The total number of alkenes possible by dehyd | vedclass

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(A) $3$-Bromo-$3$-cyclopentylhexane has three different types of $\beta$-hydrogens (not $\alpha$-hydrogens as $\alpha$ is the carbon attached to $Br$)

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(A) $3$-Bromo-$3$-cyclopentylhexane has three different types of $\beta$-hydrogens (not $\alpha$-hydrogens as $\alpha$ is the carbon attached to $Br$). These are marked as $(a)$,$(b)$,and $(c)$ in the structure. Dehydrobromination leads to the formation of three structurally isomeric alkenes: $1$. From $\beta$-hydrogen $(a)$: $CH_3CH_2CH=C(cyclopentyl)CH_2CH_3$ (exists as $E$ and $Z$ isomers). $2$. From $\beta$-hydrogen $(b)$: $CH_3CH_2CH_2C(cyclopentyl)=CHCH_3$ (exists as $E$ and $Z$ isomers). $3$. From $\beta$-hydrogen $(c)$: $CH_3CH_2CH_2C(cyclopentyl)=CHCH_2CH_3$ (this is the same as the product from $(b)$). Wait,re-evaluating the structure: The $\beta$-carbons are the two $CH_2$ groups of the hexane chain and the $CH$ group of the cyclopentyl ring. - Removal of $H$ from $CH_2$ (position $a$): $CH_3CH_2CH=C(cyclopentyl)CH_2CH_3$ ($E/Z$ isomers). - Removal of $H$ from $CH_2$ (position $b$): $CH_3CH_2CH_2C(cyclopentyl)=CHCH_3$ ($E/Z$ isomers). - Removal of $H$ from $CH$ (position $c$): $CH_3CH_2CH_2C(cyclopentyl)=C(CH_2CH_3)_2$ (this is a different alkene,no $E/Z$ isomerism). Total alkenes = $2 (E/Z) + 2 (E/Z) + 1 = 5$.

The total number of alkenes possible by dehydrobromination of $3$-bromo-$3$-cyclopentylhexane using alcoholic $KOH$ is

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