(A) With HBr, anisole forms bromobenzene and methyl alcohol.
(R ) `Br^(-)` ion will combine with smaller group to avoid steric hindrance.

Options

A. If both assertion and reason are correct and reason is correct explanation for assertion.
B. If both assertion and reason are correct but reason is not correct for assertion.
C. If assertion is correct but reason is incorrect.
D. If both assertion and reason are incorrect.

Explanation

To solve the problem, we need to analyze the reaction of anisole (methyl phenyl ether) with HBr and understand the mechanism involved.

### Step-by-Step Solution:

1. **Identify Anisole Structure**:
Anisole is represented as a benzene ring with a methoxy group (-OCH3) attached. Its structure can be depicted as:
\[ \text{C}_6\text{H}_5\text{-O-CH}_3 \]

2. **Reaction with HBr**:
When anisole reacts with HBr, the ether bond (C-O) is protonated. The oxygen atom in the methoxy group has a lone pair of electrons that can attract the hydrogen ion (H+) from HBr, leading to the formation of an oxonium ion:
\[ \text{C}_6\text{H}_5\text{-O-CH}_3 + \text{HBr} \rightarrow \text{C}_6\text{H}_5\text{-O}^+\text{-CH}_3 + \text{Br}^- \]

3. **Formation of Oxonium Ion**:
The oxonium ion structure is:
\[ \text{C}_6\text{H}_5\text{-O}^+\text{-CH}_3 \]
Here, the oxygen carries a positive charge due to the protonation.

4. **Bond Strength Consideration**:
The bond between the oxygen and the methyl group (C-O) is weaker than the bond between the oxygen and the benzene ring (C-O). This is due to resonance stabilization in the latter bond, which gives it partial double bond character.

5. **Nucleophilic Attack**:
The bromide ion (Br^-) will perform a nucleophilic attack on the weaker bond (C-O) of the methyl group. This follows the SN2 mechanism, where the nucleophile attacks the less sterically hindered site:
\[ \text{C}_6\text{H}_5\text{-O}^+\text{-CH}_3 + \text{Br}^- \rightarrow \text{C}_6\text{H}_5\text{-OH} + \text{CH}_3\text{Br} \]

6. **Products Formation**:
The products of this reaction are phenol (C6H5OH) and methyl bromide (CH3Br):
\[ \text{Products: Phenol and Methyl Bromide} \]

7. **Conclusion**:
The assertion that anisole forms bromobenzene and methyl alcohol with HBr is incorrect. The correct products are phenol and methyl bromide. The reason provided is correct as the Br^- ion attacks the smaller alkyl group to avoid steric hindrance.

### Final Answer:
- Assertion: **Incorrect** (Anisole does not form bromobenzene and methyl alcohol).
- Reason: **Correct** (Br^- ion attacks the smaller group due to steric hindrance).