MeMgBr (2 equiv:) 2) H3O + OMe OH Ph Ph
The Correct Answer and Explanation is:
Here’s the breakdown of the reaction:
- Grignard Reagent Formation:
- MeMgBr is a Grignard reagent, typically used to form carbon-carbon bonds. Grignard reagents are prepared by reacting an alkyl or aryl halide (in this case, methyl bromide) with magnesium in an anhydrous solvent like ether.
- Reaction with Ester:
- The “OMe” could represent a methoxy group attached to an ester (i.e., R-COOCH₃ or something similar), and “Ph” represents a phenyl group. The Grignard reagent (MeMgBr) would attack the carbonyl group of the ester in a nucleophilic addition.
- The methyl group (Me) from the Grignard reagent will add to the electrophilic carbonyl carbon, breaking the ester bond, forming a tetrahedral intermediate.
- Hydrolysis:
- After the nucleophilic addition, the intermediate is then hydrolyzed with H₃O⁺ (acidic aqueous workup). This protonates the alkoxide to form a hydroxyl group, resulting in an alcohol and the conversion of the ester into a secondary alcohol (in this case, with one of the groups being the methyl group from MeMgBr).
Mechanism Summary:
- The Grignard reagent (MeMgBr) attacks the carbonyl carbon of the ester, breaking the ester bond.
- The intermediate undergoes hydrolysis (with H₃O⁺) to yield a secondary alcohol.
- The OMe group remains attached to the alcohol.
Example:
If the ester involved is methyl benzoate (PhCOOCH₃), the reaction could proceed as:PhCOOCH₃+MeMgBr→H₃O⁺PhCH(OMe)CH₃\text{PhCOOCH₃} + \text{MeMgBr} \xrightarrow{\text{H₃O⁺}} \text{PhCH(OMe)CH₃}PhCOOCH₃+MeMgBrH₃O⁺PhCH(OMe)CH₃
This results in 2-methylphenylmethanol after the addition of the methyl group to the phenyl group via Grignard addition and hydrolysis.
