JEE Chemistry 2023 Question with Solution

Given: We compare the alkenes formed on dehydrohalogenation of the given bromoalkanes, excluding rearrangement.

Find: Which compound gives the maximum number of isomeric alkenes.

Dehydrohalogenation involves elimination of $\text{HBr}$ to form alkenes. The number of isomeric alkenes depends on the number of different $\beta$-hydrogens that can be removed.

For 2-Bromopentane, two different $\beta$-carbons are available, leading to the formation of multiple alkenes such as $\text{pent-1-ene}$ and $\text{pent-2-ene}$. Additionally, $\text{pent-2-ene}$ can exist as cis and trans isomers, giving a total of three isomeric alkenes.

For the other options:

• 1-Bromo-2-methylbutane and 2-Bromo-3,3-dimethylpentane have only one possible elimination product.
• 2-Bromopropane gives only one product, $\text{propene}$.

The solution states the correct option as D. Hence, following the provided the solution, the correct option is D.