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 HNO2/H2O 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:

Step 1: Empirical Formula calculation: Moles of C = (61.01/12) approx 5.08 Moles of H = (15.25/1) approx 15.25 Moles of N = (23.74/14) approx 1.69 Ratio C : H : N = (5.08/1.69) : (15.25/1.69) : (1.69/1.69) approx 3 : 9 : 1 Empirical formula = C3H9N. Step 2: Chemical tests deduction. The compound is treated with HNO2/H2O and produces a nitrogen-free compound Y. This implies X is a primary aliphatic amine (C3H7NH2), giving an alcohol Y (C3H7OH). 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 (CH3COCH3). For Y to oxidize to acetone, Y must be Isopropyl alcohol (CH3CH(OH)CH3). Therefore, X must be Isopropylamine: (CH3)2CHNH2. (CH3)2CHNH2 HNO2 (CH3)2CH-N2^+ -N2 (CH3)2CH^+ H2O (CH3)2CHOH (Y) (CH3)2CHOH Oxidation CH3COCH3 (Z) CH3COCH3 responds to the iodoform test. The structure of 'X' is Isopropylamine. Iodoform test on an oxidized alcohol means the alcohol was a secondary alcohol with a terminal methyl group. Combining this with HNO2 reacting with primary amines gives the exact functional groups. Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques Class 12 Chemistry: Amines Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids

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 HNO2/H2O produced another compound 'Y' which

Step 1: Empirical Formula calculation: Moles of C = (61.01/12) approx 5.08 Moles of H = (15.25/1) approx 15.25 Moles of N = (23.74/14) approx 1.69 Ratio C : H : N = (5.08/1.69) : (15.25/1.69) : (1.69/1.69) approx 3 : 9 : 1 Empirical formula = C3H9N. Step 2: Chemical tests deduction. The compound is treated with HNO2/H2O and produces a nitrogen-free compound Y. This implies X is a primary aliphatic amine (C3H7NH2), giving an alcohol Y (C3H7OH). Compound Y upon controlled oxidation forms Z, which

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 HNO2/H2O 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:

Step 1: Empirical Formula calculation: Moles of C = (61.01/12) approx 5.08 Moles of H = (15.25/1) approx 15.25 Moles of N = (23.74/14) approx 1.69 Ratio C : H : N = (5.08/1.69) : (15.25/1.69) : (1.69/1.69) approx 3 : 9 : 1 Empirical formula = C3H9N. Step 2: Chemical tests deduction. The compound is treated with HNO2/H2O and produces a nitrogen-free compound Y. This implies X is a primary aliphatic amine (C3H7NH2), giving an alcohol Y (C3H7OH). 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 (CH3COCH3). For Y to oxidize to acetone, Y must be Isopropyl alcohol (CH3CH(OH)CH3). Therefore, X must be Isopropylamine: (CH3)2CHNH2. (CH3)2CHNH2 HNO2 (CH3)2CH-N2^+ -N2 (CH3)2CH^+ H2O (CH3)2CHOH (Y) (CH3)2CHOH Oxidation CH3COCH3 (Z) CH3COCH3 responds to the iodoform test. The structure of 'X' is Isopropylamine. Iodoform test on an oxidized alcohol means the alcohol was a secondary alcohol with a terminal methyl group. Combining this with HNO2 reacting with primary amines gives the exact functional groups. Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques Class 12 Chemistry: Amines Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids

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 HNO2/H2O produced another compound 'Y' which

Step 1: Empirical Formula calculation: Moles of C = (61.01/12) approx 5.08 Moles of H = (15.25/1) approx 15.25 Moles of N = (23.74/14) approx 1.69 Ratio C : H : N = (5.08/1.69) : (15.25/1.69) : (1.69/1.69) approx 3 : 9 : 1 Empirical formula = C3H9N. Step 2: Chemical tests deduction. The compound is treated with HNO2/H2O and produces a nitrogen-free compound Y. This implies X is a primary aliphatic amine (C3H7NH2), giving an alcohol Y (C3H7OH). Compound Y upon controlled oxidation forms Z, which

A student performed analysis of aliphatic organic compound 'X' which on analysis gave C = 61.01%, H = 15.25%, N =…

Year	2026	2025	2024	Total
Questions	16	14	10	40

Q60 jee_main_2026_21_jan_morning Preparation of Amines

An organic compound (P) on treatment with aqueous ammonia under hot condition forms compound (Q) which on heating with

Br₂

and KOH forms compound (R) having molecular formula

C₉H₇N

. Names of P, Q and R respectively are.

A. $Benzoic acid, benzamide, aniline$
B. $Toluic acid, methylbenzamide, 2-methylaniline$
C. $Benzoic acid,4-methylbenzamide,4-methylaniline.$
D. $Phenylethanoic acid, phenylethanamide, benzamine$

Solution

Core Logic

The reaction of an amide with $Br₂$ $Br_2$ and KOH is Hoffmann bromamide degradation. It steps down the carbon chain by one carbonyl carbon to form a primary amine.

Let's analyze the options and molecular formula. The question says (R) has molecular formula $C₉H₇N$ $C_9H_7N$. Wait, looking at the standard solutions for this type of problem, aniline is $C₆H₇N$ $C_6H_7N$. The PDF says $C₉H₇N$ $C_9H_7N$ which is likely a typo in the original paper for $C₆H₇N$ $C_6H_7N$, since option (1) gives Aniline ( $C₆H₇N$ $C_6H_7N$). Let's assume the standard sequence:

$Ph-COOH NH₃, Δ Ph-CO-NH₂$ (Benzoic acid to Benzamide)
$Ph-CO-NH₂ Br₂/KOH Ph-NH₂$ (Benzamide to Aniline)

Aniline is $C₆H₅NH₂ = C₆H₇N$ $C_6H_5NH_2 = C_6H_7N$. So P is Benzoic acid, Q is Benzamide, R is Aniline.

Pattern Recognition

Reaction sequence: $Carboxylic Acid NH₃, Δ Amide Br₂/KOH Amine$ . The $Br₂/KOH$ $Br_2/KOH$ step is Hoffmann bromamide reaction.

Chapter Mix

Class 12 Chemistry: Amines Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids

Q74 jee_main_2026_21_jan_morning Electrophilic Substitution Reactions

Consider the following reaction sequence Benzene

conc. HNO₃ + conc. H₂SO₄, 333 K P 1. Sn/HCl/Δ 2. pH neutralised Q (CH₃CO)₂O R 1. conc. HNO₃ + conc. H₂SO₄ 2. pH neutralised (major product) S HCl / EtOH / Δ T

The percentage of nitrogen in product ‘T’ formed is ____%. (Nearest integer) (Given molar mass in

g mol⁻¹

H:1, C:12, N:14, O:16)

Numerical Answer. Answer: 20 to 20

Solution

Core Logic

Step 1: Nitration of benzene gives nitrobenzene (P).

Ph-H HNO₃/H₂SO₄ Ph-NO₂ (P)

Electrophilic Substitution Reactions diagram for Q74 - JEE Main 2026 Morning

Step 2: Reduction of nitrobenzene with Sn/HCl gives aniline (Q).

Ph-NO₂ Sn/HCl Ph-NH₂ (Q)

Step 3: Acetylation of aniline with acetic anhydride gives acetanilide (R). This protects the amino group to prevent oxidation and polysubstitution in the next step.

Ph-NH₂ (CH₃CO)₂O Ph-NH-CO-CH₃ (R)

Step 4: Nitration of acetanilide gives predominantly p-nitroacetanilide (S) due to steric hindrance at ortho position.

Ph-NH-CO-CH₃ HNO₃/H₂SO₄ p-NO₂-C₆H₄-NH-CO-CH₃ (S)

Step 5: Acidic hydrolysis of the amide linkage yields p-nitroaniline (T).

p-NO₂-C₆H₄-NH-CO-CH₃ HCl/EtOH/Δ p-NO₂-C₆H₄-NH₂ (T)

Molecular formula of p-nitroaniline (T) is $C₆H₆N₂O₂$ $C_6H_6N_2O_2$. Molar mass = $(6 \times 12) + (6 \times 1) + (2 \times 14) + (2 \times 16) = 72 + 6 + 28 + 32 = 138 g/mol$ . Total mass of Nitrogen = $2 \times 14 = 28 g$ $2 \times 14 = 28\text{ g}$.

Percentage of Nitrogen = $(28)/(138) \times 100 \approx 20.29%$ .

Step 1: Final Conclusion

Nearest integer is 20.

Chapter Mix

Class 12 Chemistry: Amines

Q jee_main_2026_21_jan_evening Chemical Reactions of Amines and Halogenation

Consider the above sequence of reactions. (1)

Br₂ / FeBr₃ / Δ

(2)

Sn / HCl / Δ

(3)

pH neutralisation arrow Major Product (P)

(4)

Br₂ / H₂O

(5)

NaNO₂ / HBr, 0-5°C

(6)

CuBr / NaBr

The number of bromine atom(s) in the final product (P) will be:

Reaction sequence diagram for Q53 - JEE Main 2026 Evening — Reaction sequence diagram showing starting material and reagents for synthesis of product P.

A. $(1) 1$ $(1) \ 1$
B. $(2) 6$ $(2) \ 6$
C. $(3) 5$ $(3) \ 5$
D. $(4) 3$ $(4) \ 3$

Solution

Core Logic

Tracing the steps through nitration/bromination, reduction to amine via $Sn/HCl$ $\text{Sn/HCl}$, subsequent extensive bromination with $Br₂/H₂O$ $\text{Br}_2/\text{H}_2\text{O}$, diazotization, and Sandmeyer bromination ( $CuBr/NaBr$ $\text{CuBr/NaBr}$), we get substitution at multiple positions leading to 5 bromine atoms in the final product structure.

Step 1: Final Calculation

Number of Br atoms in major product (P) = 5.

Pattern Recognition

Sees: Multi-step aromatic conversion involving halogenation and diazotization. Trap: Counting substituent groups incorrectly after Sandmeyer reaction.

Chapter Mix

Class 12 Chemistry: Amines

Q66 jee_main_2026_22_january_morning Hofmann Bromamide Degradation

'A' is a neutral organic compound (M. F :

C₈H₉ON

). On treatment with aqueous

Br₂/HO(-)

, 'A' forms a compound 'B' which is soluble in dilute acid. 'B' on treatment with aqueous

NaNO₂/HCl(0-5°C)

produces a compound 'C' which on treatment with CuCN/NaCN produces 'D' Hydrolysis of 'D' produces 'E' which is also obtainable from the hydrolysis of 'A'. 'E' on treatment with acidified

KMnO₄

produces 'F'. 'F' contains two different types of hydrogen atoms. The structure of 'A' is

A. $Structure 1$ $\text{Structure 1}$
B. $Structure 2$ $\text{Structure 2}$
C. $Structure 3$ $\text{Structure 3}$
D. $Structure 4$ $\text{Structure 4}$

Solution

Core Logic

Let's trace the sequence:

A ( $C₈H₉ON$ $C_8H_9ON$) is neutral and reacts with $Br₂/OH^-$ $Br_2/OH^-$ (Hofmann Bromamide Degradation). This means A is a primary amide.
Product B is soluble in dilute acid, meaning it is a primary amine ( $Ar-NH₂$ $Ar-NH_2$ or alkyl amine).
B reacts with $NaNO₂/HCl$ $NaNO_2/HCl$ at $0-5°C$ $0-5^{\circ}C$ to form C. Since C undergoes Sandmeyer with CuCN to form D, B must be an aromatic primary amine, and C is a diazonium salt.
D is an aryl cyanide ( $Ar-CN$ $Ar-CN$). Hydrolysis of D yields E ( $Ar-COOH$ $Ar-COOH$).
Crucially, E is also obtainable from the direct hydrolysis of A. This confirms A is an aryl amide of the form $Ar-CONH₂$ $Ar-CONH_2$.

Sequence of reactions for identifying Compound A

Let's analyze the formula $C₈H₉ON$ $C_8H_9ON$. The amide group is $-CONH₂$ $-CONH_2$. Removing $-CONH₂$ $-CONH_2$ leaves $C₇H₇$ $C_7H_7$. A benzene ring with one methyl group is a tolyl group. So A is a methylbenzamide ( $CH₃-C₆H₄-CONH₂$ $CH_3-C_6H_4-CONH_2$).
E is methylbenzoic acid ( $CH₃-C₆H₄-COOH$ $CH_3-C_6H_4-COOH$).
E is oxidized by acidified $KMnO₄$ $KMnO_4$ to F. The methyl group on the benzene ring oxidizes to $-COOH$ $-COOH$. Thus, F is a benzenedicarboxylic acid ( $HOOC-C₆H₄-COOH$ $HOOC-C_6H_4-COOH$).

The problem states that F contains two different types of hydrogen atoms. Let's check the isomers of benzenedicarboxylic acid:
Phthalic acid (ortho): Contains 2 types of aromatic hydrogens + 1 type of COOH hydrogen = 3 types.
Isophthalic acid (meta): Contains 3 types of aromatic hydrogens + 1 type of COOH hydrogen = 4 types.
Terephthalic acid (para): Due to symmetry, all 4 aromatic hydrogens are equivalent. So it contains 1 type of aromatic hydrogen + 1 type of COOH hydrogen = 2 types.

Since F has only two types of hydrogens, F must be terephthalic acid (para isomer). Thus, E is p-methylbenzoic acid, and A is p-methylbenzamide.

Step 1: Final Identification

Compound A is p-methylbenzamide. This corresponds to the structure in option (3).

Pattern Recognition

A classic sequence linking Hofmann bromamide, Sandmeyer, and side-chain oxidation. Symmetrical molecules like para-isomers minimize the number of unique proton environments (critical for NMR or simple counting).

Chapter Mix

Class 12 Chemistry: Amines Class 12 Chemistry: Aldehydes Ketones and Carboxylic Acids

Q55 jee_main_2026_22_january_evening Benzoylation and Reduction of Amides

C₆H₅NH₂ [NaOH]C₆H₅COCl [A] [H₂O]LiAlH₄ [B]

The final product [B] is:

A.
B.
C.
D.

Solution

Related Formula

Aniline + Benzoyl Chloride Schotten-Baumann Benzanilide [A]

Amide [A] LiAlH₄ Secondary Amine [B]

Core Logic

Step 1: Reaction of aniline with benzoyl chloride ( $PhCOCl$ $\text{PhCOCl}$) in basic medium yields benzanilide ( $Ph-NH-CO-Ph$ $\text{Ph-NH-CO-Ph}$) as intermediate $[A]$ $[A]$.

Step 2: Reduction of benzanilide using $LiAlH₄$ $\text{LiAlH}_4$ converts the carbonyl group $-C(=O)-$ $-\text{C}(=\text{O})-$ into a methylene group $-CH₂-$ $-\text{CH}_2-$, forming dibenzylamine ( $Ph-NH-CH₂-Ph$ $\text{Ph-NH-CH}_2\text{-Ph}$) as final product $[B]$ $[B]$.

Reaction flow of Benzoylation and reduction for Q55 - JEE Main 2026 Evening

Pattern Recognition

Sees: Acylation followed by $LiAlH₄$ $\text{LiAlH}_4$ reduction. Shortcut: Amide carbonyl group reduces directly to $-CH₂-$ $-\text{CH}_2-$, resulting in secondary amine structure (option 3).

Chapter Mix

Class 12 Chemistry: Amines

Solution & Explanation

Core Logic

Step 1: Reaction Sequence

Step 2: Final Conclusion

Pattern Recognition

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