Question

The given reaction involves benzyl chloride ($\text{C}_6\text{H}_5\text{CH}_2\text{Cl}$) reacting with methanol ($\text{CH}_3\text{OH}$). What is the product of this reaction?

Answer 1

Benzyl methyl ether

Answer 2

The reaction of benzyl chloride with methanol is a nucleophilic substitution reaction ($\text{S}_{\text{N}}1$).

1. Formation of Benzylic Carbocation: The chlorine atom leaves as a chloride ion, forming a stable benzylic carbocation:

   $$\text{C}_6\text{H}_5\text{CH}_2\text{Cl} \xrightarrow{\text{Slow}} \text{C}_6\text{H}_5\text{CH}_2^+ + \text{Cl}^-$$

   The benzylic carbocation is resonance-stabilized by the benzene ring.

2. Nucleophilic Attack: The oxygen atom of methanol attacks the positively charged benzylic carbon:

   $$\text{C}_6\text{H}_5\text{CH}_2^+ + \text{CH}_3\text{OH} \xrightarrow{\text{Fast}} \text{C}_6\text{H}_5\text{CH}_2-\text{O}^+(\text{H})\text{CH}_3$$

3. Deprotonation: The protonated ether intermediate loses a proton, yielding benzyl methyl ether:

   $$\text{C}_6\text{H}_5\text{CH}_2-\text{O}^+(\text{H})\text{CH}_3 + \text{CH}_3\text{OH} \longrightarrow \text{C}_6\text{H}_5\text{CH}_2-\text{O}-\text{CH}_3 + \text{CH}_3\text{OH}_2^+$$

The overall reaction replaces the chlorine atom with a methoxy ($\text{-OCH}_3$) group, forming benzyl methyl ether.