Q.(a) Give one example of the Sandmeyer reaction.

The Wurtz Reaction: Why Sodium Makes Alkanes Grow

Imagine you have two identical building blocks — say, two methyl groups ($CH_3-$) — and you want to join them into a longer chain ($CH_3-CH_3$). That's the core idea of the Wurtz reaction. You take an alkyl halide (like bromoethane) and treat it with sodium metal in dry ether. The sodium strips away the halogen atoms, and the two alkyl fragments couple together.

The reaction was discovered by Charles-Adolphe Wurtz in 1855, and it was one of the first methods to make carbon-carbon bonds deliberately.

The precise statement

Two molecules of an alkyl halide react with two atoms of sodium metal in anhydrous (dry) ether to give a symmetrical alkane and sodium halide.

Here $R$ is an alkyl group (methyl, ethyl, propyl, etc.) and $X$ is a halogen (usually chlorine, bromine, or iodine). The product $R-R$ is always a symmetrical alkane — meaning both halves of the new molecule are identical.

How it actually works (the mechanism)

The reaction proceeds through a free-radical intermediate. Sodium metal donates one electron to the alkyl halide, breaking the $C-X$ bond and generating an alkyl radical ($R\bullet$) and a sodium halide. Two such radicals then combine to form the $R-R$ bond.

A concrete example

Take bromoethane (ethyl bromide) with sodium:

The product is butane — a symmetrical alkane with four carbons.

The big limitation: unsymmetrical products are a mess

If you try to couple two different alkyl halides, say $CH_3Br$ and $CH_3CH_2Br$, you get a mixture of three alkanes:

• $CH_3-CH_3$ (ethane) — from two methyl radicals
• $CH_3CH_2-CH_2CH_3$ (butane) — from two ethyl radicals
• $CH_3-CH_2CH_3$ (propane) — from one methyl and one ethyl radical

This mixture is hard to separate, so the Wurtz reaction is practically useful only when both alkyl halides are identical.

Why dry ether? Why sodium? …