In the following reactions, which one is NOT correct?

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$\text{C}_6\text{H}_5\text{CONH}_2$):

This matches perfectly.

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

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:

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

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

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.

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

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:

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

Mass of product B produced:

Expressing in the specified format:

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$W_B = W_A \cdot \frac{M_B}{M_A} = 11.25 \cdot \frac{93}{112.5} = 9.3$.

Chapter Mix

Class 12 Chemistry: Amines

Related Formula

Core Logic

• Aniline with NaNO₂/HCl$\mathrm{NaNO_2/HCl}$ at 273--278~K$273\text{--}278\mathrm{~K}$ undergoes diazotization to yield benzenediazonium chloride [A] = C₆H₅N₂^+Cl^-$[\mathrm{A}] = \mathrm{C_6H_5N_2^+Cl^-}$.
• Reaction of [A]$[\mathrm{A}]$ with KI$\mathrm{KI}$ replaces the diazonium group with iodine to form iodobenzene [B] = C₆H₅I$[\mathrm{B}] = \mathrm{C_6H_5I}$.
• Treating iodobenzene [B]$[\mathrm{B}]$ with 2Na/dry ether$\mathrm{2Na/dry\ ether}$ yields biphenyl [C] = C₆H₅-C₆H₅$[\mathrm{C}] = \mathrm{C_6H_5-C_6H_5}$ via the Fittig reaction.

Step 1: Option Matching

Option (3) correctly identifies [A] = C₆H₅N₂^+Cl^-$[\mathrm{A}] = \mathrm{C_6H_5N_2^+Cl^-}$, [B] = C₆H₅I$[\mathrm{B}] = \mathrm{C_6H_5I}$, and [C] = Biphenyl$[\mathrm{C}] = \text{Biphenyl}$.

Pattern Recognition

Aniline arrow$\rightarrow$ Diazotization arrow$\rightarrow$ Iodination arrow$\rightarrow$ Fittig coupling arrow$\rightarrow$ Biphenyl.

Chapter Mix

Class 12 Chemistry: Amines