In the given reaction, 'A' can be:

Step 1: Analyze the product

Product structure shows triphenylmethanol: $\ce{(C6H5)3C-OH}$

This has:

• Three phenyl groups
• One OH group
• Central carbon

Step 2: Understand Grignard reaction with esters

Grignard + Ester → Tertiary alcohol (2 equiv of Grignard)

General reaction: $\ce{2 R-MgX + R'-COOR'' -> R2R'C-OH}$

For methyl benzoate ($\ce{C6H5-COOCH3}$): $\ce{2 R-MgX + C6H5-COOCH3 -> R2(C6H5)C-OH}$

Step 3: Determine R group

Product: $\ce{(C6H5)3C-OH}$

This means:

• R = $\ce{C6H5}$ (phenyl)
• R' = $\ce{C6H5}$ (from ester)

So: $\ce{2 C6H5-MgX + C6H5-COOCH3 -> (C6H5)3C-OH}$

Step 4: Identify compound A

Grignard reagent: $\ce{C6H5-MgX}$

Starting halide: $\ce{C6H5-X}$ = Bromobenzene ($\ce{C6H5Br}$)

Answer: (b) Bromobenzene

Key Reaction Mechanism:

1. Grignard formation: $\ce{C6H5Br + Mg ->[THF] C6H5MgBr}$

2. First addition: $\ce{C6H5MgBr + C6H5-COOCH3 -> C6H5-C(=O)-C6H5 + CH3OMgBr}$ (Ketone intermediate)

3. Second addition: $\ce{C6H5MgBr + C6H5-C(=O)-C6H5 -> (C6H5)2C(OMgBr)-C6H5}$

4. Hydrolysis: $\ce{(C6H5)3C-OMgBr + H3O+ -> (C6H5)3C-OH}$

Key Points:

• Esters react with 2 equiv of Grignard → tertiary alcohol
• Aldehydes/ketones react with 1 equiv → secondary/tertiary alcohol
• Aryl Grignards can be formed from aryl bromides