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The one giving maximum number of isomeric alkenes on dehydrohalogenation reaction is (excluding rearrangement) (1) 1-Bromo-2-methylbutane (2) 2-Bromopentane (3) 2-Bromo-3, 3-dimethylpentane (4) 2-Bromopropane

1. 1-Bromo-2-methylbutane
2. 2-Bromopentane
3. 2-Bromo-3, 3-dimethylpentane
4. 2-Bromopropane

Correct answer: 2-Bromopentane

Solution

2-Bromopentane can undergo dehydrohalogenation to form multiple alkenes due to the presence of two different β-hydrogens that can be eliminated, leading to various structural isomers. This results in a higher number of possible alkenes compared to the other options.

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