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

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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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