What will be the final product in the following reaction sequence? `CH_3CH_2CN overset(H^(+)//H_2O)to A underset(Delta)overset(NH_3)to B overset(NaOBr)to C `

Options

A. `CH_(3)CH_(2)CONH_(2)`
B. `CH_(3)CH_(2)COBr`
C. `CH_(3)CH_(2)NH_(2)`
D. `CH_(3)CH_(2)CH_(2)NH_(2)`

Answer

To solve the given reaction sequence step by step, let's analyze each part of the reaction:

### Step 1: Hydrolysis of CH₃CH₂CN
- The starting compound is **CH₃CH₂CN** (ethyl cyanide).
- When ethyl cyanide is treated with **H⁺** and **H₂O**, it undergoes hydrolysis.
- Hydrolysis of cyanides generally produces carboxylic acids.
- The product formed after hydrolysis is **propanoic acid (CH₃CH₂C(=O)OH)**.

**Product A: CH₃CH₂C(=O)OH (Propanoic acid)**

### Step 2: Reaction with Ammonia (NH₃)
- The next step involves heating the product A (propanoic acid) with **NH₃** (ammonia).
- During this reaction, the carboxylic acid reacts with ammonia to form an amide through a condensation reaction, where water is eliminated.
- The product formed is **propionamide (CH₃CH₂C(=O)NH₂)**.

**Product B: CH₃CH₂C(=O)NH₂ (Propionamide)**

### Step 3: Reaction with Sodium Hypobromite (NaOBr)
- The final step involves treating product B (propionamide) with **NaOBr**.
- Sodium hypobromite is used in the **Hofmann degradation** of amides, which results in the formation of amines with one carbon atom less than the original amide.
- In this case, propionamide will lose one carbon atom and yield **ethylamine (CH₃CH₂NH₂)**.

**Final Product C: CH₃CH₂NH₂ (Ethylamine)**

### Summary of the Reaction Sequence:
1. **CH₃CH₂CN + H⁺/H₂O → CH₃CH₂C(=O)OH (Propanoic acid)**
2. **CH₃CH₂C(=O)OH + NH₃ → CH₃CH₂C(=O)NH₂ (Propionamide)**
3. **CH₃CH₂C(=O)NH₂ + NaOBr → CH₃CH₂NH₂ (Ethylamine)**

### Final Answer:
The final product C is **CH₃CH₂NH₂ (Ethylamine)**.

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