Question

The one giving maximum number of isomeric alkenes on dehydrohalogenation reaction is (excluding rearrangement):

• A. 1-Bromo-2-methylbutane
• B. 2-Bromopropane
• C. 2-Bromopentane
• D. 2-Bromo-3,3-dimethylpentane

Hint:

Always look for a \(\beta\)-carbon that allows for double bond formation in two different directions, and check if the resulting internal alkene is substituted enough to have E/Z isomers.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Dehydrohalogenation follows Saytzeff's rule. The number of isomeric alkenes depends on the number of non-equivalent \(\beta\)-hydrogens available and whether the resulting alkenes can show cis-trans (E/Z) isomerism.

Step 2: Detailed Explanation:
1. 1-Bromo-2-methylbutane: Only one type of \(\beta\)-H. Gives only 2-methylbut-1-ene. (1 product). 2. 2-Bromopropane: Two equivalent \(\beta\)-carbons. Gives only propene. (1 product). 3. 2-Bromopentane: Two non-equivalent \(\beta\)-carbons (C1 and C3). - From C1: Pent-1-ene. - From C3: Pent-2-ene. Pent-2-ene exists as cis and trans isomers. - Total = 3 isomers. 4. 2-Bromo-3,3-dimethylpentane: C3 has no hydrogens. Only \(\beta\)-H available is on C1. Gives 3,3-dimethylpent-1-ene. (1 product).

Step 3: Final Answer
2-Bromopentane gives the maximum number of isomeric alkenes.