Answer:

To determine which alkyl chloride undergoes dehydrohalogenation in the presence of a strong base to give 2-pentene as the only alkene product, we need to consider the stability of the carbocation intermediate formed during the reaction.

Dehydrohalogenation:
Dehydrohalogenation is a type of elimination reaction where a hydrogen atom and a halogen atom are removed from an alkyl halide to form an alkene. In the presence of a strong base, such as KOH or NaOH, the halogen is replaced by a hydroxide ion (OH-) to form an alcohol intermediate. The alcohol intermediate then undergoes elimination to form the alkene product.

Stability of carbocation intermediates:
The stability of carbocations is determined by the number of alkyl groups attached to the positively charged carbon. The greater the number of alkyl groups, the more stable the carbocation.

Explanation:
Let's analyze the alkyl chlorides listed:

a) 1-chloropentane: The carbocation intermediate formed during the dehydrohalogenation would be a primary carbocation. Primary carbocations are highly unstable and would undergo rearrangement or compete with other reactions. Therefore, it is unlikely to give 2-pentene as the only alkene product.

b) 2-chloropentane: The carbocation intermediate formed during the dehydrohalogenation would be a secondary carbocation. Secondary carbocations are more stable than primary carbocations but less stable than tertiary carbocations. While it is possible for 2-chloropentane to undergo dehydrohalogenation to form 2-pentene, there is a possibility of competing reactions or side products.

c) 3-chloropentane: The carbocation intermediate formed during the dehydrohalogenation would be a tertiary carbocation. Tertiary carbocations are the most stable carbocations. They have the maximum number of alkyl groups attached to the positively charged carbon, providing greater stability. Therefore, 3-chloropentane is the most likely alkyl chloride to undergo dehydrohalogenation to give 2-pentene as the only alkene product.

d) 1-chloro-2-methylbutane: The carbocation intermediate formed during the dehydrohalogenation would be a secondary carbocation. Similar to 2-chloropentane, there is a possibility of competing reactions or side products.

Conclusion:
Among the alkyl chlorides listed, 3-chloropentane is the most likely to undergo dehydrohalogenation in the presence of a strong base to give 2-pentene as the only alkene product.