Q.(a) Write the mechanism of the following reaction: $2\,CH_3CH_2OH \xrightarrow[413\,K]{H^+} CH_3CH_2OCH_2CH_3 + H_2O$

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 …

The Kolbe Reaction: From Intuition to Mechanism

Imagine you have a benzene ring with an –OH group attached — that's phenol. Now, phenol is weakly acidic, so when you treat it with a strong base like NaOH, you get sodium phenoxide. That oxygen now carries a full negative charge. That negative charge is the key.

The oxygen's lone pairs are strongly electron-donating. They push electron density into the ring, making the ortho and para positions much more reactive toward electrophiles. The ortho position, being right next to the oxygen, gets the most activation. This is the intuition: the negatively charged oxygen "wakes up" the ring, especially the ortho carbon, making it hungry for a positive or electron-deficient species.

Now, enter carbon dioxide. CO₂ is a weak electrophile — the carbon is partially positive because of the two oxygens pulling electrons away. Under normal conditions, CO₂ is too weak to attack phenol directly. But under high pressure (about 4–7 atm) and moderate heat (125–150°C), the activated ortho position of sodium phenoxide can attack the carbon of CO₂.

The precise steps

1. Formation of sodium phenoxide

2. Carboxylation under pressure The ortho carbon of the phenoxide ion attacks CO₂. A tetrahedral intermediate forms, which then rearranges to give sodium salicylate.

3. Acidification Treating with dilute HCl gives the free carboxylic acid.

Why this reaction matters

Salicylic acid is the precursor to aspirin (acetylsalicylic acid). The Kolbe reaction is the industrial route to salicylic acid — cheap, simple, and high-yielding. Without it, aspirin would be far more expensive. …