(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 .

Question

Want to see the full answer?

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.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Students have asked these similar questions

So, the first image is what I'm trying to understand regarding my approach. The second image illustrates my teacher's method, and the third image includes my notes on the concepts behind these types of problems.

HAND DRAW

Draw a mental model for calcium chloride mixed with sodium phosphate

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.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

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.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

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.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

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

Expert Solution & Answer

Answer 11

Expert Solution & Answer

Answer 12

Want to see the full answer?

Check out a sample textbook solution

See solution

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.

Concept explainers

Videos

Want to see the full answer?

Chapter 17 Solutions