A bromoalkane ‘X’ reacts with magnesium in dry ether to form compound ‘Y’. The reaction of ‘Y’ with methanal followed by hydrolysis yield an alcohol having molecular formula, C₄H₁₀O. The compound ‘X’ is

Given:

• X is a bromoalkane.
• X + Mg/ether → Y (Grignard reagent).
• Y reacts with methanal, then hydrolysis, gives C4H10O (which is butan-1-ol or butan-2-ol).

For the alcohol C4H10O:

• If the alcohol is butan-1-ol, you'll need a propyl Grignard (from 1-bromopropane, CH3CH2CH2Br) and methanal.
• 1-bromopropane + Mg/ether gives propyl magnesium bromide (Y).
• Propyl magnesium bromide + methanal → butan-1-ol (after hydrolysis).

So, the compound X is 1-bromopropane (CH3CH2CH2Br).