Question

The organic compound A of molecular formula \[{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{4}}}{{\text{O}}_{\text{2}}}\] gives brisk effervescence with a sodium bicarbonate solution. The sodium salt of A on treatment with soda lime gives a Hydrocarbon B of molecular mass 16. It belongs to the first member of the alkane family. What are A and B and how will you prepare A from ethanol?

Answer 1

Hint:Carbon is known to have a very high degree of catenation. Catenation is the ability of an element to form bonds with itself and form a long chain. Thus, carbon can form a long chain with other carbon atoms. And due to this millions of organic compounds are found. Organic compounds are generally any chemical species having carbon-hydrogen bonds. They usually also contain other hetero-atoms such as oxygen, nitrogen, Sulphur, halogens, etc.

Complete step by step answer:
The given compound A reacts with Sodium bicarbonate to give a brisk effervescence. Sodium bicarbonate is a base and therefore the compound A must be an acid because it can react with sodium bicarbonate.
An acid when reacts with sodium bicarbonate carbon dioxide is released and therefore a brisk effervescence is observed.
It is quite clear now that compound A is an acid. The molecular formula of A is given as \[{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{4}}}{{\text{O}}_{\text{2}}}\], therefore compound A should be a carboxylic acid having two carbon atoms, probably acetic acid.
Also, the sodium salt of A given a compound B having a molecular weight of 16 when treated with Soda lime (a mixture of calcium hydroxide and sodium hydroxide). The sodium salt of carboxylic acid when treated with soda lime, the acid undergoes decarboxylation i.e., removal of carbon dioxide.
Compound B is the first member of the alkane series and having molecular mass 16. Thus, it must be methane.
Considering the above facts, compound A is acetic acid and B is methane.
The chemical reactions involved are:
${\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH + NaHC}}{{\text{O}}_{\text{3}}} \to {\text{C}}{{\text{H}}_{\text{3}}}{\text{COONa + }}{{\text{H}}_{\text{2}}}{\text{O + C}}{{\text{O}}_{\text{2}}}$
${\text{C}}{{\text{H}}_{\text{3}}}{\text{COONa + NaOH}} \to {\text{C}}{{\text{H}}_{\text{4}}}{\text{ + N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$
Acetic acid can be prepared from ethanol through oxidation in the presence of potassium dichromate. The oxidation occurs in two steps, first, the ethanol is converted to ethanal and then the latter is converted to acetic acid:
${\text{C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{OH}}\xrightarrow{{{{\text{K}}_{\text{2}}}{\text{C}}{{\text{r}}_{\text{2}}}{{\text{O}}_{{\text{7,}}}}{{\Delta }}}}{\text{C}}{{\text{H}}_{\text{3}}}{\text{CHO}}\xrightarrow{{{{\text{K}}_{\text{2}}}{\text{C}}{{\text{r}}_{\text{2}}}{{\text{O}}_{\text{7}}}{{,\Delta }}}}{\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH}}$

Note:
Carbon has a very high tendency of catenation and due to this reason millions and millions of carbon compounds are known. Compounds having carbon-hydrogen bonds are called organic compounds. And these organic compounds are so huge in number that a separate branch of chemistry was created which is completely dedicated to the study of properties, nature, and reactions of the organic compounds and to this study easier they are divided into families such as the alkane family, alkene family, alcohol family and so on.