Question 4

Analyzing each option: **(A)** 2-chloroethanol + NaCN → 3-hydroxypropanenitrile (HO–CH₂CH₂–CN); hydrolysis with HO⁻/H₂O then H₃O⁺ gives 3-hydroxypropanoic acid (β-hydroxy acid), a monocarboxylic acid – not a dicarboxylic acid. **(B)** Glucose with Br₂/H₂O: mild oxidation of only the –CHO group to –COOH, giving gluconic acid (monocarboxylic acid). The terminal CH₂OH is not oxidized (that would require HNO₃ to give saccharic acid). **(C)** Bromocyclohexane with KOH/EtOH → cyclohexene (E2 elimination). Then hot acidic KMnO₄ cleaves the C=C bond oxidatively, opening the ring to give adipic acid, HOOC–(CH₂)₄–COOH – a dicarboxylic acid. **(D)** 3-hydroxy-2,4-dimethylpentan-3-one... actually the structure is 1-hydroxy-2,4-dimethylpentan-3-one (a β-keto primary alcohol). H₂CrO₄ oxidizes the primary alcohol to a carboxylic acid: 2,4-dimethyl-3-oxopentanoic acid – a keto acid, only one COOH group. Only option (C) yields a dicarboxylic acid (adipic acid) as the major product.

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