In the given reaction
`CH_3-overset(Br)overset(|)(C )H-CH_2-COOH overset(NaOH)(rarr)(X)`.
'X' will be

Options

A. `CH_3-underset(OH)underset(|)OH(C )H-CH_2-COOH`
B. `CH_3-underset(OH)underset(|)(C )H-CH_2-COOH`
C. `CH_3-underset(OH)underset(|)(C )H-CH_2-COONa`
D. `CH_3-CH=CH-COOH`

Answer

To solve the given reaction step by step, let's analyze the process:

### Step 1: Identify the Reactants
The reactant is a compound with the structure:
`CH3-CH(Br)-CH2-COOH`.
This compound contains a bromine atom (Br) and a carboxylic acid group (COOH).

### Step 2: Recognize the Role of Sodium Hydroxide (NaOH)
Sodium hydroxide is a strong base and will participate in two types of reactions:
1. Acid-base reaction with the carboxylic acid group (COOH).
2. Nucleophilic substitution reaction where the hydroxide ion (OH-) will replace the bromine atom (Br).

### Step 3: Acid-Base Reaction
The carboxylic acid (COOH) will react with NaOH. The hydroxide ion (OH-) from NaOH will deprotonate the carboxylic acid, forming a carboxylate ion (COO-). The reaction can be represented as:
\[ \text{CH}_3-\text{CH(Br)}-\text{CH}_2-\text{COOH} + \text{NaOH} \rightarrow \text{CH}_3-\text{CH(Br)}-\text{CH}_2-\text{COO}^- + \text{H}_2\text{O} \]

### Step 4: Nucleophilic Substitution Reaction
In the next step, the hydroxide ion (OH-) acts as a nucleophile and attacks the electrophilic carbon that is bonded to bromine (Br). The bromine atom, being a good leaving group, will be replaced by the hydroxide ion. The reaction can be represented as:
\[ \text{CH}_3-\text{CH(Br)}-\text{CH}_2-\text{COO}^- + \text{OH}^- \rightarrow \text{CH}_3-\text{CH(OH)}-\text{CH}_2-\text{COO}^- + \text{Br}^- \]

### Step 5: Final Product
The final product after both reactions will be:
\[ \text{CH}_3-\text{CH(OH)}-\text{CH}_2-\text{COO}^- \]
This compound is a carboxylate ion with a hydroxyl group (OH) attached to the carbon that was originally bonded to bromine.

### Conclusion
The final product 'X' is:
\[ \text{CH}_3-\text{CH(OH)}-\text{CH}_2-\text{COO}^- \]

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