Question: For \[70\,mg\] of an organic compound on vapourisation in Victor Meyer test displaces \[22.4\,ml\] o...

Question

For $70\,mg$ of an organic compound on vapourisation in Victor Meyer test displaces $22.4\,ml$ of air measured at NTP. The compound is:
A. ${C_5}{H_{10}}$
B. ${C_4}{H_{10}}$
C. ${C_2}{H_4}$
D. $C{H_4}$

Answer 1

Solution

We know that, Victor Meyer test is used to differentiate between primary, secondary and tertiary alcohols. They can be distinguished by observing the colour obtained when they interact with Victor Meyer reagent. It is red in colour and contains a functional group i.e. $RC(N{O_2}) = NOH$

Complete answer:
First let us know about the application of Victor Meyer’s Test. It is considered as an important test for the distinction of alcohols. It is used to distinguish between primary, secondary and tertiary alcohol. The unknown alcohol interacts with Victor Meyer’s reagent, they give different colours which then are observed.
In this process, the alcohol is first treated with Iodine in presence of red phosphorus to obtain Iodoalkane. Then, Iodoalkane which is formed reacts with alcoholic silver nitrate in order to obtain nitroalkane. The nitroalkane is treated with nitrous acids (mixture of $NaN{O_2}$ and $HCl$ ) and then treated with Caustic Soda to make it alkaline.
Different types of alcohol give different colours. The primary alcohol gives blood red colour. The secondary alcohol gives the blue colour and the tertiary alcohol does not give any colour.
The following reaction occurs in the preparation of Victor Meyer’s reagent.
$RC{H_2}OH\,\, + \,\,HI \to \,RC{H_2}I$
$RC{H_2}I\,\, + \,\,AgN{O_2}\, \to \,\,RC{H_2}N{O_2}$
$\,RC{H_2}N{O_2}\,\, + \,\,\,HN{O_2} \to RC(N{O_2}) = NOH$
The final compound i.e. $RC(N{O_2}) = NOH$ is called Victor Meyer’s reagent.
In the above question,
Given mass of an organic compound = $70\,mg$ = $70\, \times \,{10^{ - 3}}\,g$
According to the given question,
$22.4\,ml$ Of air displaces $70\,mg$
So, Molecular weight of the compound = $\dfrac{{Mass\,\,of\,compound}}{{Volume\,\,at\,NTP}}\,\, \times \,22400$
Molecular weight of the compound = $\dfrac{{70\, \times \,{{10}^{ - 3}}}}{{22.4}}\,\, \times \,22400$
Molecular weight of the compound = $70$
The molecular weight of the compound is $70$ which indication that the molecular formula is ${C_5}{H_{10}}$
Hence, the correct answer is option (A).

Note:
Victor Meyer’s test is also known as the red, blue and white test. When this reagent reacts with secondary alcohol, it produces the product called pseudo nitrile. It does not react with tertiary alcohols.