Exam (elaborations) TEST BANK FOR The Art of Writing Reasonable Organic Reaction Mechanisms 3rd Edition By Robert B. Grossman (Solution Manual)-Converted

1.4. Furan has sp2 hybridization. One of the lone pairs is in a p orbital, and the other is in an sp2 orbital. Only the lone pair in the p orbital is used in resonance. 1.5. (a) No by-products. C(1–3) and C(6–9) are the keys to numbering. (b) After numbering the major product, C6 and Br25 are left over, so make a bond between them and call it the by-product. 1.6. (a) Make C4–O12, C6–C11, C9–O12. Break C4–C6, C9–C11, C11–O12. (b) Make C8–N10, C9–C13, C12–Br24. Break O5–C6, C8–C9. 1.7. PhCºCH is much more acidic than BuCºCH because the pKb of HO– is 15, PhCºCH has a pKa ≤ 23 and BuCºCH has pKa 23. H2C O CH3 H H H H B F F F sp2 sp2 sp2 sp3 sp2 sp2 sp OH Ph O H+, H2O O Ph H 1 2 3 O 4 5 6 7 8 9 10 11 12 13 12 13 10 9 8 2 1 3 4 5 7 6 11 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 HN Br OMe O OMe H Br N O O OMe Br 19 20 21 22 23 Br Br 24 25 1 Me Br 2 3 4 5 6 7 8 10 9 19 18 17 16 15 24 14 13 20 21 11 12 25 1.8. The OH is more acidic (pKa ≈ 17) than the C a to the ketone (pKa ≈ 20). Because the by-product of the reaction is H2O, there is no need to break the O–H bond to get to product, but the C–H bond a to the ketone must be broken. Answers to Chapter 1: EndofChapter Problems 1. (a) Both N and O in amides have lone pairs that can react with electrophiles. When the O reacts with an electrophile E+, a product is obtained for which two good resonance structures can be drawn. When the N reacts, only one good resonance structure can be drawn for the product. (b) Esters are lower in energy than ketones because of resonance stabilization from the O atom. Upon addition of a nucleophile to either an ester or a ketone, a tetrahedral intermediate is obtained for which resonance is not nearly as important, and therefore the tetrahedral product from the ester is nearly the same energy as the tetrahedral product from the ketone. As a result, it costs more energy to add a nucleophile to an ester than it does to add one to a ketone. (c) Exactly the same argument as in (b) can be applied to the acidity of acyl chlorides versus the acidity of esters. Note that Cl and O have the same electronegativity, so the difference in acidity between acyl chlorides and esters cannot be due to inductive effects and must be due to resonance effects. (d) A resonance structure can be drawn for 1 in which charge is separated. Normally a charge-separated structure would be a minor contributor, but in this case the two rings are made aromatic, so it is much more important than normal. (e) The difference between 3 and 4 is that the former is cyclic. Loss of an acidic H from the g -C of 3 gives a structure for which an aromatic resonance structure can be drawn. This is not true of 4. R N R R O+ E R N R R O E reaction on O reaction on N R N R R O E (f) Both imidazole and pyridine are aromatic compounds. The lone pair of the H-bearing N in imidazole is required to maintain aromaticity, so the other N, which has its lone pair in an sp2 orbital that is perpendicular to the aromatic system, is the basic one. Protonation of this N gives a compound for which two equally good aromatic resonance structures can be drawn. By contrast, protonation of pyridine gives an aromatic compound for which only one good resonance structure can be drawn. (g) The C=C π bonds of simple hydrocarbons are usually nucleophilic, not electrophilic. However, when a nucleophile attacks the exocyclic C atom of the nonaromatic compound fulvene, the electrons from the C=C π bond go to the endocyclic C and make the ring aromatic. (h) The tautomer of 2,4-cyclohexadienone, a nonaromatic compound, is phenol, an aromatic compound. (i) Carbonyl groups C=O have an important resonance contributor C + –O – . In cyclopentadienone, this resonance contributor is antiaromatic. [Common error alert: Many cume points have been lost over the years when graduate students used cyclohexadienone or cyclopentadienone as a starting material in a synthesis problem!] (j) PhOH is considerably more acidic than EtOH (pKa= 10 vs. 17) because of resonance stabilization of the conjugate base in the former. S is larger than O, so the S(p)–C(p) overlap in PhS– is much smaller O O H3C H