`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 question regarding the reaction of C₂H₅MgI (a Grignard reagent) with HCHO (formaldehyde), we will follow these steps:

### Step 1: Identify the Reactants
The reactants in this reaction are:
- C₂H₅MgI (Ethylmagnesium iodide, a Grignard reagent)
- HCHO (Formaldehyde)

### Step 2: Understand the Nature of the Grignard Reagent
Grignard reagents are nucleophiles due to the presence of the carbon atom bonded to magnesium. The carbon atom in C₂H₅MgI (C₂H₅⁻) is negatively charged and will act as a nucleophile.

### Step 3: Identify the Electrophilic Site in Formaldehyde
In formaldehyde (HCHO), the carbon atom is electrophilic because it is bonded to an oxygen atom that has a partial negative charge. This makes the carbon in HCHO susceptible to nucleophilic attack.

### Step 4: Nucleophilic Attack
The ethyl group (C₂H₅) from the Grignard reagent will attack the electrophilic carbon of formaldehyde. This results in the formation of an alkoxide intermediate.

### Step 5: Formation of the Alkoxide Intermediate
The nucleophilic attack leads to the formation of an intermediate:
- The structure of the intermediate will be:
- C₂H₅-CH₂-O⁻ (alkoxide ion)

### Step 6: Protonation of the Alkoxide
The alkoxide ion formed in the previous step will then be protonated by water or any acid (usually present in the reaction environment), leading to the formation of the final product.

### Step 7: Final Product
The final product after protonation will be:
- C₂H₅-CH₂-OH (which is ethanol)

### Summary of the Reaction
The overall reaction can be summarized as:
\[ \text{C}_2\text{H}_5\text{MgI} + \text{HCHO} \rightarrow \text{C}_2\text{H}_5\text{CH}_2\text{OH} + \text{MgI(OH)} \]

### Final Answer
The last product formed is ethanol (C₂H₅CH₂OH).

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