Consider the aggressive dehydrobromination of 3-bromo-3-cyclopentylhexane utilizing alcoholic KOH an — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

3-bromo-3-cyclopentylhexane has the structure $CH_3-CH_2-C(Br)(C_5H_9)-CH_2-CH_2-CH_3$. It possesses exactly three distinct sets of $\beta$ -hydrogens. 1) Elimination towards the cyclopentyl ring: Abstracting a proton from the tertiary ring carbon forms an exocyclic double bond, yielding 1 unique isomer. 2) Elimination towards the C2 ethyl carbon: Forms 3-cyclopentylhex-2-ene. This specific internal double bond can exhibit $E/Z$ geometrical isomerism, yielding 2 distinct isomers. 3) Elimination towards the C4 propyl carbon: Forms 3-cyclopentylhex-3-ene. This internal double bond can also exhibit $E/Z$ isomerism, yielding 2 distinct isomers. Absolute total = $1 + 2 + 2 = 5$ distinct alkene isomers.