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MCQEasyJEE 2025SN1 & SN2 Reactions

JEE Chemistry 2025 Question with Solution

Given below are two statements:

Statement (I): Alcohols are formed when alkyl chlorides are treated with aqueous potassium hydroxide by elimination reaction.

Statement (II): In alcoholic potassium hydroxide, alkyl chlorides form alkenes by abstracting the hydrogen from the β\beta-carbon.

In the light of the above statements, choose the most appropriate answer from the options given below:

  • A

    Both Statement I and Statement II are incorrect

  • B

    Statement I is incorrect but Statement II is correct

  • C

    Statement I is correct but Statement II is incorrect

  • D

    Both Statement I and Statement II are correct

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: Two statements about the reaction of alkyl chlorides with aqueous and alcoholic potassium hydroxide.

Find: Which statement is correct.

For aqueous KOH, alkyl chlorides form alcohols by nucleophilic substitution, not by elimination.

R−Cl+KOH(aq)→R−OH+KClR-\text{Cl} + \text{KOH(aq)} \rightarrow R-\text{OH} + \text{KCl}

So, Statement I is incorrect because the product is alcohol, but the reaction type mentioned is wrong.

For alcoholic KOH, alkyl chlorides undergo elimination. A hydrogen is removed from the β\beta-carbon and an alkene is formed.

R−CH2−CHCl−R′+KOH(alc)→R−CH=CH−R′+KCl+H2OR-\text{CH}_2-\text{CHCl}-R' + \text{KOH(alc)} \rightarrow R-\text{CH}=\text{CH}-R' + \text{KCl} + \text{H}_2\text{O}

So, Statement II is correct.

Therefore, the correct option is B.

Step-by-step Evaluation

Given:

  • Statement I links aqueous potassium hydroxide with alcohol formation by elimination.
  • Statement II links alcoholic potassium hydroxide with alkene formation by abstraction of hydrogen from the β\beta-carbon.

Find: The most appropriate option.

  1. Evaluate Statement I
  • Alkyl chlorides react with aqueous potassium hydroxide to give alcohols.
  • However, this occurs through a substitution reaction because OH−\text{OH}^- replaces Cl−\text{Cl}^-.
R−Cl+KOH(aq)→R−OH+KClR-\text{Cl} + \text{KOH(aq)} \rightarrow R-\text{OH} + \text{KCl}
  • Therefore, the statement is wrong because it calls the process elimination.
  1. Evaluate Statement II
  • In alcoholic potassium hydroxide, the base removes a hydrogen from the β\beta-carbon.
  • This leads to elimination of hydrogen halide and formation of an alkene.
  • This is dehydrohalogenation.
R−CH2−CHCl−R′+KOH(alc)→R−CH=CH−R′+KCl+H2OR-\text{CH}_2-\text{CHCl}-R' + \text{KOH(alc)} \rightarrow R-\text{CH}=\text{CH}-R' + \text{KCl} + \text{H}_2\text{O}
  • Therefore, Statement II is correct.

Conclude: Statement I is incorrect but Statement II is correct. Hence, the correct option is B.

Common mistakes

  • Confusing substitution with elimination in aqueous KOH. Aqueous potassium hydroxide provides OH−\text{OH}^- as a nucleophile, so alcohol formation occurs by substitution. Do not label this reaction as elimination.

  • Assuming alcoholic KOH also gives alcohols. In alcoholic medium, KOH behaves mainly as a base and removes hydrogen from the β\beta-carbon, leading to alkene formation instead.

  • Ignoring the role of the solvent. The reaction outcome depends strongly on whether KOH is aqueous or alcoholic. Always identify the medium before deciding between substitution and elimination.

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