Ethers — Williamson synthesis, cleavage by HI: Exercises

Definition First: what do "hindered / unhindered" and "

$S_{N} 1$ vs $S_{N} 2$ " actually mean? How crowded a carbon is (its substitution class) is counted by how many other carbons are directly bonded to it:

Methyl: the carbon carries $0$ other carbons ( $C H_{3} -$ ). Totally open.
Primary (1°): $1$ other carbon (e.g. $C H_{3} C H_{2} -$ ). Nearly open.
Secondary (2°): $2$ other carbons (e.g. $(C H_{3})_{2} C H -$ ). Half-blocked.
Tertiary (3°): $3$ other carbons (e.g. $(C H_{3})_{3} C -$ ). Fully walled off. So "unhindered/small" = methyl or 1°; "hindered/bulky" = 3° (2° sits in between). This is structural, not vague — just count the carbons on the reacting carbon.

$S_{N} 2$ (bimolecular, concerted): the nucleophile and substrate both appear in the rate law, $rate = k [Nu] [substrate]$ . Bond-making and bond-breaking happen in one step, with the nucleophile hitting the backside — so crowding (2°, 3°) slows or kills it. Favours methyl/1°. $S_{N} 1$ (unimolecular, stepwise): rate $= k [substrate]$ only; the leaving group departs first to make a carbocation, then the nucleophile adds. Favours carbons that give a stable cation — 3°, benzylic, allylic. The figure below sketches both mechanisms side by side as energy profiles.

Alt-text: two reaction-energy curves. Left ( $S_{N} 2$ ): a single smooth hump — one transition state with the nucleophile, carbon and leaving group all partly bonded — going straight from reactants to products. Right ( $S_{N} 1$ ): a tall first hump up to a high-energy carbocation valley, then a small second hump down to products — two steps, showing the cation intermediate.

Level 1 — Recognition

Problem 1.1

Classify each as a symmetrical or unsymmetrical (mixed) ether: (a) $C H_{3} - O - C H_{3}$ (b) $C_{2} H_{5} - O - C H_{3}$ (c) $C_{6} H_{5} - O - C_{6} H_{5}$ (d) $C_{6} H_{5} - O - C H_{3}$

Recall Solution

"Symmetrical" means the two carbon groups on oxygen are identical.

(a) $C H_{3} - O - C H_{3}$ — both sides methyl → symmetrical.
(b) $C_{2} H_{5} - O - C H_{3}$ — ethyl vs methyl → unsymmetrical.
(c) $C_{6} H_{5} - O - C_{6} H_{5}$ — both phenyl → symmetrical (diphenyl ether).
(d) $C_{6} H_{5} - O - C H_{3}$ — phenyl vs methyl → unsymmetrical (anisole).

Problem 1.2

Name the mechanism of the Williamson ether synthesis, and name the two reactants by type.

Recall Solution

Mechanism: == $S_{N} 2$ == (bimolecular nucleophilic substitution — one concerted step, rate depends on both reactants). Reactants: an alkoxide $R - O^{-}$ (the nucleophile) and an alkyl halide $R^{'} - X$ (the electrophile, carrying the leaving group $X$ ).

Problem 1.3

Fill in: In HI cleavage the first step is always ______, because the neutral ether oxygen is a ______ leaving group.

Recall Solution

First step: protonation of the ether oxygen (by $H^{+}$ from HI). The neutral ether O is a bad leaving group; once protonated it can leave as a neutral alcohol, which is a good leaving group. (See Nucleophilicity and leaving group ability.)

Level 2 — Application

Problem 2.1

Give the Williamson reagents to make ethyl methyl ether, $C_{2} H_{5} - O - C H_{3}$ . Which combination is correct and why?

Recall Solution

Both carbons are unhindered (methyl has $0$ other carbons, ethyl is 1° with $1$ ), so either could be the halide. The better choice puts the halide on the least hindered carbon — the methyl (0 carbons attached). $C_{2} H_{5} O^{-} N a^{+} + C H_{3} I ⟶ C_{2} H_{5} - O - C H_{3} + N a I$ Methyl carbon is completely open to backside $S_{N} 2$ attack. (The reverse, $C H_{3} O^{-} + C_{2} H_{5} I$ , also works but the 1° centre is slightly more hindered.)

Problem 2.2

Write the products of $C H_{3} - O - C H_{3}$ (dimethyl ether) with excess HI.

Recall Solution

Protonate O; iodide attacks a methyl carbon ( $S_{N} 2$ ) → $C H_{3} I + C H_{3} O H$ . With excess HI the methanol is itself protonated and attacked by another $I^{-}$ → a second $C H_{3} I + H_{2} O$ : $C H_{3} - O - C H_{3} 2 H I 2 C H_{3} I + H_{2} O$

Problem 2.3

Predict the products of anisole $C_{6} H_{5} - O - C H_{3}$ with HI, and identify the C–O bond that breaks.

Recall Solution

The aryl C–O bond cannot break (no $S_{N} 1/ S_{N} 2$ at aromatic carbon). So iodide must strike the methyl carbon; the aryl fragment leaves as phenol. $C_{6} H_{5} - O - C H_{3} H I C_{6} H_{5} - O H + C H_{3} - I$ Product is phenol, never iodobenzene. See Phenols — properties and reactions.

Figure — Ethers — Williamson synthesis, cleavage by HI

Alt-text: the anisole molecule with its oxygen between a benzene ring (left) and a $C H_{3}$ (right). A red "✗" blocks the aryl–O bond (aromatic carbon can't be attacked); a green arrow shows $I^{-}$ hitting the methyl carbon. Products drawn: phenol on the left, $C H_{3} I$ on the right.

Level 3 — Analysis

Problem 3.1

You must make tert-butyl methyl ether, $(C H_{3})_{3} C - O - C H_{3}$ . Two routes exist on paper: Route A: $(C H_{3})_{3} C - O^{-} + C H_{3} I$ Route B: $C H_{3} O^{-} + (C H_{3})_{3} C - B r$ Pick the correct route and explain what goes wrong with the other.

Recall Solution

Route A is correct.

Route A attacks a methyl carbon (0 other carbons → fully open) with the hindered tert-butoxide as nucleophile → clean $S_{N} 2$ ether.
Route B tries $S_{N} 2$ on a 3° carbon (3 methyls attached) — those methyls block backside attack. Worse, the strong alkoxide base $C H_{3} O^{-}$ pulls off a β-hydrogen, giving E2 elimination → isobutylene, not ether. (See E2 Elimination.) $(C H_{3})_{3} C - O^{-} + C H_{3} I ⟶ (C H_{3})_{3} C - O - C H_{3} + N a I ✓$

Alt-text: two side-by-side carbon centres. Left: a methyl carbon ringed by three small H's; a green arrow ( $R O^{-}$ ) has a clear runway to the backside, iodide leaving on the far side → "S $_{N}$ 2 ✓". Right: a tertiary carbon ringed by three bulky $C H_{3}$ groups; a red arrow ( $R O^{-}$ ) is blocked by a wall → "E2 ✗", with an alkene drawn as the by-product. Read the green arrow as the clear path, the red arrow as the walled-off one.

Problem 3.2

The same ether $(C H_{3})_{3} C - O - C H_{3}$ is now treated with HI. Which C–O bond breaks, and what are the products? Contrast this with how you built it.

Recall Solution

Cleavage runs by $S_{N} 1$ : protonate O, then the 3° C–O bond breaks to give a stable tertiary carbocation $(C H_{3})_{3} C^{+}$ , which grabs $I^{-}$ . The methyl leaves as an alcohol. $C H_{3} - O - C (C H_{3})_{3} H I C H_{3} O H + (C H_{3})_{3} C - I$ Contrast: in synthesis the unhindered (methyl) carbon was attacked; in cleavage the iodide ends up on the hindered (3°) carbon. Building follows $S_{N} 2$ (open carbon wins); cutting a 3° ether follows $S_{N} 1$ (stable cation wins). The two rules point at opposite carbons.

Level 4 — Synthesis

Problem 4.1

Design a full synthesis of anisole ( $C_{6} H_{5} - O - C H_{3}$ ) starting from phenol and methanol (plus any inorganic reagents). Show each step.

Recall Solution

Step 1 — make the alkoxide (here, phenoxide). Deprotonate phenol with NaOH (phenol is acidic enough, $p K_{a} \approx 10$ , unlike alcohols — see Alcohols — preparation and acidity): $C_{6} H_{5} O H + N a O H ⟶ C_{6} H_{5} O^{-} N a^{+} + H_{2} O$ Step 2 — make the halide from methanol. Convert $C H_{3} O H$ to $C H_{3} I$ (e.g. with HI or $P B r_{3}$ -type reagents): $C H_{3} O H H I C H_{3} I + H_{2} O$ Step 3 — Williamson $S_{N} 2$ : phenoxide (nucleophile) + methyl iodide (open carbon): $C_{6} H_{5} O^{-} N a^{+} + C H_{3} I ⟶ C_{6} H_{5} - O - C H_{3} + N a I$ Aryl must be the alkoxide (can't be the halide); methyl is the perfect $S_{N} 2$ substrate.

Problem 4.2

Propose a synthesis of benzyl ethyl ether $C_{6} H_{5} C H_{2} - O - C_{2} H_{5}$ from benzyl alcohol and ethanol. Justify the choice of which becomes the alkoxide.

Recall Solution

Both carbons are 1° (benzylic $C H_{2}$ and ethyl $C H_{2}$ each carry one other carbon), so either can be the halide. The safest is to make either alcohol into an alkoxide with Na/NaH and use the other as a 1° halide. $C_{2} H_{5} O H N a C_{2} H_{5} O^{-} N a^{+} C_{6} H_{5} C H_{2} O H H B r C_{6} H_{5} C H_{2} B r$ $C_{2} H_{5} O^{-} N a^{+} + C_{6} H_{5} C H_{2} B r ⟶ C_{6} H_{5} C H_{2} - O - C_{2} H_{5} + N a B r$ Benzyl halide is an especially good $S_{N} 2$ substrate (adjacent ring stabilises the transition state), so making it the halide is ideal. Note: do not deprotonate a phenol-type OH here — benzyl alcohol's OH is a normal, non-aromatic alcohol.

Level 5 — Mastery

Problem 5.1

An ether $R - O - R^{'}$ is treated with excess HI and yields two moles of $C H_{3} C H_{2} I$ and one mole of $H_{2} O$ per mole of ether. Identify $R - O - R^{'}$ and explain the stoichiometry.

Recall Solution

Two identical iodides means both carbon groups are ethyl, and both C–O bonds eventually break (excess HI). So the ether is diethyl ether, $C_{2} H_{5} - O - C_{2} H_{5}$ .

1st HI: protonate O, $I^{-}$ hits one 1° ethyl ( $S_{N} 2$ ) → $C_{2} H_{5} I + C_{2} H_{5} O H$ .
2nd HI: protonate ethanol's O, $I^{-}$ hits the ethyl → $C_{2} H_{5} I + H_{2} O$ . $C_{2} H_{5} - O - C_{2} H_{5} 2 H I 2 C_{2} H_{5} I + H_{2} O$ Two C–O bonds broken → 2 HI consumed → 2 iodides + 1 water. ✓

Problem 5.2

An unsymmetrical ether $C_{6} H_{5} - O - C H_{2} C H_{3}$ (phenetole) is treated with excess HI. Give all organic products and the exact number of HI equivalents consumed. Explain why the count is not two.

Recall Solution

The aryl C–O bond cannot be cleaved. So only the ethyl side reacts.

Protonate O; $I^{-}$ attacks the ethyl carbon ( $S_{N} 2$ ) → $C_{2} H_{5} I$ + phenol $C_{6} H_{5} O H$ .
Excess HI does not touch phenol's aryl C–O (aromatic), so it stops here. $C_{6} H_{5} - O - C_{2} H_{5} H I C_{6} H_{5} O H + C_{2} H_{5} I$ Only 1 equivalent of HI is consumed — not two — because the aryl–O bond is inert. Contrast diethyl ether (2 equivalents).

Problem 5.3 (rank & justify)

Rank $H I$ , $H B r$ , $H C l$ for cleaving an ether, and give the two independent reasons HI wins, backed by data.

Recall Solution

$H I > H B r > H C l$ Two reasons, both favouring HI, each with a number:

Nucleophile strength: $I^{-}$ is the largest, most polarizable halide → best nucleophile for attacking the protonated carbon. In protic solvent the nucleophilicity order is $I^{-} > B r^{-} > C l^{-}$ ; the ion's radius grows from $C l^{-} \approx 181 pm$ to $B r^{-} \approx 196 pm$ to $I^{-} \approx 220 pm$ , and its loosely-held outer electrons are easier to donate.
Acid strength: HI is the strongest of the three at protonating the ether oxygen (step 1). The gas-phase acidities run $p K_{a} (HI) \approx - 10$ , $p K_{a} (HBr) \approx - 9$ , $p K_{a} (HCl) \approx - 7$ — HI donates $H^{+}$ most readily. HCl fails on both counts ( $C l^{-}$ weakest nucleophile, HCl weakest acid), so it barely cleaves ethers. See Nucleophilicity and leaving group ability.

Problem 5.4 (boundary case — secondary ether)

Isopropyl methyl ether $(C H_{3})_{2} C H - O - C H_{3}$ is cleaved by HI. The secondary isopropyl carbon can react by either $S_{N} 1$ or $S_{N} 2$ . Predict the major products and name the C–O bond that breaks, reasoning through both possibilities.

Recall Solution

Two carbons compete for iodide:

Methyl (0 attached carbons) is a superb $S_{N} 2$ target; isopropyl is 2° (2 attached carbons) — borderline for both mechanisms, only a modestly stable secondary cation.
Because the methyl offers a clean, fast $S_{N} 2$ backside attack while the 2° cation is not especially stabilised, iodide preferentially strikes the methyl carbon. Isopropyl leaves as the alcohol. $(C H_{3})_{2} C H - O - C H_{3} H I (C H_{3})_{2} C H - O H + C H_{3} I$ With excess HI and heat the resulting isopropanol can be further converted to $(C H_{3})_{2} C H I$ , but the first, cleaner cleavage places iodide on the methyl. This is the honest edge case: 2° ethers sit between the pure- $S_{N} 2$ (methyl/1°) and pure- $S_{N} 1$ (3°) worlds, and the less hindered partner still wins the first attack.

Problem 5.5 (boundary case — vinyl ether)

Methyl vinyl ether $C H_{2} = C H - O - C H_{3}$ is treated with HI. Explain, by analogy to the aryl case, which C–O bond is inert and which breaks.

Recall Solution

The vinyl carbon ( $C H_{2} = C H -$ ) is $s p^{2}$ , just like an aryl carbon: it resists $S_{N} 2$ (poor backside geometry) and $S_{N} 1$ (a vinyl cation is very unstable). So the vinyl–O bond does not cleave by simple substitution. Iodide instead hits the methyl carbon ( $S_{N} 2$ ): $C H_{2} = C H - O - C H_{3} H I C H_{2} = C H - O H + C H_{3} I$ The fragment $C H_{2} = C H - O H$ is an enol, which tautomerises to acetaldehyde $C H_{3} C H O$ . Key parallel: like aryl ethers, a vinyl ether gives up its alkyl side as the iodide and never forms a vinyl (or aryl) iodide by this route.

Recall One-line summary to lock in

Classify each carbon by counting attached carbons (methyl 0, 1° one, 2° two, 3° three). Build with $S_{N} 2$ (hindered = alkoxide, unhindered = halide; aryl/vinyl/3° never the halide). Cut with HI (protonate O; unhindered carbon by $S_{N} 2$ , stable-cation carbon by $S_{N} 1$ ; aryl/vinyl side → phenol/enol, never their iodide). $H I > H B r > H C l$ .