ORGANIC CHEMISTRY-STUD.SOLNS.MAN+SG(LL)
ORGANIC CHEMISTRY-STUD.SOLNS.MAN+SG(LL)
4th Edition
ISBN: 9781119659587
Author: Klein
Publisher: WILEY
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Chapter 25, Problem 59PP

(a)

Interpretation Introduction

Interpretation:

The products formed in the given set of transformation need to be predicted.

Concept introduction:

α -bromination of carboxylic acid in presence of bromine and a catalytic amount of phosphorous tribromide to form α -bromocarboxylic acid is known as Hell-Volhard-Zelinsky reaction.  The formed α -bromocarboxylic acid when treated with ammonia gives α -aminocarboxylic acid as the product.

ORGANIC CHEMISTRY-STUD.SOLNS.MAN+SG(LL), Chapter 25, Problem 59PP , additional homework tip  1

(b)

Interpretation Introduction

Concept introduction:

Strecker synthesis is a process in which the aldehyde reacts with ammonium chloride and alkali cyanide to form α -aminonitrile.  The formed α -aminonitrile is undergoes hydrolysis to give the respective amino acids.  The general scheme can be shown as,

ORGANIC CHEMISTRY-STUD.SOLNS.MAN+SG(LL), Chapter 25, Problem 59PP , additional homework tip  2

Steps involved in Strecker synthesis are,

  • Protonation of carbonyl
  • Nucleophilic attack of ammonia to form imine
  • Attack of cyanide to form α -aminonitrile
  • Hydrolysis to give α -aminoacid

(c)

Interpretation Introduction

Concept introduction:

Amidomalonate synthesis is a process in which a halide is converted to an amino acid with two new carbon atoms (comes from amidomalonate).  By use of this method a new alkyl groups can be introduced.  A general scheme of this synthesis is shown below,

ORGANIC CHEMISTRY-STUD.SOLNS.MAN+SG(LL), Chapter 25, Problem 59PP , additional homework tip  3

Steps involved in amidomalonater synthesis are,

  • Deprotonation of α carbon in amidomalonate
  • Alkylation of enolate
  • Hydrolysis of ester using aqueous acid
  • Decarboxylation at high temperature

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Give the product of the bimolecular elimination from each of the isomeric halogenated compounds. Reaction A Reaction B. КОВ CH₂ HotBu +B+ ко HOIBU +Br+ Templates More QQQ Select Cv Templates More Cras QQQ One of these compounds undergoes elimination 50x faster than the other. Which one and why? Reaction A because the conformation needed for elimination places the phenyl groups and to each other Reaction A because the conformation needed for elimination places the phenyl groups gauche to each other. ◇ Reaction B because the conformation needed for elimination places the phenyl groups gach to each other. Reaction B because the conformation needed for elimination places the phenyl groups anti to each other.
Five isomeric alkenes. A through each undergo catalytic hydrogenation to give 2-methylpentane The IR spectra of these five alkenes have the key absorptions (in cm Compound Compound A –912. (§), 994 (5), 1643 (%), 3077 (1) Compound B 833 (3), 1667 (W), 3050 (weak shoulder on C-Habsorption) Compound C Compound D) –714 (5), 1665 (w), 3010 (m) 885 (3), 1650 (m), 3086 (m) 967 (5), no aharption 1600 to 1700, 3040 (m) Compound K Match each compound to the data presented. Compound A Compound B Compound C Compound D Compound

Chapter 25 Solutions

ORGANIC CHEMISTRY-STUD.SOLNS.MAN+SG(LL)

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