`R-underset(NH_(2))underset(|)(CH)-COO^(-)overset(HCI(aq))rarr`

Answer

To solve the given question, we need to analyze the reaction of the compound R-(NH2)-(CH)-COO^(-) with aqueous HCl. Let's break down the steps involved in this reaction:

### Step-by-Step Solution:

1. **Identify the Structure**:
The compound R-(NH2)-(CH)-COO^(-) represents an amino acid in its zwitterionic form. Here, R is a hydrocarbon chain, NH2 is the amino group, CH is the carbon backbone, and COO^(-) is the carboxylate group.

2. **Understanding Zwitterion Formation**:
In the zwitterionic form, the amino acid has both a positive and a negative charge. The amino group (NH2) can accept a proton (H+) to become NH3^+, while the carboxylate group (COO^(-)) has already lost a proton.

3. **Reaction with HCl**:
When this zwitterion is treated with aqueous HCl, the HCl provides protons (H+) that will react with the amino group (NH2) to form NH3^+. The carboxylate group remains negatively charged (COO^(-)).

4. **Product Formation**:
After the reaction, the amino group will be protonated, resulting in NH3^+. The carboxylate group will remain as COO^(-). Thus, the final product of the reaction can be represented as RCHCOO^(-) and NH3^+.

5. **Final Product Representation**:
The final product can be expressed as:
\[ RCHCOO^{-} \, \text{and} \, NH3^{+} \]
This indicates that we have a carboxylate anion and an ammonium cation.

### Final Answer:
The final products of the reaction of R-(NH2)-(CH)-COO^(-) with aqueous HCl are RCHCOO^(-) and NH3^+.

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