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Chapter 9: Problem 42

Which isomer of 1-bromo-3-isopropylcyclohexane reacts faster when refluxed with potassium tert-butoxide, the cis isomer or the trans isomer? Draw the structure of the expected product from the faster reacting compound.

Short Answer

Expert verified
Draw the structure of the expected product from the faster reacting compound. Answer: The trans isomer of 1-bromo-3-isopropylcyclohexane reacts faster with potassium tert-butoxide when refluxed. The expected product from the faster reacting trans isomer is 1-isopropyl-1-methylcyclohexene.

Step by step solution

01

Draw the structures of cis and trans isomers of 1-bromo-3-isopropylcyclohexane

To begin, let's draw the structures of the cis and trans isomers of 1-bromo-3-isopropylcyclohexane. For the cis isomer, the bromine atom and the isopropyl group are on the same side of the cyclohexane ring. In the trans isomer, the bromine atom and the isopropyl group are on opposite sides of the cyclohexane ring.
02

Analyze the E2 elimination reaction mechanism

In this reaction, potassium tert-butoxide (KOtBu) acts as a strong base, promoting an E2 elimination mechanism. This means that the base will abstract a β-hydrogen atom from the carbon next to the bromine-bearing carbon (the α-carbon), followed by the simultaneous formation of a C=C double bond and loss of the bromide ion (Br-). The stereochemistry of the reaction requires that the β-hydrogen and the leaving group (Br) be in an anti-periplanar orientation for the reaction to proceed.
03

Check the stability and accessibility of transition states

The rate of the E2 elimination reaction is directly affected by the stability and accessibility of the transition states. The more stable and easily accessible the transition state, the faster the reaction rate. For the trans isomer, the anti-periplanar β-hydrogen and the leaving group are easily accessible in the chair conformation, allowing for an unhindered E2 elimination reaction. In the case of the cis isomer, the β-hydrogen and the leaving group are not anti-periplanar in the chair conformation, necessitating a ring flip. This leads to a higher energy transition state, making the reaction slower compared to the trans isomer.
04

Identify the faster reacting isomer and draw the product

Since the trans isomer has a more stable and accessible transition state, it reacts faster in this E2 elimination reaction. The expected product from the faster reacting trans isomer will be formed via the abstraction of the β-hydrogen and the loss of the bromide ion, resulting in the formation of a C=C double bond. The product from the trans isomer is 1-isopropyl-1-methylcyclohexene.

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Most popular questions from this chapter

Attempts to prepare optically active iodides by nucleophilic displacement on optically active bromides using I \(\mathrm{I}^{-}\)normally produce racemic iodoalkanes. Why are the product iodoalkanes racemic?

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