Q.In Williamson synthesis the ________ reacts with sodium alkoxide and give dialkyl ether.

Williamson Ether Synthesis: The Why Before the How

Imagine you want to build a molecule with an oxygen atom bridging two carbon chains — that's an ether. The simplest way to forge that C–O–C link is to take a negatively charged oxygen (an alkoxide ion) and let it attack a carbon that carries a good leaving group. That is the entire intuition behind Williamson ether synthesis: a strong nucleophile (alkoxide) meets an electrophile (alkyl halide) in an SN2 reaction.

Why does this work so cleanly? Because the alkoxide ion is both a strong base and a strong nucleophile. It wants to donate its electron pair to an electron-deficient carbon. The alkyl halide provides that carbon, with the halogen (Cl, Br, I) acting as a leaving group. The reaction is a single-step, backside attack — the hallmark of SN2.

The product can be symmetrical (R = R') or unsymmetrical (R ≠ R'). The sodium alkoxide itself is usually made in situ by reacting an alcohol with sodium metal or sodium hydride.

The Critical Constraint: Why Only Primary Halides?

Here is where most students lose marks. The reaction is an SN2, and SN2 reactions are brutally sensitive to steric hindrance. If you use a secondary or tertiary alkyl halide, the alkoxide (a strong base) will overwhelmingly prefer to do elimination (E2) instead of substitution. You will get an alkene, not an ether.

So the rule is ironclad: the carbon bearing the leaving group must be primary (or, in special cases, methyl or allylic/benzylic). The alkoxide side can be primary, secondary, or even tertiary — that carbon is not the site of attack.

Choosing Which Side to Make the Alkoxide

For an unsymmetrical ether $\text{R–O–R'}$, you have two possible routes. Which one should you pick? The answer: make the alkoxide from the smaller or less hindered alcohol, and use the larger group as the alkyl halide. This minimises steric hindrance at the SN2 transition state.

Example: To make ethyl tert-butyl ether ($\text{CH}_3\text{CH}_2\text{–O–C(CH}_3)_3$):

• Correct: Sodium ethoxide ($\text{CH}_3\text{CH}_2\text{O}^- \text{Na}^+$) + tert-butyl bromide? No — tert-butyl is tertiary, elimination dominates.
• Correct: Sodium tert-butoxide ($(\text{CH}_3)_3\text{CO}^- \text{Na}^+$) + ethyl bromide (primary). This works because the electrophile is primary.

The Mechanism in One Step …