Questions: Williamson Ether Synthesis via SN2

Question 1

You want to synthesize tert-butyl methyl ether (CH₃OC(CH₃)₃) via Williamson synthesis. Which reagent combination is correct?

• A. tert-Butyl alcohol + NaH, then methyl iodide — the tert-butoxide attacks the primary methyl electrophile
• B. Methanol + NaH, then tert-butyl bromide — the methoxide attacks the tert-butyl electrophile
• C. tert-Butyl alcohol reacted directly with methanol under acidic conditions
• D. Sodium tert-butoxide reacted with tert-butyl bromide in DMSO

Answer: (A)

The SN2 reaction requires a primary (or methyl) electrophile. Methyl iodide is the best possible SN2 substrate — no steric hindrance whatsoever. tert-Butyl alcohol is deprotonated with NaH to form tert-butoxide, which serves as the nucleophile. Option B is the classic disconnection error: using tert-butyl bromide as the electrophile places the SN2 reaction at a tertiary carbon, where the bulky alkoxide base instead abstracts a β-proton (E2 elimination), yielding isobutene instead of the ether. Whenever one fragment is tertiary, it must be the alkoxide, never the halide.

Question 2

Why does Williamson ether synthesis fail when a tertiary alkyl halide is used as the electrophile?

• A. Tertiary alkyl halides cannot form a leaving group because the C–X bond is too strong
• B. The alkoxide is a strong base; at a tertiary substrate the backside carbon is too hindered for SN2 attack, so the alkoxide abstracts a β-proton instead, giving E2 elimination
• C. Tertiary carbons are too electronegative to be attacked by oxygen nucleophiles
• D. The reaction produces a carbocation intermediate at the tertiary center that immediately rearranges

Answer: (B)

Alkoxide ions are simultaneously strong nucleophiles and strong bases. For primary substrates, the backside is accessible and SN2 dominates. For tertiary substrates, three substituents block backside approach, making SN2 effectively impossible — but the exposed β-hydrogens are easily abstracted. The alkoxide acts as a base (E2 path) instead of a nucleophile (SN2 path), and the product is an alkene, not an ether. Option D is wrong: Williamson synthesis is strictly SN2 and does not proceed through carbocation intermediates.

Question 3

In Williamson ether synthesis, the alkoxide nucleophile should generally be derived from the more substituted (more hindered) alcohol.

• T. True
• F. False

Answer: False

This is a common misconception. The governing rule is not which alcohol is more substituted — it is that the *electrophilic alkyl halide* must be primary. The alkoxide can come from any alcohol, including tertiary ones. When making tert-butyl methyl ether, the tert-butyl group correctly becomes the alkoxide (nucleophile) and methyl becomes the halide (electrophile) — even though tert-butoxide is highly hindered — because the alternative (tert-butyl halide as electrophile) gives only elimination. Substitution level of the alkoxide is not the relevant constraint; substitution level of the electrophile is.

Question 4

Tosylates (OTs) can replace alkyl halides as the electrophilic partner in Williamson ether synthesis because the tosylate group is a competent leaving group in SN2 reactions.

• T. True
• F. False

Answer: True

Tosylates are prepared from alcohols by reaction with toluenesulfonyl chloride (TsCl). The tosylate group (–OTs) is a good leaving group — comparable to iodide — because the negative charge is stabilized by the sulfonyl group and aromatic ring. The alkoxide attacks the carbon bearing the tosylate with inversion, displacing –OTs exactly as it would displace –Br or –I. Tosylates are sometimes preferred when the corresponding alkyl halide is inconvenient to prepare or unstable.

Question 5

Explain the disconnection analysis used in planning a Williamson synthesis for an unsymmetrical ether, and identify what governs which fragment becomes the alkoxide versus the alkyl halide.

Model answer: To plan the synthesis of R–O–R', mentally break the C–O bond on one side and ask which fragment should be the nucleophile (alkoxide, RO⁻) and which should be the electrophile (alkyl halide, R'–X). The governing rule is that the alkyl halide must be primary — or methyl — to avoid E2 elimination competing with SN2. So you break the bond such that the primary carbon becomes the halide and the other fragment (which may be secondary or tertiary) becomes the alkoxide. If both sides are primary, either disconnection works. If one side is tertiary, it must become the alkoxide without exception — the tertiary group cannot serve as the electrophile.

Disconnection analysis is the key synthetic planning skill. The synthesis only works when the SN2 step occurs at an unhindered carbon. Choosing the wrong disconnection — putting a secondary or tertiary carbon on the halide side — gives elimination instead of substitution. The ability to recognize and apply the correct disconnection distinguishes students who understand the mechanism from those who have only memorized reaction names.