Identify correct statements: (A) Primary amines do not give diazonium salts when treated with NaNO₃$NaNO_{3}$ in acidic condition. (B) Aliphatic and aromatic primary amines on heating with CHCl₃$CHCl_{3}$ and ethanolic KOH form carbylamines. (C) Secondary and tertiary amines also give carbylamine test. (D) Benzenesulfonyl chloride is known as Hinsberg's reagent. (E) Tertiary amines reacts with benzenesulfonyl chloride very easily. Choose the correct answer from the options given below :

Core Logic

The reaction of an amide with Br₂$Br_2$ 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 C₉H₇N$C_9H_7N$. Wait, looking at the standard solutions for this type of problem, aniline is C₆H₇N$C_6H_7N$. The PDF says C₉H₇N$C_9H_7N$ which is likely a typo in the original paper for C₆H₇N$C_6H_7N$, since option (1) gives Aniline (C₆H₇N$C_6H_7N$). Let's assume the standard sequence:

• Ph-COOH NH₃, Δ Ph-CO-NH₂$\mathrm{Ph-COOH} \xrightarrow{\mathrm{NH}_3, \Delta} \mathrm{Ph-CO-NH}_2$ (Benzoic acid to Benzamide)
• Ph-CO-NH₂ Br₂/KOH Ph-NH₂$\mathrm{Ph-CO-NH}_2 \xrightarrow{\mathrm{Br}_2/\mathrm{KOH}} \mathrm{Ph-NH}_2$ (Benzamide to Aniline)

Aniline is C₆H₅NH₂ = C₆H₇N$C_6H_5NH_2 = C_6H_7N$. So P is Benzoic acid, Q is Benzamide, R is Aniline.

Pattern Recognition

Reaction sequence: Carboxylic Acid NH₃, Δ Amide Br₂/KOH Amine$Carboxylic\ Acid \xrightarrow{NH_3, \Delta} Amide \xrightarrow{Br_2/KOH} Amine$. The Br₂/KOH$Br_2/KOH$ step is Hoffmann bromamide reaction.

Chapter Mix

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

Core Logic

Step 1: Nitration of benzene gives nitrobenzene (P).

Step 2: Reduction of nitrobenzene with Sn/HCl gives aniline (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.

Step 4: Nitration of acetanilide gives predominantly p-nitroacetanilide (S) due to steric hindrance at ortho position.

Step 5: Acidic hydrolysis of the amide linkage yields p-nitroaniline (T).

Molecular formula of p-nitroaniline (T) is C₆H₆N₂O₂$C_6H_6N_2O_2$. Molar mass = (6 × 12) + (6 × 1) + (2 × 14) + (2 × 16) = 72 + 6 + 28 + 32 = 138 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 × 14 = 28 g$2 \times 14 = 28\text{ g}$.

Percentage of Nitrogen = (28)/(138) × 100 ≈ 20.29%$\frac{28}{138} \times 100 \approx 20.29\%$.

Step 1: Final Conclusion

Nearest integer is 20.

Chapter Mix

Class 12 Chemistry: Amines

Core Logic

Tracing the steps through nitration/bromination, reduction to amine via Sn/HCl$\text{Sn/HCl}$, subsequent extensive bromination with Br₂/H₂O$\text{Br}_2/\text{H}_2\text{O}$, diazotization, and Sandmeyer bromination (CuBr/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:

• A (C₈H₉ON$C_8H_9ON$) is neutral and reacts with Br₂/OH^-$Br_2/OH^-$ (Hofmann Bromamide Degradation). This means A is a primary amide.
• Product B is soluble in dilute acid, meaning it is a primary amine (Ar-NH₂$Ar-NH_2$ or alkyl amine).
• B reacts with NaNO₂/HCl$NaNO_2/HCl$ at 0-5°C$0-5^{\circ}C$ 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.
• D is an aryl cyanide (Ar-CN$Ar-CN$). Hydrolysis of D yields E (Ar-COOH$Ar-COOH$).
• Crucially, E is also obtainable from the direct hydrolysis of A. This confirms A is an aryl amide of the form Ar-CONH₂$Ar-CONH_2$.



• Let's analyze the formula C₈H₉ON$C_8H_9ON$. The amide group is -CONH₂$-CONH_2$. Removing -CONH₂$-CONH_2$ leaves C₇H₇$C_7H_7$. A benzene ring with one methyl group is a tolyl group. So A is a methylbenzamide (CH₃-C₆H₄-CONH₂$CH_3-C_6H_4-CONH_2$).
• E is methylbenzoic acid (CH₃-C₆H₄-COOH$CH_3-C_6H_4-COOH$).
• E is oxidized by acidified KMnO₄$KMnO_4$ to F. The methyl group on the benzene ring oxidizes to -COOH$-COOH$. Thus, F is a benzenedicarboxylic acid (HOOC-C₆H₄-COOH$HOOC-C_6H_4-COOH$).



• 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).

Chapter Mix

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

Related Formula

Core Logic

Step 1: Reaction of aniline with benzoyl chloride (PhCOCl$\text{PhCOCl}$) in basic medium yields benzanilide (Ph-NH-CO-Ph$\text{Ph-NH-CO-Ph}$) as intermediate [A]$[A]$.

Step 2: Reduction of benzanilide using LiAlH₄$\text{LiAlH}_4$ converts the carbonyl group -C(=O)-$-\text{C}(=\text{O})-$ into a methylene group -CH₂-$-\text{CH}_2-$, forming dibenzylamine (Ph-NH-CH₂-Ph$\text{Ph-NH-CH}_2\text{-Ph}$) as final product [B]$[B]$.

Pattern Recognition

Sees: Acylation followed by LiAlH₄$\text{LiAlH}_4$ reduction. Shortcut: Amide carbonyl group reduces directly to -CH₂-$-\text{CH}_2-$, resulting in secondary amine structure (option 3).

Chapter Mix

Class 12 Chemistry: Amines