Amines Previous Year Questions — JEE Main Chemistry
40 past-year Amines questions from JEE Main (Chemistry).
Q401 (2025)
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 $\mathrm{g~mol^{-1}}$ H:1, C:12, N:14, O:16, S:32
- 343
- 330
- 33
- 66
### Related Formula
$$\text{Mass (g)} = \text{Number of moles} \times \text{Molar mass } (\mathrm{g~mol^{-1}})$$
### Core Logic
Sulphanilic acid is diazotized under cold conditions ($273~\mathrm{K}$) 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:
{{SOL_IMG1}}
Nitrous acid reacts with the amine to form the diazonium compound:
{{SOL_IMG2}}
This intermediate couples with 1-naphthylamine at the para-position to give the red azo dye compound:
{{SOL_IMG3}}
### Step 1: Calculate the Molar Mass
The molecular formula of the red-azo dye formed is $\mathrm{C_{16}H_{13}N_{3}O_{3}S}$.
Let's calculate its molar mass using the given atomic masses:
$$\text{Molar mass} = 16(12) + 13(1) + 3(14) + 3(16) + 32$$
$$\text{Molar mass} = 192 + 13 + 42 + 48 + 32 = 327~\mathrm{g~mol^{-1}}$$
### Step 2: Calculate the Mass of 0.1 Mole
Using the relation for mass:
$$\text{Mass of } 0.1~\text{mole} = 0.1 \times 327 = 32.7~\mathrm{g} \approx 33~\mathrm{g}$$
Hence, the nearest option is $33~\mathrm{g}$.
### 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.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Chemistry: Amines
Q38 (2025)
The correct order of basic nature on aqueous solution for the bases $\mathrm{NH}_3$, $\mathrm{H}_2\mathrm{N}-\mathrm{NH}_2$, $\mathrm{CH}_3\mathrm{CH}_2\mathrm{NH}_2$, $(\mathrm{CH}_3\mathrm{CH}_2)_2\mathrm{NH}$ and $(\mathrm{CH}_3\mathrm{CH}_2)_3\mathrm{N}$ is:
- $(1)\ \mathrm{NH}_3 < \mathrm{H}_2\mathrm{N} - \mathrm{NH}_2 < (\mathrm{CH}_3\mathrm{CH}_2)_3\mathrm{N} < \mathrm{CH}_3\mathrm{CH}_2\mathrm{NH}_2 < (\mathrm{CH}_3\mathrm{CH}_2)_2\mathrm{NH}$
- $(2)\ \mathrm{NH}_3 < \mathrm{H}_2\mathrm{N} - \mathrm{NH}_2 < \mathrm{CH}_3\mathrm{CH}_2\mathrm{NH}_2 < (\mathrm{CH}_3\mathrm{CH}_2)_2\mathrm{NH} < (\mathrm{CH}_3\mathrm{CH}_2)_3\mathrm{N}$
- $(3)\ \mathrm{H}_2\mathrm{N} - \mathrm{NH}_2 < \mathrm{NH}_3 < (\mathrm{CH}_3\mathrm{CH}_2)_3\mathrm{N} < \mathrm{CH}_3\mathrm{CH}_2\mathrm{NH}_2 < (\mathrm{CH}_3\mathrm{CH}_2)_2\mathrm{NH}$
- $(4)\ \mathrm{NH}_2 - \mathrm{NH}_2 < \mathrm{NH}_3 < \mathrm{CH}_3\mathrm{CH}_2\mathrm{NH}_2 < (\mathrm{CH}_3\mathrm{CH}_2)_3\mathrm{N} < (\mathrm{CH}_3\mathrm{CH}_2)_2\mathrm{NH}$
### Related Formula
Basic strength in aqueous medium depends on three combined effects:
$$\text{Basic Strength} \propto \text{Inductive Effect (+I)} + \text{Solvation Energy} - \text{Steric Hindrance}$$
### 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:
$$\mathrm{(Et)_2NH > (Et)_3N > EtNH_2}$$
* Ammonia ($\mathrm{NH}_3$) is less basic than aliphatic substituted structures due to the absence of electron-donating alkyl clusters.
* Hydrazine ($\mathrm{H}_2\mathrm{N-NH_2}$) 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:
$$\mathrm{NH_2-NH_2 < NH_3 < CH_3CH_2NH_2 < (CH_3CH_2)_3N < (CH_3CH_2)_2NH}$$
### 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.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Chemistry: Amines
Q192 (2025)
The sequence from the following that would result in giving predominantly 3, 4, 5-Tribromoaniline is :
### Related Formula
Directing effects in multi-substituted benzenes:
- $-NH_2$ is a strong activating group and ortho/para director.
- $-NO_2$ is a strong deactivating, meta-directing group.
- Diazotization followed by Sandmeyer reaction replaces an $-NH_2$ 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 $\mathrm{Br}_2$ in acetic acid:
- The amino group ($-NH_2$) is a strong activator and directs to its ortho positions (positions 2 and 6).
- Positions 2 and 6 are meta to the $-NO_2$ group, which is compatible.
- This yields 2,6-dibromo-4-nitroaniline.
### Step 2: Diazotization and replacement of amino group
1. $\mathrm{NaNO}_2 + \mathrm{HCl}$ diazotizes the amino group to a diazonium salt:
$$\mathrm{R-NH_2} \rightarrow \mathrm{R-N_2^+ Cl^-}$$
2. Addition of $\mathrm{CuBr}$ (Sandmeyer reaction) replaces the diazonium group with bromine:
$$\mathrm{R-N_2^+ Cl^-} \xrightarrow{\mathrm{CuBr}} \mathrm{R-Br}$$
This yields 3,4,5-tribromonitrobenzene.
### Step 3: Reduction of nitro group
Reduction of the nitro group using $\mathrm{Sn/HCl}$ converts $-NO_2$ to $-NH_2$:
$$\mathrm{R-NO_2} \xrightarrow{\mathrm{Sn, HCl}} \mathrm{R-NH_2}$$
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 $-NH_2$ 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.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Amines
Q527 (2025)
Which of the following amine(s) show(s) positive carbylamine test?
A. {{IMG1}}
B. $(CH_3)_2NH$
C. $CH_3NH_2$
D. $(CH_3)_3N$
E. {{IMG2}}
Choose the correct answer from the options given below:
- $\text{A and E Only}$
- $\text{C Only}$
- $\text{A and C Only}$
- $\text{B, C and D Only}$
### Related Formula
$$\text{R}-\text{NH}_2 + \text{CHCl}_3 + 3\text{KOH} \rightarrow \text{R}-\text{NC} + 3\text{KCl} + 3\text{H}_2\text{O}$$
### Core Logic
Only primary ($1^\circ$) aliphatic and aromatic amines yield a positive carbylamine test (forming foul-smelling alkyl/aryl isocyanides).
- **A** is Aniline (primary aromatic amine) $\rightarrow$ Positive
- **B** is Dimethylamine (secondary aliphatic amine) $\rightarrow$ Negative
- **C** is Methylamine (primary aliphatic amine) $\rightarrow$ Positive
- **D** is Trimethylamine (tertiary aliphatic amine) $\rightarrow$ Negative
- **E** is N-Methylaniline (secondary aromatic amine) $\rightarrow$ Negative
Thus, only A and C show a positive test.
### Pattern Recognition
Shortcut: Look directly for any amine with a plain $-\text{NH}_2$ functional group. Secondary ($-\text{NH}-$) and tertiary ($-\text{N}-$) amines never react.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Chemistry: Amines
Q578 (2025)
An organic compound 'A' undergoes the following sequence of transformations:
$$\text{'A'} \xrightarrow[\text{(ii) } H_3O^+]{\text{(i) } NaOH} \text{'B'} \xrightarrow[\text{(ii) } H_2SO_4, \Delta]{\text{(i) } EtOH} \text{'C'}$$
* 'A' shows a positive Lassaigne's test for nitrogen and its molar mass is $121 \text{ g mol}^{-1}$.
* 'B' gives effervescence with aqueous $\text{NaHCO}_3$.
* 'C' gives a characteristic fruity smell.
Identify A, B, and C from the options below:
### Core Logic
Let's perform a step-by-step diagnostic analysis:
1. **Molar Mass & Nitrogen Test**: Compound 'A' has a nitrogen atom and a molar mass of $121 \text{ g mol}^{-1}$. Let's verify Benzamide ($\text{C}_6\text{H}_5\text{CONH}_2$):
$$\text{Mass} = (7 \times 12) + (7 \times 1) + 14 + 16 = 84 + 7 + 14 + 16 = 121 \text{ g mol}^{-1}$$
This matches perfectly.
2. **Alkaline Hydrolysis**: Hydrolysis of benzamide under basic conditions yields benzoic acid upon acidification:
$$\text{C}_6\text{H}_5\text{CONH}_2 \xrightarrow[H_3O^+]{NaOH} \text{C}_6\text{H}_5\text{COOH (Compound B)} + \text{NH}_3$$
Benzoic acid reactively gives effervescence with $\text{NaHCO}_3$ due to the liberation of $\text{CO}_2$ gas.
3. **Esterification**: Reaction of benzoic acid with ethanol in the presence of acid catalyst results in the creation of ethyl benzoate, an ester with a pleasant fruity smell:
$$\text{C}_6\text{H}_5\text{COOH} + \text{EtOH} \xrightarrow{H_2SO_4, \Delta} \text{C}_6\text{H}_5\text{COOEt (Compound C)} + \text{H}_2\text{O}$$
{{SOL_IMG1}}
### Pattern Recognition
"Fruity smell" is an absolute indicator for an ester product. "Effervescence with $\text{NaHCO}_3$" dictates a carboxylic acid intermediate. Basic hydrolysis converting an organo-nitrogen compound into an acid points directly to an amide or a nitrile—molar mass calculation establishes benzamide over benzonitrile ($M = 103$).
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Chemistry: Amines
Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids
