KCET 2026 Chemistry Solved Paper

Concept:In an $\mathrm{S}_{\mathrm{N}}1$ reaction, a planar carbocation intermediate is formed, and nucleophilic attack can occur from both faces.A racemic mixture is produced only when the product is chiral and both enantiomers are formed equally.Explanation:For a racemic mixture to form, the alkyl halide must generate a chiral product.2-bromobutane has a chiral carbon at the second position:$$\mathrm{CH}_3-\mathrm{CHBr}-\mathrm{CH}_2-\mathrm{CH}_3$$This carbon is attached to four different groups: $\mathrm{CH}_3$, $\mathrm{H}$, $\mathrm{Br}$, and $\mathrm{C}_2\mathrm{H}_5$.During $\mathrm{S}_{\mathrm{N}}1$ hydrolysis, the bromide ion leaves, forming a planar carbocation:$$\mathrm{CH}_3-\stackrel{+}{\mathrm{C}}\mathrm{H}-\mathrm{CH}_2-\mathrm{CH}_3$$Water can attack this planar carbocation from either side with equal probability.This attack produces both $R$ and $S$ enantiomers of butan-2-ol:$$\mathrm{CH}_3-\mathrm{CH}(\mathrm{OH})-\mathrm{CH}_2-\mathrm{CH}_3$$Thus, the product is a racemic mixture.Tertiary butyl bromide also undergoes $\mathrm{S}_{\mathrm{N}}1$, but tert-butyl alcohol is achiral and gives no racemic mixture.Isopropyl bromide and methyl bromide also yield achiral products.Therefore, the correct alkyl halide is 2-bromobutane.Answer:Option B: 2-bromobutane