A)
\[C{{H}_{3}}C{{H}_{2}}Br<{{(C{{H}_{3}})}_{3}}CCl<\] 
\[CHCl-C{{H}_{3}}\]
B) \[C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}l<C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}Br\]\[<C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}Cl\]
C) \[{{C}_{6}}{{H}_{5}}CCl{{(C{{H}_{3}})}_{2}}<{{C}_{6}}{{H}_{5}}CHClC{{H}_{3}}\]\[<{{C}_{6}}{{H}_{5}}C{{H}_{2}}Cl\]
D) \[C{{H}_{2}}=CHCl<C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}Cl<C{{H}_{2}}=\]\[CH-C{{H}_{2}}Cl\]
Correct Answer: C
Solution :
More the crowding in the transition state, lessees will be the rate of \[{{S}_{N}}2\] reaction. Hence, the correct order is\[{{C}_{6}}{{H}_{5}}CCI{{(C{{H}_{3}})}_{2}}<{{C}_{6}}{{H}_{5}}CHCIC{{H}_{3}}<{{C}_{6}}{{H}_{5}}C{{H}_{2}}CI\]
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