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

(N/A) In the Wurtz reaction,an alkyl halide $(R-X)$ reacts with sodium in the presence of dry ether to form a symmetric alkane with double the number of carbon atoms present in the alkyl group.
If we attempt to prepare an alkane with an odd number of carbon atoms by using a mixture of two different alkyl halides ($R-X$ and $R'-X$),a mixture of three different alkanes is formed: $R-R$,$R'-R'$,and $R-R'$.
For example,to prepare propane $(C_3H_8)$,a mixture of methyl bromide $(CH_3Br)$ and ethyl bromide $(C_2H_5Br)$ is used:
$2CH_3Br + 2Na \xrightarrow{\text{Dry ether}} CH_3-CH_3 + 2NaBr$
$2C_2H_5Br + 2Na \xrightarrow{\text{Dry ether}} C_2H_5-C_2H_5 + 2NaBr$
$CH_3Br + C_2H_5Br + 2Na \xrightarrow{\text{Dry ether}} CH_3-C_2H_5 + 2NaBr$
As seen,the product is a mixture of ethane,$n$-butane,and propane. Since the boiling points of these alkanes are very close,their separation is extremely difficult and costly. Therefore,the Wurtz reaction is not preferred for the synthesis of alkanes with an odd number of carbon atoms.