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 5
Q72jee_main_2024_30_jan_morningPreparation of Amines
The final product A, formed in the following multistep reaction sequence is:
The image outlines a synthesis pathway converting bromobenzene through several steps finally using Hoffmann bromamide reaction.
A.The image outlines a synthesis pathway converting bromobenzene through several steps finally using Hoffmann bromamide reaction.
B.The image outlines a synthesis pathway converting bromobenzene through several steps finally using Hoffmann bromamide reaction.
C.The image outlines a synthesis pathway converting bromobenzene through several steps finally using Hoffmann bromamide reaction.
D.The image outlines a synthesis pathway converting bromobenzene through several steps finally using Hoffmann bromamide reaction.
Solution
### Core Logic
Step 1: Bromobenzene + Mg, ether rightarrow Phenylmagnesium bromide$Mg, ether \rightarrow Phenylmagnesium bromide$ (Grignard reagent).
Step 2: Grignard + CO_2$CO_2$ followed by H^+ rightarrow Benzoic acid$H^+ \rightarrow Benzoic acid$ (C_6H_5COOH$C_6H_5COOH$).
Step 3: Benzoic acid + NH_3, Delta rightarrow Benzamide$NH_3, \Delta \rightarrow Benzamide$ (C_6H_5CONH_2$C_6H_5CONH_2$).
Step 4: Benzamide + Br_2/NaOH$Br_2/NaOH$ (Hoffmann bromamide degradation) rightarrow Aniline$\rightarrow Aniline$ (C_6H_5NH_2$C_6H_5NH_2$).
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.
### Evaluation Rubric / Model Answer
null
### 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-NH_2 xrightarrowA Ph-N_2^+Cl^- xrightarrowB textScarlet red dye$Ph-NH_2 \xrightarrow{A} Ph-N_2^+Cl^- \xrightarrow{B} \text{Scarlet red dye}$
### Core Logic
The reaction sequence represents the classic dye test for aromatic primary amines.
Step 1 (Diazotization): Aniline (Ph-NH_2$Ph-NH_2$) reacts with nitrous acid (generated in situ from NaNO_2 + HCl$NaNO_2 + HCl$) at low temperature (0-5^circ C$0-5^\circ C$) to form benzene diazonium chloride (Ph-N_2^+Cl^-$Ph-N_2^+Cl^-$).
Thus, Reagent A is NaNO_2 + HCl$NaNO_2 + HCl$ at 0-5^circ 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 beta$\beta$-naphthol in a weakly basic medium (NaOH).
Chemical Reactions of Amines solution diagram for Q78 - JEE Main 2024 Morning
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Amines
Q69jee_main_2024_31_jan_eveningReactions of Diazonium Salts
The azo-dye (Y) formed in the following reactions is
textSulphanilic acid + NaNO_2 + CH_3COOH rightarrow 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).
### Core Logic
1) Sulphanilic acid reacts with NaNO_2$NaNO_2$ and CH_3COOH$CH_3COOH$ to form a diazonium salt (X).
2) 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.
3) 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.
### Evaluation Rubric / Model Answer
null
### 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. H_2SO_4$H_2SO_4$ 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 AlCl_3$AlCl_3$ 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.text(1) Statement I is false but statement II is true$\text{(1) Statement I is false but statement II is true}$
B.text(2) Both statement I and statement II are false$\text{(2) Both statement I and statement II are false}$
C.text(3) Statement I is true but statement II is false$\text{(3) Statement I is true but statement II is false}$
D.text(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 H_2SO_4$H_2SO_4$ 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 Fe^3+$Fe^{3+}$ to form [Fe(SCN)]^2+$[Fe(SCN)]^{2+}$. Thus, Statement I is true.
Statement II: In Friedel-Crafts reactions, the Lewis acid catalyst AlCl_3$AlCl_3$ 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 -NH_2$-NH_2$ 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.
### Evaluation Rubric / Model Answer
null
### 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 108mathrm~g\,mol^-1$108\mathrm{~g\,mol^{-1}}$ undergoes acetylation to give product with molar mass 192mathrm~g\,mol^-1$192\mathrm{~g\,mol^{-1}}$. The number of amino groups in the compound (x) is ________.
Numerical Answer.Answer: 2 to 2
Solution
### Related Formula
R-NH_2 + CH_3COCl rightarrow R-NH-COCH_3 + HCl$$R-NH_2 + CH_3COCl \rightarrow R-NH-COCH_3 + HCl$$
### Core Logic
During the acetylation of an amino group, one hydrogen atom (mass = 1text g/mol$1\text{ g/mol}$) is replaced by an acetyl group (-COCH_3$-COCH_3$, mass = 43text g/mol$43\text{ g/mol}$).
Gain in molecular weight for every one -NH_2$-NH_2$ group acetylated = 43 - 1 = 42text 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 = 84text g/mol$192 - 108 = 84\text{ g/mol}$.
textNumber of amino groups = fractextTotal mass increasetextMass increase per group = frac8442 = 2$$\text{Number of amino groups} = \frac{\text{Total mass increase}}{\text{Mass increase per group}} = \frac{84}{42} = 2$$
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Amines
More Amines Questions — jee_main_2026_21_jan_morning
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