Consider the following reaction sequence Benzene xrightarrowtextconc. HNO_3 + textconc. H_2SO_4, 333text K textP xrightarrow1.text Sn/HCl/Delta quad 2.text pH neutralised textQ xrightarrow(mathrmCH_3mathrmCO)_2mathrmO textR xrightarrow1.text conc. HNO_3 + text conc. H_2SO_4 quad 2.text pH neutralised (major product) textS xrightarrowmathrmHCl / mathrmEtOH / Delta textT The percentage of nitrogen in product ‘T’ formed is ____%. (Nearest integer) (Given molar mass in mathrmg\ mol^-1 H:1, C:12, N:14, O:16)

Numerical Answer Type:
Enter a numerical value Answer: 20 to 20 +4 marks

Solution & Explanation

### Core Logic Step 1: Nitration of benzene gives nitrobenzene (P). mathrmPh-H xrightarrowHNO_3/H_2SO_4 mathrmPh-NO_2 quad text(P)
Electrophilic Substitution Reactions diagram for Q74 - JEE Main 2026 Morning
Electrophilic Substitution Reactions diagram for Q74 - JEE Main 2026 Morning
Step 2: Reduction of nitrobenzene with Sn/HCl gives aniline (Q). mathrmPh-NO_2 xrightarrowSn/HCl mathrmPh-NH_2 quad text(Q)
Electrophilic Substitution Reactions diagram for Q74 - JEE Main 2026 Morning
Electrophilic Substitution Reactions diagram for Q74 - JEE Main 2026 Morning
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. mathrmPh-NH_2 xrightarrow(CH_3CO)_2O mathrmPh-NH-CO-CH_3 quad text(R)
Electrophilic Substitution Reactions diagram for Q74 - JEE Main 2026 Morning
Electrophilic Substitution Reactions diagram for Q74 - JEE Main 2026 Morning
Step 4: Nitration of acetanilide gives predominantly p-nitroacetanilide (S) due to steric hindrance at ortho position. mathrmPh-NH-CO-CH_3 xrightarrowHNO_3/H_2SO_4 ptext-NO_2text-C_6textH_4text-NH-CO-CH_3 quad text(S) Step 5: Acidic hydrolysis of the amide linkage yields p-nitroaniline (T). ptext-NO_2text-C_6textH_4text-NH-CO-CH_3 xrightarrowHCl/EtOH/Delta ptext-NO_2text-C_6textH_4text-NH_2 quad text(T)
Electrophilic Substitution Reactions diagram for Q74 - JEE Main 2026 Morning
Electrophilic Substitution Reactions diagram for Q74 - JEE Main 2026 Morning
Molecular formula of p-nitroaniline (T) is C_6H_6N_2O_2. Molar mass = (6 times 12) + (6 times 1) + (2 times 14) + (2 times 16) = 72 + 6 + 28 + 32 = 138text g/mol. Total mass of Nitrogen = 2 times 14 = 28text g. Percentage of Nitrogen = frac28138 times 100 approx 20.29\%. ### Step 1: Final Conclusion Nearest integer is 20. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Amines

Reference Study Guides

Amines

More Amines Previous-Year Questions

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 Br_2 and KOH forms compound (R) having molecular formula C_9H_7N. Names of P, Q and R respectively are.
  • A. textBenzoic acid, benzamide, aniline
  • B. textToluic acid, methylbenzamide, 2-methylaniline
  • C. textBenzoic acid,4-methylbenzamide,4-methylaniline.
  • D. textPhenylethanoic acid, phenylethanamide, benzamine

Solution

### Core Logic The reaction of an amide with 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 C_9H_7N. Wait, looking at the standard solutions for this type of problem, aniline is C_6H_7N. The PDF says C_9H_7N which is likely a typo in the original paper for C_6H_7N, since option (1) gives Aniline (C_6H_7N). Let's assume the standard sequence: 1. mathrmPh-COOH xrightarrowmathrmNH_3, Delta mathrmPh-CO-NH_2 (Benzoic acid to Benzamide) 2. mathrmPh-CO-NH_2 xrightarrowmathrmBr_2/mathrmKOH mathrmPh-NH_2 (Benzamide to Aniline) Aniline is C_6H_5NH_2 = C_6H_7N. So P is Benzoic acid, Q is Benzamide, R is Aniline. ### Pattern Recognition Reaction sequence: Carboxylic\ Acid xrightarrowNH_3, Delta Amide xrightarrowBr_2/KOH Amine. The Br_2/KOH step is Hoffmann bromamide reaction. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Amines Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids
Q jee_main_2025_02_april_evening Diazotisation and Coupling Reactions
When a concentrated solution of sulphanilic acid and 1-naphthylamine is treated with nitrous acid (273 K) and acidified with acetic acid, the mass (g) of 0.1 mole of product formed is : Given molar mass in mathrmg~mol^-1 H:1, C:12, N:14, O:16, S:32
  • A. 343
  • B. 330
  • C. 33
  • D. 66

Solution

### Related Formula textMass (g) = textNumber of moles times textMolar mass (mathrmg~mol^-1) ### Core Logic Sulphanilic acid is diazotized under cold conditions (273~mathrmK) with nitrous acid to form a diazonium salt intermediate. This diazonium salt undergoes a coupling reaction with 1-naphthylamine to form a red azo dye. First, sulphanilic acid acts as a zwitterion and undergoes diazotization:
Azo-dye synthesis reaction from diazotized sulphanilic acid and 1-naphthylamine
Azo-dye synthesis reaction from diazotized sulphanilic acid and 1-naphthylamine
Nitrous acid reacts with the amine to form the diazonium compound:
Azo-dye synthesis reaction from diazotized sulphanilic acid and 1-naphthylamine
Azo-dye synthesis reaction from diazotized sulphanilic acid and 1-naphthylamine
This intermediate couples with 1-naphthylamine at the para-position to give the red azo dye compound:
Azo-dye synthesis reaction from diazotized sulphanilic acid and 1-naphthylamine
Azo-dye synthesis reaction from diazotized sulphanilic acid and 1-naphthylamine
### Step 1: Calculate the Molar Mass The molecular formula of the red-azo dye formed is mathrmC_16H_13N_3O_3S. Let's calculate its molar mass using the given atomic masses: textMolar mass = 16(12) + 13(1) + 3(14) + 3(16) + 32 textMolar mass = 192 + 13 + 42 + 48 + 32 = 327~mathrmg~mol^-1 ### Step 2: Calculate the Mass of 0.1 Mole Using the relation for mass: textMass of 0.1~textmole = 0.1 times 327 = 32.7~mathrmg approx 33~mathrmg Hence, the nearest option is 33~mathrmg. ### Pattern Recognition Azo coupling reactions are clean electrophilic aromatic substitution reactions. Diazotized sulphanilic acid has a highly electron-withdrawing sulphonic acid group, making it an excellent electrophile that couples selectively at the para-position of 1-naphthylamine. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Amines
Q38 jee_main_2025_02_april_morning Basic Strength of Amines
The correct order of basic nature on aqueous solution for the bases mathrmNH_3, mathrmH_2mathrmN-mathrmNH_2, mathrmCH_3mathrmCH_2mathrmNH_2, (mathrmCH_3mathrmCH_2)_2mathrmNH and (mathrmCH_3mathrmCH_2)_3mathrmN is:
  • A. (1)\ mathrmNH_3 < mathrmH_2mathrmN - mathrmNH_2 < (mathrmCH_3mathrmCH_2)_3mathrmN < mathrmCH_3mathrmCH_2mathrmNH_2 < (mathrmCH_3mathrmCH_2)_2mathrmNH
  • B. (2)\ mathrmNH_3 < mathrmH_2mathrmN - mathrmNH_2 < mathrmCH_3mathrmCH_2mathrmNH_2 < (mathrmCH_3mathrmCH_2)_2mathrmNH < (mathrmCH_3mathrmCH_2)_3mathrmN
  • C. (3)\ mathrmH_2mathrmN - mathrmNH_2 < mathrmNH_3 < (mathrmCH_3mathrmCH_2)_3mathrmN < mathrmCH_3mathrmCH_2mathrmNH_2 < (mathrmCH_3mathrmCH_2)_2mathrmNH
  • D. (4)\ mathrmNH_2 - mathrmNH_2 < mathrmNH_3 < mathrmCH_3mathrmCH_2mathrmNH_2 < (mathrmCH_3mathrmCH_2)_3mathrmN < (mathrmCH_3mathrmCH_2)_2mathrmNH

Solution

### Related Formula Basic strength in aqueous medium depends on three combined effects: textBasic Strength propto textInductive Effect (+I) + textSolvation Energy - textSteric Hindrance ### Core Logic Let's list structural elements row-by-row: * Ethyl substituted amine trends in aqueous systems uniquely align into a 2° > 3° > 1° configuration due to competing steric and hydration energies: mathrm(Et)_2NH > (Et)_3N > EtNH_2 * Ammonia (mathrmNH_3) is less basic than aliphatic substituted structures due to the absence of electron-donating alkyl clusters. * Hydrazine (mathrmH_2mathrmN-NH_2) is exceptionally weak compared to ammonia because the adjacent electronegative nitrogen creates an electron-withdrawing (-I) effect, while lone-pair repulsions reduce overall stability. ### Step 1: Ordering Assembling the fragments gives the complete verified thermodynamic order: mathrmNH_2-NH_2 < NH_3 < CH_3CH_2NH_2 < (CH_3CH_2)_3N < (CH_3CH_2)_2NH ### Pattern Recognition Remember the standard numeric rules for aliphatic basic strength order in aqueous media: - Methyl amines follow: **213** - Ethyl amines follow: **231** This simple sequence trick handles complex ranking items instantly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Amines
Q jee_main_2025_03_april_evening Aniline Reactions and Directing Effects
The sequence from the following that would result in giving predominantly 3, 4, 5-Tribromoaniline is :
  • A.
  • B.
  • C.
  • D.

Solution

### Related Formula Directing effects in multi-substituted benzenes: - -NH_2 is a strong activating group and ortho/para director. - -NO_2 is a strong deactivating, meta-directing group. - Diazotization followed by Sandmeyer reaction replaces an -NH_2 group with a halogen. ### Core Logic To prepare 3,4,5-tribromoaniline, we must introduce three bromine atoms adjacent to each other (meta to the final amino group, with one para and two meta). Let's trace the sequence in Option (3) starting from 4-nitroaniline (p-nitroaniline): ### Step 1: Bromination of p-nitroaniline Treatment of 4-nitroaniline with excess mathrmBr_2 in acetic acid: - The amino group (-NH_2) is a strong activator and directs to its ortho positions (positions 2 and 6). - Positions 2 and 6 are meta to the -NO_2 group, which is compatible. - This yields 2,6-dibromo-4-nitroaniline. ### Step 2: Diazotization and replacement of amino group 1. mathrmNaNO_2 + mathrmHCl diazotizes the amino group to a diazonium salt: mathrmR-NH_2 rightarrow mathrmR-N_2^+ Cl^- 2. Addition of mathrmCuBr (Sandmeyer reaction) replaces the diazonium group with bromine: mathrmR-N_2^+ Cl^- xrightarrowmathrmCuBr mathrmR-Br This yields 3,4,5-tribromonitrobenzene. ### Step 3: Reduction of nitro group Reduction of the nitro group using mathrmSn/HCl converts -NO_2 to -NH_2: mathrmR-NO_2 xrightarrowmathrmSn, HCl mathrmR-NH_2 This yields 3,4,5-tribromoaniline as the predominant product. Thus, Option (3) is correct. ### Pattern Recognition To brominate meta positions relative to an amino group, use a nitro precursor at the para position. The -NH_2 group activates these positions first, after which the initial amino group is replaced with a halogen, and the nitro group is subsequently reduced back to a primary amine. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Amines

More Amines Questions — jee_main_2026_21_jan_morning

Practice all Amines previous-year questions →

Rankbit System
JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%) | JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%)
YOUR FIRST PREP STEP STARTS HERE

We Map Every Repeating Question in Competitive Exams.

Say goodbye to generic mock test fatigue. RankBit uses smart analysis to group past exam questions into their foundational Repeating Question Types. Find chapter weightage, track repeating questions, and score higher with targeted practice.

Select Your Target Exam

Choose an exam track below to find formulas per chapter and patterns.

Syncing Exam Intelligence

Mapping formulas and patterns across all tracks…

PATH A — FULL LENGTH PRACTICE

Full Mock Test Hub

Simulate real NTA exam conditions with fully tracked mocks. Time yourself against past papers.

Now Live Open
PATH B — TARGETED PRACTICE

Topic-wise Practice Hub

Practice past-year questions one chapter at a time. Pick an exam → subject → chapter and get every PYQ for that topic — pulled together from all past papers — with the chapter's key formulas alongside.

Loading Questions... Browse Topics

Exam Analysis Tools

Chapter weightage trends and topic-wise practice — all mapped from past shift papers.

Year-wise PYQ

Chapter Weightage & Trends

See high-yield chapters, year-wise counts and the frequency-based chance each topic repeats next exam.

Weightage Board Open
Targeted PYQs

Topic-wise Practice Hub

Every past-year question for a chapter pulled from all papers, with its key formulas alongside.

One chapter at a time Open
Latest from the Blog
View all →

Loading articles...