JEE Main 2025 · 22 January · morning · Single correct · +4 / −1
How many different stereoisomers are possible for the given molecule?
CH₃-CH-CH=CH-CH₃
- (a) 2
- (b) 1
- (c) 3
- (d) 4
Answer and solution
Answer (d) 4
Solution
Chemistry ma’am · 4 steps
Find every stereo unit first: one chiral carbon and one C=C that can be cis or trans, so 2².
- Chiral centre. C-2 carries four different groups: H, OH, CH₃ and -CH=CH-CH₃. So it is a stereocentre and can be R or S.
- Geometrical isomerism. At the double bond, C-3 holds H and -CH(OH)CH₃, and C-4 holds H and CH₃. Each carbon has two different groups, so the double bond can be E or Z.
- Count the combinations. The two stereo units are independent, and the molecule has no symmetry, so no meso form is possible. That gives 2ⁿ with n = 2: (2R, E), (2S, E), (2R, Z) and (2S, Z).
Stereoisomers = 2ⁿ = 2² = 4
- So. Pent-3-en-2-ol has 4 stereoisomers, which is option (d).
The idea to keep. Count chiral centres and stereogenic double bonds together as n. If the molecule is unsymmetrical, the stereoisomers number 2ⁿ. Look for a meso form only when the two ends are identical.
Answer from the NTA final answer key. Solution by PrepWiser.