Which of the following amine(s) show(s) positive carbylamine test? A.

B. (CH₃)₂NH$(CH_3)_2NH$ C. CH₃NH₂$CH_3NH_2$ D. (CH₃)₃N$(CH_3)_3N$ E.

Choose the correct answer from the options given below:

Core Logic

The reaction sequence represents the classic dye test for aromatic primary amines. Step 1 (Diazotization): Aniline (Ph-NH₂$Ph-NH_2$) reacts with nitrous acid (generated in situ from NaNO₂ + HCl$NaNO_2 + HCl$) at low temperature (0-5^° C$0-5^\circ C$) to form benzene diazonium chloride (Ph-N₂^+Cl^-$Ph-N_2^+Cl^-$). Thus, Reagent A is NaNO₂ + HCl$NaNO_2 + HCl$ at 0-5^° C$0-5^\circ C$.

Step 2: Coupling Reaction

Step 2: The diazonium salt undergoes an electrophilic substitution (coupling reaction) with an electron-rich aromatic ring to form an azo dye. The formation of a 'scarlet red dye' is specifically the result of coupling benzene diazonium chloride with β$\beta$-naphthol in a weakly basic medium (NaOH).

Chapter Mix

Class 12 Chemistry: Amines

Core Logic

• Sulphanilic acid reacts with NaNO₂$NaNO_2$ and CH₃COOH$CH_3COOH$ to form a diazonium salt (X).
• The diazonium salt (X) then reacts with N,N-dimethylaniline (given in the coupling step image). The coupling takes place at the para position of the highly activated N,N-dimethylaniline ring.
• This coupling yields Methyl Orange, an azo dye. Its structure is p$p$-dimethylaminoazobenzenesulphonic acid.

Step 1: Final Identification

The final product (Y) matches option (4) structurally, containing the sulphonic acid group on one ring, the azo linkage, and the N,N-dimethylamine group on the para position of the other ring.

Chapter Mix

Class 12 Chemistry: Amines

Core Logic

Statement I: Aniline reacting with concentrated H₂SO₄$H_2SO_4$ gives anilinium hydrogensulphate, which on heating at 453-473 K produces sulphanilic acid (p-aminobenzene sulphonic acid). Because sulphanilic acid contains both Nitrogen and Sulphur, it gives a blood-red colouration in Lassaigne's test due to the formation of thiocyanate ion SCN^-$SCN^-$ which reacts with Fe³⁺$Fe^{3+}$ to form [Fe(SCN)]²⁺$[Fe(SCN)]^{2+}$. Thus, Statement I is true.

Statement II: In Friedel-Crafts reactions, the Lewis acid catalyst AlCl₃$AlCl_3$ reacts with the lone pair on the nitrogen atom of aniline to form a salt. This generates a positive charge on the nitrogen, transforming the -NH₂$-NH_2$ group from a strong activating group into a strong deactivating group, thus preventing the Friedel-Crafts reaction from occurring. Thus, Statement II is true.

Step 1: Final Conclusion

Both Statement I and Statement II are true. Option (4) is correct.

Chapter Mix

Class 12 Chemistry: Amines Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques

Related Formula

Core Logic

During the acetylation of an amino group, one hydrogen atom (mass = 1 g/mol$1\text{ g/mol}$) is replaced by an acetyl group (-COCH₃$-COCH_3$, mass = 43 g/mol$43\text{ g/mol}$). Gain in molecular weight for every one -NH₂$-NH_2$ group acetylated = 43 - 1 = 42 g/mol$43 - 1 = 42\text{ g/mol}$.

Step 1: Calculating Number of Groups

Total increase in molecular weight = Final mass - Initial mass = 192 - 108 = 84 g/mol$192 - 108 = 84\text{ g/mol}$.

Chapter Mix

Class 12 Chemistry: Amines