When neopentyl alcohol is heated with an acid | vedclass

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Question

(A) The dehydration of neopentyl alcohol $(CH_3-C(CH_3)_2-CH_2OH)$ in the presence of an acid proceeds via the formation of a primary carbocation,whic

Vedclass · Accepted Answer

$2-$methyl$-2-$butene and $2-$methyl$-1-$butene

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(A) The dehydration of neopentyl alcohol $(CH_3-C(CH_3)_2-CH_2OH)$ in the presence of an acid proceeds via the formation of a primary carbocation,which undergoes a $1,2$-methyl shift to form a more stable tertiary carbocation $(CH_3-C^+(CH_3)-CH_2-CH_3)$.From this tertiary carbocation,two alkenes can be formed by the loss of a proton $(H^+)$:$1$. Loss of a proton from the $CH_2$ group leads to $2$-methyl-$2$-butene $(CH_3-C(CH_3)=CH-CH_3)$,which is the more substituted and thus more stable alkene (major product,$85\%$).$2$. Loss of a proton from the $CH_3$ group leads to $2$-methyl-$1$-butene $(CH_3-C(CH_3)-CH=CH_2)$,which is less substituted (minor product,$15\%$).Therefore,the mixture consists of $2$-methyl-$2$-butene and $2$-methyl-$1$-butene.

When neopentyl alcohol is heated with an acid,it slowly converts into an $85:15$ mixture of alkenes $A$ and $B$,respectively. What are these alkenes $?$ (Note: $A$ is the major product and $B$ is the minor product.)

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