Consider all the structural isomers with molecular formula C5H11Br are separately treated with KOH(aq) to give respective substitution products, without any rearrangement. The number of products which can exhibit optical isomerism from these is .
Answer
3
Solution
C5H11Br has 8 structural isomers; aq. KOH replaces Br by OH (no rearrangement):
(i) CH3CH2CH2CH2CH2Br → pentan-1-ol: no chiral C
(ii) CH3CH2CH2CHBrCH3 → pentan-2-ol: C-2 chiral
(iii) CH3CH2CHBrCH2CH3 → pentan-3-ol: no chiral C
(iv) BrCH2CH(CH3)CH2CH3 → 2-methylbutan-1-ol: C-2 chiral
(v) (CH3)2CBrCH2CH3 → 2-methylbutan-2-ol: no chiral C
(vi) (CH3)2CHCHBrCH3 → 3-methylbutan-2-ol: C-2 chiral
(vii) (CH3)2CHCH2CH2Br → 3-methylbutan-1-ol: no chiral C
(viii) (CH3)3CCH2Br → 2,2-dimethylpropan-1-ol: no chiral C
products that can show optical isomerism = 3
Answer: 3