(a) Interpretation: How to carry out the given transformation is to be shown. Concept introduction: Sodium hydride is a strong base but a poor nucleophile. So it will deprotonate the alpha carbon of the carbonyl compound. This will generate a strong nucleophile as well as a strong base, the enolate ion. This enolate ion attacks the electrophilic carbon in alkyl halides to displace the halide ion in an S N 2 reaction. This will yield an alpha-alkylated aldehyde or ketone .

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Answer 1

Question

Chapter 17, Problem 17.46P

Interpretation Introduction

(a)

Interpretation:

How to carry out the given transformation is to be shown.

Concept introduction:

Sodium hydride is a strong base but a poor nucleophile. So it will deprotonate the alpha carbon of the carbonyl compound. This will generate a strong nucleophile as well as a strong base, the enolate ion. This enolate ion attacks the electrophilic carbon in alkyl halides to displace the halide ion in an SN2 reaction. This will yield an alpha-alkylated aldehyde or ketone.

Interpretation Introduction

(b)

Interpretation:

How to carry out the given transformation is to be shown.

Concept introduction:

Sodium hydride is a strong base but a poor nucleophile. So it will deprotonate the alpha carbon of the carbonyl compound. This will generate a strong nucleophile as well as a strong base, the enolate ion. This enolate ion attacks the electrophilic carbon in alkyl halides to displace the halide ion in an SN2 reaction. This will yield an alpha-alkylated aldehyde or ketone.

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(SYN) Show how to carry out each of the following transformations. (a) O (b) ? ? Ph

Answer 2

Question

Chapter 17, Problem 17.46P

Interpretation Introduction

(a)

Interpretation:

How to carry out the given transformation is to be shown.

Concept introduction:

Sodium hydride is a strong base but a poor nucleophile. So it will deprotonate the alpha carbon of the carbonyl compound. This will generate a strong nucleophile as well as a strong base, the enolate ion. This enolate ion attacks the electrophilic carbon in alkyl halides to displace the halide ion in an SN2 reaction. This will yield an alpha-alkylated aldehyde or ketone.

Interpretation Introduction

(b)

Interpretation:

How to carry out the given transformation is to be shown.

Concept introduction:

Sodium hydride is a strong base but a poor nucleophile. So it will deprotonate the alpha carbon of the carbonyl compound. This will generate a strong nucleophile as well as a strong base, the enolate ion. This enolate ion attacks the electrophilic carbon in alkyl halides to displace the halide ion in an SN2 reaction. This will yield an alpha-alkylated aldehyde or ketone.

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Answer 3

Question

Chapter 17, Problem 17.46P

Interpretation Introduction

(a)

Interpretation:

How to carry out the given transformation is to be shown.

Concept introduction:

Sodium hydride is a strong base but a poor nucleophile. So it will deprotonate the alpha carbon of the carbonyl compound. This will generate a strong nucleophile as well as a strong base, the enolate ion. This enolate ion attacks the electrophilic carbon in alkyl halides to displace the halide ion in an SN2 reaction. This will yield an alpha-alkylated aldehyde or ketone.

Interpretation Introduction

(b)

Interpretation:

How to carry out the given transformation is to be shown.

Concept introduction:

Sodium hydride is a strong base but a poor nucleophile. So it will deprotonate the alpha carbon of the carbonyl compound. This will generate a strong nucleophile as well as a strong base, the enolate ion. This enolate ion attacks the electrophilic carbon in alkyl halides to displace the halide ion in an SN2 reaction. This will yield an alpha-alkylated aldehyde or ketone.

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Answer 4

Question

Chapter 17, Problem 17.46P

Interpretation Introduction

(a)

Interpretation:

How to carry out the given transformation is to be shown.

Concept introduction:

Sodium hydride is a strong base but a poor nucleophile. So it will deprotonate the alpha carbon of the carbonyl compound. This will generate a strong nucleophile as well as a strong base, the enolate ion. This enolate ion attacks the electrophilic carbon in alkyl halides to displace the halide ion in an SN2 reaction. This will yield an alpha-alkylated aldehyde or ketone.

Interpretation Introduction

(b)

Interpretation:

How to carry out the given transformation is to be shown.

Concept introduction:

Sodium hydride is a strong base but a poor nucleophile. So it will deprotonate the alpha carbon of the carbonyl compound. This will generate a strong nucleophile as well as a strong base, the enolate ion. This enolate ion attacks the electrophilic carbon in alkyl halides to displace the halide ion in an SN2 reaction. This will yield an alpha-alkylated aldehyde or ketone.

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Answer 5

Chapter 17, Problem 17.46P

Interpretation Introduction

(a)

Interpretation:

How to carry out the given transformation is to be shown.

Concept introduction:

Sodium hydride is a strong base but a poor nucleophile. So it will deprotonate the alpha carbon of the carbonyl compound. This will generate a strong nucleophile as well as a strong base, the enolate ion. This enolate ion attacks the electrophilic carbon in alkyl halides to displace the halide ion in an SN2 reaction. This will yield an alpha-alkylated aldehyde or ketone.

Interpretation Introduction

(b)

Interpretation:

How to carry out the given transformation is to be shown.

Concept introduction:

Sodium hydride is a strong base but a poor nucleophile. So it will deprotonate the alpha carbon of the carbonyl compound. This will generate a strong nucleophile as well as a strong base, the enolate ion. This enolate ion attacks the electrophilic carbon in alkyl halides to displace the halide ion in an SN2 reaction. This will yield an alpha-alkylated aldehyde or ketone.

Answer 6

Chapter 17, Problem 17.46P

Interpretation Introduction

(a)

Interpretation:

How to carry out the given transformation is to be shown.

Concept introduction:

Sodium hydride is a strong base but a poor nucleophile. So it will deprotonate the alpha carbon of the carbonyl compound. This will generate a strong nucleophile as well as a strong base, the enolate ion. This enolate ion attacks the electrophilic carbon in alkyl halides to displace the halide ion in an SN2 reaction. This will yield an alpha-alkylated aldehyde or ketone.

Answer 7

Chapter 17, Problem 17.46P

Interpretation Introduction

(a)

Interpretation:

How to carry out the given transformation is to be shown.

Concept introduction:

Sodium hydride is a strong base but a poor nucleophile. So it will deprotonate the alpha carbon of the carbonyl compound. This will generate a strong nucleophile as well as a strong base, the enolate ion. This enolate ion attacks the electrophilic carbon in alkyl halides to displace the halide ion in an SN2 reaction. This will yield an alpha-alkylated aldehyde or ketone.

Answer 8

Interpretation Introduction

(b)

Interpretation:

How to carry out the given transformation is to be shown.

Concept introduction:

Sodium hydride is a strong base but a poor nucleophile. So it will deprotonate the alpha carbon of the carbonyl compound. This will generate a strong nucleophile as well as a strong base, the enolate ion. This enolate ion attacks the electrophilic carbon in alkyl halides to displace the halide ion in an SN2 reaction. This will yield an alpha-alkylated aldehyde or ketone.

Answer 9

Interpretation Introduction

(b)

Interpretation:

How to carry out the given transformation is to be shown.

Concept introduction:

Sodium hydride is a strong base but a poor nucleophile. So it will deprotonate the alpha carbon of the carbonyl compound. This will generate a strong nucleophile as well as a strong base, the enolate ion. This enolate ion attacks the electrophilic carbon in alkyl halides to displace the halide ion in an SN2 reaction. This will yield an alpha-alkylated aldehyde or ketone.

Answer 10

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Answer 11

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Answer 12

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Answer 13

Ch. 17 - Prob. 17.1P Ch. 17 - Prob. 17.2P Ch. 17 - Prob. 17.3P Ch. 17 - Prob. 17.4P Ch. 17 - Prob. 17.5P Ch. 17 - Prob. 17.6P Ch. 17 - Prob. 17.7P Ch. 17 - Prob. 17.8P Ch. 17 - Prob. 17.9P Ch. 17 - Prob. 17.10P

Solution Summary: The author explains how the given transformation is carried out using sodium hydride and the desired alkyl halide.

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