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COMEDK 2025 Afternoon Solved Paper

Question : 90 of 180

Marks: +1, -0

Identify the incorrect statement.

When (一)-2-bromooctane is reacted with aqueous KOH , the substitution follows $\mathrm{S}_{\mathrm{N}} 2$ mechanism and product formed is (一) -Octan-2-ol.
Benzylic halides show higher reactivity towards $S_N 1$ reaction.
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.
Optically active (-)-2-Methylbutan-1-ol on reaction with HCl yields a product $(+)$ -1-Chloro-2-methylbutane.

Solution:

Concept:

In an

\mathrm{S_N2}

reaction, the nucleophile attacks from the opposite side of the leaving group, causing inversion of configuration at the stereogenic centre.

Explanation:

The question asks for the incorrect statement among the given options.

Option A claims that (–)-2‑bromooctane reacts with aqueous KOH via an

\mathrm{S_N2}

mechanism to give (–)-octan‑2‑ol.

However, in an

\mathrm{S_N2}

reaction, the nucleophile ()-Octan‑2‑ol would have the opposite absolute configuration.

Starting from (–)-2‑bromooctane (specific rotation negative), inversion produces (+)-octan‑2‑ol, not (–)-octan‑2‑ol.

Thus, the product stated in Option A is incorrect.

All other options (B, C, D) are correct statements.

Answer:

Option A is the incorrect statement.

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