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:

### Related Formula textR-textNH_2 + textCHCl_3 + 3textKOH rightarrow textR-textNC + 3textKCl + 3textH_2textO$$\text{R}-\text{NH}_2 + \text{CHCl}_3 + 3\text{KOH} \rightarrow \text{R}-\text{NC} + 3\text{KCl} + 3\text{H}_2\text{O}$$ ### Core Logic Only primary (1^circ$1^\circ$) aliphatic and aromatic amines yield a positive carbylamine test (forming foul-smelling alkyl/aryl isocyanides). - **A** is Aniline (primary aromatic amine) rightarrow$\rightarrow$ Positive - **B** is Dimethylamine (secondary aliphatic amine) rightarrow$\rightarrow$ Negative - **C** is Methylamine (primary aliphatic amine) rightarrow$\rightarrow$ Positive - **D** is Trimethylamine (tertiary aliphatic amine) rightarrow$\rightarrow$ Negative - **E** is N-Methylaniline (secondary aromatic amine) rightarrow$\rightarrow$ Negative Thus, only A and C show a positive test. ### Pattern Recognition Shortcut: Look directly for any amine with a plain -textNH_2$-\text{NH}_2$ functional group. Secondary (-textNH-$-\text{NH}-$) and tertiary (-textN-$-\text{N}-$) amines never react. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Amines

### Core Logic Let's perform a step-by-step diagnostic analysis: 1. **Molar Mass & Nitrogen Test**: Compound 'A' has a nitrogen atom and a molar mass of 121 text g mol^-1$121 \text{ g mol}^{-1}$. Let's verify Benzamide (textC_6textH_5textCONH_2$\text{C}_6\text{H}_5\text{CONH}_2$): textMass = (7 times 12) + (7 times 1) + 14 + 16 = 84 + 7 + 14 + 16 = 121 text g mol^-1$$\text{Mass} = (7 \times 12) + (7 \times 1) + 14 + 16 = 84 + 7 + 14 + 16 = 121 \text{ g mol}^{-1}$$ This matches perfectly. 2. **Alkaline Hydrolysis**: Hydrolysis of benzamide under basic conditions yields benzoic acid upon acidification: textC_6textH_5textCONH_2 xrightarrow[H_3O^+]NaOH textC_6textH_5textCOOH (Compound B) + textNH_3$$\text{C}_6\text{H}_5\text{CONH}_2 \xrightarrow[H_3O^+]{NaOH} \text{C}_6\text{H}_5\text{COOH (Compound B)} + \text{NH}_3$$ Benzoic acid reactively gives effervescence with textNaHCO_3$\text{NaHCO}_3$ due to the liberation of textCO_2$\text{CO}_2$ gas. 3. **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: textC_6textH_5textCOOH + textEtOH xrightarrowH_2SO_4, Delta textC_6textH_5textCOOEt (Compound C) + textH_2textO$$\text{C}_6\text{H}_5\text{COOH} + \text{EtOH} \xrightarrow{H_2SO_4, \Delta} \text{C}_6\text{H}_5\text{COOEt (Compound C)} + \text{H}_2\text{O}$$

### Pattern Recognition "Fruity smell" is an absolute indicator for an ester product. "Effervescence with textNaHCO_3$\text{NaHCO}_3$" 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$M = 103$). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Amines Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids

### 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.

### 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

### 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$112.5\,\mathrm{g\,mol}^{-1}$.

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

### 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$\text{C}_6\text{H}_5-\text{C}_6\text{H}_5$).

### 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