Ethers — Williamson synthesis, cleavage by HI: Worked examples

The scenario matrix

Before solving anything, let us enumerate the whole space. An ether problem is really a choice of three switches:

Cell	Task type	Structural feature	Which "trap" it tests
A	Build (Williamson)	both groups 1°/methyl	free choice — pick either as alkoxide
B	Build (Williamson)	one 3°, one 1°/methyl	must put 3° as alkoxide (else E2)
C	Build (Williamson)	one aryl, one alkyl	aryl must be the alkoxide (phenoxide)
D	Cut (HI, 1 eq)	both 1°/methyl	$S_{N} 2$ on the smaller carbon
E	Cut (HI, 1 eq)	one 3° / benzylic	$S_{N} 1$ → stable-cation carbon takes $I^{-}$
F	Cut (HI, excess)	any dialkyl	both C–O bonds cut → two iodides + water
G	Cut (HI)	alkyl–aryl	aryl → phenol, alkyl → iodide (never aryl iodide)
H	Degenerate / limiting	symmetric ether, or HCl vs HI	tests "does it even react?"
I	Word problem	real synthesis planning	translate words → correct partners
J	Exam twist	epoxide / cyclic ether, or wrong-route detection	subtle mechanism read

Ten cells. Below, ten worked examples — one per cell. Each is labelled with its cell letter.

Cell A — Build a simple ether, free choice

Worked example

(A) Prepare ethyl methyl ether, $C H_{3} - O - C_{2} H_{5}$ , by Williamson synthesis. Forecast: Both carbons are small (methyl and 1° ethyl). Do you think there is one right answer, or two?

Step 1. Split the target at the C–O–C oxygen into two fragments: a $C H_{3}$ side and a $C_{2} H_{5}$ side. Why this step? Williamson always joins one alkoxide ( $R O^{-}$ ) to one alkyl halide ( $R^{'} - X$ ); we must decide which fragment plays which role.

Step 2. Check hindrance. Methyl carbon and 1° ethyl carbon are both open to backside attack. Why this step? $S_{N} 2$ (the Williamson mechanism, see SN1 vs SN2 mechanisms) needs a clear backside; either small carbon qualifies, so both routes work.

Step 3. Write either valid route: $C H_{3} O^{-} N a^{+} + C_{2} H_{5} B r ⟶ C H_{3} - O - C_{2} H_{5} + N a B r$ $C_{2} H_{5} O^{-} N a^{+} + C H_{3} B r ⟶ C H_{3} - O - C_{2} H_{5} + N a B r$

Verify: Count atoms — each route conserves one $C H_{3}$ , one $C_{2} H_{5}$ , one O; the halide leaves as $N a B r$ . Product mass $= 60 g/mol$ ( $C_{3} H_{8} O$ ). ✓ Free choice confirmed: cell A has two correct answers.

Cell B — One bulky group: forced choice

Worked example

(B) Prepare tert-butyl ethyl ether, $(C H_{3})_{3} C - O - C_{2} H_{5}$ , by Williamson. Forecast: One carbon is 3° (blocked), one is 1° (open). Which fragment is forced to be the alkoxide?

Step 1. Identify hindrance: $(C H_{3})_{3} C$ – is 3° (crowded), $C_{2} H_{5}$ – is 1° (open). Why this step? The carbon that gets attacked must be uncrowded for $S_{N} 2$ .

Step 2. Reject the wrong route. If we tried $C_{2} H_{5} O^{-} + (C H_{3})_{3} C - B r$ , the strong alkoxide base rips a β-H off the 3° halide → E2 elimination (E2 Elimination) giving isobutylene, not the ether. Why this step? Steric bulk + strong base = elimination beats substitution.

Step 3. Use the 3° group as the alkoxide, the 1° group as the halide: $(C H_{3})_{3} C - O^{-} N a^{+} + C_{2} H_{5} - B r ⟶ (C H_{3})_{3} C - O - C_{2} H_{5} + N a B r$

Verify: The attacked carbon is now the ethyl 1° carbon — backside is clear, no β-elimination trap. Product $C_{6} H_{14} O$ , mass $102 g/mol$ . ✓

Cell C — Aryl group present

Worked example

(C) Prepare phenetole (ethyl phenyl ether), $C_{6} H_{5} - O - C_{2} H_{5}$ . Forecast: One side is aromatic. Can the aryl carbon ever be the one attacked?

Step 1. Aryl carbon is $s p^{2}$ , part of the ring; the ring physically blocks backside approach and aryl C–X bonds do not do $S_{N} 2$ / $S_{N} 1$ . Why this step? This forbids the aryl fragment from being the halide.

Step 2. So the aryl group must be the alkoxide — here phenoxide, made from a phenol (Phenols — properties and reactions) which is acidic enough to deprotonate with $N a O H$ . Why this step? Phenol's O–H ( $p K_{a} \approx 10$ ) is deprotonated even by hydroxide, giving $C_{6} H_{5} O^{-}$ .

Step 3. Pair phenoxide with the 1° ethyl halide: $C_{6} H_{5} O^{-} N a^{+} + C_{2} H_{5} - B r ⟶ C_{6} H_{5} - O - C_{2} H_{5} + N a B r$

Verify: Attacked carbon is 1° ethyl → clean $S_{N} 2$ . Product $C_{8} H_{10} O$ , mass $122 g/mol$ . ✓

Cell D — Cut a small–small ether (SN2)

Cell E — Cut with a 3° / stable cation (SN1)

Cell F — Excess HI on a dialkyl ether

Cell G — Alkyl–aryl ether cleavage (the classic trap)

Cell H — Degenerate / limiting: "does it react at all?"

Worked example

(H) You try to cleave diethyl ether with concentrated HCl instead of HI. What happens, and why does the choice of acid matter? Forecast: Same ether, same protonation idea — will HCl finish the job?

Step 1. Protonation still occurs, but $C l^{-}$ is a poor nucleophile (small, hard, not polarizable) — see Nucleophilicity and leaving group ability. Why this step? Cleavage needs the halide to attack the carbon; a weak nucleophile stalls step 2.

Step 2. Also HCl is a weaker acid than HI, so it protonates O less effectively. Why this step? Both roles ( $H^{+}$ donor and nucleophile) are done best by HI.

Step 3. Result: essentially no cleavage with HCl. Reactivity order: $H I > H B r > H C l$ (the degenerate/limiting case: HCl ≈ no reaction.)

Verify: The order tracks two things at once — acid strength ( $H I$ strongest) and nucleophile polarizability ( $I^{-} > B r^{-} > C l^{-}$ ). Both favour HI. ✓

Cell I — Word problem (real synthesis planning)

Worked example

(I) A student in the lab has tert-butanol, methanol, sodium metal, and methyl iodide. Design a synthesis of tert-butyl methyl ether, $(C H_{3})_{3} C - O - C H_{3}$ . Which reagents do you combine, and in which order? Forecast: You have two alcohols and one halide. Guess which alcohol must be deprotonated.

Step 1. Target has a 3° carbon and a methyl carbon on either side of O. The methyl carbon (from $C H_{3} I$ ) must be the one attacked. Why this step? $C H_{3} I$ is a perfect $S_{N} 2$ substrate; the 3° side cannot be the halide (E2).

Step 2. So make the 3° alkoxide: react tert-butanol with Na metal (see Alcohols — preparation and acidity): $(C H_{3})_{3} C - O H + N a \to (C H_{3})_{3} C - O^{-} N a^{+} + \frac{1}{2} H_{2}$ Why this step? We need the bulky group as the nucleophile; Na deprotonates the alcohol.

Step 3. Add $C H_{3} I$ : $(C H_{3})_{3} C - O^{-} N a^{+} + C H_{3} I \to (C H_{3})_{3} C - O - C H_{3} + N a I$ Why this step? Alkoxide attacks methyl carbon → clean ether.

Verify: Methanol and tert-butyl bromide are not used — using $(C H_{3})_{3} C - B r$ would eliminate. The correct route attacks only the methyl carbon. Product $C_{5} H_{12} O$ , mass $88 g/mol$ . ✓

Cell J — Exam twist: wrong-route detection

Worked example

(J) An exam claims: "Make neopentyl phenyl ether, $C_{6} H_{5} - O - C H_{2} C (C H_{3})_{3}$ , via $C_{6} H_{5} - B r +^{-} O - C H_{2} C (C H_{3})_{3}$ ." Is this valid? Give the correct route. Forecast: Neopentyl is technically 1°, but is it really open to attack?

Step 1. The proposed halide is aryl ( $C_{6} H_{5} - B r$ ) — aryl carbons never do $S_{N} 2$ / $S_{N} 1$ . So the proposed route is wrong on that ground alone. Why this step? Same aryl rule as cell C.

Step 2. Swap roles: aryl must be the alkoxide (phenoxide), alkyl must be the halide. Why this step? Only legal assignment.

Step 3. But note the twist — neopentyl halide $B r C H_{2} C (C H_{3})_{3}$ is a notoriously sluggish $S_{N} 2$ substrate (the big tert-butyl next door shields the backside). It still reacts (it is 1°, no β-H arrangement for easy E2 at that carbon), just slowly. Correct route: $C_{6} H_{5} O^{-} N a^{+} + B r C H_{2} C (C H_{3})_{3} ⟶ C_{6} H_{5} - O - C H_{2} C (C H_{3})_{3} + N a B r$

Verify: Aryl is the nucleophile (legal), neopentyl 1° carbon is attacked (slow but possible). Product $C_{11} H_{16} O$ , mass $164 g/mol$ . ✓ The exam's route fails only because it made the aryl ring the electrophile.

Reveal check

The two Williamson rules

bulky/aryl group = alkoxide; small (1°/methyl) group = alkyl halide.

Which cleavage carbon takes

I^{-}

under

S_{N} 2

the smaller, less hindered carbon.

Which cleavage carbon takes

I^{-}

under

S_{N} 1

the one giving the more stable carbocation (3°/benzylic).

Anisole + HI product that is NOT formed

iodobenzene — you get phenol +

C H_{3} I

instead.

Why HCl barely cleaves ethers

C l^{-}

is a poor nucleophile and HCl a weaker acid; HI wins on both.