JEE Chemistry 2025 Question with Solution

Given: Compound (X) has molecular formula $\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}$ and is not readily oxidised. On reduction it gives compound (Y) with formula $\mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}$. Compound (Y) reacts with HBr to give bromide (Z), which forms a Grignard reagent. This Grignard reagent reacts with (X) and after hydrolysis gives 2,3-dimethylbutan-2-ol.

Find: Compounds (X), (Y) and (Z).

A compound with formula $\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}$ that is not readily oxidised is acetone (propanone), so

$$\mathrm{X} = \mathrm{CH}_{3}\mathrm{COCH}_{3}$$

Ketones on reduction give secondary alcohols, therefore

$$\mathrm{CH}_{3}\mathrm{COCH}_{3} \xrightarrow[\text{reduction}]{} \mathrm{CH}_{3}\mathrm{CH}(\mathrm{OH})\mathrm{CH}_{3}$$

So,

$$\mathrm{Y} = \mathrm{CH}_{3}\mathrm{CH}(\mathrm{OH})\mathrm{CH}_{3}$$

Now (Y) reacts with HBr to form the corresponding alkyl bromide:

$$\mathrm{CH}_{3}\mathrm{CH}(\mathrm{OH})\mathrm{CH}_{3} \xrightarrow{\mathrm{HBr}} \mathrm{CH}_{3}\mathrm{CH}(\mathrm{Br})\mathrm{CH}_{3}$$

Hence,

$$\mathrm{Z} = \mathrm{CH}_{3}\mathrm{CH}(\mathrm{Br})\mathrm{CH}_{3}$$

This bromide forms the Grignard reagent, which reacts with acetone and on hydrolysis gives the tertiary alcohol 2,3-dimethylbutan-2-ol, confirming the choice.

Therefore, the correct option is B.

1. Compound (X) has formula $\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}$ and is not readily oxidised, so it is taken as a ketone rather than an aldehyde.
2. Therefore, (X) is acetone:

$$\mathrm{CH}_{3}\mathrm{COCH}_{3}$$

3. Reduction of a ketone gives a secondary alcohol, so (Y) is isopropanol:

$$\mathrm{CH}_{3}\mathrm{CH}(\mathrm{OH})\mathrm{CH}_{3}$$

4. Reaction of isopropanol with HBr gives 2-bromopropane, so (Z) is:

$$\mathrm{CH}_{3}\mathrm{CH}(\mathrm{Br})\mathrm{CH}_{3}$$

5. The Grignard reagent formed from (Z) adds to acetone, and after hydrolysis produces 2,3-dimethylbutan-2-ol. This matches the required product.

Hence,

$$(\mathrm{X}, \mathrm{Y}, \mathrm{Z}) = (\mathrm{CH}_{3}\mathrm{COCH}_{3},\; \mathrm{CH}_{3}\mathrm{CH}(\mathrm{OH})\mathrm{CH}_{3},\; \mathrm{CH}_{3}\mathrm{CH}(\mathrm{Br})\mathrm{CH}_{3})$$

So the correct option is B.