Why is Wurtz reaction not preferred for the preparation of alkanes containing odd number of carbon atoms? Illustrate your answer by taking one example
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Wurtz reaction is not preferred for the preparation of alkanes containing odd number of carbon atoms due to the formation of side products. For example, by starting with 1 -bromopropane and 1 - bromobutane, hexane and octane are the side products besides heptane
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for product of alkane containing odd no of carbons , the reactants will be two different alkyl halides. when two different alkyl halides reacts ,then mixture of alkanes formed. thats why wurtz reaction is not preferred for the preparation of alkanes containing odd no of carbons.
example of reaction of two different alkyl halides with sodium .
3CH3I + 4C2H5I + 6Na --> CH3CH3 + C2H5CH3 + C2H5C2H5 + 6NaI
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