In a set reaction propionic acid yielded a compound `D`
`CH_(3)CH_(2)COOH overset (SOCI_(2)) to B overset (NH_(3)) to C overset (KOH)underset (Br_(2)) to D`
The structure of D would be

Options

A. (a) `CH_3CH_2CH_2NH_2`
B. (b) `CH_3CH_2CONH_2`
C. (c) `CH_3CH_2NHCH_3`
D. (d) `CH_3CH_2NH_2`

Explanation

To solve the problem, we will go through the reactions step by step, starting with propionic acid and ending with compound D.

### Step 1: Identify the starting compound
The starting compound is propionic acid, which has the structure:
\[ \text{CH}_3\text{CH}_2\text{COOH} \]

### Step 2: Reaction with SOCl₂
When propionic acid reacts with thionyl chloride (SOCl₂), the hydroxyl group (-OH) of the carboxylic acid is replaced by a chlorine atom (-Cl). This results in the formation of propionyl chloride (compound B):
\[ \text{CH}_3\text{CH}_2\text{COOH} \xrightarrow{\text{SOCl}_2} \text{CH}_3\text{CH}_2\text{COCl} \]

### Step 3: Reaction with NH₃
Next, propionyl chloride (compound B) reacts with ammonia (NH₃). In this reaction, the chlorine atom is replaced by an amine group (-NH₂), and hydrochloric acid (HCl) is released. This results in the formation of propanamide (compound C):
\[ \text{CH}_3\text{CH}_2\text{COCl} + \text{NH}_3 \rightarrow \text{CH}_3\text{CH}_2\text{CONH}_2 + \text{HCl} \]

### Step 4: Reaction with KOH and Br₂
Finally, propanamide (compound C) undergoes a reaction known as the Hofmann degradation when treated with potassium hydroxide (KOH) and bromine (Br₂). In this reaction, the amide group is converted to an amine group, and a carbon atom is lost in the process, resulting in the formation of ethylamine (compound D):
\[ \text{CH}_3\text{CH}_2\text{CONH}_2 \xrightarrow{\text{KOH}, \text{Br}_2} \text{CH}_3\text{CH}_2\text{NH}_2 + \text{CO}_2 \]

### Conclusion
The final compound D is ethylamine, which has the structure:
\[ \text{CH}_3\text{CH}_2\text{NH}_2 \]

Thus, the structure of compound D is:
**D: CH₃CH₂NH₂**