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

[prasanna]

While preparing higher alkanes by Wurtz reac¬tion only those alkanes are preferably prepared which contain even number of carbon atoms, e.g., ethane, butane, hexane containing respectively 2, 4,6 carbon atoms.

side products, e.g., by starting with 1-bromopropane and 1-bromobutane, besides the main product heptane , hexane and octane are formed as side products.