Question

How to prepare ethanal from ethyne by shortest method?

Answer 1

Hint: We should know about reactants and products. The given reactant is an alkyne and the product is an aldehyde. We have to determine the reagent. The compounds that react with the main reactant are known as a reagent. We have to break the triple bond of alkyne and attach an oxygen atom to get the aldehyde. We can use strong acid or metal salt to break the triple bond and water.

Complete step by step answer:
The ethyne is an alkyne and ethanal is an aldehyde. The ethanal can be prepared by the ethyne by treating the ethyne with mercuric sulphate in presence of acid.
The reaction equation is as follows:
\[{\text{HC}} \equiv {\text{CH}}\,\mathop {\mathop \to \limits^{{\text{HgS}}{{\text{O}}_{\text{4}}}} }\limits_{{{\text{H}}_2}{\text{S}}{{\text{O}}_{\text{4}}}} \,\,{\text{C}}{{\text{H}}_{\text{3}}}{\text{CHO}}\]
First the mercury attacks on the reactant ethyne and forms a cyclic structure. Then water attacks on this cyclic structure forming a mercury cation. Then by the removal of hydride alcohol forms.
The enol can be converted into keto form by keto-enol tautomerism. The oxygen atom of the ketone attacks on the acid and gets protonated and the diacation of mercury removes. The enol form again converts into keto form by keto-enol tautomerism.

The mechanism of the reaction of ethyne with mercuric ion and acid is as follows:

So, ethanal can be prepared from ethyne by mercuration demercuration.

Therefore, the shortest method to prepare ethanal from ethyne is mercuration demercuration.

Note: In this reaction, mercury attaches with the reactant in between the reaction mechanism and then removes so, the reaction is known as oxymercuration demercuration. In the case of a symmetrical alkyne, water in the second step can attack from any side, so the ring can open from any side. But in the case of an asymmetrical alkyne, the water attacks from the more nucleophilic side.