Hydroxylation of trans-but-2-ene results in the formation of

The correct answer is Option (4) → Racemic mixture of Butan-2, 3-diol

Hydroxylation adds an OH group to each carbon of a C=C bond. With cold dilute alkaline \( \ce{KMnO4} \) (Baeyer’s reagent) or \( \ce{OsO4} \), both OH groups add from the same face of the double bond (syn addition):

    \( \displaystyle \ce{CH3-CH=CH-CH3 + H2O + [O]} \xrightarrow{\text{cold dil. }\ce{KMnO4}} \ce{CH3-CH(OH)-CH(OH)-CH3} \)

Butane-2,3-diol has two stereocentres (C-2 and C-3). The stereochemistry of the product depends on the geometry of the alkene:

• Syn addition to cis-but-2-ene gives the meso diol, which has a plane of symmetry and is optically inactive.

• Syn addition to trans-but-2-ene can take place from the top face or the bottom face with equal probability. The two faces give the (2R,3R) and (2S,3S) diols, which are enantiomers, in equal amounts.

So trans-but-2-ene gives a racemic mixture of butane-2,3-diol.

Why the other options are incorrect

• Option (1) Butanol — Incorrect. A mono-alcohol is formed by hydration (adding H and OH), not by hydroxylation, which adds two OH groups.

• Option (2) Butan-1, 4-diol — Incorrect. The OH groups go onto the two carbons of the double bond (C-2 and C-3), not onto the end carbons.

• Option (3) Butan-2, 3-diol — Incorrect. This names the right compound but not its stereochemistry; trans-but-2-ene gives it as a racemic mixture of the two enantiomers, which Option (4) states fully.

Note: The question is from beyond the NCERT textbook.

NCERT (2026-27 edition):
Stereochemical Aspects of Nucleophilic Substitution Reactions – …