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Chapter 4: Problem 65

When an alkyl chloride (A) is treated with alcoholic potash, 2 -methyl-2-butene is obtained as the main product. (A) is (a) 1 -chloro-3-methylbutane (b) 2 -chloro- 2 -methylbutane (c) 2,3 -dichloro pentane (d) 1 -chloro-2-methylbutane

Short Answer

Expert verified
Answer: (b) 2-chloro-2-methylbutane

Step by step solution

01

Recognize the reaction

The reaction mentioned in the exercise is an elimination reaction. Alcoholic potash is a strong base that deprotonates the alkyl halide, leading to the formation of an alkene. The specific reaction type is called E2 (Elimination, bimolecular).
02

Identify the product's structure

We are given that the main product of the reaction is 2-methyl-2-butene. Its structure is as follows: ``` CH3-CH=C(CH3)-CH3 ```
03

Determine the possible structure of compound (A)

Now, we should determine the possible structure of compound (A) that would lead to the formation of 2-methyl-2-butene through an E2 reaction. To eliminate a hydrogen and a halogen, there must be a halogen (the leaving group) on a carbon and a hydrogen atom on the adjacent carbon. Considering this condition, the possible structure of compound (A) is: ``` CH3-CH(H)C(CH3)(Cl)-CH3 ```
04

Compare the possible structure with the answer choices

Now, compar the possible structure of compound (A) with the given answer choices: (a) 1-chloro-3-methylbutane (b) 2-chloro-2-methylbutane (c) 2,3-dichloro pentane (d) 1-chloro-2-methylbutane Only option (b) has the correct structure as the expected compound (A): ``` CH3-CH(H)C(CH3)(Cl)-CH3 ``` Answer: (b) 2-chloro-2-methylbutane is the correct alkyl chloride that when treated with alcoholic potash would yield 2-methyl-2-butene as the main product.

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