`C_(2)H_(5)MgI` reacts with HCHO to form last product :

Options

A. `CH_(3)CHO`
B. `C_(3)H_(7)OH`
C. `CH_(3)COCH_(3)`
D. `CH_(3)COOCH_(3)`

Answer

To solve the problem of what product is formed when C₂H₅MgI (ethyl magnesium iodide) reacts with HCHO (formaldehyde), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants**:
- The reactants are ethyl magnesium iodide (C₂H₅MgI) and formaldehyde (HCHO).

2. **Understand the Reaction**:
- Ethyl magnesium iodide is a Grignard reagent. Grignard reagents react with carbonyl compounds (like aldehydes) to form alcohols.

3. **Mechanism of Reaction**:
- The carbon atom in formaldehyde (HCHO) is electrophilic (positively charged), and the ethyl group (C₂H₅) from the Grignard reagent acts as a nucleophile (negatively charged).
- The nucleophile attacks the electrophilic carbon in formaldehyde.

4. **Formation of the Intermediate**:
- When C₂H₅MgI reacts with HCHO, the ethyl group (C₂H₅) attacks the carbon of the carbonyl group (C=O) in formaldehyde.
- This results in the formation of an intermediate alkoxide:
\[ \text{C₂H₅-CH₂-O}^-\text{MgI}^+ \]
- The structure can be represented as CH₂C₂H₅ with an alkoxide (O⁻) bonded to magnesium iodide (MgI).

5. **Hydrolysis of the Intermediate**:
- The next step involves hydrolysis, where the intermediate alkoxide is treated with water (H₂O).
- During hydrolysis, the alkoxide ion (O⁻) gets protonated by water, resulting in the formation of an alcohol.
- The reaction can be summarized as:
\[ \text{C₂H₅-CH₂-O}^-\text{MgI}^+ + \text{H₂O} \rightarrow \text{C₂H₅-CH₂-OH} + \text{MgI(OH)} \]

6. **Final Product**:
- The final product formed after hydrolysis is propyl alcohol (C₃H₇OH), also known as propanol.
- The by-product is magnesium iodide hydroxide (MgI(OH)).

### Conclusion:
The last product formed when C₂H₅MgI reacts with HCHO is propyl alcohol (C₃H₇OH).