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 & Explanation
### 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
Keywords:#Chemical Reactions of Amines#JEE Main 2024 Evening Q70#Amines JEE Main 2024#Chemical Reactions of Amines JEE Main 2024
More Amines Previous-Year Questions — Page 2
Q41jee_main_2025_29_jan_eveningDiazotization and Coupling Reactions
Which one of the following reaction sequences will give an azo dye?
(1) Nitrobenzene treated with (i) Sn/HCl$Sn/HCl$, (ii) NaNO_2/HCl$NaNO_{2}/HCl$, (iii) beta$\beta$-naphthol, NaOH$NaOH$
(2) Benzenesulfonic acid treated with (i) SOCl_2$SOCl_2$, (ii) NH_3$NH_3$, (iii) Benzyl chloride
(3) Benzonitrile treated with (i) 70\% H_2SO_4$70\% H_2SO_4$, (ii) PCl_5$PCl_5$, (iii) Aniline
(4) Aniline treated with (i) HCl/NaNO_2$HCl/NaNO_2$, (ii) Toluene
A. Reaction sequence (1)
B. Reaction sequence (2)
C. Reaction sequence (3)
D. Reaction sequence (4)
Solution
### Core Logic
Let's track sequence (1):
1) Nitrobenzene (Ph-NO_2$Ph-NO_2$) is reduced using Sn/HCl$Sn/HCl$ to form Aniline (Ph-NH_2$Ph-NH_2$).
2) Aniline undergoing diazotization with NaNO_2/HCl$NaNO_{2}/HCl$ at cold temperatures (0-5^circC$0-5^{circ}C$) creates Benzene diazonium chloride (Ph-N_2^+Cl^-$Ph-N_{2}^{+}Cl^{-}$).
3) The diazonium salt undergoes a coupling reaction with beta$\beta$-naphthol in alkaline conditions (NaOH$NaOH$) to synthesize a highly vibrant red-orange azo dye.
Diazotization and Coupling Reactions diagram for Q41 - JEE Main 2025 Evening
### Pattern Recognition
The standard sequence for azo dye preparation is: Aromatic Nitro
ightarrow$
ightarrow$ Primary Amine
ightarrow$
ightarrow$ Diazonium Salt
ightarrow$
ightarrow$ Phenol/Naphthol Coupling.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Amines
Q49jee_main_2025_28_jan_morningYield and Stoichiometric Calculations
Consider the following sequence of reactions :
The flowchart tracks a chemical conversion starting from chlorobenzene down to final compound B.
11.25 mg of chlorobenzene will produce mathrmx times 10^-1$\mathrm{x} \times 10^{-1}$ mg of product B.
(Consider the reactions result in complete conversion.)
[Given molar mass of C, H, O, N and Cl as 12, 1, 16, 14 and 35.5mathrmg\,mol^-1$35.5\mathrm{g\,mol}^{-1}$ respectively]
Numerical Answer.Answer: 93 to 93
Solution
### Core Logic
The reaction sequence details the functional conversion of chlorobenzene down to product B (aniline, with a molar mass of 93\,mathrmg\,mol^-1$93\,\mathrm{g\,mol}^{-1}$).
Following stoichiometric preservation:
textmoles of chlorobenzene = textmoles of Aniline (B)$\text{moles of chlorobenzene} = \text{moles of Aniline (B)}$
Molar mass of chlorobenzene (mathrmC_6mathrmH_5mathrmCl$\mathrm{C}_6\mathrm{H}_5\mathrm{Cl}$) = 112.5\,mathrmg\,mol^-1$112.5\,\mathrm{g\,mol}^{-1}$.
The flowchart tracks a chemical conversion starting from chlorobenzene down to final compound B.textmoles = frac11.25 times 10^-3\,mathrmg112.5\,mathrmg\,mol^-1 = 10^-4\,mathrmmol$\text{moles} = \frac{11.25 \times 10^{-3}\,\mathrm{g}}{112.5\,\mathrm{g\,mol}^{-1}} = 10^{-4}\,\mathrm{mol}$
Mass of product B produced:
textMass = 10^-4\,mathrmmol times 93\,mathrmg\,mol^-1 = 9.3 times 10^-3\,mathrmg = 9.3\,mathrmmg$\text{Mass} = 10^{-4}\,\mathrm{mol} \times 93\,\mathrm{g\,mol}^{-1} = 9.3 \times 10^{-3}\,\mathrm{g} = 9.3\,\mathrm{mg}$
Expressing in the specified format:
9.3\,mathrmmg = 93 times 10^-1\,mathrmmg Rightarrow x = 93$9.3\,\mathrm{mg} = 93 \times 10^{-1}\,\mathrm{mg} \Rightarrow x = 93$
### Pattern Recognition
Sees: Conversion sequence preserving a 1:1 mole ratio layout.
Shortcut: Directly compute target weight via W_B = W_A cdot fracM_BM_A = 11.25 cdot frac93112.5 = 9.3$W_B = W_A \cdot \frac{M_B}{M_A} = 11.25 \cdot \frac{93}{112.5} = 9.3$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Amines
Q37jee_main_2025_03_april_morningDiazonium Salts and Reactions
Identify [A], [B], and [C], respectively in the following reaction sequence:
The reaction shows Aniline reacting with sodium nitrite and hydrochloric acid to produce intermediate A, followed by treatment with potassium iodide to yield B, which then dimerizes via sodium in dry ether to form C.
A. Option (1)
B. Option (2)
C. Option (3)
D. Option (4)
Solution
### Core Logic
Let us resolve each structural step sequentially:
1. **Step 1:** Aniline undergoes diazotization when treated with textNaNO_2 + textHCl$\text{NaNO}_2 + \text{HCl}$ at 273-278text K$273-278\text{ K}$, forming benzene diazonium chloride [A]$[A]$ (textC_6textH_5textN_2^+textCl^-$\text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^-$).
2. **Step 2:** Warming benzene diazonium chloride with potassium iodide (textKI$\text{KI}$) substitutes the diazonium group with iodine, producing iodobenzene [B]$[B]$ (textC_6textH_5textI$\text{C}_6\text{H}_5\text{I}$).
3. **Step 3:** Treating iodobenzene with sodium metal in dry ether causes a Fittig coupling reaction, dimerizing two phenyl radicals into biphenyl [C]$[C]$ (textC_6textH_5-textC_6textH_5$\text{C}_6\text{H}_5-\text{C}_6\text{H}_5$). The reaction shows Aniline reacting with sodium nitrite and hydrochloric acid to produce intermediate A, followed by treatment with potassium iodide to yield B, which then dimerizes via sodium in dry ether to form C.The reaction shows Aniline reacting with sodium nitrite and hydrochloric acid to produce intermediate A, followed by treatment with potassium iodide to yield B, which then dimerizes via sodium in dry ether to form C.
### Pattern Recognition
Shortcut: Aniline
ightarrow textNaNO_2/textHCl
ightarrow$
ightarrow \text{NaNO}_2/\text{HCl}
ightarrow$ Diazonium salt
ightarrow textKI
ightarrow$
ightarrow \text{KI}
ightarrow$ Iodobenzene. The final sodium metal treatment triggers a symmetrical radical dimer homocoupling (Fittig reaction) to yield a biphenyl product.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Amines
Class 12 Chemistry: Haloalkanes and Haloarenes
Q38jee_main_2025_03_april_morningReactions of Diazonium Salts
In the following reactions, which one is NOT correct?
A. Reaction Scheme (1)
B. Reaction Scheme (2)
C. Reaction Scheme (3)
D. Reaction Scheme (4)
Solution
### Core Logic
When benzene diazonium chloride is treated with ethanol (textCH_3textCH_2textOH$\text{CH}_3\text{CH}_2\text{OH}$), it undergoes a reduction reaction (deamination). Ethanol acts as a reducing agent and gets oxidized to ethanal (textCH_3textCHO$\text{CH}_3\text{CHO}$), while the diazonium group is replaced by hydrogen to yield pure **benzene**, not phenetole (ethoxybenzene). Deamination chemical verification scheme for Q38 - JEE Main 2025 Morning
### Step 1: Review of Alternative Choices
Reactions (2), (3), and (4) show standard correct transformations: hypophosphorous acid reduction to benzene, potassium iodide substitution to iodobenzene, and cuprous cyanide substitution to benzonitrile.
### Pattern Recognition
Shortcut: Remember that textH_3textPO_2$\text{H}_3\text{PO}_2$ and textCH_3textCH_2textOH$\text{CH}_3\text{CH}_2\text{OH}$ are standard classic reducing agents that reduce textArN_2^+textCl^-$\text{ArN}_2^+\text{Cl}^-$ directly down to textArH$\text{ArH}$ (benzene). They do not undergo nucleophilic ether substitution paths.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Amines
The major product (A) formed in the following reaction sequence is:
The flowchart traces the conversion of nitrobenzene using Sn/HCl, Ac2O/pyridine, Br2/AcOH, and aqueous NaOH.
A.textProduct 1$\text{Product 1}$
B.textProduct 2$\text{Product 2}$
C.textProduct 3$\text{Product 3}$
D.textProduct 4$\text{Product 4}$
Solution
### Core Logic
Let's track the chemical transformations sequentially:
1. **Step 1 (Sn + HCl$Sn + HCl$):** Nitrobenzene is cleanly reduced to yield Aniline (C_6H_5NH_2$C_6H_5NH_2$).
2. **Step 2 (Ac_2O + textPyridine$Ac_2O + \text{Pyridine}$):** Protecting step. Aniline undergoes acetylation to form Acetanilide (C_6H_5NHCOCH_3$C_6H_5NHCOCH_3$). This tempers the highly activating -NH_2$-NH_2$ group to prevent poly-bromination.
3. **Step 3 (Br_2 + AcOH$Br_2 + AcOH$):** The -NHCOCH_3$-NHCOCH_3$ amide group safely directs electrophilic bromination to the less-hindered **para** position, yielding p-bromoacetanilide.
4. **Step 4 (NaOH_(aq)$NaOH_{(aq)}$):** Basic hydrolysis removes the protecting acetyl group, restoring the free amine function to yield the final product: **p-bromoaniline**.
### Pattern Recognition
Acetylation of aniline followed by halogenation and subsequent hydrolysis is the standard synthetic pathway to produce mono-substituted para-haloanilines.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Amines
More Amines Questions — jee_main_2024_31_jan_evening
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 "Diazotization, Benzene diazonium chloride, Iodobenzene, Wurtz-Fittig biphenyl")
 "Diazotization, Benzene diazonium chloride, Iodobenzene, Wurtz-Fittig biphenyl")
 "Diazotization, Benzene diazonium chloride, Iodobenzene, Wurtz-Fittig biphenyl")
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