Amines Previous Year Questions — JEE Main Chemistry
40 past-year Amines questions from JEE Main (Chemistry) — page 6.
Q74 (2026)
Consider the following reaction sequence
Benzene $\xrightarrow{\text{conc. } HNO_{3} + \text{conc. } H_{2}SO_{4}, 333\text{ K}} \text{P} \xrightarrow{1.\text{ Sn/HCl/}\Delta \quad 2.\text{ pH neutralised}} \text{Q} \xrightarrow{(\mathrm{CH}_{3}\mathrm{CO})_{2}\mathrm{O}} \text{R} \xrightarrow{1.\text{ conc. } HNO_{3} + \text{ conc. } H_{2}SO_{4} \quad 2.\text{ pH neutralised (major product)}} \text{S} \xrightarrow{\mathrm{HCl} / \mathrm{EtOH} / \Delta} \text{T}$
The percentage of nitrogen in product ‘T’ formed is ____%. (Nearest integer)
(Given molar mass in $\mathrm{g\ mol^{-1}}$ H:1, C:12, N:14, O:16)
### Core Logic
Step 1: Nitration of benzene gives nitrobenzene (P).
$$\mathrm{Ph-H} \xrightarrow{HNO_3/H_2SO_4} \mathrm{Ph-NO_2} \quad \text{(P)}$$
{{SOL_IMG1}}
Step 2: Reduction of nitrobenzene with Sn/HCl gives aniline (Q).
$$\mathrm{Ph-NO_2} \xrightarrow{Sn/HCl} \mathrm{Ph-NH_2} \quad \text{(Q)}$$
{{SOL_IMG2}}
Step 3: Acetylation of aniline with acetic anhydride gives acetanilide (R). This protects the amino group to prevent oxidation and polysubstitution in the next step.
$$\mathrm{Ph-NH_2} \xrightarrow{(CH_3CO)_2O} \mathrm{Ph-NH-CO-CH_3} \quad \text{(R)}$$
{{SOL_IMG3}}
Step 4: Nitration of acetanilide gives predominantly p-nitroacetanilide (S) due to steric hindrance at ortho position.
$$\mathrm{Ph-NH-CO-CH_3} \xrightarrow{HNO_3/H_2SO_4} p\text{-NO}_2\text{-C}_6\text{H}_4\text{-NH-CO-CH}_3 \quad \text{(S)}$$
Step 5: Acidic hydrolysis of the amide linkage yields p-nitroaniline (T).
$$p\text{-NO}_2\text{-C}_6\text{H}_4\text{-NH-CO-CH}_3 \xrightarrow{HCl/EtOH/\Delta} p\text{-NO}_2\text{-C}_6\text{H}_4\text{-NH}_2 \quad \text{(T)}$$
{{SOL_IMG4}}
Molecular formula of p-nitroaniline (T) is $C_6H_6N_2O_2$.
Molar mass = $(6 \times 12) + (6 \times 1) + (2 \times 14) + (2 \times 16) = 72 + 6 + 28 + 32 = 138\text{ g/mol}$.
Total mass of Nitrogen = $2 \times 14 = 28\text{ g}$.
Percentage of Nitrogen = $\frac{28}{138} \times 100 \approx 20.29\%$.
### Step 1: Final Conclusion
Nearest integer is 20.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Chemistry: Amines
Q2173 (2026)
Consider the above sequence of reactions. {{IMG1}}
(1) $\text{Br}_2 / \text{FeBr}_3 / \Delta$
(2) $\text{Sn} / \text{HCl} / \Delta$
(3) $\text{pH neutralisation} \rightarrow \text{Major Product (P)}$
(4) $\text{Br}_2 / \text{H}_2\text{O}$
(5) $\text{NaNO}_2 / \text{HBr}, 0-5^{\circ}\text{C}$
(6) $\text{CuBr} / \text{NaBr}$
The number of bromine atom(s) in the final product (P) will be:
- $(1) \ 1$
- $(2) \ 6$
- $(3) \ 5$
- $(4) \ 3$
### Core Logic
Tracing the steps through nitration/bromination, reduction to amine via $\text{Sn/HCl}$, subsequent extensive bromination with $\text{Br}_2/\text{H}_2\text{O}$, diazotization, and Sandmeyer bromination ($\text{CuBr/NaBr}$), we get substitution at multiple positions leading to 5 bromine atoms in the final product structure.
### Step 1: Final Calculation
Number of Br atoms in major product (P) = 5.
### Pattern Recognition
Sees: Multi-step aromatic conversion involving halogenation and diazotization.
Trap: Counting substituent groups incorrectly after Sandmeyer reaction.
### Chapter Mix
Class 12 Chemistry: Amines
Q66 (2026)
'A' is a neutral organic compound (M. F : $C_{8}H_{9}ON$). On treatment with aqueous $Br_{2}/HO^{(-)}$, 'A' forms a compound 'B' which is soluble in dilute acid. 'B' on treatment with aqueous $NaNO_{2}/HCl(0-5^{\circ}C)$ produces a compound 'C' which on treatment with CuCN/NaCN produces 'D' Hydrolysis of 'D' produces 'E' which is also obtainable from the hydrolysis of 'A'. 'E' on treatment with acidified $KMnO_{4}$ produces 'F'. 'F' contains two different types of hydrogen atoms. The structure of 'A' is
- $\text{Structure 1}$
- $\text{Structure 2}$
- $\text{Structure 3}$
- $\text{Structure 4}$
### Core Logic
Let's trace the sequence:
1. A ($C_8H_9ON$) is neutral and reacts with $Br_2/OH^-$ (Hofmann Bromamide Degradation). This means A is a primary amide.
2. Product B is soluble in dilute acid, meaning it is a primary amine ($Ar-NH_2$ or alkyl amine).
3. B reacts with $NaNO_2/HCl$ at $0-5^{\circ}C$ to form C. Since C undergoes Sandmeyer with CuCN to form D, B must be an aromatic primary amine, and C is a diazonium salt.
4. D is an aryl cyanide ($Ar-CN$). Hydrolysis of D yields E ($Ar-COOH$).
5. Crucially, E is also obtainable from the direct hydrolysis of A. This confirms A is an aryl amide of the form $Ar-CONH_2$.
{{SOL_IMG1}}
6. Let's analyze the formula $C_8H_9ON$. The amide group is $-CONH_2$. Removing $-CONH_2$ leaves $C_7H_7$. A benzene ring with one methyl group is a tolyl group. So A is a methylbenzamide ($CH_3-C_6H_4-CONH_2$).
7. E is methylbenzoic acid ($CH_3-C_6H_4-COOH$).
8. E is oxidized by acidified $KMnO_4$ to F. The methyl group on the benzene ring oxidizes to $-COOH$. Thus, F is a benzenedicarboxylic acid ($HOOC-C_6H_4-COOH$).
{{SOL_IMG2}}
9. The problem states that F contains two different types of hydrogen atoms. Let's check the isomers of benzenedicarboxylic acid:
- Phthalic acid (ortho): Contains 2 types of aromatic hydrogens + 1 type of COOH hydrogen = 3 types.
- Isophthalic acid (meta): Contains 3 types of aromatic hydrogens + 1 type of COOH hydrogen = 4 types.
- Terephthalic acid (para): Due to symmetry, all 4 aromatic hydrogens are equivalent. So it contains 1 type of aromatic hydrogen + 1 type of COOH hydrogen = 2 types.
{{SOL_IMG3}}
Since F has only two types of hydrogens, F must be terephthalic acid (para isomer). Thus, E is p-methylbenzoic acid, and A is p-methylbenzamide.
### Step 1: Final Identification
Compound A is p-methylbenzamide. This corresponds to the structure in option (3).
### Pattern Recognition
A classic sequence linking Hofmann bromamide, Sandmeyer, and side-chain oxidation. Symmetrical molecules like para-isomers minimize the number of unique proton environments (critical for NMR or simple counting).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Amines
Class 12 Chemistry: Aldehydes Ketones and Carboxylic Acids
Q55 (2026)
$$\mathrm{C}_6\mathrm{H}_5\mathrm{NH}_2 \xrightarrow[\mathrm{NaOH}]{\mathrm{C}_6\mathrm{H}_5\mathrm{COCl}} [\mathrm{A}] \xrightarrow[\mathrm{H}_2\mathrm{O}]{\mathrm{LiAlH}_4} [\mathrm{B}]$$
The final product [B] is:
- {{IMG1}}
- {{IMG2}}
- {{IMG3}}
- {{IMG4}}
### Related Formula
$$\text{Aniline} + \text{Benzoyl Chloride} \xrightarrow{\text{Schotten-Baumann}} \text{Benzanilide } [A]$$
$$\text{Amide } [A] \xrightarrow{\text{LiAlH}_4} \text{Secondary Amine } [B]$$
### Core Logic
Step 1: Reaction of aniline with benzoyl chloride ($\text{PhCOCl}$) in basic medium yields benzanilide ($\text{Ph-NH-CO-Ph}$) as intermediate $[A]$.
Step 2: Reduction of benzanilide using $\text{LiAlH}_4$ converts the carbonyl group $-\text{C}(=\text{O})-$ into a methylene group $-\text{CH}_2-$, forming dibenzylamine ($\text{Ph-NH-CH}_2\text{-Ph}$) as final product $[B]$.
{{SOL_IMG1}}
{{SOL_IMG2}}
{{SOL_IMG3}}
### Pattern Recognition
Sees: Acylation followed by $\text{LiAlH}_4$ reduction.
Shortcut: Amide carbonyl group reduces directly to $-\text{CH}_2-$, resulting in secondary amine structure (option 3).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Amines
Q74 (2026)
The mass of benzanilide obtained from the benzoylation reaction of $5.8\text{ g}$ of aniline, if yield of product is 82%, is ____ g (nearest integer).
(Given molar mass in $\text{g mol}^{-1}$ H:1, C:12, N:14, O:16)
### Related Formula
$$\text{Aniline } (\text{C}_6\text{H}_7\text{N}, M=93\text{ g/mol}) \xrightarrow{\text{PhCOCl}} \text{Benzanilide } (\text{C}_{13}\text{H}_{11}\text{NO}, M=197\text{ g/mol})$$
$$\text{Actual Mass} = n_{\text{theoretical}} \times \frac{\text{Yield}\%}{100} \times M_{\text{benzanilide}}$$
### Core Logic
Step 1: Calculate initial moles of aniline:
$$n_{\text{aniline}} = \frac{5.8}{93} \approx 0.06237\text{ mol}$$
Step 2: Calculate actual moles of benzanilide produced at 82% yield:
$$n_{\text{actual}} = 0.06237 \times \frac{82}{100} \approx 0.05114\text{ mol}$$
Step 3: Calculate mass of benzanilide:
$$\text{Mass} = 0.05114 \times 197 \approx 10.075\text{ g} \approx 10\text{ g}$$
{{SOL_IMG1}}
### Pattern Recognition
Sees: Amine acylation reaction with percentage yield.
Shortcut: $(5.8 / 93) \times 0.82 \times 197 = 10.07\text{ g} \approx 10\text{ g}$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Amines
