The product A formed in the following reaction is:
Diazotization and Sandmeyer Reaction diagram for Q71 - JEE Main 2024 Evening
The diagram details aniline undergoing diazotization with sodium nitrite and hydrochloric acid, followed by treatment with cuprous chloride.

Solution & Explanation

Related Formula
Ar-NH₂ NaNO₂ + HCl, 0^ Ar-N₂^+Cl^- Cu₂Cl₂ Ar-Cl
Core Logic

The schematic outlines a classic sequence:

  • Diazotization Step: Aniline reacts with nitrous acid generated in situ (NaNO₂ + HCl) at low temperature (0-5^) to form benzene diazonium chloride intermediate.
  • Sandmeyer Replacement Step: Treating the diazonium salt with cuprous chloride (Cu₂Cl₂) results in replacement of the diazo group by a chlorine atom.
Step 1: Visual Verification

Tracing structural shifts through intermediates validates the final outcome:

Diazotization and Sandmeyer Reaction solution diagram for Q71 - JEE Main 2024 Evening
The diagram details aniline undergoing diazotization with sodium nitrite and hydrochloric acid, followed by treatment with cuprous chloride.

Pattern Recognition

Aniline arrow Diazonium salt arrow Chlorobenzene via standard Sandmeyer synthesis protocols.

Chapter Mix

Class 12 Chemistry: Organic Compounds Containing Nitrogen

More Organic Compounds Containing Nitrogen Previous-Year Questions — Page 5

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. Let's verify Benzamide (C₆H₅CONH₂):
Mass = (7 × 12) + (7 × 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 due to the liberation of CO₂ 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
Esterification reaction mechanism diagram for Q28

Pattern Recognition

"Fruity smell" is an absolute indicator for an ester product. "Effervescence with NaHCO₃" 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).

Chapter Mix

Class 12 Chemistry: Amines Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids

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) is reduced using Sn/HCl to form Aniline (Ph-NH₂).
  • Aniline undergoing diazotization with NaNO₂/HCl at cold temperatures (0-5circC) creates Benzene diazonium chloride (Ph-N₂⁺Cl⁻).
  • The diazonium salt undergoes a coupling reaction with β-naphthol in alkaline conditions (NaOH) to synthesize a highly vibrant red-orange azo dye.
  • Diazotization and Coupling Reactions diagram for Q41 - JEE Main 2025 Evening
    Diazotization and Coupling Reactions diagram for Q41 - JEE Main 2025 Evening

Pattern Recognition

The standard sequence for azo dye preparation is: Aromatic Nitro arrow Primary Amine arrow Diazonium Salt arrow 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 × 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). Following stoichiometric preservation:

moles of chlorobenzene = moles of Aniline (B)

Molar mass of chlorobenzene (C₆H₅Cl) = 112.5 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 × 10⁻³ g112.5 g mol⁻¹ = 10⁻⁴ mol

Mass of product B produced:

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

Expressing in the specified format:

9.3 mg = 93 × 10⁻¹ mg ⇒ x = 93
Pattern Recognition

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

Chapter Mix

Class 12 Chemistry: Amines

Q37 jee_main_2025_03_april_morning Diazonium Salts and Reactions
Identify [A], [B], and [C], respectively in the following reaction sequence:
Reaction scheme diagram for Q37 - JEE Main 2025 Morning
Reaction sequence starting from aniline reacting with NaNO2/HCl to give [A], which reacts with KI to give [B], and with 2Na/Dry ether to give [C].
  • A. Option (1)
  • B. Option (2)
  • C. Option (3)
  • D. Option (4)

Solution

Related Formula
Ar-NH₂ NaNO₂/HCl, 273-278K Ar-N₂^+Cl^- KI Ar-I 2 Ar-I 2Na, dry ether Ar-Ar (Fittig reaction)
Core Logic
  • Aniline with NaNO₂/HCl at 273--278~K undergoes diazotization to yield benzenediazonium chloride [A] = C₆H₅N₂^+Cl^-.
  • Reaction of [A] with KI replaces the diazonium group with iodine to form iodobenzene [B] = C₆H₅I.
  • Treating iodobenzene [B] with 2Na/dry ether yields biphenyl [C] = C₆H₅-C₆H₅ via the Fittig reaction.
Step 1: Option Matching

Option (3) correctly identifies [A] = C₆H₅N₂^+Cl^-, [B] = C₆H₅I, and [C] = Biphenyl.

Pattern Recognition

Aniline arrow Diazotization arrow Iodination arrow Fittig coupling arrow Biphenyl.

Chapter Mix

Class 12 Chemistry: Amines

Q38 jee_main_2025_03_april_morning Reactions of Diazonium Salts
In the following reactions, which one is NOT correct?
  • A. Reaction (1)
  • B. Reaction (2)
  • C. Reaction (3)
  • D. Reaction (4)

Solution

Related Formula
Ar-N₂^+Cl^- + CH₃CH₂OH arrow Ar-H + N₂ + HCl + CH₃CHO
Core Logic

Reaction of benzenediazonium chloride with ethanol (EtOH) is a reduction/deamination reaction that converts the diazonium salt to benzene, while ethanol gets oxidized to ethanal (CH₃CHO).

It does NOT yield phenetole (ethoxybenzene, Ar-OEt). Therefore, Reaction (1) is incorrect.

Step 1: Verification of Other Reactions
  • Reaction with H₃PO₂/H₂O gives benzene. (Correct)
  • Reaction with KI gives iodobenzene. (Correct)
  • Reaction with CuCN/KCN gives benzonitrile. (Correct)
Pattern Recognition

Diazonium salt + Ethanol arrow Benzene (reduction), NOT Ethoxybenzene.

Chapter Mix

Class 12 Chemistry: Amines

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