JEE Main 2026 — Jan 23, Shift 1 — Chemistry solution

JEE Main2026Jan 23, Shift 1Chemistry

Consider all the structural isomers with molecular formula C $_{5}$ H $_{11}$ Br 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 ______.

Your answer

Solution

The structural isomers of C $_{5}$ H $_{11}$ Br are 8 in number (including isomers about both chain and Br position). Aqueous KOH (S $_{N}$ substitution) replaces Br with —OH on the same carbon, with no rearrangement. So the products are pentanols carrying —OH at the position where Br originally sat.

An alcohol shows optical isomerism if its carbinol carbon (or another sp $^{3}$ C) is a stereocentre — i.e. is attached to four different groups.

Counting C $_{5}$ H $_{11}$ OH products whose carbinol carbon (the one carrying —OH from substitution) is a stereocentre:

• 2-pentanol CH $_{3}$ CH(OH)CH $_{2}$ CH $_{2}$ CH $_{3}$ — C2 has —H, —OH, —CH $_{3}$ , —CH $_{2}$ CH $_{2}$ CH $_{3}$ : 4 different groups ✓

• 3-methyl-1-butanol and 2-methyl-1-butanol: only one of these has a stereocentre. 2-Methyl-1-butanol CH $_{3}$ CH $_{2}$ CH(CH $_{3}$ )CH $_{2}$ OH has a stereocentre at C2 ✓

• 3-methyl-2-butanol (CH $_{3}$ ) $_{2}$ CHCH(OH)CH $_{3}$ has a stereocentre at C2 (substituents: —H, —OH, —CH $_{3}$ , —CH(CH $_{3}$ ) $_{2}$ ) ✓

Other pentanols (1-pentanol, 3-pentanol, 3-methyl-1-butanol, 2-methyl-2-butanol, neopentyl alcohol) lack a stereocentre.

$∴$ Number of products showing optical isomerism = 3

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