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Carbylamine test is used to detect which one of the following?
A. ${C}_{6}{H}_{5}CHO$
B. ${C}_{6}{H}_{5}C{O}_{2}H$
C. ${C}_{6}{H}_{5}N{H}_{2}$
D. ${C}_{6}{H}_{5}OH$

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Last updated date: 20th Jun 2024
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Answer
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Hint: Carbylamine test is a process which involves dehydrohalogenation of chloroform in alkaline medium. In this reaction, analyte is heated with alcoholic potassium hydroxide and chloroform.

Complete step by step answer:
- Primary amines are the amines which are attached to only one carbon atom of the hydrocarbon chain.
- Secondary amines are the amines which are attached to two carbon atoms of the hydrocarbon chain.
- Tertiary amines are the amines which are attached to three carbon atoms of the hydrocarbon chain.
- The carbylamine reaction is also known as Hofmann isocyanide synthesis. This reaction is only given by primary amines because secondary and tertiary amines have more than one alkyl group attached to them and this hinders the approach of $-N{H}_{2}$ group.
- It includes a primary amine, chloroform and a base to synthesize an isocyanide.
- In carbylamine reaction, the amine is added to the intermediate that is created by the dehydrohalogenation of chloroform. This intermediate is known as dichlorocarbene. This intermediate is very important for this reaction.
- The general carbylamine test can be written as follows:
$\underset { Alkyl\quad amine }{ R-N{ H }_{ 2 } } \quad +\quad \underset { Chloroform }{ CH{ Cl }_{ 3 } } \quad +\quad \underset { Alcoholic }{ 3KOH } \quad \longrightarrow \quad \underset { Carbylamine }{ { R}-\overset { + }{ N } \equiv \overset { - }{ C } } \quad +\quad \underset { Potassium\quad chloride }{ 3KCl } \quad +\quad \underset { Water }{ 3{ H }_{ 2 }O }$
As we know, that primary amines undergo this reaction, then among the options provided to us, only option (C) fulfils this criteria. Options (A), (B), and (D) do not give this reaction because they don't have an amine group. Therefore, the correct option is option (C).
- Let us now look, how this process takes place.
Firstly, the dehydrohalogenation, which is the removal of hydrogen form a given substrate, of chloroform takes place which, as a result, gives a very reactive intermediate known as dichlorocarbene. This electrophilic dichlorocarbene is the nucleophilic nitrogen in the primary amine. After this, the base helps to form isocyanide. The reaction involved is given below.
$\underset { Aniline }{ { C }_{ 6 }{ H }_{ 5 }N{ H }_{ 2 } } \quad +\quad \underset { Chloroform }{ CH{ Cl }_{ 3 } } \quad +\quad \underset { Alcoholic }{ 3KOH } \quad \longrightarrow \quad \underset { Phenyl\quad isocyanide }{ { C }_{ 6 }{ H }_{ 5 }-\overset { + }{ N } \equiv \overset { - }{ C } } \quad +\quad \underset { Potassium\quad chloride }{ 3KCl } \quad +\quad \underset { Water }{ 3{ H }_{ 2 }O }$

Note: Carbylamine tests can be used in the synthesis of isocyanides to prepare secondary amines. This test can also be used to detect the presence of primary amines.