Amines Previous Year Questions — JEE Main Chemistry
40 past-year Amines questions from JEE Main (Chemistry) — page 2.
Q41 (2025)
Which one of the following reaction sequences will give an azo dye?
(1) Nitrobenzene treated with (i) $Sn/HCl$, (ii) $NaNO_{2}/HCl$, (iii) $\beta$-naphthol, $NaOH$
(2) Benzenesulfonic acid treated with (i) $SOCl_2$, (ii) $NH_3$, (iii) Benzyl chloride
(3) Benzonitrile treated with (i) $70\% H_2SO_4$, (ii) $PCl_5$, (iii) Aniline
(4) Aniline treated with (i) $HCl/NaNO_2$, (ii) Toluene
- Reaction sequence (1)
- Reaction sequence (2)
- Reaction sequence (3)
- Reaction sequence (4)
### Core Logic
Let's track sequence (1):
1) Nitrobenzene ($Ph-NO_2$) is reduced using $Sn/HCl$ to form Aniline ($Ph-NH_2$).
2) Aniline undergoing diazotization with $NaNO_{2}/HCl$ at cold temperatures ($0-5^{circ}C$) creates Benzene diazonium chloride ($Ph-N_{2}^{+}Cl^{-}$).
3) The diazonium salt undergoes a coupling reaction with $\beta$-naphthol in alkaline conditions ($NaOH$) to synthesize a highly vibrant red-orange azo dye.
{{SOL_IMG41}}
### Pattern Recognition
The standard sequence for azo dye preparation is: Aromatic Nitro $
ightarrow$ Primary Amine $
ightarrow$ Diazonium Salt $
ightarrow$ Phenol/Naphthol Coupling.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Chemistry: Amines
Q49 (2025)
Consider the following sequence of reactions :
{{IMG1}}
11.25 mg of chlorobenzene will produce $\mathrm{x} \times 10^{-1}$ mg of product B.
(Consider the reactions result in complete conversion.)
[Given molar mass of C, H, O, N and Cl as 12, 1, 16, 14 and $35.5\mathrm{g\,mol}^{-1}$ respectively]
### Core Logic
The reaction sequence details the functional conversion of chlorobenzene down to product B (aniline, with a molar mass of $93\,\mathrm{g\,mol}^{-1}$).
Following stoichiometric preservation:
$$\text{moles of chlorobenzene} = \text{moles of Aniline (B)}$$
Molar mass of chlorobenzene ($\mathrm{C}_6\mathrm{H}_5\mathrm{Cl}$) = $112.5\,\mathrm{g\,mol}^{-1}$.
{{SOL_IMG1}}
$$\text{moles} = \frac{11.25 \times 10^{-3}\,\mathrm{g}}{112.5\,\mathrm{g\,mol}^{-1}} = 10^{-4}\,\mathrm{mol}$$
Mass of product B produced:
$$\text{Mass} = 10^{-4}\,\mathrm{mol} \times 93\,\mathrm{g\,mol}^{-1} = 9.3 \times 10^{-3}\,\mathrm{g} = 9.3\,\mathrm{mg}$$
Expressing in the specified format:
$$9.3\,\mathrm{mg} = 93 \times 10^{-1}\,\mathrm{mg} \Rightarrow x = 93$$
### Pattern Recognition
Sees: Conversion sequence preserving a 1:1 mole ratio layout.
Shortcut: Directly compute target weight via $W_B = W_A \cdot \frac{M_B}{M_A} = 11.25 \cdot \frac{93}{112.5} = 9.3$.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Chemistry: Amines
Q37 (2025)
Identify [A], [B], and [C], respectively in the following reaction sequence: {{IMG1}}
- Option (1)
- Option (2)
- Option (3)
- Option (4)
### Related Formula
$$\mathrm{Ar-NH_2 \xrightarrow{NaNO_2/HCl, 273-278K} Ar-N_2^+Cl^- \xrightarrow{KI} Ar-I}$$
$$\mathrm{2 Ar-I \xrightarrow{2Na, \text{dry ether}} Ar-Ar \text{ (Fittig reaction)}}$$
### Core Logic
1. Aniline with $\mathrm{NaNO_2/HCl}$ at $273\text{--}278\mathrm{~K}$ undergoes diazotization to yield benzenediazonium chloride $[\mathrm{A}] = \mathrm{C_6H_5N_2^+Cl^-}$.
2. Reaction of $[\mathrm{A}]$ with $\mathrm{KI}$ replaces the diazonium group with iodine to form iodobenzene $[\mathrm{B}] = \mathrm{C_6H_5I}$.
3. Treating iodobenzene $[\mathrm{B}]$ with $\mathrm{2Na/dry\ ether}$ yields biphenyl $[\mathrm{C}] = \mathrm{C_6H_5-C_6H_5}$ via the Fittig reaction.
### Step 1: Option Matching
Option (3) correctly identifies $[\mathrm{A}] = \mathrm{C_6H_5N_2^+Cl^-}$, $[\mathrm{B}] = \mathrm{C_6H_5I}$, and $[\mathrm{C}] = \text{Biphenyl}$.
### Pattern Recognition
Aniline $\rightarrow$ Diazotization $\rightarrow$ Iodination $\rightarrow$ Fittig coupling $\rightarrow$ Biphenyl.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Chemistry: Amines
Q38 (2025)
In the following reactions, which one is NOT correct?
- Reaction (1)
- Reaction (2)
- Reaction (3)
- Reaction (4)
### Related Formula
$$\mathrm{Ar-N_2^+Cl^- + CH_3CH_2OH \rightarrow Ar-H + N_2 + HCl + CH_3CHO}$$
### Core Logic
Reaction of benzenediazonium chloride with ethanol ($\mathrm{EtOH}$) is a reduction/deamination reaction that converts the diazonium salt to benzene, while ethanol gets oxidized to ethanal ($\mathrm{CH_3CHO}$).
It does NOT yield phenetole (ethoxybenzene, $\mathrm{Ar-OEt}$). Therefore, Reaction (1) is incorrect.
### Step 1: Verification of Other Reactions
- Reaction with $\mathrm{H_3PO_2/H_2O}$ gives benzene. (Correct)
- Reaction with $\mathrm{KI}$ gives iodobenzene. (Correct)
- Reaction with $\mathrm{CuCN/KCN}$ gives benzonitrile. (Correct)
### Pattern Recognition
Diazonium salt + Ethanol $\rightarrow$ Benzene (reduction), NOT Ethoxybenzene.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Chemistry: Amines
Q935 (2025)
The major product (A) formed in the following reaction sequence is:
{{IMG1}}
### Core Logic
Let's track the chemical transformations sequentially:
1. **Step 1 ($Sn + HCl$):** Nitrobenzene is cleanly reduced to yield Aniline ($C_6H_5NH_2$).
2. **Step 2 ($Ac_2O + \text{Pyridine}$):** Protecting step. Aniline undergoes acetylation to form Acetanilide ($C_6H_5NHCOCH_3$). This tempers the highly activating $-NH_2$ group to prevent poly-bromination.
3. **Step 3 ($Br_2 + AcOH$):** The $-NHCOCH_3$ amide group safely directs electrophilic bromination to the less-hindered **para** position, yielding p-bromoacetanilide.
4. **Step 4 ($NaOH_{(aq)}$):** Basic hydrolysis removes the protecting acetyl group, restoring the free amine function to yield the final product: **p-bromoaniline**.
### Pattern Recognition
Acetylation of aniline followed by halogenation and subsequent hydrolysis is the standard synthetic pathway to produce mono-substituted para-haloanilines.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Chemistry: Amines
