1
Answers To Chapter 1 In-Chapter Problems.
1.1. The resonance structure on the right is better becau every atom has its octet.
1.2.
2+
C
O
22
2
C
222
222
N N N
N 2H 3H 3N CH 2
H 3H 3
the cond structure is hopelessly strained
1.3.
N
Ph
O–
H33H33
sp
sp2
sp2sp
2
all sp2all sp2
3
sp3sp3sp3
sp3
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O
3
B
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F
F
sp2
sp2
sp2
sp3
2
1.4. The O atom in furan has sp2 hybridization. One lone pair resides in the p orbital and is ud in resonance; the other resides in an sp2 orbital and is not ud in resonance.
1.5.
(a) No by-products. C(1–3) and C(6–9) are the keys to numbering.
H+, H2O
1
7
13
13
10
(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
uol
5
2425
Me Br
1
6
24
新托福真题
25
1.6. (a) Make C4–O12, C6–C11, C9–O1
2. Break C4–C6, C9–C11, C11–O12.
(b) Make C8–N10, C9–C13, C12–Br24. Break O5–C6, C8–C9.
chine medicine1.7. PhC≡CH is much more acidic than BuC≡CH. Becau the p K b of HO– is 15, PhC≡CH has a p K a≤23 and BuC≡CH has p K a > 23.
1.8. The OH is more acidic (p K a≈ 17) than the C α to the ketone (p K a≈ 20). Becau 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 α to the ketone must be broken.
Answers To Chapter 1 End-Of-Chapter 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.
nellie mckayO R
N E R
O R N R
R O N E R reaction on N
erar怎么读reaction on O
(b) Esters are lower in energy than ketones becau of resonance stabilization from the O atom. Upon addition of a nucleophile to either an ester or a ketone, a tetrahedral intermediate is obtained f
or 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 parated. Normally a charge-parated structure would be a minor contributor, but in this ca the two rings are made aromatic, so it is much
more important than normal.
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(e) The difference between 3 and 4 is that the former is cyclic. Loss of an acidic H from the γ C of 3gives a structure for which an aromatic resonance structure can be drawn. This is not true of 4roundtrip
.
O
O
H 3C
H
– H
+
O
H 3O
O考研数学二
H 3C
(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 sp 2 orbital that is perpen-dicular 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.
N
H N H
+
(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.
aromatic
non-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 ud cyclohexadienone or cyclopentadienone as a starting material in a synthesis problem!]
(j) PhOH is considerably more acidic than EtOH (pK a = 10 vs. 17) becau of resonance stabilization of the conjugate ba in the former. S is larger than O, so the S(p)–C(p) overlap in PhS – is much smaller than the O(p)–C(p) overlap in PhO –. The reduced overlap in PhS – leads to reduced resonance stabiliza-tion, so the prence of a Ph ring makes less of a difference for the acidity of RSH than it does for the acidity of ROH.
(k) Attack of an electrophile E + on C2 gives a carbocation for which three good resonance structures can be drawn. Attack of an electrophile E + on C3 gives a carbocation for which only two good resonance
structures can be drawn.
O
H
H
H
H E E
+
