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 solve the problem, we need to identify the compounds A, B, and C in the reaction sequence involving ethyl bromide (C2H5Br) reacting with silver cyanide (AgCN) followed by hydrolysis. Let's break it down step by step.

### Step 1: Identify Compound A
1. **Starting Material**: The reaction begins with ethyl bromide (C2H5Br).
2. **Reagent**: It reacts with AgCN, which is a covalent compound and does not dissociate into ions.
3. **Mechanism**: The reaction proceeds via a nucleophilic substitution mechanism. The nitrogen in the cyanide group (CN) acts as the nucleophile and attacks the carbon attached to the bromine in ethyl bromide.
4. **Product Formation**: The product formed from this reaction is ethyl isocyanide (C2H5NC). Therefore, **Compound A is C2H5NC (ethyl isocyanide)**.

### Step 2: Identify Compound B and C
1. **Next Step**: The next part of the reaction involves the hydrolysis of Compound A (C2H5NC).
2. **Hydrolysis Reaction**: When ethyl isocyanide undergoes hydrolysis in the presence of H3O^+, it forms an intermediate.
3. **Intermediate Formation**: The hydrolysis leads to the formation of an unstable intermediate which can be represented as CH3CH2N^+H2 (where nitrogen carries a positive charge) and a negatively charged carbon (C).
4. **Final Products**: The hydrolysis further leads to the formation of two products:
- **Compound B**: Ethylamine (C2H5NH2)
- **Compound C**: Formic acid (HCOOH)

### Conclusion
- **Compound A**: C2H5NC (ethyl isocyanide)
- **Compound B**: C2H5NH2 (ethylamine)
- **Compound C**: HCOOH (formic acid)

### Final Answer
- A: C2H5NC (ethyl isocyanide)
- B: C2H5NH2 (ethylamine)
- C: HCOOH (formic acid)

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