A student performed analysis of aliphatic organic compound 'X' which on analysis gave C = 61.01%$C = 61.01\%$, H = 15.25%$H = 15.25\%$, N = 23.74%$N = 23.74\%$.
This compound, on treatment with HNO₂/H₂O$HNO_{2}/H_{2}O$ produced another compound 'Y' which did not contain any nitrogen atom. However, the compound 'Y' upon controlled oxidation produced another compound 'Z' that responded to iodoform test.
The structure of 'X' is:
Step 1: Empirical Formula calculation:
Moles of C = (61.01)/(12) ≈ 5.08$C = \frac{61.01}{12} \approx 5.08$
Moles of H = (15.25)/(1) ≈ 15.25$H = \frac{15.25}{1} \approx 15.25$
Moles of N = (23.74)/(14) ≈ 1.69$N = \frac{23.74}{14} \approx 1.69$
Ratio C : H : N = (5.08)/(1.69) : (15.25)/(1.69) : (1.69)/(1.69) ≈ 3 : 9 : 1$\frac{5.08}{1.69} : \frac{15.25}{1.69} : \frac{1.69}{1.69} \approx 3 : 9 : 1$
Empirical formula = C₃H₉N$C_3H_9N$.
Step 2: Chemical tests deduction.
The compound is treated with HNO₂/H₂O$HNO_2/H_2O$ and produces a nitrogen-free compound Y. This implies X is a primary aliphatic amine (C₃H₇NH₂$C_3H_7NH_2$), giving an alcohol Y (C₃H₇OH$C_3H_7OH$).
Compound Y upon controlled oxidation forms Z, which gives positive iodoform test. For Z to give an iodoform test, it must be a methyl ketone. Thus Z is acetone (CH₃COCH₃$CH_3COCH_3$).
For Y to oxidize to acetone, Y must be Isopropyl alcohol (CH₃CH(OH)CH₃$CH_3CH(OH)CH_3$).
Therefore, X must be Isopropylamine: (CH₃)₂CHNH₂$(CH_3)_2CHNH_2$.
Step 1: Reaction Sequence
Reactions Of Amines diagram for Q63 - JEE Main 2026 Evening(CH₃)₂CHNH₂ HNO₂ (CH₃)₂CH-N₂^+ -N₂ (CH₃)₂CH^+ H₂O (CH₃)₂CHOH (Y)$(CH_3)_2CHNH_2 \xrightarrow{HNO_2} (CH_3)_2CH-N_2^+ \xrightarrow{-N_2} (CH_3)_2CH^+ \xrightarrow{H_2O} (CH_3)_2CHOH \text{ (Y)}$(CH₃)₂CHOH Oxidation CH₃COCH₃ (Z)$(CH_3)_2CHOH \xrightarrow{\text{Oxidation}} CH_3COCH_3 \text{ (Z)}$CH₃COCH₃$CH_3COCH_3$ responds to the iodoform test.
Step 2: Final Conclusion
The structure of 'X' is Isopropylamine.
Pattern Recognition
Iodoform test on an oxidized alcohol means the alcohol was a secondary alcohol with a terminal methyl group. Combining this with HNO₂$HNO_2$ reacting with primary amines gives the exact functional groups.
Chapter Mix
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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
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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
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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
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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
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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_28_january_evening
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.
