Question

Among the given halides, which one will give the same product in both S\(_N1\) and S\(_N2\) reactions? 

• A. (III) only
• B. (I) and (II)
• C. (III) and (IV)
• D. (I), (III), and (IV)

Hint:

Compounds that do not undergo carbocation rearrangement will give the same product in both S\(_N1\) and S\(_N2\) mechanisms. Typically, cycloalkyl halides and simple secondary alkyl halides meet this criterion.

Answer 1

The Correct Option is C

Step 1: Understanding S\(_N1\) and S\(_N2\) reaction mechanisms. - S\(_N1\) (Unimolecular Nucleophilic Substitution): This mechanism occurs via the formation of a carbocation intermediate. It is favored in tertiary and benzylic/allylic halides due to carbocation stability. - S\(_N2\) (Bimolecular Nucleophilic Substitution): This mechanism proceeds via a backside attack, leading to inversion of configuration. It is favored in primary and secondary alkyl halides with minimal steric hindrance. 
Step 2: Analyzing the given halides. 
- (III) (Cyclohexyl chloride): - This structure does not form a resonance-stabilized carbocation. - Whether the reaction follows S\(_N1\) or S\(_N2\), the product remains the same. - Satisfies the condition of giving the same product in both reactions. 
- (IV) Secondary alkyl bromide with \(\beta\) branching: - The presence of branching reduces the likelihood of rearrangement. - Whether the reaction proceeds via S\(_N1\) or S\(_N2\), the substitution product remains unaffected. - Satisfies the condition of giving the same product in both reactions. 
Step 3: Eliminating incorrect Option. 
- (I) (Branched alkyl bromide): This can undergo carbocation rearrangement in S\(_N1\), leading to different products. 
- (II) (Cyclohexyl chloride with a methyl group): The possibility of rearrangement in S\(_N1\) exists, leading to different products. 
Thus, the correct answer is (C) (III) and (IV).