Question

The order of reactivity of the following compounds towards SN2 displacement reaction is (i) \(C_{6}H_{5}CH(CH_{3})Br\) (ii) \(C_{6}H_{5}CH(C_{6}H_{5})Br\) (iii) \(C_{6}H_{5}C(CH_{3})(C_{6}H_{5})Br\) (iv) \(C_{6}H_{5}CH_{2}Br\)

• A. (ii)$>$(i)$>$(iii)$>$(iv)
• B. (iv)$>$(ii)$>$(i)$>$(iii)
• C. (ii)$>$(iii)$>$(i)$>$(iv)
• D. (i)$>$(ii)$>$(iii)$>$(iv)
• E. (iv)$>$(i)$>$(ii)$>$(iii)

Hint:

Benzylic halides are more reactive than simple alkyl halides in SN2 due to stabilization of the transition state by the phenyl ring.

Answer 1

Answer: (E)

Step 1: Understanding the Concept:
SN2 reactivity: Least steric hindrance around the carbon bearing the leaving group → fastest. Order: primary$>$secondary$>$tertiary. Benzylic halides are reactive, but bulky groups hinder.

Step 2: Detailed Explanation:
(iv) \(C_6H_5CH_2Br\): Benzylic, primary → most reactive.
(i) \(C_6H_5CH(CH_3)Br\): Secondary benzylic.
(ii) \(C_6H_5CH(C_6H_5)Br\): Secondary but with two large aryl groups → more hindered than (i).
(iii) \(C_6H_5C(CH_3)(C_6H_5)Br\): Tertiary benzylic → most hindered → least reactive.
Order: (iv)$>$(i)$>$(ii)$>$(iii).

Step 3: Final Answer:
The correct order is (iv)$>$(i)$>$(ii)$>$(iii).