Compound with chiral center and E/Z isomerism has most stereoisomers

Steps:

• Draw structures: All are hex-2-enes (CH3-CH=CH-CH2-CH2-CH3 base) with methyl branch at specified carbon.
• Assess E/Z: Requires each double bond carbon (C2, C3) to have two different substituents.
• Identify chiral centers: Carbon with four different substituents.
• Compute total stereoisomers: 2^(number of stereogenic units: E/Z + chiral centers).

Why C is correct:

• 4-methylhex-2-ene (CH3-CH=CH-CH(CH3)-CH2-CH3) has E/Z at double bond (C2: CH3/H; C3: H/CH(CH3)CH2CH3) and chiral C4 (H, CH3, CH=CHCH3, CH2CH3), so 2 × 2 = 4 stereoisomers.

Why the others are wrong:

• A: 2-methylhex-2-ene has identical CH3 groups on C2, no E/Z; no chiral centers; 1 stereoisomer.
• B: 3-methylhex-2-ene has E/Z but no chiral centers; 2 stereoisomers.
• D: 5-methylhex-2-ene has E/Z but C5 has two CH3 groups, not chiral; 2 stereoisomers.

Final answer: C