Introduction

In this CSIR‑NET style Grignard reagent problem, compound X reacts with CH₃MgBr followed by acidic workup to give tertiary alcohol P, 2‑phenyl‑2‑propanol, as the major product. Understanding how esters, ketones and aldehydes behave with Grignard reagents makes it easy to eliminate wrong options and pick the correct structures for X.

Target Alcohol P and Required Precursor

The product P shown is a tertiary alcohol whose carbon bearing OH is attached to:

• One phenyl group (C₆H₅–)
• Two methyl groups (–CH₃, –CH₃)

This structure is 2‑phenyl‑2‑propanol (also called α,α‑dimethyl benzyl alcohol).

To obtain this, the carbonyl compound must have:

• A carbonyl carbon directly bonded to the phenyl ring, and
• One methyl group already present on that carbon, because the other methyl comes from CH₃MgBr.

Therefore, the effective carbonyl intermediate just before the last Grignard addition is acetophenone (C₆H₅–CO–CH₃).

Detailed Mechanism Logic

• Reaction type – CH₃MgBr is a Grignard reagent that adds as a nucleophilic methyl group (CH₃⁻ equivalent) to a carbonyl carbon.
• Aldehydes and ketones – one equivalent of Grignard reagent adds to form an alkoxide which after protonation gives a secondary (from aldehyde) or tertiary (from ketone) alcohol.
• Esters – two equivalents of Grignard reagent attack: first to give a ketone after loss of the alkoxy group, and the second to convert that ketone to a tertiary alcohol.

So any X that is either acetophenone itself, or an ester whose acyl part is that of acetophenone (that is, that can generate acetophenone after the first addition–elimination), will give 2‑phenyl‑2‑propanol with CH₃MgBr.

Option‑wise Explanation

Option A – Methyl Benzoate (Ph–COOCH₃)

Structure: aromatic ester of benzoic acid, Ph–COOCH₃.

Step 1: CH₃MgBr adds to the ester carbonyl forming a tetrahedral intermediate; methoxide (–OCH₃) leaves, giving acetophenone (Ph–CO–CH₃) as the first product.

Step 2: the acetophenone so formed is more reactive and undergoes a second nucleophilic attack by another CH₃MgBr, followed by acidic workup, giving 2‑phenyl‑2‑propanol.

Therefore, option A is correct as X.

Option B – Anisole‑Type Ether (Ph–OCH₃)

Structure: aryl ether (anisole), not a carbonyl compound, so there is no C=O group for nucleophilic addition.

Grignard reagents generally do not add to aromatic ethers; instead they can slowly cleave certain ethers only under harsh conditions and do not give the required tertiary alcohol skeleton.

With CH₃MgBr, anisole cannot generate acetophenone or the tertiary alcohol P.

Hence, option B is incorrect.

Option C – Acetophenone (Ph–CO–CH₃)

Structure: aromatic ketone, Ph–CO–CH₃.

One equivalent of CH₃MgBr attacks the carbonyl carbon, giving a tertiary alkoxide (Ph–C(OMgBr)(CH₃)₂) which on protonation affords 2‑phenyl‑2‑propanol directly.

The substituents on the alcohol carbon become: phenyl, original methyl, and new methyl from the Grignard reagent, matching P exactly.

Therefore, option C is correct for X.

Option D – Benzaldehyde (Ph–CHO)

Structure: aromatic aldehyde, Ph–CHO.

Reaction with one equivalent of CH₃MgBr gives a secondary alcohol, Ph–CH(OH)–CH₃, which is 1‑phenylethanol, not a tertiary alcohol.

The alcohol carbon here has only one methyl and one hydrogen, so it cannot match the two‑methyl substitution pattern of P.

Thus, option D is incorrect.

Final Answer and Exam‑Style Takeaway

The major product 2-phenyl‑2‑propanol (P) is obtained when methyl benzoate (option A) or acetophenone (option C) reacts with excess CH₃MgBr followed by acidic hydrolysis, so the correct options for X are A and C.

In CSIR‑NET and similar exams, always reconstruct the carbon framework of the required alcohol and work backward to identify which carbonyl compound or ester can supply the needed groups with one or two Grignard additions.