JEE Main 202623 JanuaryMorning Shift
- Exam
- JEE Main
- Year
- 2026
- Date
- 23 January
- Shift
- Morning Shift
- Subject
- Chemistry
- Chapter
- Haloalkanes and Haloarenes
JEE Main 2026 · 23 January · Morning Shift · Chemistry · Haloalkanes and Haloarenes
Consider all the structural isomers with molecular formula C_ 5 H_...
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 .
Correct answer
3
Step-by-step solution
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.