Identify correct statements:
(A) Primary amines do not give diazonium salts when treated with NaNO_3$NaNO_{3}$ in acidic condition.
(B) Aliphatic and aromatic primary amines on heating with CHCl_3$CHCl_{3}$ and ethanolic KOH form carbylamines.
(C) Secondary and tertiary amines also give carbylamine test.
(D) Benzenesulfonyl chloride is known as Hinsberg's reagent.
(E) Tertiary amines reacts with benzenesulfonyl chloride very easily.
Choose the correct answer from the options given below :
A.(B) and (D) only
B.(A) and (B) only
C.(D) and (E) only
D.(B) and (C) only
Solution & Explanation
### Related Formula
The Carbylamine reaction is specific to primary amines:
R-NH_2 + CHCl_3 + 3KOH xrightarrowDelta R-NC + 3KCl + 3H_2O$R-NH_2 + CHCl_3 + 3KOH \xrightarrow{\Delta} R-NC + 3KCl + 3H_2O$
### Core Logic
Evaluating each amine statement:
- (A) Primary aromatic amines form stable diazonium salts with NaNO_2/HCl$NaNO_2/HCl$ at low temperatures, making this statement false.
- (B) Both aliphatic and aromatic primary amines undergo the carbylamine test to produce foul-smelling isocyanides. This is **correct**.
- (C) Secondary and tertiary amines do not undergo the carbylamine reaction, making this statement false.
- (D) Benzenesulfonyl chloride (C_6H_5SO_2Cl$C_6H_5SO_2Cl$) is the definition of Hinsberg's reagent. This is **correct**.
- (E) Tertiary amines do not possess an acidic hydrogen on nitrogen and do not react with Hinsberg's reagent under standard analytical testing conditions, making this statement false.
### Step 1: Selecting Correct Entries
Statements (B) and (D) are verified as true.
Reaction equations summary for primary amine classification
### Pattern Recognition
Hinsberg's reagent and the carbylamine test are key analytical methods used to differentiate primary, secondary, and tertiary amines. The carbylamine test is strictly positive *only* for primary (1^circ$1^\circ$) amine groups.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Amines
Keywords:#Carbylamine test primary amines#JEE Main 2025 Evening Q43#Hinsberg reagent chemical definition#Amine analytical identification steps
More Amines Previous-Year Questions — Page 2
Q412025Diazotization 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
Q492025Yield 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
Q372025Diazonium 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
Q382025Reactions 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
Q352025Aniline Reactions
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_2025_28_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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