Amines Previous Year Questions — JEE Main Chemistry
40 past-year Amines questions from JEE Main (Chemistry) — page 8.
Q59 (2026)
A student has planned to prepare acetanilide from aniline using acetic anhydride. The student has started from 9.3g of aniline. However, the student has managed to obtain 11 g of dry acetanilide.
The % yield of this reaction is :-
- 81.5%
- 97.5%
- 59.5%
- 72.5%
### Related Formula
$$\%\text{ yield} = \frac{\text{Actual Moles Produced}}{\text{Theoretical Moles Expected}} \times 100$$
### Core Logic
{{SOL_IMG1}}
Given mass of aniline ($C_6H_5NH_2$) = 9.3 gm
Molar mass of aniline ($MW$) = 93 g/mol
Moles of aniline ($n_{\text{theoretical}}$) = $\frac{9.3}{93} = 0.1 \text{ moles}$
Given mass of dry acetanilide ($C_6H_5NHCOCH_3$) obtained = 11 gm
Molar mass of acetanilide ($MW$) = 135 g/mol
Moles of acetanilide obtained ($n_{\text{actual}}$) = $\frac{11}{135} \simeq 0.08148 \text{ moles}$
### Step 1: Calculate Percentage Yield
$$\% \text{ yield} = \frac{0.08148}{0.1} \times 100 = 81.48\% \simeq 81.5\%$$
### Pattern Recognition
A 1:1 molar ratio dictates that theoretical yield in moles of product directly matches moles of starting material. Converting experimental mass to moles removes the need to calculate theoretical mass, speeding up the final ratio.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Amines
Class 11 Chemistry: Some Basic Concepts of Chemistry
Q65 (2026)
Given below are two statements :
Statement I : The dipole moment of R-CN is greater than R-NC and R-NC can undergo hydrolysis under acidic medium to produce
$$
\begin{array}{c} \mathrm{O} \\ \mathrm {\parallel} \\ \mathrm{R-C-OH.} \end{array}
$$
Statement II : R-CN hydrolyses under acidic medium to produce a compound which on treatment with $SOCl_{2}$ , followed by the addition of $NH_{3}$ gives another compound(x). This compound (x) on treatment with NaOCl/NaOH gives a product, that on treatment with $CHCl_{3}/KOH/\Delta$ produces R-NC
In the light of the above statements, choose the correct answer from the options given below :
- Both Statement I and Statement II are false
- Both Statement I and Statement II are true
- Statement I is true but Statement II is false
- Statement I is false but Statement II is true
### Core Logic
Statement I: While it is true that the dipole moment of R-CN is greater than R-NC, the hydrolysis of isocyanide (R-NC) under acidic conditions yields a primary amine and formic acid, not a carboxylic acid (R-COOH) with the same alkyl chain intact.
Reaction: $\mathrm{R-NC} \xrightarrow{\mathrm{H}_3\mathrm{O}^+} \mathrm{R-NH}_2 + \mathrm{HCOOH}$
Thus, Statement I is False.
Statement II:
1. Hydrolysis of Cyanide: $\mathrm{R-CN} \xrightarrow{\mathrm{H}_3\mathrm{O}^+} \mathrm{R-COOH}$
2. With $\mathrm{SOCl}_2$: $\mathrm{R-COOH} \xrightarrow{\mathrm{SOCl}_2} \mathrm{R-COCl}$
3. With $\mathrm{NH}_3$: $\mathrm{R-COCl} \xrightarrow{\mathrm{NH}_3} \mathrm{R-CONH}_2$ (Compound x)
4. Hoffmann Bromamide / Hypochlorite degradation: $\mathrm{R-CONH}_2 \xrightarrow{\mathrm{NaOCl/NaOH}} \mathrm{R-NH}_2$
5. Carbylamine Reaction: $\mathrm{R-NH}_2 \xrightarrow{\mathrm{CHCl}_3 + \mathrm{KOH}} \mathrm{R-NC}$
The sequence correctly produces R-NC. Thus, Statement II is True.
### Step 1: Final Conclusion
Statement I is false but Statement II is true.
### Pattern Recognition
Isocyanide hydrolysis breaks the N-C bond, leaving the R group attached to Nitrogen (forming a $1^{\circ}$ amine). The remaining C becomes formic acid.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Amines
Q63 (2026)
Consider the following reactions giving major product. Identify the correct reaction.
- {{IMG1}}
- {{IMG2}}
- {{IMG3}}
- {{IMG4}}
### Step 1: Evaluate Reaction (1)
Nitration of acetanilide produces primarily the para-isomer because the $-NHCOCH_3$ group is strongly ortho/para directing and sterically hinders the ortho position. The option incorrectly suggests the meta product. (Incorrect)
### Step 2: Evaluate Reaction (2)
Gabriel phthalimide synthesis does not work with aryl halides because aryl halides do not readily undergo nucleophilic substitution. Thus, bromobenzene cannot form aniline this way. (Incorrect)
### Step 3: Evaluate Reaction (3)
Reaction of secondary amine with $CHCl_3/KOH$ does NOT yield an isocyanide. The carbylamine test is strictly for primary amines. (Incorrect)
### Step 4: Evaluate Reaction (4)
Hoffmann bromamide degradation:
$H_3C-CH_2-CO-NH_2 + Br_2 + 4KOH \xrightarrow{\Delta} H_3C-CH_2-NH_2 + 2KBr + K_2CO_3 + 2H_2O$
This cleanly forms ethylamine, completely accurately represented. (Correct)
### Pattern Recognition
Carbylamine test = 1° amines only. Gabriel Phthalimide = aliphatic 1° amines only (no aryl halides). Hoffmann degradation strictly lowers carbon chain by 1 ($CO$ removal).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Amines
Q63 (2026)
A student performed analysis of aliphatic organic compound 'X' which on analysis gave $C = 61.01\%$, $H = 15.25\%$, $N = 23.74\%$.
This compound, on treatment with $HNO_{2}/H_{2}O$ produced another compound 'Y' which did not contain any nitrogen atom. However, the compound 'Y' upon controlled oxidation produced another compound 'Z' that responded to iodoform test.
The structure of 'X' is:
- $(1)\text{ } CH_3CH_2CH_2NH_2$
- $(2)\text{ } Ph-CH(CH_3)-NH_2$
- $(3)\text{ } (CH_3)_2CH-NH_2$
- $(4)$
### Core Logic
Step 1: Empirical Formula calculation:
Moles of $C = \frac{61.01}{12} \approx 5.08$
Moles of $H = \frac{15.25}{1} \approx 15.25$
Moles of $N = \frac{23.74}{14} \approx 1.69$
Ratio C : H : N = $\frac{5.08}{1.69} : \frac{15.25}{1.69} : \frac{1.69}{1.69} \approx 3 : 9 : 1$
Empirical formula = $C_3H_9N$.
Step 2: Chemical tests deduction.
The compound is treated with $HNO_2/H_2O$ and produces a nitrogen-free compound Y. This implies X is a primary aliphatic amine ($C_3H_7NH_2$), giving an alcohol Y ($C_3H_7OH$).
Compound Y upon controlled oxidation forms Z, which gives positive iodoform test. For Z to give an iodoform test, it must be a methyl ketone. Thus Z is acetone ($CH_3COCH_3$).
For Y to oxidize to acetone, Y must be Isopropyl alcohol ($CH_3CH(OH)CH_3$).
Therefore, X must be Isopropylamine: $(CH_3)_2CHNH_2$.
### Step 1: Reaction Sequence
{{SOL_IMG1}}
$(CH_3)_2CHNH_2 \xrightarrow{HNO_2} (CH_3)_2CH-N_2^+ \xrightarrow{-N_2} (CH_3)_2CH^+ \xrightarrow{H_2O} (CH_3)_2CHOH \text{ (Y)}$
$(CH_3)_2CHOH \xrightarrow{\text{Oxidation}} CH_3COCH_3 \text{ (Z)}$
$CH_3COCH_3$ responds to the iodoform test.
### Step 2: Final Conclusion
The structure of 'X' is Isopropylamine.
### Pattern Recognition
Iodoform test on an oxidized alcohol means the alcohol was a secondary alcohol with a terminal methyl group. Combining this with $HNO_2$ reacting with primary amines gives the exact functional groups.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
Class 12 Chemistry: Amines
Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids
Q66 (2026)
Total number of alkali insoluble solid sulphonamides obtained by reaction of given amines with Hinsberg's reagent is .....
Aniline, N-Methylaniline, Methanamine,
N, N-Dimethylmethanamine,
N-Methyl methanamine, Phenylmethanamine,
N-propylaniline, N-phenylaniline,
N, N-Dimethylaniline, Allyl amine,
Isopropyl amine
- $(1)\text{ 4}$
- $(2)\text{ 2}$
- $(3)\text{ 8}$
- $(4)\text{ 5}$
### Core Logic
Hinsberg reagent (Benzenesulfonyl chloride) reacts with:
$1^{\circ}$ Amines $\rightarrow$ Forms sulfonamide which is SOLUBLE in alkali.
$2^{\circ}$ Amines $\rightarrow$ Forms sulfonamide which is INSOLUBLE in alkali.
$3^{\circ}$ Amines $\rightarrow$ Do not react.
We need to find the total number of $2^{\circ}$ amines from the list.
1) Aniline $\rightarrow$ $1^{\circ}$
2) N-Methylaniline $\rightarrow$ $2^{\circ}$ (Reacts, insoluble) ✓
3) Methanamine $\rightarrow$ $1^{\circ}$
4) N, N-Dimethylmethanamine $\rightarrow$ $3^{\circ}$
5) N-Methyl methanamine $\rightarrow$ $2^{\circ}$ (Reacts, insoluble) ✓
6) Phenylmethanamine $\rightarrow$ $1^{\circ}$
7) N-propylaniline $\rightarrow$ $2^{\circ}$ (Reacts, insoluble) ✓
8) N-phenylaniline $\rightarrow$ $2^{\circ}$ (Reacts, insoluble) ✓
9) N, N-Dimethylaniline $\rightarrow$ $3^{\circ}$
10) Allyl amine $\rightarrow$ $1^{\circ}$
11) Isopropyl amine $\rightarrow$ $1^{\circ}$
### Step 1: Final Conclusion
There are exactly 4 secondary amines: N-Methylaniline, N-Methyl methanamine, N-propylaniline, and N-phenylaniline.
### Pattern Recognition
Alkali insoluble sulfonamide = strictly Secondary Amine. Alkali soluble = Primary Amine.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Amines
