(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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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?

The name of the parent six-membered sulfur-containing heterocycle is thiane. It is numbered beginning at sulfur. Multiple incorporation of sulfur in the ring is indicated by the prefixes di-, tri-, and so on. (a) How many methyl-substituted thianes are there? Which ones are chiral? (b) Write structural formulas for 1,4-dithiane and 1,3,5-trithiane. (c) Which dithiane isomer (1,2-, 1,3-, or 1,4-) is a disulfide?(d) Draw the two most stable conformations of the sulfoxide derived from thiane.

Chlorination of 2-butanone yields two isomeric products, each having the molecular formula C4H7ClO. (a) What are these two compounds? (b) Write structural formulas for the enol intermediates that lead to each of these compounds. (c) Using curved arrows, show the flow of electrons in the reaction of each of the enols with Cl2.

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