Determine the number of chiral products (incl | vedclass

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(B) The reaction is an $E_2$ elimination. The substrate has two distinct $\beta$-hydrogen positions available for elimination.$1$. Elimination of the

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$4$

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(B) The reaction is an $E_2$ elimination. The substrate has two distinct $\beta$-hydrogen positions available for elimination.$1$. Elimination of the $\beta$-hydrogen from the $-CH_2-$ group attached to the cyclopentyl ring leads to the formation of a double bond between the $\alpha$ and $\beta$ carbons. This alkene can exist as $E$ and $Z$ isomers. Since the molecule contains other chiral centers (the cyclopentyl and cyclohexyl rings),each of these geometric isomers will be chiral,resulting in $2$ chiral products.$2$. Elimination of the $\beta$-hydrogen from the $-CH(CH_3)-$ group leads to a different alkene. This alkene also exhibits geometric isomerism ($E$ and $Z$). Again,due to the existing chiral centers in the rings,both the $E$ and $Z$ isomers will be chiral,resulting in another $2$ chiral products.Total number of chiral products = $2 + 2 = 4$.

Determine the number of chiral products (including stereoisomers) formed upon $E_2$ elimination of the given substrate with $alc. KOH$.

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