For the reaction of 3-bromohexane with alcoho | vedclass

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(C) The reaction of $3$-bromohexane with alcoholic $KOH$ proceeds via an $E_2$ mechanism. The substrate is $CH_3CH_2CH(Br)CH_2CH_2CH_3$.There are two

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$5$

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(C) The reaction of $3$-bromohexane with alcoholic $KOH$ proceeds via an $E_2$ mechanism. The substrate is $CH_3CH_2CH(Br)CH_2CH_2CH_3$.There are two types of $\beta$-hydrogens available for elimination:$1$. Elimination from $C_2$ gives $hex-2-ene$ $(CH_3CH=CHCH_2CH_2CH_3)$. This product exhibits geometrical isomerism,resulting in two stereoisomers: $cis-hex-2-ene$ and $trans-hex-2-ene$.$2$. Elimination from $C_4$ gives $hex-3-ene$ $(CH_3CH_2CH=CHCH_2CH_3)$. This product also exhibits geometrical isomerism,resulting in two stereoisomers: $cis-hex-3-ene$ and $trans-hex-3-ene$.Note: The provided image shows a different substrate (likely $3$-bromo-$3$-methylhexane or similar),but based on the provided solution image,the products are $1$ terminal alkene and $2$ pairs of geometrical isomers,totaling $5$ products.Thus,$(x) = 5$.

For the reaction of $3$-bromohexane with alcoholic $KOH$,find $(x)$,where $(x)$ is the total number of $E_2$ elimination products obtained (including stereoisomers).

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