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

The key idea is that HI cleaves ethers via an S$_N$2 mechanism, because iodide is an excellent nucleophile and the methyl group has no steric hindrance.

Step 1: Protonation. The oxygen of methoxymethane ($\mathrm{CH_3-O-CH_3}$) is protonated by HI, forming a good leaving group (methanol) and a methyloxonium ion.

Step 2: S$_N$2 attack. The iodide ion attacks one of the methyl carbons from the back, displacing methanol in a single step.

$\mathrm{I^- + CH_3-\overset{+}{O}(H)-CH_3 \rightarrow CH_3I + CH_3OH}$ …

The reaction of an ether with hydrogen iodide is a classic example of acid-catalysed cleavage of ethers. The key idea is that HI is a strong acid and also a source of the excellent nucleophile $\ce{I-}$. For a symmetrical ether like methoxymethane ($\ce{CH3-O-CH3}$), the reaction proceeds cleanly through an S$_N$2 pathway.

Why does this work so well? The oxygen atom in the ether is basic — it gets protonated by HI, turning the poor leaving group ($\ce{-OR}$) into an excellent one ($\ce{-OHR+}$). Once the oxygen is positively charged, the carbon-oxygen bond becomes highly polarised and vulnerable to attack. Iodide is both a strong nucleophile and a weak base, making it perfect for S$_N$2 attack without causing elimination.

Let's walk through the mechanism step by step.

1. Protonation of the ether oxygen

The lone pairs on oxygen make methoxymethane a weak base. In the presence of strong acid HI, the oxygen gets protonated:

This step is fast and reversible. The protonated ether now has a positively charged oxygen, which makes the carbon-oxygen bonds much weaker — the oxygen wants to leave as a neutral molecule (methanol) rather than as an alkoxide ion.

2. S$_N$2 attack by iodide on one methyl carbon

Iodide ion, present from the first step, acts as a nucleophile. It attacks the less-hindered carbon of the protonated ether — in this case, either methyl carbon, since both are identical. The attack occurs from the back side, inverting the configuration at that carbon (though for a methyl group, this stereochemical detail is irrelevant).

As the $\ce{C-O}$ bond breaks, the oxygen leaves as a neutral methanol molecule:

At this point, we have one molecule of iodomethane and one molecule of methanol.

3. The methanol formed also reacts with HI …

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

The Williamson Ether Synthesis is a method to prepare ethers via an $S_N2$ reaction between an alkoxide ion and a primary alkyl halide.

The reverse reaction — acidic cleavage of ethers — is what happens when an ether reacts with a strong acid like HI. This is not Williamson synthesis itself, but a key related reaction often tested alongside it.

Method: Acidic Cleavage of Ethers (via $S_N2$ mechanism)

Reaction:

Step-by-Step Mechanism

Step 1: Protonation of the ether oxygen

The lone pair on oxygen attacks the proton ($\text{H}^+$) from HI, forming a protonated ether (oxonium ion).

Why? Oxygen is basic; the proton makes the $\text{C-O}$ bond a better leaving group (as neutral $\text{CH}_3\text{OH}$).

Step 2: $S_N2$ attack by iodide ion

The iodide ion ($\text{I}^-$) attacks the less hindered carbon (here, both are methyl — equally hindered) in an $S_N2$ fashion, displacing methanol.

$\text{I}^- + \text{CH}_3\text{OH}^+\text{CH}_3 \xrightarrow{\text{slow}} \text{CH}_3\text{I} + \text{CH}_3\text{OH}$ …

Here are the most common mistakes students make when tackling the Williamson Ether Synthesis mechanism (specifically the cleavage of methoxymethane with HI), along with clear strategies to avoid them.

1. Forgetting that this is Acid-Catalyzed Cleavage, not Synthesis

The biggest conceptual error is treating this as the formation of an ether (Williamson synthesis) rather than its cleavage.

• The Mistake: Students try to write an $\text{S}_\text{N}2$ mechanism where the alkoxide ion attacks an alkyl halide.
• Why it’s wrong: The reaction is $\text{HI}$ with an ether. The ether is the substrate, not the product. The $\text{H}^+$ from HI protonates the ether oxygen first, making it a good leaving group.
• How to Avoid: Always check the reagents. If you see a strong acid (HI, HBr) + an ether, you are doing ether cleavage, not synthesis. The first step is always protonation.

2. Skipping the Protonation Step

Students often jump straight to bond-breaking without showing why the oxygen becomes a leaving group.

• The Mistake: Writing $\text{CH}_3\text{OCH}_3 + \text{I}^- \rightarrow \text{CH}_3\text{I} + \text{CH}_3\text{O}^-$ directly.
• Why it’s wrong: $\text{CH}_3\text{O}^-$ is a terrible leaving group (strong base). The reaction won’t happen without protonation.
• How to Avoid: Always draw the first step as:

This creates a good leaving group (neutral methanol).

3. Choosing the Wrong Carbon for $\text{S}_\text{N}2$ Attack

Methoxymethane is a symmetrical ether, but students still get confused about which carbon the iodide attacks.

• The Mistake: Attacking the carbon that is not the one bearing the leaving group, or attacking the oxygen directly.
• Why it’s wrong: The $\text{S}_\text{N}2$ attack must occur on the carbon that was bonded to the protonated oxygen. In methoxymethane, both carbons are identical, so it doesn’t matter which one you pick — but you must pick one and show the $\text{I}^-$ attacking it.
• How to Avoid: After protonation, identify the two $\text{CH}_3$ groups. The $\text{I}^-$ will do a backside attack on one of them, displacing the neutral methanol ($\text{CH}_3\text{OH}$).

4. Forgetting the Second Equivalent of HI

The reaction produces methanol, but methanol itself reacts further with HI.

• The Mistake: Stopping the mechanism after forming $\text{CH}_3\text{I} + \text{CH}_3\text{OH}$.
• Why it’s wrong: Methanol is a primary alcohol and will undergo the same acid-catalyzed $\text{S}_\text{N}2$ reaction with excess HI to give $\text{CH}_3\text{I}$ and water.
• How to Avoid: Write the second cycle:
  1. Protonate the $\text{OH}$ of methanol.
  2. $\text{I}^-$ attacks the $\text{CH}_3$ group.
  3. Final products: 2 moles of $\text{CH}_3\text{I}$ and $\text{H}_2\text{O}$.

5. Drawing Incorrect Arrow Pushing (Curly Arrows)

Arrow pushing is the language of mechanisms — mistakes here lose marks. …