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Problem 16-1
o-, m-, and p-Bromotoluene
Problem 16-3
o-xylene: 2; p-xylene: 1; m-xylene: 3
Problem 16-4
D+ does electrophilic substitutions on the ring.
Problem 16-5
No rearrangement: (a), (b), (e)
Problem 16-6
Problem 16-7
Problem 16-8
Phenol > Toluene > Benzene > Nitrobenzene
Phenol > Benzene > Chlorobenzene > Benzoic acid
Aniline > Benzene > Bromobenzene > Benzaldehyde
Problem 16-9
o- and p-Bromonitrobenzene
o- and p-Chlorophenol
o- and p-Bromoaniline
Problem 16-10
Alkylbenzenes are more reactive than benzene itself, but acylbenzenes are less reactive.
Problem 16-11
Toluene is more reactive; the trifluoromethyl group is electron-withdrawing.
Problem 16-12
The nitrogen electrons are donated to the nearby carbonyl group by resonance and are less available to the ring.
Problem 16-13
The meta intermediate is most favored.
Problem 16-14
Ortho and para to  − OCH3
Ortho and para to  − NH2
Ortho and para to  − Cl
Problem 16-15
Reaction occurs ortho and para to the  − CH3 group.
Reaction occurs ortho and para to the  − OCH3 group.
Problem 16-16
The phenol is deprotonated by KOH to give an anion that carries out a nucleophilic acyl substitution reaction on the fluoronitrobenzene.
Problem 16-17
Only one benzyne intermediate can form from p-bromotoluene; two different benzyne intermediates can form from m-bromotoluene.
Problem 16-18
m-Nitrobenzoic acid
p-tert-Butylbenzoic acid
Problem 16-19
A benzyl radical is more stable than a primary alkyl radical by 52 kJ/mol and is similar in stability to an allyl radical.
Problem 16-20
1. CH3CH2Cl, AlCl3; 2. NBS; 3. KOH, ethanol
Problem 16-21
1. PhCOCl, AlCl3; 2. H2/Pd
Problem 16-22
1. HNO3, H2SO4; 2. Cl2, FeCl3
1. CH3COCl, AlCl3; 2. Cl2, FeCl3; 3. H2/Pd
1. CH3CH2COCl, AlCl3; 2. Cl2, FeCl3; 3. H2/Pd; 4. HNO3, H2SO4
1. CH3Cl, AlCl3; 2. Br2, FeBr3; 3. SO3, H2SO4
Problem 16-23
Friedel–Crafts acylation does not occur on a deactivated ring.
Rearrangement occurs during Friedel–Crafts alkylation with primary halides; chlorination occurs ortho to the alkyl group.
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