Question

Alcohol which gives blue coloration in Victor Meyer test is:
(A) Only $ {\text{C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{ - OH}} $
(B) Only $ {\text{C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_2}{\text{C}}{{\text{H}}_{\text{2}}}{\text{ - OH}} $
(C) All alcohols containing $ - {\text{C}}{{\text{H}}_2} - {\text{OH}} $ group
(D) All alcohols containing $ > {\text{CH - OH}} $ group

Answer 1

Hint: To answer this question, you must recall the Victor- Meyer’s test. Victor- Meyer’s test is used to distinguish between different types of alcohols. On treatment with Victor Meyer’s reagent, different colors are obtained for primary, secondary and tertiary alcohols and thus they are easily distinguishable. Blue color is obtained for secondary alcohol.

Complete step by step solution
In the question, we are given that the color obtained in the Victor- Meyer’s test is blue. Let us study the reaction taking place in the test.
It is a three step reaction. The first step involves the reaction of the alcohol with red phosphorus and iodine to give alkyl iodide as the product. It is a substitution reaction. The reaction occurring in the first step can be given as:
 $ {\text{R - OH}}\xrightarrow[{}]{{{\text{Red Phosphorous + }}{{\text{I}}_2}}}{\text{R - I}} $
In the second step, the alkyl iodide is reacted with silver nitrate to produce nitro- alkane as the product. The chemical reaction occurring between alkyl iodide and silver nitrate is represented as:
 $ {\text{R - I}}\xrightarrow[{}]{{{\text{AgN}}{{\text{O}}_2}}}{\text{R - N}}{{\text{O}}_2} $
In the third step, the nitro alkane formed in the second step of the reaction is made to react with nitrous acid, which is prepared in- situ by a mixture of sodium nitrate and hydrochloric acid. The reaction occurring between the nitro alkane and nitrous acid is given as:
 $ {\text{R - N}}{{\text{O}}_{\text{2}}}\xrightarrow[{}]{{{\text{HN}}{{\text{O}}_{\text{2}}}}}{\text{Nitrolic acid or Psedonitriol}} $
The above solution is made alkaline and then the colour obtained helps us determine the type of alcohol. If the coloration is blue, the alcohol is a secondary alcohol.
The correct answer is D.

Note
If the reaction is conducted for a primary alcohol, then we get a blood red solution. If the reaction is conducted for a tertiary alcohol, then we get a colorless solution.