ORGANIC CHEMISTRY GGC>CUSTOM<-TEXT
ORGANIC CHEMISTRY GGC>CUSTOM<-TEXT
2nd Edition
ISBN: 9781119288510
Author: Klein
Publisher: WILEY
Question
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Chapter 23.5, Problem 12PTS

(a)

Interpretation Introduction

Interpretation: Using a Gabriel synthesis, a give set of primary amine compounds have to be synthesized.

Concept Introduction: The general formula for primary amine is –NH2.  There are several methods available to prepare primary amines.  Among them, Gabriel synthesis plays a very important role for preparing it.  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 GGC>CUSTOM<-TEXT, Chapter 23.5, Problem 12PTS , 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 GGC>CUSTOM<-TEXT, Chapter 23.5, Problem 12PTS , 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 GGC>CUSTOM<-TEXT, Chapter 23.5, Problem 12PTS , additional homework tip  3

(b)

Interpretation Introduction

Interpretation: Using a Gabriel synthesis, a give set of primary amine compounds have to be synthesized.

Concept Introduction: The general formula for primary amine is –NH2.  There are several methods available to prepare primary amines.  Among them, Gabriel synthesis plays a very important role for preparing it.  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 GGC>CUSTOM<-TEXT, Chapter 23.5, Problem 12PTS , additional homework tip  4

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 GGC>CUSTOM<-TEXT, Chapter 23.5, Problem 12PTS , additional homework tip  5

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 GGC>CUSTOM<-TEXT, Chapter 23.5, Problem 12PTS , additional homework tip  6

(c)

Interpretation Introduction

Interpretation: Using a Gabriel synthesis, a give set of primary amine compounds have to be synthesized.

Concept Introduction: The general formula for primary amine is –NH2.  There are several methods available to prepare primary amines.  Among them, Gabriel synthesis plays a very important role for preparing it.  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 GGC>CUSTOM<-TEXT, Chapter 23.5, Problem 12PTS , 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 GGC>CUSTOM<-TEXT, Chapter 23.5, Problem 12PTS , 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 GGC>CUSTOM<-TEXT, Chapter 23.5, Problem 12PTS , additional homework tip  9

(d)

Interpretation Introduction

Interpretation: Using a Gabriel synthesis, a give set of primary amine compounds have to be synthesized.

Concept Introduction: The general formula for primary amine is –NH2.  There are several methods available to prepare primary amines.  Among them, Gabriel synthesis plays a very important role for preparing it.  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 GGC>CUSTOM<-TEXT, Chapter 23.5, Problem 12PTS , additional homework tip  10

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 GGC>CUSTOM<-TEXT, Chapter 23.5, Problem 12PTS , additional homework tip  11

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 GGC>CUSTOM<-TEXT, Chapter 23.5, Problem 12PTS , additional homework tip  12

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Chapter 23 Solutions

ORGANIC CHEMISTRY GGC>CUSTOM<-TEXT

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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