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(C) The reaction of $4$-bromo-$1$-pentene with alcoholic $KOH$ is a dehydrohalogenation ($E2$ elimination) reaction. The removal of $HBr$ from $C-3$ a

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$CH_2=CH-CH=CH-CH_3$

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(C) The reaction of $4$-bromo-$1$-pentene with alcoholic $KOH$ is a dehydrohalogenation ($E2$ elimination) reaction. The removal of $HBr$ from $C-3$ and $C-4$ positions results in the formation of $1,3$-pentadiene $(CH_2=CH-CH=CH-CH_3)$,which is a conjugated diene and more stable than the isolated diene ($1,4$-pentadiene) formed by elimination from $C-4$ and $C-5$. Thus,$1,3$-pentadiene is the major product.

What is the major product of the following reaction?
$CH_2=CH-CH_2-CH(Br)-CH_3 + KOH \xrightarrow[\text{heat}]{\text{alcohol}} \dots$

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What is the major product of the following reaction? $CH_2=CH-CH_2-CH(Br)-CH_3 + KOH \xrightarrow[\text{heat}]{\text{alcohol}} \dots$