Consider the following sequence of reactions :

11.25 mg of chlorobenzene will produce mathrmx times 10^-1$\mathrm{x} \times 10^{-1}$ 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 35.5mathrmg\,mol^-1$35.5\mathrm{g\,mol}^{-1}$ respectively]

### Core Logic The reaction of an amide with Br_2$Br_2$ and KOH is Hoffmann bromamide degradation. It steps down the carbon chain by one carbonyl carbon to form a primary amine. Let's analyze the options and molecular formula. The question says (R) has molecular formula C_9H_7N$C_9H_7N$. Wait, looking at the standard solutions for this type of problem, aniline is C_6H_7N$C_6H_7N$. The PDF says C_9H_7N$C_9H_7N$ which is likely a typo in the original paper for C_6H_7N$C_6H_7N$, since option (1) gives Aniline (C_6H_7N$C_6H_7N$). Let's assume the standard sequence: 1. mathrmPh-COOH xrightarrowmathrmNH_3, Delta mathrmPh-CO-NH_2$\mathrm{Ph-COOH} \xrightarrow{\mathrm{NH}_3, \Delta} \mathrm{Ph-CO-NH}_2$ (Benzoic acid to Benzamide) 2. mathrmPh-CO-NH_2 xrightarrowmathrmBr_2/mathrmKOH mathrmPh-NH_2$\mathrm{Ph-CO-NH}_2 \xrightarrow{\mathrm{Br}_2/\mathrm{KOH}} \mathrm{Ph-NH}_2$ (Benzamide to Aniline) Aniline is C_6H_5NH_2 = C_6H_7N$C_6H_5NH_2 = C_6H_7N$. So P is Benzoic acid, Q is Benzamide, R is Aniline. ### Pattern Recognition Reaction sequence: Carboxylic\ Acid xrightarrowNH_3, Delta Amide xrightarrowBr_2/KOH Amine$Carboxylic\ Acid \xrightarrow{NH_3, \Delta} Amide \xrightarrow{Br_2/KOH} Amine$. The Br_2/KOH$Br_2/KOH$ step is Hoffmann bromamide reaction. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Amines Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids

### Core Logic Step 1: Nitration of benzene gives nitrobenzene (P). mathrmPh-H xrightarrowHNO_3/H_2SO_4 mathrmPh-NO_2 quad text(P)$$\mathrm{Ph-H} \xrightarrow{HNO_3/H_2SO_4} \mathrm{Ph-NO_2} \quad \text{(P)}$$

Step 2: Reduction of nitrobenzene with Sn/HCl gives aniline (Q). mathrmPh-NO_2 xrightarrowSn/HCl mathrmPh-NH_2 quad text(Q)$$\mathrm{Ph-NO_2} \xrightarrow{Sn/HCl} \mathrm{Ph-NH_2} \quad \text{(Q)}$$

Step 3: Acetylation of aniline with acetic anhydride gives acetanilide (R). This protects the amino group to prevent oxidation and polysubstitution in the next step. mathrmPh-NH_2 xrightarrow(CH_3CO)_2O mathrmPh-NH-CO-CH_3 quad text(R)$$\mathrm{Ph-NH_2} \xrightarrow{(CH_3CO)_2O} \mathrm{Ph-NH-CO-CH_3} \quad \text{(R)}$$

Step 4: Nitration of acetanilide gives predominantly p-nitroacetanilide (S) due to steric hindrance at ortho position. mathrmPh-NH-CO-CH_3 xrightarrowHNO_3/H_2SO_4 ptext-NO_2text-C_6textH_4text-NH-CO-CH_3 quad text(S)$$\mathrm{Ph-NH-CO-CH_3} \xrightarrow{HNO_3/H_2SO_4} p\text{-NO}_2\text{-C}_6\text{H}_4\text{-NH-CO-CH}_3 \quad \text{(S)}$$ Step 5: Acidic hydrolysis of the amide linkage yields p-nitroaniline (T). ptext-NO_2text-C_6textH_4text-NH-CO-CH_3 xrightarrowHCl/EtOH/Delta ptext-NO_2text-C_6textH_4text-NH_2 quad text(T)$$p\text{-NO}_2\text{-C}_6\text{H}_4\text{-NH-CO-CH}_3 \xrightarrow{HCl/EtOH/\Delta} p\text{-NO}_2\text{-C}_6\text{H}_4\text{-NH}_2 \quad \text{(T)}$$

Molecular formula of p-nitroaniline (T) is C_6H_6N_2O_2$C_6H_6N_2O_2$. Molar mass = (6 times 12) + (6 times 1) + (2 times 14) + (2 times 16) = 72 + 6 + 28 + 32 = 138text g/mol$(6 \times 12) + (6 \times 1) + (2 \times 14) + (2 \times 16) = 72 + 6 + 28 + 32 = 138\text{ g/mol}$. Total mass of Nitrogen = 2 times 14 = 28text g$2 \times 14 = 28\text{ g}$. Percentage of Nitrogen = frac28138 times 100 approx 20.29\%$\frac{28}{138} \times 100 \approx 20.29\%$. ### Step 1: Final Conclusion Nearest integer is 20. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Amines

### 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