Organic Chemistry, Binder Ready Version
Organic Chemistry, Binder Ready Version
2nd Edition
ISBN: 9781118454312
Author: David R. Klein
Publisher: WILEY
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Chapter 23, Problem 50PP

(a)

Interpretation Introduction

Interpretation: Using a different type of reactions, 1-hexanol is to be prepared from hexyl amine, heptyl amine and pentyl amine

Concept Introduction: A number of transformations are used to prepare 1-hexanol.  Some of them are listed as follows:

  1. a) Alcohol on treatment with phosphorous tribromide gives alkyl bromide
  2. b) Alkyl bromide in azide synthesis produces primary amine
  3. c) Alkyl halide on treatment with sodium cyanide gives alkyl cyanide
  4. d) Cyanide on reduction gives alkyl amine with an increment of one carbon atom skeleton
  5. e) Alkene on ozonolysis produces carbonyl compounds
  6. f) Alkyl halides with strong base gives alkene
  7. g) Sodium cyanoborohydride is a strong reducing agent than sodium borohydride.  It reduces the carbonyl group into amine group in a rapid way.  So, it is called as reductive amination reactions.  Aldehyde or ketone group is reacted with ammonia in the presence of sodium cyanoborohydride as a reducing agent and a proton source in the reaction medium to produce primary amines.

Organic Chemistry, Binder Ready Version, Chapter 23, Problem 50PP , additional homework tip  1

Using these concepts, we can transfer 1-hexanol into the given compounds.

(b)

Interpretation Introduction

Interpretation: Using a different type of reactions, 1-hexanol is to be prepared from hexyl amine, heptyl amine and pentyl amine

Concept Introduction: A number of transformations are used to prepare 1-hexanol.  Some of them are listed as follows:

  1. a) Alcohol on treatment with phosphorous tribromide gives alkyl bromide
  2. b) Alkyl bromide in azide synthesis produces primary amine
  3. c) Alkyl halide on treatment with sodium cyanide gives alkyl cyanide
  4. d) Cyanide on reduction gives alkyl amine with an increment of one carbon atom skeleton
  5. e) Alkene on ozonolysis produces carbonyl compounds
  6. f) Alkyl halides with strong base gives alkene
  7. g) Sodium cyanoborohydride is a strong reducing agent than sodium borohydride.  It reduces the carbonyl group into amine group in a rapid way.  So, it is called as reductive amination reactions.  Aldehyde or ketone group is reacted with ammonia in the presence of sodium cyanoborohydride as a reducing agent and a proton source in the reaction medium to produce primary amines.

Organic Chemistry, Binder Ready Version, Chapter 23, Problem 50PP , additional homework tip  2

Using these concepts, we can transfer 1-hexanol into the given compounds.

(c)

Interpretation Introduction

Interpretation: Using a different type of reactions, 1-hexanol is to be prepared from hexyl amine, heptyl amine and pentyl amine

Concept Introduction: A number of transformations are used to prepare 1-hexanol.  Some of them are listed as follows:

  1. a) Alcohol on treatment with phosphorous tribromide gives alkyl bromide
  2. b) Alkyl bromide in azide synthesis produces primary amine
  3. c) Alkyl halide on treatment with sodium cyanide gives alkyl cyanide
  4. d) Cyanide on reduction gives alkyl amine with an increment of one carbon atom skeleton
  5. e) Alkene on ozonolysis produces carbonyl compounds
  6. f) Alkyl halides with strong base gives alkene
  7. g) Sodium cyanoborohydride is a strong reducing agent than sodium borohydride.  It reduces the carbonyl group into amine group in a rapid way.  So, it is called as reductive amination reactions.  Aldehyde or ketone group is reacted with ammonia in the presence of sodium cyanoborohydride as a reducing agent and a proton source in the reaction medium to produce primary amines.

Organic Chemistry, Binder Ready Version, Chapter 23, Problem 50PP , additional homework tip  3

Using these concepts, we can transfer 1-hexanol into the given compounds.

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Chapter 23 Solutions

Organic Chemistry, Binder Ready Version

Ch. 23.4 - Prob. 10CCCh. 23.4 - Prob. 11CCCh. 23.5 - Prob. 2LTSCh. 23.5 - Prob. 12PTSCh. 23.5 - Prob. 13PTSCh. 23.6 - Prob. 3LTSCh. 23.6 - Prob. 14PTSCh. 23.6 - Prob. 15ATSCh. 23.6 - Prob. 16ATSCh. 23.6 - Prob. 17ATSCh. 23.7 - Prob. 18PTSCh. 23.7 - Prob. 19PTSCh. 23.7 - Prob. 20PTSCh. 23.7 - Prob. 21ATSCh. 23.8 - Prob. 22CCCh. 23.8 - Prob. 23CCCh. 23.8 - Prob. 24CCCh. 23.9 - Prob. 5LTSCh. 23.9 - Prob. 25PTSCh. 23.9 - Prob. 26ATSCh. 23.9 - Prob. 27ATSCh. 23.9 - Prob. 28ATSCh. 23.10 - Prob. 29CCCh. 23.11 - Prob. 30CCCh. 23.11 - Prob. 6LTSCh. 23.11 - Prob. 31PTSCh. 23.11 - Prob. 32ATSCh. 23.11 - Prob. 33ATSCh. 23.12 - Prob. 34CCCh. 23.12 - Prob. 35CCCh. 23.13 - Prob. 36CCCh. 23.13 - Prob. 37CCCh. 23 - Prob. 38PPCh. 23 - Prob. 39PPCh. 23 - Prob. 40PPCh. 23 - Prob. 41PPCh. 23 - Prob. 42PPCh. 23 - Prob. 43PPCh. 23 - Prob. 44PPCh. 23 - Prob. 45PPCh. 23 - Prob. 46PPCh. 23 - Prob. 47PPCh. 23 - Prob. 48PPCh. 23 - Prob. 49PPCh. 23 - Prob. 50PPCh. 23 - Prob. 51PPCh. 23 - Prob. 52PPCh. 23 - Prob. 53PPCh. 23 - Prob. 54PPCh. 23 - Prob. 55PPCh. 23 - Prob. 56PPCh. 23 - Prob. 57PPCh. 23 - Prob. 58PPCh. 23 - Prob. 59PPCh. 23 - Prob. 60PPCh. 23 - Prob. 61PPCh. 23 - Prob. 62PPCh. 23 - Prob. 63PPCh. 23 - Prob. 64PPCh. 23 - Prob. 65PPCh. 23 - Prob. 66PPCh. 23 - Prob. 67PPCh. 23 - Prob. 68PPCh. 23 - Prob. 69PPCh. 23 - Prob. 70PPCh. 23 - Prob. 71PPCh. 23 - Prob. 72PPCh. 23 - Prob. 73PPCh. 23 - Prob. 74PPCh. 23 - Prob. 75PPCh. 23 - Prob. 76PPCh. 23 - Prob. 77IPCh. 23 - Prob. 78IPCh. 23 - Prob. 79IPCh. 23 - Prob. 80IPCh. 23 - Prob. 81IPCh. 23 - Prob. 82IPCh. 23 - Prob. 83IPCh. 23 - Prob. 84IPCh. 23 - Prob. 85IPCh. 23 - Prob. 86IPCh. 23 - Prob. 87IPCh. 23 - Prob. 88IPCh. 23 - Prob. 89IPCh. 23 - Prob. 90IPCh. 23 - Prob. 91CPCh. 23 - Prob. 92CPCh. 23 - Prob. 93CPCh. 23 - Prob. 94CPCh. 23 - Prob. 95CPCh. 23 - Prob. 96CPCh. 23 - Prob. 97CPCh. 23 - Prob. 98CP
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