Answer
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Hint: Wurtz reaction not preferred for the preparation of alkane containing odd number of carbon atoms due to formation of number of products i.e. mixture of alkane.
Complete answer:
Wurtz reaction not preferred for the preparation of alkane containing an odd number of carbon atoms due to the formation of a number of products. When two different types of alkyl halides are taken, a mixture of three alkanes with odd and even numbers of carbon atoms are obtained.
For example: \[1\]- chloroethane and \[1\]- chloropropane react with Na in the presence of dry ether. It gives butane, hexane, pentane.
By the reaction
\[
\mathop {C{H_3}C{H_2}Cl}\limits_{\left( {Chloropropane} \right)} + Na + Cl - \mathop {C{H_2}C{H_2}C{H_3}}\limits_{\left( {Chloropropane} \right)} \xrightarrow{{dry{\text{ }}ether}}\mathop {{\text{C}}{{\text{H}}_3}C{H_2}C{H_3}}\limits_{\left( {Bu\tan e} \right)} + \mathop {C{H_3}C{H_2}C{H_2}C{H_2}C{H_3}}\limits_{\left( {Pen\tan e} \right)} + \mathop {C{H_3}C{H_2}C{H_2}C{H_2}C{H_2}C{H_3}}\limits_{\left( {hexane} \right)} \\
mixtures{\text{ }}of{\text{ }}alkane \\
\]
This reaction is preceded by an ionic and free radical mechanism.Pure even numbers of carbon atoms of alkane are not formed.
Wurtz reaction is a coupling reaction that must have at least two carbon atoms in alkyl halide and only primary halides are used but not secondary and tertiary halide. In secondary and tertiary halides, alkene is also formed by this reaction. Wurtz reaction generally fails when tertiary alkyl halides are used because tertiary alkyl halides include elimination reaction as side reactions. It happens due to steric – hindrance.
Note:The reason why we we add dry ether to the wurtz reaction is that we have used the Na metal in the reaction which is highly reactive so we need to add a solvent that does not react with the metal.Moreover here if we do not add dry ether to the reaction then sodium reacts with water to form oxides and hydroxides and disturbs the reaction,thus we use dry ether to the reaction.
Complete answer:
Wurtz reaction not preferred for the preparation of alkane containing an odd number of carbon atoms due to the formation of a number of products. When two different types of alkyl halides are taken, a mixture of three alkanes with odd and even numbers of carbon atoms are obtained.
For example: \[1\]- chloroethane and \[1\]- chloropropane react with Na in the presence of dry ether. It gives butane, hexane, pentane.
By the reaction
\[
\mathop {C{H_3}C{H_2}Cl}\limits_{\left( {Chloropropane} \right)} + Na + Cl - \mathop {C{H_2}C{H_2}C{H_3}}\limits_{\left( {Chloropropane} \right)} \xrightarrow{{dry{\text{ }}ether}}\mathop {{\text{C}}{{\text{H}}_3}C{H_2}C{H_3}}\limits_{\left( {Bu\tan e} \right)} + \mathop {C{H_3}C{H_2}C{H_2}C{H_2}C{H_3}}\limits_{\left( {Pen\tan e} \right)} + \mathop {C{H_3}C{H_2}C{H_2}C{H_2}C{H_2}C{H_3}}\limits_{\left( {hexane} \right)} \\
mixtures{\text{ }}of{\text{ }}alkane \\
\]
This reaction is preceded by an ionic and free radical mechanism.Pure even numbers of carbon atoms of alkane are not formed.
Wurtz reaction is a coupling reaction that must have at least two carbon atoms in alkyl halide and only primary halides are used but not secondary and tertiary halide. In secondary and tertiary halides, alkene is also formed by this reaction. Wurtz reaction generally fails when tertiary alkyl halides are used because tertiary alkyl halides include elimination reaction as side reactions. It happens due to steric – hindrance.
Note:The reason why we we add dry ether to the wurtz reaction is that we have used the Na metal in the reaction which is highly reactive so we need to add a solvent that does not react with the metal.Moreover here if we do not add dry ether to the reaction then sodium reacts with water to form oxides and hydroxides and disturbs the reaction,thus we use dry ether to the reaction.
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