A) \[C{{H}_{3}}Cl>{{(C{{H}_{3}})}_{2}}CHCl>C{{H}_{3}}C{{H}_{2}}Cl>{{(C{{H}_{3}})}_{3}}CCl\]
B) \[C{{H}_{3}}Cl>C{{H}_{3}}C{{H}_{2}}Cl>{{(C{{H}_{3}})}_{2}}CHCl>{{(C{{H}_{3}})}_{3}}CCl\]
C) \[C{{H}_{3}}C{{H}_{2}}Cl>C{{H}_{3}}Cl>{{(C{{H}_{3}})}_{2}}CHCl>{{(C{{H}_{3}})}_{3}}CCl\]
D) \[{{(C{{H}_{3}})}_{2}}CHCl>C{{H}_{3}}C{{H}_{2}}Cl>C{{H}_{3}}Cl>{{(C{{H}_{3}})}_{3}}CCl\]
Correct Answer: B
Solution :
[b] Steric hindrance around the carbon atom undergoing the inversion process will slow down the \[{{S}_{N}}2\] reaction, hence less hindrance faster will the reaction. So, the order is \[C{{H}_{3}}Cl>C{{H}_{3}}C{{H}_{2}}-Cl>{{(C{{H}_{3}})}_{2}}CH-Cl>{{(C{{H}_{3}})}_{3}}CCl\]You need to login to perform this action.
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