`PhCOCHBr_(2)overset(OH^(-))toAoverset(OH^(-))toBoverset(H^(+))toC`
The compound 'C' is

Options

A. PhCH(OH)CHO
B. PhCH(OH)COOH
C. `PhCOunderset(H)underset(|)CBr_2`
D. `Ph-undersetunderset(O)(||)(O)-CH_2-OH`

Explanation

To determine the compound 'C' from the given reaction sequence, let's break down the steps involved in the reaction.

### Step 1: Identify the Starting Compound
The starting compound is PhCOCHBr₂, which can be represented as:
- Ph = Phenyl group (C₆H₅)
- CO = Carbonyl group (C=O)
- CHBr₂ = A carbon atom bonded to two bromine atoms.

### Step 2: Reaction with OH⁻
The first reaction involves the nucleophilic attack by hydroxide ion (OH⁻) on the carbon atom that is bonded to the bromine atoms (SN2 mechanism). The bromine atoms are good leaving groups, and one of them will leave.

The product 'A' after this step will be:
- PhC(OH)CHBr (where one Br has been replaced by OH)

### Step 3: Elimination of Br⁻
In the next step, the compound 'A' reacts again with OH⁻. The hydroxide ion will attack the carbon that is still bonded to Br, leading to the elimination of Br⁻.

The product 'B' after this step will be:
- PhC(OH)CHO (where the second Br has been replaced by a carbonyl group, forming an aldehyde)

### Step 4: Reaction with OH⁻ (Canizzaro Reaction)
The compound 'B' undergoes a Canizzaro reaction, which occurs in the presence of a strong base (OH⁻) when there is no alpha-hydrogen. This reaction results in the formation of an alcohol and a carboxylic acid.

The product 'C' after this step will be:
- PhC(OH)COOH (where the aldehyde is oxidized to a carboxylic acid)

### Step 5: Final Product
The final product 'C' can be represented as:
- PhC(OH)COOH, which is a phenyl group attached to a carbon that has a hydroxyl group and a carboxylic acid.

### Conclusion
The compound 'C' is:
- **C₆H₅C(OH)COOH** (or simply, **salicylic acid**)

### Summary of Steps:
1. Start with PhCOCHBr₂.
2. React with OH⁻ to form PhC(OH)CHBr (product A).
3. Eliminate Br⁻ to form PhC(OH)CHO (product B).
4. Undergo Canizzaro reaction to form PhC(OH)COOH (product C).