Which of the following amine(s) show(s) positive carbamylamine test? 
The carbamylamine test is used to detect primary amines. It results in the formation of an isocyanide (carbamylamine).
Option A \( \text{NH}_2 \) (Phenylamine) is a primary amine and will give a positive result.
Option B \( \text{(CH}_3)_2\text{NH} \) (Dimethylamine) is a secondary amine and does not give a positive result.
Option C \( \text{CH}_3\text{NH}_2 \) (Methylamine) is a primary amine and will give a positive result.
Option D \( \text{(CH}_3)_3\text{N} \) (Trimethylamine) is a tertiary amine and does not give a positive result.
Option E \( \text{H}\text{NCH}_3 \) (Methylamine attached to a benzene ring) is a primary amine and will give a positive result.
Thus, the correct options are A and C.
The carbamylamine test is used to detect the presence of primary amines (1° amines).
Reaction Mechanism:
- When a primary amine (either aliphatic or aromatic) is treated with chloroform (CHCl₃) in the presence of an alkali (such as potassium hydroxide, KOH), it undergoes a reaction to form an isocyanate.
- This is a characteristic test for primary amines, as secondary and tertiary amines do not undergo this reaction.
The reaction can be represented as follows:
\[ \text{RNH}_2 \xrightarrow{\text{CHCl}_3 + \text{KOH}} \text{RNC} \] Where:
- RNH₂ represents the primary amine group.
- CHCl₃ is chloroform, which reacts with the amine in the presence of an alkaline medium.
- KOH is potassium hydroxide, which is used to generate the basic condition needed for the reaction.
- RNC represents the isocyanate product formed.
Note:
- The reaction is specific to primary amines (1° amines), both aliphatic and aromatic. Secondary (2°) and tertiary (3°) amines do not respond to this test. Thus, if an isocyanate (RNC) is formed, it indicates the presence of a primary amine.
This test is useful in organic chemistry to distinguish primary amines from other types of amines.
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)

Cobalt chloride when dissolved in water forms pink colored complex $X$ which has octahedral geometry. This solution on treating with cone $HCl$ forms deep blue complex, $\underline{Y}$ which has a $\underline{Z}$ geometry $X, Y$ and $Z$, respectively, are


What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)