The major product (A) formed in the following reaction sequence is:

Q: The major product (A) formed in the following reaction sequence is:

### Core Logic Let's track the chemical transformations sequentially: 1. **Step 1 (Sn + HCl):** Nitrobenzene is cleanly reduced to yield Aniline (C6H5NH2). 2. **Step 2 (Ac2O + Pyridine):** Protecting step. Aniline undergoes acetylation to form Acetanilide (C6H5NHCOCH3). This tempers the highly activating -NH2 group to prevent poly-bromination. 3. **Step 3 (Br2 + AcOH):** The -NHCOCH3 amide group safely directs electrophilic bromination to the less-hindered **para** position, yielding p-bromoacet

Year	2026	2025	2024	Total
Questions	16	14	10	40

More Amines Previous-Year Questions — Page 6

Q36 jee_main_2025_07_april_evening Basicity of Amines

The descending order of basicity of following amines is: (A) Aniline (B) p-Methoxyaniline (C) p-Nitroaniline (D)

CH₃NH₂

(E)

(CH₃)₂NH

Choose the correct answer from the options given below:

Basicity of Amines diagram for Q36 - JEE Main 2025 Evening — The images show structural configurations of the aromatic amines: aniline, p-methoxyaniline, and p-nitroaniline.

A. $B > E > D > A > C$
B. $E > D > B > A > C$
C. $E > D > A > B > C$
D. $E > A > D > C > B$

Solution

Related Formula

Basicity \propto Availability of lone pair of electrons on Nitrogen atom

Basicity \propto +I, +M groups Basicity \propto 1-I, -M groups

Core Logic

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.
Among aliphatic amines (aqueous standard configurations implicit): Secondary amine $(CH₃)₂NH$ $(\text{CH}_3)_2\text{NH}$ is a stronger base than primary $CH₃NH₂$ $\text{CH}_3\text{NH}_2$ due to combined inductive effect ( $+I$ $+\text{I}$) and solvation fields. Hence, $E > D$ $\text{E} > \text{D}$.
Among substituted aromatic amines:
(B) p-Methoxyaniline: $-OCH₃$ $-\text{OCH}_3$ exerts a strong electron-donating resonance effect ( $+M$ $+\text{M}$), maximizing ring density and electronic availability on N.
(A) Aniline: Baseline reference value with no extra substitutions.
(C) p-Nitroaniline: $-NO₂$ $-\text{NO}_2$ acts as an intensive electron-withdrawing field ( $-M, -I$ $-\text{M}, -\text{I}$), pulling electron clouds heavily and quenching basicity.

Step 1: Consolidating Rankings

Combining both structural domains cleanly provides:

E > D > B > A > C

Pattern Recognition

Basicity hierarchy shortcut: Aliphatic secondary $>$ $>$ Aliphatic primary $>$ $>$ Aromatic with EDG ( $+M$ $+\text{M}$) $>$ $>$ Unsubstituted aniline $>$ $>$ Aromatic with EWG ( $-M$ $-\text{M}$). This immediately gives $E > D > B > A > C$ without deep arithmetic.

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Class 12 Chemistry: Amines Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques

Q28 jee_main_2025_24_jan_evening Chemical Reactions of Aniline

For reaction

Chemical Reactions of Aniline diagram for Q28 - JEE Main 2025 Evening — The image shows a multi-step organic conversion starting with sulfanilic acid intermediate structure mapping to a selectively brominated major product.

The correct order of set of reagents for the above conversion is :

A. $Br2 | FeBr3, H2O(Δ), NaOH$
B. $H2SO4, Ac2O, Br2, H2O(Δ), NaOH$
C. $Ac2O, Br2, H2O(Δ), NaOH$
D. $Ac2O, H2SO₄, Br2, NaOH$

Solution

Core Logic

To direct selective monobromination ortho to the amino functionality while utilizing the masking capability of the sulfonic acid group:

Treating Aniline with conc. $H2SO₄$ $\mathrm{H}{2}SO_{4}$ at high temperature ( $453-473 K$ $453-473\text{ K}$) yields Sulfanilic acid due to para sulfonating preference.
Acetylation with $Ac₂O$ $\mathrm{Ac}_{2}O$ protects the amine as an acetanilide functionality to moderate activation power and prevent over-bromination.
Electrophilic substitution using $Br₂$ $\mathrm{Br}_{2}$ selectively places bromine at the position ortho to the protected acetamido group (the only available activated site since para is occupied).
Acidic/thermal desulfonation via $H₂O(Δ)$ $\mathrm{H}_{2}O(\Delta)$ cleaves the para-sulfonic acid group.
Alkaline hydrolysis with $NaOH$ $\mathrm{NaOH}$ 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:

Chemical Reactions of Aniline solution diagram for Q28 - JEE Main 2025 Evening — The image shows a multi-step organic conversion starting with sulfanilic acid intermediate structure mapping to a selectively brominated major product.

Pattern Recognition

When dealing with aniline conversions requiring blocked para positions followed by a removal step, look for the sequence tracking: Sulfonation $arrow$ $\rightarrow$ Protection $arrow$ $\rightarrow$ Halogenation $arrow$ $\rightarrow$ Desulfonation $arrow$ $\rightarrow$ Deprotection.

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Class 12 Chemistry: Amines

Q43 jee_main_2025_28_jan_evening Chemical Reactions of Amines

Identify correct statements: (A) Primary amines do not give diazonium salts when treated with

NaNO₃

in acidic condition. (B) Aliphatic and aromatic primary amines on heating with

CHCl₃

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 :

A. (B) and (D) only
B. (A) and (B) only
C. (D) and (E) only
D. (B) and (C) only

Solution

Related Formula

The Carbylamine reaction is specific to primary amines:

R-NH₂ + CHCl₃ + 3KOH Δ R-NC + 3KCl + 3H₂O

Core Logic

Evaluating each amine statement:

(A) Primary aromatic amines form stable diazonium salts with $NaNO₂/HCl$ $NaNO_2/HCl$ 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 ( $C₆H₅SO₂Cl$ $C_6H_5SO_2Cl$) 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.

Reaction equations summary for primary amine classification

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 ( $1^°$ $1^\circ$) amine groups.

Chapter Mix

Class 12 Chemistry: Amines

Q jee_main_2025_29_jan_morning Basic Character of Amines

Given below are some nitrogen containing compounds.

Nitrogen containing compounds profiles for Q46 - JEE Main 2025 Morning — Four amine candidates are indexed to find the optimal basic compound for reaction calculations.

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)

Numerical Answer. Answer: 341 to 341

Solution

Related Formula

Moles = MassMolar Mass

Mass of HCl consumed = namine \cdot MHCl

Core Logic

Step 1: Identify the most basic amine Benzylamine ( $C₆H₅CH₂NH₂$ ) 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

C₆H₅CH₂NH₂ + HCl arrow C₆H₅CH₂NH₃^+ Cl^-

Molar Mass of Benzylamine ( $C₇H₉N$ $\mathrm{C}_7\mathrm{H}_9\mathrm{N}$):

M = (7 \cdot 12) + (9 \cdot 1) + 14 = 84 + 9 + 14 = 107 g/mol

Moles of Benzylamine in $1.0 g$ $1.0\text{ g}$ :

n = (1.0)/(107) 0.009346 mol

Since 1 mole of benzylamine reacts with 1 mole of $HCl$ $\mathrm{HCl}$ :

Moles of HCl consumed = 0.009346 mol

Mass of HCl = 0.009346 \cdot 36.5 = 0.3411 g = 341.1 mg arrow 341

Pattern Recognition

Aliphatic localized clusters (like the $-CH₂NH₂$ $-\mathrm{CH}_2\mathrm{NH}_2$ segment in benzylamine) always show higher basicity than aromatic ring-conjugated arrays.

Chapter Mix

Class 12 Chemistry: Amines

The major product (A) formed in the following reaction sequence is:
The flowchart traces the conversion of nitrobenzene using Sn/HCl, Ac2O/pyridine, Br2/AcOH, and aqueous NaOH.

Solution & Explanation

Core Logic

Pattern Recognition

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More Amines Previous-Year Questions — Page 6

Solution

Related Formula

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Step 1: Consolidating Rankings

Pattern Recognition

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Solution

Core Logic

Step-by-Step Mechanism

Pattern Recognition

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Solution

Related Formula

Core Logic

Step 1: Selecting Correct Entries

Pattern Recognition

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Related Formula

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The major product (A) formed in the following reaction sequence is: The flowchart traces the conversion of nitrobenzene using Sn/HCl, Ac2O/pyridine, Br2/AcOH, and aqueous NaOH.

Solution & Explanation

Core Logic

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More Amines Previous-Year Questions — Page 6

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Related Formula

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Related Formula

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The major product (A) formed in the following reaction sequence is:
The flowchart traces the conversion of nitrobenzene using Sn/HCl, Ac2O/pyridine, Br2/AcOH, and aqueous NaOH.