Amines Previous Year Questions — JEE Main Chemistry
40 past-year Amines questions from JEE Main (Chemistry) — page 3.
Q36 (2025)
The descending order of basicity of following amines is:
(A) Aniline
(B) p-Methoxyaniline
(C) p-Nitroaniline
(D) $\text{CH}_3\text{NH}_2$
(E) $(\text{CH}_3)_2\text{NH}$
Choose the correct answer from the options given below:
- $\text{B} > \text{E} > \text{D} > \text{A} > \text{C}$
- $\text{E} > \text{D} > \text{B} > \text{A} > \text{C}$
- $\text{E} > \text{D} > \text{A} > \text{B} > \text{C}$
- $\text{E} > \text{A} > \text{D} > \text{C} > \text{B}$
### Related Formula
$$\text{Basicity} \propto \text{Availability of lone pair of electrons on Nitrogen atom}$$
$$\text{Basicity} \propto +\text{I}, +\text{M groups} \quad \text{Basicity} \propto \frac{1}{-\text{I}, -\text{M groups}}$$
### Core Logic
1. Aliphatic amines vs Aromatic amines: In aromatic amines (A, B, C), the lone pair on nitrogen is delocalized into the benzene ring via resonance, decreasing basicity compared to aliphatic amines (D, E) where electron pairs are localized.
2. Among aliphatic amines (aqueous standard configurations implicit): Secondary amine $(\text{CH}_3)_2\text{NH}$ is a stronger base than primary $\text{CH}_3\text{NH}_2$ due to combined inductive effect ($+\text{I}$) and solvation fields. Hence, $\text{E} > \text{D}$.
3. Among substituted aromatic amines:
- **(B) p-Methoxyaniline**: $-\text{OCH}_3$ exerts a strong electron-donating resonance effect ($+\text{M}$), maximizing ring density and electronic availability on N.
- **(A) Aniline**: Baseline reference value with no extra substitutions.
- **(C) p-Nitroaniline**: $-\text{NO}_2$ acts as an intensive electron-withdrawing field ($-\text{M}, -\text{I}$), pulling electron clouds heavily and quenching basicity.
### Step 1: Consolidating Rankings
Combining both structural domains cleanly provides:
$$\text{E} > \text{D} > \text{B} > \text{A} > \text{C} $$
### Pattern Recognition
Basicity hierarchy shortcut: Aliphatic secondary $>$ Aliphatic primary $>$ Aromatic with EDG ($+\text{M}$) $>$ Unsubstituted aniline $>$ Aromatic with EWG ($-\text{M}$). This immediately gives $\text{E} > \text{D} > \text{B} > \text{A} > \text{C}$ without deep arithmetic.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Amines
Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
Q28 (2025)
For reaction
{{IMG1}}
The correct order of set of reagents for the above conversion is :
- $\mathrm{Br}{2} | \mathrm{FeBr}{3}, \mathrm{H}{2}O(\Delta), \mathrm{NaOH}$
- $\mathrm{H}{2}SO{4}, \mathrm{Ac}{2}O, \mathrm{Br}{2}, \mathrm{H}{2}O(\Delta), \mathrm{NaOH}$
- $\mathrm{Ac}{2}O, \mathrm{Br}{2}, \mathrm{H}{2}O(\Delta), \mathrm{NaOH}$
- $\mathrm{Ac}{2}O, \mathrm{H}{2}SO_{4}, \mathrm{Br}{2}, \mathrm{NaOH}$
### Core Logic
To direct selective monobromination ortho to the amino functionality while utilizing the masking capability of the sulfonic acid group:
1. Treating Aniline with conc. $\mathrm{H}{2}SO_{4}$ at high temperature ($453-473\text{ K}$) yields Sulfanilic acid due to para sulfonating preference.
2. Acetylation with $\mathrm{Ac}_{2}O$ protects the amine as an acetanilide functionality to moderate activation power and prevent over-bromination.
3. Electrophilic substitution using $\mathrm{Br}_{2}$ selectively places bromine at the position ortho to the protected acetamido group (the only available activated site since para is occupied).
4. Acidic/thermal desulfonation via $\mathrm{H}_{2}O(\Delta)$ cleaves the para-sulfonic acid group.
5. Alkaline hydrolysis with $\mathrm{NaOH}$ removes the acetyl protecting group to regenerate the pristine primary amine structure yielding ortho-bromoaniline.
### Step-by-Step Mechanism
The reaction mechanism progresses linearly through the designated strategic intermediates:
{{SOL_IMG1}}
{{SOL_IMG2}}
{{SOL_IMG3}}
### Pattern Recognition
When dealing with aniline conversions requiring blocked para positions followed by a removal step, look for the sequence tracking: Sulfonation $
ightarrow$ Protection $
ightarrow$ Halogenation $
ightarrow$ Desulfonation $
ightarrow$ Deprotection.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Amines
Q43 (2025)
Identify correct statements:
(A) Primary amines do not give diazonium salts when treated with $NaNO_{3}$ in acidic condition.
(B) Aliphatic and aromatic primary amines on heating with $CHCl_{3}$ and ethanolic KOH form carbylamines.
(C) Secondary and tertiary amines also give carbylamine test.
(D) Benzenesulfonyl chloride is known as Hinsberg's reagent.
(E) Tertiary amines reacts with benzenesulfonyl chloride very easily.
Choose the correct answer from the options given below :
- (B) and (D) only
- (A) and (B) only
- (D) and (E) only
- (B) and (C) only
### Related Formula
The Carbylamine reaction is specific to primary amines:
$$R-NH_2 + CHCl_3 + 3KOH \xrightarrow{\Delta} R-NC + 3KCl + 3H_2O$$
### Core Logic
Evaluating each amine statement:
- (A) Primary aromatic amines form stable diazonium salts with $NaNO_2/HCl$ at low temperatures, making this statement false.
- (B) Both aliphatic and aromatic primary amines undergo the carbylamine test to produce foul-smelling isocyanides. This is **correct**.
- (C) Secondary and tertiary amines do not undergo the carbylamine reaction, making this statement false.
- (D) Benzenesulfonyl chloride ($C_6H_5SO_2Cl$) is the definition of Hinsberg's reagent. This is **correct**.
- (E) Tertiary amines do not possess an acidic hydrogen on nitrogen and do not react with Hinsberg's reagent under standard analytical testing conditions, making this statement false.
### Step 1: Selecting Correct Entries
Statements (B) and (D) are verified as true.
{{SOL_IMG1}}
### Pattern Recognition
Hinsberg's reagent and the carbylamine test are key analytical methods used to differentiate primary, secondary, and tertiary amines. The carbylamine test is strictly positive *only* for primary ($1^\circ$) amine groups.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Amines
Q1321 (2025)
Given below are some nitrogen containing compounds.
{{IMG1}}
Each of them is treated with HCl separately.
1.0 g of the most basic compound will consume ________ mg of HCl.
(Given molar mass in g mol $^{-1}$ C:12, H : 1, O : 16, Cl : 35.5)
### Related Formula
$$\text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}}$$
$$\text{Mass of HCl consumed} = n_{\text{amine}} \cdot M_{\text{HCl}}$$
### Core Logic
Step 1: Identify the most basic amine
Benzylamine ($\mathrm{C}_6\mathrm{H}_5\mathrm{CH}_2\mathrm{NH}_2$) is the most basic compound here because its nitrogen lone pair is localized and not involved in aromatic resonance. This stands in contrast to aniline or amides, which delocalize their lone pairs into the ring or carbonyl group .
Step 2: Neutralization Stoichiometry
$$\mathrm{C_6H_5CH_2NH_2} + \mathrm{HCl}
ightarrow \mathrm{C_6H_5CH_2NH_3^+ Cl^-}$$
Molar Mass of Benzylamine ($\mathrm{C}_7\mathrm{H}_9\mathrm{N}$):
$$M = (7 \cdot 12) + (9 \cdot 1) + 14 = 84 + 9 + 14 = 107 \, \mathrm{g/mol}$$
Moles of Benzylamine in $1.0\text{ g}$ :
$$n = \frac{1.0}{107} \simeq 0.009346 \, \mathrm{mol}$$
Since 1 mole of benzylamine reacts with 1 mole of $\mathrm{HCl}$ :
$$\text{Moles of HCl consumed} = 0.009346 \, \mathrm{mol}$$
$$\text{Mass of HCl} = 0.009346 \cdot 36.5 = 0.3411 \, \mathrm{g} = 341.1 \, \mathrm{mg}
ightarrow 341$$
### Pattern Recognition
Aliphatic localized clusters (like the $-\mathrm{CH}_2\mathrm{NH}_2$ segment in benzylamine) always show higher basicity than aromatic ring-conjugated arrays.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Amines
Q70 (2024)
Given below are two statements:
Statement (I) : Aminobenzene and aniline are same organic compounds.
Statement (II) : Aminobenzene and aniline are different organic compounds.
In the light of the above statements, choose the most appropriate answer from the options given below:
- $\text{Both Statement I and Statement II are correct}$
- $\text{Statement I is correct but Statement II is incorrect}$
- $\text{Statement I is incorrect but Statement II is correct}$
- $\text{Both Statement I and Statement II are incorrect}$
### Core Logic
Aniline is the common name for the simplest aromatic amine, which consists of a phenyl group attached to an amino group ($C_6H_5NH_2$).
According to IUPAC nomenclature, the amino group attached to a benzene ring can also be called aminobenzene.
### Step 1: Statement Validation
Statement I: True. Aminobenzene is just the systematic IUPAC name for aniline.
Statement II: False. They refer to the exact same molecule.
### Pattern Recognition
Common names for simple aromatic compounds are often accepted as IUPAC names. Aniline = Benzenamine = Aminobenzene.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Amines
