1.1
Answers To Chapter 1 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.
N E R
R N R
R O N R
E R reaction on N
reaction on O
(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 γ C of 3 gives a structure for which an aromatic resonance structure can be drawn.  This is not true of 4.
H3
+
H3
O
H3
(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 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.
H
+
HN
(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.
Nu
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 used cyclohexadienone or cyclopentadienone as a starting material in a synthesis problem!]
(j)  PhOH is considerably more acidic than EtOH (pK a= 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 than
the O(p)–C(p) overlap in PhO –.  The reduced overlap in PhS – leads to reduced resonance stabilization, so the presence 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.
E +
E
2.  (a)
Cl 3inductive electron-withdrawing effect of F is greater than Cl
(b)
H
In general, AH + is more acidic than AH
3
Ketones are more
acidic than esters
(d)
Deprotonation of 5-membered
ring gives aromatic anion;
deprotonation of 7-membered
ring gives anti-aromatic anion.
(e)
2
The N(sp2) lone pair derived from deprotonation
of pyridine is in lower energy orbital, hence more
stable, than the N(sp3) lone pair derived from
deprotonation of piperidine.
(f)
2
Acidity increases as you move down a
column in the periodic table due to
increasing atomic size and hence worse
overlap in the A–H bond
(g)
2Et
The anion of phenylacetate
is stabilized by resonance
into the phenyl ring.
(h)
EtO2C
CO2Et
Anions of 1,3-dicarbonyl
compounds are stabilized
by resonance into two
carbonyl groups
O2
The anion of 4-nitrophenol is stabilized
by resonance directly into the nitro
group.  The anion of 3-nitrophenol can't
do this.  Draw resonance structures to
convince yourself of this.
N
–O
–O
(j)
H3NH2
More electronegative atoms are more
acidic than less electronegative atoms
in the same row of the periodic table
(k)
CH3
Ph
C(sp) is more acidic than C(sp3),
even when the anion of the latter can
be delocalized into a Ph ring.
(l
)
The anion of the latter cannot overlap
with the C=O π bond, hence cannot
delocalize, hence is not made acidic by
the carbonyl group.
(m
)
The C(sp2)–H bond on the upper atom is the plane of the paper,react with
orthogonal to the p orbitals of the C=O bond, so the C=O bond
provides no acidifying influence.  The C(sp3)–H bonds on the
lower atom are in and out of the plane of the paper, so there is
overlap with the C=O orbitals.
3.
(a)Free-radical.  (Catalytic peroxide tips you off.)

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