Question bank — Ethers — Williamson synthesis, cleavage by HI

False. Williamson is [[SN1 vs SN2 mechanisms|$S_{N}2$]]; a 3° halide is sterically blocked, and the strong alkoxide base drives E2 to give an alkene instead of the ether.

False. Product stability doesn't set the mechanism's rate; $S_{N}2$ cares about steric access to the carbon being attacked, and a 3° carbon is the most crowded, so it's the worst choice for the halide.

False. The aryl C–O bond cannot undergo [[SN1 vs SN2 mechanisms|$S_{N}1$ or $S_{N}2$]], so $I^{-}$ hits the methyl carbon: products are phenol ($C_6{H}_{5}OH$) + $C{H}_{3}I$, never iodobenzene.

True. A leaving $R{O}^{-}$ is a strong, high-energy base, so it won't depart; you must first protonate the O to convert it into a neutral alcohol (a good leaving group) before any substitution happens.

False. HI wins because $I^{-}$ is the most polarizable nucleophile and HI is the stronger acid (better at protonating O); HCl barely reacts. Order is $HI>HBr>HCl$.

True. Both are strong enough bases; Na reduces $H^{+}$ to $H_2$ gas, NaH grabs the proton to give $H_2$ — either way you get $R{O}^{-}N{a}^{+}$ ready for substitution.

False (it's symmetric). Both carbons are equivalent 1° centres, so either C–O can break; with excess HI both ends eventually become $C_2{H}_{5}I$ and the oxygen leaves as $H_{2}O$.

True. The phenol O still bears a lone pair and negative charge; it attacks a 1°/methyl halide fine — the aryl group just can't be on the halide side.

Error: the 3° halide. The methoxide base will trigger E2 on $(C{H}_3{)}_{3}C\text{–}Br$ (→ isobutylene). Swap roles: use $(C{H}_3{)}_{3}C{O}^{-}$ (alkoxide) + $C{H}_{3}I$ (halide).

Error: aryl halide. Aryl carbons are sp² and ring-shielded, so no $S_{N}2$ at $C_6{H}_5\text{–}Br$. Put the aryl on the alkoxide: $C_6{H}_5{O}^{-}+C{H}_{3}I$.

Error: wrong side got the iodide. The 3° carbon forms a stable [[SN1 vs SN2 mechanisms|$S_{N}1$]] carbocation and grabs $I^{-}$; a methyl cation is far too unstable. Correct: $C{H}_{3}OH+(C{H}_3{)}_{3}CI$.

Error: wrong atom. $I^{-}$ is a carbon nucleophile here; it attacks the electrophilic carbon, pushing the neutral alcohol off the C–O bond. Oxygen is never the attack site.

Error: no cation. Williamson is concerted $S_{N}2$ — bond-making and bond-breaking happen in one step with no carbocation. Cation stability is irrelevant to it (that logic belongs to $S_{N}1$ cleavage).

Error: stoichiometry. One HI can only cleave once, giving $C_2{H}_{5}I+C_2{H}_{5}OH$. You need excess (2 eq) HI to also convert the ethanol to a second $C_2{H}_{5}I$ and release $H_{2}O$.

Because $S_{N}2$ needs an open backside approach to the $C^{\delta +}$; a bulky carbon blocks the incoming alkoxide, so we keep the attacked carbon small (methyl/1°).

It converts a departing $R{O}^{-}$ (a strong, unstable base) into a neutral $R\text{–}OH$ molecule, which is stable and happy to leave — the essence of turning a bad LG into a good one.

Iodide is large and highly polarizable, so its electron cloud distorts easily to form the new C–I bond; smaller, harder $C{l}^{-}$ is far less nucleophilic in these conditions.

The aryl C–O bond can't undergo nucleophilic substitution, so cleavage happens only at the alkyl carbon; the oxygen stays with the ring, giving phenol.

The 3° carbon is too crowded for backside attack, but it can ionize to a stable tertiary carbocation, so the reaction takes the $S_N1$ route and the cation captures $I^{-}$.

The alkoxide carries a full negative charge (higher electron density on O), making it a much stronger nucleophile than the neutral, lone-pair-only alcohol.

A tetrahedral-like $s{p}^3$ oxygen (two bonds + two lone pairs) gives an angle near $109.5^{\circ }$; the observed $\sim 110^{\circ }$ matches, confirming $s{p}^3$ hybridisation like water.

The first step gives phenol + $C{H}_{3}I$; phenol's aryl C–O still can't be cleaved, so no further iodide forms on the ring — you stop at $C_6{H}_{5}OH+C{H}_{3}I$.

The less hindered carbon (methyl) is attacked fastest in $S_{N}2$, so with 1 eq HI the major products are $C{H}_{3}I+C_2{H}_{5}OH$.

It fails — aryl halides don't do $S_{N}2$. The synthesis only works if the aryl group is on the alkoxide (phenoxide) side.

Elimination competes: the alkoxide can act as a base removing a $\beta$-H to give an alkene, lowering ether yield — favour 1°/methyl halides to suppress this.

Since both sides are identical, you always get the same alkyl iodide + the same alcohol (e.g. $C_2{H}_{5}I+C_2{H}_{5}OH$); no ambiguity, unlike unsymmetrical ethers.

The benzylic carbon forms a resonance-stabilised cation ($S_{N}1$) and grabs $I^{-}$, giving $C_6{H}_{5}C{H}_{2}I+C{H}_{3}OH$ — here the aryl ring is not directly bonded to O, so the C is cleavable.