Q40 / 180 · COMEDK 2025 (10 May Shift 2) · Chemistry

Identify the incorrect statement.

  • A
    The reaction of $\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2-\mathrm{CHCl}-\mathrm{C}_6 \mathrm{H}_5$ with alc. KOH on heating yields two products.
  • Correct answer
    B
    When (-)-2-bromooctane is reacted with aqueous KOH , the substitution follows $\mathrm{S}_{\mathrm{N}} 2$ mechanism and product formed is $(-)$-Octan-2-ol.
  • C
    Benzylic halides show higher reactivity towards $S_N 1$ reaction.
  • D
    Optically active (-)-2-Methylbutan-1-ol on reaction with HCl yields a product $(+)$-1-Chloro-2-methylbutane.

Solution

(1) Correct: Dehydrohalogenation of $C_6H_5CH_2-CHCl-C_6H_5$ with alc. KOH gives stilbene ($C_6H_5-CH=CH-C_6H_5$), which exists as two geometric isomers (cis and trans) — hence two products.

(2) Incorrect: $S_N2$ involves Walden inversion. Reaction of $(-)$-2-bromooctane with aqueous KOH via $S_N2$ gives $(+)$-octan-2-ol (inverted configuration), not $(-)$-octan-2-ol.

(3) Correct: Benzylic halides readily undergo $S_N1$ because the benzylic carbocation intermediate is stabilised by resonance with the benzene ring.

(4) Correct: Reaction of $(-)$-2-methylbutan-1-ol with HCl replaces $-OH$ at the primary carbon $C_1$. No bond at the chiral centre $C_2$ is broken, so configuration is retained. However the sign of optical rotation changes, giving $(+)$-1-chloro-2-methylbutane.

The incorrect statement is (2).

Hence, the correct answer is option: 'B'.

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