In the following reaction CH_3-C(CH_3)_2-CH=C | vedclass

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Question

(A) The reaction proceeds via a carbocation intermediate.First,the addition of $H^{+}$ to $CH_3-C(CH_3)_2-CH=CH_2$ follows Markovnikov's rule to form

Vedclass · Accepted Answer

$CH_3-C(OH)(CH_3)-CH(CH_3)-CH_3$

Vedclass · Answer

(A) The reaction proceeds via a carbocation intermediate.First,the addition of $H^{+}$ to $CH_3-C(CH_3)_2-CH=CH_2$ follows Markovnikov's rule to form a secondary carbocation: $CH_3-C(CH_3)_2-C^{+}H-CH_3$.Next,a $1,2$-methyl shift occurs to convert the secondary carbocation into a more stable tertiary carbocation: $CH_3-C^{+}(CH_3)-CH(CH_3)-CH_3$.Finally,the attack of $H_2O$ on this tertiary carbocation followed by deprotonation yields $2,3$-dimethyl-$2$-butanol,which is $CH_3-C(OH)(CH_3)-CH(CH_3)-CH_3$,as the major product $A$.

In the following reaction $CH_3-C(CH_3)_2-CH=CH_2 \xrightarrow{H_2O/H^{+}} A \text{ (major)} + B \text{ (minor)}$,the major product $A$ is:

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