Which of the following compound is most reactive in the nucleophilic aromatic substitution reaction with NaOH ?

Options

A. `p-` nitrofluorobenezene
B. `p-`nitrochlorobenzene
C. `p-`nitrobromobenzene
D. `p-`nitroiodobenzene

Explanation

To determine which compound is most reactive in a nucleophilic aromatic substitution reaction with NaOH, we need to consider the reactivity order of different halogens in aromatic compounds. Here’s a step-by-step solution:

### Step 1: Understand Nucleophilic Aromatic Substitution
Nucleophilic aromatic substitution occurs when a nucleophile (like OH⁻ from NaOH) attacks an aromatic compound, replacing a leaving group (like a halogen). The reactivity of the aromatic compound depends on the nature of the leaving group.

### Step 2: Identify the Leaving Groups
In the context of nucleophilic aromatic substitution, the halogens (F, Cl, Br, I) are common leaving groups. Their reactivity in this type of reaction generally follows the order:
- **Fluorine (F) > Chlorine (Cl) > Bromine (Br) > Iodine (I)**

### Step 3: Analyze the Given Compounds
Assuming the compounds in question are substituted aromatic compounds with different halogens, we can analyze them based on the reactivity order established in Step 2. The compound with fluorine as the leaving group will be the most reactive, followed by chlorine, bromine, and iodine.

### Step 4: Conclusion
Based on the reactivity order of the halogens in nucleophilic aromatic substitution, the compound with fluorine as the leaving group will be the most reactive when treated with NaOH.

### Final Answer
Thus, the compound that is most reactive in the nucleophilic aromatic substitution reaction with NaOH is the one containing **Fluorine (F)** as the leaving group.

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