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

Hinsberg reagent (Benzenesulfonyl chloride) reacts with: 1°$1^{\circ}$ Amines arrow$\rightarrow$ Forms sulfonamide which is SOLUBLE in alkali. 2°$2^{\circ}$ Amines arrow$\rightarrow$ Forms sulfonamide which is INSOLUBLE in alkali. 3°$3^{\circ}$ Amines arrow$\rightarrow$ Do not react.

We need to find the total number of 2°$2^{\circ}$ amines from the list.

• Aniline arrow$\rightarrow$ 1°$1^{\circ}$
• N-Methylaniline arrow$\rightarrow$ 2°$2^{\circ}$ (Reacts, insoluble) ✓
• Methanamine arrow$\rightarrow$ 1°$1^{\circ}$
• N, N-Dimethylmethanamine arrow$\rightarrow$ 3°$3^{\circ}$
• N-Methyl methanamine arrow$\rightarrow$ 2°$2^{\circ}$ (Reacts, insoluble) ✓
• Phenylmethanamine arrow$\rightarrow$ 1°$1^{\circ}$
• N-propylaniline arrow$\rightarrow$ 2°$2^{\circ}$ (Reacts, insoluble) ✓
• N-phenylaniline arrow$\rightarrow$ 2°$2^{\circ}$ (Reacts, insoluble) ✓
• N, N-Dimethylaniline arrow$\rightarrow$ 3°$3^{\circ}$
• Allyl amine arrow$\rightarrow$ 1°$1^{\circ}$
• Isopropyl amine arrow$\rightarrow$ 1°$1^{\circ}$

Step 1: Final Conclusion

There are exactly 4 secondary amines: N-Methylaniline, N-Methyl methanamine, N-propylaniline, and N-phenylaniline.

Pattern Recognition

Alkali insoluble sulfonamide = strictly Secondary Amine. Alkali soluble = Primary Amine.

Chapter Mix

Class 12 Chemistry: Amines

Related Formula

Core Logic

Sulphanilic acid is diazotized under cold conditions (273~K$273~\mathrm{K}$) with nitrous acid to form a diazonium salt intermediate. This diazonium salt undergoes a coupling reaction with 1-naphthylamine to form a red azo dye.

First, sulphanilic acid acts as a zwitterion and undergoes diazotization:

Nitrous acid reacts with the amine to form the diazonium compound:

This intermediate couples with 1-naphthylamine at the para-position to give the red azo dye compound:

Step 1: Calculate the Molar Mass

The molecular formula of the red-azo dye formed is C₁₆H₁₃N₃O₃S$\mathrm{C_{16}H_{13}N_{3}O_{3}S}$. Let's calculate its molar mass using the given atomic masses:

Step 2: Calculate the Mass of 0.1 Mole

Using the relation for mass:

Hence, the nearest option is 33~g$33~\mathrm{g}$.

Pattern Recognition

Azo coupling reactions are clean electrophilic aromatic substitution reactions. Diazotized sulphanilic acid has a highly electron-withdrawing sulphonic acid group, making it an excellent electrophile that couples selectively at the para-position of 1-naphthylamine.

Chapter Mix

Class 12 Chemistry: Amines

Related Formula

Basic strength in aqueous medium depends on three combined effects:

Core Logic

Let's list structural elements row-by-row:

• Ethyl substituted amine trends in aqueous systems uniquely align into a 2° > 3° > 1° configuration due to competing steric and hydration energies:

• Ammonia (NH₃$\mathrm{NH}_3$) is less basic than aliphatic substituted structures due to the absence of electron-donating alkyl clusters.
• Hydrazine (H₂N-NH₂$\mathrm{H}_2\mathrm{N-NH_2}$) is exceptionally weak compared to ammonia because the adjacent electronegative nitrogen creates an electron-withdrawing (-I) effect, while lone-pair repulsions reduce overall stability.

Step 1: Ordering

Assembling the fragments gives the complete verified thermodynamic order:

Pattern Recognition

Remember the standard numeric rules for aliphatic basic strength order in aqueous media:

• Methyl amines follow: 213
• Ethyl amines follow: 231

This simple sequence trick handles complex ranking items instantly.

Chapter Mix

Class 12 Chemistry: Amines

Related Formula

Directing effects in multi-substituted benzenes:

• -NH₂$-NH_2$ is a strong activating group and ortho/para director.
• -NO₂$-NO_2$ is a strong deactivating, meta-directing group.
• Diazotization followed by Sandmeyer reaction replaces an -NH₂$-NH_2$ group with a halogen.

Core Logic

To prepare 3,4,5-tribromoaniline, we must introduce three bromine atoms adjacent to each other (meta to the final amino group, with one para and two meta). Let's trace the sequence in Option (3) starting from 4-nitroaniline (p-nitroaniline):

Step 1: Bromination of p-nitroaniline

Treatment of 4-nitroaniline with excess Br₂$\mathrm{Br}_2$ in acetic acid:

• The amino group (-NH₂$-NH_2$) is a strong activator and directs to its ortho positions (positions 2 and 6).
• Positions 2 and 6 are meta to the -NO₂$-NO_2$ group, which is compatible.
• This yields 2,6-dibromo-4-nitroaniline.

Step 2: Diazotization and replacement of amino group

• NaNO₂ + HCl$\mathrm{NaNO}_2 + \mathrm{HCl}$ diazotizes the amino group to a diazonium salt:

• Addition of CuBr$\mathrm{CuBr}$ (Sandmeyer reaction) replaces the diazonium group with bromine:

This yields 3,4,5-tribromonitrobenzene.

Step 3: Reduction of nitro group

Reduction of the nitro group using Sn/HCl$\mathrm{Sn/HCl}$ converts -NO₂$-NO_2$ to -NH₂$-NH_2$:

This yields 3,4,5-tribromoaniline as the predominant product. Thus, Option (3) is correct.

Pattern Recognition

To brominate meta positions relative to an amino group, use a nitro precursor at the para position. The -NH₂$-NH_2$ group activates these positions first, after which the initial amino group is replaced with a halogen, and the nitro group is subsequently reduced back to a primary amine.

Chapter Mix

Class 12 Chemistry: Amines