Chiral secondary chloride yielding two alkenes via E2

Steps:

• Identify structures with a chiral carbon (four different groups including Cl) for optical isomers.
• Eliminate A: Cl on primary carbon with two H atoms, achiral.
• Confirm B and C represent the same 2-chlorobutane structure, chiral at C2.
• Verify D's structure has a chiral secondary carbon with Cl, H, methyl, and isopropyl-like chain fitting C4H9Cl.
• Assess E2 elimination: beta carbons must allow exactly two distinct alkene products.

Why D is correct:

• D features a chiral carbon bonded to Cl, H, CH3, and CH2CH3, enabling two optical isomers; E2 removes beta-H from either adjacent methyl or methylene, forming only two alkenes (1-butene and 2-butene) per Zaitsev/Saytzeff rule.

Why the others are wrong:

• A: Achiral (no optical isomers); E2 yields one alkene, 2-methylpropene.
• B: Identical to C (2-chlorobutane); but question specifies unique fit for two alkenes considering stereoisomerism in products.
• C: Same as B, forms 1-butene and 2-butene (with cis/trans, but structural count exceeds if stereo separated).

Final answer: D