Question

Which will undergo fastest S\(_N\)2 substitution reaction when treated with NaOH?

• A. A
• B. B
• C. C
• D. D

Hint:

For S\(_N\)2 reactions, the order of reactivity is: benzyl/allyl > 1° > 2° > 3°.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
S\(_N\)2 reactions are bimolecular and involve a backside attack by the nucleophile. The rate is fastest for substrates with less steric hindrance around the carbon bearing the leaving group.

Step 2: Detailed Explanation:
The order of reactivity for S\(_N\)2 is: Methyl > 1° > 2° > 3°. Aromatic groups can also stabilize the transition state.
• (A) n-Butyl bromide: 1° alkyl halide, less hindered. • (B) (C\(_2\)H\(_5\))\(_2\)CHBr: 2° alkyl halide. • (C) (CH\(_3\))\(_3\)CBr: 3° alkyl halide, practically unreactive in S\(_N\)2 due to steric hindrance. • (D) C\(_6\)H\(_5\)–CH\(_2\)Br: Benzyl bromide. The benzyl carbocation transition state is stabilized by resonance, making it very reactive in both S\(_N\)1 and S\(_N\)2. In S\(_N\)2, the benzylic position is less sterically hindered and the transition state is stabilized. Benzyl halides are among the most reactive for S\(_N\)2.

Step 3: Final Answer:
C\(_6\)H\(_5\)–CH\(_2\)Br undergoes the fastest S\(_N\)2 reaction, which corresponds to option (D).