Chiral C4H9Cl yielding two alkenes via elimination

Steps:

• C4H9Cl isomers include n-butyl, isobutyl, sec-butyl, and tert-butyl chlorides; optical isomers require a tetrahedral carbon with four different substituents.
• Sec-butyl chloride (CH3CH2CHClCH3) has a chiral carbon at C2, existing as a pair of enantiomers.
• Ethanol/NaOH induces E2 elimination; from sec-butyl chloride, β-hydrogen abstraction yields but-1-ene (from C1-C2) and but-2-ene (from C2-C3).
• This produces a mixture of two alkenes, matching the reaction outcome.

Why C is correct:

• CH3CH2CHClCH3 has a stereocenter (C2 attached to H, Cl, CH3, CH2CH3), forming two optical isomers, and E2 gives two constitutional alkene isomers per Zaitsev/Saytzeff rule.

Why the others are wrong:

• A: (CH3)2CHCH2Cl lacks a chiral center (CH2Cl has two H's) and yields only one alkene, 2-methylpropene.
• B: CH3CH2CHCH3 is not a C4H9Cl formula (missing Cl), invalid structure.
• D: CH3CH2CH2CH2Cl has no chiral center and yields mainly one alkene, but-1-ene.

Final answer: C