Q.(a) What happens when chlorobenzene reacts with:

The Core Idea

Imagine you have two different Lego bricks — one is an aromatic ring with a halogen attached (aryl halide), and the other is a simple carbon chain with a halogen (alkyl halide). You want to snap them together to make a single molecule where the alkyl chain is directly attached to the benzene ring. That's exactly what the Wurtz-Fittig reaction does.

The reaction is a coupling — it takes two different organic halides and joins them at the carbon that used to hold the halogen, using metallic sodium as the "glue."

The Precise Statement

Here, Ar is an aryl group (like phenyl, tolyl, etc.), R is an alkyl group (methyl, ethyl, etc.), and X is a halogen (usually Br or I). The reaction is carried out in dry ether (anhydrous diethyl ether) because sodium reacts violently with water.

The product is an alkyl-substituted aromatic hydrocarbon — for example, bromobenzene and methyl bromide give toluene:

How It Actually Works (The Mechanism)

The reaction proceeds through a free radical mechanism, similar to the classic Wurtz reaction (which couples two alkyl halides).

Step 1: Sodium metal donates an electron to the halogen of the aryl halide, breaking the C–X bond and generating an aryl radical and NaX.

Step 2: Simultaneously, the same happens with the alkyl halide, giving an alkyl radical.

Step 3: The two radicals combine to form the coupled product.

Why Not Just Use Friedel-Crafts Alkylation?

You might ask: why not simply alkylate benzene using Friedel-Crafts? The Wurtz-Fittig reaction is useful when the alkyl group you want to attach is primary and you want to avoid rearrangements. Friedel-Crafts alkylation often rearranges carbocations (e.g., n-propyl chloride gives isopropylbenzene). The Wurtz-Fittig reaction gives the straight-chain product without rearrangement.

The Catch: Side Products

Because you have two different halides, the radicals can also couple with themselves:

• Two aryl radicals → biaryl (e.g., biphenyl from bromobenzene)
• Two alkyl radicals → alkane (e.g., ethane from methyl bromide)

So the desired Ar–R product is always mixed with Ar–Ar and R–R. This is why the reaction is most useful when the alkyl halide is cheap and the aryl halide is valuable — you can use an excess of the alkyl halide to favour the cross-coupling. …