Q.(a) How do you convert the following:

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 Intuition: Why Would Chloroform Add an Aldehyde?

Imagine you have a phenol molecule — a benzene ring with an –OH group. That –OH is not just sitting there; it's a powerful electron-donating group. It pushes electron density into the ring, especially onto the ortho and para positions. This makes those positions unusually reactive toward electrophiles (species that love electrons).

Now, chloroform ($\text{CHCl}_3$) in the presence of a strong base like aqueous $\text{NaOH}$ does something dramatic. The base pulls off a proton from chloroform, generating a highly reactive species called dichlorocarbene ($:\text{CCl}_2$). This carbene is a fierce electrophile — it has a sextet of electrons and desperately wants two more.

The phenol's ortho position, rich in electrons, is the perfect target. The carbene attacks there, and a cascade of reactions follows, ultimately converting that –CHCl₂ group into an aldehyde (–CHO). The product is salicylaldehyde (2-hydroxybenzaldehyde).

The Precise Statement

Reimer–Tiemann Reaction:

When phenol is treated with chloroform ($\text{CHCl}_3$) and aqueous sodium hydroxide ($\text{NaOH}$) at about 60–70 °C, followed by acidification, the formyl group (–CHO) is introduced at the ortho position relative to the –OH group. The major product is salicylaldehyde.

Step-by-Step Mechanism (Why It Works)

1. Generation of dichlorocarbene $\text{NaOH}$ deprotonates chloroform:

The trichloromethyl anion loses a chloride ion to form the electrophilic carbene:

2. Attack on phenoxide ion

   Phenol first reacts with $\text{NaOH}$ to form the more nucleophilic phenoxide ion ($\ce{C6H5O-}$). The carbene attacks the ortho carbon of the phenoxide ring.

3. Rearrangement and hydrolysis

   The intermediate undergoes ring-opening to a dichloromethyl phenol derivative, which then hydrolyzes under basic conditions to give the aldehyde.

4. Acidification

   After the reaction, adding dilute acid converts the sodium salt of salicylaldehyde back to the free aldehyde.

Key Points for Exams

• Reagents: Phenol + $\text{CHCl}_3$ + aqueous $\text{NaOH}$ (not alcoholic $\text{NaOH}$ — that would give a different reaction).
• Temperature: ~60–70 °C (reflux). Too low, the carbene doesn't form; too high, side reactions dominate.
• Product: Salicylaldehyde (ortho-hydroxybenzaldehyde). A small amount of para-hydroxybenzaldehyde may also form, but ortho is the major product. …