JEE Main 2026 — Haloalkanes and Haloarenes Question with Solution
JEE Main 2026 (23 January Shift 1)
Question
Consider all the structural isomers with molecular formula are separately treated with to give respective substitution products, without any rearrangement. The number of products which can exhibit optical isomerism from these is .
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Show full solutionCorrect answer: 3
Correct answer
3
Step-by-step explanation
For C₅H₁₁Br structural isomers to show optical isomerism after KOH substitution (Br→OH), the resulting product must have a chiral center (asymmetric carbon with four different groups).
Possible carbon skeletons: n-pentane, 2-methylbutane, 2,2-dimethylpropane.
Analysis of isomers:
2-bromopentane → CH₃CHOHCH₂CH₂CH₃: C2 bonded to H, OH, CH₃, and CH₂CH₂CH₃ (chiral).
3-bromopentane → CH₃CH₂CHOHCH₂CH₃: C3 bonded to H, OH, CH₂CH₃, CH₂CH₃ (NOT chiral).
3-bromo-2-methylbutane → CH₃CH(CH₃)CHOHCH₃: Chiral center present.
1-bromo-3-methylbutane → HOCH₂CH(CH₃)CH₂CH₃: C2 bonded to H, OH, CH₃, and CH₂CH₃ (chiral).
2,2-dimethylpropane derivatives: No chiral centers possible.
Compounds giving chiral products: 2-bromopentane, 3-bromo-2-methylbutane, and 1-bromo-3-methylbutane. Total = 3.
Possible carbon skeletons: n-pentane, 2-methylbutane, 2,2-dimethylpropane.
Analysis of isomers:
2-bromopentane → CH₃CHOHCH₂CH₂CH₃: C2 bonded to H, OH, CH₃, and CH₂CH₂CH₃ (chiral).
3-bromopentane → CH₃CH₂CHOHCH₂CH₃: C3 bonded to H, OH, CH₂CH₃, CH₂CH₃ (NOT chiral).
3-bromo-2-methylbutane → CH₃CH(CH₃)CHOHCH₃: Chiral center present.
1-bromo-3-methylbutane → HOCH₂CH(CH₃)CH₂CH₃: C2 bonded to H, OH, CH₃, and CH₂CH₃ (chiral).
2,2-dimethylpropane derivatives: No chiral centers possible.
Compounds giving chiral products: 2-bromopentane, 3-bromo-2-methylbutane, and 1-bromo-3-methylbutane. Total = 3.
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