A : When `C_(2) H_(5) - O- CH_(3)` is reacted with oen mole of Hl then `C_(2)H_(5)OH& CH_(3) l` is formed .
R : It is `S_(N)1` reaction

Options

A. If both Assertion & Reason are true and the reason is the correct explanation of the assertion, then mark (1)
B. If both Assertion & Reason are true but the reason is not the correct explanation of the assertion, then mark (2)
C. If Assertion is true statement but Reason is false , then mark (3)
D. If both Assertion and Reason are false statements , then mark (4)

Answer

To solve the question, we need to analyze the assertion and reasoning provided:

### Assertion (A):
When \( C_2H_5 - O - CH_3 \) is reacted with one mole of HI, then \( C_2H_5OH \) and \( CH_3I \) are formed.

### Reason (R):
It is an \( S_N1 \) reaction.

### Step-by-Step Solution:

1. **Identify the Reactants**:
The reactant is an ether, specifically ethyl methyl ether, represented as \( C_2H_5 - O - CH_3 \).

2. **Reaction with HI**:
When this ether reacts with hydroiodic acid (HI), the ether undergoes cleavage. The reaction can be represented as:
\[ C_2H_5 - O - CH_3 + HI \rightarrow C_2H_5OH + CH_3I \]
Here, the ether is cleaved to form ethanol (\( C_2H_5OH \)) and methyl iodide (\( CH_3I \)).

3. **Mechanism of the Reaction**:
- The oxygen atom in the ether has lone pairs and can protonate when it reacts with \( H^+ \) from HI, leading to the formation of a positively charged intermediate.
- The bond between the oxygen and the methyl group breaks, leading to the formation of a methyl cation (\( CH_3^+ \)).
- However, the methyl cation is unstable, and therefore, the iodide ion (\( I^- \)) attacks the methyl group from the backside, resulting in the formation of methyl iodide (\( CH_3I \)).
- The ethyl group remains as ethanol (\( C_2H_5OH \)).

4. **Determine the Type of Reaction**:
- The reaction mechanism here is not \( S_N1 \) because \( S_N1 \) involves the formation of a stable carbocation, which is not the case for the methyl group.
- Instead, this reaction follows an \( S_N2 \) mechanism, where the nucleophile (iodide ion) attacks the electrophile (methyl group) directly.

5. **Conclusion**:
- The assertion is true as the products \( C_2H_5OH \) and \( CH_3I \) are indeed formed.
- The reason given is false because the reaction follows an \( S_N2 \) mechanism, not \( S_N1 \).

### Final Answer:
- Assertion (A) is true.
- Reason (R) is false.