`{:(CH_(2)-CONH_(2)),("|"),(CH_(2)-CONH_(2)):} overset(P_(2)O_(5))underset(Delta)to(P) overset(H_(3)O^(o+))underset(Delta)to(Q)`
The product (Q) is

Options

A. `CH_(3)-CH_(2)-COOH`
B. `{:(CH_(2)-COOH),("|"),(CH_(2)-COOH):} `
C.
D. `CH_(3)-COOH`

Explanation

To solve the problem step by step, we will analyze the reaction of the compound with P₂O₅ and then with H₃O⁺ under heat.

### Step 1: Identify the starting compound
The starting compound is given as CH₂CONH₂, which is an amide.

### Step 2: Reaction with P₂O₅
When CH₂CONH₂ is treated with P₂O₅ in the presence of heat, P₂O₅ acts as a dehydrating agent. This means it will remove water molecules from the compound.

- **Dehydration Reaction**: The amide group (CONH₂) will lose two water molecules (H₂O) upon heating with P₂O₅, converting the amide into a nitrile (CN group).

The reaction can be summarized as:
\[ \text{CH}_2\text{CONH}_2 \xrightarrow{\Delta, \text{P}_2\text{O}_5} \text{CH}_2\text{CN} \]

### Step 3: Formation of Product P
The product formed after this dehydration step is a nitrile, specifically CH₂CN.

### Step 4: Hydrolysis of the Nitrile
Next, the nitrile (CH₂CN) undergoes hydrolysis in the presence of H₃O⁺ (acidic conditions) and heat.

- **Hydrolysis Reaction**: The nitrile group (C≡N) will be converted into a carboxylic acid (COOH) through hydrolysis.

The reaction can be summarized as:
\[ \text{CH}_2\text{CN} \xrightarrow{\Delta, \text{H}_3\text{O}^+} \text{CH}_2\text{COOH} \]

### Step 5: Final Product Q
After hydrolysis, the product formed is CH₂COOH, which is a carboxylic acid.

### Step 6: Identify the final product
However, if we consider the reaction conditions and the possibility of further dehydration, we can also consider that the carboxylic acid can lose a water molecule upon heating, forming an anhydride.

The final product Q can be represented as:
\[ \text{CH}_2\text{COO}\text{CH}_2 \]

### Conclusion
Thus, the product Q is:
\[ \text{CH}_2\text{COO}\text{CH}_2 \]