My book says that Wurtz reaction can not be used for preparation of alkanes with odd number of carbon atoms. Why is it so?


It's always good to think of what would probably go on in the reaction vessel. Here's a mechanism of the Wurtz reaction:

First, one electron from the metal is transferred to the halogen to produce a metal halide and an alkyl radical.

$\ce{R–X + M → R. + M+X−}$

The alkyl radical then accepts an electron from another metal atom to form an alkyl anion and the metal becomes cationic. This intermediate has been isolated in several cases.

$\ce{R. + M → R^-M+}$

The last step is a $\ce{S_N2}$ reaction; the nucleophilic carbon of the alkyl anion displaces the halide, forming a new carbon-carbon covalent bond

$\ce{R−M+ + R–X → R–R + M+X−}$

Simple. Now, let's take a look at an example.

If we start with a single alkyl halide, in this case a methyl halide, $\ce{2 CH3-X + Na ->[][-2NaX] CH3-CH3 }$

Now, let's consider two dissimilar alkyl halides, ethyl and methyl halide: 3 possible reactions take place, giving a statistical distribution of products.

$\ce{2CH3-X + Na ->[][-2NaX] X-CH3}$

$\ce{2CH3-CH2-X + Na ->[][-2NaX] CH3-CH2-CH2-CH3}$

$\ce{CH3-X + X-CH2-CH3 + Na ->[][-2NaX] CH3-CH2-CH3}$

Sure, you've produced an alkane with an odd number of carbon atoms, but at the same time you have a mixture of alkanes (this is not really a selective reaction), that you will have a hard time separating.

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