Q.Compound (X) reacts with magnesium in the presence of dry ether to form a Grignard reagent, which is then treated with compound (Y) containing 3 carbon atoms to form an adduct which on hydrolysis gives 2-methylbutan-2-ol. Write the chemical reactions involved and name the compounds (X) and (Y).

Ethylmagnesium bromide (from bromoethane + Mg) adds across acetone's carbonyl; hydrolysis of the adduct gives 2-methylbutan-2-ol.

The target alcohol, 2-methylbutan-2-ol, is $CH_3-C(OH)(CH_3)-CH_2-CH_3$ — a tertiary alcohol with an ethyl group and two methyl groups attached to the carbinol (C–OH) carbon. A tertiary alcohol of this type is made by adding a Grignard reagent, $R-MgX$, across a ketone $R'COR''$: the carbanion-like R group of the Grignard attacks the electrophilic carbonyl carbon, and after hydrolysis the three groups R, R', R'' (plus OH) end up on that carbon.

Working backwards: the carbinol carbon of the product carries $-CH_3$, $-CH_3$, and $-C_2H_5$. The ketone (Y) supplies two of these groups still joined together as its own $R'-CO-R''$, and the Grignard reagent supplies the third group.

• Compound X (2 carbons): bromoethane, $CH_3CH_2Br$. Reacting with magnesium turnings in dry ether forms the Grignard reagent, ethylmagnesium bromide: $CH_3CH_2Br + Mg \xrightarrow{\text{dry ether}} CH_3CH_2MgBr$.
• Compound Y (3 carbons): propan-2-one (acetone), $CH_3-CO-CH_3$ — it already carries the two methyl groups needed on the final carbinol carbon. …