Q.Give the major products that are formed by heating each of the following ethers with HI.

The key idea is that HI cleaves ethers: when both alkyl groups are primary/methyl, $I^-$ attacks the less hindered carbon ($S_N2$); a tertiary or benzylic side instead forms a stable carbocation ($S_N1$) and takes the iodine directly, while the oxygen leaves with the other side as the alcohol.

(i) $CH_3-CH_2-CH(CH_3)-CH_2-O-CH_2-CH_3$

The less hindered side is the ethyl group ($-CH_2CH_3$). $S_N2$ attack by $\ce{I-}$ gives ethyl iodide and 2-methylbutan-1-ol. The alcohol does not react further (primary alcohol).

(ii) $CH_3-CH_2-CH_2-O-C(CH_3)_2-CH_2CH_3$

The carbon on the right of the oxygen is tertiary, so cleavage follows $S_N1$: the tertiary carbocation forms and captures $I^-$, giving 2-iodo-2-methylbutane, while the oxygen leaves with the propyl group as propan-1-ol. Being primary, propan-1-ol reacts only sluggishly with HI and is isolated as the alcohol.

(iii) $C_6H_5-CH_2-O-C_6H_5$ …

Concept & Intuition

Heating an ether with concentrated HI is the classic acidic cleavage reaction. The mechanism is a two-step nucleophilic substitution: first the ether oxygen is protonated by HI, making it a good leaving group (as a neutral alcohol molecule). Then the iodide ion ($I^-$) attacks one of the carbon atoms adjacent to the oxygen.

The critical decision is which C–O bond breaks. This depends entirely on the structure of the alkyl groups attached to the oxygen.

• If both groups are primary or methyl, the reaction follows $S_N2$: the iodide attacks the less hindered (less substituted) carbon.
• If one group is tertiary, benzylic, or allylic, that carbon can form a relatively stable carbocation, so the reaction follows $S_N1$: the C–O bond breaks to give that carbocation, which is then trapped by iodide. In this case, the more substituted (or resonance-stabilised) carbon gets the iodine.

(i) $CH_3-CH_2-CH(CH_3)-CH_2-O-CH_2-CH_3$

Step 1: Identify the alkyl groups.

The ether is:

Left side: $CH_3-CH_2-CH(CH_3)-CH_2-$ — this is a primary carbon (the carbon directly attached to oxygen is a $CH_2$ group, even though the chain has a branch further away).

Right side: $CH_3-CH_2-$ — this is also primary (ethyl group).

Step 2: Decide the mechanism.

Both groups are primary. No tertiary, benzylic, or allylic carbons. So the reaction proceeds via $S_N2$.

Step 3: Which bond breaks?

In $S_N2$, the iodide attacks the less hindered primary carbon. The right-side ethyl carbon is less hindered than the left-side carbon (which has a branched chain nearby), so the iodide attacks the ethyl carbon, breaking the $CH_3CH_2–O$ bond.

Step 4: Write the products.

The oxygen stays with the more substituted fragment (the branched chain) as an alcohol, and the ethyl group leaves as ethyl iodide.

The resulting alcohol, 2-methylbutan-1-ol, is primary — primary alcohols react only slowly with HI (the substitution is comparatively sluggish), so under the conditions that cleave the ether it is normally isolated as the free alcohol rather than being converted on to a second iodide.

Final products:

$CH_3CH_2CH(CH_3)CH_2OH$ (2-methylbutan-1-ol) and $CH_3CH_2I$ (iodoethane).

(ii) $CH_3-CH_2-CH_2-O-C(CH_3)_2-CH_2CH_3$

Step 1: Identify the alkyl groups.

Left side: $CH_3CH_2CH_2-$ — this is a primary carbon (propyl).

Right side: $–C(CH_3)_2–CH_2CH_3$ — the carbon directly attached to oxygen is a tertiary carbon (it has three other carbon substituents: two methyls and one ethyl).

Step 2: Decide the mechanism.

One group is tertiary. The tertiary carbon can form a stable tertiary carbocation. So the reaction follows $S_N1$: the C–O bond breaks to give the tertiary carbocation, which is then attacked by iodide.

Step 3: Which bond breaks?

The bond that breaks is the one that gives the more stable carbocation — the tertiary carbon–oxygen bond. So the oxygen stays with the primary propyl group (as an alcohol), and the tertiary group becomes the iodide.

Step 4: Write the products.

First cleavage:

$CH_3CH_2CH_2–O–C(CH_3)_2CH_2CH_3 + HI \rightarrow CH_3CH_2CH_2OH + (CH_3)_2C(I)CH_2CH_3$

The resulting propan-1-ol is primary — like 2-methylbutan-1-ol in part (i), it reacts only sluggishly with HI under these conditions, so it is isolated as the free alcohol rather than being converted on to a second iodide.

Final products:

$CH_3CH_2CH_2OH$ (propan-1-ol) and $(CH_3)_2C(I)CH_2CH_3$ (2-iodo-2-methylbutane). …

Method: Acid-Catalyzed Cleavage of Ethers (via S$_N$1 / S$_N$2 Mechanism)

This is not Williamson Ether Synthesis — it is the reverse reaction: ether cleavage using HI (a strong acid). The method is based on protonation of the ether oxygen, followed by nucleophilic attack by $\ce{I-}$.

General Principle

• Step 1: Ether oxygen is protonated by HI, making it a good leaving group (as $\ce{ROH}$ or $\ce{ROH2+}$).
• Step 2: $\ce{I-}$ (a strong nucleophile) attacks the less hindered carbon (S$_N$2) or the carbon that can form a stable carbocation (S$_N$1).
• Step 3: The products are alkyl iodides and alcohols (which may further react with excess HI to give more alkyl iodide).

(i) $CH_3-CH_2-CH(CH_3)-CH_2-O-CH_2-CH_3$

Name of ether: 1-ethoxy-2-methylbutane (unsymmetrical)

Step-by-step reasoning:

1. Protonate the oxygen → $\ce{R-O+(H)-R'}$
2. Two possible cleavage sites:
   • Path A (S$_N$2 at less hindered carbon): $\ce{I-}$ attacks the ethyl group ($\ce{-CH2CH3}$) — primary carbon → gives $\ce{CH3CH2I}$ (ethyl iodide) and $\ce{CH3CH2CH(CH3)CH2OH}$ (2-methylbutan-1-ol)
   • Path B (S$_N$2 at more hindered carbon): $\ce{I-}$ attacks the 2-methylbutyl group — also a primary carbon, but crowded by the adjacent branch → slower, less favored
3. The alcohol formed can react with excess HI to give the corresponding alkyl iodide.

Major products:

• $\mathbf{CH_3CH_2I}$ (ethyl iodide)
• $\mathbf{CH_3CH_2CH(CH_3)CH_2OH}$ (2-methylbutan-1-ol) (with excess HI, this alcohol converts to $\ce{CH3CH2CH(CH3)CH2I}$)

(ii) $CH_3-CH_2-CH_2-O-C(CH_3)_2-CH_2CH_3$

Name of ether: 2-methyl-2-propoxybutane (unsymmetrical, one side tertiary)

Step-by-step reasoning:

1. Protonate oxygen.
2. Two possible cleavages:
   • Path A (S$_N$1 at tertiary carbon): The $\ce{-C(CH3)2CH2CH3}$ group can form a tertiary carbocation (very stable) → $\ce{I-}$ attacks → $\ce{(CH3)2C(I)CH2CH3}$ (2-iodo-2-methylbutane) and $\ce{CH3CH2CH2OH}$ (propan-1-ol)
   • Path B (S$_N$2 at primary carbon): $\ce{I-}$ attacks the propyl group → $\ce{CH3CH2CH2I}$ (1-iodopropane) and $\ce{(CH3)2C(OH)CH2CH3}$ (2-methylbutan-2-ol) — but tertiary alcohol is less stable under acidic conditions
3. S$_N$1 path dominates because tertiary carbocation is highly stabilized.

Major products:

• $\mathbf{(CH_3)_2C(I)CH_2CH_3}$ (2-iodo-2-methylbutane)
• $\mathbf{CH_3CH_2CH_2OH}$ (propan-1-ol) (with excess HI, propan-1-ol → $\ce{CH3CH2CH2I}$)

(iii) $C_6H_5-CH_2-O-C_6H_5$ …

Here are the common mistakes students make in acidic cleavage of ethers with HI problems and how to avoid each.

Mistake 1: Forgetting the Mechanism (S$_N$1 vs S$_N$2)

Students often guess the products without checking whether an alkyl group can form a stable carbocation.

• The rule: HI cleaves ethers via S$_N$2 on the less hindered carbon unless a stable carbocation (3° or benzylic) can form — then it switches to S$_N$1, and the carbocation side takes the iodine while the oxygen leaves with the other side as the alcohol.
• How to avoid: Always check the substitution pattern of each carbon attached to oxygen. If one side is 3° or benzylic, that side becomes the alkyl iodide (via its carbocation), not the alcohol.

Example (ii):

$CH_3-CH_2-CH_2-O-C(CH_3)_2-CH_2CH_3$

• Left side: 1° — the oxygen stays here → propan-1-ol
• Right side: 3° (S$_N$1) → carbocation → captured by $I^-$ Correct: $(CH_3)_2C(I)CH_2CH_3$ + $CH_3CH_2CH_2OH$ Common wrong answer: $CH_3CH_2CH_2I$ + $(CH_3)_2C(OH)CH_2CH_3$ (the split reversed).

Mistake 2: Ignoring the Less Hindered Side for S$_N$2

Even when both sides are primary, students sometimes pick the wrong carbon for iodide attack.

• The rule: In S$_N$2, the iodide ion attacks the less sterically hindered carbon — that carbon gets the iodine; the bulkier fragment keeps the oxygen as the alcohol.
• How to avoid: Draw the ether and compare the two carbons bonded to oxygen. The smaller/less crowded one becomes the alkyl iodide.

Example (i):

$CH_3-CH_2-CH(CH_3)-CH_2-O-CH_2-CH_3$

• Left side: 1° carbon, but crowded by the adjacent branch
• Right side: 1° ethyl carbon — less hindered → gets the iodine Correct: $CH_3CH_2I$ + $CH_3CH_2CH(CH_3)CH_2OH$ (2-methylbutan-1-ol) Common wrong answer: $CH_3CH_2CH(CH_3)CH_2I$ + $CH_3CH_2OH$ (the split reversed).

Mistake 3: Forgetting the Phenol Exception with Aryl Ethers

Students treat aryl groups like alkyl groups and apply S$_N$2 directly.

• The rule: Aryl–O bonds are very strong (partial double-bond character). HI cannot break the C–O bond on the aromatic ring. Only the alkyl/benzylic side is cleaved.
• How to avoid: If one side is aryl (C$_6$H$_5$–), that bond never breaks — the ring keeps the oxygen and ends up as phenol.

Example (iii):

$C_6H_5-CH_2-O-C_6H_5$

• Left side: benzylic (stable carbocation → S$_N$1) → benzyl iodide
• Right side: aryl (does not break) → phenol Correct: $C_6H_5CH_2I$ + $C_6H_5OH$ Common wrong answer: $C_6H_5I$ + $C_6H_5CH_2OH$ (impossible — the aryl–O bond is not cleaved).

Mistake 4: Missing Carbocation Rearrangement in the S$_N$1 path

When a carbocation forms, students sometimes forget it can rearrange (hydride or alkyl shift) to a more stable one.

• The rule: If the carbocation can rearrange to a more stable one (e.g., 2° → 3°), the alkyl iodide comes from the rearranged carbocation — rearrangement changes the iodide product, because the carbocation is what the iodide attacks.
• How to avoid: Draw the carbocation formed after C–O cleavage and check for a 1,2-hydride or alkyl shift before writing the iodide.

Example (ii) revisited:

The 3° carbocation $(CH_3)_2C^+CH_2CH_3$ is already the most stable — no rearrangement needed. But if the ether had a 2° group that could become 3° via a shift, many students miss that.

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