NEETChemistryHaloalkanes and Haloarenes
Match List-I with List-II. List-I List-II (A) n -Butyl bromide (I) Undergoes predominantly S _ N 1 reaction (B) tert-Butyl bromide (II) Undergoes predominantly S _ N 2 reaction (C) Benzyl bromide (III) Unreactive towards nucleophilic substitution (D) Vinyl bromide (IV) Highly reactive in both S _ N 1 and S _ N 2 reactions Choose the correct answer from the options given below:
Options
- A(A)-(II), (B)-(I), (C)-(IV), (D)-(III)
- B(A)-(II), (B)-(I), (C)-(I), (D)-(IV)
- C(A)-(I), (B)-(II), (C)-(IV), (D)-(III)
- D(A)-(II), (B)-(IV), (C)-(I), (D)-(III)
Correct answer
A. (A)-(II), (B)-(I), (C)-(IV), (D)-(III)
Step-by-step solution
(A) n -Butyl bromide is a primary ( 1^ ) alkyl halide with minimal steric hindrance, so it undergoes predominantly S _ N 2 reaction. (B) tert-Butyl bromide is a tertiary ( 3^ ) alkyl halide. Due to severe steric hindrance, it does not undergo S _ N 2 , but readily undergoes S _ N 1 reaction by forming a stable 3^ carbocation. (C) Benzyl bromide is highly reactive in both S _ N 1 and S _ N 2 reactions. It forms a resonance-stabilized benzyl carbocation (favouring S _ N 1 ) and has low steric hindrance at the -carbon