An organic compound (P) on treatment with aqueous ammonia under hot condition forms compound (Q) which on heating with Br_2$Br_{2}$ and KOH forms compound (R) having molecular formula C_9H_7N$C_{9}H_{7}N$. Names of P, Q and R respectively are.
### Core Logic
The reaction of an amide with Br_2$Br_2$ 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$C_9H_7N$. Wait, looking at the standard solutions for this type of problem, aniline is C_6H_7N$C_6H_7N$. The PDF says C_9H_7N$C_9H_7N$ which is likely a typo in the original paper for C_6H_7N$C_6H_7N$, since option (1) gives Aniline (C_6H_7N$C_6H_7N$). Let's assume the standard sequence:
1. mathrmPh-COOH xrightarrowmathrmNH_3, Delta mathrmPh-CO-NH_2$\mathrm{Ph-COOH} \xrightarrow{\mathrm{NH}_3, \Delta} \mathrm{Ph-CO-NH}_2$ (Benzoic acid to Benzamide)
2. mathrmPh-CO-NH_2 xrightarrowmathrmBr_2/mathrmKOH mathrmPh-NH_2$\mathrm{Ph-CO-NH}_2 \xrightarrow{\mathrm{Br}_2/\mathrm{KOH}} \mathrm{Ph-NH}_2$ (Benzamide to Aniline)
Aniline is C_6H_5NH_2 = C_6H_7N$C_6H_5NH_2 = C_6H_7N$. 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$Carboxylic\ Acid \xrightarrow{NH_3, \Delta} Amide \xrightarrow{Br_2/KOH} Amine$. The Br_2/KOH$Br_2/KOH$ step is Hoffmann bromamide reaction.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Amines
Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids
Keywords:#Hoffmann bromamide degradation#JEE Main 2026 Morning Q60#Amines JEE Main 2026#Preparation of Amines JEE Main 2026
More Amines Previous-Year Questions — Page 4
Qjee_main_2025_29_jan_morningBasic Character of Amines
Given below are some nitrogen containing compounds.
Four amine candidates are indexed to find the optimal basic compound for reaction calculations.
Each of them is treated with HCl separately.
1.0 g of the most basic compound will consume ________ mg of HCl.
(Given molar mass in g mol ^-1$^{-1}$ C:12, H : 1, O : 16, Cl : 35.5)
Four amine candidates are indexed to find the optimal basic compound for reaction calculations.
Numerical Answer.Answer: 341 to 341
Solution
### Related Formula
textMoles = fractextMasstextMolar Mass$$\text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}}$$textMass of HCl consumed = n_textamine cdot M_textHCl$$\text{Mass of HCl consumed} = n_{\text{amine}} \cdot M_{\text{HCl}}$$
### Core Logic
Step 1: Identify the most basic amine
Benzylamine (mathrmC_6mathrmH_5mathrmCH_2mathrmNH_2$\mathrm{C}_6\mathrm{H}_5\mathrm{CH}_2\mathrm{NH}_2$) is the most basic compound here because its nitrogen lone pair is localized and not involved in aromatic resonance. This stands in contrast to aniline or amides, which delocalize their lone pairs into the ring or carbonyl group .
Step 2: Neutralization Stoichiometry
mathrmC_6H_5CH_2NH_2 + mathrmHCl
ightarrow mathrmC_6H_5CH_2NH_3^+ Cl^-$$\mathrm{C_6H_5CH_2NH_2} + \mathrm{HCl}
ightarrow \mathrm{C_6H_5CH_2NH_3^+ Cl^-}$$
Molar Mass of Benzylamine (mathrmC_7mathrmH_9mathrmN$\mathrm{C}_7\mathrm{H}_9\mathrm{N}$):
M = (7 cdot 12) + (9 cdot 1) + 14 = 84 + 9 + 14 = 107 \, mathrmg/mol$$M = (7 \cdot 12) + (9 \cdot 1) + 14 = 84 + 9 + 14 = 107 \, \mathrm{g/mol}$$
Moles of Benzylamine in 1.0text g$1.0\text{ g}$ :
n = frac1.0107 simeq 0.009346 \, mathrmmol$$n = \frac{1.0}{107} \simeq 0.009346 \, \mathrm{mol}$$
Since 1 mole of benzylamine reacts with 1 mole of mathrmHCl$\mathrm{HCl}$ :
textMoles of HCl consumed = 0.009346 \, mathrmmol$$\text{Moles of HCl consumed} = 0.009346 \, \mathrm{mol}$$textMass of HCl = 0.009346 cdot 36.5 = 0.3411 \, mathrmg = 341.1 \, mathrmmg
ightarrow 341$$\text{Mass of HCl} = 0.009346 \cdot 36.5 = 0.3411 \, \mathrm{g} = 341.1 \, \mathrm{mg}
ightarrow 341$$
### Pattern Recognition
Aliphatic localized clusters (like the -mathrmCH_2mathrmNH_2$-\mathrm{CH}_2\mathrm{NH}_2$ segment in benzylamine) always show higher basicity than aromatic ring-conjugated arrays.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Amines
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.textBoth Statement I and Statement II are correct$\text{Both Statement I and Statement II are correct}$
B.textStatement I is correct but Statement II is incorrect$\text{Statement I is correct but Statement II is incorrect}$
C.textStatement I is incorrect but Statement II is correct$\text{Statement I is incorrect but Statement II is correct}$
D.textBoth 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 (C_6H_5NH_2$C_6H_5NH_2$).
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.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Amines
Given below are two statements:
Statement (I): The NH_2$NH_2$ 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.textBoth Statement I and Statement II are correct$\text{Both Statement I and Statement II are correct}$
B.textBoth Statement I and Statement II are incorrect$\text{Both Statement I and Statement II are incorrect}$
C.textStatement I is incorrect but Statement II is correct.$\text{Statement I is incorrect but Statement II is correct.}$
D.textStatement I is correct but Statement II is incorrect$\text{Statement I is correct but Statement II is incorrect}$
Solution
### Core Logic
Statement (I): The -NH_2$-NH_2$ 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 AlCl_3$AlCl_3$. Aniline is a Lewis base (due to the lone pair on N) and reacts with the Lewis acid AlCl_3$AlCl_3$ to form a stable salt/complex (C_6H_5overset+NH_2-AlCl_3^-$C_6H_5\overset{+}{N}H_2-AlCl_3^-$). This removes the lone pair from resonance and converts the -NH_2$-NH_2$ 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 (NH_2$NH_2$) reacts with the catalyst (AlCl_3$AlCl_3$) before the reaction can proceed.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Amines
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 -NH_2$-NH_2$ group is a strongly activating group and directs incoming electrophiles to the ortho and para positions due to resonance electron donation (+M$M$ effect).
### Step 1: Identifying Sigma Complexes (Arenium Ions)
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 -NH_2$-NH_2$ 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 -NH_2$-NH_2$ 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 -NH_2$-NH_2$ 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 -NH_2$-NH_2$ substituted carbon).
Electrophilic Substitution in Amines diagram for Q68 - JEE Main 2024 Morning
Since -NH_2$-NH_2$ is ortho/para directing, the meta-arenium ion will not be formed.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Amines
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. HNO_3$HNO_3$ and conc. H_2SO_4$H_2SO_4$ gives nitrobenzene.
Step 2: Reduction of nitrobenzene with Sn/HCl$Sn/HCl$ yields aniline.
Step 3: Aniline reacts with NaNO_2/HCl$NaNO_2/HCl$ at 0-5^circ 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.
### Pattern Recognition
Nitration rightarrow$\rightarrow$ Reduction rightarrow$\rightarrow$ Diazotization rightarrow$\rightarrow$ Coupling (Azo Dye Test).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Amines
More Amines Questions — jee_main_2026_21_jan_morning
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