Skip to ContentGo to accessibility page

Chapter 24

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.

Problem 24-25
(CH3)3CCOCH3 ​→ ​(CH3)3CCH(NH2)CH3
Citation/Attribution
Reuse and redistribution of this content in digital or print format:
  • This book may not be used in the training of large language models or otherwise be ingested into large language models or generative AI offerings without OpenStax's prior written permission.
  • This book uses the Creative Commons Attribution-NonCommercial-ShareAlike License, which means that you can reuse and modify the material only for noncommercial purposes, must attribute OpenStax, and must distribute any derivative works under the same license.
  • Any commercial printing of this textbook, including using a local or custom printer, must be approved by OpenStax, and proper citation provided.
  • OpenStax-copyrighted images, activities, assessments, and similar components of this book are subject to the same licensing – CC-BY-NC-SA. They can be used for noncommercial purposes with attribution. Commercial use requires permission.
  • Permission requests: Anyone who intends to incorporate this content (including text, images, and other components) into large language models, use it in AI offerings, use it commercially (including in print), and/or has questions about another use case is welcome to complete our reuse request form.
Attribution information
  • If you are redistributing all or part of this book in a noncommercial print format, then you must include on every physical page the following attribution:

    Access for free at https://openstax.org/books/organic-chemistry/pages/1-why-this-chapter

  • If you are redistributing all or part of this book in a noncommercial digital format, then for every page that includes OpenStax content, you must license the derivative work under the same CC-BY-NC-SA license as the original, and include on every digital page view the following attribution:

    Access for free at https://openstax.org/books/organic-chemistry/pages/1-why-this-chapter

Citation information

The information below includes the information needed to generate citations in most major styles (APA, MLA, etc.); you must reformat and organize the information as needed to fit the requirements of the style. Use the information below to generate a citation. We recommend using a citation tool such as this one.

© Jul 1, 2026 OpenStax. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike License. The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo, and Rice University name, and Rice University logo trademarks, or wordmarks are not subject to the Creative Commons license and may not be reproduced without the prior and express written consent of Rice University.