Testing Your Organic Chemistry Knowledge to Reinforce ...

Testing Your Organic Chemistry Knowledge to Reinforce Comprehension and 

Understanding using Web-based Multiple Choice Questions 

Andrew Boa and Jason Eames* 

Department of Chemistry, University of Hull 

a.n.boa@hull.ac.uk; j.eames@hull.ac.uk 

This database was generated in Respondus ®®®® (http://www.respondus.com/) and can be 

exported in a variety of formats – please contact the authors for futher details. 

Index Pages 

1. Physical properties of organic molecules 2-87 

85 Questions and Answers (65 with feedback) 

2. Charges 88-112 


3. Functional groups in organic chemistry 113-143 


4. Nomenclature of organic compounds 144-171 


5. Structural isomers in organic chemistry 172-184 


6. Conformation 185-219 


7. Chirality 220-293 


8. Acids and bases 294-363 


9. Aromaticity in Organic Chemistry 364-426 


10. Substitution and Elimination Reactions 427-520 


11. Chemistry of the Carbonyl Group 521-620 


12. Organic Reactions 621-666 


653 Questions and Answers (511 with feedback)






Physical properties of organic molecules 


exported in a variety of formats – please contact the authors for futher details.






1. Dipole Moment 

Pick out the molecule which has a non-zero dipole moment (µ ≠ 0). 

a. A 

b. B 

c. C 

*d. D 

e. E 

Correct Answer Reply: 

Dipole moments are vector quantities which mean that they have both magnitude and 

direction. Considering the polarization of each individual C-Cl bonds in molecule D, each 

of which provides a component of the overall dipole moment, we see that these bonds are 

all generally pointing downwards (as represented). The C-H bond, pointing upwards, is 

essentially unpolarized. In all the other cases, any polar bond has an equal an opposite 

dipole to cancel it out.






2. Dipole Moment 2 

Which of the following molecules has a zero dipole moment? 

a. A 

b. B 

c. C 

d. D 

*e. None of the above. 



direction. Considering the polarization of each individual C-X bonds in molecules A, B, 

C and D, each of which provides a component of the overall dipole moment for the given 

example; you can see that all of these molecules have a dipole.







Which of the following molecules are polarised? 

a. A and B 

b. B and C 

c. C and D 

d. A and C 

*e. B and D 


The polarization of a bond is determined by the relative electronegativity of each 

particular atom. The fluorine atom is more electronegative than a hydrogen atom and 

bromine atom, so both molecules B and D have polarized bonds, whereas, symmetrical 

molecules A and C have no polarization.







Pick out the molecule which has the correct dipole orientation. 

a. A 

*b. B 

c. C 

d. D 

e. E 


The polarization of a bond is determined by the relative electronegativity of each 

particular atom. As Cl is more electronegative than a hydrogen atom, so the dipole 

moment points towards the Cl atom along the H-Cl bond.







Pick out the molecule which has a zero dipole moment. 

a. CH2Br2 

b. NCl3 

*c. trans-HFC=CHF 

d. cis-HBrC=CFH 

e. fluoromethane 



direction. Drawing out the structure (c) in full you can see that the dipole moments 

associated with each polar C-F bond are of equal magnitude and point in opposite 

directions leading to an overall zero dipole.






6. Bond Length 

Which of the following molecules has the longest carbon-carbon bond length? 

a. 

*b. 

c. 

d. All the same 


An alkane has a longer carbon-carbon bond (C-C) than an alkene (C=C), which in turn is 

longer than an alkyne (C≡C).






7. Bond Length 2 

Put the following molecules in the order of increasing carbon-carbon bond length: 

a. A, B, C 

*b. C, A, B 

c. B, A, C 

d. C, B, A 

e. They are all the same bond length 


For carbon-carbon bonds, as the bond order decreases the bond length increases.







Which of the following carbon-carbon bonds would you expect to be the shortest? 

*a. A 

b. B 

c. C 

d. D 

e. They are all the same length. 


For carbon-carbon bonds, as the bond order increases the bond length decreases. The 

triple bond is therefore the shortest bond.







Put the following molecules in order of increasing carbon-carbon bond length: 

*a. A > B > C 

b. B > A > C 

c. C > A > B 

d. A > C > B 

e. C > B > A 


For carbon-carbon bonds, as the bond order decreases, the bond length increases. Due to 

the delocalization of the pi electrons in benzene, the carbon-carbon bonds are neither 

single nor double bonds, but have a bond order of 1.5.







Which of these cyclic molecules has the shortest carbon-carbon bond length? 

*a. A 

b. B 

c. C 

d. D 

e. E






11. Boiling Point 

Which of these molecules has the lowest boiling point? 

a. A 

b. B 

c. C 

d. D 

*e. E 


All the molecules have hydrogen bonding capability. For compounds with the same 

functional group the lower molecular weight compounds generally have a lower boiling 

point due to the reduced opportunities for intermolecular bonding through van der Waals 

interactions.



12. Boiling Point 2 

Which amine has the highest boiling point? 




a. A 

*b. B 

c. C 

d. D 

e. They all contain nitrogen, they must have the same boiling point. 


The primary amines A and B can donate two hydrogen bonds and accept one (due to the 

lone pair on N). Generally, for compounds with the same functional group, the higher 

molecular weight compounds have a higher boiling point due to increased opportunity for 

intermolecular van der Waals interactions.







Pick out the oxygen-containing molecule which has the highest boiling point. 

*a. A 

b. B 

c. C 

d. D 

e. E 


The diol A has more intermolecular hydrogen bonding possibilities, meaning that more 

energy has to be put into change the state of the molecule.







Pick out the oxygen-containing molecule with the lowest boiling point. 

a. A 

b. B 

c. C 

d. D 

*e. E 


Neither 1,2-dimethoxypropane B and dioxane E can form intermolecular H-bonds. 

However, dioxane E has weaker intermolecular bonding interactions as the molecule is 

more compact. The decreased surface-surface contact weakens the intermolecular forces 

(e.g., dipole-dipole or Van der Waals) and so less thermal energy is needed to overcome 

them.







Which of the following molecules have the highest boiling point? 

a. A 

b. B 

c. C 

d. D 

*e. E 


Only the alcohol E can participate in hydrogen bonding. These intermolecular 

interactions are stronger than any dipole or van der Waals forces that may be present in 

the other examples.




Which amine has the highest boiling point? 




a. A 

b. B 

c. C 

*d. D 

e. They all contain nitrogen; they must have the same boiling point. 


Generally, for compounds with the same functional group the higher molecular weight 

compounds have a higher boiling point. Within this series, the amines have the same 

molecular weight, but the primary amine E can donate two H-bonds and accept one (due 

to the lone pair on N).






17. Heat of Combustion 

Which of the following molecules have the highest heat of combustion per CH2? 

*a. A 

b. B 

c. C 

d. D 

e. All have the same value 


Due to the angle strain, deviating most from the ideal angle of 109°, as found in 

cyclohexane C, cyclobutane A has the highest heat of combustion per CH2.






18. Heat of Combustion 2 

Which of the following molecules have the highest heat of combustion? 

a. 

b. 

c. 

d. 

*e. 


Combustion forms CO2, H2O and releases energy. Pentane has the most carbon-carbon 

and carbon-hydrogen bonds (i.e., greatest sum of all bond energies), so as the process of 

combustion breaks them, more energy is released.







Which of these molecules have the highest heat of combustion? 

a. A 

b. B 

*c. C 

d. D 

e. All have the same value 


Ethyne C has the very strong carbon-carbon triple bond, and will release the most energy 

when combusted.







Which cycloalkane has the lowest heat of combustion per CH2 group? 

a. cyclopropane 

b. cyclobutane 

c. cyclopentane 

*d. cyclohexane 

e. cycloheptane 


Molecules with more energy due to ring strain will release more energy per CH2. 

Considering the C-C-C bond angles, and their deviation from the ideal tetrahedral value, 

it can be seen that cyclohexane has the lowest heat of combustion per CH2 as it has the 

ideal C-C-C angles of 109°.



21. Bonding 

The carbon-carbon bonds in benzene have: 




a. different lengths due to the double-single-double bonding nature 

b. the same length as those in ethane 

c. the same length as those in ethene 

d. the same length as those in ethyne 

*e. the same length, somewhere between a single and double bond 


The delocalization in benzene makes the six C-C bonds the same. Six carbon-carbon 

bonds with 18 bonding electrons gives each C-C bond three electrons on average. This is 

compared to an ethane sigma C-C bond which has two, and an ethene C=C bond which 

has four electrons.






22. Bonding 2 

How many sigma bonds are there in CH2=CH-CH2-CH=CH2? 

a. 4 

b. 6 

c. 9 

*d. 12 

e. 14 


Multiple bonds are made up of one sigma-bond plus one (for alkenes) or two (for 

alkynes) pi-bonds. Therefore, four C-C sigma-bonds and eight C-H bonds gives an 

overall total of 12.



23. Bonding 3 

How many pi-bonds are there in 

a. 4 

b. 5 

*c. 6 

d. 7 

e. 8 

? 






alkynes) pi bonds.



24. Bonding 4 

How many sigma-bonds are there in: 

? 

a. 6 

b. 8 

c. 10 

d. 12 

*e. 14 






alkynes) pi bonds. Along with the eight CH bonds, the sigma-bond total is 14.






25. Bonding 5 

Identify the orbitals involved with the C-C sigma-bonding framework in ethyne. 

a. (2s, 1s) 

b. (2p, 2p) 

c. (2sp 2 , 2sp 2 ) 

d. (2sp 3 , 2sp 3 ) 

*e. (2sp, 2sp) 


Sigma bonds are made with end-on overlap of orbitals, such that the electron density is 

predominantly situated on the interatomic axis. In this case, the alkyne carbon is sp 

hybridized and so the orbitals involved are sp. Pi-bonds, on the hand, result from side on 

overlap of orbitals, p-orbitals in this case, with a node on the interatomic axis.






26. Bonding 6 

The angle ____ is commonly associated with tetrahedral molecules. 

a. 60º 

b. 90º 

*c. 109.5º 

d. 120º 

e. 180º






27. Bonding 7 

Which molecule has the shortest carbon-carbon bond(s)? 

a. Ethane 

b. Propene 

*c. Acetylene 

d. Allene 

e. Propane 


For carbon-carbon bonds, as the bond order decreases, the bond length increases. 

Acetylene is the trivial name for ethyne which contains a carbon-carbon triple bond.






28. Bonding 8 

The hybridisation state of an alkene carbon is: 

a. s 

b. p 

c. sp 

*d. sp 2 

e. sp 3 


Each C=C bond has one pi-bond made from overlapping of p-orbitals. This leaves the 

rest (2s, 2px and 2py) to form three sp 2 hybrids for the sigma-framework.






29. Bonding 9 

The hybridisation state of an alkyne carbon is: 

a. s 

b. p 

*c. sp 

d. sp 2 

e. sp 3 


Each carbon-carbon triple bond has two pi-bonds made from overlapping of p-orbitals. 

This leaves the others (2s and 2px) to form two sp-hybrids for the sigma-framework.






30. Bonding 10 

The hybridisation state and shape of a carbocation is: 

a. sp and linear 

*b. sp 2 and trigonal planar 

c. sp 2 and tetrahedral 

d. sp 3 and trigonal planar 

e. sp 3 and tetrahedral 


There are three bonding electron pairs around the carbon atom, and these can be placed as 

far apart as possible in an sp 2 trigonal planar framework. The empty orbital, which results 

in the charge, is the 2pz atomic (unhybridized) orbital.







The hybridisation state of a tetrahedral carbon is: 

a. s 

b. p 

c. sp 

d. sp 2 

*e. sp 3 


There are no carbon-carbon multiple bonds in tetrahedral molecules (which would be 

made from overlapping p orbitals). So all the orbitals (2s, 2pX, 2pY and 2pZ) are available 

to form four sp 3 hybrids for the sigma-framework.







The pZ-orbital of a methyl carbocation CH3 + contains _____ electrons. 

*a. 0 

b. 1 

c. 2 

d. 3 

e. None of the above 


There are three bonding electron pairs around the carbon atom, and these can be placed as 

far apart as possible in an sp 2 trigonal planar framework. The empty orbital is therefore 

the 2pZ atomic (unhybridised) atomic orbital.







The t-butyl carbocation is stabilised by hyperconjugation of the positive charge through 

the _________ orbitals. 

a. s/p 

b. p/p 

c. sp 2 /sp 3 

d. sp 3 /s 

*e. p/sp 3 


The empty orbital, which results in the overall positive charge, is the 2pZ atomic 

(unhybridised) orbital on the C + . This orbital can be aligned with a neighbouring sp 3 

hybrid involved in any one of the nine C-H bonds to assist stabilization through 

hyperconjugation.







Which of the following molecules contains sp-hybridisation? 

a. A 

*b. A and B 

c. A and C 

d. B and C 

e. All of them 


Benzyne, C, has a very weak second pi-bond, but this is from side on overlap of 

neighboring sp 2 -orbitals. The central carbon atom of B is involved in two pi bonds so 

must be sp-hybridised.







Which of the following molecules contain conjugation? 

a. A and B 

b. C and D 

c. A, B and C 

*d. A, C and D 

e. B, C and D 


Conjugation involves the overlapping of two or more pi systems, or overlap of lone pairs 

of electrons into such pi-systems. Bicyclic amide B is so arranged such that the nitrogen 

lone pair is almost perpendicular to the pi system of the C=O and therefore is not 

conjugated with it.







How many formal double bonds are there in: 

a. four 

b. five 

c. six 

*d. seven 

e. eight 

? 


The C-C bond order in benzene and the N-O bond order in the nitro groups are 1.5 each. 

However, when a particular canonical form is drawn out in full, they can be considered to 

have 3 and 1 formal double bonds respectively.








? 

a. zero 

b. one 

*c. two 

d. three 

e. four 


When presented with a resonance hybrid one should draw a representative canonical 

structure in full. From this the number of formal double bonds can be deduced.







Which of the following molecules can form hydrogen bonds? 

a. A and B 

b. B and C 

c. B and D 

d. C and D 

*e. All of them 


A molecule may be a hydrogen bond donor if it contains an H atom attached to an 

electronegative atom, such as an O and N atoms, or a hydrogen bond acceptor, if it 

contains a lone pair. It may of course be both a donor and acceptor of hydrogen bonds.







Put the following molecules in the order of increasing H-X-H bond angle. 

a. A > B > C 

b. B > C > A 

c. C > A > B 

*d. A > C > B 

e. B > A > C 


Lone pairs are held closest to the nucleus of the central atom and so cause a reduction on 

the H-X-H angle through increased repulsion of the bonding pairs.








? 

a. four 

*b. five 

c. six 

d. seven 

e. eight 


Unless you are familiar with a functional group, it is always worthwhile to draw it out in 

full, making sure each atom has the correct valency.







Pick out the molecules which can act as a hydrogen bond donor: 

a. A and B 

b. B and C 

*c. B and D 

d. C and D 




electronegative atom, such as either an O atom or an N atom, or a hydrogen bond 

acceptor (if it contains a lone pair). It may of course be both a donor and acceptor. If in 

doubt, draw a Lewis dot structure to identify any lone pairs.







Pick out the molecules which can act as a hydrogen bond acceptor: 

a. A and B 

b. B and C 

c. B and D 

d. C and D 




electronegative atom, such as either an O atom or an N atom, or a hydrogen bond 

acceptor (if it contains a lone pair). It may of course be both a donor and acceptor. If in 

doubt, draw a Lewis dot structure to identify any lone pairs.



43. Aromaticity 

Pick out which molecule is aromatic: 

a. A 

*b. B 

c. C 

d. D 

e. None of them 





Only molecule B obeys Hückel's rules of aromaticity. It must be cyclic, planar and have a 

continuously conjugated system containing 4n+2 pi-electrons.






44. Aromaticity 2 

Pick out which of the following molecules is not aromatic: 

a. A 

b. B 

c. C 

*d. D 

e. E 


Hückel's rules of aromaticity state that a molecule is aromatic if it is cyclic, planar, and 

has a continuously conjugated system containing 4n+2 pi-electrons. All molecules are 

planar and have a cyclic conjugated system, but molecule D does not have 4n+2 pielectrons.







Pick out which of the following molecules is aromatic: 

a. A 

b. B 

c. C 

d. D 



All of these molecules obey Hückel's rules of aromaticity. Hückel's rules of aromaticity 

state that a molecule is aromatic if it is cyclic, planar, and has a continuously conjugated 

system containing 4n+2 pi-electrons.







Which of the following would you expect to be aromatic? 

a. A 

b. B 

c. C 

d. D 

*e. None of the above 


All of these molecules obey some of Hückel's rules of aromaticity. Hückel's rules of 

aromaticity state that a molecule is aromatic if it is cyclic, planar, and has a continuously 

conjugated system containing 4n+2 pi-electrons. 

None of these examples obeys all of these rules.






47. Resonance 

Which of the following are not resonance forms of one another? 

*a. A 

b. B 

c. C 

d. D 

e. E 


There are no lone pairs or pi bonds in carbocation A which may be used to delocalize the 

positive charge. In all other examples, the radical or electron pair can be delocalized 

using a neighboring pi-bond. 

Note: a resonance structure only involves the movement of pi- or non-bonded electrons.



48. Resonance 2 

Which of these is an example of resonance? 

a. A 

b. B 

c. C 

d. B and C 






If you can draw curly arrows to show how electrons (lone or bonding pairs or a radical) 

may be delocalized into a neighbouring pi-bond or empty orbital then you have a 

resonance form. All of these examples show how electrons or charges may be delocalized 

using pi bonds.







Which of the following representations are resonance structures? 

a. A 

b. B 

c. C 

d. A and C 

*e. None of them 


Resonance is delocalization of electrons into neighbouring pi bond(s) or empty orbital(s) 

but which does not involve movement of atoms. In delocalizing electrons valency rules 

still apply! 

Note: a resonance structure only involves the movement of pi- or non-bonded electrons.



50. Radicals 

An allyl radical has ___ pi-electrons. 

a. 0 

b. 1 

c. 2 

*d. 3 

e. 5 



a.n.boa@hull.ac.uk; j.eames@hull.ac.uk






51. Stability 

Which of the following free radicals is most stable? 

a. A 

b. B 

*c. C 

d. D 

e. E 


In E, the radical can be delocalized using the neighbouring pi bond; you must also 

consider also the inductive effect of the substituents on the stabilization of the radical. 

The delocalized radical in C has the electron density shared between a secondary and 

primary carbon atom.






52. Stability 2 

Pick out which primary free radical is the least stable. 

a. A 

b. B 

*c. C 

d. D 

e. E 


All radicals shown are primary and there are no cases of resonance stabilization; even 

molecule E where the pi-system is not conjugated to the singly occupied orbital. This 

leaves only the inductive effect which is least significant in the case of the single methyl 

substituent in molecule C.







Pick out the carbocation which is most stable. 

a. A 

*b. B 

c. C 

d. D 

e. E 


Stabilization of a carbocation can come from either inductive or resonance effects or a 

combination of both. Cations A, C and E are not stabilized by delocalization as the pisystem 

is not conjugated to the unoccupied orbital. This leaves the inductive effect which 

is most significant in the case of the molecule B where the charge can be delocalized 

between a tertiary (as shown) and a secondary centre.







Which of the following molecules is most stable? 

a. A 

b. B 

*c. C 

d. D 

e. E 


Stability can come from a variety of factors. Conjugation of pi-systems leads to a 

stabilization, whereas, increased angle strain leads to a destabilization. The double bonds 

in molecule C are conjugated and there is no strained bond angles associated with the 

four sp 2 hybridised carbon atoms in a cycloctene ring system.




Which of these carbocations is most stable? 

a. A 

*b. B 

c. C 

d. D 

e. E 





Greatest stabilization of carbocations comes when the degree of resonance delocalization 

is greatest; generally the more resonance forms that can be drawn, the more stable it is. 

Remember that if an sp 2 orbital is empty, but is perpendicular to the pi-system it cannot 

overlap with it. Cations A, C and E have such empty sp 2 orbitals, but in the case of the 

cation B the carbocation is delocalized around both rings. In cation D, only the fivemembered 

ring is conjugated to the carbocation.




Pick out which free radical is most stable: 

a. A 

b. B 

c. C 

*d. D 

e. E 





Stabilization of a cation can come from either inductive or resonance effects or a 

combination of both. Resonance, however, is the most important. 

In radical D, the pi-system is conjugated to the unoccupied orbital, but this is not the case 

for radical E. The inductive effect is a much less significant factor, so radicals A, B and 

C can be discounted.






57. Hydrogenation 

Pick out the diene which has the highest heat of hydrogenation? 

a. A 

b. B 

c. C 

d. D 

*e. E 


Heat released in the hydrogenation of an alkene is a measure of its thermodynamic 

stability. 

Factors contributing to stability of alkenes include conjugation and double bond 

geometry. Diene E has both a conjugated system and the more stable (E,E) alkene 

configuration.






58. Atomic Orbitals 

Select the most electronegative atomic orbital. 

*a. 2s 

b. 2p 

c. 2sp 

d. 2sp 2 

e. 2sp 3






59. Atomic Orbitals 2 

Which of the following molecules contain an empty atomic p-orbital? 

a. BF3 and NH3 

b. Me3C + and HCl 

*c. BF3 and Me3C + 

d. NH3 and HCl 

e. None of them







Which of the following is a 3p atomic orbital? 

a. A 

b. B 

c. C 

*d. D 

e. E







Which of the following molecules have the same number of pi-electrons? 

a. A and B 

b. C and D 

*c. A, B and C 

d. A, C and D 

e. B, C and D







Select the most electronegative atomic orbital. 

*a. A 

b. B 

c. C 

d. D 

e. E







An electron in which of the following orbitals is most stable? 

a. A 

*b. B 

c. C 

d. D 

e. E







Which of the following is a 2s atomic orbital? 

a. A 

b. B 

*c. C 

d. D 

e. E







Which of the following is a 2p atomic orbital? 

a. A 

b. B 

c. C 

*d. D 

e. E






66. Forces 

Which of the following terms is not an intermolecular attractive force? 

b. H-bonding 

c. Dispersion forces 

d. Dipole-dipole interactions 

*e. Resonance structures 

f. Ion-ion interactions






67. Valency 

Which of the following representations of nitromethane have incorrect valency? 

a. A 

*b. B 

c. C 

d. A and C 

e. B and C






68. Valency 2 

For the incomplete Lewis structure drawn, assign the charges to the O and S atoms of 

dimethylsulphoxide (DMSO) (overall charge on the molecule is neutral). 

a. S 1- and O 1+ 

b. S 0 and O 0 

*c. S 1+ and O 1- 

d. S δ+ 

and Oδe. 

S δ- and O δ+






69. Valency 3 

For the incomplete Lewis structure drawn, assign the charges to the O and N atoms of 

nitromethane (overall charge on the molecule is neutral). 

a. N 1- and O 1+ 

b. N 0 and O 0 

*c. N 1+ and O 1- 

d. N δ+ and O δ- 

e. N δ- and O δ+






70. Valency 4 

For the incomplete Lewis structure drawn, assign the charges to the C and O atoms of 

acetone (overall charge on the molecule is neutral). 

a. C 1- and O 1+ 

b. C 0 and O 0 

*c. C 1+ and O 1- 

d. C δ+ and O δ- 

e. C δ- and O δ+






71. Valency 5 

For the incomplete Lewis structure drawn, assign the charges to the N atoms of azide 

(overall the molecule has a single negative charge). 

a. a N 0 , b N 1- and c N 0 

b. a N 1- , b N 0 and c N 0 

*c. a N 1- , b N 1+ and c N 1- 

d. a N 1/2+ , b N 0 and c N 1/2- 

e. a N 0 , b N 2- and c N 1+






72. Valency 6 

For the incomplete Lewis structure drawn, how many lone pairs of electrons are present 

in the following molecule? 

a. 0 

b. 1 

c. 2 

d. 3 

*e. 4






73. Valency 7 

How many lone pairs of electrons are present in the following molecule? 

a. 0 

*b. 1 

c. 2 

d. 3 

e. 4 


The nitrogen lone pair is in an sp 2 orbital and is therefore perpendicular to (and not 

conjugated to) the pi-system. It is therefore truly a lone pair.






74. Valency 8 


*a. 0 

b. 1 

c. 2 

d. 3 

e. 4 


The seemingly “lone” pair of electrons on the NH nitrogen is in fact delocalized into the 

pi-system, and as such, is not directly available, e.g. for protonation.






75. Valency 9 


a. 0 

*b. 1 

c. 2 

d. 3 

e. 4 


The seemingly “lone” pair electrons on the NH nitrogen is in fact delocalized into the pisystem, 

and as such is not available, e.g. for protonation. However, the other nitrogen 

atom has a lone pair in an sp 2 orbital perpendicular to the pi-system and is available for 

protonation.






76. Acids 

Which of the following molecules are Lewis acids? 

a. BF3 and H2O 

b. NH3 and Me3C + 


d. NH3 and H2O 



A Lewis acid is an electron pair acceptor by virtue of it having an empty orbital. Both 

BH3 and CH3 + are electron deficient, and therefore are Lewis acids.






77. Acids 2 

Which of the following molecules are Bronsted acids? 

a. BH3 and NH3 

b. CH4 and H2O 

c. BH3 and CH4 

d. NH3 and H2O 



A Brønsted acid is a proton donor by virtue of having a hydrogen atom attached to 

another, usually an electronegative, atom. 

All molecules which contain a hydrogen atom are potential Brønsted acids irrespective 

of their ability to donate a hydrogen atom as a proton.






78. Acids 3 

Which of the following molecules are Lewis bases? 

a. BF3 and H2O 

b. NH3 and Me3C + 

c. BF3 and Me3C + 

*d. NH3, H2O and BF3 



A Lewis base is an electron pair donor by virtue of having a lone pair of electrons. BH3 

and CH3 + are electron deficient molecules and are therefore Lewis acids.






79. Acids 4 

Which of the following molecules are Bronsted bases? 

a. BH3 and NH3 

b. CH4 and H2O 

c. BH3 and CH4 

*d. NH3 and H2O 



A Brønsted base is a proton acceptor by virtue of having a lone pair of electrons. BH3 is 

electron deficient at boron, and CH4 has covalently shared outer electrons; both 

molecules do not contain a lone pair of electrons.






80. Molecular Weight 

Which of the following molecules have approximately a molecular weight of 156? 

a. A 

b. B 

c. C 

d. A and C 



Compounds of the same molecular formula must have the same molecular weight. Those 

with atoms which have weights which are multiples of lighter atoms (e.g., O is 16 and S 

is 32) may have the same mass approximately. If measured to several decimal places they 

can however easily be distinguished.






81. Spectroscopy 

Pick out the molecules which contain an IR stretching frequency of 3500-3000 cm -1 . 

(I) amines; 

(II) carboxylic acids; 

(III) phenols; 

(IV) alcohols. 

a. (I) and (II) 

b. (II) and (III) 

c. (II) and (IV) 

d. (III) and (IV) 



The range 3500-3000cm -1 is typically associated with stretches between heteroatoms and 

a hydrogen atom, for example NH and OH bonds. They may appear broad, intense and 

jagged (COOH), broad and smooth (OH) or less intense and sharper (NH).






82. Spectroscopy 2 

Pick out the molecules which contain an IR stretching frequency of 1720 cm -1 . 

(I) amines; 

(II) carboxylic acids; 

(III) phenols; 

(IV) alcohols. 

*a. (I) and (II) 

b. (II) and (III) 

c. (II) and (IV) 

d. (III) and (IV) 

e. All of them






83. Electrophile Recognition 

Pick out the electrophiles from the following molecules: 

a. BF3 and NH3 

b. Me3C + and HCl 


d. NH3 and HCl 

e. None of them






84. Electrons 


a. 3 

b. 4 

c. 5 

*d. 6 

e. 7 


Amines are Brønsted bases and as such their lone pair of electrons is ‘tied up’ upon 

protonation to form the corresponding ammonium salt. Therefore, only the number of 

lone pairs for the oxygen atom and chloride ion needs to be considered.






85. Electrons 2 

Pick out the molecules which have six pi-electrons. 

a. A 

b. B 

c. C 

d. A and B 

*e. A, B and C 


Lone pairs of electrons in p-orbitals (on the NH of pyrrole B and imidazole C) may 

overlap with the conjugated system and contribute to the total, but those in orthogonal sp 2 

orbitals on nitrogen (in pyridine A and pyrrole C) are clearly perpendicular to the pisystem, 

and therefore cannot be delocalized.






Charges 








86. Molecules and Ions 

Which of the following structures bear a positive charge? 

a. A and B 

b. C and D 

c. E and A 

*d. B and C 

e. D and E 


In this example the number of lone and bonding pairs of electrons needs to be 

determined. The normal valency of the central atom can then be used be determined 

whether the charge is positive (B and C), negative (E) or zero (A and D).






87. Shape 

Which of the following structures are tetrahedral? 

a. A and C 

b. C and E 

c. A and E 

*d. B and E 

e. D and B 


A molecule will be tetrahedral if it has four sigma bonding pairs of electrons. If the 

groups are identical then the tetrahedron will be regular (e.g. CCl4). If the groups are 

different (e.g. CHCl3) then not all bond angles will be absolutely identical.






88. Shape 2 

Which of the following structures is trigonal planar? 

*a. A 

b. B 

c. C 

d. D 

e. E 


A trigonal planar arrangement can be found in two types of molecules where: 

(i) no multiple bonds are involved, the central atom has a maximum of three lone or 

bonding pairs of electrons. 

(ii) three sigma bonding pairs of electrons are used on the central atom, but a fourth 

bonding pair of electrons is involved in a pi-bond - for example, in alkenes. 

In these examples, C contains a lone pair of electrons so is pyramidal not trigonal planar.






89. Shape 3 

Which of the following structures are trigonal planar? 

*a. A and B 

b. B and C 

c. D and A 

d. E 

e. All of the above 


A trigonal planar arrangement can be found in two types of molecules where: 

(i) no multiple bonds are involved, the central atom has a maximum of three lone or 

bonding pairs of electrons. 

(ii) three sigma bonding pairs of electrons are used on the central atom, but a fourth 

bonding pair of electrons is involved in a pi-bond - for example, in alkenes. 

In these examples, the number of lone and bonding pairs of electrons need to be 

determined in order to see if it fits the first situation.






90. Shape 4 

Which of the following structures could not be considered as planar? 

a. A 

b. B 

*c. C 

d. D 

e. All of the above






91. Charge 

What is the formal charge on the oxygen atom in the trimethyl oxonium ion? 

a. +2 

*b. +1 

c. 0 

d. -1 

e. -2 


Being in Group 16, oxygen normally forms two bonds with other atoms and has two lone 

pairs. In this case the oxygen has three bonds and one lone pair. So overall, it has a share 

in five electrons (1 from each lone electron, half from each shared electron) and so the 

charge must be +1.






92. Charge 2 

Which of the following structures contain an oxygen atom which has a formal charge of 

+1? 

a. A and C 

b. B and D 

c. E and A 

*d. B and E 

e. C and D 


In this example the number of lone and bonding pairs of electrons on oxygen will be 4. 

The normal valency of the atom is 2 and this can then be used be determined whether the 

charge is therefore positive (three bonds, B and E), negative (one bond, A and D) or zero 

(two bonds, C).






93. Charge 3 

Which of the following structures have a central atom with a formal charge of -1? 

a. A 

*b. B 

c. C 

d. D 

e. E 


The number of lone and bonding pairs of electrons on nitrogen will be 4. The normal 

valency of the nitrogen atom is 3 and this can then be used be determined whether the 

charge is therefore positive (four bonds, E), negative (two bonds, B) or zero (three bonds, 

A, C and D). (This of course assumes that the total number of electrons around the 

nitrogen atom is 8, the balance made up with lone pairs which may or may not be 

explicitly indicated).






94. Charge 4 

For the incomplete Lewis structure drawn, assign the charges to the C, N and O atoms of 

an isocyanate (overall the molecule has a single negative charge). 

a. a O -1 , b C 0 and c N 0 

b. a O 0 , b C -1 and c N 0 

*c. a O 0 , b C 0 and c N -1 

d. a O +1 , b C -1 and c N -1 

e. a O -1 , b C 0 and c N -1 


The normal valency of carbon is 4, oxygen is 2 and nitrogen 3. Adding electrons to make 

a full octet means adding two lone pairs to both oxygen and nitrogen atoms. For a -1 

charge this must therefore be situated on N as it only has two bonding pairs of electrons 

instead of three.






95. Charge 5 

What is the formal charge on the carbon atom of structure A? 

a. +2 

b. +1 

*c. 0 

d. -1 

e. -2 


Being in Group 14, carbon normally bonds four times (with other atoms) and has no lone 

pairs. In this case, the carbon atom has two bonds and one lone pair. So overall, the 

carbon atom has a share in four electrons (1 from each lone electron, half from each 

shared electron) and so the charge must be 0.






96. Charge 6 


a. +2 

b. +1 

c. 0 

d. -1 

*e. -2 


Being in Group 14, carbon normally forms four bonds (with other atoms) and has no lone 

pairs. In this case, the carbon atom has two bonds and two lone pairs. So overall, the 

carbon atom has a share in six electrons (1 from each lone electron, half from each shared 

electron) and so the charge must be -2.






97. Charge 7 

What is the formal charge on the central carbon atom of structure A? 

a. +2 

*b. +1 

c. 0 

d. -1 

e. -2 



pairs. In this case, the carbon atom has three bonds and no lone pairs. So overall, the 

carbon atom has a share in three electrons (half from each shared electron) and so the 

charge must be +1.






98. Charge 8 

What is the formal charge on the sulfur atom of structure A? 

a. +2 

*b. +1 

c. 0 

d. -1 

e. -2 


Sulfur being in Group 16 normally forms two bonds with other atoms and has two lone 

pairs. In this case, the sulfur has three bonds and one lone pair. So overall, it has a share 


charge must be +1.






99. Charge 9 

What is the formal charge on the nitrogen atom of structure A? 

a. +2 

b. +1 

*c. 0 

d. -1 

e. -2 


Being in Group 15, nitrogen normally forms three bonds (with other atoms) and has one 

lone pair. In this case, the nitrogen has three bonds and one lone pair. So overall, it has a 

share in five electrons (1 from each lone electron, half from each shared electron) and so 

the charge must be 0.






100. Charge 10 

What is the formal charge on the nitrogen atom of structure A? 

a. +2 

*b. +1 

c. 0 

d. -1 

e. -2 


Being in Group 15, nitrogen normally forms three bonds (with other atoms) and has one 

lone pair. In this case, the nitrogen has four bonds and no lone pairs. So overall, it has a 

share in four electrons (half from each shared electron) and so the charge must be +1.








*a. +2 

b. +1 

c. 0 

d. -1 

e. -2 



pairs. In this case, the carbon has two bonds and no lone pairs. So overall, it has a share 

in two electrons (half from each shared electron) and so the charge must be +2.







What is the formal charge on the boron atom of structure A? 

a. +2 

b. +1 

c. 0 

*d. -1 

e. -2 


Being in Group 3, boron normally forms three bonds (with other atoms) and has no lone 

pairs. In this case, B has two bonds and one lone pairs. So overall, it has a share in four 

electrons (1 from each lone electron, half from each shared electron) and so the charge 

must be -1.







What is the formal charge on the beryllium atom of structure A? 

a. +2 

b. +1 

*c. 0 

d. -1 

e. -2 


Being in Group 2, beryllium normally forms two bonds (with other atoms) and has no 

lone pairs. In this case, Be has two bonds and no lone pairs. So overall, it has a share in 

two electrons (half from each shared electron) and so the charge must be 0.







What is the formal charge on the oxygen atom of structure A? 

a. +2 

*b. +1 

c. 0 

d. -1 

e. -2 


Being in Group 16, oxygen normally forms two bonds with other atoms and has two lone 

pairs. In this case, the oxygen has three bonds and one lone pair. So overall, it has a share 


charge must be +1.







What is the formal charge on the phosphorus atom of structure A? 

a. +2 

b. +1 

*c. 0 

d. -1 

e. -2 


Being in Group 15, phosphorus, as found in phosphines (PR3), can form three bonds 

(with other atoms) and has one lone pair. In this case, the phosphorus has four bonds and 

no lone pairs. This phosphorus ylide is therefore in a higher oxidation state (+4) (for this 

it must involve the use of d orbitals) compared to the phosphine (+3). However, overall it 

still has a share in five electrons (half from each shared electron) and so the charge like in 

the phosphine is 0.







What is the formal charge on the sulfur atom of structure A? 

a. +2 

b. +1 

*c. 0 

d. -1 

e. -2 


Being in Group 16, sulfur, as found in thioethers (R-S-R) can form two bonds (with other 

atoms) and has two lone pairs. In this case, the sulfur has four bonds and one lone pair. 

The sulfoxide is in a higher oxidation state (+4) (for this it must involve the use of d 

orbitals) compared to the thioether (+2). However, overall it still has a share in six 

electrons (1 from each lone electron, half from each shared electron) and so the formal 

charge will be zero (like a sulfide).



107. Structure 

Pick out the structure of nitric acid (HNO3): 

a. A 

b. B 

*c. C 

d. D 

e. None of these 






108. Structure 2 

Pick out the structure of a diazonium salt: 

a. A 

*b. B 

c. C 

d. D 

e. E 






109. Structure 3 

Pick out the structure of a nitrite ion: 

*a. A 

b. B 

c. C 

d. D 

e. None of these 









Functional groups in organic chemistry 








110. Cyclics 

Which of the following molecules is a bicyclohexane? 

a. A 

*b. B 

c. C 

d. D 

e. E 


A cyclohexane will always have only six carbon atoms in the ring(s) irrespective of the 

presence or absence of the prefix bicyclo.






111. Chemical Tests 

Which of the following reagents would be a simple test to distinguish between benzene 

and ethene? 

a. A 

*b. B 

c. C 

d. D 

e. E 


The decolourization of bromine by alkenes is a classic test for the presence of carboncarbon 

double bonds; ethene adds bromine to form a 1,2-dibromoethane. However, with 

benzene, a Lewis acid catalyst is needed before a (indirect substitution) reaction can take 

place.






112. Formation 

Which of the following compounds is capable of forming a delta-lactone? 

a. A 

*b. B 

c. C 

d. D 

e. E 


Lactone is an alternative name for a cyclic ester and contains the CO(C=O)C unit within 

the ring. The smallest lactone is a three membered ring and is called an alpha-lactone. A 

delta-lactone would therefore contain a six membered ring.



113. Aldehydes 

Which of the following is an aldehyde? 

a. A 

b. B 

c. C 

*d. D 

e. E 






114. Ketones 

Which compound is a ketone? 

a. A 

*b. B 

c. C 

d. D 

e. E 






115. Esters 

Which compound is an acyclic ester? 

a. A 

b. B 

*c. C 

d. D 

e. E 





A cyclic ester, also known as a lactone, contains the CO(C=O)C unit within the ring. 

Molecule E is a lactone, and molecule C is an acyclic ester.



116. Esters 2 

Which compound is a cyclic ester? 

a. A 

b. B 

c. C 

d. D 

*e. E 





A lactone is an alternative name for a cyclic ester and contains the CO(C=O)C unit 

within the ring. Therefore, molecule C is not a lactone, but an ester, even though there is 

a ring present.



117. Esters 3 

Which compound is a lactone? 

a. A 

b. B 

c. C 

*d. D 






A lactone is an alternative name for a cyclic ester and contains the CO(C=O)C unit 

within the ring.



118. Alcohols 

Which compound is an alcohol? 

a. A and D 

b. A and B 

c. C and E 

*d. B 

e. D 





In hemi-aminal A and enol B, the presence of adjacent atoms / functionality adjacent to 

the OH group means their chemistry is not the same as a simple alcohol, such as alcohol 

B. Therefore, molecules A and B cannot be classed as alcohols.






119. Alcohols 2 

Which of the following compounds is a secondary alcohol? 

a. A and D 

b. B and E 

c. C 

*d. D 



A secondary alcohol, such as molecule D, is one in which the OH group is attached to a 

secondary carbon atom (which bears only one C-H bond).



120. Alcohols 3 

Which compound is an alcohol? 

a. A 

b. B 

*c. C 

d. D 

e. E 





The presence of adjacent atoms/functionality adjacent to the OH group means the 

chemistry of molecules A, D and E is not the same as a simple alcohol like molecule C. 

Molecule B is the closest in chemistry to molecule C, but even in this case, the aromatic 

ring alters the properties enough for it to be classed as a phenol rather than an alcohol.



121. Amides 

Which compound is a lactam? 

*a. A 

b. B 

c. C 

d. D 

e. E 





A lactam is an alternative name for a cyclic amide and contains the CN(C=O)C unit 

within the ring.



122. Amides 2 

Which compound is an amide? 

*a. A 

b. B 

c. C 

d. D 

e. E 






123. Amides 3 

Which compound is a tertiary amide? 

a. A 

b. B 

c. C 

d. D 

*e. E 






124. Amines 

Which compound is a secondary amine? 

a. A and D 

b. B 

c. C 

d. E 






A secondary amine has nitrogen atom bearing only one N-H bond and two alkyl or aryl 

groups. It is often confused with a primary amine (R2CHNH2) in which the nitrogen is 

attached to a secondary carbon atom.



125. Amines 2 

Which compound is a tertiary amine? 

a. A and C 

*b. B and D 

c. C and E 

d. A and B 

e. D and E 





A secondary amine has nitrogen atom bearing three alkyl or aryl groups. It is often 

confused with primary amines (R3CNH2) in which the nitrogen atom is attached to a 

tertiary carbon atom. The chemistry of the nitrogen atom in enamine A is influenced by 

the adjacent double bond (into which the lone pair of electrons on the nitrogen atom is 

delocalized) so it cannot be classed as an amine.



126. Amines 3 


a. A, B and C 

b. A, C and E 

*c. B, C and D 

d. C, D and E 






The chemistry of the nitrogen atom in enamine E is influenced by the adjacent double 

bond, where the lone pair of electrons on the nitrogen atom is delocalized into; so it 

cannot be classed as an amine. The presence of a C=N double bond also alters the 

reactivity on imine A, which is much closer to a ketone than an amine.



127. Amines 4 


*a. A 

b. B 

c. C 

d. D 






A secondary amine has nitrogen atom bearing only one N-H bond and two alkyl or aryl 

groups. It is often confused with a primary amine (R2CHNH2) in which the nitrogen is 

attached to a secondary carbon atom.






128. Ethers 

Which of the following compounds is an ether? 

*a. A 

b. A and B 

c. A and C 

d. A and D 

e. A and E 


An ether must have a C-O-C sub-unit, which in the carbon atoms may only bear other H 

atoms or C substituents.






129. Ethers 2 

Which of the following is not true about ethers? 

a. A 

b. B 

*c. C 

d. D 

e. E 


An ether will generally be less soluble in water than alcohol of similar molecular weight 

as the OH group in the alcohol can act as a hydrogen bond donor as well as acceptor.



130. Ethers 3 

Which compound is an acetal? 

a. A 

b. B 

*c. C 

d. D 

e. E 









131. Dienes 

Select the structures which are conjugated dienes. 

a. A and C 

*b. B and D 

c. C and E 

d. A and B 

e. C, D and E



132. Bifunctional Chemistry 

Which compound is an aminal? 

*a. A 

b. B 

c. C 

d. D 

e. E 






133. Bifunctional Chemistry 2 

Which compound is an enol? 

a. A 

b. B 

c. C 

*d. D 

e. E 







Which compound is an hemiacetal? 

a. A 

*b. B 

c. C 

d. D 

e. E 







Which compound is a ketal? 

a. A 

b. B 

*c. C 

d. D 

e. E 






136. Naming 

Propan-2-one is commonly known as: 

a. A 

b. B 

c. A and C 

*d. D 

e. E 






137. Naming 2 

Acetic acid is commonly known as: 

a. A 

*b. B 

c. A and C 

d. D 

e. E 






138. Naming 3 

Ethyl acetate is commonly known as: 

*a. A 

b. B 

c. C 

d. D 

e. E 









139. Naming 4 

Which of the following compounds are phenols? 

a. A 

b. A and B 

c. A and C 

d. A and D 

*e. A and E






Nomenclature of organic compounds 





140. IUPAC Naming 

What is the IUPAC name for: 




a. 2-methyl-5-isobutylheptane 

*b. 2,7-dimethyl-4-ethyloctane 

c. 2,7-dimethyl-5-ethyloctane 

d. 2-methyl-5-(2-methylpropyl)heptane 

e. 2,7,7-trimethyl-4-ethylheptane 


The longest chain length determines the basic alkane name; for this particular case it is 

octane. 

Locants (substituent position numbers) are assigned such that the lowest possible 

numbers are used [i.e., answer (b) NOT (c)]. 

In the final stage of naming a compound, the order of substituents is alphabetic (dimethyl 

comes before ethyl).



141. IUPAC Naming 2 





*a. 5-bromo-3-(bromomethyl)pent-1-ene 

b. 3-(1-bromopropyl)-4-bromobut-1-ene 

c. 1,4-dibromo-3-ethenylbutane 

d. 1,4-dibromo-2-ethenylbutane 

e. 1,4-dibromo-2-ethenylbutane 


The longest chain length determines the base part of the name; for this particular case it is 

a pentene and NOT a butene. Therefore, this is a 5-bromopent-1-ene and not a 1bromopent-5-ene 

as the priority order of the functional groups; an alkene has higher 

priority than a bromo-substituent.








*a. 2,3,4,5,6,7,8-heptamethylnonane 

b. 1,2,3,4,5-pentamethyl-2,5-isopropylpentane 

c. 1,1',2,3,5,6,7,7'-nonamethylheptane 

d. di-(1,2,3-trimethylbutyl)ethane 



The longest chain length determines the basic alkane name; for this particular case it is a 

nonane. 


numbers are used. In this case it would not matter at which end you assigned as carbon 1 

due to the symmetry of the structure.








a. 1-ethyl-2,3-dimethylcyclohexane 

b. 1,2-dimethyl-3-ethylcyclohexane 

c. 1,6-dimethyl-5-ethylcyclohexane 

d. 1-ethyl-5,6-dimethylcyclohexane 

*e. 2,3-dimethyl-1-ethylcyclohexane 


Cycloalkanes are named like their alicyclic counterparts, based on their number of atoms 

in the ring adding the prefix cyclo to indicate the ring; for this particular case it is a 

cyclohexane. 


numbers are used where possible. In this case, the ethyl substituent is the bigger and so is 

given the locant 1 (1-ethyl). The rest follow using the lowest numbers (i.e., 2,3-dimethyl 

and not 5,6-dimethyl) with the overall order of substituents presented in alphabetical 

order.




Select the systematic name for: 

a. cis-1,4-dibromocyclopentane 

b. trans-1,4-dibromocyclopentane 

*c. cis-1,3-dibromocyclopentane 

d. trans-1,3-dibromocyclopentane 

e. dibromopentane 





Cyclopentane is not flat but has rather an (envelope) shape in which one carbon atom is 

out of the plane made by the other four carbon atoms. In a 2D representation, the 

cyclopentane ring can be considered to be flat-ish. Whichever carbon atom is out of the 

plane is unimportant, however, the orientation of the substituents around this ring is 

important; they can either be generally pointing towards or away from the observer. 

If groups are on the same side they are referred to as being cis-, whereas, on opposite 

sides they are referred to as trans-.




Select the systematic name for: 




*a. trans-1-bromo-3-methylcyclopentane 

b. trans-1-bromo-5-methycyclopentane 

c. cis-1-bromo-3-methylcyclopentane 

d. trans-1-methyl-3-bromocyclopentane 

e. cis-1-methyl-3-bromocyclopentane 


Cyclopentane is not flat but has rather an (envelope) shape in which one carbon atom is 

out of the plane made by the other four carbon atoms. In a 2D representation, the 

cyclopentane ring can be considered to be flat-ish. Whichever carbon atom is out of the 

plane is unimportant, however, the orientation of the substituents around this ring is 

important; they can either be generally pointing towards or away from the observer. 

If groups are on the same side they are referred to as being cis-, whereas, on opposite 

sides they are referred to as trans-. 

Remember the substituents are always given the smallest locants possible.







Which of the following systematic names is best for: 

a. 3-bromo-4-methyl-penta-1,4-diene 

*b. 3-bromo-2-methyl-penta-1,4-diene 

c. 3-bromo-4-methylene-pentene 

d. 3-bromo-2-methylene-pent-4-ene 



The longest chain length determines the basic alkane name; for this particular case it has 

five carbon atoms and has the pent prefix. As there are two C5 chains, but alkene groups 

at both ends of only one these, it means this chain takes priority; this molecule is a 

substituted penta-1,4-diene. The locants are then assigned such that the lowest possible 

numbers are used [(b) not (a)].







The correct IUPAC name for the following compound is: 

a. 2-bromo-4-ethylene-pent-4-ene 

b. 4-methylene-2-bromohexane 

*c. 2-bromo-4-methylene-hexane 

d. 5-bromo-3-methylene-hexane 

e. 4-bromo-2-ethylpent-1-ene 


The longest chain length determines the basic alkane name; for this particular case it has 

six carbon atoms and is a hexane. 


numbers are used [i.e. answer (c) not (d)]. 

In the final stage of naming a compound the order of substituents is alphabetic [bromo 

comes before methylene; so (b) not (c)].







The correct IUPAC name for the following molecule is: 

a. (E)-1-bromo-4-(bromomethylene)-pentane 

b. (Z)-1-bromo-4-(bromomethylene)-pentane 

*c. (E)-1,5-dibromo-2-methylpent-1-ene 

d. (Z)-1,5-dibromo-2-methylpent-1-ene 



Where there is more than one choice, the longest chain containing the most functional 

groups takes priority.








a. cis-5-ethyl-3,4-dimethylhex-2-ene 

*b. (E)-3,4,5-trimethylhept-2-ene 

c. (Z)-3,4,5-trimethylhept-2-ene 

d. cis-3,4,5-trimethylhept-2-ene 

e. trans-3,4,5-trimethylhept-2-ene 


(E)- and (Z)- (and NOT cis- and trans-) and are the correct terms in IUPAC nomenclature 

for naming geometric isomers.







Which of the following molecules is 5-bromo-4-ethylhex-1-ene? 

*a. A 

b. B 

c. C 

d. D 

e. E







Which of the following molecules is 5-bromo-3-propylhex-1-ene? 

a. A 

b. B 

*c. C 

d. D 

e. E







Pick out the structure(s) for 5-chloro-4-methyl-2-hydroxy-hex-3-ene: 

a. A 

b. B 

*c. A and C 

d. B and D 

e. E 


A and C are geometric (Z-/E-) isomers, as are B and D. However, only A and C match 

the IUPAC name given.








a. trifluorobutene 

*b. 4,4,4,-trifluorobut-1-ene 

c. 1,1,1-trifluorobut-3-ene 

d. 1,1,1-trifluorobut-4-ene 

e. None of the above







The correct IUPAC name for the following molecule is: 

a. 3-formyl-but-1-ene 

b. 3-methylbut-2-enal 

*c. 2-methylbut-3-enal 

d. 2-vinylpropanal 



Formyl and vinyl are the trivial names for aldehyde and ethenyl groups, respectively. 

Although commonly used they are not part of the IUPAC system.




Which of the following is an anhydride? 

a. A 

b. B 

*c. C 

d. D 

e. E 










Which of the following molecules is toluene? 

*a. A 

b. B 

c. C 

d. D 

e. E







Which of the following molecules is an amide? 

a. A 

b. B 

c. C 

d. D 

*e. E







Which of the following molecules is a carboxylic acid? 

a. A 

b. B 

c. C 

*d. D 

e. E







Which of the following molecules are alkenes? 

a. A and C 

b. B and D 

c. C and D 

*d. A, B and E 



Isolated carbon-carbon double bonds may be referred to as alkenes, but if the C=C is 

conjugated to a C=O, as in C and D, then its chemistry is dramatically altered. It cannot 

therefore be grouped together with the other simple alkenes.







Which of the following molecules is an aldehyde? 

a. A 

b. B 

c. C 

*d. D 

e. E







Which of the following molecules is a ketone? 

a. A and B 

b. B and C 

c. C and D 

d. E 

*e. None of them







Which of the following molecules is an amide? 

a. A 

b. B 

c. C 

*d. D 

e. E







Which of the following molecules is a peroxy acid? 

a. A 

*b. B 

c. C 

d. D 

e. E







Which of the following molecules is an ether? 

*a. A 

b. B 

c. C 

d. D 

e. E






165. Empirical Formula 

What is the empirical formula for cyclopentane? 

a. CH 

*b. CH2 

c. C2H4 

d. C5H5 

e. C5H10






166. Empirical Formula 2 

Molecule A contains only carbon and chlorine, and has a molecular weight of 285 g 

mol -1 . What is the molecular formula for this compound? 

a. C2Cl2 

b. C3Cl3 

c. C4Cl4 

d. C5Cl5 

*e. C6Cl6






Structural isomers in organic chemistry 








167. Structural Isomers 

Which of the following pairs of compounds represent pairs of structural isomers? 

a. A 

b. B 

c. C 

d. D 

*e. E 


Structural (constitutional) isomers have the same molecular formula but a different 

arrangement of atom connectivity. 

Structural isomers can come in three sub-classes: 

(i) chain isomers (variation in the basic carbon skeleton); (ii) position isomers (position 

of a functional group in a chain), and; (iii) functional group isomers (where the rearrangement 

of atoms also leads to a change in the functional group).






168. Structural Isomers 2 

How many structural isomers are there for hexane, C6H14? 

a. 2 

b. 3 

c. 4 

*d. 5 

e. 6 


A pen and paper is very useful for such questions. 

In re-arranging carbon atoms you must be careful not to “double count” isomers. Always 

check the length of the longest carbon chain, and therefore the atom number for any 

branch points.






169. Structural Isomers 3 

How many different difluorotoluenes are possible? 

a. 4 

b. 6 

c. 8 

d. 9 

*e. 10 


A pen and paper is very useful for such questions. In considering the basic carbon 

skeleton of toluene you should immediately note that there is a plane of symmetry 

through the methyl group and “para CH”. This helps to identify arrangements of atoms 

that are the same, but seem different due to translation or rotation.







Pick out the conformation of 1,2-dibromoethane which has a zero dipole moment: 

a. A and B 

*b. C 

c. D 

d. All have a dipole 

e. 1,2-dibromoethane always has a dipole moment 


The carbon-bromine bonds are polar due to the difference in electronegativity of a carbon 

atom and a bromine atom. These individual dipole moments cancel each other out only in 

the anti-periplanar arrangement as shown in C.






171. Cis-Trans Isomers 

Pick out the molecules which exist as (E)- and (Z)- isomers: 

a. A 

b. A and B 

*c. A and C 

d. A and D 

e. B and C 


Geometric isomers (E-/Z-) exist in molecules where there is restricted rotation about a 

multiple bond (e.g., C=C) and where there is no plane or axis of symmetry about the 

double bond. Terminal alkenes, those with the double bond at the end of a carbon chain, 

do not exhibit geometric isomerism.






172. Cis-Trans Isomers 2 

Which one of the following can exhibit geometric isomerism? 

a. A 

b. B 

c. C 

d. D 

*e. E 


Geometric isomers (E-/Z-) exist in molecules where there is restricted rotation about a 

multiple bond (e.g., C=C) and where there is no plane or axis of symmetry about the 

double bond. Terminal alkenes, those with the double bond at the end of a carbon chain, 

do not exhibit geometric isomerism.






173. Isomers 

Which of the following represent the same compound: 

a. A and B 

b. B and C 

c. A and C 

*d. B and D 



In naming compounds the numbering system can read from left to right or the reverse, 

but always such that the functional group is given the lowest locant possible. Thus, 

molecules B and D are both hex-1-ene.






174. Plane Of Symmetry 

Pick out the following compounds which do NOT possess a plane of symmetry: 

a. A 

b. B 

c. C 

*d. D 

e. E 


Larger saturated cyclic systems are not normally planar and may adopt several different 

conformations, such as in the boat and chair forms of cyclohexane. In problems such as 

this one, you can treat the ring as being planar on average, with the substituents either 

placed above or below this plane. Cis-isomers will have both substituents on the same 

side of the ring (either up or down), whereas, the trans-isomers will have will have both 

substituents on the opposite side of the ring.







Pick out the most stable isomer of hexa-2,4-diene? 

a. A 

b. B 

*c. C 

d. D 

e. E 


The trans-isomer of an alkene is, normally, more stable than the corresponding cisisomer. 

In such diene and higher polyene systems, the ‘all trans’ isomer will be similarly 

more stable than an isomer containing one, several or all cis bonds.






176. Functional Group Isomers 

Pick out the functional group isomer for ketone Z: 

a. A and C 

*b. B, D and E 

c. C and E 

d. A, B and C 

e. B only 


Functional group isomers are a subset of structural isomers and the term is used where 

the re-arrangement of atoms leads to a change in the functional group(s) present. Z is a 

ketone. 

A and C are ketones (positional isomers) but B is an aldehyde (change in functional 

group). Tautomers are a special subset of functional group isomers as they exist in 

equilibrium. Thus, D and E are enol tautomers of ketones B and A, respectively, but can 

be considered as functional group isomers of Z.






177. Functional Group Isomers 2 

Pick out the functional group isomer for enol Z: 

a. A and B 

b. C and E 

c. D 

d. A, B and C 

*e. A, B, D and E 


Functional group isomers are a subset of structural isomers where the re-arrangement of 

atoms leads to a change in the functional group(s). Tautomers are a further subset of 

functional group isomers as they can exist in equilibrium with another form (arrangement 

of atoms). Ketones, A and B, are in fact the same molecule, and are tautomers of enol Z. 

Enol C is a positional isomer of Z, whereas, and D and E are more standard functional 

group isomers of Z.






178. Functional Group Isomers 3 

Pick out the functional group isomer for carboxylic acid Z: 

a. A 

b. B 

c. C 

d. D 

*e. All of the above 


Functional group isomers are a subset of structural (constitutional) isomers where the rearrangement 

of atoms leads to a change in the functional group(s) present. All of the 

molecules above have the same molecular formula but only Z is a carboxylic acid, so 

molecules A-D are all functional group isomers of Z.






Conformation 

This database was generated in Respondus®®®® (http://www.respondus.com/) and can be 







179. Strain 

What type of strain influences the stability of small ring compounds? 

a. A 

b. B 

*c. C 

d. D 

e. E 


Deviation from the ideal tetrahedral angle of 109.5 degrees, which is found in 

cyclohexane, increases the ring strain for smaller cyclic molecules. In larger molecules, 

the conformational flexibility of these rings means that the ideal angle can also be 

achieved, but cross ring torsional strain comes into consideration.






180. Stable Conformers 

Which is the most stable conformation of cyclohexane? 

a. A 

b. B 

c. C 

*d. D 

e. E 


In the chair and boat conformation, the C-C-C bond angle is ideal tetrahedral 109.5 

degrees, but only in the chair are the C-H bonds all staggered. In the boat conformation 

some C-H bonds are eclipsed and so the torsional strain means the energy is higher than 

the chair conformation.






181. Stable Conformers 2 

Which of the following Newman projections is the most stable conformation of 1,2diphenylethane? 

a. A 

b. B 

*c. C 

d. D 



An all-staggered arrangement with the two bulky phenyl rings antiperiplanar is only 

found in C, which therefore is the most stable conformation.







The preferred conformation of cis-1-tert-butyl-3-methylcyclohexane is the one in which: 

a. A 

b. B 

c. C 

*d. D 

e. E 


Conformations in which most and /or bulkiest substituents are equatorial are the most 

stable conformation as there is more space than in the axial positions. In 1,3-disubstituted 

cyclohexanes, both groups can adopt equatorial positions if the groups are cis.







What structure represents the most stable conformation of cis-1,3-difluorocyclohexane? 

a. A 

*b. B 

c. C 

d. D 



In 1,3-disubstituted cyclohexanes both groups can adopt equatorial positions in the stable 

chair conformation if the groups are cis-. This is favoured as equatorial groups have more 

space than in the axial positions, where torsional strain factors come into consideration.







Which is the most stable conformation of cyclohexane? 

a. A 

*b. B 

c. C 

d. D 

e. E 


In the chair conformation B, the C-C-C bond angles are the ideal tetrahedral 109.5 

degrees, and all of the C-H bonds are staggered. In the boat conformation A, some C-H 

bonds are eclipsed and so the energy is higher. This eclipsing is also seen in C, D and E 

where also the bond angles are not ideal.







Pick out the conformation for butane which has the lowest potential energy: 

a. A 

*b. B 

c. C 

d. D 



An all-staggered arrangement in which the two bulkier methyl groups are antiperiplanar 

is found only in B. This therefore is the most stable conformation.






186. Conformers 

What structure represents the most stable conformation of syn-1,2-dibromocyclohexane? 

a. A 

b. B 

c. C 

d. D 

*e. A and B 


In syn 1,2-disubstituted cyclohexanes if one group adopts an equatorial position the other 

must be axial. In this case, as the substituents are the same, chairs A and B have the same 

energy. It does not matter which bromine is in the favoured equatorial position. Boat 

conformations C and D are less stable due to the eclipsing bromine atoms (as well as the 

less stable boat conformation).






187. Conformers 2 

Which conformation for ethane has the lowest potential energy? 

a. A 

b. B 

c. C 

d. D 

*e. E







Pick out the gauche conformation for butane: 

a. A 

b. B 

c. C 

d. D 

*e. A and C 


The term gauche conformation means that the substituents are staggered and synclinal to 

one another.







Pick out the chiral conformation(s) for butane: 

a. A and C 

b. B and D 

c. A, B and C 

d. A and D 

*e. A, C and D 


The conformations A, C and D do not possess a mirror plane, unlike B, and so are chiral 

conformations. However, conformers can in most cases rapidly interchange, here through 

rotation about the central C-C bond, and so naturally butane is not chiral overall.







Pick out the staggered conformation(s) for butane: 

a. A 

b. B 

c. C 

d. D 

*e. All of the above







Pick out the eclipsed conformation(s) for butane: 

a. A and C 

b. B and C 

c. A, B and D 


e. All of the above






192. Projections 

For the following linear structure Z, pick out its perspective projection: 

a. A 

*b. B 

c. C 

d. D 



D is like a partial perspective projection, in which the viewpoint is a side on to the C-C 

bond. It is referred to as a sawhorse projection and is an alternative to the Newman 

projection, C, which views the same C-C bond but end on.






193. Projections 2 


a. A 

*b. B 

c. C 

d. D 



A is bit like a partial perspective projection, in which the viewpoint is a side on to the C- 

C bond. It is referred to as a sawhorse projection and is an alternative to the Newman 

projection, D, which views the same C-C bond but end on.








a. A 

b. B 

c. C 

*d. D 



A is like a partial perspective projection, in which the viewpoint is a side on to the C-C 

bond. It is referred to as a sawhorse projection and is an alternative to the Newman 

projection, C, which views the same C-C bond but from one end.







For the following linear structure Z, pick out its Newman projection: 

a. A 

b. B 

*c. C 

d. D 









a. A 

b. B 

c. C 

*d. D 









a. A 

b. B 

*c. C 

d. D 








For the following linear structure Z, pick out its sawhorse projection: 

a. A 

b. B 

c. C 

*d. D 








For the following linear structure Z, pick out its sawhorse projection: 

*a. A 

b. B 

c. C 

d. D 









a. A 

b. B 

*c. C 

d. D 








For the following linear structure Z, pick out its skeletal representation: 

*a. A 

b. B 

c. C 

d. D 









a. A 

b. B 

*c. C 

d. D 









a. A 

*b. B 

c. C 

d. D 








For the following perspective projection Z, pick out its Newman projection (viewed 

along the C2-C3 bond from the right hand side): 

a. A 

*b. B 

c. C 

d. D 



Note the relative displacement of similar groups. The two methyls are antiperiplanar and 

the two chlorines are synclinal. It is then easy to spot B as the answer.








along the C2-C3 bond): 

a. A 

b. B 

c. C 

d. D 



Note the relative displacement of similar groups. The two methyls are synclinal and the 

two chlorines are antiperiplanar. It is then relatively easy to spot that none of the 

projections are correct.







For the following Newman projection Z, pick out its perspective projection: 

a. A 

b. B 

*c. C 

d. D 



Note the relative displacement of similar groups. The two methyls and the two hydroxyls 

are antiperiplanar. It is then relatively easy to spot that projection C is correct.







For the following Newman projection Z, pick out its perspective projection: 

a. A 

b. B 

c. C 

*d. D 



Note the relative displacement of similar groups. The two methyls and the two hydroxyls 

are synclinal. It is then relatively easy to spot that projection D is correct.








along the C2-C3 bond from the right hand side): 

a. A 

*b. B 

c. C 

d. D 



A tip here is to draw first the “Y shape” made by the display of the substituents on the 

nearer atom; add the substituents, then the circle and only lastly the partially hidden 

inverted Y, which should be in a staggered arrangement to the first Y.







What structure represents the most stable conformation of compound Z? 

a. A 

b. B 

*c. C 

d. D 



In A and D, the bulk tert-butyl group is axial, resulting in unfavourable interactions with 

the parallel CH bonds two atoms away (so called 1,3-diaxial interactions). These 

conformations are therefore much less stable than B, where the smaller hydroxyl group is 

axial. Molecule C is the most stable conformation as both substituents are equatorial.








a. A 

b. B 

c. C 

*d. D 



Molecule D has the most stable arrangement with both substituents equatorial. In 

molecules A, B or C, at least one methyl group must be axial which results in 

unfavourable interactions with the parallel CH bonds two atoms away (so called 1,3diaxial 

interactions).








*a. A and B 

b. B and C 

c. C and D 

d. D and A 



Molecule A (and B, which is simply the mirror image of A) has the more stable 

arrangement with diequatorial substituents. In molecule C (where D is its mirror image), 

the diaxial position of the groups results in unfavourable interactions with the parallel CH 

and C-R bonds two atoms away (so called 1,3-diaxial interactions).








*a. A and B 

b. B and C 

c. C and D 

d. D and A 



Molecule A (and B, which is simply the mirror image of A) has the more stable 

arrangement with diequatorial substituents. In molecule C (where D is its mirror image) 

the diaxial position of the groups results in unfavourable interactions with the parallel CH 

bonds two atoms away (so called 1,3-diaxial interactions).






Chirality 








213. Reaction Configuration 

Pick out the following reactions which involve racemisation: 

a. A 

*b. B 

c. C 

d. D 

e. none of the above 


Reaction A involves turning a chiral material into an achiral product, and reaction C 

involves turning an achiral material into a racemate. Reaction D is an example of a 

stereospecific process where one enantiomer of starting material leads to one enantiomer 

of product. Only in reaction B, is there conversion of a single enantiomer into a mixture 

of enantiomers (racemate); i.e., racemization has taken place.






214. Chiral 

Pick out the following molecules which can be made optically active: 

a. A and B 

b. B and C 

c. C and D 

d. A and C 

*e. A and D 


Both structures A and D could be made optically active as they contain a chiral centre. 

Aldehyde A is quite sensitive and is prone to racemisation via the enol tautomer.



215. Chiral 2 

Pick out which molecules are achiral: 

a. A and B 

b. B and C 

c. C and D 

d. A and C 

*e. B and D 





A is a single enantiomer, but C is a racemic mixture. However neither are achiral as they 

both have a chiral centre.



216. Chiral 3 

Pick out which molecules are achiral: 

a. A and B 

b. B and C 

c. C and D 

*d. A and C 

e. B and D 





A has an internal mirror plane and is superimposable in its mirror image. It is a mesocompound. 

Molecule C has no chiral centre, so is achiral.






217. Chiral 4 

Pick out the physical property that is different for a pair of enantiomers: 

a. A 

*b. B 

c. C 

d. D 

e. E 


The optical rotation is the only physical property that distinguishes enantiomers.






218. Chiral 5 

For a carbon centre to be chiral, it must contain an equal pair of side chains. 

a. True 

*b. False






219. Chiral 6 

For a carbon centre to be a chiral centre, it must be bonded to four dissimilar groups. 

*a. True 

b. False






220. Chiral 7 

A mixture of compound A ([α] 25 

D = +20.00) and its enantiomer B ([α] 25 

D = -20.00) has a 

specific rotation of +10.00. What is the composition? 

a. 25% A, 75% B 

b. 50% A, 50% B 

*c. 75% A, 25% B 

d. 50% A, 0% B 


The specific rotation of a 50:50 (racemic) mixture is an average of the single enantiomer 

specific rotations. As these are of the same magnitude but of opposite sign the value is 

zero. The specific rotation of a mixture has a linear relationship with the composition of 

that mixture, so if the two specific rotations of the pure enantiomers are available, the 

composition can be easily calculated.






221. Chiral 8 

The prefix, (R)-, before the name of a compound indicates that it is a racemic mixture. 

a. True 

*b. False






222. Chiral 9 

For an (S)-enantiomer; which of the following statements is true? 

a. A 

b. B 

c. C 

d. D 

*e. E






223. Chiral 10 

Pick out the (R)-configuration of the following molecules: 

*a. A, B and C 

b. B, C and D 

c. A, C and D 

d. none of them 

e. all of them 


In each case assign first the priority of the groups attached to the chiral centre based on 

the atomic number (S>O>N>C>H). If two atoms are the same, then the next level must 

be considered. If an atom is multiply bonded this is considered equivalent to the number 

of equivalent single bonds. If the lowest priority group happens to be pointing towards 

you, you can still easily work out the order of the higher priority groups, but remember to 

reverse your answer.







Which of the following molecules are achiral? 

a. A and B 

b. B and C 

c. D and E 

d. A and C 

*e. A and D 


Molecule A has an internal mirror plane and is superimposable on its mirror image. The 

same holds for molecule D, but you must consider the conformation where the two 

bromines are syn-periplanar to see this clearly. It is a meso-compound.







Which of the following molecules are chiral? 

a. A and B 

b. B and C 

c. D and E 

*d. A and C 

e. A and D 


Molecules, B and D have no chiral centres. In molecule D, there is a mirror plane in the 

ring intersecting the two oxygen atoms.







Pick out the following molecules which can be made optically active at room 

temperature: 

*a. A and B 

b. B and C 

c. C and D 

d. A and C 

e. A and D 


An atom bonded to three substituents in a pyramidal geometry and possessing one 

unshared electron pair may undergo an inversion of configuration. The energy barrier to 

this inversion may dictate whether such pyramidal structures, if chiral, can be resolved. 

Both structures C and D can undergo facile inversion of configuration at room 

temperature by changing the hybridization about carbon or nitrogen. The bonding orbitals 

about the central atom in this transition state are sp 2 (rather than sp 3 orbitals in the 

pyramidal structure) and the lone pair is in a p-orbital (the case of carbanion C is 

however further complicated by counterion effects). In the phosphine B, the C-P-C bond 

angles are only just above 90º indicating a large degree of p-character for the bonding 

orbitals, and therefore s-character for the lone pair. This means the inversion mechanism, 

in which the lone pair needs to be in a p-orbital, would involve more energy. Thus 

phosphines, such as B, need more energy to invert configuration and can as a result be 

resolved.







Which of the following molecules is optically active? 

*a. A 

b. B 

c. C 

d. D 



Molecule A is a single enantiomer, as indicated by the hatched bond, whereas, molecule 

C is a racemic mixture as indicated by the wavy bond.




Which of the following molecules is meso? 

a. A 

b. B 

c. C 

*d. D 






A meso-compound has chiral centres, but also has an internal mirror plane (or a point of 

symmetry) and is superimposable on its mirror image. Molecule D has an internal mirror 

plane as can be seen more readily in the conformation where the C-Br bonds are placed 

syn-periplanar.







Pick out the following molecules which are chiral: 

a. A and B 

b. B and C 

c. C and D 

*d. A and C 

e. B and D 


Molecular models are useful for answering these questions. The mirror images of 

molecules B and D can be superimposed upon the original by a combination of 

translation and rotation. This cannot be achieved with chiral molecules like A and C.







Pick out the following molecules which do not rotate the plane of plane polarized light: 

a. A + B 

b. B + C 

c. C + D 

d. D + A 

*e. All of these 


Rotation of plane polarized light is indicative of a chiral molecule. If not then the 

compounds must be meso or achiral. A is achiral - two different representations of the 

methyl ester group have been used. C has an internal mirror plane and B and D are 

superimposable on their mirror image.







What is the minimum number of chiral centres a meso-compound must have? 

a. 0 

b. 1 

*c. 2 

d. 3 

e. > 4






232. Epimers 

For molecule A, which molecules are diastereoisomers? 

a. A + B, and C + D 

b. A + C, and B + D 

c. A + D, and B + C 

d. A + B, and A + C 

*e. A + C, and A + D 


Diastereoisomers are non-superimpossible, non-mirror image stereoisomers. Molecules A 

and B are mirror images, and therefore are enantiomers, whereas, molecules C and D are 

not mirror images.






233. Epimers 2 

For molecule A, which molecules are epimers? 

a. A + B 

b. A + C 

c. A + D 

*d. A + B, and A + C 

e. A + C, and A + D 


An epimer is a stereoisomer which has two or more chiral (stereogenic) centres, but a 

different configuration at one chiral (stereogenic) centre. By comparing molecules A and 

B you can see that they have three common stereocentres, with only one which is of the 

opposite configuration. The same is true for molecules A and C. Therefore, molecule A is 

an epimer of molecule C, and molecule A is an epimer of molecule B, but molecule B is 

not an epimer of molecule C as there are two differences in configuration at chiral 

centres.



234. Stereoisomers 

How many stereoisomers can be drawn for: 

? 

a. 0 

b. 1 

c. 2 

*d. 3 

e. 4 





A molecule with n-chiral centres has a maximum of 2 n stereoisomers, except when there 

is a meso-form which will reduce this number. The given compound can have (R,R), (S,S) 

and (R,S) stereoisomers. The (S,R)-isomer is the same as (R,S); they are the meso-forms 

(superimposable mirror image forms).






235. Chiral Reaction 

Which of the following molecules are produced from the reaction below? 

a. A 

b. B 

c. C 

*d. A + B 

e. A + C 


Compounds A and B are formed by addition of H + to give the more stable secondary 

carbocation. This species is planar and so addition of the resulting bromide ion can occur 

from either face giving rise to an epimeric mixture at the new chiral centre. Compound C 

results from anti-Markovnikov addition of HBr, which is possible but only occurs in the 

presence of peroxides (via a radical mechanism).



236. Chiral Isomers 

The following two molecules are: 

*a. Identical 

b. Structural isomers 

c. Enantiomers 

d. Conformers 

e. Diastereoisomers 





Structural isomers have different bond connectivity for the same molecular formula. 

Enantiomers are non-superimposable mirror images of one another. Conformers are any 

one of an infinite number of 3D arrangements a molecule can adopt by rotation about a 

single bond. Diastereoisomers are pairs of stereoisomers which are non-superimposable 

non-mirror images. 

The given representations can be superimposed by simple translation / rotation so are 

identical.



237. Chiral Isomers 2 


a. identical 

b. structural isomers 

c. enantiomers 

d. conformers 

*e. diastereoisomers 










The given representations are non-superimposable non-mirror images stereoisomers.





a. non isomeric 



d. conformers 

*e. diastereoisomers 










The given representations are non-superimposable non-mirror images stereoisomers.





a. non isomeric 



*d. conformers 

e. diastereoisomers 










The given representations are conformeric.



240. Enantiomers 

Enantiomers are: 




a. molecules that have chiral carbon atoms 

b. molecules that have a mirror image 

*c. molecules that have a non-superimposable mirror image 

d. molecules that have a non-superimposable non-mirror image 

e. non-superimposable molecules






241. Enantiomers 2 

Which is the enantiomer of the following molecule? 

a. A 

b. B 

c. C 

d. D 

*e. E 


This molecule is achiral as it has two identical subsitutents (CH3) on the central carbon 

atom.







Which of the following is/are enantiomeric to the molecule shown? 

a. A 

b. B 

c. C 

d. A and C 

*e. B and D 


Enantiomers are non-superimposable mirror images of one another. They may however 

be presented from a variety of different perspectives / conformers. Representations A and 

C are simply different projections (conformers) of the same enantiomer.




Which pair of molecules is enantiomeric? 

*a. A + B, and C + D 

b. A + C, and B + D 

c. A + D, and B + C 

d. A + B, and A + C 

e. A + C, and A + D 





The enantiomer of any given molecule with multiple chiral centres can be identified 

without mental gymnastics if the configuration of each centre is known. 

For a pair to be enantiomeric, every chiral centre in one molecule must have the opposite 

configuration(s) in the second molecule. It is easy to see in the representation shown of 

molecule A that each stereocentre has the opposite configuration to the corresponding 

centre in B. The same can be seen in molecules C and D.




Diastereoisomers are: 




a. stereoisomers that have chiral carbon atoms 

b. stereoisomers that have a mirror image 

c. stereoisomers that have a non-superimposable mirror image 

d. stereoisomers that have a superimposable mirror image 

*e. stereoisomers that have a non-superimposable non-mirror image







Pick out the following molecules which are achiral and have no chiral centres: 

a. A 

b. B 

c. C 

d. D 








Pick out the following molecules which are chiral and have chiral centres: 

a. A and B 

b. B and C 

c. C and D 

*d. D and A 

e. all of the above 


Sulfoxides, and also the related sulfinic acids and esters, are chiral as the sulfur atom is in 

a tetrahedral environment to accommodate the lone pair of electrons which are easily 

identified from the Lewis structure.







Pick out the following molecules which are achiral but have chiral centres: 

a. A and B 

*b. B and C 

c. C and D 

d. D and A 

e. all of the above 


Meso-compounds have chiral centres but are superimposable on their mirror image. 

Molecule C has an internal mirror plane as seen in the conformation where the bromine 

atoms are synperiplanar. Molecule B has a centre of inversion, which is another identifier 

of meso-compounds; this is also the case for molecule C.




Assign the configuration of molecule A: 

a. (R)- 

*b. (S)- 

c. (E)- 

d. (Z)- 

e. (rac)- 





The hydroxyl is the highest priority group due to its atomic number of O being 8 

compared to 6 for carbon. Hydrogen is nearly always, but not exclusively, the lowest 

priority group and in this example is conveniently placed pointing away from the 

direction of view. The ethyl group is higher priority than the methyl due to the additional 

carbon atom.





a. (R)- 

*b. (S)- 

c. (E)- 

d. (Z)- 

e. (rac)- 





The C=O is higher priority than the CH2OH as the double bond is equivalent to two 

single bonds to oxygen.





a. (R)- 

*b. (S)- 

c. (E)- 

d. (Z)- 

e. (rac)- 





In acid chlorides it is easy to overlook that the Cl (atomic number = 17) bonded to a 

carbon is more important than the two bonds to oxygen.





*a. (R)- 

b. (S)- 

c. (E)- 

d. (Z)- 

e. (rac)- 





When applying the CIP rules, you need to know that the atomic number of F is 9.







Assign the configuration of the following molecule: 

a. (R)- 

*b. (S)- 

c. (E)- 

d. (Z)- 

e. (rac)- 


Applying the CIP rules to differentiate the priority of the alkyne and phenyl groups you 

will need to consider the third level of atom substitution.








a. (R)- and (Z)- 

b. (S)- and (Z)- 

c. (R)- and (E)- 

*d. (S)- and (E)- 

e. (rac)-







Which of the following molecules have the same conformation as molecule Z? 

a. A and B 

b. B and C 

c. C and D 

*d. A and C 

e. B and D 


In this type of question, it is important to note the relationship between the C-Cl bond and 

the furthest methyl group; in this case, molecule Z has an s-cis conformation about the 

C2-C3 C-C bond. 

Representations B and D can then be easily eliminated. Also representations having the 

same conformation need not necessarily be conformers (conformational isomers of the 

same molecule).







Which of the following molecules have the same configuration as molecule Z? 

a. A and B 

b. B and C 

*c. C and D 

d. A and C 

e. B and D 


You can either apply the CIP rules to work out the configuration in each case, or else 

apply a mixture of translation and bond rotation. It is quite easy to see that 

representations A and B are conformers of the one another, as are representations C and 

D. Once you can match or mismatch Z to one of these four then the answer is quickly 

revealed.







Which of the following molecules have the same configurations as molecule Z? 

a. A and B 

b. B and C 

c. C and D 

d. A and C 

*e. B and D 


It is quite easy to see that representations Z, B and D are trans-isomers about the C=C 

double bond, whereas, representations A and C are cis-isomers. Representation Z can be 

simply rotated clockwise by 120º to superimpose on D. It can also be rotated by 180º 

about a vertical axis to superimpose the chiral centre, followed by rotation about the C2- 

C3 bond to map the alkene fragment.







Which of the following molecules have the same conformation as molecule Z? 

a. A and B 

b. B and C 

c. C and D 

d. A, C and D 

*e. none of the above combinations 


In this type of question, it is important to note the relationship between the C-OH bond 

and the alkene fragment; in this case, molecule Z has an s-cis conformation about the C2- 

C3 C-C bond. 

Representations A and C can then be easily eliminated. Representation B is the s-trans 

form, whereas, only representation D is s-cis.







Which of the following molecules have the same conformation and configuration as 

molecule Z? 

a. A and B 

b. B and C 

c. C and D 

d. A, C and D 

*e. none of them 


After consideration of the configuration of the alkene C=C and the chiral centre, 

conformational isomers can be most easily identified by considering the relationship 

between the OH and double bond.







Predict the number of stereoisomers for molecule A: 

a. 0 

b. 1 

c. 2 

d. 3 

*e. 4 


A molecule with n-chiral centres has a maximum of 2 n stereoisomers, but watch out for 

meso-compounds (those with internal mirror planes or centres of inversion) which may 

reduce this number. Compound A can exist as four stereoisomers [i.e., (R,R), (S,S), (S,R) 

and (R,S)].








a. 0 

b. 1 

*c. 2 

d. 3 

e. 4 


This is an allene (C=C=C) in which the central atom is sp-hybridized and linear in 

arrangement. The end carbon atoms are sp 2 -hybridised and trigonal planar just like 

alkenes. Allenes can thus give rise to (R)- and (S)-isomers just like tetrahedral carbons. 

Molecule A is chiral.








a. 0 

b. 1 

c. 2 

d. 3 

*e. 4 


A molecule with n-chiral centres has a maximum of 2 n stereoisomers, but watch out for 


reduce this number. Compound A can exist as four stereoisomers [i.e., (R,R), (S,S), (S,R) 

and (R,S)] as the sulfur atom is tetrahedral, and therefore chiral due to the presence of a 

lone pair.








a. 0 

b. 1 

*c. 2 

d. 3 

e. 4 


A molecule with n chiral centres has a maximum of 2 n stereoisomers, but watch out for 


reduce this number. Compound A can exist as two stereoisomers [i.e., (R)- and (S)-] as it 

has only one chiral centre.







From the following molecules, pick out those with a plane of symmetry: 

a. A and B 

b. B and C 

c. C and D 

*d. A, B and C 

e. B, C and D 


Even though saturated cyclic systems are not normally planar they may adopt several 

different conformations, such as in the boat and chair forms of cyclohexane. In problems 

such as this one you can however treat the ring as being planar on average, but the 

substituents must be either placed either above or below this plane.







From the following molecules, pick out TWO with a centre of symmetry: 

a. A and B 

b. B and C 

c. C and D 

*d. A and C 

e. B and D 


The very centre of the ring is the centre of inversion in benzene C, and midway between 

the double bond C=C in (E)-alkene A.








a. (R)- and (Z)- 

*b. (S)- and (Z)- 

c. (R)- and (E)- 

d. (S)- and (E)- 

e. (rac)-







Pick out the following molecules which are chiral but have no chiral centres: 

a. A 

b. B 

c. C 

d. D 

*e. all of the above





a. (R)- 

b. (S)- 

*c. (E)- 

d. (Z)- 

e. (rac)- 








a. (R)- 

b. (S)- 

c. (E)- 

d. (Z)- 

*e. none of the above 








a. (R)- 

b. (S)- 

*c. (E)- 

d. (Z)- 






(E)- and (Z)-oximes, unlike alkenes, may interconvert reasonably easily and form an 

mixture at equilbrium. This interconversion occurs through the nitroso tautomer and not 

by breaking of the C=N pi-bond.





a. (R)- 

b. (S)- 

c. (E)- 

d. (Z)- 

*e. none of the above 








a. (R)- 

b. (S)- 

*c. (E)- 

d. (Z)- 











Which of the following molecules is an alternative projection of both conformation and 

configuration of that shown in Z? 

a. A 

b. B 

*c. C 

d. D 

e. All are the same







What type of isomers are present in the molecule Z: 

a. A 

*b. B 

c. C 

d. D 

e. E







From the following molecules, pick out those with a plane of symmetry: 

a. A and B 

b. B and C 

c. C and D 

d. None of them 



Even though saturated cyclic systems are not normally planar they may adopt several 

different conformations, such as in the boat and chair forms of cyclohexane. In problems, 

such as this one, you can however treat the ring as being planar on average, but the 

substituents must always be placed either above or below this plane. All of these 

examples therefore have at least one plane of symmetry.






275. Isomers 2 

How are the following molecules, CH3CH3CHClCH3 and CH3CH2CH2CH2Cl related? 

a. enantiomers 

b. non-superimposable mirror images 

c. diastereoisomers 

*d. constitutional isomers 

e. not isomeric






276. Isomers 2 2 

How are the following molecules, (Z)-CHBr=CBrCHBrCH3 and (E)- 

CHBr=CBrCH2CH2Br related? 

a. enantiomers 


c. diastereoisomers 

*d. constitutional isomers 

e. not isomeric







How are the following molecules, (2R,3R)-CH3CHBrCHBrCH3 and (2S,3R)- 

CH3CHBrCHBrCH3 related? 

a. structural isomers 


*c. diastereoisomers 

d. enantiomers 

e. none of the above




A meso-compound is defined as: 

a. A 

b. B 

c. C 

d. D 

*e. E 







The definition of a diastereoisomer is: 

a. A 

b. B 

c. C 

d. D 

*e. E 







The definition of an enantiomer is: 

a. A 

b. B 

*c. C 

d. D 

e. E 







For a molecule to be chiral it must have: 

a. A 

b. B 

*c. C 

d. D 

e. E 









282. Newman 

Which of the following representations is a Newman projection of ethane? 

*a. A 

b. B 

c. C 

d. D 

e. E






283. Cahn-Ingold-Prelog 

Use the Cahn-Ingold-Prelog (CIP) rules to arrange the following groups in order of 

decreasing priority (highest priority at the left): 

a. A > B > C > D 

b. D > C > B > A 

c. A > C > B > D 

*d. D > B > C > A 

e. A = B = C = D 


To assign priority in the CIP rules, you need to identify the atomic number of each atom 

bonded directly to the chiral centre. The highest atomic number takes highest priority. In 

cases which are not clear cut the number of bonds to the highest priority atom identified 

needs to considered. In the case of multiple bonds, this is considered as equivalent to the 

same number of single bonds to the atom concerned. The carbon in carboxylic acid D is 

bonded three times to oxygen, in acetal B and aldehyde C only twice, and in alcohol A 

only once. However, acetal B is bonded to two oxygen atoms (via two C-O sigma bonds), 

whereas, aldehyde C is only bonded to one oxygen atom (in the C=O bond) and therefore 

has higher priority.






284. Cahn-Ingold-Prelog 2 



a. A > B > C > D 

*b. D > C > B > A 

c. A > C > B > D 

d. D > B > C > A 

e. A = B = C = D 


To assign priority in the CIP rules, you need to identify the atomic number of each atom 

bonded directly to the chiral centre. The highest atomic number takes highest priority. In 

cases which are not clear cut the number of bonds to the highest priority atom identified 

needs to considered. In the case of multiple bonds, this is considered as equivalent to the 

same number of single bonds to the atom concerned. 

A has lowest priority (fourth) due to the nitrogen atom being substituted with H only. The 

nitrogen atoms in B, C and D are all bonded to carbon, but at the next level B can be 

placed third. To distinguish C and D, note that the carbon is bonded to oxygen in D 

making this the higher priority and take precedence over C in which the third level of 

substitution is with carbon atoms.






285. Cahn-Ingold-Prelog 3 



a. A > B > C > D 

b. D > C > B > A 

*c. A > C > B > D 

d. D > B > C > A 

e. A = B = C = D 


To assign priority in the CIP rules you need to identify the atomic mass of each atom 

bonded directly to the chiral centre. A has highest priority as bromine is the heaviest of 

the first level of substituents. Following this S is heavier than O which is in turn heavier 

lighter than D, the lowest priority group.






Acids and Bases 








286. Strongest Acid 

Which of the following organic compounds is the strongest acid? 

a. A, KA = 10 -45 

b. B, KA = 10 -42 

c. C, KA = 10 -18 

*d. D, KA = 1 

e. They all have the same acidity. 


In an acid base equilibrium, AH A - + H + , the further the equilibrium lies to the right 

the stronger the acid. This gives largest equilibrium constant, KA, which can be calculated 

from the equation KA = ([H + ]×[A - ])/[HA]. Thus D is the strongest acid as the KA is the 

largest. 

KEYWORD: CONJUGATION






287. Strongest Acid 2 

Which of the following alcohols would you expect to be the strongest acid? 

*a. Cl3COH 

b. Cl2CHOH 

c. ClCH2OH 

d. CH3OH 

e. Alcohols have equal acidity. 


The strongest acid is found in the molecule with the most stable conjugate base. For this 

particular case, the trichloromethoxide is the more stable conjugate base by virtue of the 

three electron withdrawing chlorine atoms. 

KEYWORD: INDUCTIVE







Which of the following would have the largest KA? 

a. A 

b. B 

c. C 

d. D 

*e. E 


In an acid base equilibrium, AH A- + H+, the largest KA value is associated with the 

strongest acid where the equilibrium lies furthest to the right. The strongest acid is 

associated with the most stable conjugate base. Phenol E is most acidic due to the 

electron withdrawing para-nitro group (note -I and –M effects) which can stabilize the 

resulting oxyanion through resonance delocalization. 

KEYWORD: RESONANCE




Pick out the molecule which is most acidic: 

*a. A 

b. B 

c. C 

d. D 

e. E 





The strongest acid is associated with the most stable conjugate base. Carboxylic acid A is 

most acidic due to the electronegative chloro-substituent stabilizing the resulting 

carboxylate anion through inductive effects. Note that fluorinated B is an aldehyde and 

not a carboxylic acid! 

KEYWORD: INDUCTIVE




Pick out the molecule which is most acidic: 

a. A 

b. B 

c. C 

d. D 

*e. E 





The strongest acid is associated with the most stable conjugate base. Phenol E is most 

acidic due to the stabilization of the carboxylate anion through the inductive effect of 

three electron withdrawing (-I) chloro-substituents on the aromatic ring. 

KEYWORD: INDUCTIVE







Which of the following acids have the largest pKA? 

*a. A 

b. B 

c. C 

d. D 

e. E 


In an acid base equilibrium, AH A- + H+, the largest pKA value is associated with the 

weakest acid. The unhalogenated carboxylic acid A is the weakest acid. In the other 

cases, the electron withdrawing (-I) halogen substituent stabilize the carboxylate 

conjugate base making the parent acid more acidic. 

KEYWORD: INDUCTIVE







Which of the following compounds is the strongest acid? 

a. A 

b. B 

c. C 

*d. D 

e. E 


Carboxylic acid D is a stronger acid than alcohol A, esters C and E and alkane B. The 

resulting conjugate base, a carboxylate anion (for D), can be readily stabilized through 

delocalization (resonance) of the negative charge between both electronegative oxygen 

atoms. 

KEYWORDS: RESONANCE VERSUS INDUCTIVE




Pick out the strongest acid: 

a. A 

b. B 

*c. C 

d. D 

e. E 





Simple hydrocarbons are, in relation to other organic compounds, weakly acidic. 

However, carbanions can be formed and their stability depends heavily on the 

hybridization state of the orbital containing the lone pair of electrons. Alkyne C is most 

acidic as the lone pair of the conjugate base is held in a sp-orbital, rather than sp 3 (for A) 

or sp 2 orbital (for B, D and E). By having a greater proportion of “s character”, the 

electron density in an sp-orbital is held closer to the positively charged nucleus and 

therefore is electrostatically more stable. 

KEYWORD: HYBRIDIZATION




Pick out the strongest acid: 

a. A 

b. B 

c. C 

*d. D 

e. E 





Molecule C does not contain any acidic protons, and the mono anions B and E are more 

difficult to ionize than neutral A and D due to their existing negative charge. In maleic 

acid, D, the cis-configuration means that there is intramolecular hydrogen bonding which 

is not possible in its trans-isomer, fumaric acid A. This H-bonding means that the 

ionization of the OH bond is more facile as the intramolecular H-bond stabilizes the 

incipient carboxylate. 

KEYWORDS: HYDROGEN BONDING




Which is more acidic, A or B? 

*a. A 

b. B 





Deprotonation in molecule A, between both carbonyl groups, leads to a carbanion where 

the pair of electrons can be delocalized over both carbonyl groups. This greater degree of 

delocalization means the conjugate base of A is more stable compared to that of B which 

has the charge delocalized to a less extent. Molecule A is therefore more acidic. 

KEYWORD: RESONANCE





*a. A 

b. B 


KEYWORD: RESONANCE 








a. A 

*b. B 










a. A 

*b. B 





KEYWORDS: RESONANCE VERSUS CONFORMATION





*a. A 

b. B 





KEYWORDS: RESONANCE VERSUS HETEROATOM





a. A 

*b. B 





KEYWORDS: HETEROATOM AND PERIODICITY





a. A 

*b. B 










a. A 

*b. B 










*a. A 

b. B 





Deprotonation of carbocation A occurs on the CH2 position alpha to the carbon bearing a 

positive charge. The resulting conjugate base is benzene. For carbocation B, 

deprotonation leads to the less stable conjugate base, cyclohexadiene. 

The conjugate base derived from A is more stable that that from B due to the stability 

gained through being aromatic and therefore the parent acid A is more acidic. 

KEYWORDS: AROMATICITY VERSUS RESONANCE





*a. A 

b. B 


KEYWORDS: CHARGE EFFECTS 








a. A 

*b. B 


KEYWORDS: RESONANCE 








a. A 

*b. B 





KEYWORDS: HETEROATOM, PERIODICITY AND RESONANCE





a. A 

*b. B 





KEYWORDS: HETEROATOM, PERIODICITY AND RESONANCE





*a. A 

b. B 










*a. A 

b. B 










a. A 

*b. B 


KEYWORDS: CHARGE EFFECTS 








*a. A 

b. B 










*a. A 

b. B 











313. Strongest Conjugate Base 

Which of the following organic compounds would have the strongest conjugate base? 

*a. A, KA = 10 -45 

b. B, KA = 10 -42 

c. C, KA = 10 -18 

d. D, KA = 1 

e. They would all produce equally strong conjugate bases 


In an acid base equilibrium, AH A - + H + , the weakest acid is associated with the least 

stable (i.e., strongest) conjugate base. The weakest acid has the smallest KA value and is 

least dissociated. Molecule A is least dissociated and has the strongest conjugate base. 

KEYWORDS: HETEROATOM VERSUS RESONANCE






314. Strongest Conjugate Base 2 

Which of the following organic compounds would have the strongest conjugate base? 

a. A, KA = 1.76 x 10 -5 

b. B, KA = 5140 x 10 -5 

*c. C, KA = 10 -18 

d. D, KA = 90000 x 10 -5 

e. They would all produce equally strong conjugate bases 


In an acid base equilibrium, AH A - 

+ H + 

, the weakest acid is associated with the least 

stable (i.e. strongest) conjugate base. The weakest acid has the smallest KA value and is 

least dissociated. Molecule C is least dissociated and has the strongest conjugate base. 

Alternatively, by looking at the structure of the molecules; you should immediately note 

that A, B and D are carboxylic acids, whereas, C is an alcohol. 

KEYWORDS: HETEROATOM VERSUS RESONANCE






315. Acidity 

Which of the following correctly lists the compounds in order of decreasing acidity? 

a. A > B > C > D 

*b. A > B > D > C 

c. B > D > A > C 

d. C > A > D > C 

e. C > D > B > A 


The lone pairs of the conjugate bases of molecules A, C and D are all held in related sp 3 

hybrid orbitals. Consequently, the electronegativity of the atom bearing the negative 

charge tells us about their relative stability (and by inference the acidity of the acid). 

Taking this into consideration, the relative acidic of these molecules are A > D > C. 

Inspection of the answers, this leaves only options (b) and (d) as the correct answers. 

Molecule B generates a carbanion where the lone pair is held in an sp-orbital. The 

electrons are somewhat stabilized by being held closer to the nucleus, but not as much as 

if being located on the electronegative oxygen (in A) and so (b) must be the answer by 

the process of elimination. 

KEYWORDS: HETEROATOM AND HYBRIDIZATION






316. Acidity 2 

Which of the following statements is true when ethane, ethene and ethyne are compared 

with one another? 

a. A 

b. B 

c. C 

*d. D 

e. E 


The stability of carbanions derived from simple hydrocarbons depends on the 

hybridization state of the orbital containing the lone pair of electrons. 

Alkynes are most acidic as the lone pair of the conjugate base is held in a sp orbital, 

rather than sp 2 (alkene) or sp 3 (alkane). Having a greater proportion “s character” the 

electron density in an sp-orbital is held closer to the positively charged nucleus and so is 

more stable. Therefore ethyne is the strongest acid and also has the shortest C-H bond 

length. 

KEYWORD: HYBRIDIZATION







Cyclopentadiene is unusually acidic for a hydrocarbon. Select the explanation: 

a. A 

*b. B 

c. C 

d. D 

e. E 


In an acid base equilibrium, AH A - + H + , a strong acid is associated with the stable 

conjugate base. The conjugate base of cyclopentadiene, the cyclopentadienyl anion, is a 

planar, cyclic continuously conjugated system containing six pi-electrons which means it 

is stabilized through being aromatic. Statements A, C and D are clearly incorrect, but 

only statement B explains why cyclopentadiene is unusually acidic. 








Which of the following would have the largest pKA? 

a. A 

b. B 

*c. C 

d. D 

e. E 


In an acid base equilibrium, AH A - + H + , the largest pKA value is associated with the 

weakest acid, where the equilibrium lies furthest to the left. The weakest acid is clearly 

associated with the least stable conjugate base. 

Phenol C is the least acidic due to its electron donating (+I) para-methyl group which 

destabilizes the delocalized oxyanion in contrast to the electron withdrawing F, Cl and 

NO2 groups present in B, D and E respectively. 

KEYWORDS: INDUCTIVE VERSUS RESONANCE







Which of the following compounds are more acidic than ethanol? 

*a. A 

b. B 

c. C 

d. A and B only 

e. A and C only 


Enolates of simple monocarbonyl species are generally less stable than alkoxides. 

Within this context, ethanol is more acidic than ketone B and ester C. However, when 

ethyl acetoacetate A is deprotonated, the resulting negative charge (and pair of electrons) 

is located between and delocalized over both carbonyl groups. This greater degree of 

delocalization (resonance) ensures the conjugate base of A is relatively stable. Ethyl 

acetoacetate A is more acidic than ethanol. 

KEYWORD: RESONANCE







The pKA of phenol in water is 10. The pKA of phenol in dichloromethane is less than 10. 

Is this statement true or false? 

a. True 

*b. False 


The acidity of phenol in a given solvent is related to the stability of the conjugate base, 

i.e., the phenate ion. 

In relative terms, the greater the stability of the phenate ion, the more acidic phenol will 

be. Due to hydrogen bonding, the phenate ion is more stable in water than 

dichloromethane and so phenol is more acidic (smaller pKA) in water than 

dichloromethane. 

KEYWORD: SOLVATION







The pKA of acetic acid in water is 5. The pKA of acetic acid in hexane is greater than 5. Is 

this statement true or false? 

*a. True 

b. False 


The acidity of acetic acid in a given solvent is related to the stability of the conjugate 

base, i.e., the acetate ion. 

In relative terms, the greater the stability the of acetate ion, the more acidic acetic acid 

will be. Due to hydrogen bonding, the acetate ion is more stable in water than hexane, 

and so acetic acid is less acidic (larger pKA) in hexane than water. 

KEYWORD: SOLVATION







An aqueous solution of 4-nitrophenol of concentration, 0.1 mol/dm 3 , has a pH = 4.1. 

Estimate the pKA of this phenol. 

a. 5.2 

b. 6.2 

*c. 7.2 

d. 8.2 

e. 9.2 


In the ionization of acid HA to give H + and A - 

, the equilibrium constant KA is defined KA 

= ([H + ]×[A - ])/[HA]. As pH = -log[H + ] this value reveals both the [H + ] and also the [A - ] 

concentration which must be equal to [H + ]. Thus the pKA (=-log10KA) can be calculated. 

KEYWORD: EQUILIBRIA







For the following molecules, order them with decreasing acidity (the most acidic to the 

left). 

a. A > B > C > D 

b. D > C > B > A 

c. B > A > D > C 

*d. D > B > C > A 

e. A = B = C = D 


The strength of the H-X bond is weakest when X is I as the 5sp 3 orbital on iodine is very 

different in size and shape to the 1s orbital of the hydrogen atom. Overlap is therefore 

poor, and so the H-I bond is very weak. Therefore, HI is the most acidic. The acidity of 

the others then follows a simple periodic group trend based on the same argument. 

Interestingly, the most acidic molecule contains the least electronegative halogen. 









left). 

a. A > B > C > D 

b. D > C > B > A 

c. C > D > A > B 

d. B > D > A > C 

*e. A > C > B > D 


The strength of the H-XH bond is weakest when X is Te as the 5sp 3 orbital on tellurium 

is very different in size and shape to the 1s orbital of the hydrogen atom. Overlap is 

therefore poor, and so the H-Te bond is very weak. Therefore, H-TeH is the most acidic. 

The acidity of the others then follows a simple periodic group trend based on the same 

argument. Interestingly, the most acidic molecule contains the least electronegative 

chalcogen. 









left). 

a. A > B > C > D 

b. D > C > B > A 

c. C > D > A > B 

*d. B > D > A > C 

e. A > C > B > D 


In this particular case, the stability of the conjugate base (and therefore the acidity of the 

acid) increases with the electronegativity of the central atom. However, the strength of 

the H-X bond can be deceiving as it increases with an increase with electronegativity. 

H-F is most acidic due to the stability of the fluoride ion through electronegativity and 

solvation effects. The acidity of the others then follows a simple periodic group trend 

based on the same argument. 





Which hydrogen is most acidic? 

a. A 

b. B 

c. C 

d. D 

*e. E 





In order to answer such questions, the stability of each resulting conjugate base should be 

considered; the more stable this is, the stronger the parent acid. 

For enolates, the greater the degree of delocalization of charge, (i.e., the more resonance 

structures that can be drawn), the stronger the acid. Therefore, proton C is more acidic 

than B. However, the carboxylate arising from loss of E is the most stable as the charge is 

shared between two electronegative oxygen atoms (of the CO2- group). Proton E is the 

strongest acid. 






a. A 

b. B 

*c. C 

d. D 

e. E 







For enolates, the greater the degree of delocalization of charge (i.e., the more resonance 

structures that can be drawn) the stronger the acid. Proton A and C can be shared over 

five atoms each, but in C, two of these atoms are the more electronegative oxygen 

compared to just one for A. Proton C is therefore the most acidic. 






a. A 

*b. B 

c. C 

d. D 

e. E 







For enolates, the greater the degree of delocalization of charge (i.e. the more resonance 

structures that can be drawn) the stronger the acid. Proton HB is most acidic due to the 

charge (and lone pair of electrons) on the conjugate base being delocalized over both 

carbonyl groups and the benzene ring. 









left). 

a. A > B > C > D > E 

*b. D > A > B > E > C 

c. E > B > C > A > D 

d. D > A > B > C > E 

e. B > C > D > E > A 


C is clearly the least acidic since it does not contain any obviously acidic protons. Mono 

anions B and E are more difficult to ionize than neutral dicarboxylic acids A and D due 

to their existing negative charge. 

For maleic acid, D, the cis-configuration promotes efficient intramolecular hydrogen 

bonding which is not possible for its trans-isomeric form, fumaric acid A. 

D is more acidic than A and, by similar arguments, B is more acidic than E as the second 

acidic proton is tightly held by H-bonding. 

KEYWORDS: INTRAMOLECULAR HYDROGEN BONDING



330. Basicity 

Pick out the strongest base: 

a. A 

b. B 

*c. C 

d. D 

e. E 





Basicity can be influenced by the electronegativity and the nature of the orbital the 

electron pair resides in. 

For this particular case, the stability of the conjugate base (and therefore the acidity of the 

acid) increases with an increase in the electronegativity of the central atom. However, the 

strength of the H-X bond can be deceiving as it increases with an increase with 

electronegativity. 

The lone pair of electrons in A, C and D are all held in sp 3 orbitals (and therefore further 

away from the positive nucleus) and are less stable than if they were in an sp 2 or sporbital. 

In C, the atom bearing the charge is carbon, which is less able to stabilize the 

charge compared to the more electronegative oxygen (A) and nitrogen (D). Molecule C is 

the strongest base. 

KEYWORDS: HYBRIDIZATION VERSUS HETEROATOM






331. Basicity 2 

For the following molecules in the GAS PHASE, order them with decreasing basicity 

(the most basic to the left). 

a. A > B > C > D 

b. B > C > D > A 

c. C > D > A > B 

*d. A > C > B > D 

e. B > D > A > C 


Basicity is greater if the conjugate acid is stabilized. 

Protonation of amines leads to ammonium ions. The positive charges on the nitrogen 

atom in the conjugate acids are increasingly stabilized with increasing alkyl substitution 

due to their +I inductive effect. 

Note: in the gas phase there are no solvation issues to be considered. 

KEYWORDS: INDUCTIVE EFFECTS






332. Basicity 3 

For the following molecules in WATER, order them with decreasing basicity (the most 

basic to the left). 

a. A > B > C > D 

b. B > C > D > A 

c. C > D > A > B 

d. A > D > B > C 

*e. C > A > B > D 


Basicity is greater if the conjugate acid is stabilized. 

Protonation of amines leads to ammonium ions. The positive charges on the nitrogen 

atom in the conjugate acids are increasingly stabilized with increasing alkyl substitution 

due to their +I inductive effect. 

This would imply the relative basicity (and the answer) would be: A > C > B > D. 

However, in water, solvation through intermolecular H- bonding is possible. Amine C 

becomes more basic than A due to the resulting dimethylammonium ion being able to 

form two hydrogen bonds to water, whereas, the related trimethylammonium ion can only 

form one. 

KEYWORD: SOLVATION



333. Strongest Base 


a. A 

*b. B 

c. C 

d. D 

e. E 





In these examples, the strongest base can be identified by considering the delocalization, 

or not, of the nitrogen lone pair of the amino group. In all examples, except for the B, the 

lone pair is stabilized by conjugation with an unsaturated system - the aromatic rings in 

molecules C, D and E, and the carbonyl group in A. 

KEYWORD: RESONANCE



334. Strongest Base 2 


a. A 

*b. B 

c. C 

d. D 

e. E 





Compound B is the strongest base due to the charged hydroxide counter anion. All other 

compounds A, C and D are neutral amine bases, except for amide E, which is not basic as 

the lone pair is conjugated to the carbonyl group. 

KEYWORD: HETEROATOM





a. A 

b. B 

*c. C 

d. D 

e. E 





Carboxylates B and E are stabilized through inter- and intramolecular H-bonding 

respectively. The dicarboxylate C (which is coincidentally the conjugate acid of B), is 

therefore the strongest base due to charge repulsion. 

KEYWORDS: CHARGE EFFECTS VERSUS HYDROGEN BONDING




Which is more basic, A or B? 

a. A 

*b. B 


KEYWORD: HYBRIDIZATION 








a. A 

*b. B 










*a. A 

b. B 










a. A 

*b. B 










a. A 

*b. B 


KEYWORD: INDUCTIVE 








a. A 

*b. B 





KEYWORDS: CONFORMATION VERSUS RESONANCE





*a. A 

b. B 










*a. A 

b. B 












In aqueous solution, which is more basic, A or B? 

*a. A 

b. B 


KEYWORD: SOLVATION





a. A 

*b. B 





KEYWORDS: HYBRIDIZATION AND LEWIS STRUCTURES





*a. A 

b. B 





KEYWORD: RESONANCE VERSUS CONFORMATION





a. A 

*b. B 





KEYWORD: HYBRIDIZATION VERSUS AROMATICITY





*a. A 

b. B 





KEYWORD: HYBRIDIZATION VERSUS AROMATICITY






349. Weakest Base 

Which compound would you expect to be the weakest base? 

a. A 

b. B 

c. C 

*d. D 

e. E 


The lone pair on nitrogen in pyrrole, D, is delocalized into the conjugated system and 

therefore contributes to pyrrole’s aromaticity. 

This lone pair is not available for protonation, and clearly pyrrole, D, is the weakest base. 

Note: in pyridine, B, the lone pair is in a sp 2 orbital perpendicular to the ring system and 

is therefore not delocalized (or part of pyridine’s aromatic system). 

KEYWORDS: AROMATICITY VERSUS HYBRIDIZATION



350. Weakest Base 2 

Which is less basic, A or B? 

*a. A 

b. B 











351. pH 

The pKA of acetic acid is 5. At what pH is the concentration of acetic acid equal to that of 

its conjugate base, acetate? 

a. 3 

*b. 5 

c. 7 

d. 9 

e. 11 


The Henderson-Hasselbalch equation linking the pH and pKA gives pH = pKA + log([A - 

]/[AH]). When [A - ]=[HA] the second term in the equation becomes zero, and so the pH = 

pKA 

KEYWORDS: HENDERSON-HASSELBALCH EQUATION






352. pH 2 

The pKA of acetic acid is 5. What is the pH at which the concentration of acetic acid will 

be 10 3 that of its conjugate base, acetate? 

*a. 2 

b. 3 

c. 5 

d. 7 

e. 8 



]/[AH]). When [A - ]/[HA] =10 -3 the second term on the right is -3, and so the pH = pKA - 

3 = 2. 







353. pH 3 

The pKA of acetic acid is 5. What is the pH at which the concentration of acetic acid will 

be 10 -6 that of its conjugate base, acetate? 

a. -1 

b. 1 

c. 7 

*d. 11 

e. 12 



]/[AH]). When [A - ]/[HA] =10 6 the second term on the right is +6, and so the pH = pKA + 

6 = 11 







354. pKa 

If an acid, HA, has a pKA of 5 in water. What is the pKB of its conjugate base? 

a. 5 

b. 7 

*c. 9 

d. 11 



In water, the pKA (acid) + pKB (conjugate base) = 14. 

So simple maths gives the answer as C (14 - 5 = 9). 

KEYWORD: HENDERSON-HASSELBALCH EQUATION






Chemistry of Aromatic Molecules 









Which of the following molecules would you expect to have resonance stabilization? 

a. A 

b. B 

c. C 

*d. All of the above 



All the molecules have conjugated systems in which electron density can be delocalized 

across the pi-bonds. This provides the molecule with extra stabilization, called resonance 

stabilization, compared to similar examples where there is no conjugation. All these 

molecules are aromatic.






356. Resonance 2 2 

Which of the following structures would NOT contribute to the resonance hybrid of a 

benzyl cation? 

a. A 

b. B 

*c. C 

d. D 

e. They all contribute equally. 


All canonical (resonance) forms must be valid Lewis structures and only the position of 

the pi- or non-bonded electrons can be changed – atoms and sigma-bonds cannot move. 

Canonical forms can be deduced, starting from one structure, moving sequentially a 

series of curly arrows to indicate the delocalization of electrons.







Pick out the major contributing form of the resonance hybrid for the carbocation shown: 

a. A 

b. B 

c. C 

*d. D 



The positive inductive (+I) methyl group means more electron density is pushed towards 

the ring carbon bearing the charge in canonical C compared to those in canonicals A, B 

and D.







Which of the following do NOT contribute to the resonance hybrid of the intermediate 

formed from para-chlorination of bromobenzene: 

a. A 

b. B 

c. C 

d. D 



Upon chlorination the initial charge is found ortho to the newly introduced group (in 

structures A or D). All subsequent canonical (resonance) forms can be deduced using 

curly arrows to indicate the delocalization of electrons. 

Note that in all cases the new position of the charge is two bonds away (on the third 

atom) from where the charge was originally placed.







How many different canonical forms can be drawn for the imidazolate anion? 

a. 2 

b. 3 

c. 4 

*d. 5 

e. 6 


All canonical (resonance) forms can be deduced using curly arrows to indicate the 

delocalization of electrons. 

Note, that in all cases, the new position of the charge is two bonds away (on the third 

atom) from where the charge was originally placed. Counting round the ring, you will 

find that the charge is located back on the original nitrogen (N) atom after four other 

canonical forms.







Which of the following molecules would you expect to have the largest resonance 

energy? 

a. A 

*b. B 

c. C 

d. D 

e. E 


In general, the greater the degree of electron delocalization, the greater stability a 

particular molecule has. In all these cases, there are six pi-electrons in cyclic systems, but 

only in molecule B is the molecule entirely symmetrical meaning that the electron density 

is evenly spread out. In the other cases, the electron density will be situated to a greater 

extent on the more electronegative heteroatom - i.e., more localization of charge.






361. Orbitals 

In the molecular orbital model of benzene, the six atomic p-orbitals combine to form how 

many molecular orbitals? 

a. 1 

b. 3 

c. 4 

*d. 6 

e. 12 


n Atomic orbitals combine to form n-molecular orbitals. These will be a mixture of 

bonding and anti-bonding orbitals (non-bonding orbitals in some cases), but in most cases 

only the bonding or non-bonding orbitals will be filled with electrons.






362. Orbitals 2 

How many molecular orbitals are filled with pi-electrons for benzene? 

a. 1 

*b. 3 

c. 4 

d. 6 

e. 12 


n Atomic orbitals combine to form n molecular orbitals (MOs). Within benzene this will 

mean six MOs are formed, comprising of three bonding and three anti-bonding MOs. 

Each p orbital contributes one electron, so the 3 electron pairs go into the three lower 

energy bonding orbitals.







Pick out the following compound which is not aromatic: 

a. A 

b. B 

*c. C 

d. D 

e. E 


In all cases, except for borocycle C, there are six electrons in a cyclic, planar, 

continuously conjugated pi-system. In pyrrole B, thiophene D and furan E, the 

heteroatom can provide an electron pair in a p-orbital to this total, but Group 2, boron (B) 

cannot. It therefore only has four pi-electrons, and therefore is anti-aromatic.






364. Aromaticity 2 2 

Which of the following statements regarding the cyclopentadiene radical is correct? 

a. It is aromatic 

b. It obeys Hückel's rule 

c. It undergoes reactions characteristic of benzene 

d. It has a closed shell of 4 pi-electrons 

*e. None of the above statements are correct 


As this molecule does not have 4n+2 pi electrons (it has 5 pi electrons, two from double 

bonds and one single electron), it cannot comply with Hückel’s rule for aromaticity. This 

means it does not behave like benzene, but rather like a conjugated radical, such as the 

allyl radical.







For the following canonical structures, pick out the most stable form: 

*a. A 

b. B 

c. C 

d. D 

e. E 


Structures A, B and E are resonance forms of the molecule shown. Structure C is a 

invalid Lewis structure and structure D is an carbocation which is not stabilized by 

delocalization (it is in an sp 2 orbital perpendicular to the pi-system). Only in structure A 

is the aromatic ring of benzene intact; in structures B and E, the aromaticity is lost when 

the continuous cyclic conjugation is disrupted. This means the lower energy resonance 

form A is the major contributor to the hybrid.




Pick out the molecules which are aromatic: 

a. A 

b. B 

c. C 

d. All of these 

*e. None of these 





To be aromatic a molecule must possess 4n+2 pi-electrons in a planar, continuously 

conjugated cyclic system. Structures B and C are therefore ruled out. In molecule A, the 

trans-double bond means that the ring can only adopt a puckered conformation (due to 

the C-H clashes), so disrupting the planarity. In addition the double bonds are not 

continuously conjugated and so A is not aromatic either.







Pick out the molecules which are anti-aromatic: 

a. A 

b. B 

c. C 

d. All of these 

*e. None of these 


To be anti-aromatic, a molecule must possess 4n pi-electrons in a planar, continuously 

(cyclic) conjugated system. In structures A and B, the electron count is wrong, whereas, 

molecule C does not have a planar structure. The internal angle at each vertex of a 

regular octagon is 135º which is not possible when the trigonal planar angle of an alkene 

is ideally 120º (make a model to prove this).




For a molecule to aromatic it must have: 

a. A 

b. B 

c. C 

d. D 

*e. E 





To be aromatic a molecule must possess 4n+2 pi-electrons as well as them being in a 

planar, continuously conjugated cyclic system.




For a molecule to aromatic it must have: 

a. A 

b. B 

c. C 

*d. D 

e. E 







Which of the following statements is true? 

a. A 

b. B 

c. C 

*d. D 

e. E 







Which of the following statements are true? 

a. A and B 

b. B and C 

c. C and D 

d. A and C 

*e. B and D 







Which of the following statements is true? 

a. A 

b. B 

*c. C 

d. A and B 

e. All the statements are false 









373. Reactivity 

Pick out the following compound which is most reactive towards ring bromination: 

a. A 

b. B 

c. C 

*d. D 

e. E 


The most reactive phenyl rings to electrophilic substitution are those which are electron 

rich. In aniline, D, the nitrogen lone pair is delocalized into the aromatic ring to making it 

more reactive (i.e., a stronger nucleophile). In acetanilide, E, the lone pair on nitrogen 

can also be delocalized into the ring, but is also delocalized into the more electronwithdrawing 

carbonyl group, towards the electronegative oxygen. In acetophenone C, the 

partial positive charge on the carbonyl carbon means the group is an electron 

withdrawing substituent.






374. Reactivity 2 

Pick out the following compound which you would give the greatest amount of metaproduct 

when subjected to ring nitration: 

a. A 

b. B 

*c. C 

d. D 

e. E 


Substituents with a -M effect direct substitution to the meta position; these can be easily 

identified as groups which have a positive or partial positive charge (or an empty orbital) 

on the atom attached directly to the ring. These groups destabilize in particular the 

intermediate carbocation formed when ortho or para attack occurs, so directing the 

preferred attack to the meta position. You should be able to draw the resonance structures 

to show this.







Pick out the compound which would be the most reactive towards electrophilic 

substitution: 

a. A 

*b. B 

c. C 

d. D 

e. E 


The most reactive rings to electrophilic substitution are electron rich. In phenol, B, the 

oxygen lone pair is delocalized into the aromatic ring to making it more 

reactive/nucleophilic (it is a +M group). In benzaldehyde C and nitrobenzene D the 

(partial) positive charge on the atom attached to the ring means the group is an electron 

withdrawing (-M) substituent, and is therefore meta directing. The Br in bromobenzene E 

is ortho-para directing (+M), but as the bromine is electronegative (-I) the compound is 

less reactive than phenol B.







Which of the following compounds would be least reactive to ring bromination? 

a. A 

b. B 

c. C 

d. D 

*e. E 



identified as groups which have a positive (or an empty orbital) on the atom attached 

directly to the ring. These groups destabilize the intermediate carbocation formed upon 

attack of an electrophile which means the Eact for this process is much greater than that 

compared with benzene. A bigger Eact means a slower reaction (under given conditions).







Pick out the following compounds which does NOT have a meta-directing substituent: 

a. A 

b. B 

c. C 

*d. D 

e. E 



identified as groups which have a positive (or an empty orbital) on the atom attached 

directly to the ring (e.g., molecules A, B, C and E). Ortho-para directors on the other 

hand can be identified by having a lone pair of electrons on the atom attached directly to 

the ring (e.g., molecule D).







Order the following aromatic systems in order of decreasing nucleophilic (electron rich) 

character of the aromatic ring (most nucleophilic first): 

a. A > B > C > D 

b. B > C > D > A 

c. C > D > A > B 

d. A > C > B > D 

*e. B > D > A > C 


In anisole, B, the oxygen lone pair is delocalized into the aromatic ring making it more 

nucleophilic. The +I effect of the methyl group in toluene D is less effective than the +M 

methoxy group at making the ring nucleophilic, whereas, the strongly electronegative -I 

chlorine makes the chlorobenzene C less nucleophilic than benzene A.









a. A > B > C > D 

b. B > C > D > A 

*c. C > D > A > B 

d. A > C > B > D 

e. B > D > A > C 


In anisole, B, the oxygen lone pair is delocalized into the aromatic ring making it more 

nucleophilic than benzene A. The weak +I effect of the CH2 group in ether D is less 

effective than the +M methoxy group at making the ring nucleophilic, but is still greater 

compared to benzene. The electronegative -I and -M nitro group makes the ring electron 

deficient and more resistant to electrophilic attack.









a. A > B > C > D 

b. A > C > B > D 

*c. C > A > D > B 

d. B > D > A > C 

e. D > A > C > B 


In N-methylaniline, C, the nitrogen lone pair is delocalized into the aromatic ring making 

it more nucleophilic than benzene A. The -I / -M effect of the CHO group makes the ring 

electron deficient compared to benzene and so more resistant to electrophilic attack. The 

very electronegative -I and -M nitro group is the least electron rich, and therefore the 

least nucleophilic.







For a series of mono-substituted benzenes, pick out the substituent which has a positive 

inductive effect: 

a. A 

b. B 

c. C 

*d. D 



Groups which have a positive charge (or a partial positive charge orbital) on the atom 

attached directly to the ring are easily identified as -I groups. This also includes atoms 

attached directly to the aromatic ring which are more electronegative than carbon. 

Simple, unsubstituted alkyl groups always have a +I effect.







For a series of mono-substituted benzenes, pick out the substituents which has a negative 

inductive effect: 

a. A and B 

b. B and C 

c. C and D 

*d. A, B and C 

e. A, C and D 


Groups which have a positive charge (or a partial positive charge) on the atom attached 

directly to the ring are easily identified as -I groups. This also includes atoms attached 

directly to the aromatic ring which are more electronegative than carbon. Simple, 

unsubstituted alkyl groups always have a +I effect.







For a series of mono-substituted benzenes, pick out the substituent which has a negative 

resonance (mesomeric) effect: 

a. A 

*b. B 

c. C 

d. D 

e. A and B 


Substituents which have a (partial) positive (or an empty orbital) on the atom attached 

directly to the ring are easily identified as electron withdrawing groups. If these also 

contain a multiple bond conjugated to the ring then they can be identified as having an - 

M effect.







For a series of mono-substituted benzenes, pick out the substituents which has a positive 

resonance (mesomeric) effect: 

a. A and B 

b. B and C 

c. C and D 

*d. A and C 

e. B and D 


Substituents with a +M effect can be identified as those in which the atom attached 

directly to the ring bears a lone pair of electrons.







Pick out the molecule which reacts faster with sodium methoxide (NaOMe): 

a. A 

b. B 

c. C 

d. D 

*e. E 


Nucleophilic attack by sodium methoxide results in the formation of an intermediate 

carbanion which is resonance delocalized in the pi-system. Electron withdrawing groups, 

especially those with a negative mesomeric effect, can help stabilize further this charge - 

such as the nitro group in molecule E. In general, electron deficient halobenzenes 

undergo nucleophilic substitution more readily, whereas electron rich benzenes undergo 

electrophilic substitution more readily.







Pick out the molecule which reacts faster with sodium methoxide (NaOMe): 

a. A 

b. B 

c. C 

*d. D 

e. E 


Nucleophilic attack by sodium methoxide results in the formation of an intermediate 

carbanion which is resonance delocalized in the pi-system. Electron withdrawing groups, 

especially those with a negative mesomeric effect, can further stabilize this charge - such 

as the nitro groups in D and E. However, this mesomeric effect can only be effective if 

the charge can be delocalized onto the nitro substituent and this can only take place when 

the chlorine and nitro groups are ortho or para to one another; i.e., in D only.







Place the following aromatic systems in order of decreasing reaction rate with sodium 

methoxide (NaOMe)(most reactive at the left): 

a. A > B > C > D 

b. B > C > D > A 

c. C > D > A > B 

*d. A > C > B > D 

e. B > D > A > C 


Nucleophilic attack by sodium methoxide at the C-Hal carbon results in the formation of 

an intermediate carbanion which is resonance delocalized in the pi-system and the para 

nitro-substituent. The second step is loss of the halogen, and the rate of this is determined 

by the strength of the C-Hal bond (c.f., rate of SN2 reactions in C-Hal systems). As this 

step is not the rate-determining step, addition of sodium methoxide to halobenzene is 

dependent on the electronegativity of the halogen; the more electronegative the halogen, 

the more electrophilic the halobenzene and consequently the faster the addition processes. 

Note: this process does not involve a leaving group effect.






388. Reactions 

Which reagent(s) would be most suitable to carry out the following transformation? 

a. A 

b. B 

*c. C 

d. D 

e. E






389. Reactions 2 


*a. A 

b. B 

c. C 

d. D 

e. E 


This transformation is a side-chain chlorination reaction and leaves the aromatic ring 

intact; it proceeds efficiently via a free radical mechanism as the intermediate benzylic 

radical formed is stabilized by resonance.








a. A 

*b. B 

c. C 

d. D 

e. E 


This is an electrophilic ring transformation rather than a side-chain chlorination reaction; 

it proceeds efficiently via electrophilic chlorination (“Cl+”) of the aromatic ring, which 

normally involves chorine with a suitable Lewis acid, such as transition metal chloride.







Which of the following reactions would give isopropylbenzene Z as the major product? 

a. A 

b. B 

c. C 

d. D 

*e. all of the above 


This is Friedel Crafts alkylation reaction, and as such, reactions proceed via a carbocation 

electrophile. From the product Z, we can deduce that this must have been the isopropyl 

cation (Me2CH + ), and this can be generated in many ways; e.g., protonation of the alkene 

in B, or ionization of the secondary alcohol or chloride in C and D. Process A can also 

generate the product as the initially formed primary cation can undergo a hydride shift to 

give the more stable secondary carbocation before reacting with the ring.







Which of the following reactions would be used to synthesise tert-butylbenzene Z? 

a. A 

b. B 

c. C 

*d. All of the above 



This is Friedel Crafts alkylation reaction, and as such, reactions proceed via a carbocation 

electrophile. From the product Z, we can deduce that this must have been the tert-butyl 

cation (Me3C + ), and this can be generated in many ways; e.g., protonation of the alkene 

in A, to give the more stable tertiary carbocation, or ionization of the tertiary alcohol or 

chloride in B and C.








a. A 

b. B 

c. C 

*d. D 

e. E 


Reaction A would generate para-nitro tert-butylbenzene; B would generate 1,4-di(tertbutyl)benzene; 

in reaction C, sulfurous acid is oxidized by peroxides to sulfuric acid, but 

in the absence of other reagents would not react with tert-butylbenzene. 

Only reaction D would generate the product(s) shown. In fact, the para isomer would be 

far the greater of the two (probably formed exclusively) due to steric effects of both 

electrophile and substituent.







What would you expect to be the major product obtained from the following reaction? 

a. A 

b. B 

*c. C 

d. D 



The first step is to decide which of the two aromatic rings is the more electron rich, and 

therefore will react the fastest. In this case, the right hand ring is an anilide (derivative of 

aniline - with a +M amino group), whereas, the left ring is a derivative of benzoic acid 

(with a -M CO2H group). This analysis reveals that the anilide ring is the more reactive 

(nucleophilic) and +M (ortho-para) directing, leading to required product C.







Which of the following reactions would give the product(s) indicated? 

a. A 

b. B 

c. C 

d. All of the above 



Reaction A would generate para-nitrochlorobenzene; B would generate acetanilide by 

acetylation of the nitrogen atom. In reaction C, a bromoanisole would be produced, but as 

the OMe group is +M and ortho-para directing, the meta-product shown would not be 

formed.







Indicate the correct product, if any, of the following reaction: 

a. A 

b. B 

c. C 

d. D 

*e. There is no reaction 


HBr is a strong Brønsted acid. For bromination of phenol, molecular bromine, Br2, is 

normally used.







Which would be the major product of the following reaction? 

a. A 

b. B 

*c. C 

d. D 

e. E 


An OH group, which has a +M effect is ortho-para directing. Due to the steric bulk of the 

bromine atom, the fastest forming product is the para isomer, para-bromophenol. Once 

brominated, however, the aromatic ring now becomes less electron rich due to the 

electron-withdrawing (-I) Br atom, and so less susceptible to further bromination. 

Therefore, with one equivalent of molecular bromine, C is the major product.







The most efficient synthesis of p-nitroacetophenone would be: 

a. A 

b. B 

c. C 

*d. D 

e. None of the Above 


Route A would not work as the nitro group is a -M meta director, and similarly route D 

would also fail as the acetyl group is a meta director too. Both routes would produce the 

same 1,3-disubituted product. 

Route B would not work as the acid chloride reacts with the Grignard reagent to form a 

ketone, which would further react more readily with the remaining Grignard to form a 

tertiary alcohol. 

In the case of route D, the lithium carboxylate salt can still react with the second 

equivalent of the very reactive methyl lithium to produce a stable dialkoxide which only 

forms the ketone on aqueous work up of the reaction.







Select the final product from the following reaction sequence: 

*a. A 

b. B 

c. C 

d. D 



Step 1i forms the diazonium salt from the amine which is decomposed with CuCl (step 

1ii) to form the corresponding chlorobenzene. In step 2i, Sn/HCl reduces the nitro group 

to the anilinium hydrochloride, which is then neutralized with NaOH in step 2ii. Step 3i 

sees a new diazonium salt produced from the newly formed amino group and this 

decomposed with hypophosphorous acid to give the Ar-H substituent seen in A.







The most efficient synthesis of 3,5-dibromotoluene would involve the following reagents: 

a. A 

*b. B 

c. C 

d. D 

e. E 


Analysis of the target reveals that both bromine and the methyl group are ortho-para 

directing groups but the target molecule is in an all meta arrangement; no straightforward 

sequence is therefore available. 

This quickly rules out option A which forms 2,4-dibromotoluene, option C which would 

initially produces 2,4-dinitrotoluene, option D which produces 2,4-dibromotoluene (steric 

hindrance would disfavour 2,6-dibromotoluene) and option E which would produce a 

mixture of 3-bromo-4-nitrotoluene and 5-bromo-2-nitrotoluene. This leaves only option 

B, where the activating o/p-directing amino group leads to the formation of 4-amino-3,5dibromotoluene. 

This amino group can be removed using the very useful diazonium salt 

chemistry.







The most efficient synthesis of 1,3-dibromobenzene would involve the following 

reagents: 

a. A 

*b. B 

c. C 

d. D 

e. E 


Analysis of the target reveals that both ortho-para directing bromine atoms are in a meta 

arrangement; no straightforward sequence is available. 

Option A would first form 4-nitrobromobenzene, option C would produce 1,4bromobenzene 

(steric hindrance disfavouring the 1,2-isomer). 

Option D would seem to be sensible at first glance, as aniline can produce 2,4dibromoaniline. 

However, as aniline is very reactive it readily tribrominates in a reaction 

difficult to stop at the intermediate stages. Only Option B therefore works. The first metadirecting 

nitro group directs the second nitro group meta. Both can be reduced (Fe/H + ) to 

the amine, and then diazotized and can then be converted to the dibromide using CuBr








a. A 

b. B 

c. C 

d. D 



The nitro group is a -M, -I meta director. Introduction of a second group therefore must 

be directed 1,3- to the first group. None of these examples are meta-dinitrobenzene.








a. A 

b. B 

*c. C 

d. D 



Electrophilic sulphonation of naphthalene leads to the rapid formation of the 1-isomer B. 

This process is under kinetic control and is reversible at higher temperatures. Above 

160 o C, this kinetic product rearranges to the more stable thermodynamic product, 2isomer 

C








a. A 

b. B 

c. C 

d. D 



In this Friedel Crafts reaction, the primary carbocation formed first from ionization of the 

halide readily re-arranges to the more stable tertiary butyl carbocation through a hydride 

shift. Option D is incorrect as the chlorine is ortho-para directing, and the bulky 

electrophile would react preferentially at the para-position and NOT the meta-position.








a. A 

b. B 

*c. C 

d. D 



In this Friedel Crafts reaction, the primary carbocation formed first from ionization of the 

halide readily re-arranges to the more stable tertiary butyl carbocation through a hydride 

shift. Option C is correct as the methoxy group is ortho-para directing, and the bulky 

electrophile would react at the para position preferentially.








a. A 

b. B 

c. C 

d. D 

*e. E 


The Birch reduction occurs via two single electron transfer-proton exchange steps. 

Addition of an electron to methoxybenzene leads to the formation of the MORE stable 

radical anion 1, protonation of which (using ethanol) gives radical 3. Further electron 

addition, followed by protonation (using ethanol) leads to the product E.








a. A 

b. B 

*c. C 

d. D 



The Birch reduction occurs via two single electron transfer-proton exchange steps. 

Addition of an electron to nitrobenzene leads to the formation of the MORE stable radical 

anion 2, protonation of which (using ethanol) gives radical 3. Further electron addition, 

followed by KINETIC protonation (using ethanol) leads to the product C.







What would you expect to be the major product obtained from the following reaction 

under thermodynamic control? 

a. A 

b. B 

c. C 

*d. D 

e. E 


Correct. Friedel-Crafts alkylations occur via carbocation electrophiles. In this case, the 

primary cation 1 readily rearranges to the more stable secondary cation 2 which then 

reacts with benzene to give A. This first alkyl group introduced acts as a bulky para 

director and reacts to give B. As both groups are o/p-directors the next product formed 

can only be C, but this is sterically hindered process due to the isopropyl groups. Under 

thermodynamic control C can isomerise under acid to give the more stable product D.







Pick out the product from the following reaction: 

*a. A 

b. B 

c. C 

d. D 



Under these reaction conditions, nitration occurs NOT sulfonation. The products can 

either be A or B BUT NOT C or D. The NMe2 group is an activating and normally an 

ortho/para-director, so one might presume that B is the correct answer. However in this 

case the steric clash of the four methyl groups (make a model!) mean that the NMe2 

group is twisted from the plane of the ring and the nitrogen lone pair is no longer 

conjugated with the ring. Therefore, it cannot act as a +M group and is like a -I only 

substituent. This means the methyl are dominant in directing the intro group in a position 

ortho-para to the methyls but meta to the NMe2.







Protonation of pyrrole leads to either A or B? 

a. A 

*b. B 


Resonance stabilization of the conjugate acid explains why structure B is the correct 

answer here. The charge on structure A cannot be delocalized into the pi-system as the 

nitrogen atom bearing the charge is already bonded with four sigma bonds. In structure B 

on the other hand, the charge can be delocalized over the pi-system meaning it is more 

stable. (Conversely A is a stronger acid than B).



411. Acid and Base 


*a. A 

b. B 





Pyridinium ion B is still aromatic despite protonation, but protonation of the nitrogen in 

A leads to loss of aromaticity. This is regained upon loss of the H + , therefore, A is the 

more acidic.



412. Acid and Base 2 


*a. A 

b. B 





Molecule B is still aromatic despite protonation, but protonation of the nitrogen in A 

leads to loss of aromaticity. This is regained upon loss of the H + , therefore, A is the more 

acidic.





a. A 

*b. B 





Cyclopentadiene B is the more acidic as loss of a proton from the methylene (CH2) leads 

to an aromatic conjugate base.





a. A 

*b. B 





Cyclopentadiene B is the more acidic as loss of a proton from the methylene (CH2) leads 

to an aromatic conjugate base. Pent-1,4-diene A produces a resonance stabilized anion, 

but clearly it is NOT aromatic.





*a. A 

b. B 





Molecule A is the more acidic as loss of a proton from the methylene (CH2) leads to an 

aromatic conjugate base - benzene! Molecule B produces anti-aromatic cyclobutadiene.







Pick out the most basic carbon atom within the following molecule: 

a. A 

b. B 

*c. C 

d. None of them are basic 

e. All the carbons have equal basicity 


Carbon A is non-basic (as it does not contain a pair of non-bonded electrons). Protonation 

of carbon B leads to a secondary carbocation, whereas, protonation of carbon C leads to 

the more stable tertiary carbocation. Therefore, carbon C is more basic than B because 

the conjugate acid (carbocation) is more stable.






Substitution and Elimination Reactions 








417. Major Product 

Pick out the structure which is the major product formed in the following reaction: 

*a. A 

b. B 

c. C 

d. D 

e. E 


This is a free-radical photobromination, and proceeds preferentially via the more stable 

tertiary radical which gives rise to product A.






418. Major Product 2 

Pick out the structure which is the major product formed in the following reaction: 

a. A 

*b. B 

c. C 

d. D 

e. E 


The alkyl bromide ionizes to give a stabilized tertiary carbocation. In the presence of a 

strong base, KOH, abstraction of a proton resulting in a net elimination of HBr produces 

the more stable, more substituted alkene B.






419. Rate 

For the following SN2 reaction; what would be the relative rate if the concentrations of 

the bromide and hydroxide ions were doubled? 

a. Increase the rate six times 

*b. Increase the rate four times 

c. Increase the rate three times 

d. Increase the rate two times 

e. no effect 


In this SN2 reaction the rate is proportional to the concentration of both reactants 

(rate=k[RBr]×[OH]). If both concentrations were doubled, the rate would therefore 

increase four-fold.






420. Rate 2 

Pick out the most reactive alkyl halide for an SN1 reaction: 

*a. A 

b. B 

c. C 

d. D 

e. E 


In an SN1 process, the alkyl bromide ionizes to give a carbocation. Inductive effects 

stablize this cation and so the tertiary cation arising from A is the most stable.






421. Rate 3 

Pick out the second order reaction from the following rate laws: 

a. A 

*b. B 

c. C 

d. D 

e. E 


An SN2 reaction is typical of a second order reaction where the rate is equally 

proportional to the concentration of both reactants.






422. Rate 4 

Pick out the compound which reacts fastest in the presence of AgNO3: 

*a. A 

b. B 

c. C 

d. D 

e. E 


The silver halide assists ionization of an alkyl chloride and forms insoluble AgCl. This 

ionization is easier if the resulting carbocation is more stable, and this happens to be the 

case for the tertiary alkyl chloride A.






423. Rate 5 

From the following, pick out the explanation for why molecule Y hydrolyses faster than 

molecule Z: 

a. A 

b. B 

c. C 

*d. D 

e. E 


With molecule Y, the carbocation is stabilized by resonance delocalization with the 

neighbouring aromatic ring (as well as inductive effects seen with molecule Z).






424. Rate 6 

From the following, pick out the explanation for why molecule Y hydrolyses faster than 

molecule Z: 

a. A 

b. B 

*c. C 

d. D 

e. E 


With both molecules Y and Z, the carbocation is tertiary and stabilized by inductive 

effects. In molecule Y, however, the neighbouring phenyl ring can stablize the 

carbocation by forming a spirocyclopropyl cation.






425. Rate 7 

Which of the following molecules will react faster with sodium hydroxide? 

a. A 

*b. B 


In an SN2 process the bromide is displaced by the nucleophile in a concerted process via 

rear-side attack. This process is faster with less steric hindrance and so the least hindered 

primary halide E reacts fastest.






426. Rate 8 


*a. A 

b. B






427. Rate 9 


a. A 

*b. B






428. Rate 10 


*a. A 

b. B






429. Yield 

Pick out the most reactive alkyl halide for an SN2 reaction: 

a. A 

b. B 

c. C 

d. D 

*e. E 


In an SN2 process the bromide is displaced by the nucleophile in a concerted process via 

rear-side attack. This process is faster with less steric hindrance and so the least hindered 

primary halide E reacts fastest.






430. Sn1 Reactions 

Which of the following statements is/are true for SN1 reactions? 

*a. A 

b. B 

c. C 

d. A and B 

e. A, B and C 


In an SN1 process the ionization of the alkyl halide to form a cation is the ratedetermining 

step and is independent of the nucleophile. Non-polar solvents cannot 

stabilize the cation, and so lowers the rate of reaction.






431. Sn1 Reactions 2 

Pick out the nucleophile which has the greatest effect on the rate of an SN1 reaction: 

a. A 

b. B 

c. C 

d. D 

*e. All behave the same 


In an SN1 process, the ionization of the electrophile is the rate-determining step; it is 

independent of the nucleophile.







From the following, pick out the potential energy profile for an SN1 reaction: 

a. A 

b. B 

*c. C 

d. D 



In all reactions the product is more stable than the starting material (so obviously not 

profile D), and the SN1 proceeds via a discrete carbocation intermediate (so not profile 

A). The intermediate cation is also much less stable than the product or starting material 

and so profile C can be the only answer.







Which of the following electrophiles react slowest in an SN1 reaction: 

a. A 

b. B 

c. C 

d. D 

*e. E 


SN1 substitution proceeds via a carbocation, and the less stable the cation, the slower the 

reaction (higher Eact). The -M para-nitro group destabilizes the cation arising from 

bromide E through resonance effects, and so is less stable than the cations arising from 

all the other examples given.







Which of the following electrophiles react fastest in an SN1 reaction: 

a. A 

*b. B 

c. C 

d. D 

e. E 


SN1 substitution proceeds via a carbocation, and the more stable the cation the faster the 

reaction (lower Eact). The +M para-methoxy group stabilizes the cation arising from 

bromide B through resonance effects, and so is more stable than the cations arising from 

all the other examples given.







Which of the following electrophiles react slowest in an SN1 reaction: 

*a. A 

b. B 

c. C 

d. D 

e. All of the above have the same rate of reaction 


SN1 substitution proceeds via a carbocation resulting from ionization of the R-Hal bond. 

The stronger the R-Hal bond, the more difficult the ionization, and therefore the slower 

the reaction rate.







Pick out the strongest electrophile for an SN1 reaction: 

a. A 

b. B 

c. C 

*d. D 

e. E 


SN1 substitution proceeds via a carbocation resulting from ionization of the R-Br bond in 

the rate determining step. The more stable cation is the more substituted case, which is 

derived from bromide D, which in-turn has the lowest Eact and fastest reaction rate.








a. A 

b. B 

c. C 

d. D 

*e. E 


SN1 substitution proceeds via a carbocation resulting from ionization of the R-Br bond in 

the rate determining step. The more stable the cation, the faster the reaction rate. For 

bromide E, the resulting carbocation can be resonance stabilized by the +M effect from 

the neighbouring sulfur atom.







Pick out the following factor(s) which promote an SN1 reaction: 

a. A and B 

b. B and C 

c. C and D 

*d. A and C 



A change in temperature affects all reactions, but only a change in concentration of the 

alkyl halide affects the rate of SN1 reactions as the rate determining step (RDS) involves 

this molecule only. The polarity of the solvent can lower the activation energy for the 

RDS through better solvation of the cation intermediate.



439. Sn2 Reactions 

For the following: 

a. A 

*b. B 

c. C 

d. D 

e. E 





Factors A and D hinder SN2 processes, and the solvent (factor C) may affect the rate 

depending on the reagents. As the reaction is a one-step concerted process, the 

nucleophile is involved in the rate determining step and so factor B favours an increase in 

the SN2 reaction rate.







What product(s) would you expect from the following SN2 reaction? 

a. A 

*b. B 

c. C 

d. Equimolar mixture of A and B 

e. Non-equimolar mixture of A and B 


SN2 reactions proceed with inversion of configuration, so molecule A cannot be formed 

in this case. Molecule C is the product from elimination.







Pick out the strongest nucleophile for an SN2 reaction: 

a. A 

b. B 

c. C 

*d. D 

e. E 


Hydroxide is the strongest nucleophile as (i) the charge is not delocalized (as in C and E) 

and (ii) the O is negatively charged (compared to A and B).







Pick out the weakest nucleophile for an SN2 reaction: 

*a. A 

b. B 

c. C 

d. D 

e. E 


In molecule A, the charge is delocalized and so carboxylates in general are weaker 

nucleophiles compared to hydroxide and alkoxides.







Pick out the strongest nucleophile for an SN2 reaction: 

*a. A 

b. B 

c. C 

d. D 

e. E 


The carboxylate C is a weaker nucleophile as it has a delocalized charge - like the 

phenolate E. In the selenide ion A, the outer electrons are held further from a less 

electronegative nucleus and so A is more polarizable than the smaller ions in the group. 

This means the charge is “more available” and so is the strongest nucleophile of those 

shown.







From the following, pick out the potential energy profile for an SN2 reaction: 

*a. A 

b. B 

c. C 

d. D 



An SN2 reaction is a concerted process and does not involve intermediates (ruling out 

profile C and D). Also, reaction transition states are higher energy than the starting 

material leaving profile A as the only option.








*a. A 

b. B 

c. C 

d. D 

e. E 


In an SN2 process, the bromide is displaced by the nucleophile in a concerted process via 

rear-side attack. This process is faster with less steric hindrance, and so the least hindered 

primary halide A reacts fastest.








a. A 

b. B 

*c. C 

d. D 

e. All of react at the same rate







Pick out the following factor(s) which promote an SN2 reaction: 

a. A and B 

b. B and C 

c. B and D 

d. A and C 








Pick out the alkyl bromide which proceeds with retention of configuration in an SN2 

reaction: 

a. A 

b. B 

*c. C 

d. D 

e. All of react at the same rate 


Neighbouring group participation takes place due to the nucleophilic sulfur atom. This 

can easily displace the bromide to form a 5-membered cyclic sulfonium ion with 

inversion. The nucleophile (MeOH) can then ring open the sulfonium ion with inversion 

to form the product - a net retention of configuration overall (by double inversion).







Rearrangements are likely to occur in which of the following reaction types? 

a. SN1 reactions 

b. SN2 reactions 

c. E1 reactions 

d. E2 reactions 

*e. Both SN1 and E1 reactions 


Carbocations undergo rearrangements and these are the intermediates formed in SN1 and 

E1 reactions.







Pick out the major product from the following reaction: 

a. A 

*b. B 

c. C 

d. D 

e. E 


The alkyl halide can ionize to form a stabilized tertiary cation. In the presence of the 

strong base elimination (reaction B) is preferred to the substitution pathway (reaction A).







Pick out the nucleophile in the following reaction: 

a. A 

*b. B 

c. C 

d. D 

e. E 


It is a common student misconception that water ionizes to hydroxide prior to substitution 

forming an alcohol. However it is water that acts as the nucleophile and then the resulting 

‘onium’ ion subsequently loses a proton to form the product alcohol.







Pick out the leaving group in the following reaction: 

a. A 

b. B 

c. C 

*d. D 

e. E




Pick out the weakest nucleophile: 

*a. A 

b. B 

c. C 

d. D 

e. All behave the same 





Fluoride has a high charge density (small atom, outer electrons close to nucleus) and so is 

less polarizable than the larger ions in the group. This means the charge is “less 

available” to act as a nucleophile, unlike in iodide which is the strongest nucleophile of 

those shown.







From the following, pick out the descriptor that best describes a mechanism which 

involves separated ion-pairs: 

a. A and B 

b. B and C 

c. C and D 

*d. A and C 

e. B and C 


E2 and SN2 process are concerted and so do not involve distinct ionic intermediates, 

which is the opposite of the stepwise E1 and SN1 reactions.







From the following, pick out the descriptor that best describes a mechanism that involves 

separated ion-pairs: 

a. A and B 

*b. A and C 

c. B and C 

d. C and D 

e. E 


E2 and SN2 process are concerted and so do not involve distinct ionic intermediates, 

which is the opposite of the stepwise E1cb and SN1 reactions.







From the following, pick out the reaction(s) which form PhCH2NH2: 

a. A 

b. B 

c. C 

d. D 



All of these processes form the amine indicated, but the route outline in A is less 

commonly used as secondary and tertiary amines may be formed as by products when the 

amine produced reacts further with the starting alkyl bromide.







From the following, pick out the alcohol which reacts most readily with HCl: 

a. A 

b. B 

c. C 

*d. D 

e. E 


The oxonium ion derived from protonation of tertiary alcohol D loses water most readily. 

The resulting carbocation is stabilized by inductive effects to a greater degree than the 

cations arising from alcohols A, B, C and E.







For the following reaction, pick out the following statement which is correct: 

a. A 

*b. B 

c. C 

d. D 

e. E 


Primary alcohols do not readily lose water on protonation as the resulting cation is not 

stabilized greatly by inductive effects. 

However, the protonated alcohol is a substrate for SN2-type substitution by a second 

molecule of the alcohol to form a dialkyl ether as the product.







For the following reaction, pick out the term which best describes its mechanism: 

a. A 

b. B 

*c. C 

d. D 

e. E 


The neighbouring group participation is due to the nucleophilic sulfur atom which can 

displace the bromide to form a 5-membered cyclic sulfonium ion. The nucleophile 

(MeOH) can then ring open the sulfonium ion to form he product.







Pick out the process which proceeds with inversion of configuration: 

a. A 

*b. B 

c. C 

d. D 

e. E







Pick out the process which proceeds with retention of configuration: 

a. A 

*b. B 

c. C 

d. D 

e. E 


SN1 processes, proceeding via a planar carbocation, may form products in which the 

configuration is the same as the starting material, but only as a mixture along with 

inverted product.







Pick out the process which does not involve carbocation formation: 

a. A 

b. B 

*c. C 

d. D 



Reaction C is a free radical substitution, as indicated by the need for light (hv) to form 

the bromine atoms to initiate this process.







For the following reaction, pick out the following statement(s) which is/are true: 

a. A 

b. B 

c. C 

*d. A and B are correct 

e. A, B and C are correct 


This is an SN2 process as indicated by the inversion of configuration in the product. The 

statement B must be true as for SN2 reactions, the rate=k[OH]×[RBr].







Which of the following molecules is more nucleophilic? 

*a. A 

b. B








a. A 

*b. B








*a. A 

b. B








*a. A 

b. B




The following reaction proceeds with: 

*a. A 

b. B 








*a. A 

b. B 








a. A 

*b. B 










Which of the following alkyl halides is most electrophilic? 

a. A 

b. B 

*c. C







Which of the following alkyl halides is least electrophilic? 

*a. A 

b. B 

c. C







From the following, pick out which is a protic solvent. 

a. A 

b. B 

c. C 

*d. D 

e. E 


A protic solvent is one which can readily liberate a proton (has an acidic hydrogen atom). 

Such solvents include water, alcohols, simple carboxylic acids (formic / acetic acid) etc.







From the following, pick out which are aprotic solvents. 

a. A and B 

b. A and C 

c. B and D 


e. B, C and D 


A protic solvent is one which can readily liberate a proton (has an acidic hydrogen atom), 

whereas, an aprotic solvent does not have an acidic hydrogen atom.








a. A 

b. B 

c. C 

*d. D 

e. E 


A ketone readily forms enolates reversibly in the presence of hydroxide. When the 

enolate is formed alpha to the epoxide, it can trigger a ring opening reaction to give the 

product shown. The enolate formation is rapid compared to the ring opening step and so 

this process is a stepwise E1cb process rather than a concerted E2-type elimination.








a. A 

b. B 

c. C 

d. D 

*e. E 


This is a classic E2 elimination involving an alkyl halide and a moderately strong base.








a. A 

b. B 

*c. C 

d. D 

e. E 


Tertiary (hindered) alkyl halides are the classic substrates for E1 eliminations with strong 

bases and they proceed via the stabilized tertiary carbocation.








a. A 

b. B 

c. C 

*d. D 

e. E 


The presence of a –M anion-stabilizing group beta to a leaving group makes for a good 

substrate for an E1cb process. Deprotonation of the substrate is rapid and reversible due to 

the –M sulfone, and the slower rate determining elimination of the leaving group then 

forms the alkenic product.








a. A 

*b. B 

c. C 

d. D 

e. E 


This substitution reaction proceeds with retention of configuration so cannot be either an 

SN2 (with inversion) or SN1 (produces a mixture) process. Radical processes often 

involve formation of a mixture of configurations in the product.







Pick out the strongest base in an aprotic solvent: 

*a. A 

b. B 

c. C 

d. D 

e. They all behave the same 


Fluorine is the most electronegative atom, and in aprotic solvents there is no hydrogen 

bonding to reduce the basicity of the fluoride ion.







From the following, pick out the descriptor(s) that best describes a mechanism which 

involves a concerted process: 

a. A and B 

b. B and C 

c. C and D 

d. A and C 

*e. B and D 


SN1 and E1 reactions are stepwise reactions proceeding via carbocation intermediates, 

whereas E2 and SN2 are one-step (concerted) processes.







Pick out the processes which proceed via a non-concerted pathway: 

a. A 

*b. B 

c. C 

d. D 

e. E







Pick out the process which preferentially proceeds via front side attack: 

a. A 

*b. B 

c. C 

d. D 

e. E







Pick out the process which preferentially proceeds via front side attack: 

a. A and C 

*b. C and E 

c. D and A 

d. E and B 

e. None of them







Pick out the processes which involve electrophiles and nucleophiles: 

a. A and C 

*b. A, B and C 

c. D and E 

d. A, B, C and D 



Substitutions involve electrophiles and nucleophiles, whereas, eliminations involve acids 

and bases.







Pick out the processes which involve acids and bases: 

a. A and C 

b. A, B and C 

*c. D and E 

d. A, B, C and D 



Eliminations involve acids and bases, whereas, substitutions involve electrophiles and 

nucleophiles.






487. Elimination 


*a. A 

b. B 

c. C 

d. Equimolar amount of A and B 



E2 eliminations are stereospecific reactions in which the C-H bond and C-Cl must be 

antiperiplanar during the concerted elimination. In most cases the more substituted alkene 

is produced as the major product, but only product A can be made from the required 

conformation. Less substituted alkene C would be a minor product.






488. Elimination 2 

Pick out the more reactive substrate towards an E2 elimination: 

*a. A 

b. B








a. A 

b. B 

*c. C 

d. Equimolar amount of A and B 



Hoffman-type elimination, which occurs with bulky leaving groups, produces the least 

substituted alkene as the major product.







Pick out the more reactive substrate towards an E2 elimination: 

a. A 

*b. B








*a. A 

b. B 

c. C 

d. Equimolar amount of B and C 



In cis-alkenyl bromides, E2 elimination produces an alkyne as the bond antiperiplanar to 

the bromine is a C-H bond (and not a C-C bond as in the trans-isomer). This reaction 

takes place in preference to elimination to form an allene (C=C=C) as the alkenyl CH 

bond is weaker than an alkyl CH bond.








a. A 

b. B 

c. C 

*d. Equimolar amount of B and C 



As this process is an E2 elimination it produces, like simple alkyl halide substrates, a 

mixture of cis- and trans- isomers. It cannot form an alkyne as the bond antiperiplanar to 

the bromine is a C-C and not a C-H bond.








a. A 

b. B 

c. C 

*d. Non-equimolar amount of A and B 



This syn-periplanar concerted elimination occurs when the oxyanion of the amine Noxide 

abstracts a beta-hydrogen. The reaction is stereospecific, but as the starting material 

exists as a mixture of isomers, both A and B may be formed. Less substituted alkene C 

would be a minor product.








*a. A 

b. B 

c. C 

d. Non-equimolar amount of A and B 



This syn-periplanar concerted elimination occurs when the oxyanion of the amine Noxide 

abstracts a beta-hydrogen. As the reaction is stereospecific, only alkene isomer A is 

formed, and not alkene B, from the single isomer of the starting material. The less 

substituted alkene isomer C would be a minor product.







Pick out the reaction(s) which proceeds via an E1cb mechanism: 

*a. A 

b. B 

c. C 

d. A and C 

e. B and C 


In the stepwise E1cb mechanism deprotonation, forming an anion, precedes the loss of the 

leaving group. Elimination of A, via the readily formed enolate, is typical of the type of 

substrate which undergoes this type of reaction.







From the following, pick out the base which could give preferential E2 elimination for 

bromohexane: 

*a. A 

b. B 

c. C 

d. D 

e. E 


Ethoxide is the strongest base as the negative charge is not delocalized (as in B and D) 

and as oxygen is more electronegative than sulfur resulting in a higher charge density on 

the charged atom.







From the following, pick out the base which would favour E2 elimination in preference 

to SN2 substitution: 

a. A 

*b. B 

c. C 

d. D 

e. E 


An E2 reaction is preferred to an SN2 process when the reaction occurs with a strong 

base. Due to the large caesium ion, CsOEt, has the weakest metal-oxygen bond and is 

therefore the strongest base.




E2 eliminations proceed via: 

a. A and C 

*b. B 

c. B and D 

d. A and D 

e. E 





An E2 reaction is a concerted process where the bonds to the beta hydrogen and leaving 

group are broken simultaneously. This occurs preferentially in the lowest energy 

conformation making these processes extremely efficient.







What product(s) would you expect from the following E2 reaction? 

a. A 

b. B 

c. C 

d. D 

*e. Equimolar mixture of C and D 


An E2 reaction of the bromide produces an alkene by net elimination of HBr, where the 

H + comes from an atom beta to the leaving group. This process produces a racemic 

mixture of alkenes (C and D).




Pick out the rate law for an E1 elimination: 

*a. A 

b. B 

c. C 

d. D 

e. E 





An E1 reaction is typical of a first order reaction, where the rate is proportional to the 

concentration of the electrophile only.




Pick out the rate law for an E2 elimination: 

a. A 

*b. B 

c. C 

d. D 

e. E 





An E2 reaction is typical of a second order reaction, where the rate is equally 

proportional to the concentration of both reactants, RX and hydroxide.







Pick out the strongest base for an E2 reaction: 

a. A 

b. B 

c. C 

*d. D 

e. E 


Hydroxide is the strongest base as (i) the charge is not delocalized (as in molecules C and 

E) and (ii) the O is charged compared to molecules A and B.







Pick out the base which has the greatest effect on the rate of an E1 reaction: 

a. A 

b. B 

c. C 

d. D 

*e. They all behave the same 


In an E1 process the ionization of the electrophile is the rate determining step; it is 

independent of the base.







Pick out the strongest base for an E2 reaction: 

a. A 

b. B 

c. C 

*d. D 

e. E 


Hydroxide is the strongest base as (i) the charge is not delocalized (as in molecules C and 

E) and (ii) O is more electronegative than those elements further down Group 16 (S or 

Se) in molecules A and B.







From the following, pick out the molecule which would react preferentially by E1 

elimination: 

a. A 

*b. B 

c. C 

d. D 

e. E 


E1 elimination proceeds via a carbocation, and the more stable the cation, the faster the 

elimination. The +M para-methoxy group can stabilize the planar cation arising from 

bromide B through resonance and is therefore more stable than the cations derived from 

the other bromides.







From the following, pick out the potential energy profile for an E2 reaction: 

*a. A 

b. B 

c. C 

d. D 



In all reactions, the product is more stable than the starting material (so it clearly cannot 

be profile D), and the E2 elimination proceeds via a converted process (therefore, no 

intermediates so it cannot be profile B, C or D). Profile A can therefore be the only 

answer.







From the following, pick out the potential energy profile for an E1 reaction: 

a. A 

b. B 

*c. C 

d. D 



In all reactions the product is more stable than the starting material (so it clearly cannot 

be profile D), and the El elimination proceeds via a carbocation intermediate (therefore, it 

is not concerted - so it cannot be profile A). The intermediate cation is also much less 

stable than the product or starting material, therefore, profile C can be the only answer.







Which of the following substrate(s) allow E2 elimination? 

a. A 

b. B 

c. C 

d. D 



An E2 elimination requires the presence of a hydrogen atom beta to the leaving group. 

So, none of these examples can form an alkene by any mechanism.







Pick out the substrate which does NOT allow E1 or E2 eliminations to occur: 

a. A 

b. B 

*c. C 

d. D 

e. E 


Whether by an E1 or E2 process, elimination requires the presence of a hydrogen atom 

beta to the leaving group. Only molecule C has no beta carbon-hydrogen bond.






Chemistry of the Carbonyl Group 








510. Product 

For the following reaction, pick out the product(s) Z: 

a. A 

b. B 

*c. C 

d. An equimolar amount of A and B 

e. An equimolar amount of A and C 


This is an example of the Cannizzaro reaction in which disproportionation of an aromatic 

aldehyde produces an alcohol and a carboxylate.






511. Product 2 

For the following reaction, pick out the product Z: 

a. A 

b. B 

*c. C 

d. D 

e. E 


Hemiacetal B is formed as an intermediate to the acetal C. In a reverse process aqueous 

acid can convert the acetal C back to the aldehyde, PhCHO.








a. A 

*b. B 

c. C 

d. D 

e. E 


Hemiacetal B is formed under basic condition by methoxide addition to the electrophilic 

carbonyl group. Further conversion of B to the dimethyl acetal C requires the use of acid.







Compound Z can be formed by dissolving benzaldehyde in ethanol. Pick out the structure 

of this compound: 

a. A 

b. B 

c. C 

*d. D 

e. E 


Hemiacetal D is formed by addition of ethanol to the electrophilic carbonyl group. As the 

strong double bond is broken in this process, elimination back to the aldehyde group is 

also facile. Further conversion of D to the diethyl acetal E requires the use of acid.








a. A 

b. B 

c. C 

*d. D 

e. E 


This is an example of a crossed Cannizzaro reaction. This reaction proceeds via oxidation 

of formaldehyde to formic acid, with benzaldehyde being reduced to the alcohol. The 

deuterium label is not required but has been used to show that in the mechanism 

transferal of a deuteride ion from the formaldehyde to the benzaldehyde takes place.








a. A 

b. B 

c. C 

*d. D 

e. E 


The base, sodium deuteroxide, forms reversibly the enolate of the ketone by abstraction 

of the proton alpha to the carbonyl group. In the presence of D2O only the alpha-H is 

replaced by deuterium giving product D. This process is under equilibrating conditions.








a. Equimolar amount of A and B 

b. Equimolar amount of A and C 

c. Equimolar amount of A and D 

d. Equimolar amount of B and C 

*e. Equimolar amount of C and D 


The base, sodium deuteroxide, forms reversibly the planar enolate of the ketone by 

abstraction of the alpha-proton. In the presence of D2O the alpha-H is replaced by 

deuterium but as the chiral centre is racemized during enolization both enantiomeric 

forms of the deuterated ketone (C and D) are formed.







Pick out from the following reactions an appropriate synthesis of ethyl acetate (ethyl 

ethanoate): 

a. A 

b. B 

c. C 

d. D 



Reaction D is an example of an SN2 reaction. Reaction A and C, using an acid chloride 

and an anhydride, are examples of SNAc addition-elimination substitution reactions. 

Reaction B is an example of a classical alcohol + acid catalyst esterification.








a. A 

b. B 

*c. C 

d. D 

e. E 


The reducing agent sodium borodeuteride delivers a deuteride ion (D - ) to the 

electrophilic carbonyl carbon. On the addition of water in step 2, protonation of the 

alkoxide intermediate with water gives the alcohol.








a. A 

b. B 

c. C 

d. D 

*e. E 


The reducing agent lithium aluminium deuteride delivers two deuteride ions (D - ) to the 

electrophilic carbonyl carbon. The reaction proceeds via the intermediate aldehyde C, but 

on final addition of water (in step 2 of the mechanism), protonation of the alkoxide 

intermediate gives the alcohol E.








a. A 

b. B 

c. C 

*d. D 

e. E 


This is an example of the Claisen condensation. In the presence of methoxide base, the 

ester is in equilibrium with the ester enolate (by simple deprotonated-reprotonation at the 

alpha-carbon). A condensation of the enolate with the ester takes place, and the final 

product from this reaction is the beta-ketoester D.







For the following reaction, pick out the major product Z: 

a. A 

*b. B 

c. C 

d. D 

e. E 


This is an example of a crossed Claisen condensation. In the presence of methoxide base, 

the acetic ester is in equilibrium with its enolate (by simple deprotonated-reprotonation at 

the alpha-carbon). The benzoate does not form an enolate as it does not have an acidic 

alpha-hydrogen, but is however a more reactive electrophile than the acetic ester. A 

crossed condensation of the acetic enolate with the benzoate takes place and the final 

product from this reaction is the beta-ketoester B.








a. A 

*b. B 

c. C 

d. D 

e. E 


This is an example of a crossed aldol condensation. In the presence of base, the ketone is 

in equilibrium with its enolate (by simple deprotonation-reprotonation at the alphacarbon). 

The aldehyde does not form an enolate as it does not have an acidic alphahydrogen, 

but is however a more reactive electrophile than the ketone. A crossed 

condensation of the enolate with the aldehyde takes place forming C but the final 

product, after dehydration, from this reaction is the alpha,beta-unsaturated ketone B.








*a. A 

b. B 

c. C 

d. D 

e. E 


In this acid-catalysed esterification, the only isotopically labelled oxygen atom is in the 

alcohol attached via the CH3-O bond. This bond is unbroken in the reaction, and is 

retained in the final product, A, with loss of unlabelled water as a byproduct.








a. A 

*b. B 

c. C 

d. D 

e. E 


Carboxylic acids can readily exchange their acidic proton in a tautomerisation-like 

conversion. In this acid-catalysed process the isotopically labelled oxygen atom in the 

water can exchange with the OH group in the acid in a reaction mechanistically identical 

to an esterification. As a result of these two steps both oxygen atoms can be exchange for 

an isotopically labeled oxygen and B is the isolated product Z.







Compound Z can be formed by dissolving benzaldehyde in aqueous sodium hydroxide. 

Pick out the structure of this compound: 

a. A 

*b. B 

c. C 

d. D 



This is an example of the Cannizzaro reaction in which disproportionation of an aromatic 

aldehyde produces an alcohol and a carboxylate.







Pick out the product Z from the following reaction: 

a. A 

b. B 

*c. C 

d. D 



This is an example of a crossed Cannizzaro reaction in which disproportionation of an 

aromatic aldehyde produces an alcohol and a carboxylate.







Pick out the product, Z, from the following reaction: 

a. A 

*b. B 

c. C 

d. D 

e. E 


Protonation of the C=O bond makes the carbonyl group of the aldehyde a stronger 

electrophile and so the cyanide anion, acting as a nucleophile, can readily add to give a 

cyanohydrin B.








a. A 

b. B 

c. C 

d. D 

*e. E 


This is an example of the benzoin condensation. Initially the aldehyde adds HCN to 

produce cyanohydrin B. Excess - CN then can act as a base to deprotonate alpha to the 

CN group in B, and this anion adds to another molecule of aldehyde to produce the 

carbon-skeleton similar to E. The cyanohydrin portion can then lose HCN to reform the 

more stable carbonyl group and thus forms product E.








a. A 

b. B 

c. C 

*d. D 

e. E 


Cyanide ion can react readily but reversibly with the C=O to form the kinetic direct (1,2-) 

addition product E. An alternative slower and irreversible pathway, but forming the more 

stable (thermodynamic) product, is the conjugate (1,4-) addition product D. This can be 

eventually formed due to the reversibility of the direct addition.








*a. A 

b. B 

c. C 

d. D 

e. E 


This is an example of a conjugate (1,4-) addition reaction giving product A. The cuprate 

reagent, typical of the reagents used for conjugate additions, will not add directly to the 

carbonyl group.








a. A 

b. B 

c. C 

d. D 

*e. E 


This is an example of a direct (1,2-) addition reaction giving product E. The Grignard 

reagent is a strong enough nucleophile to add directly to the carbonyl group, but this 

process is not reversible and so then alternative thermodynamic conjugate addition 

product B cannot form.








a. A 

b. B 

c. C 

*d. D 

e. E 


Lithium aluminium hydride (LiAlH4) is a powerful reducing agent. With alpha, betaunsaturated 

ketones, the reduction can give a product where both the C=C double bond 

and the carbonyl group are reduced by virtue of a mechanism which sees hydride add in 

the carbonyl group, followed by intramolecular (tethered) reduction of the carbon-carbon 

double bond of the intermediate allylic alkoxide.








a. A 

b. B 

c. C 

d. D 

*e. E 


Reduction of alpha, beta-unsaturated ketones can give a product where either the C=C 

double bond or the carbonyl group has been reduced. In this particular reaction (known as 

a Luche reduction), utilizes cerium trichloride (CeCl3) and sodium borohydride, and 

results in high yields of the direct reduction product E only. 

CeCl3, a Lewis acid, catalyses the formation of sodium methoxyborohydrides (from 

methanol and sodium borohydride) and promotes 1,2-addition by binding to the C=O of 

the enone. This is a “harder” reducing agent than the parent borohydride (according to 

HSAB principles), and therefore results in high selectivity for the 1,2-reduction.








a. A 

b. B 

*c. C 

d. D 

e. E 


Secondary amines, such as morpholine, react with aldehydes and ketones expelling water 

as a by product. Instead of forming an imine (C=NR), as is formed with primary amines 

(H2NR), the enamine product C is formed instead. This tautomer is preferred 

energetically to the more substituted, but sterically demanding, double bond in enamine 

D.








a. A 

*b. B 

c. C 

d. D 



Primary amines, such as methylamine, react with aldehydes and ketones by nucleophilic 

attack at the carbonyl carbon to produce the hydroxyamine A. This can loses water to 

form the imine (C=NR) B by reforming the strong C=N double bond. Unless the imine 

itself is particularly stable and due to the reversibility of this process, a successful 

synthesis usually requires the water to be removed from the reaction as it is produced (by 

using a Dean-Stark apparatus or a drying agent such as magnesium sulfate).







Pick out the product from the following reaction: 

a. A 

b. B 

*c. C 

d. D 

e. E 


Grignard (organomagnesium) reagents, such as phenyl magnesium bromide, react with 

aldehydes by nucleophilic attack at the carbonyl carbon to produce a new carbon-carbon 

bond and an intermediate magnesium alkoxide. The corresponding free alcohol, C, is 

produced upon work-up with dilute acid.







Pick out the minor product, Z, from the following reaction: 

a. A 

*b. B 

c. C 

d. D 

e. E 


Grignard (organomagnesium) reagents normally react with aldehydes and ketones by 

nucleophilic attack at the carbonyl carbon. However, in certain substrates with 

particularly acidic alpha-hydrogens, such as this example with an alpha-phenyl 

substituent, the Grignard reagent can act as a base. Deprotonation of the aldehyde results 

in the formation of a magnesium enolate (which is conjugated with the adjacent Ph ring). 

When this is quenched (at the end of the reaction) with acid, the enol is produced, but this 

rapidly tautomerises back to the aldehyde - i.e., the starting material.







Pick out the products from the following reaction: 

a. A and C 

b. B and D 

c. C and D 

*d. B and E 

e. C and A 


Grignard (organomagnesium) reagents, such as ethyl magnesium bromide, react with 

ketones by nucleophilic attack at the carbonyl carbon to produce a new carbon-carbon 

bond and an intermediate magnesium alkoxide. The corresponding free alcohol, E, is 

produced upon work-up with dilute acid. 

As a minor side reaction the Grignard reagent can act as a base. This results in the 

formation of a magnesium enolate which when this is quenched with acid produces an 

enol. This rapidly tautomerises back to the ketone - i.e., the starting material.







Pick out all the products from the following reaction: 

a. A, B and C 

b. B, C and D 

c. C, D and E 

d. D, E and A 

*e. E, A and B 


Grignard reagents are most commonly known as nucleophiles which add to the C=O 

bond of ketones to form alcohols (E). 

Also possible with certain substrates is when they act as base to form a magnesium 

enolate, which upon work-up forms an enol and then tautomerizes back to the starting 

ketone (B). 

In certain circumstances, however, products can also arise when they behave as a 

reducing agent through acting as a beta-hydride reducing source (to give alcohols like A).








a. A 

b. B 

*c. C 

d. D 

e. E 


Reaction of a Grignard reagent with carbon dioxide results in an addition to the C=O 

bond of carbon dioxide, forming an intermediate magnesium carboxylate. Quenching 

with acid at the end of the reaction produces the carboxylic acid. This method is a useful 

way of chain homologation - increasing the carbon chain length by one carbon atom.








*a. A 

b. B 

c. C 

d. D 

e. E 


Grignard (organomagnesium) reagents, such as ethyl magnesium bromide, react with 

aldehydes (here it is formaldehyde) by nucleophilic attack at the carbonyl carbon to 

produce a new carbon-carbon bond and an intermediate magnesium alkoxide. The 

corresponding free alcohol, A, is produced on addition of acid (in the work-up 

procedure).








a. A 

b. B 

c. C 

*d. D 

e. E 


Grignard (organomagnesium) reagents, such as butyl magnesium bromide, react with 

nitriles by nucleophilic attack at the carbon of the CN triple bond. A stable intermediate, 

the N-magnesium salt of an imine (C=N-Mg) is produced. Addition of acid (on work-up) 

leads to the corresponding free imine D.








a. A 

b. B 

c. C 

d. D 

*e. E 


Grignard (organomagnesium) reagents can be considered as simple carbanions. 

Protonation, or as in this case deuteration, adds a single proton (deuteron, D + ) to the 

unfilled valency of the charged carbon atom to produce an alkane.






544. Molecule Choice 

Pick out the compound which would have greatest enol content: 

a. A 

b. B 

*c. C 

d. D 

e. E 


To form an enol, a carbonyl compound must possess a carbon bearing at least one 

hydrogen atom next to (alpha to) the carbonyl carbon. In these examples only C can form 

an enol.






545. Molecule Choice 2 


a. A 

*b. B 

c. C 

d. D 

e. E 



hydrogen atom next to (alpha to) the carbonyl carbon. Clearly, molecule D cannot form 

an enol. 

Enolization is easier for aldehydes than for ketones > esters > acids. This reflects the 

increasing stability of the carbonyl C=O bond which is “broken” in the enol tautomer. 

Therefore, aldehyde B has the greatest enol content.








a. A 

*b. B 

c. C 

d. D 

e. E 



hydrogen atom next to (alpha to) the carbonyl carbon. Enolization is easier for aldehydes 

than for ketones > esters > acids. This reflects the increasing stability of the carbonyl 

C=O bond which is “broken” in the enol tautomer. 

However, 1,3-dicarbonyl compounds have even greater enol content due to the resulting 

enol C=C bond being able to conjugate with the neighbouring C=O bond. Additionally, 

the new OH of the enol is suitably positioned for intramolecular hydrogen bond to this 

second C=O group. These factors mean that B has the greatest enol content.








a. A 

b. B 

c. C 

d. D 

*e. E 


All of these compounds are 1,3-dicarbonyl species, but only E is cyclic. The 

conformational restriction of a ring in E means that the degree of conjugation 

(stabilization) between the C=C of the enol and C=O is maximized, and so E has the 

greatest enol content. In addition, there is weaker conjugation present in the O-C=O motif 

due to the conformation of the internal ester (lactone) which destabilizes the starting 

material.







Pick out the compound with acetal functionality: 

*a. A 

b. B 

c. C 

d. D 

e. E 


An acetal is formed from an aldehyde and two molecules of an alcohol. Acetals can be 

described as mixed if the alcohol does not have the same alkyl chain. The mixed acetal A 

in this case is made up from PhCHO, MeOH and EtOH.







Pick out the compound with ketal functionality: 

a. A 

*b. B 

c. C 

d. D 

e. E 


A ketal is formed from a ketone and two molecules of an alcohol (PhCOMe and MeOH 

form B in this case).







Pick out the compound with orthoformate functionality: 

*a. A 

b. B 

c. C 

d. D 

e. E 


The orthoformate A is to the ester PhCO2Me as the ketal B is to the ketone PhCOMe.







Pick out the compound with hemiketal functionality: 

a. A 

b. B 

c. C 

*d. D 

e. E 


A hemiketal is formed from a ketone by addition of a molar equivalent of an alcohol, 

normally in the form of an alkoxide (PhCOMe and MeO - in MeOH form D in this case).







Pick out the compound with hemiacetal functionality: 

a. A 

b. B 

*c. C 

d. D 

e. E 


A hemiacetal is formed from an aldehyde by addition of a molar equivalent of an alcohol, 

normally in the form of an alkoxide (MeCHO and MeO - in MeOH form C in this case).







Pick out the compound with hydrate functionality: 

a. A 

b. B 

c. C 

d. D 

*e. E 


A hydrate is formed from an aldehyde by addition of a molar equivalent of water.







Pick out the molecule which is most electrophilic: 

a. A 

*b. B 

c. C 

d. D 

e. E 


The electrophilic atom in these derivatives is the carbonyl carbon due to the 

electronegative carbonyl oxygen. This electrophilicity is also influenced by the other 

substituents attached to the carbonyl carbon. 

The chlorine atom in B is highly electronegative, and also least likely to donate electrons 

into the C=O double bond, so B is the strongest electrophile. Interesting, the oxygen atom 

in A is the most electronegative atom, but “back-donation” to the C=O bond is 

particularly efficient making it less electrophilic.







Pick out the molecule which is least electrophilic: 

a. A 

b. B 

c. C 

d. D 

*e. E 



electronegative carbonyl oxygen. This electrophilicity is also influenced by the other 

substituents attached to the carbonyl carbon. 

The nitrogen atom in E is moderately electronegative but most readily donates electrons 

into the C=O double bond (through resonance). This fact is reflected in the restricted 

rotation about the OC-N bond. Amide E is the least electrophilic.








a. A 

*b. B 

c. C 

d. D 

e. All have the same electrophilicity 


The electrophilic carbonyl carbons in these molecules vary in the attachment of just one 

substituent. Of these the A, C and D are lone-pair bearing atoms (O and N) which can 

delocalize electrons into the neighbouring C=O as so stabilize the functional group and 

lower their electrophilicity. This leaves B as the most electrophilic.








a. A 

*b. B 

c. C 

d. D 

e. E 


The electrophilic carbon in B is attached to two electronegative atoms in chlorine (sp 3 ) 

and oxygen (sp 2 ), so B is the most reactive towards nucleophiles.







Pick out the molecule which is least electrophilic: 

a. A 

b. B 

*c. C 

d. D 

e. E 


All molecules contain an electrophilic C-Cl bond. Efficient displacement of chloride for 

A, D and E can occur via an SN2 reaction. For the remaining molecules, B and C, the 

corresponding displacement can occur via an addition-elimination reaction. This process 

is extremely disfavoured for vinyl chlorides like C. So C is the least reactive towards 

nucleophiles, and therefore least electrophilic.








a. A 

*b. B 

c. C 

d. D 

e. E 



electronegative carbonyl oxygen. This electrophilicity is dependent on the substitution 

pattern and the substituents present. 

The chlorine atom in B is highly electronegative, and also least likely to donate electrons 

into the C=O double bond, so B is the strongest electrophile. Interesting, the oxygen atom 

in A and D is the most electronegative atom, but “back-donation” to the C=O bond is 

reasonably efficient making it less electrophilic.







Pick out from the following reactions an appropriate synthesis of phenylmethyl ketone 

(PhCOMe): 

a. A and B 

b. B and C 

c. C and D 

d. A, B and C 



In equation A the first product formed is a ketone, and this may react again with the 

Grignard reagent and go on to form an alcohol. Hence this method is the least attractive 

but can work if the Grignard reagent is added slowly to the acid chloride (and in not the 

reverse order). 

In the case of equation D the Grignard reagent adds to the nitrile to form the stable 

magnesium salt of the imine, Ph(Me)C=NMgBr. Only upon addition of aqueous acid 

during work up is the double bond hydrolysed to give the ketone. 

In equation B the carboxylic acid reacts with the first equivalent of MeLi to produce a 

lithium salt, which then reacts with a second equivalent of the highly reactive MeLi to 

produce a stable dilithium dialkoxide. Upon acidic work up, the ketone hydrate formed 

then loses water to produce the ketone. 

Equation C is a standard Friedel-Crafts acylation reaction.







From the following pairs of molecules, pick out the tautomeric pair: 

a. A 

*b. B 

c. C 

d. A and B 

e. A, B and C 


In an enol, the tautomeric form of a ketone, the hydroxyl is always attached to a carbon 

atom which is part of the double bond.







From the following pairs of molecules, pick out the keto-enol tautomeric pair: 

a. A 

*b. B 

c. C 

d. None of the above 



In enol-keto tautomerism, the carbonyl group is replaced with a hydroxyl attached to a 

carbon atom which forms part of a C=C double bond.







From the following pairs of molecules, pick out the keto-enol tautomeric pair: 

a. A and B 

*b. C and D 

c. A and C 

d. B and D 



Tautomers have the same molecular formula, different structures but are in equilibrium 

with one another. In enol-keto tautomerism, the carbonyl group is replaced with a 

hydroxyl attached to a carbon atom which forms part of a C=C double bond which can be 

seen in C and D. In A though, the enolate (negatively charged) is closely related to the 

enol, but is not a tautomer of the ketone.







Pick out from the following reaction(s) which gives benzaldehyde as the product: 

a. A 

b. B 

c. C 

d. A and C 



Potassium dichromate, in A, oxidizes the alcohol to the aldehyde (this only goes on to the 

acid in the presence of water). In solution, the hydrate B readily loses water to form the 

aldehyde, and the acetal C can also form the aldehyde in a similar process but requires an 

acid catalyst.







Pick out from the following reaction(s) which give benzyl alcohol as the product: 

a. A 

*b. B 

c. C 

d. A and C 

e. A, B and C 


Potassium permanganate, in A, is an oxidant and so cannot produce benzyl alcohol from 

benzoic acid (a reduction). 

Hemiacetal C will be converted to the benzaldehyde with acidic water, but as there is no 

reducing agent present, so no further reaction will take place. 

Hydrate B is converted in water to benzaldehyde and in the presence of sodium 

borohydride will be reduced to the required alcohol.







Pick out from the following reactions an appropriate synthesis of compound Z: 

a. A 

b. B 

c. C 

d. A and C 



In general a secondary alcohol can be produced from either combination of aldehyde and 

Grignard reagent (A or C) or by reduction of the corresponding ketone B.







Pick out from the following reactions an appropriate synthesis of compound Z: 

a. A 

b. B 

c. C 

d. A and B 



Addition of methyl magnesium bromide to the carbonyl group of ketone A, followed an 

acidic work-up leads to the tertiary alcohol Z. 

Addition of methyl magnesium bromide to the carbonyl group of ester B, followed by 

elimination of magnesium ethoxide (BrMgOEt) leads to the intermediate formation of 

ketone A. This in-turn is converted to tertiary alcohol Z by the process outlined about. 

Double addition of phenyl magnesium bromide to the carbonyl group of ester E leads to 

the formation of related diphenylmethyl tertiary alcohol. 

Both carbonyl derivatives, A and B, lead to the required tertiary alcohol Z.







Pick out the compound which could be prepared using a Mannich reaction: 

*a. A 

b. B 

c. C 

d. D 

e. E 


The Mannich electrophile, Me2N + =CH2, is an iminium salt produced from the reaction of 

a secondary amine (Me2NH) and a non-enolizable aldehyde (formaldehyde CH2O). This 

electrophile is readily attacked by enols to produce beta-amino carbonyl compounds, 

such as A.







Pick out the compound which could be prepared using an Aldol condensation under 

acidic conditions: 

a. A 

b. B 

*c. C 

d. D 

e. E 


The aldol reaction is a condensation of an enol (or enolate if using base) with a ketone or 

aldehyde to produce a beta-hydroxy carbonyl compound, such as B. Under acidic 

conditions, this readily dehydrates to produce a conjugated enone C.







Pick out the compound which could be prepared using an Aldol reaction: 

a. A 

*b. B 

c. C 

d. D 

e. E 


The aldol reaction is a condensation of an enol (or enolate if using base) with a ketone or 

aldehyde to produce a beta-hydroxy carbonyl compound B.







Order the following molecules with respect to their ability to form hydrates: 

a. A > B > C > D 

*b. B > C > A > D 

c. C > B > D > A 

d. D > A > C > B 

e. B > D > A > C 


The ability to form a hydrate (by addition of water across the C=O bond) can be seen as 

measure of the stability of a carbonyl group. 

Aldehydes, like B, are most reactive (unstable) as they are least sterically hindered and 

possess only one +I (stabilizing) substituent attached to the electrophilic carbonyl carbon. 

The ketone C has two +I (stabilizing) substituents. 

The ester A, followed by the amide A, are more stable (less electrophilic) due to 

delocalization of the lone pair of electrons on the heteroatom onto the carbonyl pi-bond.








a. A > D > B > C 

b. D > B > C > A 

*c. C > D > B > A 

d. B > A > C > D 

e. D > C > A > B 


The ability to form a hydrate (by addition of water across the C=O bond) can be seen as 


All these compounds are ketones which have two +I (stabilizing) substituents attached to 

the electrophilic carbonyl carbon, but steric effects means that the smallest groups (Me) 

hinder hydrate formation the least, and the biggest (Bu) the most. In addition, hydrate 

formation is promoted by having less hydrophobic substituents present.








a. A > D > B > C 

b. B > A > C > D 

c. C > B > D > A 

*d. D > A > C > B 

e. B > A > D > C 


The ability to form a hydrate (addition of water across the C=O bond) can be seen as 


Factors affecting this are the electronic and steric effects of substitutents attached to the 

electrophilic carbonyl carbon. Hydrogen bond acceptors also aid hydrate formation. 

The central C=O in D is flanked by electron withdrawing (activating) CO groups and so 

is the most unstable (most reactive). Stability of the hydrate is promoted through 

intramolecular hydrogen bonding between the central C=O bond and the hydrate motif. 

Formaldehyde A has no stablizing groups and is unhindered. Intermolecular hydrogen 

bonding (solvation) aids stabilization of the hydrate. 

Ketones B and D look similar in terms of steric and electronic effects, but the 

cyclohexanone is the least reactive as the hydrate formed is less stable due to increased 

1,3-diaxial interactions in the cyclohexyl ring. In addition, ketone C, being less 

hydrophobic, has better solvation through intermolecular hydrogen bonding.








a. A > D > B > C 

b. B > C > D > A 

c. C > B > A > D 

d. D > A > C > B 

*e. B > A > D > C 






Strained cyclopropanone B is the most ready to form a hydrate as the ring bond angle is 

far from the ideal trigonal planar angle. This strain is relieved somewhat upon formation 

of a near tetrahedral hydrate. 

Formaldehyde A is more reactive (unstable) than aldehyde D which in turn is more 

reactive than ketone C. Across this sequence more +I methyl groups are being added to 

the electrophilic C=O carbon atom. In addition, the less hydrophobic the carbonyl 

compound (and the corresponding hydrate), the better the solvation through 

intermolecular hydrogen bonding (i.e., A > D > C).








a. A > D > B > C 

b. B > C > D > A 

c. C > B > A > D 

d. D > C > A > B 

*e. D > A > B > C 






Trichloraldehyde D, is most reactive (unstable), whereas, the hydrate is most stable due 

to intramolecular H-bonds being formed from the OH of the hydrate and the Cl atoms. 

Formaldehyde A is more reactive (unstable) than aldehyde B which in turn is more 

reactive than ketone C. Across this sequence more +I methyl groups are being added to 

the electrophilic C=O carbon atom. In addition, the less hydrophobic the carbonyl 

compound (and the corresponding hydrate), the better the solvation through 

intermolecular hydrogen bonding (i.e., A > B > C).







Order the following molecules with respect to their ability to form enols: 

a. A > D > B > C 

b. B > A > D > C 

c. C > B > A > D 

*d. D > C > B > A 

e. D > A > B > C 


A cannot form an enol at all (having no acidic H atoms on a carbon alpha to the C=O), 

and B is least likely of the rest to form an enol as this would mean forming a highly 

strained cyclopropene ring. 

During enolization, the C=O bond is broken and so if there is a strong C=O bond this 

process is energetically less favourable than if the bond is weaker. Therefore, the 

aldehyde D has a greater enol content compared to the ketone C due to its lower C=O 

stability.








a. A > D > B > C 

*b. B > C > A > D 

c. C > B > A > D 

d. D > C > B > A 

e. (D = A) > (B = C) 


During enolization the C=O bond is broken and so if there is a strong C=O bond this 

process is energetically less favourable than if the bond is weaker (i.e. less enolization). 

Aldehyde B > ketone C as there is only one +I methyl groups stabilizing the electron 

deficient C=O in the aldehyde compared to the ketone. 

Ester A > amide D has lower enol content because the lone pair on the more 

electronegative oxygen is less readily delocalized into the C=O compared to the nitrogen.







Pick out the molecule which has highest enol content: 

a. A 

*b. B 

c. C 

d. D 

e. E 


Formaldehyde A and aldehyde D cannot form enols since they do not have any acidic H 

atoms on a carbon alpha to the C=O. The cyclopropanone enol is highly strained due to 

the three membered ring. 


process is energetically less favourable than if the bond is weaker. Therefore, aldehyde B 

has a greater enol content compared to the ketone C.







Pick out the molecule which has highest enol content: 

a. A 

b. B 

*c. C 

d. D 

e. E 


Formaldehyde A and diketone D cannot form enols since they do not have any acidic H 

atoms on a carbon alpha to the C=O. 


process is energetically less favourable than if the bond is weaker. Therefore, the ketone 

C has a greater enol content compared to the ester E and the amide B due to the absence 

of conjugation.








a. A > B > C 

b. B > A > C 

c. C > A > B 

*d. A > C > B 

e. B > C > A 


The enol of ketone A is phenol which benefits greatly from aromatic stability. The enol 

of C produces a conjugated diene-like structure, but ketone B produces a simple enol 

with no particular stabilizing features.








*a. A > B > C 

b. B > C > A 

c. C > A > B 

d. C > B > A 

e. A > C > B 


The enol of ketone A is phenol which benefits greatly from aromatic stability. The enol 

of B produces a stabilized conjugated triene structure. The enol of C produces a 1,3,5,7cycloctatetraene. 

This molecule however cannot adopt a planar form as the bond angles 

would be too strained and deviated from the preferred trigonal 120º. It therefore adopts an 

approximately U shaped structure and any conjugation of the double bonds is disrupted. 

Anti-aromatic enols (of C) by their very nature are very unstable.








a. A > B > C 

b. B > C > A 

c. C > A > B 

d. A > C > B 

*e. C > B > A 


Compounds B and C are more ready to form enols compared to A due to the existing 

C=C bond to which the enol C=C will be conjugated. Compound C possesses aring 

oxygen bearing a lone pair of electrons in an sp2 orbital. The electron pair is conjugated 

to both the C=C bonds within the ring and so this tautomer is aromatic (like furan) and 

will exist completely in the enol form.








*a. A > B > C 

b. B > C > A 

c. C > A > B 

d. B > A > C 

e. C > B > A 


These three molecules are all 1,3-dicarbonyl compounds and so can form an enol in 

which the new C=C bond is conjugated to the remaining C=O bond and the enol OH 

bond can be additional stabilized through intramolecular hydrogen bonding. 

The differences between these compounds are the number of ester versus ketone groups. 

The lone pair of the oxygen atom (within the ester C-O motif) can be conjugated into the 

C=O bond, thus making it more stable than that of a ketone. This makes enolization of an 

ester energetically less favourable than for the related ketones.







Pick out the most electrophilic carbonyl derivative: 

a. A 

b. B 

c. C 

d. D 

*e. E 


Electrophilicity is determined by both electronic and steric factors. Aldehydes are 

generally more reactive than ketones and esters. However, the cylopropanone E is highly 

strained as the ring angle (approx. 60 degrees) is far from the trigonal planar angle of 120 

degrees for an acyclic ketone. Upon nucleophilic addition, this strain is partially relieved 

upon adoption of a near tetrahedral arrangement (109 degrees).







Pick out the most nucleophilic carbonyl group: 

*a. A 

b. B 

c. C 

d. D 

e. E 


The ester has the most nucleophilic C=O group as the neighbouring oxygen has a lone 

pair of electrons delocalized into the C=O bond. This increases the electron density on 

the oxygen C=O atom.







Order the following molecules with respect to their nucleophilic character: 

a. A > D > B > C 

b. B > C > A > D 

c. C > B > A > D 

*d. D > A > C > B 

e. (B = C) > (D = A) 


The amide has the most nucleophilic C=O group, followed by the ester, as the 

neighbouring heteroatom has a lone pair of electrons delocalized into the C=O bond; 

amide D > ester A as oxygen is more electronegative than nitrogen. 

The ketone is more nucleophilic than the aldehyde (C>B) as the extra +I methyl group 

reduces the electron deficiency of the carbonyl carbon. This means the more 

electronegative oxygen has a greater share of the electron density in the C=O bond.







Pick out the ester which is most easily hydrolyzed in acidic solution: 

a. A 

b. B 

c. C 

*d. D 

e. E 


Under acidic conditions, esters are hydrolysed via nucleophilic attack of water at the 

carbonyl carbon of the protonated double bond, C=O + H. The hindered tert-butyl ester, 

however, upon protonation of the C=O, preferably fragments by loss of the tert-butyl 

carbocation and are thus most easily hydrolysed under acidic conditions.






588. Reagents 

Pick out the most appropriate reagent, Z, for the following interconversion: 

*a. A 

b. B 

c. C 

d. D 



In A, the organocadmium reagent is sufficiently nucelophilic to react with the more 

electrophilic acid chloride but not the less electrophilic ketone. This reaction therefore 

stops after formation of the corresponding ketone. However, under similar conditions, the 

Grignard reagent B will react readily with this ketone and so, unless you are very careful, 

a tertiary alcohol will be produced. Methyl lithium, C, can only be used to produce the 

tertiary alcohol, and the borane D will not react.






589. Reagents 2 

Pick out the appropriate reagents, Z, for the following interconversion: 

a. A 

b. B 

c. C 

d. D 



Grignard reagents in A, B and C are formed by addition of magnesium (Mg) to an alkyl 

or aryl halide (e.g., PhBr or EtBr). 

A simple, unsymmetrical tertiary alcohol can in principal be made from any combination 

of a ketone and Grignard reagent (A, B and C). The hydration of an alkene may present 

issues of regioselectivity, but here this reaction proceeds via a highly stabilized tertiary 

cation conjugated to the phenyl ring and so only the product shown is produced.







Pick out the appropriate reagents for the following interconversion: 

a. A 

b. B 

c. C 

d. A and B 



Of these esterification methods, A is the most general. Method B can be used if the alkyl 

halide is sufficiently electrophilic, and the diazoalkane method C is mostly restricted 

towards making methyl esters on a small scale.








*a. A 

b. B 

c. C 

d. D 

e. E 


In A, the carboxylic acid reacts with the first equivalent of MeLi to produce a lithium 

carboxylate salt, which then reacts at the C=O with a second equivalent of the highly 

reactive MeLi to produce a stable dilithium dialkoxide. Upon acidic work up, the 

(ketone) hydrate is formed which subsequently loses water to produce the ketone, 

acetophenone.








a. A 

b. B 

c. C 

d. D 

*e. E 


PCl5 can be used to make the acid chloride (from the carboxylic acid directly), but NOT 

via the ester as in D.







Pick out the most appropriate reagents for the following interconversion: 

*a. A 

b. B 

c. C 

d. D 

e. E 


Reagents B will hydrolyse the ester to the carboxylate salt but this is inert towards the 

iodobenzene. Reagents D will produce a phenyl ester, and Lewis acid triphenyl borane E 

will simply coordinate to the C=O oxygen. The Grignard reagent C will also react readily 

with first formed ketone product and so, unless you are very careful, a tertiary alcohol 

will be produced. 

Using reagent A, however, the organocadmium reagent is sufficiently nucleophilic 

(reactive) to react with the more electrophilic acid chloride but not the less electrophiic 

ketone. This reaction therefore stops at the ketone stage.








a. A 

b. B 

*c. C 

d. D 

e. E 


The Grignard reagent A will react more readily with product so producing a tertiary 

alcohol. Reagents B will hydrolyse the ester to the salt and then reform the methyl ester 

starting material. The cuprate D is not reactive towards the ester, and Lewis acid 

trimethyl borane E will simply coordinate to the C=O oxygen. 

With reagents C, the first-formed carboxylic acid reacts with the first equivalent of MeLi 

to produce a lithium salt, which then reacts at the C=O with a second equivalent of the 

highly reactive MeLi to produce a stable dilithium dialkoxide. Upon addition of acid 

(acidic work up), the ketone hydrate is formed which loses water to produce the more 

stable ketone.








a. A 

b. B 

c. C 

*d. D 

e. E 


This reaction is a reduction but sodium hydride, E, is a base and hydrogenation, C, 

cannot be used to reduce the ester. 

The other reducing agents have differing strengths where only DIBAL can be used to 

reduce esters to aldehydes; sodium borohydride is a weaker reducing agent and converts 

aldehydes to alcohols and the powerful lithium aluminium hydride, B, reduces esters all 

the way to the alcohols (via the corresponding aldehyde).








a. A 

b. B 

*c. C 

d. D 

e. E 


This particular reduction of acid chlorides is known as the Rosenmund reduction.








*a. A 

b. B 

c. C 

d. D 

e. E 


Reagents B and C are not nucleophilic enough to react the chosen ketone, acetophenone. 

Reagent E produces the corresponding hemiketal, whereas, reagent C is inert. 

Grignard reagents, like A, react efficiently with ketones, such as acetophenone, to give 

the corresponding tertiary alcohol.








*a. A 

b. B 

c. C 

d. D 

e. E 


The Grignard reagent A adds to the nitrile to form the stable magnesium salt of the imine, 

Ph(Me)C=NMgBr. Upon addition of water - during work up procedure - the double bond 

is hydrolysed to the ketone. None of the other reagents (on their own) are nucleophilic 

enough to add to the CN triple bond.








a. A 

*b. B 

c. C 

d. D 

e. E 


This reaction is a reduction but sodium hydride, E, is a base and hydrogenation, C, 

cannot be used to reduce the ester. The other reducing agents have differing strengths 

where DIBAL reduces esters preferentially to aldehydes, sodium borohydride reduces 

aldehydes to alcohols and only lithium aluminium hydride, B, being able to reduce esters 

to alcohols.








a. A 

*b. B 

c. C 

d. D 

e. E 


This reaction is an oxidation but reagents in C and D are reducing agents and hydroxide, 

E, is a base. Potassium dichromate, A, can only oxidize the aldehyde on to the acid in the 

presence of water. This process occurs via formation of an intermediate hydrate (due to 

the presence of water) which in turn is oxidized to give the corresponding carboxylic 

acid.








a. A 

*b. B 

c. C 

d. D 

e. E 


This reaction is an oxidation but reagents in C and D are reducing agents and hydroxide, 

E, is a base. Potassium dichromate, A, can only oxidize the intermediate aldehyde on to 

the acid in the presence of water. This process occurs via formation of an intermediate 

aldehyde and its corresponding hydrate (due to the presence of water) which in turn is 

oxidized to give the corresponding carboxylic acid.








a. A 

b. B 

c. C 

d. D 

*e. E 


The reaction starts with condensation of the aldehyde with hydroxylamine to form the 

oxime (PhCH=NOH). Thionyl chloride then causes the dehydration of the oxime to give 

the required nitrile.








a. A 

b. B 

c. C 

d. D 

*e. E 


The hydrazine condenses with the ketone to form a hydrazone (Ph2C=N-NH2). The 

strong base then causes a tautomerisation to form Ph2CH-N=NH which then can 

thermally eliminate the very stable nitrogen molecule and form the hydrocarbon. This is 

an example of the Wolff-Kishner reduction.








a. A 

b. B 

c. C 

*d. D 

e. E 


This is an example of the iodoform reaction which proceeds via iodination of the ketone 

enolate to form the phenyl tri-iodomethylketone. Hydrolysis of this product forms 

iodoform (CHI3) and sodium benzoate which gives the benzoic acid upon addition of 

HCl.






605. Eq Constant 

Pick out the approximate equilibrium constant for the following process: 

a. 10 -2 

*b. 10 -4 

c. 10 -8 

d. 10 -12 

e. 10 -16 


The pKA values shown lead to the equilibrium constants for the ionization of the two 

acids acetone and ethanol. 

As pKA = -log10KA these KA values are 10 -20 and 10 -16 respectively. Thus the overall KA 

values is 10 -20 /10 -16 = 10 -4 .






606. Eq Constant 2 


a. 10 -3 

b. 10 -6 

*c. 10 -9 

d. 10 -12 

e. 10 -15 



acids ethyl acetate and ethanol. 


values is 10 -25 /10 -16 = 10 -9 .








a. 10 2 

b. 10 4 

c. 10 6 

*d. 10 8 

e. 10 10 



acids ethyl acetate and diisopropylamine. 


values is 10 -25 /10 -33 = 10 8 .








a. 10 3 

b. 10 4 

c. 10 5 

d. 10 6 

*e. 10 7 


The pKA values shown lead to the equilibrium constants for the ionization of 

the two acids acetyl acetone and ethanol. 


values is 10 -9 /10 -16 = 10 7 .






Organic Reactions 










a. A 

*b. B 

c. C 

d. D 

e. E 


Protonation of the alkene gives preferentially the more stable tertiary carbocation, which 

in turn gives rise to the more substituted alcohol B. Possible minor by-products are the 

alcohols C and D which will be formed from the less stable secondary cation.








a. A 

b. B 

c. C 

*d. D 

e. E 


Borane adds in a syn-fashion across the carbon-carbon double bond, with boron adding to 

less substituted end, to give an alkylborane compound. This derivative with anti 

stereochemistry is then oxidized in a separate step to the alcohol D, giving a net anti- 

Markovnikov addition of water across the double bond.








a. A 

b. B 

c. C 

*d. D 

e. E 


The action of UV light on chlorine results in Cl-Cl homolysis and gives rise to chlorine 

free radicals. These abstract a hydrogen atom at the benzylic CH2 position to give a 

resonance stabilized radical. This reacts with chlorine to give benzyl chloride D and a 

new chlorine atom to perpetuate the chain reaction.








a. A 

*b. B 

c. C 

d. D 

e. E 


Addition of a proton to the alkyne occurs at the unsubstituted end to give a cation, which 

is resonance stabilized by the neighbouring phenyl ring. This in-turn captures a chloride 

ion to form 1-chloro-1-phenylethene. The second HCl addition occurs in the same 

orientation due to the same factors above, which is also favoured by the +M effect of the 

first chlorine.








*a. A 

b. B 

c. C 

d. D 

e. C and D 


The action of UV light on bromine produces bromine free radicals via homolysis of the 

Br-Br bond. A bromine radical in-turn can abstract a proton from the allylic carbon to 

produce a resonance-stabilised radical. This reacts readily with bromine to produce the 

allylic bromide A, and another bromine radical to perpetuate the chain reaction.








*a. A 

b. B 

c. C 

d. D 

e. E 


Reduction of the alkyne by sodium metal goes through radical anion and radical 

intermediates formed by sequential addition of electrons to the multiple bond. The anions 

formed can abstract a proton from the ammonia solvent leading to the intermediate transbut-2-en-2-yl 

anion, protonation of which gives the required trans-but-2-ene.








a. A 

b. B 

c. C 

*d. D 

e. E 


Deprotonation of chloroform with a strong base forms dichlorocarbene (:CCl2). This can 

add to alkenes to produce dichlorocyclopropanes. If the alkene is constrained by being 

endocyclic then this must produce a cis-fused ring junction.








a. A 

*b. B 

c. C 

d. D 

e. E 


Bromine is polarised by the C=C double bond, and “Br + ” adds to the double bond to form 

the cyclic bromonium ion E. The liberated bromide can attack the C-Br bond with 

inversion to produce the trans-1,2-dibromide B.








*a. A 

b. B 

c. C 

d. D 

e. E 


Protonation of the double bond produces the more stable tertiary carboation (c.f. 

Markovnikov addition). The liberated bromide can then attack this cation to produce the 

tertiary bromide A.








a. A 

*b. B 

c. C 

d. D 

e. E 


Step 1: Friedel Crafts acylation using AlCl3 produces acetophenone (methyl phenyl 

ketone). 

Step 2. The amine condenses with the ketone to form an N-methylimine 

[Ph(Me)C=NMe)and water as the by-product. 

Step 3. Borohydride reduces the imine to give the secondary amine B - the final product.








*a. A 

b. B 

c. C 

d. D 

e. E 


This is an example of an intramolecular (the two ketone groups are in the same molecule) 

acid-catalyzed aldol reaction. This reaction proceeds via single enol formation and 

intramolecular carbonyl addition to give a beta hydroxyl ketone. This readily dehydrates 

under acid conditions to form the conjugated unsaturated ketone A. As the ketone 

components in the starting molecule are equivalent, it does not matter from which enol 

instigates the reaction.








*a. A 

b. B 

c. C 

d. D 

e. E 


This is an example of a base-catalyzed crossed aldol condensation in which the acetone 

enolate reacts with the benzaldehyde electrophile. The aldehyde has no acidic alphahydrogens 

so cannot form an enolate. This crossed reaction takes place in preference to a 

self condensation as the aldehyde electrophile is more reactive than the ketone. This 

reaction initially forms molecule B, but this dehydrates to form the stable, conjugated 

enone A as the final product.








a. A 

*b. B 

c. C 

d. D 

e. E 


Lithium diisopropyl amide (LiNi-Pr2; LDA) is a well known sterically demanding strong 

base. It can readily deprotonate a ketone to form the less substituted lithium enolate (the 

kinetic enolate). This in-turn can react with an alkyl halide (electrophile) to produce the 

required dimethyl ketone B.








*a. A 

b. B 

c. C 

d. D 

e. E 


The first step is an example of the Strecker reaction. The ammonia condenses with the 

given aldehyde to produce an imine (RCH=NH). Cyanide then attacks the C=N bond to 

produce the cyano amine D. The final step, involving aqueous acid, hydrolyses the nitrile 

to the carboxylic acid producing product A. This sequence is commonly used to convert 

an aldehyde into an alpha-amino acid.







For the following reaction, pick out the major product Z: 

*a. A 

b. B 

c. C 

d. D 



Br-Cl is a polarized molecule in which the more electronegative Cl atom bears a partial 

negative charge and the Br atom a partial positive charge. The Br atom is therefore 

electrophilic. When this reacts with an alkene, a cyclic bromonium ion is formed. The 

remaining chloride ion can then attack from the opposite side to the C-Br bond(s) to 

produce the anti-isomer of 1,2-bromochloroalkane (rather than the syn). As the alkene is 

unsymmetrical, the positive charge on the bromonium ion is preferentially centered on 

the more substitution tertiary centre so isomer A is produced in preference to isomer B.








*a. A 

b. B 

c. C 

d. D 

e. E 


Hydroxylamine attacks the C=O group using the more nucleophilic nitrogen atom. An 

intermediate in this process is the hydroxyamine D, but this eliminates water and so the 

more stable oxime A is produced.






625. Reagents1 

Pick out the appropriate sequence of reagents for the following interconversion: 

*a. A 

b. B 

c. C 

d. D 



The product is an anti-1,2-diol. The ring opening of an epoxide is a good synthetic 

approach for making this type of molecule. 

NaOEt is a base which assists elimination of the chloride to an alkene. The thiophenolate 

(in sequence C) on the other hand is an excellent nucleophile, and so an SN2 process will 

occur to give a sulfide. m-CPBA (meta-chloroperoxybenzoic acid) produces epoxides 

from alkenes, and then hydroxide (HO-) can open this epoxide to give the required anti- 

1,2-diol. In sequence B, osium tetraoxide (OsO4) produces a syn-1,2-diol from an alkene. 

In sequence D, an alcohol is produced.







Pick out the appropriate reagent(s), Z, for the following interconversion: 

a. A 

b. B 

*c. C 

d. D 

e. E 


Addition of molecular hydrogen across an alkene to produce an alkane can only take 

place in the presence of a catalyst. Palladium (Pd) is one example commonly used and 

this is most often supported on powdered graphite (carbon) (“Pd-on-C” or “Pd-C”).







Pick out from the following options an appropriate synthesis of molecule Z: 

*a. A 

b. B 

c. C 

d. D 



The terminal proton in an alkyne is relatively acidic (pKA = 25), and therefore can be 

deprotonated with sodium amide. Alkylation with an alkyl halide then produces the 

required di-substituted alkyne. Reagents in reaction B produces an alkene, as do the 

reagents in answer C. In this case, an initial organometallic is formed, but the leaving 

group on the beta-carbon is lost in an E1cb-type mechanism. In answer D, the initial 

acetylide forms, but addition of an alcohol results in simple deprotonation of the alcohol 

to give back the alkyne and sodium ethoxide.







Pick out from the following options an appropriate synthesis of molecule Z: 

a. A 

b. B 

c. C 

d. D 



In answer A: The terminal proton of the alkyne is deprotonated with sodium amide. 

Alkylation with an ethyl iodide produces the required product. 

In answer B: An initial organometallic, RZnBr, is formed by insertion in the weak C-Br 

bond. However, the leaving group (bromide) on the beta-carbon is then lost in an E1cbtype 

mechanism to give the corresponding alkene. This process happens again, but this 

time involving the C-Cl bonds, to install the new pi-bond in alkyne component. 

In answer C: An initial organometallic, RZnBr, is formed by insertion in the weak C-Br 

bond. E1cb-type mechanism of the remaining bromide leads to the required alkyne. 

In answer D: This is an example of an E2 elimination. Concerted deprotonation of the 

bromoalkene using the strong base sodium amide, and subsequent loss of bromide, leads 

to the required alkyne.








a. A 

b. B 

c. C 

d. D 

*e. E 


Sodium borohydride is a weak reducing agent, and will therefore normally react with 

moderately electrophilic species, such as aldehydes and ketones, leaving the less 

electrophilic ester functional group intact.








*a. A 

b. B 

c. C 

d. D 

e. E 


Lithium aluminium hydride is a powerful reducing agent and can reduce all carbonyl 

containing groups, such as aldehydes, esters, ketones, acids, amides etc.








a. A 

*b. B 

c. C 

d. D 

e. E 


DIBAL is an electrophilic source of hydride. By its very nature, it preferentially reduces 

nucleophilic carbonyl motifs, such as esters and amides, in the presence of aldehydes and 

ketones. 

Note: the nucleophilic component of a carbonyl group is the oxygen atom of the C=O 

bond.








*a. A 

b. B 

c. C 

d. D 

e. E 


The aluminium-catalyzed hydride shift from the α-carbon of the propan-2-ol to the 

carbonyl carbon the ketone, which proceeds via a six-membered transition state, is named 

Meerwein-Ponndorf-Verley (MPV) reduction. 

The other product of this disproportionation is propanone. This reaction is also known as 

the Oppenauer Oxidation, if viewed from the perspective of the alcohol component. Nonenolizable 

ketones can serve as the carbonyl component in this case.








a. A 

b. B 

c. C 

d. D 

*e. E 


This reaction proceeds via bromination of the enol tautomer which is formed by acid 

catalysis. 

If a base is used instead of acid (as with process D), the increased acidity of the alphaproton 

in the product (due to the electron withdrawing halogen) allows further 

bromination to occur until the 1,1,1-tribrormopropanone is produced. This particular 

product in-turn can be hydrolysed with hydroxide to give bromoform and a carboxylic 

acid (c.f., the bromoform reaction).








a. A 

b. B 

c. C 

*d. D 

e. E 


This is the bromoform reaction. It proceeds via sequential bromination of an enol 

tautomer until 1,1,1-tribromopropanone is produced. The CBr3 group is sufficiently 

electron withdrawing for the hydroxide to hydrolyse the O=C-CBr3 bond to form the acid 

(like hydrolysis of an acid chloride) and bromoform (CHBr3).








a. A 

b. B 

c. C 

d. D 

*e. E 


Addition of the methyl Grignard reagent (CH3MgBr) to the ketone produces a tertiary 

alcohol with the correct number of carbon atoms. Elimination of the water to form the 

alkene is then readily achieved with acidic conditions (process E) as opposed to basic 

conditions (process D).








a. A 

*b. B 

c. C 

d. D 

e. E 


The ethoxide readily deprotonates alpha to the ester C=O, aided by the electron 

withdrawing chlorine substitutent. The resulting enolate undergoes an addition to the 

ketone C=O to form an intermediate alkoxide which then can trigger nucleophilic 

displacement of the chlorine to form an epoxide. This reaction is known as the Darzens’ 

reaction. In processes A or C, without the chlorine atom, the ethoxide base would cause a 

self-condensation (aldol) reaction of the ketone.








a. A 

b. B 

c. C 

d. D 



These methods are all successful for the given transformation. It is interesting to note, a 

variety of conditions are used, such as acid, base and hydrogenation. Therefore, 

compatibility with other functional groups within different substrates may direct use of 

one particular set of reagents over another.








a. A 

b. B 

c. C 

*d. D 

e. E








a. A 

b. B 

*c. C 

d. D 



In this reaction, there is a substitution with inversion of configuration at the chiral 

(stereogenic) centre. It is well documented that SN2 processes result in inversion of 

configuration. In process C, a sulfonate ester leaving group is initially formed with no 

change (retention) of configuration. Cyanide displacement proceeds with inversion of 

configuration and results with the loss of the sulfonate ester leaving group.








a. A 

b. B 

*c. C 

d. D 



In this reaction, there is a substitution with net retention of configuration at the chiral 

(stereogenic) centre. An SN2 process leads to a single inversion of configuration, 

whereas, two consecutive SN2 processes will result in the required net retention (double 

inversion). 

In process A, a sulfonate ester leaving group is initially formed (with no change in 

configuration) and then cyanide displaces the leaving group with inversion; i.e., the 

enantiomer of the required product is formed. In process C, the alcohol is converted by 

addition of thionyl chloride (SOCl2) to the alkyl chloride with inversion of conversion. 

Displacement of the chloride by the cyanide anion (with another inversion of 

configuration) leads to the required product.






641. Statement 

For the acid-catalysed dehydration of tertiary alcohols, which of the following statements 

are true: 

a. A 

b. B 

c. C 

d. D 








Pick out the alcohol which would most easily dehydrate under acidic conditions: 

a. A 

*b. B 

c. C 

d. D 

e. E 


Protonation of the hydroxyl leads to loss of water and formation of a carbocation (E1 

mechanism). This cation is stabilized in the case of alcohol B by the inductive effects of 

the three methyl groups through hyperconjugation. At the other extreme, the electron 

withdrawing NO2 group in alcohol E will destabilizes any cation formed, and so this 

alcohol will eliminate the slowest. Alcohol C cannot eliminate water as it has no beta 

proton!







Pick out the alcohol which would most easily dehydrate under basic conditions: 

a. A 

b. B 

c. C 

d. D 

*e. E 


Base-catalyzed elimination takes place via a concerted E2 mechanism and abstraction of 

a proton beta to the leaving group (OH). Alcohol C cannot eliminate water as it has no 

beta proton! Unlike the other cases, in alcohol E the transition state is stabilized by 

conjugation of the incipient pi-bond with the nitro group. This lowers the energy of the 

transition state leading to a lower activation energy barrier, and therefore results in a 

faster elimination process.







Pick out the alcohol which would most readily dehydrate under acidic conditions: 

a. A 

*b. B 

c. C 

d. D 

e. E 


Protonation of the hydroxyl group and subsequent loss of water leads the formation of a 

carbocation via an E1 mechanism. For alcohol C, the resulting carbocation is highly 

conjugated, and therefore stable, but there is no beta proton to be lost to form an alkene. 

At the other extreme, the electron withdrawing CF3 in alcohol D destabilizes any cation 

formed. Alcohol E most readily dehydrates as this leads to a stable cation which has a 

beta-hydrogen for the all important deprotonation step.







Order the following molecules with respect to their ability to form carbocations on 

treatment with acid: 

a. A > B > C 

b. B > C > A 

*c. C > A > B 

d. A > C > B 

e. B > A > C 


Protonation of the hydroxyl group and subsequent loss of water leads the formation of a 

carbocation via an E1 mechanism. For alcohol C, the resulting carbocation is highly 

conjugated, and therefore most stable. In both alcohols B and A, a tertiary cation is 

formed, however, alcohol B cannot form a trigonal planar shape due to the restricted 

flexibility of the bicyclic system. The opportunity for hyperconjugation to stabilize this 

carbocation (derived from alcohol B) is greatly reduced.







Pick out the reaction which occurs with anti-addition to an alkene: 

*a. A 

b. B 

c. C 

d. D 

e. E 


Reagent A gives a 1,2-dibromide via bromide anion attack on the resulting bromonium 

ion. This addition process occurs on the opposite face of the bromonium ion leading the 

formation of an anti-1,2-dibromide. Reagent B produces an epoxide, reagent D a 1,2-diol, 

and reagent E an alkane, all via syn-addition mechanisms. Reagent C produces an alcohol 

via a trigonal planar cation.







Pick out the reactions which occur with syn-addition to an alkene: 

a. A, B and C 

b. C, D and E 

c. A, C and E 

*d. B, D and E 

e. B, C and D 


Reagent A produce a 1,2-dibromide from an alkene, reagent B produces an epoxide, 

reagent C an alcohol, reagent D a 1,2-diol, and reagent E an alkane.







Pick out the molecule which would have the lowest heat of hydrogenation: 

a. A 

b. B 

c. C 

d. D 

*e. E 


The heat released in the hydrogenation of an alkene is a measure of its thermodynamic 

stability. Factors contributing to stability of an alkene are conjugation, double bond 

geometry, whether cyclic or not and, in this example, the degree of substitution. Alkene 

E, with an exocyclic double bond, is the least stable, and therefore has the lowest heat of 

hydrogenation.







For a Diels-Alder reaction, pick out the least reactive diene: 

a. A 

b. B 

c. C 

d. D 

*e. E 


In a Diels Alder reaction the diene must be in the s-cis conformation; i.e., the two double 

bonds must be in a synperiplanar conformation. In all cases, except for diene E, may 

adopt the correct conformation by rotation about a C-C bond. However, diene E can 

never adopt the correct conformation as the C-C bond between the alkenes is constrained 

within the ring. Therefore, E is unreactive in a Diels-Alder reaction.







For a Diels-Alder reaction, pick out the most reactive diene: 

a. A 

b. B 

c. C 

*d. D 

e. E 


In a Diels Alder reaction the diene must be in the s-cis conformation; i.e., the two double 

bonds must be in a synperiplanar conformation. 

Evidently, this rules out the conformation associated with dienes A, C and E as the 

answers. Diene D is more reactive than compared to diene B as the cyclic structure holds 

the diene in the most reactive conformer.






651. Reactant 1 

Ozonolysis of compound Z gives the following products. From this information, deduce 

the structure of compound Z: 

a. A 

b. B 

c. C 

*d. D 

e. E 


Ozonolysis of alkenes cleaves the C=C double bond and produces two carbonyl 

compounds, and these may be acids, ketones or aldehydes depending on the reaction 

“work-up” conditions. In this particular sequence, dimethyl sulfide is used, so the 

aldehyde produced is not oxidised to the corresponding acid. The two-carbon aldehyde 

and four-carbon ketone must come from a trisubstituted alkene, and only alkene D has 

this arrangement.







Ozonolysis of compound Z gives the following products. From this information, deduce 

the structure of compound Z: 

*a. A and B 

b. B and C 

c. C and D 

d. D and A 



Ozonolysis of alkenes cleaves the C=C double bond and produces two carbonyl 

compounds, and these may be acids, ketones or aldehydes depending on the reaction 

“work-up” conditions. In this particular sequence, dimethyl sulfide is used, so the 

aldehyde produced is not oxidised to the corresponding acid. 

The four-carbon and five-carbon ketones must come from a tetrasubstituted alkene. 

However, the reaction occurs irrespective of the alkene geometry so either the transisomer 

A or its cis-isomer B will produce the same ketones.







Pick out from the following reactions which would be exothermic: 

a. A 

b. B 

*c. C 

d. D 



In all cases, bar reaction C, one bond is made (exothermic) and another is broken 

(endothermic) and an unstable radical is produced. However, in reaction C, only bond 

formation occurs, and this process is very exothermic overall.

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