Following `1^@` amine has chiral carbon as indicated
`H-undersetunderset(C_2H_5)|oversetoverset(CH_3)|(C^(***))-NH_2`
This on reaction with (`NaNO_2` + HCI) forms

Options

A. `1^(@) ` alcohol with retention of configuration
B. `2^(@) ` alcohol with inverted configuration
C. Racemic mixture of `2^@ ` alcohol
D. race mixture of `1^@` alcohol

Answer

To solve the problem, we need to analyze the given structure of the amine and its reaction with sodium nitrite (NaNO2) in the presence of hydrochloric acid (HCl).

### Step-by-Step Solution:

1. **Identify the Structure of the Amine**:
The amine is represented as:
```
H
|
C2H5-C-CH3
|
NH2
```
Here, the carbon atom attached to the `NH2` group is a chiral center because it is bonded to four different groups: `H`, `C2H5`, `CH3`, and `NH2`.

2. **Reaction with NaNO2 and HCl**:
When a primary amine reacts with sodium nitrite (NaNO2) in the presence of hydrochloric acid (HCl), it undergoes diazotization. The reaction can be summarized as follows:
- The amine (R-NH2) reacts with HCl to form the ammonium salt (R-NH3^+ Cl^-).
- The ammonium salt then reacts with NaNO2 to form a diazonium salt (R-N2^+ Cl^-).

3. **Formation of the Diazotization Product**:
The diazonium salt formed is:
```
C2H5-C-CH3
|
N2^+ Cl^-
```
This is a stable compound at low temperatures and can be used for further reactions.

4. **Conclusion**:
The final product of the reaction of the given chiral amine with NaNO2 and HCl is the corresponding diazonium salt.

### Final Answer:
The reaction of the given chiral amine with NaNO2 and HCl forms a diazonium salt.

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