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

We need to find which compound gives the maximum number of isomeric alkenes on dehydrohalogenation (excluding rearrangement).

1-Bromo-2-methylbutane: CH$$_2$$Br-CH(CH$$_3$$)-CH$$_2$$-CH$$_3$$

The only $$\beta$$-carbon is C2. Elimination gives: 2-methylbut-1-ene.

Number of isomeric alkenes = 1

2-Bromopropane: CH$$_3$$-CHBr-CH$$_3$$

$$\beta$$-hydrogens are on C1 and C3 (equivalent). Elimination gives only: propene.

Number of isomeric alkenes = 1

2-Bromopentane: CH$$_3$$-CHBr-CH$$_2$$-CH$$_2$$-CH$$_3$$

$$\beta$$-hydrogens at C1 and C3:

Elimination from C1: gives pent-1-ene

Elimination from C3: gives pent-2-ene, which exists as both (E)-pent-2-ene and (Z)-pent-2-ene

Number of isomeric alkenes = 3 (maximum)

2-Bromo-3,3-dimethylpentane: CH$$_3$$-CHBr-C(CH$$_3$$)$$_2$$-CH$$_2$$-CH$$_3$$

C3 is quaternary (no $$\beta$$-H). Only $$\beta$$-H at C1.

Elimination from C1: gives 3,3-dimethylpent-1-ene

Number of isomeric alkenes = 1

2-Bromopentane gives the maximum number of isomeric alkenes (3).

The correct answer is Option 3: 2-Bromopentane.