Benzene cannot be iodinated with `I_(2)` directly. However, in presence of oxidants such as `HNO_(3)`, iodination is possible. The electrophiles formed in the case is

Options

A. `[I^(+)]`
B. `[I^(-)]`
C. `[overset(+delta)(I)......overset(+delta)(OH_(2))]`
D. `[overset(-delta)(I)......overset(-delta)(OH_(2))]`

Answer

To solve the problem regarding the iodination of benzene in the presence of oxidants like nitric acid (HNO3), we need to identify the electrophile formed during this reaction. Here’s a step-by-step solution:

### Step 1: Understand the Reaction Conditions
Benzene (C6H6) does not react with iodine (I2) directly under normal conditions. However, in the presence of an oxidizing agent like nitric acid (HNO3), the iodination of benzene can occur.

**Hint:** Remember that electrophilic aromatic substitution reactions require a strong electrophile.

### Step 2: Identify the Role of Nitric Acid
Nitric acid acts as an oxidizing agent and helps in generating a more reactive species from iodine. When HNO3 is mixed with iodine, it facilitates the formation of an electrophile.

**Hint:** Consider how nitric acid can influence the reactivity of iodine.

### Step 3: Formation of the Electrophile
When nitric acid reacts with iodine, it generates the electrophile I⁺ (iodonium ion) along with other products. The reaction can be represented as follows:

\[ \text{HNO}_3 + \text{I}_2 \rightarrow \text{I}^+ + \text{NO}_2 + \text{H}_2\text{O} \]

Here, I⁺ is the electrophile that will attack the benzene ring.

**Hint:** Look for the species that can donate a positive charge and act as an electrophile.

### Step 4: Electrophilic Attack on Benzene
The generated I⁺ (iodonium ion) can then react with benzene through an electrophilic aromatic substitution mechanism, resulting in the formation of iodobenzene (C6H5I).

**Hint:** Recall the mechanism of electrophilic aromatic substitution and how the electrophile interacts with the benzene ring.

### Conclusion
The electrophile formed in the reaction of benzene with iodine in the presence of nitric acid is I⁺ (iodonium ion). Therefore, the correct answer is I⁺.

**Final Answer:** The electrophile formed is I⁺ (iodonium ion).