'A' is a neutral organic compound (M. F : C₈H₉ON$C_{8}H_{9}ON$). On treatment with aqueous Br₂/HO(-)$Br_{2}/HO^{(-)}$, 'A' forms a compound 'B' which is soluble in dilute acid. 'B' on treatment with aqueous NaNO₂/HCl(0-5°C)$NaNO_{2}/HCl(0-5^{\circ}C)$ 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$KMnO_{4}$ produces 'F'. 'F' contains two different types of hydrogen atoms. The structure of 'A' is
A.Structure 1$\text{Structure 1}$
Hofmann Bromamide Degradation
B.Structure 2$\text{Structure 2}$
Hofmann Bromamide Degradation
C.Structure 3$\text{Structure 3}$
Hofmann Bromamide Degradation
D.Structure 4$\text{Structure 4}$
Hofmann Bromamide Degradation
Solution & Explanation
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$HOOC-C_6H_4-COOH$).
Sequence of reactions for identifying Compound A
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.
Sequence of reactions for identifying Compound A
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
The image outlines a synthesis pathway converting bromobenzene through several steps finally using Hoffmann bromamide reaction.
Step 1: Tracing the product
The final product 'A' is Aniline.
Chapter Mix
Class 12 Chemistry: Amines
Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids
Class 12 Chemistry: Haloalkanes and Haloarenes
Q78jee_main_2024_30_jan_morningChemical Reactions of Amines
Following is a confirmatory test for aromatic primary amines. Identify reagent (A) and (B)
Ph-NH₂ A Ph-N₂^+Cl^- B Scarlet red dye$Ph-NH_2 \xrightarrow{A} Ph-N_2^+Cl^- \xrightarrow{B} \text{Scarlet red dye}$
The reaction sequence represents the classic dye test for aromatic primary amines.
Step 1 (Diazotization): Aniline (Ph-NH₂$Ph-NH_2$) reacts with nitrous acid (generated in situ from NaNO₂ + HCl$NaNO_2 + HCl$) at low temperature (0-5^° C$0-5^\circ C$) to form benzene diazonium chloride (Ph-N₂^+Cl^-$Ph-N_2^+Cl^-$).
Thus, Reagent A is NaNO₂ + HCl$NaNO_2 + HCl$ at 0-5^° C$0-5^\circ C$.
Step 2: Coupling Reaction
Step 2: The diazonium salt undergoes an electrophilic substitution (coupling reaction) with an electron-rich aromatic ring to form an azo dye.
The formation of a 'scarlet red dye' is specifically the result of coupling benzene diazonium chloride with β$\beta$-naphthol in a weakly basic medium (NaOH).
Chemical Reactions of Amines solution diagram for Q78 - JEE Main 2024 Morning
Chapter Mix
Class 12 Chemistry: Amines
Qjee_main_2024_31_jan_eveningReactions of Diazonium Salts
The azo-dye (Y) formed in the following reactions is
Sulphanilic acid + NaNO₂ + CH₃COOH arrow X$\text{Sulphanilic acid } + NaNO_2 + CH_3COOH \rightarrow X$The image shows the reaction of intermediate X with N,N-dimethylaniline to form an azo dye (Y).
A.
B.
C.
D.
Solution
Core Logic
Sulphanilic acid reacts with NaNO₂$NaNO_2$ and CH₃COOH$CH_3COOH$ to form a diazonium salt (X).
The diazonium salt (X) then reacts with N,N-dimethylaniline (given in the coupling step image). The coupling takes place at the para position of the highly activated N,N-dimethylaniline ring.
This coupling yields Methyl Orange, an azo dye. Its structure is p$p$-dimethylaminoazobenzenesulphonic acid.
The image shows the reaction of intermediate X with N,N-dimethylaniline to form an azo dye (Y).
Step 1: Final Identification
The final product (Y) matches option (4) structurally, containing the sulphonic acid group on one ring, the azo linkage, and the N,N-dimethylamine group on the para position of the other ring.
Chapter Mix
Class 12 Chemistry: Amines
Q70jee_main_2024_31_jan_eveningChemical Reactions of Amines
Given below are two statements:
Statement I: Aniline reacts with con. H₂SO₄$H_2SO_4$ followed by heating at 453-473 K gives p-aminobenzene sulphonic acid, which gives blood red colour in the 'Lassaigne's test'.
Statement II: In Friedel-Crafts alkylation and acylation reactions, aniline forms salt with the AlCl₃$AlCl_3$ catalyst. Due to this, nitrogen of aniline acquires a positive charge and acts as deactivating group.
In the light of the above statements, choose the correct answer from the options given below:
A.(1) Statement I is false but statement II is true$\text{(1) Statement I is false but statement II is true}$
B.(2) Both statement I and statement II are false$\text{(2) Both statement I and statement II are false}$
C.(3) Statement I is true but statement II is false$\text{(3) Statement I is true but statement II is false}$
D.(4) Both statement I and statement II are true$\text{(4) Both statement I and statement II are true}$
Solution
Core Logic
Statement I: Aniline reacting with concentrated H₂SO₄$H_2SO_4$ gives anilinium hydrogensulphate, which on heating at 453-473 K produces sulphanilic acid (p-aminobenzene sulphonic acid). Because sulphanilic acid contains both Nitrogen and Sulphur, it gives a blood-red colouration in Lassaigne's test due to the formation of thiocyanate ion SCN^-$SCN^-$ which reacts with Fe³⁺$Fe^{3+}$ to form [Fe(SCN)]²⁺$[Fe(SCN)]^{2+}$. Thus, Statement I is true.
Statement II: In Friedel-Crafts reactions, the Lewis acid catalyst AlCl₃$AlCl_3$ reacts with the lone pair on the nitrogen atom of aniline to form a salt. This generates a positive charge on the nitrogen, transforming the -NH₂$-NH_2$ group from a strong activating group into a strong deactivating group, thus preventing the Friedel-Crafts reaction from occurring. Thus, Statement II is true.
Chemical Reactions of Amines diagram for Q70 - JEE Main 2024 Evening
Step 1: Final Conclusion
Both Statement I and Statement II are true. Option (4) is correct.
Chapter Mix
Class 12 Chemistry: Amines
Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
Q83jee_main_2024_31_jan_eveningAcylation of Amines
A compound (x) with molar mass 108 ~g mol⁻¹$108\mathrm{~g\,mol^{-1}}$ undergoes acetylation to give product with molar mass 192 ~g mol⁻¹$192\mathrm{~g\,mol^{-1}}$. The number of amino groups in the compound (x) is ________.
During the acetylation of an amino group, one hydrogen atom (mass = 1 g/mol$1\text{ g/mol}$) is replaced by an acetyl group (-COCH₃$-COCH_3$, mass = 43 g/mol$43\text{ g/mol}$).
Gain in molecular weight for every one -NH₂$-NH_2$ group acetylated = 43 - 1 = 42 g/mol$43 - 1 = 42\text{ g/mol}$.
Step 1: Calculating Number of Groups
Total increase in molecular weight = Final mass - Initial mass = 192 - 108 = 84 g/mol$192 - 108 = 84\text{ g/mol}$.
Number of amino groups = Total mass increaseMass increase per group = (84)/(42) = 2$$\text{Number of amino groups} = \frac{\text{Total mass increase}}{\text{Mass increase per group}} = \frac{84}{42} = 2$$
Chapter Mix
Class 12 Chemistry: Amines
More Amines Questions — jee_main_2026_22_january_morning
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.
