Reaction of propanone with methylmagnesium bromide followed by hydrolysis gives :-

Options

A. primary alcohol
B. secondary alcohol
C. tertiary alcohol
D. carboxylic acid

Answer

To solve the question regarding the reaction of propanone with methylmagnesium bromide followed by hydrolysis, we can break it down into clear steps:

### Step 1: Identify the Reactants
- The reactants in this reaction are propanone (a ketone) and methylmagnesium bromide (a Grignard reagent).

### Step 2: Understand the Structure of Propanone
- Propanone, also known as acetone, has the following structure:
\[ CH_3COCH_3 \]
It consists of a carbonyl group (C=O) flanked by two methyl groups (CH₃).

### Step 3: Reaction with Methylmagnesium Bromide
- Methylmagnesium bromide (CH₃MgBr) acts as a nucleophile. In the presence of the carbonyl group in propanone, the nucleophile attacks the electrophilic carbon of the carbonyl group.
- This results in a nucleophilic addition reaction, forming an intermediate alkoxide:
\[ CH_3COCH_3 + CH_3MgBr \rightarrow CH_3C(OMgBr)(CH_3) \]

### Step 4: Hydrolysis of the Alkoxide
- The next step is hydrolysis, where the alkoxide intermediate reacts with water (or an acid) to form the corresponding alcohol.
- During hydrolysis, the alkoxide is protonated, leading to the formation of a tertiary alcohol:
\[ CH_3C(OMgBr)(CH_3) + H_2O \rightarrow CH_3C(OH)(CH_3) + Mg(OH)Br \]
- The product formed is:
\[ (CH_3)_3COH \]
This is tert-butanol, which is a tertiary alcohol.

### Step 5: Conclusion
- Therefore, the final product of the reaction of propanone with methylmagnesium bromide followed by hydrolysis is a tertiary alcohol.

### Final Answer
The correct answer is **tertiary alcohol** (Option 3).

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