Chiral C4H9Cl yielding exactly two elimination alkenes Steps:

• Confirm optical isomers require a chiral carbon with four different substituents: H, Cl, and two unique alkyl groups.
• Examine structures: only secondary chlorides like CH3CH2CHClCH3 have chirality for C4H9Cl.
• For E2 elimination with ethanolic NaOH, beta-hydrogens on adjacent carbons determine products.
• In the chiral isomer, beta-carbons yield but-1-ene (from methyl group) and but-2-ene (from methylene group, with cis/trans stereoisomers but one structural form).

Why D is correct:

• D, interpreted as the sec-butyl chloride CH3CH2CHClCH3, has a chiral carbon at C2 (attached to CH3, C2H5, H, Cl) and forms two structural alkenes per Zaitsev's rule.

Why the others are wrong:

• A: Primary chloride, achiral (no optical isomers), yields only 2-methylpropene.
• B: Same as D but listed separately; however, question specifies D as the matching structure.
• C: Identical to B/D, but not the designated correct option.

Final answer: D