Amines Previous Year Questions — JEE Main Chemistry
40 past-year Amines questions from JEE Main (Chemistry) — page 5.
Q78 (2024)
Following is a confirmatory test for aromatic primary amines. Identify reagent (A) and (B)
$Ph-NH_2 \xrightarrow{A} Ph-N_2^+Cl^- \xrightarrow{B} \text{Scarlet red dye}$
- $A=HNO_3/H_2SO_4$; $B=\beta-\text{naphthol}$
- $A=NaNO_2+HCl, 0-5^{\circ}C$; $B=\text{phenol}$
- $A=NaNO_2+HCl, 0-5^{\circ}C$; $B=\alpha-\text{naphthol}$
- $A=NaNO_2+HCl, 0-5^{\circ}C$; $B=\beta-\text{naphthol, NaOH}$
### Core Logic
The reaction sequence represents the classic dye test for aromatic primary amines.
Step 1 (Diazotization): Aniline ($Ph-NH_2$) reacts with nitrous acid (generated in situ from $NaNO_2 + HCl$) at low temperature ($0-5^\circ C$) to form benzene diazonium chloride ($Ph-N_2^+Cl^-$).
Thus, Reagent A is $NaNO_2 + HCl$ at $0-5^\circ C$.
### Step 2: Coupling Reaction
Step 2: The diazonium salt undergoes an electrophilic substitution (coupling reaction) with an electron-rich aromatic ring to form an azo dye.
The formation of a 'scarlet red dye' is specifically the result of coupling benzene diazonium chloride with $\beta$-naphthol in a weakly basic medium (NaOH).
{{SOL_IMG1}}
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Chemistry: Amines
Q1929 (2024)
The azo-dye (Y) formed in the following reactions is
$\text{Sulphanilic acid } + NaNO_2 + CH_3COOH \rightarrow X$
{{IMG1}}
### Core Logic
1) Sulphanilic acid reacts with $NaNO_2$ and $CH_3COOH$ to form a diazonium salt (X).
2) The diazonium salt (X) then reacts with N,N-dimethylaniline (given in the coupling step image). The coupling takes place at the para position of the highly activated N,N-dimethylaniline ring.
3) This coupling yields Methyl Orange, an azo dye. Its structure is $p$-dimethylaminoazobenzenesulphonic acid.
{{SOL_IMG1}}
### Step 1: Final Identification
The final product (Y) matches option (4) structurally, containing the sulphonic acid group on one ring, the azo linkage, and the N,N-dimethylamine group on the para position of the other ring.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Chemistry: Amines
Q70 (2024)
Given below are two statements:
Statement I: Aniline reacts with con. $H_2SO_4$ followed by heating at 453-473 K gives p-aminobenzene sulphonic acid, which gives blood red colour in the 'Lassaigne's test'.
Statement II: In Friedel-Crafts alkylation and acylation reactions, aniline forms salt with the $AlCl_3$ catalyst. Due to this, nitrogen of aniline acquires a positive charge and acts as deactivating group.
In the light of the above statements, choose the correct answer from the options given below:
- $\text{(1) Statement I is false but statement II is true}$
- $\text{(2) Both statement I and statement II are false}$
- $\text{(3) Statement I is true but statement II is false}$
- $\text{(4) Both statement I and statement II are true}$
### Core Logic
Statement I: Aniline reacting with concentrated $H_2SO_4$ gives anilinium hydrogensulphate, which on heating at 453-473 K produces sulphanilic acid (p-aminobenzene sulphonic acid). Because sulphanilic acid contains both Nitrogen and Sulphur, it gives a blood-red colouration in Lassaigne's test due to the formation of thiocyanate ion $SCN^-$ which reacts with $Fe^{3+}$ to form $[Fe(SCN)]^{2+}$. Thus, Statement I is true.
Statement II: In Friedel-Crafts reactions, the Lewis acid catalyst $AlCl_3$ reacts with the lone pair on the nitrogen atom of aniline to form a salt. This generates a positive charge on the nitrogen, transforming the $-NH_2$ group from a strong activating group into a strong deactivating group, thus preventing the Friedel-Crafts reaction from occurring. Thus, Statement II is true.
{{SOL_IMG1}}
### Step 1: Final Conclusion
Both Statement I and Statement II are true. Option (4) is correct.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Chemistry: Amines
Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
Q83 (2024)
A compound (x) with molar mass $108\mathrm{~g\,mol^{-1}}$ undergoes acetylation to give product with molar mass $192\mathrm{~g\,mol^{-1}}$. The number of amino groups in the compound (x) is ________.
### Related Formula
$$R-NH_2 + CH_3COCl \rightarrow R-NH-COCH_3 + HCl$$
### Core Logic
During the acetylation of an amino group, one hydrogen atom (mass = $1\text{ g/mol}$) is replaced by an acetyl group ($-COCH_3$, mass = $43\text{ g/mol}$).
Gain in molecular weight for every one $-NH_2$ group acetylated = $43 - 1 = 42\text{ g/mol}$.
### Step 1: Calculating Number of Groups
Total increase in molecular weight = Final mass - Initial mass = $192 - 108 = 84\text{ g/mol}$.
$$\text{Number of amino groups} = \frac{\text{Total mass increase}}{\text{Mass increase per group}} = \frac{84}{42} = 2$$
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Chemistry: Amines
Q60 (2026)
An organic compound (P) on treatment with aqueous ammonia under hot condition forms compound (Q) which on heating with $Br_{2}$ and KOH forms compound (R) having molecular formula $C_{9}H_{7}N$. Names of P, Q and R respectively are.
- $\text{Benzoic acid, benzamide, aniline}$
- $\text{Toluic acid, methylbenzamide, 2-methylaniline}$
- $\text{Benzoic acid,4-methylbenzamide,4-methylaniline.}$
- $\text{Phenylethanoic acid, phenylethanamide, benzamine}$
### Core Logic
The reaction of an amide with $Br_2$ and KOH is Hoffmann bromamide degradation. It steps down the carbon chain by one carbonyl carbon to form a primary amine.
Let's analyze the options and molecular formula. The question says (R) has molecular formula $C_9H_7N$. Wait, looking at the standard solutions for this type of problem, aniline is $C_6H_7N$. The PDF says $C_9H_7N$ which is likely a typo in the original paper for $C_6H_7N$, since option (1) gives Aniline ($C_6H_7N$). Let's assume the standard sequence:
1. $\mathrm{Ph-COOH} \xrightarrow{\mathrm{NH}_3, \Delta} \mathrm{Ph-CO-NH}_2$ (Benzoic acid to Benzamide)
2. $\mathrm{Ph-CO-NH}_2 \xrightarrow{\mathrm{Br}_2/\mathrm{KOH}} \mathrm{Ph-NH}_2$ (Benzamide to Aniline)
Aniline is $C_6H_5NH_2 = C_6H_7N$. So P is Benzoic acid, Q is Benzamide, R is Aniline.
### Pattern Recognition
Reaction sequence: $Carboxylic\ Acid \xrightarrow{NH_3, \Delta} Amide \xrightarrow{Br_2/KOH} Amine$. The $Br_2/KOH$ step is Hoffmann bromamide reaction.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Chemistry: Amines
Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids
