Amines Previous Year Questions — JEE Main Chemistry
40 past-year Amines questions from JEE Main (Chemistry) — page 7.
Q57 (2026)
Consider the following sequence of reactions.
{{IMG1}}
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)
- $\text{301}$
- $\text{146}$
- $\text{228}$
- $\text{208}$
### Core Logic
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.
### Step 1: Reaction Sequence Analysis
Reaction 1: Reduction of 4-nitrotoluene using $Sn, HCl / \Delta$ followed by pH neutralization converts the $-NO_2$ group to an $-NH_2$ group, forming 4-methylaniline (p-toluidine).
{{SOL_IMG1}}
### Step 2: Acetylation
Reaction 2 (forming A): The aniline group is protected by reaction with acetic anhydride $(CH_3CO)_2O$, forming an acetanilide derivative (N-(4-methylphenyl)acetamide).
{{SOL_IMG2}}
### Step 3: Bromination
Reaction 3 (forming B): Bromination using $Br_2 / AcOH$. The $-NHCOCH_3$ group is strongly activating and ortho-directing, while the $-CH_3$ group is weakly activating and ortho-directing. The incoming $-Br$ will substitute ortho to the $-NHCOCH_3$ group due to dominant directing influence.
{{SOL_IMG3}}
### Step 4: Stoichiometric Calculation
Molar mass of 4-nitrotoluene ($C_7H_7NO_2$) = $137 \text{ g/mol}$.
Molar mass of product B ($C_9H_{10}BrNO$) = $228 \text{ g/mol}$.
Moles of 4-nitrotoluene = $\frac{137 \times 10^{-3} \text{ g}}{137 \text{ g/mol}} = 0.001 \text{ mol}$.
Since the stoichiometry is 1:1 and yields are 100%, moles of B = $0.001 \text{ mol}$.
Mass of B = $0.001 \text{ mol} \times 228 \text{ g/mol} = 0.228 \text{ g} = 228 \text{ mg}$.
### Pattern Recognition
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.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Chemistry: Amines
Q68 (2026)
Compound 'P' undergoes the following sequence of reactions :
$P \xrightarrow[\text{(ii)}\Delta]{\text{(i)}\text{NH}_3} Q \xrightarrow[\text{(ii)}\text{CHCl}_3,\text{KOH (alc)},\Delta]{\text{(i)}\text{KOH, Br}_2} \text{Cyclohexyl isocyanide}$
'P' is :
- $(1)$
- $(2)$
- $(3)$
- $(4)$
### Core Logic
Work backwards from the final product, cyclohexyl isocyanide.
### Step 1: Carbylamine Reaction
The step $(ii) \text{ } CHCl_3, KOH (alc), \Delta$ is the Carbylamine reaction. It converts a primary amine into an isocyanide. Therefore, the intermediate formed right before this step must be Cyclohexylamine ($Cyclohexyl-NH_2$).
{{SOL_IMG1}}
### Step 2: Hoffmann Bromamide Degradation
The step $(i) \text{ } KOH, Br_2$ is Hoffmann Bromamide Degradation. It converts an amide into a primary amine with one carbon less. Since the amine is Cyclohexylamine, the precursor 'Q' must be Cyclohexanecarboxamide.
### Step 3: Finding 'P'
The reaction $P \xrightarrow{NH_3, \Delta} Q$ converts a carboxylic acid into an amide. Therefore, 'P' must be Cyclohexanecarboxylic acid.
### Pattern Recognition
Isocyanide ($-NC$) final product $\rightarrow$ Primary amine precursor $\rightarrow$ Amide precursor via Hoffmann Bromamide $\rightarrow$ Carboxylic acid via $NH_3/\Delta$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Amines
Class 12 Chemistry: Aldehydes Ketones and Carboxylic Acids
Q69 (2026)
Given below are two statements:
Statement I: {{IMG1}} can be synthesized from {{IMG2}} in the order i) Acidic $KMnO_4$, ii) Ammonia, iii) Bromine and alkali
Statement II: {{IMG3}} {{IMG4}}
In the light of the above statements, choose the correct answer from the options given below
- $\text{Both Statement I and Statement II are false}$
- $\text{Statement I is true but Statement II is false}$
- $\text{Both Statement I and Statement II are true}$
- $\text{Statement I is false but Statement II is true}$
### Core Logic
Statement I analysis:
The target is to convert an ethyl side chain (propylbenzene) into an aniline derivative.
Step 1: Treatment with acidic $KMnO_4$ oxidizes any alkyl side chain with at least one benzylic hydrogen fully into a carboxylic acid group (benzoic acid derivative).
Step 2: Reaction with Ammonia ($NH_3$) converts the carboxylic acid into an amide.
Step 3: Bromine and alkali ($Br_2 + KOH$) trigger a Hoffmann bromamide degradation, chopping off the carbonyl carbon and leaving an amine ($NH_2$). This correctly yields the target structure. Statement I is True.
### Step 1: Statement II Analysis
Statement II analysis:
The goal is to convert p-toluidine to 3,5-dibromotoluene.
Step 1: Bromine water ($Br_2 + H_2O$) performs electrophilic aromatic substitution on the highly activated ring due to the $-NH_2$ group. It polybrominates the ortho and para positions relative to $-NH_2$. Since para is blocked by the methyl group, it brominates both ortho positions, yielding 2,6-dibromo-4-methylaniline.
Step 2: Diazotization with $NaNO_2 + HCl$ at $0-5^{\circ}\text{C}$ converts the $-NH_2$ group into a diazonium salt ($-N_2^+Cl^-$).
Step 3: Aqueous $H_3PO_2$ (hypophosphorous acid) acts as a reducing agent, replacing the diazonium group with a Hydrogen atom. This successfully yields 3,5-dibromotoluene (since the numbering shifts). Statement II is True.
### Pattern Recognition
Hoffmann bromamide degradation strictly steps down an amide to an amine by eliminating the carbonyl $C$. Deamination of aniline derivatives via diazonium salt followed by reduction ($H_3PO_2$ or EtOH) is the standard method for synthesizing specific meta-substituted benzenes where direct electrophilic meta-substitution fails.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Amines
Q70 (2026)
A student has been given a compound "x" of molecular formula -$C_{6}H_{7}N$. 'x' is sparingly soluble in water. However, on addition of dilute mineral acid, 'x' becomes soluble in water. 'x' when treated with $CHCl_{3}$ and KOH (alc.) 'y' is produced. 'y' has a specific unpleasant smell. On treatment with benzenesulphonyl chloride, 'x' gives a compound 'z' which is soluble in alkali. The number of different "H" atoms present in 'z' is:-
- $5$
- $8$
- $4$
- $7$
### Core Logic
Identify compound 'x':
Molecular formula $C_6H_7N$ indicates a high degree of unsaturation (Degree of Unsaturation = $6 - 7/2 + 1/2 + 1 = 4$), characteristic of a benzene ring. Given it dissolves in dilute mineral acid (due to protonation forming a salt), it's a basic amine. Since it produces an unpleasant-smelling compound 'y' with $CHCl_3$ and alc. KOH, it gives a positive carbylamine test, meaning 'x' is a primary amine. Thus, 'x' is Aniline ($Ph-NH_2$).
Compound 'y' is phenyl isocyanide ($Ph-NC$).
### Step 1: Identifying compound 'z'
Reaction with benzenesulphonyl chloride (Hinsberg's reagent):
Aniline reacts with $Ph-SO_2Cl$ to form $N$-phenylbenzenesulfonamide, which is 'z'.
Structure of 'z': $Ph-NH-SO_2-Ph$.
Because 'z' has an acidic proton on the nitrogen, it is soluble in alkali, confirming our deduction.
### Step 2: Counting different "H" atoms
Let's map the types of Hydrogen atoms (chemically distinct environments by symmetry) on the molecule 'z' ($Ph-NH-SO_2-Ph$):
1. The single hydrogen attached to the Nitrogen ($N-H$): 1 type.
2. The first phenyl ring (from aniline) has symmetry down the 1,4-axis. It has ortho-H, meta-H, and para-H. That yields 3 distinct types of H.
3. The second phenyl ring (from sulfonyl chloride) also has symmetry down its 1,4-axis. It has ortho-H, meta-H, and para-H. That yields 3 distinct types of H.
Total number of different H atoms = $1 + 3 + 3 = 7$.
### Pattern Recognition
Whenever you see $CHCl_3$ + alc. KOH producing an foul smell, it's definitively the Carbylamine test marking a primary amine. Soluble in alkali after Hinsberg's reagent confirms it's a $1^{\circ}$ amine.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Amines
Q57 (2026)
The correct stability order of the following diazonium salts is
(A) {{IMG1}}
(B) {{IMG2}}
(C) {{IMG3}}
(D) {{IMG4}}
- $\text{A} > \text{B} > \text{C} > \text{D}$
- $\text{C} > \text{D} > \text{B} > \text{A}$
- $\text{A} > \text{C} > \text{D} > \text{B}$
- $\text{C} > \text{A} > \text{D} > \text{B}$
### Core Logic
The stability of arenediazonium salts is dictated by the ability of substituents on the benzene ring to disperse the positive charge on the diazonium group ($\mathrm{-N_2^+}$).
Electron-donating groups (EDG) increase the stability by resonance (+M effect) and inductive (+I) effects, dispersing the positive charge.
Electron-withdrawing groups (EWG) decrease the stability by intensifying the positive charge (-M, -I effects).
Let's evaluate the given substituents at the para position:
(A) $\mathrm{-OCH_3}$: Strong +M effect. Highly stabilizing.
(C) Plain benzene ring: Neutral baseline.
(D) $\mathrm{-CN}$: Strong -M and -I effect. Destabilizing.
(B) $\mathrm{-NO_2}$: Very strong -M and -I effect. Most destabilizing.
### Step 1: Final Conclusion
Arranging them in decreasing order of stability:
(A) $\mathrm{-OCH_3}$ ($+M$) > (C) Baseline > (D) $\mathrm{-CN}$ ($-M$) > (B) $\mathrm{-NO_2}$ (Stronger $-M$).
Thus, the correct order is $\text{A} > \text{C} > \text{D} > \text{B}$.
### Pattern Recognition
Stability of carbocations and diazonium ions follows: $+M > +I > \text{no effect} > -I > -M$. Look directly at the para group to rank.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Chemistry: Amines
Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
