Question

Compound $X$ gives alcohol $P$ as the major product for the reaction: \;i) CH$_3$MgBr ii) H$_3$O$^+$. Suitable option(s) for $X$ is/are. ($P$ is the benzylic tertiary alcohol bearing two CH$_3$ groups) 

• A. A
• B. B
• C. C
• D. D

Hint:

To predict Grignard outcomes, sketch the target alcohol’s central carbon: count how many alkyl/aryl groups it bears. Work backwards to the carbonyl type—tertiary alcohols generally arise from adding RMgX to a ketone (or two additions to acyl derivatives).

Answer 1

The Correct Option is B, C

Requirement from the product. The product \(P\) is a tertiary benzylic alcohol of the form Ph–C(OH)(CH\(_3\))\(_2\). Therefore, after adding CH\(_3^-\) (from CH\(_3\)MgBr) and protonation, the carbonyl precursor must deliver a central carbon attached to Ph and two CH\(_3\) groups.

Functional groups that work.
• Aryl methyl ketone (acetophenone type): Ph–CO–CH\(_3\) \(\xrightarrow{\;\text{CH}_3\text{MgBr}\;}\) Ph–C(OMgBr)(CH\(_3\))\(_2\) \(\xrightarrow{\;\text{H}_3\text{O}^+\;}\) Ph–C(OH)(CH\(_3\))\(_2\) \(\Rightarrow\) matches \(P\) \(\Rightarrow\) (B) correct.
• Acyl derivatives that convert to that ketone then add again (e.g., acid chloride of benzoic acid): first addition gives acetophenone, a second methyl addition occurs under standard conditions, affording the same tertiary alcohol on work-up \(\Rightarrow\) (C) correct.

Why the others are not suitable.
Options that are aldehydes (Ph–CHO) give only a secondary alcohol (Ph–CH(OH)–CH\(_3\)), not tertiary. Esters/ether-like variants that would place an extra –OR group on the benzylic carbon either require different stoichiometry or give a different carbon skeleton than \(P\). Hence (A) and (D) do not furnish \(P\) as the major product under the given conditions.