Identify [A], [B], and [C], respectively in the following reaction sequence:

Q: Identify [A], [B], and [C], respectively in the following reaction sequence:

### Core Logic Let us resolve each structural step sequentially: 1. **Step 1:** Aniline undergoes diazotization when treated with NaNO2 + HCl at 273-278 K, forming benzene diazonium chloride [A] (C6H5N2^+Cl^-). 2. **Step 2:** Warming benzene diazonium chloride with potassium iodide (KI) substitutes the diazonium group with iodine, producing iodobenzene [B] (C6H5I). 3. **Step 3:** Treating iodobenzene with sodium metal in dry ether causes a Fittig coupling reaction, dimerizing two phenyl radicals

Year	2026	2025	2024	Total
Questions	16	14	10	40

Q66 jee_main_2026_28_january_evening Hinsberg Test

Total number of alkali insoluble solid sulphonamides obtained by reaction of given amines with Hinsberg's reagent is ..... Aniline, N-Methylaniline, Methanamine, N, N-Dimethylmethanamine, N-Methyl methanamine, Phenylmethanamine, N-propylaniline, N-phenylaniline, N, N-Dimethylaniline, Allyl amine, Isopropyl amine

A. $(1) 4$ $(1)\text{ 4}$
B. $(2) 2$ $(2)\text{ 2}$
C. $(3) 8$ $(3)\text{ 8}$
D. $(4) 5$ $(4)\text{ 5}$

Solution

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

Q jee_main_2025_02_april_evening Diazotisation and Coupling Reactions

When a concentrated solution of sulphanilic acid and 1-naphthylamine is treated with nitrous acid (273 K) and acidified with acetic acid, the mass (g) of 0.1 mole of product formed is : Given molar mass in

g~mol⁻¹

H:1, C:12, N:14, O:16, S:32

A. 343
B. 330
C. 33
D. 66

Solution

Related Formula

Mass (g) = Number of moles \times Molar mass ( g~mol⁻¹)

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:

Azo-dye synthesis reaction from diazotized sulphanilic acid and 1-naphthylamine

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:

Molar mass = 16(12) + 13(1) + 3(14) + 3(16) + 32

Molar mass = 192 + 13 + 42 + 48 + 32 = 327~ g~mol⁻¹

Step 2: Calculate the Mass of 0.1 Mole

Using the relation for mass:

Mass of 0.1~mole = 0.1 \times 327 = 32.7~g \approx 33~g

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

Q38 jee_main_2025_02_april_morning Basic Strength of Amines

The correct order of basic nature on aqueous solution for the bases

NH₃

,

H₂N-NH₂

,

CH₃CH₂NH₂

,

(CH₃CH₂)₂NH

and

(CH₃CH₂)₃N

is:

A. $(1) NH₃ < H₂N - NH₂ < (CH₃CH₂)₃N < CH₃CH₂NH₂ < (CH₃CH₂)₂NH$
B. $(2) NH₃ < H₂N - NH₂ < CH₃CH₂NH₂ < (CH₃CH₂)₂NH < (CH₃CH₂)₃N$
C. $(3) H₂N - NH₂ < NH₃ < (CH₃CH₂)₃N < CH₃CH₂NH₂ < (CH₃CH₂)₂NH$
D. $(4) NH₂ - NH₂ < NH₃ < CH₃CH₂NH₂ < (CH₃CH₂)₃N < (CH₃CH₂)₂NH$

Solution

Related Formula

Basic strength in aqueous medium depends on three combined effects:

Basic Strength \propto Inductive Effect (+I) + Solvation Energy - Steric Hindrance

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:

(Et)₂NH > (Et)₃N > EtNH₂

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:

NH₂-NH₂ < NH₃ < CH₃CH₂NH₂ < (CH₃CH₂)₃N < (CH₃CH₂)₂NH

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

Q jee_main_2025_03_april_evening Aniline Reactions and Directing Effects

The sequence from the following that would result in giving predominantly 3, 4, 5-Tribromoaniline is :

A.
B.
C.
D.

Solution

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:

R-NH₂ arrow R-N₂^+ Cl^-

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

R-N₂^+ Cl^- CuBr R-Br

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 $-NH₂$ $-NH_2$:

R-NO₂ Sn, HCl R-NH₂

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

Q jee_main_2025_07_april_morning Carbylamine Reaction

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

Carbylamine reactant aniline diagram for Q27 - JEE Main 2025 — The structures depict Aniline (A) and N-Methylaniline (E) to distinguish primary and secondary aromatic amines.

B.

(CH₃)₂NH

C.

CH₃NH₂

D.

(CH₃)₃N

E.

Choose the correct answer from the options given below:

A. $A and E Only$ $\text{A and E Only}$
B. $C Only$ $\text{C Only}$
C. $A and C Only$ $\text{A and C Only}$
D. $B, C and D Only$ $\text{B, C and D Only}$

Solution

Related Formula

R-NH₂ + CHCl₃ + 3KOH arrow R-NC + 3KCl + 3H₂O

Core Logic

Only primary ( $1^°$ $1^\circ$) aliphatic and aromatic amines yield a positive carbylamine test (forming foul-smelling alkyl/aryl isocyanides).

A is Aniline (primary aromatic amine) $arrow$ $\rightarrow$ Positive
B is Dimethylamine (secondary aliphatic amine) $arrow$ $\rightarrow$ Negative
C is Methylamine (primary aliphatic amine) $arrow$ $\rightarrow$ Positive
D is Trimethylamine (tertiary aliphatic amine) $arrow$ $\rightarrow$ Negative
E is N-Methylaniline (secondary aromatic amine) $arrow$ $\rightarrow$ Negative

Thus, only A and C show a positive test.

Pattern Recognition

Shortcut: Look directly for any amine with a plain $-NH₂$ $-\text{NH}_2$ functional group. Secondary ( $-NH-$ $-\text{NH}-$) and tertiary ( $-N-$ $-\text{N}-$) amines never react.

Chapter Mix

Class 12 Chemistry: Amines

Identify [A], [B], and [C], respectively in the following reaction sequence: The reaction shows Aniline reacting with sodium nitrite and hydrochloric acid to produce intermediate A, followed by treatment with potassium iodide to yield B, which then dimerizes via sodium in dry ether to form C.

Solution & Explanation

Core Logic

Pattern Recognition

Evaluation Rubric / Model Answer

Chapter Mix

More Amines Previous-Year Questions — Page 4

Solution

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate the Molar Mass

Step 2: Calculate the Mass of 0.1 Mole

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Ordering

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Bromination of p-nitroaniline

Step 2: Diazotization and replacement of amino group

Step 3: Reduction of nitro group

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

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Identify [A], [B], and [C], respectively in the following reaction sequence:
The reaction shows Aniline reacting with sodium nitrite and hydrochloric acid to produce intermediate A, followed by treatment with potassium iodide to yield B, which then dimerizes via sodium in dry ether to form C.