The products formed in the following reaction
`C_6H_5 - O - CH_3 + HI overset("heat")to` 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_3 and HOI`
D. `C_6H_6 and CH_3OI`

Explanation

To determine the products formed in the reaction of anisole (C₆H₅-O-CH₃) with hydroiodic acid (HI) upon heating, we can follow these steps:

### Step 1: Identify the Reactants
The reactants in this reaction are anisole (C₆H₅-O-CH₃) and hydroiodic acid (HI). Anisole is an ether, and HI is a strong acid.

### Step 2: Understand the Reaction Mechanism
When anisole reacts with HI, the ether bond (C-O) can undergo cleavage. The oxygen atom in the ether has lone pairs and can protonate in the presence of HI, leading to the formation of a more reactive species.

### Step 3: Protonation of the Ether
The oxygen atom in anisole will donate its lone pair to the hydrogen ion (H⁺) from HI, resulting in the formation of a protonated ether. This makes the ether bond (C-O) more susceptible to cleavage.

### Step 4: Cleavage of the Ether Bond
Once the ether is protonated, the C-O bond can break. This cleavage will produce two products: phenol (C₆H₅-OH) and methyl iodide (CH₃I). The methyl group (CH₃) will bond with the iodide ion (I⁻) from HI.

### Step 5: Write the Final Products
Thus, the final products of the reaction are:
1. Phenol (C₆H₅-OH)
2. Methyl iodide (CH₃I)

### Final Answer:
The products formed in the reaction of C₆H₅-O-CH₃ with HI upon heating are phenol (C₆H₅-OH) and methyl iodide (CH₃I).

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