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Ethyl ester overset(MeMgBr) underset(exc...

Ethyl ester `overset(MeMgBr) underset(excess)rarr P`. The product `P` will be :

A

B

C

D

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To solve the problem of what product \( P \) is formed when ethyl ester reacts with excess methyl magnesium bromide (\( MeMgBr \)), we can follow these steps: ### Step 1: Identify the Reactants The reactant is ethyl ester, which can be represented as \( RCOOCH_2CH_3 \). Here, \( R \) is an alkyl group that is not specified. ### Step 2: Understand Grignard Reagent Reaction Methyl magnesium bromide (\( MeMgBr \)) is a Grignard reagent. Grignard reagents are nucleophiles and react with electrophiles. In this case, the carbonyl carbon of the ester is electrophilic. ### Step 3: First Nucleophilic Attack When \( MeMgBr \) reacts with the ethyl ester, the nucleophilic methyl group (\( Me^- \)) attacks the carbonyl carbon of the ester. This leads to the formation of a tetrahedral intermediate: \[ RCOOCH_2CH_3 + MeMgBr \rightarrow R-C(OMgBr)(CH_3)(CH_2CH_3) \] ### Step 4: Elimination of Leaving Group The tetrahedral intermediate collapses, resulting in the elimination of the leaving group (ethyl group, \( C_2H_5 \)). This forms a ketone: \[ R-C(=O)(CH_3) + C_2H_5OMgBr \] ### Step 5: Second Nucleophilic Attack Since we have excess \( MeMgBr \), another molecule of \( MeMgBr \) will react with the newly formed ketone. The methyl group will again attack the carbonyl carbon: \[ R-C(=O)(CH_3) + MeMgBr \rightarrow R-C(OMgBr)(CH_3)(CH_3) \] ### Step 6: Final Product Formation After the second nucleophilic attack, we have another tetrahedral intermediate. Upon hydrolysis (using acidic workup), we will convert the magnesium alkoxide to an alcohol: \[ R-C(OH)(CH_3)(CH_3) \] ### Step 7: Identify the Product The final product \( P \) is a tertiary alcohol with the structure \( R-C(OH)(CH_3)(CH_3) \). If \( R \) is also a methyl group, the product is 2-methyl-2-propanol (tert-butyl alcohol). ### Conclusion Thus, the product \( P \) formed from the reaction of ethyl ester with excess methyl magnesium bromide is a tertiary alcohol.

To solve the problem of what product \( P \) is formed when ethyl ester reacts with excess methyl magnesium bromide (\( MeMgBr \)), we can follow these steps: ### Step 1: Identify the Reactants The reactant is ethyl ester, which can be represented as \( RCOOCH_2CH_3 \). Here, \( R \) is an alkyl group that is not specified. ### Step 2: Understand Grignard Reagent Reaction Methyl magnesium bromide (\( MeMgBr \)) is a Grignard reagent. Grignard reagents are nucleophiles and react with electrophiles. In this case, the carbonyl carbon of the ester is electrophilic. ...
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