The products formed in the following reaction
`C_(6)H_(5)-O-CH_(3)+HI overset(heat)rarr` are

Options

A. `C_6H_5-I and CH_3-OH`
B. `C_6H_5-OH and CH_3-I`
C. `C_6H_5-CH_3and HOl`
D. `C_6H_6and CH_3Ol`

Answer

To solve the reaction of `C6H5-O-CH3` with `HI` upon heating, we can follow these steps:

### Step 1: Identify the Reactants
The reactants in this reaction are:
- `C6H5-O-CH3` (which is anisole, a methoxy derivative of phenol)
- `HI` (hydroiodic acid)

### Step 2: Understand the Reaction Mechanism
When anisole is treated with hydroiodic acid (HI) and heated, the ether bond (C-O) is cleaved. Hydroiodic acid provides iodine and a proton (H+), which can lead to the formation of phenol and methyl iodide.

### Step 3: Break the Ether Bond
The ether bond between the phenyl group (`C6H5`) and the methoxy group (`O-CH3`) is broken. The oxygen atom from the ether bond will bond with a hydrogen ion (H+) from HI, forming phenol (`C6H5OH`).

### Step 4: Form the Products
After breaking the ether bond, the following products are formed:
1. **Phenol**: `C6H5OH`
2. **Methyl Iodide**: `CH3I` (the methoxy group is converted into methyl iodide)

### Step 5: Write the Balanced Equation
The balanced equation for the reaction can be written as:
\[ C6H5-O-CH3 + HI \xrightarrow{heat} C6H5OH + CH3I \]

### Final Products
The products formed from the reaction are:
- **Phenol**: `C6H5OH`
- **Methyl Iodide**: `CH3I`

### Conclusion
Thus, the products formed in the reaction of `C6H5-O-CH3` with `HI` upon heating are `C6H5OH` and `CH3I`.