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(B) The reaction involves the elimination of $HI$ from $3-iodo-2,3-dimethylpentane$ in the presence of $t-BuOH$ and heat,which proceeds via an $E1$ me

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$2,3-$dimethylpent$-2-$ene

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(B) The reaction involves the elimination of $HI$ from $3-iodo-2,3-dimethylpentane$ in the presence of $t-BuOH$ and heat,which proceeds via an $E1$ mechanism.$1$. The leaving group $I^-$ departs to form a secondary carbocation.$2$. This carbocation undergoes a $1,2-hydride$ shift to form a more stable tertiary carbocation.$3$. Further rearrangement via a $1,2-methyl$ shift leads to an even more stable tertiary carbocation.$4$. Finally,the loss of a proton from the adjacent carbon atom results in the formation of the most substituted alkene,which is the major product.$5$. The final product is $2,3-dimethylpent-2-ene$.

The major product in the following reaction is:

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