(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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Choose the Lewis structure for the compound below: H2CCHOCH2CH(CH3)2 HH H :d H H H C. Η H H HH H H H H. H H H HH H H H H H- H H H C-H H H HHHH

Each of the highlighted carbon atoms is connected to hydrogen atoms.

く Complete the reaction in the drawing area below by adding the major products to the right-hand side. If there won't be any products, because nothing will happen under these reaction conditions, check the box under the drawing area instead. Note: if the products contain one or more pairs of enantiomers, don't worry about drawing each enantiomer with dash and wedge bonds. Just draw one molecule to represent each pair of enantiomers, using line bonds at the chiral center. More... No reaction. Explanation Check O + G 1. Na O Me Click and drag to start drawing a structure. 2. H + 2025 McGraw Hill LLC. All Rights Reserved. Terms of Use | Privacy Center | Accessibility 000 Ar P

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 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