Which will form lactone on treatment with NaOH?

Options

A. `delta`- bromo acid
B. `beta`-bromo acid
C. `beta` - hydroxy acid
D. `alpha` - bromo acid

Answer

To determine which compound will form a lactone upon treatment with NaOH, we need to analyze the options provided: Delta Bromo Acid, Beta Bromo Acid, Beta Hydroxy Acid, and Alpha Bromo Acid.

### Step-by-Step Solution:

1. **Understanding Lactones**:
- Lactones are cyclic esters formed from the reaction of a hydroxy acid where the hydroxyl group and the carboxylic acid group react to form a ring.
- The stability of the lactone depends on the size of the ring formed during the reaction.

2. **Analyzing Delta Bromo Acid**:
- Delta Bromo Acid has the structure:
\[ \text{Br-CH}_2-\text{CH}_2-\text{C(=O)OH} \]
- Upon treatment with NaOH, the hydroxide ion (OH-) will deprotonate the carboxylic acid group, leading to the formation of a nucleophile.
- The nucleophile (O-) can then perform an intramolecular nucleophilic substitution (SN2) reaction, attacking the delta carbon (the carbon adjacent to the carbonyl carbon) and forming a 6-membered lactone ring.

3. **Analyzing Beta Bromo Acid**:
- Beta Bromo Acid has the structure:
\[ \text{Br-CH}_2-\text{C(=O)OH}-\text{CH}_2 \]
- Similar to Delta Bromo Acid, treatment with NaOH will lead to deprotonation and formation of a nucleophile.
- The O- can attack the beta carbon, resulting in a 4-membered ring. However, 4-membered rings are unstable due to angle strain.

4. **Analyzing Beta Hydroxy Acid**:
- Beta Hydroxy Acid has the structure:
\[ \text{OH-CH}_2-\text{C(=O)OH}-\text{CH}_2 \]
- Upon treatment with NaOH, the hydroxyl group is not a good leaving group, so the nucleophile cannot effectively form a lactone. Therefore, no lactone is formed.

5. **Analyzing Alpha Bromo Acid**:
- Alpha Bromo Acid has the structure:
\[ \text{Br-CH}_2-\text{C(=O)OH} \]
- Similar to Beta Bromo Acid, the nucleophile can attack the alpha carbon, leading to a 3-membered ring. However, 3-membered rings are also unstable due to significant angle strain.

6. **Conclusion**:
- The only compound that forms a stable lactone upon treatment with NaOH is **Delta Bromo Acid**, which forms a stable 6-membered ring lactone.

### Final Answer:
The correct answer is **A. Delta Bromo Acid**.