Reaction between dry ice and methyl magnesium bromide gives an addition compound which acidification yields:

Options

A. ethanoic acid
B. propanoic acid
C. ethanal
D. ethanol

Explanation

To solve the problem of what product is formed when dry ice reacts with methyl magnesium bromide and is then acidified, we can follow these steps:

### Step 1: Identify the Reactants
- Dry ice is solid carbon dioxide (CO₂).
- Methyl magnesium bromide is a Grignard reagent, represented as CH₃MgBr.

### Step 2: Understand the Reaction Mechanism
- The Grignard reagent (CH₃MgBr) reacts with carbon dioxide (CO₂) to form an addition compound.
- In this reaction, the nucleophilic methyl group (CH₃⁻) from the Grignard reagent attacks the electrophilic carbon of carbon dioxide.

### Step 3: Formation of the Addition Compound
- When CH₃⁻ attacks CO₂, it forms a carbanion intermediate:
\[ \text{CO}_2 + \text{CH}_3^- \rightarrow \text{(CH}_3\text{COO)}^- \]
- This intermediate is a carboxylate ion (specifically, the acetate ion).

### Step 4: Acidification of the Addition Compound
- Upon acidification (usually with a dilute acid like HCl), the carboxylate ion is protonated:
\[ \text{(CH}_3\text{COO)}^- + \text{H}^+ \rightarrow \text{CH}_3\text{COOH} \]
- This results in the formation of acetic acid (ethanoic acid).

### Final Product
- The final product of the reaction between dry ice and methyl magnesium bromide, followed by acidification, is **acetic acid (ethanoic acid)**.

### Summary of the Reaction
1. Dry ice (CO₂) reacts with methyl magnesium bromide (CH₃MgBr).
2. An addition compound is formed, leading to the carboxylate ion.
3. Acidification of this compound yields acetic acid (CH₃COOH).