14CL Final Practice Key

Chemistry 14CL
Final Exam Practice -- Key
1. The stoichiometric equation for the aldol condensation of benzaldehyde with acetone
is given below.
O
2
H +
O NaOH/H2O C2H5OH MW: 102g/mol 58g/mol 234g/mol
(a) If a reaction of 1.00 mL of benzaldehyde (density = 1.0415 g/mL) is reacted with
0.360 mL of acetone (density =0.7899 g/mL), what is the theoretical yield (in grams)
of product?
1.0415 g
102 g/mol = 1.02 X 10−2 mol benzaldehyde, which will give 5.1 X 10 -3 mol product
0.360 mL X 0.7899 g/mL
58 g/mol
= 4.9 X 10 −3 mol acetone
Since two moles of benzaldehyde are required for each mole of product, acetone is
the limiting reagent and the theoretical yield is (4.9 X 10 -3 mol) (234 g/moL) = 1.15 g
(b) If the actual experimental yield for the reaction was 89%, how many mL of acetone
would be required to produce 4.00 grams of product?
If the yield were 89% then in the previous example, 0.360 mL of acetone would give
(1.15 g) (0.89) = 1.02 g of product.
1.02 g product
0.360 mL
X = 1.41 mL
= 4.00 g product
X mL
O

2. The chromium oxidation of compound I below yields a mixture of products (II – V).
Identify the order of elution that would occur if you used an alumina stationary phase
with diethyl ether to separate the mixture of products (II, III, IV, V) by column
chromatography. (Assume there is no starting material in the products.)
Starting material
Products:
O
C 6H 5-C-C(CH 3) 3
OH
C 6 H 5 -CH-C(CH 3 ) 3
I
CH 3
CH 3-C-OH
CH 3
C 6H 5C
O
H
C 6H 5C
II III IV V
First eluted ______ II _______
Reason: The ketone is the least polar of the oxygen functional groups; in addition
there are two nonpolar hydrocarbon group—an alkyl and an aryl group.
Second eluted ____IV_______
Reason: The aldehyde is slightly more polar than the ketone and this compound
has only one hydrocarbon group—a polarizable aryl group.
Third eluted _____III________
Reason: The alcohol is more polar than the aldehyde or ketone; there is only one
alkyl group.
Fourth eluted ________V___________
Reason: The two oxygens of the acid group make this the most polar compound
of the set. In addition the one hydrocarbon group is a polarizable aryl group.
O
OH
2

3. The molecular structure and the UV spectrum of caffeine are given below.
(a) If the solution used for the spectrum had a concentration of 1mg/40mL of water, what
is the molar extinction coefficient of caffeine at 275nm? (Assume a 1cm diameter
UV cell).
H3C N
C
O
O
C
N
CH3 Molecular formula: C8H10N4O2 (determined from the structure)
Molecular weight = 194 g/mol
Concentration of solution =
A = εbc
C
C
1 mg
40 mL •
CH3 N
N
1mol
CH
194 g/mol = 1.29 X 10−4 M
0.84 = ε(1 cm)(1.29 X 10 -4 M)
ε = 6512 M -1 cm -1
(b) Briefly indicate what feature(s) of caffeine give(s) rise to the UV spectrum.
This absorption is due to a π – π* transition of the chromophore of the
conjugated diene and ketone with auxochromes in the adjacent nitrogen atoms.
3

100
50
120
110
100
0
120
110
100
90
80
70
60
50
40
30
20
10
0
90
80
70
60
50
40
30
20
10
0
4. The 13 C spectrum of caffeine is given below. The signals for each carbon have been
identified and numbered. Draw the DEPT 90 and DEPT 135 caffeine spectra on the
charts below the 13 C spectrum.
156.31(3)
159.24(2)
155.86(4)
142.33(5)
104.08(1)
29.45(8)
32.84(6)
27.90(7)
170 160 150 140 130 120 110 100 90 80 70 60 50 40 30 20
Only
one CH
H3C 7
DEPT 90
170 160 150 140 130 120 110 100 90 80 70 60 50 40 30 20
DEPT 135
170 160 150 140 130 120 110 100 90 80 70 60 50 40 30 20
O
One CH and
3 CH3’s
N
3
O
4
N
CH 3
8
1
2
CH3 6
N
N
5
4

5. (a) The infra red spectrum of caffeine is given below. Identify the functional group or
groups that give rise to the absorption(s) at the specified wavelengths and the type of
vibrations involved.
(i) 2960 cm -1 __________C-H stretch________________
(ii) 1740cm -1 or 1692 cm -1 ___________C=O stretch_______________
(iii) 1240 cm -1 ____C-C or C-N stretches_______________
5

5. (b) The mass spectrum of caffeine is given below. Identify the species that give rise to
the m/z signals at
(i) 194 ___________molecular ion___________________
(ii) 109 Fragmentation occurs next to a carbonyl groups. In this case the
compound breaks on either side of the carbonyl carbons leaving the five-membered
ring and one nitrogen with a methyl group (C8 in the NMR problem) attached
(iii) 15 _______________methyl__________________________
6. A buffer solution is 0.100 M in NaOAc and 0.200 M in HOAc. To 1.00L of this
buffer, 25.0 mL of 0.500 M NaOH is added. Calculate the change in pH that occurs.
Ka of acetic acid = 1.8 X 10 -5
Original buffer: 1.8 X 10 −5 = [H+ ](0.1)
(0.2)
[H + ] = 3.6 X 10 -5 pH = 4.44
Moles of base added = (0.500M)(0.025L) = 0.0125 mol
After addition of base:
[H+] [OAc-] [HOAc]
moles x 0.1 + 0.0125 mol 0.200-0.0125 mol
=0.1125 =0.1875
molarities X 0.1125mol/1.025L 0.1875mol/1.025L
[H + ] = (1.8 X 10−5 )(0.1875)
(0.1125)
= 3.0X10 −5
pH = 4.53 pH change = 0.09
6

7. The internal pH of a muscle cell is 6.8. Calculate the [H2PO4 - ]/[HPO4 2- ] ratio in the
cell. The second dissociation constant of phosphoric acid is 6.31 x 10 -8 .
[H2PO4 - ]/[HPO4 2- ] = [H + ]/Ka = 1.58 x 10 -7 /6.31 x 10 -8 = 2.5/1 = 2.5
8. (a) Hexane (bp = 68 o C) and octane (bp = 125 o C ) form an ideal solution. Draw the
liquid-vapor phase diagram for this system. Label the axes and the phases in each area.
Indicate the temperature at which a 60% octane solution will begin to boil. What is the
composition of the vapor in equilibrium with the solution? What is the composition of
the distillate if there are two theoretical plates in the distillation apparatus?
140
T
E
M
P
E 120
R
A
T
U
100
R
E
O C
80
60
solution
0 0.2 0.4 0.6 0.8 1.0
MOLE FRACTION OF HEXANE
vapor
Initial boiling point
Vapor ~65%hexane
Temp ~ 95 o C
Distillate after two
theoretical plates
~79% hexane
7

(b) Sketch the gas chromatograms that would result if the original mixture and the
distillate were separated by gas chromatography on a non polar column.
Areas 40 60
Original mixture
Distillate
Injection time hexane octane time
Areas 79 21
Injection time hexane octane time
9. The pKa’s for valine are 2.3 and 9.6.
(a) What is the pI (the isoelectronic point pH) for valine?
The isoelectronic point for this amino acid is the average of the pK’s that is, 5.95 ~6
(b) What species are present at pH 7 and in what proportion?
A pH of 7 is sufficiently higher than the isolectronic point that the only two species
present are the zwitterion (major component) and the anion (minor component).
The ratio is 400/1
10. Based on their molecular structures, explain why vitamin C is water soluble and
vitamin A is lipid soluble.
Vitamin C (C5H7O6) has four alcohol groups, and a carbonyl. This many polar
groups with few carbon atoms make the molecule very polar and soluble in water.
Vitamin A (C18H30O)) has only one alcohol group and 18 carbon atoms. The nonpolar
portion of the molecule is too large to be pulled into an aqueous solution by
one OH group.
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