In the given reaction sequences the final product 'C' is
`CH_3-CH_2-OH underset((ii)H_2O^(+))overset((i)KMnO_4//overset(-)(OH)//Delta)(rarr)(A)underset((ii)NH_3)overset((i)SOCl_2)(rarr)(B) overset(Br_(2)//KOH)(rarr)(C )`

Options

A. Methylamine
B. Ethylamine
C. Propylamine
D. Acetamide

Answer

To solve the given reaction sequence and determine the final product 'C', we will break down the reactions step by step.

### Step 1: Oxidation of Ethanol
The first step involves treating ethanol (CH₃-CH₂-OH) with hot alkaline potassium permanganate (KMnO₄ in the presence of OH⁻).

**Reaction:**
\[ \text{CH}_3\text{CH}_2\text{OH} \xrightarrow{\text{KMnO}_4/\text{OH}^-} \text{CH}_3\text{C(=O)O}^- \]

**Explanation:**
- KMnO₄ is a strong oxidizing agent. In alkaline conditions, it oxidizes the alcohol (ethanol) to a carboxylate ion (in this case, acetate ion).
- The hydroxyl group (-OH) of ethanol is converted to a carboxylate ion (-COO⁻) because the oxidation leads to the formation of acetic acid (CH₃COOH) after hydrolysis.

### Step 2: Hydrolysis to Acetic Acid
Next, we treat the carboxylate ion with hydronium ions (H₃O⁺) to hydrolyze it into acetic acid.

**Reaction:**
\[ \text{CH}_3\text{C(=O)O}^- + \text{H}_3\text{O}^+ \rightarrow \text{CH}_3\text{C(=O)OH} \]

**Explanation:**
- The carboxylate ion accepts a proton (H⁺) from the hydronium ion, resulting in the formation of acetic acid (CH₃COOH).

### Step 3: Conversion to Acyl Chloride
Now, we treat acetic acid with thionyl chloride (SOCl₂) to convert it into an acyl chloride.

**Reaction:**
\[ \text{CH}_3\text{C(=O)OH} \xrightarrow{\text{SOCl}_2} \text{CH}_3\text{C(=O)Cl} + \text{SO}_2 + \text{HCl} \]

**Explanation:**
- In this reaction, the hydroxyl group (-OH) of acetic acid is replaced by a chlorine atom (-Cl), forming acetyl chloride (CH₃C(=O)Cl).

### Step 4: Reaction with Ammonia
Next, we treat the acyl chloride with ammonia (NH₃).

**Reaction:**
\[ \text{CH}_3\text{C(=O)Cl} + \text{NH}_3 \rightarrow \text{CH}_3\text{C(=O)NH}_2 + \text{HCl} \]

**Explanation:**
- Ammonia acts as a nucleophile and attacks the carbonyl carbon of the acyl chloride, leading to the formation of an amide (acetamide, CH₃C(=O)NH₂) and the release of HCl gas.

### Step 5: Hofmann Degradation
Finally, we treat the amide with bromine (Br₂) in the presence of a strong base (KOH), which leads to the Hofmann degradation reaction.

**Reaction:**
\[ \text{CH}_3\text{C(=O)NH}_2 + \text{Br}_2/\text{KOH} \rightarrow \text{CH}_3\text{NH}_2 + \text{CO}_2 + \text{HBr} \]

**Explanation:**
- The Hofmann degradation reaction removes the carbonyl carbon (C=O) from the amide, resulting in the formation of methylamine (CH₃NH₂).

### Final Product
Thus, the final product 'C' is methylamine (CH₃NH₂).

### Summary of Steps
1. Ethanol is oxidized to acetic acid.
2. Acetic acid is converted to acetyl chloride.
3. Acetyl chloride reacts with ammonia to form acetamide.
4. Acetamide undergoes Hofmann degradation to yield methylamine.

### Final Answer
The final product 'C' is **methylamine**.