Identify the major product formed in the foll | vedclass

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Question

(A) The reaction of $3-$bromo$-2-$methylpentane with alcoholic $KOH$ is a dehydrohalogenation reaction ($E2$ mechanism). According to Saytzeff's rule,

Vedclass · Accepted Answer

$2-$Methylpent$-2-$ene

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(A) The reaction of $3-$bromo$-2-$methylpentane with alcoholic $KOH$ is a dehydrohalogenation reaction ($E2$ mechanism). According to Saytzeff's rule,the major product is the more substituted alkene. The starting material is $CH_3-CH(CH_3)-CH(Br)-CH_2-CH_3$. Removal of $H$ from $C-2$ gives $CH_3-C(CH_3)=CH-CH_2-CH_3$ ($2-$methylpent$-2-$ene). Removal of $H$ from $C-4$ gives $CH_3-CH(CH_3)-CH=CH-CH_3$ ($4-$methylpent$-2-$ene). $2-$Methylpent$-2-$ene is more substituted (trisubstituted) compared to $4-$methylpent$-2-$ene (disubstituted),hence it is the major product.

Identify the major product formed in the following reaction: $3-$Bromo$-2-$methylpentane $\xrightarrow{Alc. KOH, \Delta}$ Major product

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