Consider the following sequence of reactions :

11.25 mg of chlorobenzene will produce x × 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 35.5g mol⁻¹$35.5\mathrm{g\,mol}^{-1}$ respectively]

Related Formula

Core Logic

Step 1: Calculate initial moles of aniline:

Step 2: Calculate actual moles of benzanilide produced at 82% yield:

Step 3: Calculate mass of benzanilide:

Pattern Recognition

Sees: Amine acylation reaction with percentage yield. Shortcut: (5.8 / 93) × 0.82 × 197 = 10.07 g ≈ 10 g$(5.8 / 93) \times 0.82 \times 197 = 10.07\text{ g} \approx 10\text{ g}$.

Chapter Mix

Class 12 Chemistry: Amines

Core Logic

Follow the sequence of chemical transformations to determine the structure of product B, then apply stoichiometry based on moles of the starting material to find its final mass.

Step 1: Reaction Sequence Analysis

Reaction 1: Reduction of 4-nitrotoluene using Sn, HCl / Δ$Sn, HCl / \Delta$ followed by pH neutralization converts the -NO₂$-NO_2$ group to an -NH₂$-NH_2$ group, forming 4-methylaniline (p-toluidine).

Step 2: Acetylation

Reaction 2 (forming A): The aniline group is protected by reaction with acetic anhydride (CH₃CO)₂O$(CH_3CO)_2O$, forming an acetanilide derivative (N-(4-methylphenyl)acetamide).

Step 3: Bromination

Reaction 3 (forming B): Bromination using Br₂ / AcOH$Br_2 / AcOH$. The -NHCOCH₃$-NHCOCH_3$ group is strongly activating and ortho-directing, while the -CH₃$-CH_3$ group is weakly activating and ortho-directing. The incoming -Br$-Br$ will substitute ortho to the -NHCOCH₃$-NHCOCH_3$ group due to dominant directing influence.

Step 4: Stoichiometric Calculation

Molar mass of 4-nitrotoluene (C₇H₇NO₂$C_7H_7NO_2$) = 137 g/mol$137 \text{ g/mol}$. Molar mass of product B (C₉H₁₀BrNO$C_9H_{10}BrNO$) = 228 g/mol$228 \text{ g/mol}$.

Moles of 4-nitrotoluene = 137 × 10⁻³ g137 g/mol = 0.001 mol$\frac{137 \times 10^{-3} \text{ g}}{137 \text{ g/mol}} = 0.001 \text{ mol}$.

Since the stoichiometry is 1:1 and yields are 100%, moles of B = 0.001 mol$0.001 \text{ mol}$. Mass of B = 0.001 mol × 228 g/mol = 0.228 g = 228 mg$0.001 \text{ mol} \times 228 \text{ g/mol} = 0.228 \text{ g} = 228 \text{ mg}$.

Pattern Recognition

When dealing with multi-step synthesis yield questions, immediately check the initial and final molar masses if the reaction achieves 100% completion. Often, 1 mole of reactant gives 1 mole of product.

Chapter Mix

Class 12 Chemistry: Amines

Core Logic

Work backwards from the final product, cyclohexyl isocyanide.

Step 1: Carbylamine Reaction

The step (ii) CHCl₃, KOH (alc), Δ$(ii) \text{ } CHCl_3, KOH (alc), \Delta$ is the Carbylamine reaction. It converts a primary amine into an isocyanide. Therefore, the intermediate formed right before this step must be Cyclohexylamine (Cyclohexyl-NH₂$Cyclohexyl-NH_2$).

Step 2: Hoffmann Bromamide Degradation

The step (i) KOH, Br₂$(i) \text{ } KOH, Br_2$ is Hoffmann Bromamide Degradation. It converts an amide into a primary amine with one carbon less. Since the amine is Cyclohexylamine, the precursor 'Q' must be Cyclohexanecarboxamide.

Step 3: Finding 'P'

The reaction P NH₃, Δ Q$P \xrightarrow{NH_3, \Delta} Q$ converts a carboxylic acid into an amide. Therefore, 'P' must be Cyclohexanecarboxylic acid.

Pattern Recognition

Isocyanide (-NC$-NC$) final product arrow$\rightarrow$ Primary amine precursor arrow$\rightarrow$ Amide precursor via Hoffmann Bromamide arrow$\rightarrow$ Carboxylic acid via NH₃/Δ$NH_3/\Delta$.

Chapter Mix

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

Core Logic

Statement I analysis: The target is to convert an ethyl side chain (propylbenzene) into an aniline derivative. Step 1: Treatment with acidic KMnO₄$KMnO_4$ oxidizes any alkyl side chain with at least one benzylic hydrogen fully into a carboxylic acid group (benzoic acid derivative). Step 2: Reaction with Ammonia (NH₃$NH_3$) converts the carboxylic acid into an amide. Step 3: Bromine and alkali (Br₂ + KOH$Br_2 + KOH$) trigger a Hoffmann bromamide degradation, chopping off the carbonyl carbon and leaving an amine (NH₂$NH_2$). This correctly yields the target structure. Statement I is True.

Step 1: Statement II Analysis

Statement II analysis: The goal is to convert p-toluidine to 3,5-dibromotoluene. Step 1: Bromine water (Br₂ + H₂O$Br_2 + H_2O$) performs electrophilic aromatic substitution on the highly activated ring due to the -NH₂$-NH_2$ group. It polybrominates the ortho and para positions relative to -NH₂$-NH_2$. Since para is blocked by the methyl group, it brominates both ortho positions, yielding 2,6-dibromo-4-methylaniline. Step 2: Diazotization with NaNO₂ + HCl$NaNO_2 + HCl$ at 0-5°C$0-5^{\circ}\text{C}$ converts the -NH₂$-NH_2$ group into a diazonium salt (-N₂^+Cl^-$-N_2^+Cl^-$). Step 3: Aqueous H₃PO₂$H_3PO_2$ (hypophosphorous acid) acts as a reducing agent, replacing the diazonium group with a Hydrogen atom. This successfully yields 3,5-dibromotoluene (since the numbering shifts). Statement II is True.

Pattern Recognition

Hoffmann bromamide degradation strictly steps down an amide to an amine by eliminating the carbonyl C$C$. Deamination of aniline derivatives via diazonium salt followed by reduction (H₃PO₂$H_3PO_2$ or EtOH) is the standard method for synthesizing specific meta-substituted benzenes where direct electrophilic meta-substitution fails.

Chapter Mix

Class 12 Chemistry: Amines

Core Logic

Identify compound 'x': Molecular formula C₆H₇N$C_6H_7N$ indicates a high degree of unsaturation (Degree of Unsaturation = 6 - 7/2 + 1/2 + 1 = 4$6 - 7/2 + 1/2 + 1 = 4$), characteristic of a benzene ring. Given it dissolves in dilute mineral acid (due to protonation forming a salt), it's a basic amine. Since it produces an unpleasant-smelling compound 'y' with CHCl₃$CHCl_3$ and alc. KOH, it gives a positive carbylamine test, meaning 'x' is a primary amine. Thus, 'x' is Aniline (Ph-NH₂$Ph-NH_2$). Compound 'y' is phenyl isocyanide (Ph-NC$Ph-NC$).

Step 1: Identifying compound 'z'

Reaction with benzenesulphonyl chloride (Hinsberg's reagent): Aniline reacts with Ph-SO₂Cl$Ph-SO_2Cl$ to form N$N$-phenylbenzenesulfonamide, which is 'z'. Structure of 'z': Ph-NH-SO₂-Ph$Ph-NH-SO_2-Ph$. Because 'z' has an acidic proton on the nitrogen, it is soluble in alkali, confirming our deduction.

Step 2: Counting different "H" atoms

Let's map the types of Hydrogen atoms (chemically distinct environments by symmetry) on the molecule 'z' (Ph-NH-SO₂-Ph$Ph-NH-SO_2-Ph$):

• The single hydrogen attached to the Nitrogen (N-H$N-H$): 1 type.
• The first phenyl ring (from aniline) has symmetry down the 1,4-axis. It has ortho-H, meta-H, and para-H. That yields 3 distinct types of H.
• The second phenyl ring (from sulfonyl chloride) also has symmetry down its 1,4-axis. It has ortho-H, meta-H, and para-H. That yields 3 distinct types of H.

Total number of different H atoms = 1 + 3 + 3 = 7$1 + 3 + 3 = 7$.

Pattern Recognition

Whenever you see CHCl₃$CHCl_3$ + alc. KOH producing an foul smell, it's definitively the Carbylamine test marking a primary amine. Soluble in alkali after Hinsberg's reagent confirms it's a 1°$1^{\circ}$ amine.

Chapter Mix

Class 12 Chemistry: Amines