Question

How will you convert nitromethane to dimethylamine?

Answer 1

Hint:The formula for nitromethane is $\text{C}{{\text{H}}_{\text{3}}}\text{N}{{\text{O}}_{\text{2}}}$and that of dimethylamine is$\text{C}{{\text{H}}_{\text{3}}}\text{NHC}{{\text{H}}_{\text{3}}}$. There are different steps that can lead to this conversion, as nitromethane can be reduced to aminomethane which further needs to be treated to get the product.

Complete answer:
The steps that lead to the conversion to nitromethane to dimethylamine are as follows:
As first nitromethane is converted to aminomethane by reducing it with tin or zinc along with concentrated hydrochloric acid as follows:
$\text{C}{{\text{H}}_{\text{3}}}\text{N}{{\text{O}}_{\text{2}}}\xrightarrow{\text{Conc}\text{. HCl + Sn/ Zn/ Fe}}\text{C}{{\text{H}}_{\text{3}}}\text{N}{{\text{H}}_{\text{2}}}$
The aminomethane is then converted to methyl isocyanide in the presence of chloroform and potassium hydroxide as per the following reaction:
\[\text{C}{{\text{H}}_{\text{3}}}\text{N}{{\text{H}}_{\text{2}}}\xrightarrow{\text{CHC}{{\text{l}}_{\text{3}}}\text{/KOH}}\text{C}{{\text{H}}_{\text{3}}}\text{NC}\]
The methyl isocyanide so formed is then treated with sodium metal in presence of ethanol when it is reduced to form dimethylamine, as per the following reaction:
\[\text{C}{{\text{H}}_{\text{3}}}\text{NC}\xrightarrow{\text{Na/}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}\text{OH}}\text{C}{{\text{H}}_{\text{3}}}\text{NHC}{{\text{H}}_{\text{3}}}\]
Hence, the conversion of nitromethane to dimethylamine is done as above.

Note:
The sodium and lithium metals act as mild reducing agents in the presence of ethyl alcohol and together they reduce compounds that contain unsaturated bonds with dissimilar atoms such as the carbonyl bond and the isocyanide bond. This reduction is very similar to an organic reduction reaction called the “Bouveault- Blanc” reduction in which and ester is reduced to alcohols in presence of sodium and ethanol as the reducing agents. In this reaction, sodium metal is a one electron reducing agent source while the ethanol acts as the source of protons. To reduce one ester molecule, four atoms of sodium are required and the reaction produces sodium alkoxides as the products.