Q.Why is Wurtz reaction not preferred for the preparation of alkanes containing odd number of carbon atoms? Illustrate your answer by taking one example.

The key idea is that the Wurtz reaction couples two alkyl halides using sodium metal, so the product alkane always contains an even number of carbon atoms — unless a mixture of two different halides is used, which then gives a statistical mixture of three alkanes.

Reasoning:

1. In the Wurtz reaction, two alkyl radicals combine:

   $2R\text{-}X + 2Na \rightarrow R\text{-}R + 2NaX$

   The product is a symmetrical alkane with twice the number of carbon atoms of the starting halide.

2. If you start with a single alkyl halide having $n$ carbons, the product has $2n$ carbons — always even.

3. To get an odd-numbered alkane, you must mix two different halides (e.g., $C_2H_5Br$ and $CH_3Br$). This yields three products:

   $C_2H_5\text{-}C_2H_5$ (even), $CH_3\text{-}CH_3$ (even), and $C_2H_5\text{-}CH_3$ (odd).

   The desired odd alkane is contaminated with two even alkanes, making separation difficult and yield poor.

Example: …

The Wurtz reaction is a classic method for making alkanes by treating alkyl halides with sodium metal in dry ether. The reaction proceeds via a free-radical mechanism, where two alkyl radicals combine. The key limitation is that when you start with two different alkyl halides, you get a statistical mixture of three possible coupling products — not just the one you want.

Let’s understand why this makes the preparation of odd-numbered alkanes impractical.

Why the Wurtz reaction fails for odd-carbon alkanes

1. The basic reaction When a single alkyl halide (say, ethyl bromide, $C_2H_5Br$) is used, the product is a symmetrical alkane with an even number of carbons:

This works cleanly because only one type of radical ($C_2H_5^\cdot$) is formed, so only one coupling product is possible.

2. The problem with two different halides

   To get an odd-carbon alkane, you must use two different alkyl halides — one with an even number of carbons and one with an odd number. For example, to prepare propane ($C_3H_8$), you might try mixing methyl bromide ($CH_3Br$) and ethyl bromide ($C_2H_5Br$).

   But the reaction mixture now contains three possible radical species: $CH_3^\cdot$, $C_2H_5^\cdot$, and the sodium surface. These radicals couple randomly, giving three products:

   • $CH_3-CH_3$ (ethane) — from two methyl radicals
   • $C_2H_5-C_2H_5$ (butane) — from two ethyl radicals
   • $CH_3-C_2H_5$ (propane) — from one methyl and one ethyl radical

3. The yield problem

   The desired odd-carbon alkane (propane) is only one of three products. Statistically, if the two halides are equally reactive, the product ratio is roughly 1:2:1 (ethane : propane : butane). The yield of the desired product is low, and separating propane from ethane and butane is difficult because their boiling points are close.

4. Why odd-carbon alkanes are especially problematic If you want an even-carbon alkane, you can simply use a single alkyl halide (e.g., $C_2H_5Br$ gives $C_4H_{10}$). But for an odd-carbon alkane, you must use two different halides, which inevitably produces a mixture. There is no way around this with the Wurtz reaction.

Illustrated example: Attempted preparation of propane

Let’s take the specific case of trying to make propane ($C_3H_8$) using the Wurtz reaction.

Reactants: Methyl bromide ($CH_3Br$) and ethyl bromide ($C_2H_5Br$) with sodium metal in dry ether.

Reaction:

Products formed: