
Reagents needed along with alcohol in Victor Mayer’s test is:
A. red phosphorous and iodide
B. silver nitrate
C. nitrous acid
D. caustic soda
Answer
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Hint:: To determine the answer we should know what Victor Mayer’s test is. The test is used to distinguish alcohol. The alcohol is converted into iodide and then into nitroalkane. By further reaction nitroalkane form nitrolic acid, or nitrol depends upon the type of alcohol.
Complete step by step solution:
Victor Mayer’s test distinguished primary, secondary and tertiary alcohols.
First the alcohol is treated with phosphorus and iodine so the alkyl iodide forms.
The formation of iodide for all types of alcohol is shown as follows:
${\text{RC}}{{\text{H}}_2}{\text{OH}}\,\,\,{\text{ + }}\,{\text{P + }}{{\text{I}}_{\text{2}}}\, \to {\text{RC}}{{\text{H}}_2}{\text{I}}$ (Primary alcohol)
${{\text{R}}_2}{\text{CHOH}}\,\,\,{\text{ + }}\,{\text{P + }}{{\text{I}}_{\text{2}}}\, \to {{\text{R}}_2}{\text{CHI}}$(Secondary alcohol)
${{\text{R}}_3}{\text{COH}}\,\,\,{\text{ + }}\,{\text{P + }}{{\text{I}}_{\text{2}}}\, \to {{\text{R}}_3}{\text{CI}}$(Tertiary alcohol)
These alkyl iodides are treated with silver nitrate. By the reaction of alkyl iodide with silver nitrate nitro alkane forms.
The formation of nitroalkane for all types of alcohol is shown as follows:
${\text{RC}}{{\text{H}}_2}{\text{I}}\,\, + \,{\text{AgN}}{{\text{O}}_{\text{2}}}\, \to {\text{RC}}{{\text{H}}_2}{\text{N}}{{\text{O}}_{\text{2}}}\, + \,{\text{AgI}}$ (Primary alcohol)
${{\text{R}}_2}{\text{CHOH}}\,\,\, + \,{\text{AgN}}{{\text{O}}_{\text{2}}}\, \to
{{\text{R}}_2}{\text{CHN}}{{\text{O}}_{\text{2}}}\, + \,{\text{AgI}}$(Secondary alcohol)
${{\text{R}}_3}{\text{COH}}\,\, + \,{\text{AgN}}{{\text{O}}_{\text{2}}}\, \to {{\text{R}}_3}{\text{CN}}{{\text{O}}_{\text{2}}}\, + \,{\text{AgI}}$(Tertiary alcohol)
The nitro alkanes then are treated with nitrous acid. The primary alcohol gives nitrolic acid on reaction with nitrous acid. Secondary alcohol gives nitrole and tertiary alcohol does not react with nitric acid.
The reaction of nitroalkanes with nitrous acid is shown as follows:
${\text{RC}}{{\text{H}}_2}{\text{N}}{{\text{O}}_{\text{2}}}\, + \,{\text{HN}}{{\text{O}}_{\text{2}}}\, \to \,\mathop {{\text{RC( = NOH)N}}{{\text{O}}_{\text{2}}}}\limits_{{\text{nitrolic acid}}} + \,{{\text{H}}_2}{\text{O}}$ (Primary alcohol)
${{\text{R}}_2}{\text{CHN}}{{\text{O}}_{\text{2}}}\, + \,{\text{HN}}{{\text{O}}_{\text{2}}}\, \to \,\mathop {{{\text{R}}_2}{\text{C( - NO)N}}{{\text{O}}_{\text{2}}}}\limits_{{\text{pseudo nitrol}}} + \,{{\text{H}}_2}{\text{O}}$(Secondary alcohol)
${{\text{R}}_3}{\text{COH}}\,\, + \,{\text{HN}}{{\text{O}}_{\text{2}}}\, \to \,{\text{no reaction}}$(Tertiary alcohol)
So, red phosphorus and iodide and silver nitrate reagents are needed along with alcohol in Victor Mayer’s test.
Therefore, option (A), red phosphorous and iodide (B) silver nitrate, are correct.
Note: The nitrolic acid gives blood red colour in alkali solution and pseudo nitrol gives blue colour in alkali solution whereas tertiary nitroalkane does not react. So, the alcohol gets differentiated. The other test used for the differentiation of alcohol is Lucas test. Lucas tests differentiate on the basis of reactivity of alcohol towards hydrochloric acid.
Complete step by step solution:
Victor Mayer’s test distinguished primary, secondary and tertiary alcohols.
First the alcohol is treated with phosphorus and iodine so the alkyl iodide forms.
The formation of iodide for all types of alcohol is shown as follows:
${\text{RC}}{{\text{H}}_2}{\text{OH}}\,\,\,{\text{ + }}\,{\text{P + }}{{\text{I}}_{\text{2}}}\, \to {\text{RC}}{{\text{H}}_2}{\text{I}}$ (Primary alcohol)
${{\text{R}}_2}{\text{CHOH}}\,\,\,{\text{ + }}\,{\text{P + }}{{\text{I}}_{\text{2}}}\, \to {{\text{R}}_2}{\text{CHI}}$(Secondary alcohol)
${{\text{R}}_3}{\text{COH}}\,\,\,{\text{ + }}\,{\text{P + }}{{\text{I}}_{\text{2}}}\, \to {{\text{R}}_3}{\text{CI}}$(Tertiary alcohol)
These alkyl iodides are treated with silver nitrate. By the reaction of alkyl iodide with silver nitrate nitro alkane forms.
The formation of nitroalkane for all types of alcohol is shown as follows:
${\text{RC}}{{\text{H}}_2}{\text{I}}\,\, + \,{\text{AgN}}{{\text{O}}_{\text{2}}}\, \to {\text{RC}}{{\text{H}}_2}{\text{N}}{{\text{O}}_{\text{2}}}\, + \,{\text{AgI}}$ (Primary alcohol)
${{\text{R}}_2}{\text{CHOH}}\,\,\, + \,{\text{AgN}}{{\text{O}}_{\text{2}}}\, \to
{{\text{R}}_2}{\text{CHN}}{{\text{O}}_{\text{2}}}\, + \,{\text{AgI}}$(Secondary alcohol)
${{\text{R}}_3}{\text{COH}}\,\, + \,{\text{AgN}}{{\text{O}}_{\text{2}}}\, \to {{\text{R}}_3}{\text{CN}}{{\text{O}}_{\text{2}}}\, + \,{\text{AgI}}$(Tertiary alcohol)
The nitro alkanes then are treated with nitrous acid. The primary alcohol gives nitrolic acid on reaction with nitrous acid. Secondary alcohol gives nitrole and tertiary alcohol does not react with nitric acid.
The reaction of nitroalkanes with nitrous acid is shown as follows:
${\text{RC}}{{\text{H}}_2}{\text{N}}{{\text{O}}_{\text{2}}}\, + \,{\text{HN}}{{\text{O}}_{\text{2}}}\, \to \,\mathop {{\text{RC( = NOH)N}}{{\text{O}}_{\text{2}}}}\limits_{{\text{nitrolic acid}}} + \,{{\text{H}}_2}{\text{O}}$ (Primary alcohol)
${{\text{R}}_2}{\text{CHN}}{{\text{O}}_{\text{2}}}\, + \,{\text{HN}}{{\text{O}}_{\text{2}}}\, \to \,\mathop {{{\text{R}}_2}{\text{C( - NO)N}}{{\text{O}}_{\text{2}}}}\limits_{{\text{pseudo nitrol}}} + \,{{\text{H}}_2}{\text{O}}$(Secondary alcohol)
${{\text{R}}_3}{\text{COH}}\,\, + \,{\text{HN}}{{\text{O}}_{\text{2}}}\, \to \,{\text{no reaction}}$(Tertiary alcohol)
So, red phosphorus and iodide and silver nitrate reagents are needed along with alcohol in Victor Mayer’s test.
Therefore, option (A), red phosphorous and iodide (B) silver nitrate, are correct.
Note: The nitrolic acid gives blood red colour in alkali solution and pseudo nitrol gives blue colour in alkali solution whereas tertiary nitroalkane does not react. So, the alcohol gets differentiated. The other test used for the differentiation of alcohol is Lucas test. Lucas tests differentiate on the basis of reactivity of alcohol towards hydrochloric acid.
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