### Related Formula textMass (g) = textNumber of moles times textMolar mass (mathrmg~mol^-1)$\text{Mass (g)} = \text{Number of moles} \times \text{Molar mass } (\mathrm{g~mol^{-1}})$ ### Core Logic Sulphanilic acid is diazotized under cold conditions (273~mathrmK$273~\mathrm{K}$) with nitrous acid to form a diazonium salt intermediate. This diazonium salt undergoes a coupling reaction with 1-naphthylamine to form a red azo dye. First, sulphanilic acid acts as a zwitterion and undergoes diazotization:

Nitrous acid reacts with the amine to form the diazonium compound:

This intermediate couples with 1-naphthylamine at the para-position to give the red azo dye compound:

### Step 1: Calculate the Molar Mass The molecular formula of the red-azo dye formed is mathrmC_16H_13N_3O_3S$\mathrm{C_{16}H_{13}N_{3}O_{3}S}$. Let's calculate its molar mass using the given atomic masses: textMolar mass = 16(12) + 13(1) + 3(14) + 3(16) + 32$\text{Molar mass} = 16(12) + 13(1) + 3(14) + 3(16) + 32$ textMolar mass = 192 + 13 + 42 + 48 + 32 = 327~mathrmg~mol^-1$\text{Molar mass} = 192 + 13 + 42 + 48 + 32 = 327~\mathrm{g~mol^{-1}}$ ### Step 2: Calculate the Mass of 0.1 Mole Using the relation for mass: textMass of 0.1~textmole = 0.1 times 327 = 32.7~mathrmg approx 33~mathrmg$\text{Mass of } 0.1~\text{mole} = 0.1 \times 327 = 32.7~\mathrm{g} \approx 33~\mathrm{g}$ Hence, the nearest option is 33~mathrmg$33~\mathrm{g}$. ### Pattern Recognition Azo coupling reactions are clean electrophilic aromatic substitution reactions. Diazotized sulphanilic acid has a highly electron-withdrawing sulphonic acid group, making it an excellent electrophile that couples selectively at the para-position of 1-naphthylamine. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Amines

### Related Formula Basic strength in aqueous medium depends on three combined effects: textBasic Strength propto textInductive Effect (+I) + textSolvation Energy - textSteric Hindrance$\text{Basic Strength} \propto \text{Inductive Effect (+I)} + \text{Solvation Energy} - \text{Steric Hindrance}$ ### Core Logic Let's list structural elements row-by-row: * Ethyl substituted amine trends in aqueous systems uniquely align into a 2° > 3° > 1° configuration due to competing steric and hydration energies: mathrm(Et)_2NH > (Et)_3N > EtNH_2$\mathrm{(Et)_2NH > (Et)_3N > EtNH_2}$ * Ammonia (mathrmNH_3$\mathrm{NH}_3$) is less basic than aliphatic substituted structures due to the absence of electron-donating alkyl clusters. * Hydrazine (mathrmH_2mathrmN-NH_2$\mathrm{H}_2\mathrm{N-NH_2}$) is exceptionally weak compared to ammonia because the adjacent electronegative nitrogen creates an electron-withdrawing (-I) effect, while lone-pair repulsions reduce overall stability. ### Step 1: Ordering Assembling the fragments gives the complete verified thermodynamic order: mathrmNH_2-NH_2 < NH_3 < CH_3CH_2NH_2 < (CH_3CH_2)_3N < (CH_3CH_2)_2NH$\mathrm{NH_2-NH_2 < NH_3 < CH_3CH_2NH_2 < (CH_3CH_2)_3N < (CH_3CH_2)_2NH}$ ### Pattern Recognition Remember the standard numeric rules for aliphatic basic strength order in aqueous media: - Methyl amines follow: **213** - Ethyl amines follow: **231** This simple sequence trick handles complex ranking items instantly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Amines

### Related Formula Directing effects in multi-substituted benzenes: - -NH_2$-NH_2$ is a strong activating group and ortho/para director. - -NO_2$-NO_2$ is a strong deactivating, meta-directing group. - Diazotization followed by Sandmeyer reaction replaces an -NH_2$-NH_2$ group with a halogen. ### Core Logic To prepare 3,4,5-tribromoaniline, we must introduce three bromine atoms adjacent to each other (meta to the final amino group, with one para and two meta). Let's trace the sequence in Option (3) starting from 4-nitroaniline (p-nitroaniline): ### Step 1: Bromination of p-nitroaniline Treatment of 4-nitroaniline with excess mathrmBr_2$\mathrm{Br}_2$ in acetic acid: - The amino group (-NH_2$-NH_2$) is a strong activator and directs to its ortho positions (positions 2 and 6). - Positions 2 and 6 are meta to the -NO_2$-NO_2$ group, which is compatible. - This yields 2,6-dibromo-4-nitroaniline. ### Step 2: Diazotization and replacement of amino group 1. mathrmNaNO_2 + mathrmHCl$\mathrm{NaNO}_2 + \mathrm{HCl}$ diazotizes the amino group to a diazonium salt: mathrmR-NH_2 rightarrow mathrmR-N_2^+ Cl^-$\mathrm{R-NH_2} \rightarrow \mathrm{R-N_2^+ Cl^-}$ 2. Addition of mathrmCuBr$\mathrm{CuBr}$ (Sandmeyer reaction) replaces the diazonium group with bromine: mathrmR-N_2^+ Cl^- xrightarrowmathrmCuBr mathrmR-Br$\mathrm{R-N_2^+ Cl^-} \xrightarrow{\mathrm{CuBr}} \mathrm{R-Br}$ This yields 3,4,5-tribromonitrobenzene. ### Step 3: Reduction of nitro group Reduction of the nitro group using mathrmSn/HCl$\mathrm{Sn/HCl}$ converts -NO_2$-NO_2$ to -NH_2$-NH_2$: mathrmR-NO_2 xrightarrowmathrmSn, HCl mathrmR-NH_2$\mathrm{R-NO_2} \xrightarrow{\mathrm{Sn, HCl}} \mathrm{R-NH_2}$ This yields 3,4,5-tribromoaniline as the predominant product. Thus, Option (3) is correct. ### Pattern Recognition To brominate meta positions relative to an amino group, use a nitro precursor at the para position. The -NH_2$-NH_2$ group activates these positions first, after which the initial amino group is replaced with a halogen, and the nitro group is subsequently reduced back to a primary amine. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Amines

### 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