Organic Chemistry
Organic Chemistry
3rd Edition
ISBN: 9781119338352
Author: Klein
Publisher: WILEY
Question
Book Icon
Chapter 22.11, Problem 26CC

 (a)

Interpretation Introduction

Interpretation: An efficient synthesis for the given transformations has to be proposed.

Concept Introduction:

Friedel-Crafts Alkylation: The Friedel-Crafts alkylation involves the electrophilic substitution of alkyl groups on aromatic rings when arenes are treated with alkyl halides in presence of Lewis acids.  This reaction is catalyzed by Lewis acids like anhydrous AlCl3, FeX3, ZnCl2, BF3 etc.

Organic Chemistry, Chapter 22.11, Problem 26CC , additional homework tip  1

Sandmeyer reaction: Sandmeyer reactions use copper salts as the reagents.  Here, aryldiazonium salt is converted into aryl halides or aryl cyanides by using copper halides or copper cyanides.

Organic Chemistry, Chapter 22.11, Problem 26CC , additional homework tip  2

A compound containing an amino group is treated with sodium nitrite and HCl leading to the formation of diazonium salt.

Organic Chemistry, Chapter 22.11, Problem 26CC , additional homework tip  3

To find: Propose an efficient synthesis for the given transformation (a)

Decide the transformation

Organic Chemistry, Chapter 22.11, Problem 26CC , additional homework tip  4

(b)

Interpretation Introduction

Interpretation: An efficient synthesis for the given transformations has to be proposed.

Concept Introduction:

Electrophilic aromatic substitution: Electrophilic aromatic substitution is a reaction in which the hydrogen atom of an aromatic ring is replaced as a result of an electrophilic attack on the aromatic ring.  Here are three general steps to an electrophilic aromatic substitution.

  1. 1) Generation of an electrophile
  2. 2) Attack of the electrophile on the aromatic ring, creating a resonance-stabilized carbocation called an arenium ion.   It loses aromaticity in this step, so the energy of activation is high. Furthermore, this is the rate-determining step of the reaction because of the disruption of aromaticity.  The arenium ion is a hybrid resonance structure. There are three general resonance contributors of an arenium ion.
  3. 3) Deprotonation of the arenium ion by a weak base to regain aromaticity.

Sandmeyer reaction: Sandmeyer reactions use copper salts as the reagents.  Here, aryldiazonium salt is converted into aryl halides or aryl cyanides by using copper halides or copper cyanides.

Organic Chemistry, Chapter 22.11, Problem 26CC , additional homework tip  5

A compound containing an amino group is treated with sodium nitrite and HCl leading to the formation of diazonium salt.

Organic Chemistry, Chapter 22.11, Problem 26CC , additional homework tip  6

To find: Propose an efficient synthesis for the given transformation (b)

Apply a retrosynthetic analysis

Organic Chemistry, Chapter 22.11, Problem 26CC , additional homework tip  7

Organic Chemistry, Chapter 22.11, Problem 26CC , additional homework tip  8

(c)

Interpretation Introduction

Interpretation: An efficient synthesis for the given transformations has to be proposed.

Concept Introduction:

Friedel–Crafts Acylation: The Friedel–Crafts acylation is the reaction of an arene with acyl chlorides or anhydrides using a strong Lewis acid catalyst. This reaction proceeds via electrophilic aromatic substitution to form monoacylated products.

Organic Chemistry, Chapter 22.11, Problem 26CC , additional homework tip  9

A compound containing an amino group is treated with sodium nitrite and HCl leading to the formation of diazonium salt.

Organic Chemistry, Chapter 22.11, Problem 26CC , additional homework tip  10

To find: Propose an efficient synthesis for the given transformation (c)

Apply a retrosynthetic analysis

Organic Chemistry, Chapter 22.11, Problem 26CC , additional homework tip  11

(d)

Interpretation Introduction

Interpretation: An efficient synthesis for the given transformations has to be proposed.

Concept Introduction:

Friedel-Crafts Alkylation: The Friedel-Crafts alkylation involves the electrophilic substitution of alkyl groups on aromatic rings when arenes are treated with alkyl halides in presence of Lewis acids.  This reaction is catalyzed by Lewis acids like anhydrous AlCl3, FeX3, ZnCl2, BF3 etc.

Organic Chemistry, Chapter 22.11, Problem 26CC , additional homework tip  12

Sandmeyer reaction: Sandmeyer reactions use copper salts as the reagents.  Here, aryldiazonium salt is converted into aryl halides or aryl cyanides by using copper halides or copper cyanides.

Organic Chemistry, Chapter 22.11, Problem 26CC , additional homework tip  13

A compound containing an amino group is treated with sodium nitrite and HCl leading to the formation of diazonium salt.

Organic Chemistry, Chapter 22.11, Problem 26CC , additional homework tip  14

To find: Propose an efficient synthesis for the given transformation (d)

Apply a retrosynthetic analysis

Organic Chemistry, Chapter 22.11, Problem 26CC , additional homework tip  15

(e)

Interpretation Introduction

Interpretation: An efficient synthesis for the given transformations has to be proposed.

Concept Introduction:

Electrophilic aromatic substitution: Electrophilic aromatic substitution is a reaction in which the hydrogen atom of an aromatic ring is replaced as a result of an electrophilic attack on the aromatic ring.  Here are three general steps to an electrophilic aromatic substitution.

  1. 1) Generation of an electrophile
  2. 2) Attack of the electrophile on the aromatic ring, creating a resonance-stabilized carbocation called an arenium ion.   It loses aromaticity in this step, so the energy of activation is high. Furthermore, this is the rate-determining step of the reaction because of the disruption of aromaticity.  The arenium ion is a hybrid resonance structure. There are three general resonance contributors of an arenium ion.
  3. 3) Deprotonation of the arenium ion by a weak base to regain aromaticity.

Schiemann reaction: It is the reaction in which diazonium salt is converted into fluoro group using fluoroboric acid (HBF4).

Organic Chemistry, Chapter 22.11, Problem 26CC , additional homework tip  16

A compound containing an amino group is treated with sodium nitrite and HCl leading to the formation of diazonium salt.

Organic Chemistry, Chapter 22.11, Problem 26CC , additional homework tip  17

To find: Propose an efficient synthesis for the given transformation (e)

Apply a retrosynthetic analysis

Organic Chemistry, Chapter 22.11, Problem 26CC , additional homework tip  18

(f)

Interpretation Introduction

Interpretation: An efficient synthesis for the given transformations has to be proposed.

Concept Introduction:

Sandmeyer reaction: Sandmeyer reactions use copper salts as the reagents.  Here, aryldiazonium salt is converted into aryl halides or aryl cyanides by using copper halides or copper cyanides.

Organic Chemistry, Chapter 22.11, Problem 26CC , additional homework tip  19

A compound containing an amino group is treated with sodium nitrite and HCl leading to the formation of diazonium salt.

Organic Chemistry, Chapter 22.11, Problem 26CC , additional homework tip  20

To find: Propose an efficient synthesis for the given transformation (f)

Perform elimination-addition process

Organic Chemistry, Chapter 22.11, Problem 26CC , additional homework tip  21

Blurred answer

Chapter 22 Solutions

Organic Chemistry

Ch. 22.4 - Prob. 10CCCh. 22.4 - Prob. 11CCCh. 22.5 - Prob. 2LTSCh. 22.5 - Prob. 12PTSCh. 22.5 - Prob. 13ATSCh. 22.6 - Prob. 3LTSCh. 22.6 - Prob. 14PTSCh. 22.6 - Prob. 15ATSCh. 22.7 - Prob. 4LTSCh. 22.7 - Prob. 16PTSCh. 22.7 - Prob. 17PTSCh. 22.7 - Prob. 18PTSCh. 22.7 - Prob. 19ATSCh. 22.8 - Prob. 20CCCh. 22.8 - Prob. 21CCCh. 22.8 - Prob. 22CCCh. 22.9 - Prob. 5LTSCh. 22.9 - Prob. 23PTSCh. 22.9 - Prob. 24ATSCh. 22.10 - Prob. 25CCCh. 22.11 - Prob. 26CCCh. 22.11 - Prob. 6LTSCh. 22.11 - Prob. 27PTSCh. 22.11 - Prob. 28ATSCh. 22.12 - Prob. 29CCCh. 22.12 - Prob. 30CCCh. 22.13 - Prob. 31CCCh. 22.13 - Prob. 32CCCh. 22 - Prob. 33PPCh. 22 - Prob. 34PPCh. 22 - Prob. 35PPCh. 22 - Prob. 36PPCh. 22 - Prob. 37PPCh. 22 - Prob. 38PPCh. 22 - Prob. 39PPCh. 22 - Prob. 40PPCh. 22 - Prob. 41PPCh. 22 - Prob. 42PPCh. 22 - Prob. 43PPCh. 22 - Prob. 44PPCh. 22 - Prob. 45PPCh. 22 - Prob. 46PPCh. 22 - Prob. 47PPCh. 22 - Prob. 48PPCh. 22 - Prob. 49PPCh. 22 - Prob. 50PPCh. 22 - Prob. 51PPCh. 22 - Prob. 52PPCh. 22 - Prob. 53PPCh. 22 - Prob. 54PPCh. 22 - Prob. 55PPCh. 22 - Prob. 56PPCh. 22 - Prob. 57PPCh. 22 - Prob. 58PPCh. 22 - Prob. 59PPCh. 22 - Prob. 60PPCh. 22 - Prob. 61PPCh. 22 - Prob. 62PPCh. 22 - Prob. 63PPCh. 22 - Prob. 64PPCh. 22 - Prob. 65PPCh. 22 - Prob. 66PPCh. 22 - Prob. 67PPCh. 22 - Prob. 68PPCh. 22 - Prob. 69PPCh. 22 - Prob. 70PPCh. 22 - Prob. 71PPCh. 22 - Prob. 72PPCh. 22 - Prob. 73IPCh. 22 - Prob. 74IPCh. 22 - Prob. 75IPCh. 22 - Prob. 76IPCh. 22 - Prob. 77IPCh. 22 - Prob. 78IPCh. 22 - Prob. 79IPCh. 22 - Prob. 80IPCh. 22 - Prob. 81IPCh. 22 - Prob. 82IPCh. 22 - Prob. 83IPCh. 22 - Prob. 84IPCh. 22 - Prob. 85IPCh. 22 - Prob. 86IPCh. 22 - Prob. 87IPCh. 22 - Prob. 88IPCh. 22 - Prob. 89IPCh. 22 - Prob. 90IPCh. 22 - Prob. 91IPCh. 22 - Prob. 92IPCh. 22 - Prob. 93IPCh. 22 - Prob. 94IPCh. 22 - Prob. 95IPCh. 22 - Prob. 96CPCh. 22 - Prob. 97CPCh. 22 - Prob. 98CPCh. 22 - Prob. 99CP
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