Carbylamine Reaction Mechanism - Overview, Example, Formula, Features, FAQs

Carbylamine Reaction

The Carbylamine Reaction is a characteristic chemical test used to identify primary amines (–NH₂), both aliphatic and aromatic. In this reaction, a primary amine reacts with chloroform (CHCl₃) and alcoholic potassium hydroxide (KOH) on heating to form an isocyanide (carbylamine), which has a highly unpleasant, foul odor.

General Reaction

$\mathrm{R}-\mathrm{NH}_2+\mathrm{CHCl}_3+3 \mathrm{KOH} \rightarrow \mathrm{R}-\mathrm{NC}+3 \mathrm{KCl}+3 \mathrm{H}_2 \mathrm{O}$

Where $\mathbf{R}=$ alkyl or aryl group
R-NC = isocyanide (carbylamine)

Reaction Conditions

• Reagents: Chloroform $\left(\mathrm{CHCl}_3\right)$, alcoholic KOH
• Condition: Heating
• Medium: Alcoholic (ethanol)

Hofmann's Isocyanide Test

Hofmann's Isocyanide Test is a qualitative test used to identify primary amines (both aliphatic and aromatic). In this test, a primary amine is heated with chloroform ( $\mathbf{C H C l}_{\mathbf{3}}$ ) and alcoholic potassium hydroxide $(\mathrm{KOH})$ to produce an isocyanide (carbylamine), which has a highly unpleasant smell.

$\mathrm{RNH}_2+\mathrm{CHCl}_3+3 \mathrm{KOH} \rightarrow \mathrm{RNC}+3 \mathrm{KCl}+3 \mathrm{H}_2 \mathrm{O}$

where:

• $\mathrm{RNH}_2=$ Primary amine
• $R N C=$ Isocyanide (carbylamine)

Example

$\mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_2+\mathrm{CHCl}_3+3 \mathrm{KOH} \rightarrow \mathrm{C}_6 \mathrm{H}_5 \mathrm{NC}+3 \mathrm{KCl}+3 \mathrm{H}_2 \mathrm{O}$

(Aniline forms phenyl isocyanide.)
Observation
Formation of an extremely foul-smelling isocyanide confirms the presence of a primary amine.

Mechanism

1. Chloroform reacts with alcoholic KOH to form dichlorocarbene $\left(: \mathrm{CCl}_2\right)$.

2. The primary amine attacks the carbene.

3. Subsequent elimination leads to the formation of isocyanide.

Why Secondary and Tertiary Amines Do Not Give Carbylamine Test?

The Carbylamine reaction (Hofmann’s isocyanide test) works only with primary amines because of structural and mechanistic requirements that secondary and tertiary amines do not fulfill.

Reason the Mechanism Fails in Secondary and Tertiary Amines

1. Absence of Required N–H Hydrogen Atoms

• Primary amines $\left(\mathrm{R}-\mathrm{NH}_2\right)$ have two hydrogen atoms attached to nitrogen.
• These N–H hydrogens are essential for: Formation of the reaction intermediate and Subsequent elimination steps leading to isocyanide (R–NC)
• Secondary amines $\left(\mathrm{R}_2-\mathrm{NH}\right)$ have only one N–H hydrogen
• Tertiary amines$\left(\mathrm{R}_3-\mathrm{N}\right)$ have no N–H hydrogen

Hence, the necessary elimination of HCl in the mechanism cannot occur.

2. Mechanism Requires Dehydrohalogenation

• After the primary amine attacks dichlorocarbene $\left(: \mathrm{CCl}_2\right)$, the intermediate must lose: Hydrogen from nitrogen and Chloride ion
• This dehydrohalogenation step is impossible in secondary and tertiary amines due to the lack of sufficient N–H bonds.

3. Steric Hindrance (Additional Factor)

• Secondary and especially tertiary amines are more sterically crowded.
• This reduces effective interaction with the highly reactive dichlorocarbene, further preventing the reaction.

Also read :

Isocyanide Synthesis

1. From Formamides

Isocyanides are commonly combined with dehydration. The reaction occurs in such a way that formamide is dehydrated with toluenesulfonyl chloride, phosphorus oxychloride, phosgene, diphosgene, or Burgess reagent before the base, such as triethylamine or pyridine.

The reaction is as follows:

$\mathrm{RNHC}(\mathrm{O}) \mathrm{H}+\mathrm{ArSO}_2 \mathrm{Cl}+2 \mathrm{C}_5 \mathrm{H}_5 \mathrm{~N} \rightarrow \mathrm{RNC}+\left[\mathrm{C}_5 \mathrm{H}_5 \mathrm{NH}\right]+\left[\mathrm{ArSO}_3\right]-+\left[\mathrm{C}_5 \mathrm{H}_5 \mathrm{NH}\right]+\mathrm{Cl}$

2. From Dichlorocarbon

In the reaction of carbylamine (the Alkyl isocyanide test is given by the Hofmann isocyanide test), the soluble base responds with chloroform to deliver dichlorocarbene.

Carbene then converts the four main into isocyanides.

Demonstrating the incorporation of tert-butyl isocyanide from tert-butylamine in the presence of a binding amount of catalyst transfer phase called benzyl triethylammonium chloride.

$\mathrm{Me}_3 \mathrm{CNH}_2+\mathrm{CHCl}_3(\mathrm{l})+3 \mathrm{NaOH}(\mathrm{aq}) \rightarrow \mathrm{Me}_3 \mathrm{CNC}+3 \mathrm{NaCl}(\mathrm{s})+3 \mathrm{H}_2 \mathrm{O}(\mathrm{l})$

The above reaction occurs in the azo Dye test of amines. This test is also helpful in producing secondary amines.

3. Silver Cyanide Test

The first isocyanide, called allyl isocyanide, was prepared by the reaction of allyl iodide with silver cyanide. The reaction was as follows:

$\mathrm{RI}+\mathrm{AgCN} \rightarrow \mathrm{RNC}+\mathrm{AgI}$

4. By deduction

Isocyanides include the separation of oxazoles and benzoxazoles in the second place. The following compound of organolithium exists in the synthetic balance with 2-isocyano-phenolate, which can be trapped by an electrophile as an acidic chloride.

Also read -

Some Solved Examples

Question 1: Compound ' A' reacts with nitrous acid to form p - p-nitroso N-N dimethyl aniline. Compound ' A ' is:

1) N - nitrosoaniline

2) p-nitroso N-methyl aniline

3) (correct) N - N dimethyl aniline

4) N - phenyl - N - methyl methanamine

Solution:

As we learn:
Carbylamine reaction -
The product is isocyanide & this reaction is used for the detection of primary amines.

- wherein

N - N dimethyl aniline

Hence, the answer is option (3).

Question 2: This compound does not respond to the carbylamine reaction:

1) Isopropylamine

2) (correct) Diethylamine

3) t-Butylamine

4) sec.Butylamine

Solution:

As we learn
carbylamine Reaction -
Product is isocyanide & this reaction is used for the detection of primary amines.
- wherein

$\mathrm{RNH}_2+\mathrm{CHCl}_3+3 \mathrm{KOH} \rightarrow \mathrm{R}-\mathrm{NC}+3 \mathrm{KCl}+3 \mathrm{H}_2 \mathrm{O}$

an isocyanide (foul-smelling)
Hence, the answer is option (2).

Question 3: Which of the following species are involved in the carbylamine test?
(i) $R-N C$
(ii) $\mathrm{CHCl}_3$
(iii) $\mathrm{COCl}_2$
(iv) $\mathrm{NaNO}_2+\mathrm{HCl}$
1) (correct) (i) and (ii)
2) (iii) and (iv)
3) (ii) and (iii)
4) (ii) and (iv)

Solution:

Option (i) and (ii) are the correct answers.
Explanation: Amine, when it reacts with a mixture of $\mathrm{CHCl}_3$ and KOH, gives out alkyl isocyanate. This reaction is called a Carbylamine reaction.
Here,

$\mathrm{R}-\mathrm{NH}_2+\mathrm{CHCl}+3 \mathrm{KOH} \rightarrow \mathrm{RNC}+3 \mathrm{KCl}+3 \mathrm{H}_2 \mathrm{O}$

only RNC and $\mathrm{CHCl}_3$ are involved in carbylamine reaction.

Hence, the answer is option (1).

Question 4: The peptide that gives positive ceric ammonium nitrate and carbylamine tests is:

1) (correct) Ser-Lys

2) Gln-Asp

3) Lys-Asp

4) Asp-Gln

Solution:

As we have learnt,

Red coloration with Ceric ammonium nitrate confirms the presence of alcohols

Alcohol + Ceric Ammonium Nitrate → Red Solution

Foul-smelling isocyanide obtained in the Carbylamine Reaction indicates the presence of Primary amines

$\mathrm{RNH}_2+\mathrm{CHCl}_3+3 \mathrm{KOH} \rightarrow \mathrm{R}-\mathrm{NC}+3 \mathrm{KCl}+3 \mathrm{H}_2 \mathrm{O}$

Thus,

Positive ceric ammonium nitrate -OH group

Positive carbylamine tests -NH₂ group

(1) Ser-Lys ( Serine + Lysine )

Both amine as well as alcohol groups are present so it will give positive tests for both tests.

Hence, the answer is the option (1).

Practice More Question with the link given below