Consider the following reaction}
Statement I:} In the above reaction, product formed will be a mixture of benzyl alcohol and iodobenzene.
Statement II:} In the above reaction, the \(-O-CH_2-\) bond is cleaved to give the product.
Choose the correct answer.
Step 1: Understanding the Question:
The question asks about the products and mechanism of the cleavage of an ether (Benzyl phenyl ether) by Hydroiodic acid (HI). Step 2: Key Formula or Approach:
Ether cleavage with HI follows \(S_{N}1\) or \(S_{N}2\) mechanisms depending on the nature of the alkyl groups.
If one group can form a very stable carbocation (like benzyl or tertiary butyl), the reaction follows the \(S_{N}1\) pathway at that side. Step 3: Detailed Explanation:
The reactant is Benzyl phenyl ether: \(C_{6}H_{5}-O-CH_{2}-C_{6}H_{5}\).
1. Protonation: The oxygen atom is protonated by \(H^{+}\) from HI.
2. Bond Cleavage: We have two potential bonds to break: \(O-Phenyl\) or \(O-Benzyl\).
- The \(O-Phenyl\) bond has partial double bond character due to resonance with the benzene ring and is very strong. It is not cleaved.
- The \(O-Benzyl\) bond is cleaved. The benzyl group can form a highly stable benzyl carbocation (\(C_{6}H_{5}CH_{2}^{+}\)). Thus, the reaction follows an \(S_{N}1\) mechanism at the benzyl side.
3. Products:
- The oxygen stays with the phenyl ring to form Phenol (\(C_{6}H_{5}OH\)).
- The benzyl cation reacts with \(I^{-}\) to form Benzyl iodide (\(C_{6}H_{5}CH_{2}I\)).
Evaluating Statements:
- Statement I says products are benzyl alcohol and iodobenzene. This is incorrect. (Phenol and benzyl iodide are formed).
- Statement II says the \(-O-CH_{2}-\) bond is cleaved. This is correct as it is the benzyl-oxygen bond. Step 4: Final Answer:
Statement I is false, Statement II is true.
