Question

An organic compound with formula ${{C}_{2}}{{H}_{6}}O$ undergoes oxidation with ${{K}_{2}}C{{r}_{2}}{{O}_{7}}/{{H}_{2}}S{{O}_{4}}$ to produce ‘X’ which contains 40% carbon, 6.7% hydrogen and 53.3% oxygen. The molecular formula of the compound is:
[A]$C{{H}_{2}}O$
[B]${{C}_{2}}{{H}_{4}}{{O}_{2}}$
[C]${{C}_{2}}{{H}_{4}}O$
[D]${{C}_{2}}{{H}_{6}}{{O}_{2}}$

Answer 1

Hint: The given compound is ethanol. When ethanol is oxidised, it forms carboxylic acid. If the percentage of carbon, hydrogen and oxygen atoms in carboxylic acid match with the given percentages of these atoms in ‘X’ then, carboxylic acid will be the oxidised product.

Complete answer:
In the question, it is given to us that the compound ${{C}_{2}}{{H}_{6}}O$ undergoes oxidation to give us a compound ‘X’. We can write this as a reaction as-
     \[{{C}_{2}}{{H}_{6}}O\xrightarrow{{{K}_{2}}C{{r}_{2}}{{O}_{7}}/{{H}_{2}}S{{O}_{4}}}X\]
We can write${{C}_{2}}{{H}_{6}}O$in two isomeric forms i.e.$C{{H}_{3}}-O-C{{H}_{3}}\text{ and }{{\text{C}}_{2}}{{H}_{5}}OH$.
As we know, ethers do not undergo oxidation, therefore the only feasible oxidation reaction of ${{C}_{2}}{{H}_{6}}O$ will be-
     \[{{C}_{2}}{{H}_{5}}OH\xrightarrow{{{K}_{2}}C{{r}_{2}}{{O}_{7}}/{{H}_{2}}S{{O}_{4}}}C{{H}_{3}}COOH\]
Therefore, we can assume that ‘X’ is\[C{{H}_{3}}COOH\]. To check whether it is correct or not we will compare it to the given percentages of carbon, oxygen and hydrogen.
Molecular mass of $C{{H}_{3}}COOH$ will be=$\left( 12\times 2 \right)+\left( 4\times 1 \right)+\left( 16\times 2 \right)$,as it contains 2 atoms of carbon, 4 atoms of hydrogen and 2 atoms of oxygen.
Therefore, the molecular mass = 60.
Now, we can calculate the percentage of carbon, hydrogen and oxygen atoms-
Percentage of carbon atoms =$\dfrac{no.of\text{ }C-atoms\times atomic\text{ mass of carbon}}{molecular\text{ mass of C}{{\text{H}}_{3}}COOH}\times 100=\dfrac{2\times 12}{60}\times 100=40%$

Similarly, percentage of hydrogen =$\dfrac{4\times 1}{60}\times 100=6.666%\simeq 6.67%$
Percentage of oxygen=$\dfrac{2\times 16}{60}\times 100=53.33%$
As the above percentages match with the given composition of ‘X’ which is, 40% carbon, 6.7% hydrogen and 53.3% oxygen, therefore we can say that X is $C{{H}_{3}}COOH$ which can also be written as${{C}_{2}}{{H}_{4}}{{O}_{2}}$.
So, the correct answer is “Option B”.

Note: In this case, the reaction was common and known to us so we used this method to solve the question. But if the oxidation product was unknown to us, we could have calculated the percentage of each atom in all of the given options and whichever matched with the given composition of ‘X’, that would be the correct answer.