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(B) The reaction of $3-$bromo$-2-$methylpentane with alcoholic $KOH$ is a dehydrohalogenation reaction ($\beta-$elimination).In this reaction,a hydrog

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$2-$Methylpent$-2-$ene

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(B) The reaction of $3-$bromo$-2-$methylpentane with alcoholic $KOH$ is a dehydrohalogenation reaction ($\beta-$elimination).In this reaction,a hydrogen atom is removed from the $\beta-$carbon and a bromine atom is removed from the $\alpha-$carbon to form an alkene.According to Saytzeff's rule,the major product is the more substituted alkene.For $3-$bromo$-2-$methylpentane,elimination can occur at $C-2$ or $C-4$.Elimination at $C-2$ gives $2-$methylpent$-2-$ene (a trisubstituted alkene),while elimination at $C-4$ gives $3-$methylpent$-2-$ene (a disubstituted alkene).Thus,$2-$methylpent$-2-$ene is the major product.

Identify the major product $A$ in the following reaction.
$3-$Bromo$-2-$methylpentane $\xrightarrow{alc. KOH} A$

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Identify the major product $A$ in the following reaction.$3-$Bromo$-2-$methylpentane $\xrightarrow{alc. KOH} A$