32.4.23: Chapter 24

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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 24-2
    (a) The structure of triisopropyl amine. It comprises three isopropyl groups linked to a nitrogen atom.
    (b) The structure of triallylamine. It has a nitrogen connected to three groups with structural formula C H 2 C H C H 2 (double bond between second and third carbons).
    (c) The structure of N-methylaniline. It comprises a benzene ring with an N H group linked to a methyl group.
    (d) The structure of N-ethyl-N-methylcyclopentylamine. It comprises nitrogen with ethyl, methyl, and cyclopentyl substituents.
    (e) The structure of N-isopropylcyclohexylamine. It comprises cyclohexane linked to an N H linked to a C H group further connected to two methyl groups at C 1.
    (f) A five-membered ring, one member of which is nitrogen. There are double bonds at C 2 and C 4, and nitrogen has an additional ethyl substituent.
    Problem 24-3
    (a) A benzene ring fused to another five-membered ring, one member of which is nitrogen. There is a double bond between non-fusion carbon atoms. Benzene has a methoxy substituent.
    (b) A five-membered ring, one member of which is nitrogen. There are double bonds at C 2 and C 4, methyl at C 3, and nitrogen has an additional methyl substituent.
    (c) A six-membered ring, one member of which is nitrogen, with alternating single and double bonds. Opposite the nitrogen is the substituent N with two methyl groups.
    (d) A six-membered ring, two members of which are nitrogen, with one carbon between them. There are alternating single and double bonds. C 5 has an amino substituent.
    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 24-10 4-(2-Bromoethyl)benzene-1,2-diol reacts with ammonia or 4-(bromomethyl)benzene-1,2-diol reacts with sodium cyanide, then lithium aluminum hydride.
    Problem 24-11
    (a) Ethylamine + acetone, or isopropylamine + acetaldehyde
    (b) Aniline + acetaldehyde
    (c) Cyclopentylamine + formaldehyde, or methylamine + cyclopentanone
    Problem 24-12 A benzene ring with aldehyde group on C 1 and methyl on C 3 reacts with dimethylamine in the presence of sodium borohydride. The product is not mentioned.
    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 H2CCHCH2CH2CH2N(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 24-20 Orbital diagram of thiazole involving nitrogen, sulfur, and three hydrogens, all s p 2 hybridized. N and S carry lone pairs of electrons; six total electrons in the pi system.
    Problem 24-21 4.1% protonated
    Problem 24-22 The attack of electrophile at C 2, C 3, and C 4 of pyridine. Mechanisms showing attack at C 2 and C 4 are labeled unfavorable.
    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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