Which of the following is the correct final product when $\mathrm{PhMgBr}$ reacts with $\mathrm{acetone}$ followed by acidic workup in a Grignard reaction?

• A. $\mathrm{Ph}-\mathrm{CH}{2}_{}-\mathrm{Me}$
• B. $\mathrm{Ph}-C\left(\mathrm{OH}\right)-\mathrm{Me}-\mathrm{Me}$
• C. $\mathrm{Ph}-O-\mathrm{Me}$
• D. $\mathrm{Ph}-\mathrm{CO}-\mathrm{Me}$

Answer: (B)

Verified step by step guidance

1. Identify the reactants: PhMgBr is a Grignard reagent (phenylmagnesium bromide), and acetone is a ketone with the structure (CH3)2C=O.
2. Recall the general mechanism of a Grignard reaction with a ketone: the nucleophilic carbon in the Grignard reagent attacks the electrophilic carbonyl carbon of the ketone, breaking the C=O double bond and forming a tetrahedral alkoxide intermediate.
3. Write the nucleophilic attack step: the phenyl group (Ph-) from PhMgBr attacks the carbonyl carbon of acetone, resulting in an alkoxide intermediate with the structure Ph-C(O^-)-Me-Me.
4. Consider the acidic workup step: protonation of the alkoxide intermediate by acid (H3O+) converts the alkoxide into an alcohol, yielding a tertiary alcohol with the structure Ph-C(OH)-Me-Me.
5. Conclude that the final product is a tertiary alcohol where the phenyl group is attached to the former carbonyl carbon, now bearing an OH group, and two methyl groups remain attached to that carbon.