`PhCHO+(CH_3CO)_2Ooverset(CH_3COONa)toAoverset(HBr)toB`
The product B is

Options

A. `PhCH=CHCH_2Br`
B. `Phundersetunderset(Br)(|)CH-CH_2-COOH`
C. `PhCH_2CH(Br)COOH`
D. PhCH=CH-COBr

Explanation

To solve the problem, we will follow the steps of the reaction sequence provided in the question.

### Step 1: Reaction of Benzaldehyde with Acetic Anhydride
- **Reactants**: Benzaldehyde (PhCHO) and Acetic Anhydride ((CH₃CO)₂O) in the presence of Sodium Acetate (CH₃COONa).
- **Mechanism**: The sodium acetate acts as a catalyst for the reaction, facilitating the formation of an intermediate product through a condensation reaction. This leads to the formation of an enol acetate.

The reaction can be represented as:
\[ \text{PhCHO} + \text{(CH}_3\text{CO)}_2\text{O} \xrightarrow{\text{CH}_3\text{COONa}} \text{A} \]

The product A is likely to be Cinnamic Acid (C₆H₅-CH=CH-COOH).

### Step 2: Hydrolysis of the Intermediate
- **Hydrolysis**: The intermediate product A undergoes hydrolysis in the presence of water, leading to the formation of Cinnamic Acid.

The hydrolysis can be represented as:
\[ \text{A} + \text{H}_2\text{O} \rightarrow \text{Cinnamic Acid (C₆H₅-CH=CH-COOH)} \]

### Step 3: Addition of HBr to Cinnamic Acid
- **Reactants**: Cinnamic Acid reacts with HBr.
- **Mechanism**: The double bond in Cinnamic Acid acts as a nucleophile and attacks the electrophilic hydrogen (H⁺) from HBr. This leads to the formation of a carbocation. The bromide ion (Br⁻) then attacks the carbocation to form the final product B.

The reaction can be represented as:
\[ \text{Cinnamic Acid} + \text{HBr} \rightarrow \text{B} \]

### Step 4: Identify Product B
- The product B formed after the addition of HBr to Cinnamic Acid is likely to be 3-Bromocinnamic Acid (C₆H₅-CHBr-CH=COOH).

### Final Answer
The product B is **3-Bromocinnamic Acid (C₆H₅-CHBr-CH=COOH)**.

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