Consider the following reaction ,
`C_(6)H_(5)NH_(2)+CHCl_(3)+KOHoverset(Delta)rarr(A)overset(H^(+)//H_(2)O)rarr(B)+(C)`
The compounds (B) and (C) are :-

Options

A. `C_(6)H_(5)COOH` and `NH_(3)` respectively
B. `C_(6)H_(5)NH_(2)` and `HCOOH` respectively
C. `C_(6)H_(5)NH_(2)` and `H_(2)O` respectively
D. None of these

Answer

To solve the given reaction step by step, we will analyze the reactants and the processes involved.

### Step 1: Identify the reactants
The reactants provided are:
- Aniline (C₆H₅NH₂)
- Chloroform (CHCl₃)
- Potassium hydroxide (KOH)

### Step 2: Understand the reaction conditions
The reaction is conducted under heating conditions (Δ), and KOH is used in alcoholic form. This indicates that we are likely dealing with a carbylamine reaction.

### Step 3: Carbylamine reaction
When aniline reacts with chloroform and KOH, it undergoes a carbylamine reaction. The product formed in this step is phenylisocyanide (or phenylcarbamine), which is represented as (A):
- **A** = C₆H₅N=C=O (Phenylisocyanide)

### Step 4: Hydrolysis of the product
Next, the product (A) undergoes hydrolysis in the presence of H⁺ and H₂O. The hydrolysis of phenylisocyanide leads to the formation of two products:
1. Aniline (C₆H₅NH₂)
2. Formic acid (HCOOH)

### Step 5: Identify the final products
Thus, the compounds (B) and (C) formed from the hydrolysis of (A) are:
- **B** = C₆H₅NH₂ (Aniline)
- **C** = HCOOH (Formic acid)

### Final Answer
The compounds (B) and (C) are:
- **B** = Aniline (C₆H₅NH₂)
- **C** = Formic acid (HCOOH)

---