Q.Match the ethers given in Column I with the products of their reaction with HI given in Column II.
Column I:

The key idea is that ethers cleave with HI via an $S_N1$ or $S_N2$ mechanism depending on the alkyl group. The smaller alkyl group (or the one that can form a stable carbocation) ends up as the alkyl iodide.

Step 1: For $CH_3OCH_3$, both groups are primary — $S_N2$ attack by $I^-$ on either methyl gives $CH_3I + CH_3OH$.

Step 2: For $(CH_3)_2CHOCH_3$, the isopropyl group is secondary but the methyl is primary — $S_N2$ occurs at the less hindered methyl, giving $CH_3I + (CH_3)_2CHOH$. …

The Core Idea: Williamson Ether Synthesis in Reverse

When an ether reacts with HI, the bond breaks at the oxygen. The question is: which C–O bond breaks, and which alkyl group ends up as the alkyl iodide?

The answer depends entirely on the nature of the alkyl groups attached to oxygen. HI is a strong acid — it protonates the ether oxygen first, making it a good leaving group. Then the iodide ion ($I^-$) attacks one of the carbon atoms. The mechanism that follows is either S$_N$2 or S$_N$1, depending on whether that carbon is primary, secondary, tertiary, or benzylic.

Let’s apply this rule to each ether.

1. $CH_3OCH_3$ — dimethyl ether

Both alkyl groups are identical — methyl, which is primary. There is no ambiguity. The protonated ether is $CH_3\overset{+}{O}(H)CH_3$. Iodide attacks either carbon via S$_N$2 (both are equally accessible). The products are methyl iodide and methanol.

This matches option (d).

2. $(CH_3)_2CHOCH_3$ — isopropyl methyl ether

Here we have two different alkyl groups: isopropyl (secondary) and methyl (primary). The protonated ether is $(CH_3)_2\overset{+}{O}(H)CH_3$.

Iodide can attack either carbon. But the S$_N$2 attack on the methyl carbon is much faster — methyl is unhindered and primary. Attack on the isopropyl carbon would be S$_N$2 as well, but secondary carbons are slower due to steric hindrance. So the dominant pathway is:

This gives isopropyl alcohol and methyl iodide — option (c).

3. tert-butyl methyl ether

Now we have a tertiary group (tert-butyl) and a primary group (methyl). This is the classic case where the mechanism switches.

The protonated ether is $(CH_3)_3C\overset{+}{O}(H)CH_3$. The tert-butyl group can form a very stable tertiary carbocation ($(CH_3)_3C^+$). So instead of S$_N$2, the reaction follows an S$_N$1 pathway: the C–O bond to the tert-butyl group breaks first, giving the tert-butyl carbocation and $CH_3OH$. The carbocation is then trapped by $I^-$ to give tert-butyl iodide.

This matches option (b). …

Method: SN1 vs SN2 Pathway in Acidic Cleavage of Ethers (HI Reaction)

This problem uses the mechanistic rule for ether cleavage by HI:

In acidic cleavage of unsymmetrical ethers, the smaller alkyl group leaves as the alkyl halide (via SN2), while the bulkier or more substituted group forms the alcohol (or, if tertiary, the halide via SN1).

Steps to solve:

1. Identify the two alkyl groups attached to oxygen in each ether.

2. Determine which group is smaller/less hindered (will undergo SN2 with I⁻) and which is bulkier/more substituted (will form the alcohol, or if tertiary, the carbocation → halide).

3. Apply the rule:

   • Primary or methyl → forms alkyl iodide (SN2).
   • Secondary → usually forms alcohol (unless special cases).
   • Tertiary → forms alkyl iodide via SN1 (carbocation intermediate).

Applying to each ether:

(i) $CH_3OCH_3$ (dimethyl ether)

• Both groups are methyl (identical).
• One methyl becomes $CH_3I$, the other becomes $CH_3OH$.
• Product: $CH_3I + CH_3OH$ → matches (d)

(ii) $(CH_3)_2CHOCH_3$ (isopropyl methyl ether)

• Groups: isopropyl (secondary, bulkier) and methyl (small).
• Methyl → $CH_3I$ (SN2).
• Isopropyl → $(CH_3)_2CHOH$ (alcohol).
• Product: $CH_3I + (CH_3)_2CHOH$ → matches (c)

(iii) tert-butyl methyl ether

• Groups: tert-butyl (tertiary, very bulky) and methyl (small).
• Methyl → $CH_3I$ (SN2).
• Tert-butyl → forms tert-butyl carbocation → $(CH_3)_3C^+$ → combines with I⁻ to give $(CH_3)_3CI$ (SN1).
• Product: $(CH_3)_3CI + CH_3OH$ → matches (b)

(iv) $C_6H_5OCH_3$ (anisole, phenyl methyl ether) …

✗ Mistake 1: Forgetting the mechanism — assuming all C–O bonds break equally

Why it happens:

Students treat all ethers the same and guess products based only on which alkyl group is “bigger.”

How to avoid:

Recall that HI cleaves ethers via an $S_N2$ mechanism (for primary alkyl groups) or $S_N1$ mechanism (for tertiary alkyl groups).

• The smaller alkyl group usually leaves as the alkyl halide in $S_N2$.
• The more stable carbocation forms in $S_N1$ (tertiary > secondary > primary > methyl).

Example:

For $(CH_3)_2CHOCH_3$ (isopropyl methyl ether):

• Methyl is smaller → $CH_3I$ forms via $S_N2$ on methyl.
• Isopropyl group stays as alcohol: $(CH_3)_2CHOH$. ✓ Correct match: (ii) → (c)

✗ Mistake 2: Confusing tert-butyl methyl ether with isopropyl methyl ether

Why it happens:

Both have a branched alkyl group, so students think the same rule applies.

How to avoid:

Tert-butyl is tertiary — it forms a stable carbocation.

• HI protonates the ether oxygen.
• The tert-butyl group leaves as a carbocation → forms $(CH_3)_3C^+$ → reacts with $I^-$ to give $(CH_3)_3CI$.
• The methyl group stays as $CH_3OH$.

✓ Correct match: (iii) → (b)

✗ Mistake 3: Ignoring aryl alkyl ethers — thinking the aryl group can also leave

Why it happens:

Students try to apply the same $S_N2$ logic to $C_6H_5OCH_3$.

How to avoid:

Aryl groups cannot undergo $S_N2$ or $S_N1$ because the C–O bond has partial double-bond character (resonance).

• Only the alkyl group (methyl) is cleaved.
• Product: $C_6H_5OH$ (phenol) + $CH_3I$.

✓ Correct match: (iv) → (a)

✗ Mistake 4: Mixing up which product is the alcohol and which is the alkyl halide

Why it happens:

Students write the products but swap the alcohol and halide.

How to avoid:

Remember:

• The oxygen stays with the larger / more substituted group (or the aryl group). …