Question

In Victor Mayer’s test, which compound do we get for secondary alcohol?
(a)- Iodoalkane
(b)- Nitroalkane
(c)- Nitrolic acid
(d)- Pseudonitrole

Answer 1

Victor Mayer's test is used to differentiate between the primary alcohol that gives blood-red coloration, secondary alcohol that gives blue coloration, and tertiary alcohol that remains colorless.

Complete step by step answer:
Victor Mayer's test consists of the following steps:
(i)- The given alcohol is first converted into its alkyl iodide by treating it with phosphorus and iodine ($P/{{I}_{2}}$).
(ii)- The alkyl iodide is now converted into its corresponding nitroalkane by treating it with silver nitrate ($AgN{{O}_{2}}$).
(iii)- Now this nitroalkane is then treated with nitrous acid ($HN{{O}_{2}}$), which is a mixture of sodium nitrite and hydrochloric acid ($NaN{{O}_{2}}+HCl$).
(iv)- The resulting solution we get is finally made alkaline with an aqueous solution of sodium hydroxide or potassium hydroxide.
If the color appears to be blood red, then it is a primary alcohol.
If the color appears to be blue, then it is a secondary alcohol.
If the solution doesn't change and remains colorless, then it is a tertiary alcohol.
The reaction of secondary alcohol following the above steps is given below:
${{R}_{2}}CHOH\xrightarrow{P+{{I}_{2}}}{{R}_{2}}CH-I$
${{R}_{2}}CH-I\xrightarrow{AgN{{O}_{2}}}{{R}_{2}}CH-N{{O}_{2}}$
${{R}_{2}}CH-N{{O}_{2}}\xrightarrow{HONO}{{R}_{2}}C(N=O)-N{{O}_{2}}$
${{R}_{2}}C(N=O)-N{{O}_{2}}\xrightarrow{NaOH}Blue\text{ }coloration$
The compound ${{R}_{2}}C(N=O)-N{{O}_{2}}$ is called Pseudo Nitrol which is responsible for the blue coloration.

Therefore, the correct answer is an option (d)- Pseudonitrole.

Note: The reaction of primary alcohol following the above steps is given below:
$RC{{H}_{2}}OH\xrightarrow{P+{{I}_{2}}}RC{{H}_{2}}-I$
$RC{{H}_{2}}-I\xrightarrow{AgN{{O}_{2}}}RC{{H}_{2}}-N{{O}_{2}}$
$RC{{H}_{2}}-N{{O}_{2}}\xrightarrow{HONO}RC(NOH)-N{{O}_{2}}$
$RC(NOH)-N{{O}_{2}}\xrightarrow{NaOH}Blood\text{ }red\text{ }coloration$
$RC(NOH)-N{{O}_{2}}$ is called nitrolic acid and is responsible for blood-red color.
The reaction of tertiary alcohol following the above steps is given below:
${{R}_{3}}COH\xrightarrow{P+{{I}_{2}}}{{R}_{3}}C-I$
${{R}_{3}}C-I\xrightarrow{AgN{{O}_{2}}}{{R}_{3}}C-N{{O}_{2}}$
${{R}_{3}}C-N{{O}_{2}}\xrightarrow{HONO}\text{ no reaction (colorless)}$