Chapter 7, Problem 43P

Interpretation Introduction

Interpretation:

The compound with the molecular formula ${\text{C}}_{\text{7}}{\text{H}}_{\text{13}}\text{Br}$ thatyields the indicated alkene as the exclusive product of $\text{E2}$ elimination is to be predicted.

Concept introduction:

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The alkyl halides, in the presence of a strong base, undergo dehydrohalogenation and formcorresponding alkenes.

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In dehydrohalogenation $\text{H}$ and $\text{X}$ (halogen atom) areeliminated from adjacent carbons.

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The elimination of primary alkyl halides proceeds through $\text{E2}$ mechanism.

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In $\text{E2}$ mechanism, the attack of base and detachment of leaving group occurs simultaneously, and there is no carbocation formation step.

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According to Zaitsev rule, the most substituted alkene is the major product.

Answer to Problem 43P

Solution:

The compounds of molecular formula ${\text{C}}_{\text{7}}{\text{H}}_{\text{13}}\text{Br}$ that gives each alkene shown as the exclusive product of $\text{E2}$ elimination are as follows:

a)

b)

c)

d)

e)

f)

Explanation of Solution

a) The structure of the given alkene as the product of $\text{E2}$ elimination is shown below:

In $\text{E2}$ mechanism, $\text{H}$ and $\text{Br}$ are eliminated from adjacent carbons and the double bond forms between those carbons to form an alkene.

Hence, the structure of alkyl bromide is as follows:

b) The structure of the given alkene as the product of $\text{E2}$ elimination is shown below:

In $\text{E2}$ mechanism, $\text{H}$ and $\text{Br}$ are eliminated from adjacent carbons and the double bond forms between those carbons to form an alkene.

So, there are two possible structures of alkyl bromide which are as follows:

The primary alkyl bromide forms the desired product but the tertiary alkyl bromide can form two alkenes with trisubstituted alkene as the major product. The reaction is as follows:

Hence, the structure of the compound that forms the given alkene as an exclusive product of $\text{E2}$ elimination is as follows:

c)The structure of given alkene as the product of $\text{E2}$ elimination is shown below:

In $\text{E2}$ mechanism, $\text{H}$ and $\text{Br}$ are eliminated from adjacent carbons and the double bond forms between those carbons to form an alkene.

So, there are two possible structures of alkyl bromide as follows:

The alkyl bromide in which $\text{Br}$ is placed at $\text{C-4}$ with respect to ${\text{CH}}_{\text{3}}$, forms the desired product but the alkyl bromide in which $\text{Br}$ is placed at $\text{C-3}$ with respect to ${\text{CH}}_{\text{3}}$, forms two products.The reaction is as follows:

Hence, the structure of the compound that forms the given alkene as an exclusive product of $\text{E2}$ elimination is:

d) The structure of the given alkene as the product of $\text{E2}$ elimination is shown below:

In $\text{E2}$ mechanism, $\text{H}$ and $\text{Br}$ are eliminated from adjacent carbons and the double bond forms between those carbons to form an alkene.

So, there are two possible structures of alkyl bromide as follows:

The primary alkyl bromide forms the desired product but the secondary alkyl bromide can form two alkenes with trisubstituted alkene as the major product.

Hence, the structure of the compound that forms the given alkene as an exclusive product of $\text{E2}$ elimination is as follows:

e) The structural formula of the given alkene is shown below:

In $\text{E2}$ mechanism, $\text{H}$ and $\text{Br}$ are eliminated from the adjacent carbons and the double bond forms between those carbons to form an alkene.

So, there are two possible structures of alkyl bromide as follows:

The alkyl bromide in which $\text{Br}$ is placed at $\text{C-3}$ with respect to the isopropyl group forms the desired product but the alkyl bromide in which $\text{Br}$ is placed at $\text{C-2}$ with respect to isopropyl forms two products.

Hence, the structure of the compound that forms the given alkene as an exclusive product of $\text{E2}$ elimination is

f) The structural formula of the given alkene is shown below:

In $\text{E2}$ mechanism, $\text{H}$ and $\text{Br}$ are eliminated from adjacent carbons and the double bond forms between those carbons to form an alkene.

So, there are two possible structures of alkyl bromide as follows:

The secondary alkyl bromide forms two alkenesbut the tertiary alkyl bromide forms only the desired product as shown in following equations:

Hence, the structure of the compound that forms the given alkene as an exclusive product of $\text{E2}$ elimination is as follows: