Consider the following sequence of reactions. Assuming that the reaction proceeds to completion, then 137 mg of 4-nitrotoluene will produce ___mg of B. (Given molar mass in g mol-1 H : 1, C : 12, N : 14, O : 16, Br : 80)

Follow the sequence of chemical transformations to determine the structure of product B, then apply stoichiometry based on moles of the starting material to find its final mass. Reaction 1: Reduction of 4-nitrotoluene using Sn, HCl / Delta followed by pH neutralization converts the -NO2 group to an -NH2 group, forming 4-methylaniline (p-toluidine). Reaction 2 (forming A): The aniline group is protected by reaction with acetic anhydride (CH3CO)2O, forming an acetanilide derivative (N-(4-methylphenyl)acetamide). Reaction 3 (forming B): Bromination using Br2 / AcOH. The -NHCOCH3 group is strongly activating and ortho-directing, while the -CH3 group is weakly activating and ortho-directing. The incoming -Br will substitute ortho to the -NHCOCH3 group due to dominant directing influence. Molar mass of 4-nitrotoluene (C7H7NO2) = 137 g/mol. Molar mass of product B (C9H10BrNO) = 228 g/mol. Moles of 4-nitrotoluene = 137 x 10-3 g137 g/mol = 0.001 mol. Since the stoichiometry is 1:1 and yields are 100%, moles of B = 0.001 mol. Mass of B = 0.001 mol x 228 g/mol = 0.228 g = 228 mg. When dealing with multi-step synthesis yield questions, immediately check the initial and final molar masses if the reaction achieves 100% completion. Often, 1 mole of reactant gives 1 mole of product. Class 12 Chemistry: Amines

Consider the following sequence of reactions. Assuming that the reaction proceeds to completion, then 137 mg of 4-nitrotoluene will produce ___mg of B. (Given molar mass in g mol-1 H : 1, C : 12, N :

Follow the sequence of chemical transformations to determine the structure of product B, then apply stoichiometry based on moles of the starting material to find its final mass. Reaction 1: Reduction of 4-nitrotoluene using Sn, HCl / Delta followed by pH neutralization converts the -NO2 group to an -NH2 group, forming 4-methylaniline (p-toluidine). Reaction 2 (forming A): The aniline group is protected by reaction with acetic anhydride (CH3CO)2O, forming an acetanilide derivative (N-(4-methylphe

Consider the following sequence of reactions. Assuming that the reaction proceeds to completion, then 137 mg of 4-nitrotoluene will produce ___mg of B. (Given molar mass in g mol-1 H : 1, C : 12, N : 14, O : 16, Br : 80)

Follow the sequence of chemical transformations to determine the structure of product B, then apply stoichiometry based on moles of the starting material to find its final mass. Reaction 1: Reduction of 4-nitrotoluene using Sn, HCl / Delta followed by pH neutralization converts the -NO2 group to an -NH2 group, forming 4-methylaniline (p-toluidine). Reaction 2 (forming A): The aniline group is protected by reaction with acetic anhydride (CH3CO)2O, forming an acetanilide derivative (N-(4-methylphenyl)acetamide). Reaction 3 (forming B): Bromination using Br2 / AcOH. The -NHCOCH3 group is strongly activating and ortho-directing, while the -CH3 group is weakly activating and ortho-directing. The incoming -Br will substitute ortho to the -NHCOCH3 group due to dominant directing influence. Molar mass of 4-nitrotoluene (C7H7NO2) = 137 g/mol. Molar mass of product B (C9H10BrNO) = 228 g/mol. Moles of 4-nitrotoluene = 137 x 10-3 g137 g/mol = 0.001 mol. Since the stoichiometry is 1:1 and yields are 100%, moles of B = 0.001 mol. Mass of B = 0.001 mol x 228 g/mol = 0.228 g = 228 mg. When dealing with multi-step synthesis yield questions, immediately check the initial and final molar masses if the reaction achieves 100% completion. Often, 1 mole of reactant gives 1 mole of product. Class 12 Chemistry: Amines

Consider the following sequence of reactions. Assuming that the reaction proceeds to completion, then 137 mg of 4-nitrotoluene will produce ___mg of B. (Given molar mass in g mol-1 H : 1, C : 12, N :

Follow the sequence of chemical transformations to determine the structure of product B, then apply stoichiometry based on moles of the starting material to find its final mass. Reaction 1: Reduction of 4-nitrotoluene using Sn, HCl / Delta followed by pH neutralization converts the -NO2 group to an -NH2 group, forming 4-methylaniline (p-toluidine). Reaction 2 (forming A): The aniline group is protected by reaction with acetic anhydride (CH3CO)2O, forming an acetanilide derivative (N-(4-methylphe

Consider the following sequence of reactions. Assuming that the reaction proceeds to completion, then 137 mg of…

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

Solution & Explanation

Core Logic

Step 1: Reaction Sequence Analysis

Step 2: Acetylation

Step 3: Bromination

Step 4: Stoichiometric Calculation

Pattern Recognition

Chapter Mix

Reference Study Guides

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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