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Problem 24-1
(a)
N-Methylethylamine
(b)
Tricyclohexylamine
(c)
N-Ethyl-N-methylcyclohexylamine
(d)
N-Methylpyrrolidine
(e)
Diisopropylamine
(f)
1,3-Butanediamine
Problem 24-4
(a)
CH3CH2NH2
(b)
NaOH
(c)
CH3NHCH3
Problem 24-5
Propylamine is stronger; benzylamine pKb = 4.67; propylamine pKb = 3.29
Problem 24-6
(a)
p-Nitroaniline < p-Aminobenzaldehyde < p-Bromoaniline
(b)
p-Aminoacetophenone < p-Chloroaniline < p-Methylaniline
(c)
p-(Trifluoromethyl)aniline < p-(Fluoromethyl)aniline < p-Methylaniline
Problem 24-7
Pyrimidine is essentially 100% neutral (unprotonated).
Problem 24-8
(a)
Propanenitrile or propanamide
(b)
N-Propylpropanamide
(c)
Benzonitrile or benzamide
(d)
N-Phenylacetamide
Problem 24-9
The reaction takes place by two nucleophilic acyl substitution reactions.
Problem 24-11
(a)
Ethylamine + acetone, or isopropylamine + acetaldehyde
(b)
Aniline + acetaldehyde
(c)
Cyclopentylamine + formaldehyde, or methylamine + cyclopentanone
Problem 24-13
(a)
4,4-Dimethylpentanamide or 4,4-dimethylpentanoyl azide
(b)
p-Methylbenzamide or p-methylbenzoyl azide
Problem 24-14
(a)
3-Octene and 4-octene
(b)
Cyclohexene
(c)
3-Heptene
(d)
Ethylene and cyclohexene
Problem 24-15
H2C CHCH2CH2CH2N(CH3)2
Problem 24-16
1. HNO3, H2SO4; 2. H2/PtO2; 3. (CH3CO)2O; 4. HOSO2Cl; 5. aminothiazole; 6. H2O, NaOH
Problem 24-17
(a)
1. HNO3, H2SO4; 2. H2/PtO2; 3. 2 CH3Br
(b)
1. HNO3, H2SO4; 2. H2/PtO2; 3. (CH3CO)2O; 4. Cl2; 5. H2O, NaOH
(c)
1. HNO3, H2SO4; 2. Cl2, FeCl3; 3. SnCl2
(d)
1. HNO3, H2SO4; 2. H2/PtO2; 3. (CH3CO)2O; 4. 2 CH3Cl, AlCl3; 5. H2O, NaOH
Problem 24-18
(a)
1. CH3Cl, AlCl3; 2. HNO3, H2SO4; 3. SnCl2; 4. NaNO2, H2SO4; 5. CuBr; 6. KMnO4, H2O
(b)
1. HNO3, H2SO4; 2. Br2, FeBr3; 3. SnCl2, H3O+; 4. NaNO2, H2SO4; 5. CuCN; 6. H3O+
(c)
1. HNO3, H2SO4; 2. Cl2, FeCl3; 3. SnCl2; 4. NaNO2, H2SO4; 5. CuBr
(d)
1. CH3Cl, AlCl3; 2. HNO3, H2SO4; 3. SnCl2; 4. NaNO2, H2SO4; 5. CuCN; 6. H3O+
(e)
1. HNO3, H2SO4; 2. H2/PtO2; 3. (CH3CO)2O; 4. 2 Br2; 5. H2O, NaOH; 6. NaNO2, H2SO4; 7. CuBr
Problem 24-19
1. HNO3, H2SO4; 2. SnCl2; 3a. 2 equiv. CH3I; 3b. NaNO2, H2SO4; 4. product of 3a + product of 3b
Problem 24-21
4.1% protonated
Problem 24-23
The side-chain nitrogen is more basic than the ring nitrogen.
Problem 24-24

Reaction at C2 is disfavored because the aromaticity of the benzene ring is lost.

Structure of indole with arrows showing C 3 attacking electrophile. Resulting resonance structures show loss of aromaticity due to 8 electrons in pi system.
Problem 24-25
(CH3)3CCOCH3 ​→ ​(CH3)3CCH(NH2)CH3
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