Acid-Base & Electron Energy Relationships- 5 3NH H NHCNThe new lone pair of e-'s created by loss of proton a is assigned to an sp 2 orbital on a negative nitrogen atom while lone pair electrons forthe other bases are assigned to either an sp3 orbital on a negative nitrogen atom or an sp3 orbital on a negative carbon atom. (See diagrambelow) Again, since, all other effects being equal, the effective nuclear charge experienced by e-'s is proportional to the % s-character, e-'sin an sp 2 orbital experience more effective nuclear charge than those in sp 3 orbitals. Further, e-'s on negative nitrogen atoms willexperience more effective nuclear charge than those on negative carbon atoms, so the protons of the sp 3 hybridized carbon atoms in thiscompound should be the least acidic protons. As we have seen in activity 9, higher effective nuclear charge yields an increase in attractiveforce experienced by e-'s, lowering their energies. So we predict that the base created by loss of proton a should have the lowest energynew e- pair. Consequently, proton a should be the most acidic proton.aNH H bHNCNLoss of any other protonwould create a lone pair ofe - 's on a negative sp 3 carbon.All ions would have HigherEnergy(B)(A)N: N:-HNHH H : N:-NCCNLone pair of e - 'son a negative sp 2nitrogenLowest Energy Base.Isolated lone pair of e - 's ona negative sp 3 nitrogen.Intermediately Energy Base.Base A has the lowest energy! PROTON a IS MOST ACIDIC
4 Acid-Base & Electron Energy Relationships- 52. For each of the following pairs of equilibrium controlled reactions, predict which reaction should produce more product at equilibrium. Provide awarrant based on the theory developed in Acid-Base-5.a.CH 2OC+-CC-CH 2OCCCCH 2OC+HC-CH-CH 2O HC CHGeneral approach:At equilibrium, the lower energy side of the reaction (reactants or products) predominates. So the most product will be produced bythe reaction with the most negative (or least positive) ΔG˚. The relative Effect Assumption leads us to conclude that ΔG˚ can beestimated from differences in the environments of the HEE in the reactants and the products. So we first identify the HEE in thereactants and products of each reaction and then compare their relative energies.CCH 2CH 2OCOC-+ : C Cnegativesp carbon-+ HC Cnegativesp 2 carbon:Hnegativesp 3 oxygen-CH 2O-CH 2CCCO HC CHIn this case, the highest energy e - 's in the products are in the same environments (sp 3 O-), but those in the reactants are in differentenvironments. Thus the larger difference in energy occurs in the reactants. The highest energy e - 's in the reactants of the first reactionare on a negative sp hybrid carbon. The highest energy e - 's in the reactants of the second reaction are on a negative sp 2 hybrid carbon. SoC
Acid-Base & Electron Energy Relationships- 5 5HEE of the reactants are in an sp orbital with more s-character, more attractive force and lower energy than the HEE of the reactants inthe second reaction (In an sp 2 orbital).G2 nd rxn1 st rxn!G 2Consequently, the second reaction should have a lower ΔG˚ and soproduce more product at equilibrium than does the first reaction.!G 1b.NCRxn. Coord.+HEE-O-HHEE-NHOCHCNH+HEE-O-HHHEE-NOHCHIn this case, the highest energy e - 's in the reactants are in the same environments (sp 3 orbital on an O - ), but those in the products are indifferent environments. Thus the larger difference in energy occurs in the products. So, in this case, the reaction that producesmore product at equilibrium would be the reaction with the lower energy products. Since the highest energy e - 's in the products of thefirst reaction are in a lower energy environment (sp 2 hybrid orbital on nitrogen, more effective nuclear charge and more attractiveforces on the e - ‘s) than those in the second reaction (sp 3 hybrid orbital on nitrogen, less effective nuclear charge and less attractiveforces on the e - ‘s), the products of the first reaction should have lower energy than those of the second.GHigher !G 2(more positive)reactantsproductsRxn 2Rxn 1Lower !G 1(less positive)Consequently, the first reaction should have a lower ΔG˚ and soproduce more product at equilibrium than does the secondreaction.Rxn. Coord.
6 Acid-Base & Electron Energy Relationships- 5C. Nomenclature of Alkenes and Alkynes3. Applicationsa. Name the following:Br4-ethyl-2-methyl-2-heptene5-bromo-1,3-dimethylcyclohexene 5-ethyl-2-heptyne 3-ethyl-1,5-dimethylcyclopenteneb. Draw structural formulas for the following compounds:BrCl5-chloro-1-cyclopentyl-1-pentyne !2,3,5-trimethylcycloheptene 4-bromo-3-ethoxy-2-methyl-2-penteneO I O!8-butoxy-6-ethyl-3-iodo-4-nonyne