`CH_(3)-CO OH overset(Br_(2)//P)to(A) overset(NaCN)to(B) overset(HOH//H^(+))to(C)`
The product (C) will be

Options

A.
B.
C.
D.

Explanation

To solve the problem step by step, we will analyze the given reactions involving acetic acid (CH₃COOH) and identify the products formed at each stage.

### Step 1: Reaction of Acetic Acid with Bromine and Phosphorus
The first step involves the reaction of acetic acid (CH₃COOH) with bromine (Br₂) in the presence of phosphorus (P). This is known as the Hell-Volhard-Zelinsky (HVZ) reaction, where an alpha hydrogen atom is replaced by a bromine atom.

- **Reaction**:
\[ \text{CH}_3\text{COOH} + \text{Br}_2/\text{P} \rightarrow \text{BrCH}_2\text{COOH} \]

- **Product (A)**: The product formed is **alpha-bromoacetic acid (BrCH₂COOH)**.

### Step 2: Reaction of Alpha-Bromoacetic Acid with Sodium Cyanide
In the second step, we treat the alpha-bromoacetic acid with sodium cyanide (NaCN). This reaction involves nucleophilic substitution, where the bromine atom is replaced by a cyanide group (CN).

- **Reaction**:
\[ \text{BrCH}_2\text{COOH} + \text{NaCN} \rightarrow \text{CH}_2\text{(CN)COOH} + \text{NaBr} \]

- **Product (B)**: The product formed is **sinoacetic acid (CH₂(CN)COOH)**.

### Step 3: Hydrolysis of Sinoacetic Acid
In the final step, we hydrolyze the sinoacetic acid using water (H₂O) in the presence of an acid (H⁺). This hydrolysis converts the nitrile group (CN) into a carboxylic acid (COOH).

- **Reaction**:
\[ \text{CH}_2\text{(CN)COOH} + \text{H}_2\text{O}/\text{H}^+ \rightarrow \text{CH}_2\text{(COOH)}_2 \]

- **Product (C)**: The product formed is **malonic acid (CH₂(COOH)₂)**.

### Final Answer
The product (C) will be **malonic acid (CH₂(COOH)₂)**.

---