In the given reaction sequence
`CH_3CHO underset((ii)H_2O^+)overset((i) NaCN//HCl)(rarr)(A) underset("reagent")overset("Fenton")(rarr)(B)`, B will be

Options

A. Acetic acid
B. Oxalic acid
C. Pyruvic acid
D. Citric acid

Answer

To solve the given reaction sequence, we will break down the steps involved in the transformation of the reactant to the final product.

### Step-by-Step Solution:

1. **Identify the Starting Material:**
The starting material is **CH₃CHO**, which is acetaldehyde (an aldehyde).

2. **Nucleophilic Addition Reaction:**
The first step involves the reaction of acetaldehyde with **NaCN** in the presence of **HCl**. Here, the cyanide ion (CN⁻) acts as a nucleophile and attacks the electrophilic carbon of the carbonyl group in acetaldehyde.

- The reaction can be represented as:
\[ CH₃CHO + CN⁻ \rightarrow CH₃C(OH)(CN) \]
- This forms a cyanohydrin intermediate, which is **2-hydroxy-2-methylpropanenitrile**.

3. **Protonation:**
The intermediate formed is then protonated by **H₂O** (from HCl), leading to the formation of **lactic acid** (A):
\[ CH₃C(OH)(CN) + H₂O \rightarrow CH₃C(OH)(COOH) \]
- This compound is **lactic acid** (2-hydroxypropanoic acid).

4. **Oxidation with Fenton's Reagent:**
Next, lactic acid (A) is treated with **Fenton's reagent**, which is a combination of hydrogen peroxide (H₂O₂) and ferrous iron (Fe²⁺). This reagent is known for its oxidizing properties.

- The oxidation of lactic acid will convert the hydroxyl group (-OH) into a carbonyl group (C=O), resulting in the formation of **pyruvic acid** (B):
\[ CH₃C(OH)(COOH) \xrightarrow{Fenton's} CH₃C(=O)COOH \]
- The product formed is **pyruvic acid** (2-oxopropanoic acid).

### Final Answer:
Thus, the compound B formed in the reaction sequence is **pyruvic acid**.