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

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Question

(C) The reaction proceeds via a carbocation intermediate. $1.$ Protonation of $CH_3-C(CH_3)_2-CH=CH_2$ gives a secondary carbocation $CH_3-C(CH_3)_2-C

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) The reaction proceeds via a carbocation intermediate. $1.$ Protonation of $CH_3-C(CH_3)_2-CH=CH_2$ gives a secondary carbocation $CH_3-C(CH_3)_2-C^{+}H-CH_3$. $2.$ $A$ $1,2$-methyl shift occurs to form a more stable tertiary carbocation $CH_3-C^{+}(CH_3)-CH(CH_3)-CH_3$. $3.$ Nucleophilic attack by $H_2O$ on the tertiary carbocation followed by deprotonation yields the major product $CH_3-C(OH)(CH_3)-CH(CH_3)-CH_3$ ($2,3$-dimethyl-$2$-butanol).

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

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