Q.Ethers can be prepared by Williamson synthesis in which an alkyl halide is reacted with sodium alkoxide. Di-tert-butyl ether cannot be prepared by this method. Explain.

The key idea is that Williamson Ether Synthesis requires an $S_N2$ mechanism, which fails with bulky substrates.

Reasoning:

1. The reaction is: $R\text{-}O^- Na^+ + R'\text{-}X \rightarrow R\text{-}O\text{-}R' + NaX$. It proceeds via $S_N2$ attack of the alkoxide on the alkyl halide.
2. To make di-tert-butyl ether, you would need to react tert-butyl halide with tert-butoxide.
3. tert-Butyl halide is a tertiary substrate — it cannot undergo $S_N2$ due to extreme steric hindrance. Instead, it eliminates (E2) to give isobutene. …

Williamson ether synthesis is one of the most reliable methods for preparing ethers. The reaction is simple: a sodium alkoxide (a strong nucleophile) attacks an alkyl halide, displacing the halide ion to form an ether. But this simplicity hides a critical constraint — the reaction proceeds through an $S_N2$ mechanism.

Why $S_N2$ matters here

The $S_N2$ mechanism demands a backside attack by the nucleophile on the carbon bearing the leaving group. This means the carbon must be sterically accessible — primary alkyl halides work beautifully, secondary ones work with some difficulty, and tertiary alkyl halides are essentially unreactive toward $S_N2$ because three bulky alkyl groups block the approach.

But there's a second problem with tertiary halides: even if you could force an $S_N2$ attack, the strong base (alkoxide) would instead pull off a proton from the tertiary halide, triggering elimination (E2) to give an alkene. So you get no ether at all.

Applying this to di-tert-butyl ether

Di-tert-butyl ether has the structure $(CH_3)_3C-O-C(CH_3)_3$. To make it via Williamson synthesis, you would need to combine:

1. A tert-butyl halide — say, $(CH_3)_3C-Br$ — as the alkyl halide.
2. Sodium tert-butoxide — $(CH_3)_3C-O^-Na^+$ — as the alkoxide.

Now look at what happens:

Step 1: The alkoxide is bulky. Sodium tert-butoxide is a very hindered base. It's a strong base but a poor nucleophile because the three methyl groups crowd the oxygen. Even if it could attack, it would struggle to reach the carbon.

Step 2: The alkyl halide is tertiary. The carbon bearing the bromine is surrounded by three methyl groups. An $S_N2$ attack here is geometrically impossible — the nucleophile cannot approach from the back without crashing into the alkyl groups.

Step 3: Elimination dominates. Instead of substitution, the tert-butoxide abstracts a proton from the tert-butyl halide, producing isobutylene (2-methylpropene) and tert-butyl alcohol. This E2 elimination is fast and irreversible. …

Williamson Ether Synthesis — Method & Limitation

Method: Williamson Ether Synthesis

This is an SN2 reaction between an alkoxide ion and an alkyl halide.

General reaction:

Steps:

1. Deprotonate the alcohol using a strong base (e.g., NaH, Na metal) to form the alkoxide ion ($RO^-$).
2. Alkylate the alkoxide by reacting it with a primary alkyl halide ($R'X$) via an SN2 mechanism.

Why Di-tert-butyl Ether Cannot Be Prepared

Core reason: Both the alkoxide and the alkyl halide are tertiary — the reaction fails because:

• Tertiary alkyl halides (like tert-butyl bromide) cannot undergo SN2 due to extreme steric hindrance.
• Instead, they undergo E2 elimination (forming an alkene) in the presence of a strong base like alkoxide.

Attempted reaction:

$(\text{CH}_3)_3\text{C—O}^- Na^+ + (\text{CH}_3)_3\text{C—Br} \xrightarrow{\text{not SN2}} \text{alkene} + \text{alcohol}$ …

Common Mistakes in Williamson Ether Synthesis (Di‑tert‑butyl ether case)

Students often lose marks on this question because they miss the mechanistic reason behind the failure. Here are the most frequent errors and how to avoid them.

Mistake 1: Thinking “any alkyl halide + any alkoxide works”

• What students write: “Di‑tert‑butyl ether cannot be prepared because tert‑butyl groups are bulky.”
• Why it’s wrong: Bulky groups alone don’t explain the failure — many bulky ethers can be made if the mechanism allows.
• How to avoid: Always check which carbon the halide is on. Williamson synthesis is an S$_N$2 reaction. The alkyl halide must be primary (or methyl) for a good yield. Tertiary halides undergo elimination (E2) instead.

Mistake 2: Forgetting the alkoxide can also be bulky

• What students write: “The alkoxide is too bulky to attack.”
• Why it’s wrong: The alkoxide is the nucleophile — it attacks the alkyl halide. Bulky alkoxides (like tert‑butoxide) are actually strong bases and favour elimination, not substitution.
• How to avoid: Remember: tertiary alkyl halide + any alkoxide → elimination (alkene). The alkoxide’s bulk makes E2 even faster.

Mistake 3: Confusing the two possible routes

• What students write: “We could use tert‑butyl bromide and sodium tert‑butoxide.”
• Why it’s wrong: That’s exactly the attempted synthesis — and it fails because both reactants are tertiary.
• How to avoid: Know the correct pairing:
  • To make an ether R–O–R', use primary alkyl halide + alkoxide of the other group.
  • For di‑tert‑butyl ether, you would need tert‑butyl halide (tertiary) + tert‑butoxide (tertiary) → E2 elimination dominates, giving isobutene.

Mistake 4: Ignoring the E2 elimination product

• What students write: “No reaction occurs.”
• Why it’s wrong: A reaction does occur — just not the desired one. The major product is 2‑methylpropene (isobutene).
• How to avoid: Always predict the elimination product when a tertiary halide is used. Write the alkene structure.

Mistake 5: Not explaining the “why” in terms of mechanism

• What students write: “It cannot be prepared because of steric hindrance.”
• Why it’s wrong: Vague. Examiners want the S$_N$2 vs E2 reasoning.
• How to avoid: Use this structure in your answer: …