Compounds A, B and C in the following reaction sequence are:
` C_(2) H_(5)Br overset("AgCN")(rarr) A overset(H3O^(+))(rarr)` B + C

Options

A. `C_(2) H_(5)CN, C_(2)H_(5)"COOH", NH_(3)`
B. `C_(2)H_(5)"NC", C_(2) H_(5) NH_(2),"HCOOH"`
C. `C_(2)H_(5)"NC", C_(2) H_(5) "NHCH"_(3),"HCOOH"`
D. none of these

Answer

To determine the compounds A, B, and C in the reaction sequence given, we can follow these steps:

### Step 1: Identify the Reaction
The reaction starts with ethyl bromide (C₂H₅Br) reacting with silver cyanide (AgCN). This is a nucleophilic substitution reaction.

### Step 2: Determine Compound A
In this reaction, the cyanide ion (CN⁻) acts as a nucleophile and attacks the ethyl bromide. However, since AgCN does not dissociate into ions, the nucleophile will be the nitrogen atom from the AgCN. The product formed from this reaction is ethyl isocyanide (C₂H₅NC). Therefore, compound A is:

**A = C₂H₅NC (ethyl isocyanide)**

### Step 3: Hydrolysis of Compound A
Next, we perform hydrolysis on compound A (ethyl isocyanide). Hydrolysis involves the addition of water (H₂O) and typically requires an acid (H⁺) to catalyze the reaction.

### Step 4: Determine Compound B and C
During hydrolysis, the ethyl isocyanide reacts with water and H⁺ to form two products:
1. Ethylamine (C₂H₅NH₂) - compound B
2. Formic acid (HCOOH) - compound C

Thus, we have:
**B = C₂H₅NH₂ (ethylamine)**
**C = HCOOH (formic acid)**

### Final Summary of Compounds
- **A = C₂H₅NC (ethyl isocyanide)**
- **B = C₂H₅NH₂ (ethylamine)**
- **C = HCOOH (formic acid)**