Q.Write the mechanism of the reaction of HI with methoxybenzene.

The key idea is that HI cleaves the ether bond in methoxybenzene via an SN2 mechanism at the methyl carbon, not at the aromatic ring.

Step 1 – Protonation: The oxygen of methoxybenzene is protonated by HI, forming a good leaving group (methanol-like group attached to the ring).

Step 2 – Nucleophilic attack: The iodide ion ($\text{I}^-$) attacks the methyl carbon in an SN2 fashion. This is possible because the methyl carbon is primary and sterically unhindered. …

The reaction of hydrogen iodide with methoxybenzene (anisole) is a classic example of ether cleavage under acidic conditions. Most students instinctively think of electrophilic aromatic substitution when they see an aromatic ring with a strong acid, but that’s a trap here. The real action happens at the methoxy group, not on the ring.

Why does this happen? The oxygen in anisole has lone pairs that are basic. In the presence of a strong acid like HI, the oxygen gets protonated first. This turns the –OCH₃ group into a much better leaving group — instead of an alkoxide (which is a poor leaving group), you now have a neutral methanol molecule attached to the ring. You might object that methanol itself isn’t a great leaving group either — the real trick is that iodide ion (I⁻) is an excellent nucleophile and a weak base, so it attacks the methyl carbon in an SN2 fashion. The aromatic ring is too bulky and too stable to be attacked directly — the methyl group is the soft target.

Let’s walk through the mechanism step by step.

1. Protonation of the oxygen The lone pair on the oxygen of methoxybenzene abstracts a proton from HI. This gives a protonated anisole intermediate — an oxonium ion.

The oxygen now bears a positive charge, making the C–O bond much more polarised and weaker.

2. Nucleophilic attack by iodide on the methyl carbon The iodide ion, present in solution, acts as a strong nucleophile. It attacks the methyl carbon from the back side (SN2 mechanism). The C–O bond breaks heterolytically, with the electron pair going to the oxygen (neutralising its charge).

This step is the key cleavage event. The products are phenol and iodomethane.

3. Proton transfer (if needed) The phenol formed is already neutral, so no further deprotonation is required. The reaction stops here. …

Concept: Williamson Ether Synthesis (and its Reverse — Ether Cleavage)

The Williamson ether synthesis is a method to prepare ethers via an SN2 reaction between an alkoxide ion and a primary alkyl halide.

However, the question here asks for the reverse reaction: the acidic cleavage of an ether using HI.

The reaction of HI with methoxybenzene (anisole) is not a Williamson synthesis — it is an ether cleavage under acidic conditions. The mechanism follows an SN1 or SN2 pathway depending on the alkyl group.

Method: Acid-Catalyzed Ether Cleavage (SN2 Mechanism for Methyl Aryl Ethers)

Why this method?

• Methoxybenzene has a methyl group attached to oxygen.
• Methyl groups undergo SN2 easily (no steric hindrance).
• The aryl–O bond is strong (resonance stabilization), so cleavage occurs at the alkyl–O bond.

Step-by-Step Mechanism

Step 1: Protonation of the ether oxygen

The lone pair on oxygen attacks a proton from HI, forming a protonated ether (oxonium ion).

Step 2: SN2 attack by iodide ion

The iodide ion ($\ce{I-}$) attacks the methyl carbon from the back, displacing the oxygen.

This step breaks the C–O bond on the methyl side, not the aryl side. …

Common Mistakes in Williamson Ether Synthesis (HI + Methoxybenzene)

This question tests your understanding of ether cleavage — specifically, how HI reacts with an aryl methyl ether (methoxybenzene). Students often confuse this with the formation of ethers (the actual Williamson Ether Synthesis). Let's break down the key errors.

Mistake 1: Confusing Ether Formation with Ether Cleavage

The error: Students write the mechanism for making an ether (e.g., $CH_3ONa + C_6H_5Br \rightarrow C_6H_5OCH_3$) instead of breaking it.

Why it happens: The question says "Williamson Ether Synthesis" in the heading, but the actual task is the reverse reaction — acid-catalyzed cleavage.

How to avoid:

• Read carefully: "Reaction of HI with methoxybenzene" means cleavage, not synthesis.
• Remember: HI (a strong acid) protonates the ether oxygen, then $I^-$ attacks.
• The product is phenol ($C_6H_5OH$) and iodomethane ($CH_3I$), not an ether.

Mistake 2: Forgetting the Aromatic Ring Doesn't Undergo $S_N2$

The error: Students show $I^-$ attacking the aryl carbon (the carbon attached to the benzene ring), writing $C_6H_5I$ as a product.

Why it happens: They treat the aryl-oxygen bond like an alkyl-oxygen bond.

How to avoid:

• Aryl halides do not form easily via $S_N2$ — the $sp^2$ carbon is too hindered and the $\pi$ system resists backside attack.
• The correct attack is on the methyl carbon ($CH_3$), which is $sp^3$ and unhindered.
• Result: $C_6H_5OH$ (phenol) + $CH_3I$ (iodomethane).

Mistake 3: Writing the Wrong Mechanism Steps

The error: Students skip protonation or show direct $S_N2$ on the neutral ether.

Why it happens: They forget that HI provides both $H^+$ (to activate the ether) and $I^-$ (the nucleophile).

How to avoid — write these steps in order:

1. Protonation of the ether oxygen by $H^+$ from HI:

2. Nucleophilic attack by $I^-$ on the methyl carbon (not the aryl carbon):

3. Deprotonation (if needed): The phenol product may lose $H^+$ in basic workup, but in acidic HI, it remains as phenol.

Mistake 4: Ignoring the Role of the Aromatic Ring

The error: Students treat methoxybenzene like a simple aliphatic ether (e.g., $CH_3OCH_3$).

Why it happens: They don't consider resonance stabilization of the aryl-oxygen bond.

How to avoid:

• The lone pair on oxygen is delocalized into the benzene ring, making the $C_6H_5-O$ bond stronger than a typical $C-O$ bond.
• This is why $I^-$ attacks the methyl group — breaking the $CH_3-O$ bond is easier than breaking the $C_6H_5-O$ bond. …