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 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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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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Solution

The correct option is D $C H_{3} C l > C H_{3} C H_{2} C l > (C H_{3})_{2} C H C l > (C H_{3})_{3} C C l$
Rate of $S_{N} 2$ reaction $\propto \frac{1}{s t e r i c o v e r c r o w d i n g i n t r a n s i t i o n s t a t e}$

So, the reactivity of $C H_{3} C l$ will be the most.

And as the steric hinderance increases rate of ${S_{N}}^{2}$ will decrease. And hinderance increases with crowding on the central atom.

Hence the correct order is option D.

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Similar questions

Which compound in each of the following pairs will react faster in S_N2 reaction with OH⁻?

(i) CH₃Br or CH₃I

(ii) (CH₃)₃CCl or CH₃Cl

S N_{1}

(C H_{3})_{3} C C l

C H_{3} C l

C H_{3} C l, C H_{3} C H_{2} C l, C H_{3} (C H_{2})_{2} C l, C H_{3} (C H_{2})_{3} C l

(C H_{3})_{3} C C l + H_{2} O \to (C H_{3})_{3} C O H + H C l (\frac{d x}{d t}) = k [(C H_{3})_{3} C C l]

{C H}_{3} C \equiv \oplus C N a + {({C H}_{3})}_{2} C H C l ⟶

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