'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
Keywords:#Hofmann Bromamide Degradation#JEE Main 2026 Morning Q66#Amines JEE Main 2026#aqueous Br2/HO
More Amines Previous-Year Questions — Page 7
Q70jee_main_2024_01_february_morningNomenclature of Amines
Given below are two statements:
Statement (I) : Aminobenzene and aniline are same organic compounds.
Statement (II) : Aminobenzene and aniline are different organic compounds.
In the light of the above statements, choose the most appropriate answer from the options given below:
A.Both Statement I and Statement II are correct$\text{Both Statement I and Statement II are correct}$
B.Statement I is correct but Statement II is incorrect$\text{Statement I is correct but Statement II is incorrect}$
C.Statement I is incorrect but Statement II is correct$\text{Statement I is incorrect but Statement II is correct}$
D.Both Statement I and Statement II are incorrect$\text{Both Statement I and Statement II are incorrect}$
Solution
Core Logic
Aniline is the common name for the simplest aromatic amine, which consists of a phenyl group attached to an amino group (C₆H₅NH₂$C_6H_5NH_2$).
According to IUPAC nomenclature, the amino group attached to a benzene ring can also be called aminobenzene.
Step 1: Statement Validation
Statement I: True. Aminobenzene is just the systematic IUPAC name for aniline.
Statement II: False. They refer to the exact same molecule.
Pattern Recognition
Common names for simple aromatic compounds are often accepted as IUPAC names. Aniline = Benzenamine = Aminobenzene.
Given below are two statements:
Statement (I): The NH₂$NH_2$ group in Aniline is ortho and para directing and a powerful activating group.
Statement (II): Aniline does not undergo Friedel-Craft's reaction (alkylation and acylation).
In the light of the above statements, choose the most appropriate answer from the options given below:
A.Both Statement I and Statement II are correct$\text{Both Statement I and Statement II are correct}$
B.Both Statement I and Statement II are incorrect$\text{Both Statement I and Statement II are incorrect}$
C.Statement I is incorrect but Statement II is correct.$\text{Statement I is incorrect but Statement II is correct.}$
D.Statement I is correct but Statement II is incorrect$\text{Statement I is correct but Statement II is incorrect}$
Solution
Core Logic
Statement (I): The -NH₂$-NH_2$ group has a lone pair of electrons on nitrogen, which undergoes resonance with the benzene ring (strong +M effect). This strongly activates the ring towards electrophilic substitution and directs incoming electrophiles to the ortho and para positions.
Statement (II): Friedel-Crafts alkylation and acylation require a Lewis acid catalyst like anhydrous AlCl₃$AlCl_3$. Aniline is a Lewis base (due to the lone pair on N) and reacts with the Lewis acid AlCl₃$AlCl_3$ to form a stable salt/complex (C₆H₅ +NH₂-AlCl₃^-$C_6H_5\overset{+}{N}H_2-AlCl_3^-$). This removes the lone pair from resonance and converts the -NH₂$-NH_2$ group into a strongly deactivating group, thereby halting the Friedel-Crafts reaction.
Step 1: Evaluate Statements
Statement I is correct.
Statement II is correct.
Pattern Recognition
Aniline NEVER undergoes Friedel-Crafts because the base (NH₂$NH_2$) reacts with the catalyst (AlCl₃$AlCl_3$) before the reaction can proceed.
Chapter Mix
Class 12 Chemistry: Amines
Qjee_main_2024_29_january_eveningDiazotization and Sandmeyer Reaction
The product A formed in the following reaction is:
The diagram details aniline undergoing diazotization with sodium nitrite and hydrochloric acid, followed by treatment with cuprous chloride.
Diazotization Step: Aniline reacts with nitrous acid generated in situ (NaNO₂ + HCl$\text{NaNO}_2 + \text{HCl}$) at low temperature (0-5^$0-5^\circ\text{C}$) to form benzene diazonium chloride intermediate.
Sandmeyer Replacement Step: Treating the diazonium salt with cuprous chloride (Cu₂Cl₂$\text{Cu}_2\text{Cl}_2$) results in replacement of the diazo group by a chlorine atom.
Step 1: Visual Verification
Tracing structural shifts through intermediates validates the final outcome:
The diagram details aniline undergoing diazotization with sodium nitrite and hydrochloric acid, followed by treatment with cuprous chloride.
Pattern Recognition
Aniline arrow$\rightarrow$ Diazonium salt arrow$\rightarrow$ Chlorobenzene via standard Sandmeyer synthesis protocols.
Chapter Mix
Class 12 Chemistry: Organic Compounds Containing Nitrogen
Q68jee_main_2024_29_jan_morningElectrophilic Substitution in Amines
The arenium ion which is not involved in the bromination of Aniline is.
A. ""
B. ""
C. ""
D. ""
Solution
Core Logic
Aniline undergoes electrophilic aromatic substitution (like bromination). The -NH₂$-NH_2$ group is a strongly activating group and directs incoming electrophiles to the ortho and para positions due to resonance electron donation (+M$M$ effect).
When an electrophile (Br^+$Br^+$) attacks the ring, an intermediate arenium ion (sigma complex) is formed.
If attack occurs at the ortho or para position, the positive charge is delocalized onto the carbon atom bearing the -NH₂$-NH_2$ group. The lone pair on nitrogen can then stabilize this positive charge via resonance, forming a highly stable resonance structure (an octet-complete intermediate).
If attack occurs at the meta position, the positive charge delocalizes only over the remaining ring carbons and never rests on the carbon bearing the -NH₂$-NH_2$ group. Thus, it misses the extra stabilization provided by the nitrogen lone pair.
Because the meta attack intermediate is less stable compared to ortho/para attack, and the -NH₂$-NH_2$ is strictly o/p directing, the meta-arenium ion is NOT a primary intermediate involved in standard bromination pathways of neutral aniline.
Step 2: Conclusion
Option 3 displays the arenium ion resulting from a meta-attack (positive charge skips the -NH₂$-NH_2$ substituted carbon).
Electrophilic Substitution in Amines diagram for Q68 - JEE Main 2024 Morning
Since -NH₂$-NH_2$ is ortho/para directing, the meta-arenium ion will not be formed.
Chapter Mix
Class 12 Chemistry: Amines
Qjee_main_2024_30_january_eveningDiazonium Salts
The products A and B formed in the following reaction scheme are respectively
The diagram shows a reaction pathway for synthesizing product A and B.
A.
B.
C.
D.
Solution
Core Logic
Step 1: Nitration of benzene using conc. HNO₃$HNO_3$ and conc. H₂SO₄$H_2SO_4$ gives nitrobenzene.
Step 2: Reduction of nitrobenzene with Sn/HCl$Sn/HCl$ yields aniline.
Step 3: Aniline reacts with NaNO₂/HCl$NaNO_2/HCl$ at 0-5^° C$0-5^\circ C$ to form benzene diazonium chloride (Product A).
Step 4: Benzene diazonium chloride undergoes a coupling reaction with phenol (typically in a mildly alkaline medium) to form p-hydroxyazobenzene, an orange dye (Product B).
The diagram shows a reaction pathway for synthesizing product A and B.
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