A student has planned to prepare acetanilide from aniline using acetic anhydride. The student has started from 9.3g of aniline. However, the student has managed to obtain 11 g of dry acetanilide. The % yield of this reaction is :-

### 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

### Core Logic Tracing the steps through nitration/bromination, reduction to amine via textSn/HCl$\text{Sn/HCl}$, subsequent extensive bromination with textBr_2/textH_2textO$\text{Br}_2/\text{H}_2\text{O}$, diazotization, and Sandmeyer bromination (textCuBr/NaBr$\text{CuBr/NaBr}$), we get substitution at multiple positions leading to 5 bromine atoms in the final product structure. ### Step 1: Final Calculation Number of Br atoms in major product (P) = 5. ### Pattern Recognition Sees: Multi-step aromatic conversion involving halogenation and diazotization. Trap: Counting substituent groups incorrectly after Sandmeyer reaction. ### Chapter Mix Class 12 Chemistry: Amines

### Core Logic Let's trace the sequence: 1. A (C_8H_9ON$C_8H_9ON$) is neutral and reacts with Br_2/OH^-$Br_2/OH^-$ (Hofmann Bromamide Degradation). This means A is a primary amide. 2. Product B is soluble in dilute acid, meaning it is a primary amine (Ar-NH_2$Ar-NH_2$ or alkyl amine). 3. B reacts with NaNO_2/HCl$NaNO_2/HCl$ at 0-5^circC$0-5^{\circ}C$ to form C. Since C undergoes Sandmeyer with CuCN to form D, B must be an aromatic primary amine, and C is a diazonium salt. 4. D is an aryl cyanide (Ar-CN$Ar-CN$). Hydrolysis of D yields E (Ar-COOH$Ar-COOH$). 5. Crucially, E is also obtainable from the direct hydrolysis of A. This confirms A is an aryl amide of the form Ar-CONH_2$Ar-CONH_2$.

6. Let's analyze the formula C_8H_9ON$C_8H_9ON$. The amide group is -CONH_2$-CONH_2$. Removing -CONH_2$-CONH_2$ leaves C_7H_7$C_7H_7$. A benzene ring with one methyl group is a tolyl group. So A is a methylbenzamide (CH_3-C_6H_4-CONH_2$CH_3-C_6H_4-CONH_2$). 7. E is methylbenzoic acid (CH_3-C_6H_4-COOH$CH_3-C_6H_4-COOH$). 8. E is oxidized by acidified KMnO_4$KMnO_4$ to F. The methyl group on the benzene ring oxidizes to -COOH$-COOH$. Thus, F is a benzenedicarboxylic acid (HOOC-C_6H_4-COOH$HOOC-C_6H_4-COOH$).

9. The problem states that F contains two different types of hydrogen atoms. Let's check the isomers of benzenedicarboxylic acid: - Phthalic acid (ortho): Contains 2 types of aromatic hydrogens + 1 type of COOH hydrogen = 3 types. - Isophthalic acid (meta): Contains 3 types of aromatic hydrogens + 1 type of COOH hydrogen = 4 types. - Terephthalic acid (para): Due to symmetry, all 4 aromatic hydrogens are equivalent. So it contains 1 type of aromatic hydrogen + 1 type of COOH hydrogen = 2 types.

Since F has only two types of hydrogens, F must be terephthalic acid (para isomer). Thus, E is p-methylbenzoic acid, and A is p-methylbenzamide. ### Step 1: Final Identification Compound A is p-methylbenzamide. This corresponds to the structure in option (3). ### Pattern Recognition A classic sequence linking Hofmann bromamide, Sandmeyer, and side-chain oxidation. Symmetrical molecules like para-isomers minimize the number of unique proton environments (critical for NMR or simple counting). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Amines Class 12 Chemistry: Aldehydes Ketones and Carboxylic Acids

### Related Formula textAniline + textBenzoyl Chloride xrightarrowtextSchotten-Baumann textBenzanilide [A]$$\text{Aniline} + \text{Benzoyl Chloride} \xrightarrow{\text{Schotten-Baumann}} \text{Benzanilide } [A]$$ textAmide [A] xrightarrowtextLiAlH_4 textSecondary Amine [B]$$\text{Amide } [A] \xrightarrow{\text{LiAlH}_4} \text{Secondary Amine } [B]$$ ### Core Logic Step 1: Reaction of aniline with benzoyl chloride (textPhCOCl$\text{PhCOCl}$) in basic medium yields benzanilide (textPh-NH-CO-Ph$\text{Ph-NH-CO-Ph}$) as intermediate [A]$[A]$. Step 2: Reduction of benzanilide using textLiAlH_4$\text{LiAlH}_4$ converts the carbonyl group -textC(=textO)-$-\text{C}(=\text{O})-$ into a methylene group -textCH_2-$-\text{CH}_2-$, forming dibenzylamine (textPh-NH-CH_2text-Ph$\text{Ph-NH-CH}_2\text{-Ph}$) as final product [B]$[B]$.

### Pattern Recognition Sees: Acylation followed by textLiAlH_4$\text{LiAlH}_4$ reduction. Shortcut: Amide carbonyl group reduces directly to -textCH_2-$-\text{CH}_2-$, resulting in secondary amine structure (option 3). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Amines