Organic Chemistry
Organic Chemistry
2nd Edition
ISBN: 9781118452288
Author: David R. Klein
Publisher: WILEY
Question
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Chapter 23.7, Problem 19PTS

(a)

Interpretation Introduction

Interpretation: For a given compound set of compounds, the given amines are to be prepared via Gabriel synthesis, reductive amination starting from potassium phthalimide

Concept Introduction: Gabriel synthesis plays a very important role for preparing amines.  In this method, secondary and tertiary amines are not formed as side products.  It involves in three steps.

Step-1:  Formation of potassium phthalimide (deprotonation)

Potassium phthalimide in alkaline KOH acts as the reagent which has negatively charged phthalimide.  It is formed by the reaction between phthalimide and potassium hydroxide.

Organic Chemistry, Chapter 23.7, Problem 19PTS , additional homework tip  1

Step-2: Formation of R−N bond by SN2 nucleophilic substitution

The negative charged nitrogen atom in phthalimide can easily attract the positive side of R−X.  In primary alkyl halides (R−X), R and X get positive and negative charges, respectively when they ionize.  As a result, a bond between nitrogen of phthalimide and carbon of R is formed.  This is SN2 nucleophilic substitution reaction.  Halogen atom is going away as halide anion.

Organic Chemistry, Chapter 23.7, Problem 19PTS , additional homework tip  2

Step-3: Formation of primary amine by hydrolysis

The resultant product further goes for hydrolysis using hydrazine as the reagent.  This reaction also follows nucleophilic substitution reaction.  Finally, primary amine is formed with a side product of hydrazine derivative.

Organic Chemistry, Chapter 23.7, Problem 19PTS , additional homework tip  3

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, Chapter 23.7, Problem 19PTS , additional homework tip  4

Aldehyde or ketone group is reacted with primary amine in the presence of sodium cyanoborohydride as a reducing agent and a proton source in the reaction medium to produce secondary amines.

Organic Chemistry, Chapter 23.7, Problem 19PTS , additional homework tip  5

Aldehyde or ketone group is reacted with secondary amine in the presence of sodium cyanoborohydride as a reducing agent and a proton source in the reaction medium to produce tertiary amines.

Organic Chemistry, Chapter 23.7, Problem 19PTS , additional homework tip  6

(b)

Interpretation Introduction

Interpretation: For a given compound set of compounds, the given amines are to be prepared via Gabriel synthesis, reductive amination starting from potassium phthalimide

Concept Introduction: Gabriel synthesis plays a very important role for preparing amines.  In this method, secondary and tertiary amines are not formed as side products.  It involves in three steps.

Step-1:  Formation of potassium phthalimide (deprotonation)

Potassium phthalimide in alkaline KOH acts as the reagent which has negatively charged phthalimide.  It is formed by the reaction between phthalimide and potassium hydroxide.

Organic Chemistry, Chapter 23.7, Problem 19PTS , additional homework tip  7

Step-2: Formation of R−N bond by SN2 nucleophilic substitution

The negative charged nitrogen atom in phthalimide can easily attract the positive side of R−X.  In primary alkyl halides (R−X), R and X get positive and negative charges, respectively when they ionize.  As a result, a bond between nitrogen of phthalimide and carbon of R is formed.  This is SN2 nucleophilic substitution reaction.  Halogen atom is going away as halide anion.

Organic Chemistry, Chapter 23.7, Problem 19PTS , additional homework tip  8

Step-3: Formation of primary amine by hydrolysis

The resultant product further goes for hydrolysis using hydrazine as the reagent.  This reaction also follows nucleophilic substitution reaction.  Finally, primary amine is formed with a side product of hydrazine derivative.

Organic Chemistry, Chapter 23.7, Problem 19PTS , additional homework tip  9

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, Chapter 23.7, Problem 19PTS , additional homework tip  10

Aldehyde or ketone group is reacted with primary amine in the presence of sodium cyanoborohydride as a reducing agent and a proton source in the reaction medium to produce secondary amines.

Organic Chemistry, Chapter 23.7, Problem 19PTS , additional homework tip  11

Aldehyde or ketone group is reacted with secondary amine in the presence of sodium cyanoborohydride as a reducing agent and a proton source in the reaction medium to produce tertiary amines.

Organic Chemistry, Chapter 23.7, Problem 19PTS , additional homework tip  12

(c)

Interpretation Introduction

Interpretation: For a given compound set of compounds, the given amines are to be prepared via Gabriel synthesis, reductive amination starting from potassium phthalimide

Concept Introduction: Gabriel synthesis plays a very important role for preparing amines.  In this method, secondary and tertiary amines are not formed as side products.  It involves in three steps.

Step-1:  Formation of potassium phthalimide (deprotonation)

Potassium phthalimide in alkaline KOH acts as the reagent which has negatively charged phthalimide.  It is formed by the reaction between phthalimide and potassium hydroxide.

Organic Chemistry, Chapter 23.7, Problem 19PTS , additional homework tip  13

Step-2: Formation of R−N bond by SN2 nucleophilic substitution

The negative charged nitrogen atom in phthalimide can easily attract the positive side of R−X.  In primary alkyl halides (R−X), R and X get positive and negative charges, respectively when they ionize.  As a result, a bond between nitrogen of phthalimide and carbon of R is formed.  This is SN2 nucleophilic substitution reaction.  Halogen atom is going away as halide anion.

Organic Chemistry, Chapter 23.7, Problem 19PTS , additional homework tip  14

Step-3: Formation of primary amine by hydrolysis

The resultant product further goes for hydrolysis using hydrazine as the reagent.  This reaction also follows nucleophilic substitution reaction.  Finally, primary amine is formed with a side product of hydrazine derivative.

Organic Chemistry, Chapter 23.7, Problem 19PTS , additional homework tip  15

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, Chapter 23.7, Problem 19PTS , additional homework tip  16

Aldehyde or ketone group is reacted with primary amine in the presence of sodium cyanoborohydride as a reducing agent and a proton source in the reaction medium to produce secondary amines.

Organic Chemistry, Chapter 23.7, Problem 19PTS , additional homework tip  17

Aldehyde or ketone group is reacted with secondary amine in the presence of sodium cyanoborohydride as a reducing agent and a proton source in the reaction medium to produce tertiary amines.

Organic Chemistry, Chapter 23.7, Problem 19PTS , additional homework tip  18

(d)

Interpretation Introduction

Interpretation: For a given compound set of compounds, the given amines are to be prepared via Gabriel synthesis, reductive amination starting from potassium phthalimide

Concept Introduction: Gabriel synthesis plays a very important role for preparing amines.  In this method, secondary and tertiary amines are not formed as side products.  It involves in three steps.

Step-1:  Formation of potassium phthalimide (deprotonation)

Potassium phthalimide in alkaline KOH acts as the reagent which has negatively charged phthalimide.  It is formed by the reaction between phthalimide and potassium hydroxide.

Organic Chemistry, Chapter 23.7, Problem 19PTS , additional homework tip  19

Step-2: Formation of R−N bond by SN2 nucleophilic substitution

The negative charged nitrogen atom in phthalimide can easily attract the positive side of R−X.  In primary alkyl halides (R−X), R and X get positive and negative charges, respectively when they ionize.  As a result, a bond between nitrogen of phthalimide and carbon of R is formed.  This is SN2 nucleophilic substitution reaction.  Halogen atom is going away as halide anion.

Organic Chemistry, Chapter 23.7, Problem 19PTS , additional homework tip  20

Step-3: Formation of primary amine by hydrolysis

The resultant product further goes for hydrolysis using hydrazine as the reagent.  This reaction also follows nucleophilic substitution reaction.  Finally, primary amine is formed with a side product of hydrazine derivative.

Organic Chemistry, Chapter 23.7, Problem 19PTS , additional homework tip  21

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, Chapter 23.7, Problem 19PTS , additional homework tip  22

Aldehyde or ketone group is reacted with primary amine in the presence of sodium cyanoborohydride as a reducing agent and a proton source in the reaction medium to produce secondary amines.

Organic Chemistry, Chapter 23.7, Problem 19PTS , additional homework tip  23

Aldehyde or ketone group is reacted with secondary amine in the presence of sodium cyanoborohydride as a reducing agent and a proton source in the reaction medium to produce tertiary amines.

Organic Chemistry, Chapter 23.7, Problem 19PTS , additional homework tip  24

(e)

Interpretation Introduction

Interpretation: For a given compound set of compounds, the given amines are to be prepared via Gabriel synthesis, reductive amination starting from potassium phthalimide

Concept Introduction: Gabriel synthesis plays a very important role for preparing amines.  In this method, secondary and tertiary amines are not formed as side products.  It involves in three steps.

Step-1:  Formation of potassium phthalimide (deprotonation)

Potassium phthalimide in alkaline KOH acts as the reagent which has negatively charged phthalimide.  It is formed by the reaction between phthalimide and potassium hydroxide.

Organic Chemistry, Chapter 23.7, Problem 19PTS , additional homework tip  25

Step-2: Formation of R−N bond by SN2 nucleophilic substitution

The negative charged nitrogen atom in phthalimide can easily attract the positive side of R−X.  In primary alkyl halides (R−X), R and X get positive and negative charges, respectively when they ionize.  As a result, a bond between nitrogen of phthalimide and carbon of R is formed.  This is SN2 nucleophilic substitution reaction.  Halogen atom is going away as halide anion.

Organic Chemistry, Chapter 23.7, Problem 19PTS , additional homework tip  26

Step-3: Formation of primary amine by hydrolysis

The resultant product further goes for hydrolysis using hydrazine as the reagent.  This reaction also follows nucleophilic substitution reaction.  Finally, primary amine is formed with a side product of hydrazine derivative.

Organic Chemistry, Chapter 23.7, Problem 19PTS , additional homework tip  27

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, Chapter 23.7, Problem 19PTS , additional homework tip  28

Aldehyde or ketone group is reacted with primary amine in the presence of sodium cyanoborohydride as a reducing agent and a proton source in the reaction medium to produce secondary amines.

Organic Chemistry, Chapter 23.7, Problem 19PTS , additional homework tip  29

Aldehyde or ketone group is reacted with secondary amine in the presence of sodium cyanoborohydride as a reducing agent and a proton source in the reaction medium to produce tertiary amines.

Organic Chemistry, Chapter 23.7, Problem 19PTS , additional homework tip  30

(f)

Interpretation Introduction

Interpretation: For a given compound set of compounds, the given amines are to be prepared via Gabriel synthesis, reductive amination starting from potassium phthalimide

Concept Introduction: Gabriel synthesis plays a very important role for preparing amines.  In this method, secondary and tertiary amines are not formed as side products.  It involves in three steps.

Step-1:  Formation of potassium phthalimide (deprotonation)

Potassium phthalimide in alkaline KOH acts as the reagent which has negatively charged phthalimide.  It is formed by the reaction between phthalimide and potassium hydroxide.

Organic Chemistry, Chapter 23.7, Problem 19PTS , additional homework tip  31

Step-2: Formation of R−N bond by SN2 nucleophilic substitution

The negative charged nitrogen atom in phthalimide can easily attract the positive side of R−X.  In primary alkyl halides (R−X), R and X get positive and negative charges, respectively when they ionize.  As a result, a bond between nitrogen of phthalimide and carbon of R is formed.  This is SN2 nucleophilic substitution reaction.  Halogen atom is going away as halide anion.

Organic Chemistry, Chapter 23.7, Problem 19PTS , additional homework tip  32

Step-3: Formation of primary amine by hydrolysis

The resultant product further goes for hydrolysis using hydrazine as the reagent.  This reaction also follows nucleophilic substitution reaction.  Finally, primary amine is formed with a side product of hydrazine derivative.

Organic Chemistry, Chapter 23.7, Problem 19PTS , additional homework tip  33

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, Chapter 23.7, Problem 19PTS , additional homework tip  34

Aldehyde or ketone group is reacted with primary amine in the presence of sodium cyanoborohydride as a reducing agent and a proton source in the reaction medium to produce secondary amines.

Organic Chemistry, Chapter 23.7, Problem 19PTS , additional homework tip  35

Aldehyde or ketone group is reacted with secondary amine in the presence of sodium cyanoborohydride as a reducing agent and a proton source in the reaction medium to produce tertiary amines.

Organic Chemistry, Chapter 23.7, Problem 19PTS , additional homework tip  36

Blurred answer

Chapter 23 Solutions

Organic Chemistry

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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