Question

Which one of the following is the major product formed when the given reaction occurs?
\[ \text{C}_6H_5\text{OCH}_2\text{CH}_3 + \text{excess conc. HCl} \xrightarrow{373K} \]

• A. A
• B. B
• C. C
• D. D

Hint:

When an alcohol undergoes an SN1 reaction with a strong acid, the hydroxyl group (OH) is replaced by a halide ion e.g., I\(^{−}\), Cl\(^{−}\) to form an alkyl halide.

Answer 1

The Correct Option is D

Step 1: Identify the reactants and reaction conditions.
We are given the reaction of an alcohol \( C_6H_5OCH_2CH_3 \) (phenylethanol) with excess concentrated HCl at 373 K. This reaction involves the substitution of the hydroxyl group (-OH) by a halogen atom, specifically an iodine (I), under acidic conditions.

Step 2: Understand the reaction mechanism.
The reaction is a nucleophilic substitution reaction (SN1 or SN2), but since excess concentrated HCl is used and the temperature is 373 K (which favors the formation of a carbocation), this is an SN1 mechanism. In the SN1 mechanism, the first step is the departure of the leaving group (OH\(^-\)), which forms a carbocation. The carbocation is then attacked by the chloride ion (Cl\(^-\)), replacing the hydroxyl group.

Step 3: Analyze the possible products.
In this case, the alcohol \( C_6H_5OCH_2CH_3 \) is likely to undergo substitution, where the hydroxyl group (-OH) is replaced by an iodine atom (I), and the major product will be \( \text{CH}_3\text{I} \). This forms the final product:
\[ \text{CH}_3\text{I} \]

Step 4: Conclusion.
The correct major product formed in this reaction is \( \text{CH}_3\text{I} \), and the correct answer is option (D).