Which of the following will react with CHCl3 + alc. KOH?
(a) Adenine and lysine
(b) Thymine and prol

Q: Why don't secondary amines give the Carbylamine test?

The formation of an isocyanide ($\text{R-N}\equiv\text{C}$) requires the nitrogen atom of the amine to lose two hydrogen atoms during the reaction mechanism. Secondary amines ($\text{R}_2\text{NH}$) only have one hydrogen atom attached to the nitrogen, so they cannot form the triple bond required for the isocyanide product.

Q: Is the amino group in Adenine considered aliphatic or aromatic?

The $-\text{NH}_2$ group in Adenine is attached directly to the aromatic purine ring, making it an aromatic primary amine. Fortunately, the Carbylamine reaction is given by both aliphatic and aromatic primary amines, so Adenine will give a positive test.

Q: Why is Proline considered a secondary amine?

In Proline, the nitrogen atom is part of a five-membered pyrrolidine ring. It is bonded to the $\alpha$-carbon and the terminal carbon of the side chain. Because it is bonded to two carbon atoms, it is classified as a secondary amine.

Q: What exactly is the foul-smelling compound formed in this reaction?

The foul-smelling compound is an isocyanide (also called a carbylamine), which has the general structure $\text{R-N}\equiv\text{C}$. The intense, unpleasant odor is the primary indicator of a positive test.

Q: Why do we use alcoholic KOH instead of aqueous KOH?

Alcoholic KOH acts as a strong base and favors elimination reactions. In the Carbylamine mechanism, it helps generate the highly reactive intermediate, dichlorocarbene ($\text{:CCl}_2$), from chloroform. Aqueous KOH would instead favor nucleophilic substitution, leading to different products.

Question

Which of the following will react with $	ext{CHCl}_3 + 	ext{alc. KOH}$?

Options:

(A) Adenine and lysine

(B) Thymine and proline

(C) Adenine and thymine

(D) Adenine and proline

Sigma Percentile · Accepted Answer

### The Magic of the Carbylamine Reaction

Imagine you are working in a chemistry lab, and you need a quick, definitive way to identify if a compound contains a primary amine. You reach for chloroform ($	ext{CHCl}_3$) and alcoholic potassium hydroxide ($	ext{KOH}$). When you mix these with a primary amine and apply a little heat, an unmistakable, intensely foul smell fills the room. This is the **Carbylamine Reaction**, also known as the Isocyanide test.

The general equation for this transformation is:

$$	ext{R-NH}_2 + 	ext{CHCl}_3 + 3	ext{KOH} \xrightarrow{\Delta} 	ext{R-N}\equiv	ext{C} + 3	ext{KCl} + 3	ext{H}_2	ext{O}$$

This reaction is a favorite among examiners because it is highly specific. It acts as a chemical bouncer, strictly allowing only **primary ($1^\circ$) amines** to pass through and form the foul-smelling isocyanide. Secondary ($2^\circ$) and tertiary ($3^\circ$) amines are turned away at the door.

### Analyzing the Biomolecules

In our problem, we are presented with four biological molecules: Adenine, Lysine, Thymine, and Proline. To solve this, we must act as molecular detectives and inspect the degree of the amine groups in each structure.

**1. Adenine:** 
Adenine is a purine nucleobase found in DNA and RNA. If you look closely at its structure, you will see an amino group ($-	ext{NH}_2$) attached to the 6th carbon of the purine ring. Because this nitrogen is attached to only one carbon atom and holds two hydrogen atoms, it is a classic **primary amine**.

**2. Lysine:** 
Lysine is an essential amino acid. Its structure is a linear aliphatic chain terminating in a carboxylic acid. Crucially, it possesses two amino groups: one at the $\alpha$-carbon and another at the $\epsilon$-carbon at the end of the chain. Both of these are $-	ext{NH}_2$ groups, making them **primary amines**.

**3. Thymine:** 
Thymine is a pyrimidine nucleobase. When we examine its ring, we find nitrogen atoms incorporated directly into the cyclic structure. These nitrogens are bonded to two adjacent carbon atoms within the ring, leaving them with only one hydrogen atom ($-	ext{NH}-$). Therefore, they are **secondary amines**.

**4. Proline:** 
Proline is a unique amino acid. Unlike the others, its side chain loops back and bonds with the backbone nitrogen, forming a five-membered pyrrolidine ring. Because the nitrogen is bonded to two carbon atoms (the $\alpha$-carbon and the side chain carbon), it is a **secondary amine**.

### The Final Verdict

Now, let's bring our logic together. The Carbylamine reaction demands a primary amine.

Thymine and Proline, being secondary amines, fail the test. Why? Because the mechanism of forming an isocyanide ($-	ext{N}\equiv	ext{C}$) requires the nitrogen atom to lose *two* protons during the reaction. Secondary amines only have one proton to give, so the reaction stalls.

Adenine and Lysine, however, proudly wear their primary $-	ext{NH}_2$ groups. They have the necessary two protons to lose and will readily react with $	ext{CHCl}_3$ and alc. $	ext{KOH}$ to produce the characteristic foul-smelling isocyanides.

**Therefore, the correct pair that will react is Adenine and Lysine.**

Animated Solution for Chemistry - Organic Chemistry: Which of the following will react with $CHCl_{3} + alc. KOH$ ?

Select Answer:

Visualized Solution

$B i o m o l ec u l es & A min es$

$C a r b y l amin e R e a c t i o n$

$S p ec i f i c i t y o f t h e T es t$

$A na l y z in g A d e nin e & L y s in e$

$A na l y z in g T h y min e & P r o l in e$

$C o n c l u s i o n$

$T h e W a y F or w a r d$

The Sigma Insight: Amines

The Magic of the Carbylamine Reaction

Analyzing the Biomolecules

The Final Verdict

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Animated Solution for Chemistry - Organic Chemistry: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Which of the following will react with CHCl3​+alc. KOH?