In $S_{N} 2$ reactions, the correct order of reactivity for the following compounds:
$C H_{3} C l, C H_{3} C H_{2} C l, (C H_{3})_{2} C H C l$ and $(C H_{3})_{3} C C l$ is:

C H_{3} C l > (C H_{3})_{2} C H C l > C H_{3} C H_{2} C l > (C H_{3})_{3} C C l

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C H_{3} C l > C H_{3} C H_{2} C l > (C H_{3})_{2} C H C l > (C H_{3})_{3} C C l

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C H_{3} C H_{2} C l > C H_{3} C l > (C H_{3})_{2} C H C l > (C H_{3})_{3} C C l

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(C H_{3})_{2} C H C l > C H_{3} C H_{2} C l > C H_{3} C l > (C H_{3})_{3} C C l

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Solution

The correct option is B $C H_{3} C l > C H_{3} C H_{2} C l > (C H_{3})_{2} C H C l > (C H_{3})_{3} C C l$
Steric congestion around the carbon atom undergoing the inversion process will slow down the $S_{N} 2$ reaction. hence the less congestion faster will the reaction. So, the order is
$C H_{3} C l > C H_{3} C H_{2} C l > (C H_{3})_{2} C H C l > (C H_{3})_{3} C C l$

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