Question

Given,\[{C_2}{H_2}\](\[20\% {H_2}S{O_4},{\text{ }}1\% H{g^{2 + }}\]) =\[A\]. A (\[Zn - Pb/HCl\](concentrate)) =\[B\]. Identify the ‘\[A\]’ & ‘\[B\]’?

Answer 1

Hint: In organic chemistry, hydrocarbons are an important topic. Hydrocarbons are majorly classified into three groups. There are alkane, alkene and alkyne. The alkane means carbon-carbon single bond. The alkene having a carbon-carbon double bond. The alkyne means carbon-carbon having a triple bond in the molecule.

Complete answer:
The chemical name of \[{C_2}{H_2}\] is ethyne.
The molecular formula of ethyne is \[{C_2}{H_2}\].
The chemical structure of \[{C_2}{H_2}\] is

The ethyne is reacted with \[20\% {H_2}S{O_4},{\text{ }}1\% H{g^{2 + }}\] to form the product of ethanal.
The chemical reaction is given below,

The ethanal is reacted with \[Zn - Pb/HCl\] to form the product of ethanol.
The chemical reaction for the above is given below,

So, molecule \[A\] is ethanal and molecule \[B\] is ethanol.
According to the above, we conclude \[{C_2}{H_2}\] (\[20\% {H_2}S{O_4},{\text{ }}1\% H{g^{2 + }}\]) =. \[A\]\[A\] (\[Zn - Pb/HCl\] (concentrate)) = \[B\], molecule \[A\] is ethanal and molecule \[B\] is ethanol.

Note:
The conversion of one type of hydrocarbon to other hydrocarbons by oxidation and reduction. The oxidation of alkane gives an alkene. The oxidation of alkene gives alkyne. The reduction of alkyne to give an alkene. The reduction of alkene to give alkane. It having some general formula. \[{{\text{C}}_{\text{n}}}{{\text{H}}_{{\text{2n + 2}}}}\] is the general formula of an alkane. \[{{\text{C}}_{\text{n}}}{{\text{H}}_{{\text{2n}}}}\] is the general formula of an alkene. \[{{\text{C}}_{\text{n}}}{{\text{H}}_{{\text{2n - 2}}}}\] is the general formula of the alkyne. Oxidation means the addition of oxygen or removal of hydrogen or the loss of electrons in the reactant in the chemical reaction. The reduction means the removal of hydrogen or the addition of oxygen or the gain of electrons in the reaction in a chemical reaction.