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Question

(C) $S_N^1$ reaction proceeds via the formation of a carbocation intermediate. The rate of $S_N^1$ reaction is directly proportional to the stability

Vedclass · Accepted Answer

$CH_3-C(CH_3)_2-Cl$

Vedclass · Answer

(C) $S_N^1$ reaction proceeds via the formation of a carbocation intermediate. The rate of $S_N^1$ reaction is directly proportional to the stability of the carbocation formed. The stability order of carbocations is $3^\circ > 2^\circ > 1^\circ > CH_3^+$.In the given options:$(a)$ $CH_3-CH_2-Cl$ forms a $1^\circ$ carbocation $(CH_3-CH_2^+)$.$(b)$ $(CH_3)_2CH-Cl$ forms a $2^\circ$ carbocation $((CH_3)_2CH^+)$.$(c)$ $CH_3-C(CH_3)_2-Cl$ forms a $3^\circ$ carbocation $((CH_3)_3C^+)$.$(d)$ $CH_3-CH(CH_3)-CH_2-Cl$ forms a $1^\circ$ carbocation $(CH_3-CH(CH_3)-CH_2^+)$.Since the $3^\circ$ carbocation is the most stable,the $S_N^1$ reaction is fastest in $CH_3-C(CH_3)_2-Cl$.

$S_N^1$ reaction is faster in

Similar Questions

The correct reactivity order of the following compounds towards alkaline hydrolysis is:-
$I. \ p-CH_3O-C_6H_4-CH_2Cl$
$II. \ C_6H_5-CH_2Cl$
$III. \ p-CH_3-C_6H_4-CH_2Cl$
$IV. \ m-CH_3-C_6H_4-CH_2Cl$

Identify the product $A$ in the following reaction: $C_2H_5Br + \text{aq. } KOH \rightarrow A + KBr$

The major product $P$ of the above reaction is:

In which case is the formation of $butanenitrile$ possible?

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