The sequence from the following that would result in giving predominantly 3, 4, 5-Tribromoaniline is :

Solution & Explanation

### 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
Aniline Reactions and Directing Effects
Aniline Reactions and Directing Effects

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
Q74 jee_main_2026_21_jan_morning Electrophilic Substitution Reactions
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. Answer: 20 to 20

Solution

### 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
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
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%)