Which of following shows highest rate of decarboxylation in presence of soda lime :

Options

A. `underset(F)underset(|)(C)H_(2)-COOH`
B. `underset(Cl)underset(|)CH_(2)-overset(O)overset(||)C-OH`
C. `underset(Br)underset(|)CH_(2)-overset(O)overset(||)C-OH`
D. `underset("I ")underset(|)CH_(2)-overset(O)overset(||)C-OH`

Answer

To determine which compound shows the highest rate of decarboxylation in the presence of soda lime, we need to analyze the factors that influence the decarboxylation reaction.

### Step-by-Step Solution:

1. **Understanding Decarboxylation**:
- Decarboxylation is the process of removing a carboxyl group (-COOH) from a carboxylic acid, resulting in the formation of an alkane. This reaction typically occurs in the presence of soda lime (a mixture of sodium hydroxide and calcium oxide).

2. **Role of Soda Lime**:
- Soda lime acts as a dehydrating agent and provides a strong base (NaOH) that facilitates the decarboxylation process. The presence of quick lime (CaO) helps to keep the reaction conditions dry, preventing the reverse reaction.

3. **Mechanism of Decarboxylation**:
- The mechanism involves the formation of a tetrahedral intermediate when the carboxylic acid reacts with the hydroxide ion. The carbonyl carbon becomes more electrophilic due to the presence of electron-withdrawing groups, which stabilize the positive charge during the reaction.

4. **Influence of Electron-Withdrawing Groups**:
- The rate of decarboxylation is influenced by the presence of electron-withdrawing groups (EWGs) attached to the carbon chain. EWGs increase the positive charge on the carbonyl carbon, making it more susceptible to nucleophilic attack.

5. **Comparing Compounds**:
- To find which compound shows the highest rate of decarboxylation, we need to evaluate the given options based on the presence of electron-withdrawing groups. The more electronegative the substituent, the stronger its electron-withdrawing effect.

6. **Identifying the Correct Option**:
- Among the options, if one compound has fluorine (the most electronegative element), it will exert a strong electron-withdrawing effect compared to other halogens like bromine or iodine. This will enhance the positive charge on the carbonyl carbon, leading to a faster rate of decarboxylation.

7. **Conclusion**:
- Therefore, the compound with fluorine (Option A) will show the highest rate of decarboxylation in the presence of soda lime due to the strong electron-withdrawing effect of fluorine.

### Final Answer:
**Option A** shows the highest rate of decarboxylation in the presence of soda lime.

---