question_answer 1)

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

Answer:

In order to prepare aIkane containing odd number of carbon atoms, a mixture of two different aloalkanes are needed. Since, two haloalkanes can react in three different ways, therefore, a mixture of three alkanes instead of the desired alkane would be formed. For example, Wurtz reaction between bromoethane and1-bromopropane gives a mixture of three alkanes, i.e., butane, pentane and hexane as shown below: \[\underset{Bromoethane}{\mathop{C{{H}_{3}}C{{H}_{2}}Br}}\,+2Na+\underset{Bromoethane}{\mathop{BrC{{H}_{2}}C{{H}_{3}}}}\,\xrightarrow[\Delta ]{Dry\,ether}\]\[\underset{\text{Butane}}{\mathop{C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}C{{H}_{3}}}}\,+2NaBr\] \[\underset{Bromoethane}{\mathop{C{{H}_{3}}C{{H}_{2}}Br}}\,+2Na+B\underset{1-Bromopropane}{\mathop{rC{{H}_{2}}C{{H}_{2}}C{{H}_{3}}}}\,\xrightarrow[\Delta ]{Dry\,ether}\] \[\underset{\text{Pentane}}{\mathop{C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}C{{H}_{2}}C{{H}_{3}}}}\,+2NaBr\] \[\underset{1-Bromopropane}{\mathop{C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}Br}}\,+2Na+\underset{1-Bromopropane}{\mathop{BrC{{H}_{2}}C{{H}_{2}}C{{H}_{3}}}}\,\xrightarrow[\Delta ]{Dry\,ether}\]\[\underset{Hexane}{\mathop{C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}C{{H}_{2}}C{{H}_{2}}C{{H}_{3}}}}\,+2NaBr\]