When iodobenzene is treated with sodium in dry ether the product is

Options

A. DDT
B. Triphenyl
C. Diphenyl
D. Dichlorobenzene

Answer

To determine the product formed when iodobenzene is treated with sodium in dry ether, we can follow these steps:

### Step 1: Identify the Reactants
We have iodobenzene (C6H5I) and sodium (Na) in the presence of dry ether. Iodobenzene consists of a benzene ring with an iodine atom attached.

**Hint:** Recognize the components of the reaction: iodobenzene and sodium metal.

### Step 2: Understand the Reaction Mechanism
When sodium is added to iodobenzene, it can donate an electron to the iodine atom. This process involves breaking the bond between the iodine and the benzene ring.

**Hint:** Consider how sodium interacts with the halogen (iodine) in iodobenzene.

### Step 3: Formation of Radicals
The reaction leads to the formation of a phenyl radical (C6H5•) and sodium iodide (NaI). The bond between the iodine and the benzene ring breaks, resulting in a positively charged benzene radical and negatively charged iodide ion.

**Hint:** Remember that breaking the bond creates a radical and an ion.

### Step 4: Combine the Radicals
Two phenyl radicals can combine to form biphenyl (C12H10). This occurs because the two radicals can bond together, resulting in a new carbon-carbon bond between the two benzene rings.

**Hint:** Think about how radicals can react with each other to form stable products.

### Step 5: Write the Overall Reaction
The overall reaction can be summarized as:
\[ 2 \, C_6H_5I + 2 \, Na \rightarrow C_{12}H_{10} + 2 \, NaI \]
This indicates that two moles of iodobenzene react with two moles of sodium to produce one mole of biphenyl and two moles of sodium iodide.

**Hint:** Ensure you balance the reaction correctly, reflecting the stoichiometry of the reactants and products.

### Conclusion
The product formed when iodobenzene is treated with sodium in dry ether is biphenyl (C12H10).

**Final Answer:** Biphenyl (C12H10)