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
... [Show More] 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 [Show Less]