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

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Question

(A) The acid-catalyzed hydration of $3,3-	ext{dimethyl}-1-	ext{butene}$ proceeds via a carbocation intermediate.$1$. Protonation of the alkene follo

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The acid-catalyzed hydration of $3,3-	ext{dimethyl}-1-	ext{butene}$ proceeds via a carbocation intermediate.$1$. Protonation of the alkene follows Markovnikov's rule to form a secondary carbocation: $CH_3-C(CH_3)_2-CH^{+}-CH_3$.$2$. $A$ $1,2-	ext{methyl shift}$ occurs to convert the secondary carbocation into 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,$2,3-	ext{dimethyl}-2-	ext{butanol}$: $CH_3-C(OH)(CH_3)-CH(CH_3)-CH_3$.

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

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