Given below are some nitrogen containing compounds.

Each of them is treated with HCl separately. 1.0 g of the most basic compound will consume ________ mg of HCl. (Given molar mass in g mol ⁻¹$^{-1}$ C:12, H : 1, O : 16, Cl : 35.5)

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