Which of the following amine(s) show(s) positive carbylamine test? A. B. (CH3)2NH C. CH3NH2 D. (CH3)3N E. Choose the…

Q: Which of the following amine(s) show(s) positive carbylamine test? A. B. (CH3)2NH C. CH3NH2 D. (CH3)3N E. Choose the correct answer from the options given below:

### Related Formula ### Core Logic Only primary (1^) aliphatic and aromatic amines yield a positive carbylamine test (forming foul-smelling alkyl/aryl isocyanides). - **A** is Aniline (primary aromatic amine) arrow Positive - **B** is Dimethylamine (secondary aliphatic amine) arrow Negative - **C** is Methylamine (primary aliphatic amine) arrow Positive - **D** is Trimethylamine (tertiary aliphatic amine) arrow Negative - **E** is N-Methylaniline (secondary aromatic amine) arrow Negative Thus, o

Year	2026	2025	2024	Total
Questions	16	14	9	39

Q jee_main_2025_08_april_evening Functional Group Analysis and Identification

An organic compound 'A' undergoes the following sequence of transformations:

'A' [(ii) H₃O^+](i) NaOH 'B' [(ii) H₂SO₄, Δ](i) EtOH 'C'

* 'A' shows a positive Lassaigne's test for nitrogen and its molar mass is

121 g mol⁻¹

. * 'B' gives effervescence with aqueous

NaHCO₃

. * 'C' gives a characteristic fruity smell. Identify A, B, and C from the options below:

A.
B.
C.
D.

Solution

Core Logic

Let's perform a step-by-step diagnostic analysis:

Molar Mass & Nitrogen Test: Compound 'A' has a nitrogen atom and a molar mass of $121 g mol⁻¹$ $121 \text{ g mol}^{-1}$. Let's verify Benzamide ( $C₆H₅CONH₂$ $\text{C}_6\text{H}_5\text{CONH}_2$):

Mass = (7 \times 12) + (7 \times 1) + 14 + 16 = 84 + 7 + 14 + 16 = 121 g mol⁻¹

This matches perfectly.

Alkaline Hydrolysis: Hydrolysis of benzamide under basic conditions yields benzoic acid upon acidification:

C₆H₅CONH₂ [H₃O^+]NaOH C₆H₅COOH (Compound B) + NH₃

Benzoic acid reactively gives effervescence with $NaHCO₃$ $\text{NaHCO}_3$ due to the liberation of $CO₂$ $\text{CO}_2$ gas.

Esterification: Reaction of benzoic acid with ethanol in the presence of acid catalyst results in the creation of ethyl benzoate, an ester with a pleasant fruity smell:

C₆H₅COOH + EtOH H₂SO₄, Δ C₆H₅COOEt (Compound C) + H₂O

Esterification reaction mechanism diagram for Q28

Pattern Recognition

"Fruity smell" is an absolute indicator for an ester product. "Effervescence with $NaHCO₃$ $\text{NaHCO}_3$" dictates a carboxylic acid intermediate. Basic hydrolysis converting an organo-nitrogen compound into an acid points directly to an amide or a nitrile—molar mass calculation establishes benzamide over benzonitrile ( $M = 103$ $M = 103$).

Chapter Mix

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Q41 jee_main_2025_29_jan_evening Diazotization and Coupling Reactions

Which one of the following reaction sequences will give an azo dye? (1) Nitrobenzene treated with (i)

Sn/HCl

, (ii)

NaNO₂/HCl

, (iii)

β

-naphthol,

NaOH

(2) Benzenesulfonic acid treated with (i)

SOCl₂

, (ii)

NH₃

, (iii) Benzyl chloride (3) Benzonitrile treated with (i)

70% H₂SO₄

, (ii)

PCl₅

, (iii) Aniline (4) Aniline treated with (i)

HCl/NaNO₂

, (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):

Nitrobenzene ( $Ph-NO₂$ $Ph-NO_2$) is reduced using $Sn/HCl$ $Sn/HCl$ to form Aniline ( $Ph-NH₂$ $Ph-NH_2$).
Aniline undergoing diazotization with $NaNO₂/HCl$ $NaNO_{2}/HCl$ at cold temperatures ( $0-5circC$ $0-5^{circ}C$) creates Benzene diazonium chloride ( $Ph-N₂⁺Cl⁻$ $Ph-N_{2}^{+}Cl^{-}$).
The diazonium salt undergoes a coupling reaction with $β$ $\beta$-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 $arrow$ $\rightarrow$ Primary Amine $arrow$ $\rightarrow$ Diazonium Salt $arrow$ $\rightarrow$ Phenol/Naphthol Coupling.

Chapter Mix

Class 12 Chemistry: Amines

Q49 jee_main_2025_28_jan_morning Yield and Stoichiometric Calculations

Consider the following sequence of reactions :

Reaction flow pathway for Q49 - JEE Main 2025 Morning — The flowchart tracks a chemical conversion starting from chlorobenzene down to final compound B.

11.25 mg of chlorobenzene will produce

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

35.5g 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 $93 g mol⁻¹$ $93\,\mathrm{g\,mol}^{-1}$). Following stoichiometric preservation:

moles of chlorobenzene = moles of Aniline (B)

Molar mass of chlorobenzene ( $C₆H₅Cl$ $\mathrm{C}_6\mathrm{H}_5\mathrm{Cl}$) = $112.5 g mol⁻¹$ $112.5\,\mathrm{g\,mol}^{-1}$.

Molar stoichiometry relation graph for Q49 - JEE Main 2025 Morning — The flowchart tracks a chemical conversion starting from chlorobenzene down to final compound B.

moles = 11.25 \times 10⁻³ g112.5 g mol⁻¹ = 10⁻⁴ mol

Mass of product B produced:

Mass = 10⁻⁴ mol \times 93 g mol⁻¹ = 9.3 \times 10⁻³ g = 9.3 mg

Expressing in the specified format:

9.3 mg = 93 \times 10⁻¹ mg \Rightarrow x = 93

Pattern Recognition

Sees: Conversion sequence preserving a 1:1 mole ratio layout. Shortcut: Directly compute target weight via $WB = WA \cdot (MB)/(MA) = 11.25 \cdot (93)/(112.5) = 9.3$ .

Chapter Mix

Class 12 Chemistry: Amines

Q jee_main_2025_03_april_morning Diazonium Salts and Reactions

Identify [A], [B], and [C], respectively in the following reaction sequence:

Organic aromatic reaction sequence diagram for Q37 - JEE Main 2025 Morning — 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.
B.
C.
D.

Solution

Core Logic

Let us resolve each structural step sequentially:

Step 1: Aniline undergoes diazotization when treated with $NaNO₂ + HCl$ $\text{NaNO}_2 + \text{HCl}$ at $273-278 K$ $273-278\text{ K}$, forming benzene diazonium chloride $[A]$ $[A]$ ( $C₆H₅N₂^+Cl^-$ ).
Step 2: Warming benzene diazonium chloride with potassium iodide ( $KI$ $\text{KI}$) substitutes the diazonium group with iodine, producing iodobenzene $[B]$ $[B]$ ( $C₆H₅I$ $\text{C}_6\text{H}_5\text{I}$).
Step 3: Treating iodobenzene with sodium metal in dry ether causes a Fittig coupling reaction, dimerizing two phenyl radicals into biphenyl $[C]$ $[C]$ ( $C₆H₅-C₆H₅$ ).
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 $arrow NaNO₂/HCl arrow$ Diazonium salt $arrow KI arrow$ $\rightarrow \text{KI} \rightarrow$ Iodobenzene. The final sodium metal treatment triggers a symmetrical radical dimer homocoupling (Fittig reaction) to yield a biphenyl product.

Evaluation Rubric / Model Answer

Option (C)

Chapter Mix

Class 12 Chemistry: Amines Class 12 Chemistry: Haloalkanes and Haloarenes

Q jee_main_2025_03_april_morning Reactions of Diazonium Salts

In the following reactions, which one is NOT correct?

A.
B.
C.
D.

Solution

Core Logic

When benzene diazonium chloride is treated with ethanol ( $CH₃CH₂OH$ $\text{CH}_3\text{CH}_2\text{OH}$), it undergoes a reduction reaction (deamination). Ethanol acts as a reducing agent and gets oxidized to ethanal ( $CH₃CHO$ $\text{CH}_3\text{CHO}$), 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 $H₃PO₂$ $\text{H}_3\text{PO}_2$ and $CH₃CH₂OH$ $\text{CH}_3\text{CH}_2\text{OH}$ are standard classic reducing agents that reduce $ArN₂^+Cl^-$ $\text{ArN}_2^+\text{Cl}^-$ directly down to $ArH$ $\text{ArH}$ (benzene). They do not undergo nucleophilic ether substitution paths.

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