(a) Interpretation: The mechanism for the given reaction is to be drawn, and the major product is to be predicted. 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 ketone or aldehyde .

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Answer 1

Question

Chapter 17, Problem 17.45P

Interpretation Introduction

(a)

Interpretation:

The mechanism for the given reaction is to be drawn, and the major product is to be predicted.

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 ketone or aldehyde.

Interpretation Introduction

(b)

Interpretation:

The mechanism for the given reaction is to be drawn, and the major product is to be predicted for the same reaction.

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 ketone or aldehyde.

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CS2(g) →CS(g) + S(g) The rate law is Rate = k[CS2] where k = 1.6 × 10−6 s−¹. S What is the concentration of CS2 after 5 hours if the initial concentration is 0.25 M?

Answer 2

Question

Chapter 17, Problem 17.45P

Interpretation Introduction

(a)

Interpretation:

The mechanism for the given reaction is to be drawn, and the major product is to be predicted.

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 ketone or aldehyde.

Interpretation Introduction

(b)

Interpretation:

The mechanism for the given reaction is to be drawn, and the major product is to be predicted for the same reaction.

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 ketone or aldehyde.

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Answer 3

Question

Chapter 17, Problem 17.45P

Interpretation Introduction

(a)

Interpretation:

The mechanism for the given reaction is to be drawn, and the major product is to be predicted.

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 ketone or aldehyde.

Interpretation Introduction

(b)

Interpretation:

The mechanism for the given reaction is to be drawn, and the major product is to be predicted for the same reaction.

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 ketone or aldehyde.

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Answer 4

Question

Chapter 17, Problem 17.45P

Interpretation Introduction

(a)

Interpretation:

The mechanism for the given reaction is to be drawn, and the major product is to be predicted.

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 ketone or aldehyde.

Interpretation Introduction

(b)

Interpretation:

The mechanism for the given reaction is to be drawn, and the major product is to be predicted for the same reaction.

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 ketone or aldehyde.

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Answer 5

Chapter 17, Problem 17.45P

Interpretation Introduction

(a)

Interpretation:

The mechanism for the given reaction is to be drawn, and the major product is to be predicted.

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 ketone or aldehyde.

Interpretation Introduction

(b)

Interpretation:

The mechanism for the given reaction is to be drawn, and the major product is to be predicted for the same reaction.

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 ketone or aldehyde.

Answer 6

Chapter 17, Problem 17.45P

Interpretation Introduction

(a)

Interpretation:

The mechanism for the given reaction is to be drawn, and the major product is to be predicted.

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 ketone or aldehyde.

Answer 7

Chapter 17, Problem 17.45P

Interpretation Introduction

(a)

Interpretation:

The mechanism for the given reaction is to be drawn, and the major product is to be predicted.

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 ketone or aldehyde.

Answer 8

Interpretation Introduction

(b)

Interpretation:

The mechanism for the given reaction is to be drawn, and the major product is to be predicted for the same reaction.

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 ketone or aldehyde.

Answer 9

Interpretation Introduction

(b)

Interpretation:

The mechanism for the given reaction is to be drawn, and the major product is to be predicted for the same reaction.

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 ketone or aldehyde.

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 that Sodium hydride is a strong base but poor nucleophile. It deprotonates the alpha carbon of the carbonyl compound.

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Chapter 17 Solutions