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 of the given acids will form a lactone upon treatment with NaOH, we need to analyze each option step by step. The acids in question are Delta Bromo Acid, Beta Bromo Acid, Beta Hydroxy Acid, and Alpha Bromo Acid.

### Step 1: Understanding the Structure of Each Acid
1. **Delta Bromo Acid**: This acid has a bromine atom attached to the delta carbon (the fourth carbon in the chain from the carboxylic acid group).
2. **Beta Bromo Acid**: This acid has a bromine atom on the beta carbon (the second carbon from the carboxylic acid group).
3. **Beta Hydroxy Acid**: This acid has a hydroxyl (-OH) group on the beta carbon.
4. **Alpha Bromo Acid**: This acid has a bromine atom on the alpha carbon (the first carbon next to the carboxylic acid group).

### Step 2: Reaction with NaOH
When these acids are treated with NaOH, the hydroxide ion (OH-) acts as a nucleophile and can abstract the most acidic hydrogen from the carboxylic acid group.

### Step 3: Analyzing Delta Bromo Acid
- Upon treatment with NaOH, the OH- will abstract the acidic hydrogen from the carboxylic acid group.
- This will generate a carboxylate ion, which can then undergo an intramolecular nucleophilic attack by the oxygen atom on the carbon that has the bromine attached (the delta carbon).
- This results in the formation of a six-membered lactone ring, which is stable.

### Step 4: Analyzing Beta Bromo Acid
- Similar to the Delta Bromo Acid, the OH- will abstract the acidic hydrogen.
- However, the intramolecular attack will occur between the carboxylate ion and the beta carbon, forming a four-membered ring.
- Four-membered rings are highly unstable due to angle strain, so this will not lead to lactone formation.

### Step 5: Analyzing Beta Hydroxy Acid
- The OH- will also abstract the acidic hydrogen here.
- The intramolecular attack would occur, but the hydroxyl group is not a good leaving group compared to bromine. Therefore, the reaction will not proceed to form a lactone.

### Step 6: Analyzing Alpha Bromo Acid
- The OH- will abstract the acidic hydrogen.
- An intramolecular attack would lead to a three-membered ring, which is highly unstable and cannot form.

### Conclusion
The only acid that can successfully form a stable lactone upon treatment with NaOH is **Delta Bromo Acid**, as it forms a six-membered lactone ring.

### Final Answer
**Delta Bromo Acid will form a lactone on treatment with NaOH.**