Question

Give a balanced chemical equation to convert methyl cyanide to ethyl alcohol.

Answer 1

Hint: We can convert methyl cyanide into a primary amine. Using a mixture of sodium nitrite and hydrochloric acid and then hydrolysing the primary amine will give us ethyl alcohol or ethanol.

Complete step by step answer:
To convert methyl cyanide to ethyl alcohol, or ethanol, we can start by reducing the cyanide to primary amine using a strong reducing agent. From that primary amine we will form a diazonium salt which we can use differently depending upon the required product. Here, the required product is ethanol which is an alcohol. We know that alcohols contain –OH functional groups, therefore we can add water to this diazonium salt and it will give us ethanol.

We can write the reactions for this process step wise as-
For conversion of methyl cyanide to a primary amine, we will need sodium metal and an alcohol-
     \[MeCN+4\left[ H \right]\xrightarrow{Na/{{C}_{2}}{{H}_{5}}OH}{{C}_{2}}{{H}_{5}}N{{H}_{2}}\]

To this primary amine, we can use a mixture of sodium nitrite and hydrochloric acid. In addition to the mixture, the amine will undergo diazotization and give us an intermediate salt. Upon addition of water to this diazonium salt we will end up on our required product which is ethyl alcohol-

     \[{{C}_{2}}{{H}_{5}}N{{H}_{2}}\xrightarrow{NaN{{O}_{2}}/HCl}\left[ C{{H}_{3}}C{{H}_{2}}{{N}_{2}}^{+}C{{l}^{-}} \right]\xrightarrow{{{H}_{2}}O}{{C}_{2}}{{H}_{5}}OH\]

Therefore, we can write the balanced equation as-
     \[MeCN+4\left[ H \right]\xrightarrow{Na/{{C}_{2}}{{H}_{5}}OH}{{C}_{2}}{{H}_{5}}N{{H}_{2}}\xrightarrow{NaN{{O}_{2}}/HCl}\left[ C{{H}_{3}}C{{H}_{2}}{{N}_{2}}^{+}C{{l}^{-}} \right]\xrightarrow{{{H}_{2}}O}{{C}_{2}}{{H}_{5}}OH\]

Note: There is one more procedure of conversion of methyl cyanide to alcohols which we can use for this conversion. It is by hydrolysing methyl cyanide in an acidic medium to give us carboxylic acid and then reduction of carboxylic acid will give us alcohol. There is formation of an aldehyde as an intermediate in this reaction.

     \[\begin{align}
  & C{{H}_{3}}CN\xrightarrow{{{H}_{2}}O/{{H}^{+}}}C{{H}_{3}}COOH \\
 & C{{H}_{3}}COOH\xrightarrow{LiAl{{H}_{4}}/{{H}_{3}}{{O}^{+}}}C{{H}_{3}}C{{H}_{2}}OH \\
\end{align}\]