(a) Interpretation: The alkyl halide and alkene required to produce derivative of cyclopropane are to be predicted. Concept introduction: Alkenes on reaction with alkyl halide in the presence of a strong base give derivative of cyclopropane. The C = C bond of alkene breaks and a single three membered cyclic ring is formed. Methylene ( :CH 2 ) is formed as a reactive intermediate during this reaction.

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Answer 1

Question

Chapter 9, Problem 9.35P

Interpretation Introduction

(a)

Interpretation:

The alkyl halide and alkene required to produce derivative of cyclopropane are to be predicted.

Concept introduction:

Alkenes on reaction with alkyl halide in the presence of a strong base give derivative of cyclopropane. The C=C bond of alkene breaks and a single three membered cyclic ring is formed. Methylene (:CH2) is formed as a reactive intermediate during this reaction.

Interpretation Introduction

(b)

Interpretation:

The alkyl halide and alkene required to produce derivative of cyclopropane are to be predicted.

Concept introduction:

Alkenes on reaction with alkyl halide in the presence of a strong base give derivative of cyclopropane. The C=C bond of alkene breaks and a single three membered cyclic ring is formed. Methylene (:CH2) is formed as a reactive intermediate during this reaction.

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Which of the following is true for the reactions of alkyl halides? (a) The characteristic reactions of alkyl halides are oxidation and reduction.(b) The characteristic reactions of alkyl halides are elimination and substitutionc. The characteristic reactions of alkyl halides are addition and substitutiond. Characteristic reactions of alkyl halides are addition and elimination.

Disiamylborane adds only once to alkynes by virtue of its two bulky secondary isoamylgroups. Disiamylborane is prepared by the reaction of BH3 # THF with an alkene.(a) Draw the structural formulas of the reagents and the products in the preparation ofdisiamylborane.(b) Explain why the reaction in part (a) goes only as far as the dialkylborane. Why isSia3B not formed?

1) The carbon-oxygen double bond present in aldehydes and ketones is very polar. What does this mean and how does it arise? 2) The carbon-oxygen double bond is readily attacked by nucleophiles like cyanide ions or ammonia. (i) What do you understand by the term nucleophile? (ii) Which part of the carbon-oxygen double bond is attractive to nucleophiles? 3) Why is there a difference between aldehydes and ketones in their response to oxidizing agents such as potassium dichromate(VI) solution acidified with dilute sulfuric acid?

Answer 2

Question

Chapter 9, Problem 9.35P

Interpretation Introduction

(a)

Interpretation:

The alkyl halide and alkene required to produce derivative of cyclopropane are to be predicted.

Concept introduction:

Alkenes on reaction with alkyl halide in the presence of a strong base give derivative of cyclopropane. The C=C bond of alkene breaks and a single three membered cyclic ring is formed. Methylene (:CH2) is formed as a reactive intermediate during this reaction.

Interpretation Introduction

(b)

Interpretation:

The alkyl halide and alkene required to produce derivative of cyclopropane are to be predicted.

Concept introduction:

Alkenes on reaction with alkyl halide in the presence of a strong base give derivative of cyclopropane. The C=C bond of alkene breaks and a single three membered cyclic ring is formed. Methylene (:CH2) is formed as a reactive intermediate during this reaction.

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Answer 3

Question

Chapter 9, Problem 9.35P

Interpretation Introduction

(a)

Interpretation:

The alkyl halide and alkene required to produce derivative of cyclopropane are to be predicted.

Concept introduction:

Alkenes on reaction with alkyl halide in the presence of a strong base give derivative of cyclopropane. The C=C bond of alkene breaks and a single three membered cyclic ring is formed. Methylene (:CH2) is formed as a reactive intermediate during this reaction.

Interpretation Introduction

(b)

Interpretation:

The alkyl halide and alkene required to produce derivative of cyclopropane are to be predicted.

Concept introduction:

Alkenes on reaction with alkyl halide in the presence of a strong base give derivative of cyclopropane. The C=C bond of alkene breaks and a single three membered cyclic ring is formed. Methylene (:CH2) is formed as a reactive intermediate during this reaction.

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Answer 4

Question

Chapter 9, Problem 9.35P

Interpretation Introduction

(a)

Interpretation:

The alkyl halide and alkene required to produce derivative of cyclopropane are to be predicted.

Concept introduction:

Alkenes on reaction with alkyl halide in the presence of a strong base give derivative of cyclopropane. The C=C bond of alkene breaks and a single three membered cyclic ring is formed. Methylene (:CH2) is formed as a reactive intermediate during this reaction.

Interpretation Introduction

(b)

Interpretation:

The alkyl halide and alkene required to produce derivative of cyclopropane are to be predicted.

Concept introduction:

Alkenes on reaction with alkyl halide in the presence of a strong base give derivative of cyclopropane. The C=C bond of alkene breaks and a single three membered cyclic ring is formed. Methylene (:CH2) is formed as a reactive intermediate during this reaction.

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Answer 5

Chapter 9, Problem 9.35P

Interpretation Introduction

(a)

Interpretation:

The alkyl halide and alkene required to produce derivative of cyclopropane are to be predicted.

Concept introduction:

Alkenes on reaction with alkyl halide in the presence of a strong base give derivative of cyclopropane. The C=C bond of alkene breaks and a single three membered cyclic ring is formed. Methylene (:CH2) is formed as a reactive intermediate during this reaction.

Interpretation Introduction

(b)

Interpretation:

The alkyl halide and alkene required to produce derivative of cyclopropane are to be predicted.

Concept introduction:

Alkenes on reaction with alkyl halide in the presence of a strong base give derivative of cyclopropane. The C=C bond of alkene breaks and a single three membered cyclic ring is formed. Methylene (:CH2) is formed as a reactive intermediate during this reaction.

Answer 6

Chapter 9, Problem 9.35P

Interpretation Introduction

(a)

Interpretation:

The alkyl halide and alkene required to produce derivative of cyclopropane are to be predicted.

Concept introduction:

Alkenes on reaction with alkyl halide in the presence of a strong base give derivative of cyclopropane. The C=C bond of alkene breaks and a single three membered cyclic ring is formed. Methylene (:CH2) is formed as a reactive intermediate during this reaction.

Answer 7

Chapter 9, Problem 9.35P

Interpretation Introduction

(a)

Interpretation:

The alkyl halide and alkene required to produce derivative of cyclopropane are to be predicted.

Concept introduction:

Alkenes on reaction with alkyl halide in the presence of a strong base give derivative of cyclopropane. The C=C bond of alkene breaks and a single three membered cyclic ring is formed. Methylene (:CH2) is formed as a reactive intermediate during this reaction.

Answer 8

Interpretation Introduction

(b)

Interpretation:

The alkyl halide and alkene required to produce derivative of cyclopropane are to be predicted.

Concept introduction:

Alkenes on reaction with alkyl halide in the presence of a strong base give derivative of cyclopropane. The C=C bond of alkene breaks and a single three membered cyclic ring is formed. Methylene (:CH2) is formed as a reactive intermediate during this reaction.

Answer 9

Interpretation Introduction

(b)

Interpretation:

The alkyl halide and alkene required to produce derivative of cyclopropane are to be predicted.

Concept introduction:

Alkenes on reaction with alkyl halide in the presence of a strong base give derivative of cyclopropane. The C=C bond of alkene breaks and a single three membered cyclic ring is formed. Methylene (:CH2) is formed as a reactive intermediate during this reaction.

Answer 10

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Answer 11

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Answer 12

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Answer 13

Ch. 9 - Prob. 9.1P Ch. 9 - Prob. 9.2P Ch. 9 - Prob. 9.3P Ch. 9 - Prob. 9.4P Ch. 9 - Prob. 9.5P Ch. 9 - Prob. 9.6P Ch. 9 - Prob. 9.7P Ch. 9 - Prob. 9.8P Ch. 9 - Prob. 9.9P Ch. 9 - Prob. 9.10P

Solution Summary: The author explains that the alkyl halide and alkene required to produce derivative of cyclopropane are shown in Figure 1.

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Chapter 9 Solutions