The reaction of`CH_(3)CH_(2)MgBr` with formaldehyde after acidification gives

Options

A. an aldehyde
B. a primary alcohol
C. a ketone
D. secondary alcohol

Explanation

To solve the question regarding the reaction of `CH₃CH₂MgBr` (ethylmagnesium bromide) with formaldehyde, we can follow these steps:

### Step 1: Identify the Reactants
The reactants in this reaction are:
- Ethylmagnesium bromide (`CH₃CH₂MgBr`)
- Formaldehyde (`HCHO`)

### Step 2: Understand the Reaction Mechanism
Ethylmagnesium bromide is a Grignard reagent, which acts as a nucleophile. It will react with the carbonyl carbon of formaldehyde.

### Step 3: Nucleophilic Attack
The nucleophilic carbon of the ethyl group (`CH₃CH₂-`) will attack the electrophilic carbon of the formaldehyde:
- The reaction can be represented as:
\[ CH₃CH₂MgBr + HCHO \rightarrow CH₃CH₂-CH₂OH \]
- This forms a tetrahedral intermediate where the oxygen of the carbonyl becomes negatively charged.

### Step 4: Protonation
After the nucleophilic attack, the intermediate is treated with an acid (usually water or dilute acid) to protonate the oxygen:
- The final product after acidification is:
\[ CH₃CH₂-CH₂OH \]
- This compound is 1-propanol, which is a primary alcohol.

### Step 5: Conclusion
Thus, the reaction of `CH₃CH₂MgBr` with formaldehyde after acidification gives a primary alcohol, specifically 1-propanol.

### Final Answer
The correct answer is **B: Primary Alcohol**.

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