Match List-I with List-II -

List-I (Separation of)	List-II (Separation Technique)
(A) Aniline from aniline-water mixture	(I) Simple distillation
(B) Glycerol from spent-lye in soap industry	(II) Fractional distillation
(C) Different fractions of crude oil in petroleum industry	(III) Distillation at reduced pressure
(D) Chloroform-Aniline mixture	(IV) Steam distillation

Choose the correct answer from the options given below:

### Related Formula According to IUPAC rules for nomenclature of aromatic compounds: - Principal functional group has highest priority: -mathrmCOOH > -mathrmOH$$-\mathrm{COOH} > -\mathrm{OH}$$ - The principal functional group carbon is designated as Carbon-1, and numbering is directed to give substituents the lowest possible locants. ### Core Logic Assign priority and number the ring: - Carbon-1: -mathrmCOOH$-\mathrm{COOH}$ (Carboxyl carbon, parent name 'benzoic acid') - Carbon-2: -mathrmOH$-\mathrm{OH}$ (Hydroxyl substituent) - Carbon-3: -mathrmBr$-\mathrm{Br}$ (Bromo substituent) - Carbon-5: -mathrmNO_2$-\mathrm{NO}_2$ (Nitro substituent) This numbering yields substituent locants at positions 2, 3, and 5. ### Step 1: Arrange alphabetically List the substituents alphabetically with locants: - 3-Bromo - 2-Hydroxy - 5-Nitro Combining these names: text3-Bromo-2-hydroxy-5-nitrobenzoic acid$$\text{3-Bromo-2-hydroxy-5-nitrobenzoic acid}$$ This matches Option (1). ### Pattern Recognition Carboxylic acid always dictates position 1 in ring numbering over alcohol. Numbering clockwise gives 2-hydroxy, 3-bromo, and 5-nitro, whereas counterclockwise numbering would yield much higher locants (2-nitro, 4-bromo, 5-hydroxy) which violates the lowest-locant rule. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids

### Related Formula Balanced reaction for nitration of nitrobenzene: mathrmC_6H_5NO_2 + mathrmHNO_3 rightarrow mathrmC_6H_4(NO_2)_2 + mathrmH_2O$$\mathrm{C_6H_5NO_2} + \mathrm{HNO_3} \rightarrow \mathrm{C_6H_4(NO_2)_2} + \mathrm{H_2O}$$ textMoles = fractextMasstextMolar Mass$$\text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}}$$ ### Core Logic From the balanced stoichiometry: - 1 mole of nitrobenzene yields 1 mole of m-dinitrobenzene. ### Step 1: Determine molar masses - Molar mass of Nitrobenzene (mathrmC_6H_5NO_2$\mathrm{C_6H_5NO_2}$): M_1 = 6(12) + 5(1) + 14 + 2(16) = 72 + 5 + 14 + 32 = 123mathrm~g/mol$$M_1 = 6(12) + 5(1) + 14 + 2(16) = 72 + 5 + 14 + 32 = 123\mathrm{~g/mol}$$ - Molar mass of m-Dinitrobenzene (mathrmC_6H_4(NO_2)_2$\mathrm{C_6H_4(NO_2)_2}$): M_2 = 6(12) + 4(1) + 2(14) + 4(16) = 72 + 4 + 28 + 64 = 168mathrm~g/mol$$M_2 = 6(12) + 4(1) + 2(14) + 4(16) = 72 + 4 + 28 + 64 = 168\mathrm{~g/mol}$$

### Step 2: Calculate moles and find X Moles of m-dinitrobenzene produced: $

### Step 2: Calculate moles and find X Moles of m-dinitrobenzene produced: $n = frac4.2mathrm~g168mathrm~g/mol = 0.025mathrm~mol$n = \frac{4.2\mathrm{~g}}{168\mathrm{~g/mol}} = 0.025\mathrm{~mol}$ Since stoichiometry is $ Since stoichiometry is $1:1, the moles of nitrobenzene required is also $, the moles of nitrobenzene required is also $0.025\mathrm{~mol}: $: $textMass of nitrobenzene X = 0.025mathrm~mol times 123mathrm~g/mol = 3.075mathrm~g$\text{Mass of nitrobenzene } X = 0.025\mathrm{~mol} \times 123\mathrm{~g/mol} = 3.075\mathrm{~g}$ Rounding to the nearest integer gives $ Rounding to the nearest integer gives $3$. ### Pattern Recognition Electrophilic aromatic substitution stoichiometry is straightforward: each aromatic precursor ring converts to exactly one product ring. Finding moles from the heavier substituted product and converting back using the reactant's molecular weight quickly yields the answer. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques Class 12 Chemistry: Amines

### Related Formula Degrees of Unsaturation (Double Bond Equivalents, DBE): mathrmDBE = C + 1 - fracH2 + fracN2$$\mathrm{DBE} = C + 1 - \frac{H}{2} + \frac{N}{2}$$ For formula mathrmC_9H_12$\mathrm{C_9H_{12}}$: mathrmDBE = 9 + 1 - frac122 = 4$$\mathrm{DBE} = 9 + 1 - \frac{12}{2} = 4$$ These 4 degrees of unsaturation match a benzene ring exactly (one ring + three double bonds). ### Core Logic Since the question specifies 'substituted benzene derivatives', we must keep the benzene core (mathrmC_6H_5-$\mathrm{C_6H_5-}$ or similar) intact. This leaves 3 carbon atoms to be distributed as alkyl substituents. ### Step 1: Categorize by substitution patterns 1. **Mono-substituted benzene** (one propyl group containing 3 carbons): - n-Propylbenzene: mathrmC_6H_5-CH_2-CH_2-CH_3$\mathrm{C_6H_5-CH_2-CH_2-CH_3}$ (Isomer 1) - Isopropylbenzene (Cumene): mathrmC_6H_5-CH(CH_3)_2$\mathrm{C_6H_5-CH(CH_3)_2}$ (Isomer 2) 2. **Di-substituted benzene** (one ethyl group and one methyl group): - 1-Ethyl-2-methylbenzene (ortho-ethylmethylbenzene) (Isomer 3) - 1-Ethyl-3-methylbenzene (meta-ethylmethylbenzene) (Isomer 4) - 1-Ethyl-4-methylbenzene (para-ethylmethylbenzene) (Isomer 5) ### Step 2: Tri-substituted benzenes 3. **Tri-substituted benzene** (three methyl groups): - 1,2,3-Trimethylbenzene (Hemimellitene) (Isomer 6) - 1,2,4-Trimethylbenzene (Pseudocumene) (Isomer 7) - 1,3,5-Trimethylbenzene (Mesitylene) (Isomer 8) ### Step 3: Total Count Summing all options: textTotal structural isomers = 2 + 3 + 3 = 8$$\text{Total structural isomers} = 2 + 3 + 3 = 8$$ ### Pattern Recognition For alkyl benzenes with N$N$ extra carbons, systematically group them as single chain substituents down to multiple methyl substituents. This hierarchical sorting prevents duplicates or missing patterns. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques Class 11 Chemistry: Hydrocarbons

### Related Formula Pressure of dry nitrogen gas: P_mathrmN_2 = P_texttotal - textAqueous tension$$P_{\mathrm{N}_2} = P_{\text{total}} - \text{Aqueous tension}$$ Using Ideal Gas Law: n_mathrmN_2 = fracP_mathrmN_2 VR T$$n_{\mathrm{N}_2} = \frac{P_{\mathrm{N}_2} V}{R T}$$ \%mathrmN = fractextMass of nitrogentextMass of organic compound times 100$$\%\mathrm{N} = \frac{\text{Mass of nitrogen}}{\text{Mass of organic compound}} \times 100$$ ### Core Logic Given parameters: - Mass of compound m = 0.4mathrm~g$m = 0.4\mathrm{~g}$ - Volume of nitrogen V = 60mathrm~mL = 0.060mathrm~L$V = 60\mathrm{~mL} = 0.060\mathrm{~L}$ - Total pressure P_texttotal = 715mathrm~mm~Hg$P_{\text{total}} = 715\mathrm{~mm~Hg}$ - Temperature T = 300mathrm~K$T = 300\mathrm{~K}$ - Aqueous tension = 15mathrm~mm~Hg$= 15\mathrm{~mm~Hg}$ ### Step 1: Calculate dry nitrogen pressure P_mathrmN_2 = 715mathrm~mm~Hg - 15mathrm~mm~Hg = 700mathrm~mm~Hg$$P_{\mathrm{N}_2} = 715\mathrm{~mm~Hg} - 15\mathrm{~mm~Hg} = 700\mathrm{~mm~Hg}$$ P_mathrmN_2 = frac700760mathrm~atm approx 0.921mathrm~atm$$P_{\mathrm{N}_2} = \frac{700}{760}\mathrm{~atm} \approx 0.921\mathrm{~atm}$$ ### Step 2: Calculate moles of nitrogen gas Using $ ### Step 2: Calculate moles of nitrogen gas Using $R = 0.0821\mathrm{~L\cdot atm\cdot K^{-1}\cdot mol^{-1}}: $: $n_mathrmN_2 = fracleft(frac700760right) times 0.0600.0821 times 300 = frac0.0552624.63 approx 2.2436 times 10^-3mathrm~mol$n_{\mathrm{N}_2} = \frac{\left(\frac{700}{760}\right) \times 0.060}{0.0821 \times 300} = \frac{0.05526}{24.63} \approx 2.2436 \times 10^{-3}\mathrm{~mol}$ Mass of $ Mass of $\mathrm{N}_2 gas: $ gas: $textMass = 2.2436 times 10^-3 times 28mathrm~g approx 0.06282mathrm~g$\text{Mass} = 2.2436 \times 10^{-3} \times 28\mathrm{~g} \approx 0.06282\mathrm{~g}$ ### Step 3: Calculate percentage of Nitrogen $$ ### Step 3: Calculate percentage of Nitrogen $$\%\mathrm{N} = \frac{0.06282\mathrm{~g}}{0.4\mathrm{~g}} \times 100 \approx 15.71\%$$ This matches Option (1). ### Pattern Recognition In Dumas' method calculations, always subtract the aqueous tension to obtain the pressure of dry nitrogen gas. Do not use the raw moist gas pressure, as doing so will overestimate the nitrogen content. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques

### Related Formula The number of hyperconjugation structures is directly related to the count of alpha$\alpha$-hydrogen atoms: textNumber of hyperconjugative structures = textNumber of alphatext-hydrogens$$\text{Number of hyperconjugative structures} = \text{Number of } \alpha\text{-hydrogens}$$ ### Core Logic Statement I Analysis: - Hyperconjugation (no-bond resonance) involves the delocalization of sigma$\sigma$ electrons of mathrmC-H$\mathrm{C-H}$ bonds of an alkyl group directly attached to an atom of unsaturated system or a positively charged carbon atom. This is a permanent ground-state electronic effect, not dependent on external reagents. Thus, Statement I is False. ### Step 1: Analyze Statement II - Statement II states that more alkyl groups attached to a carbocation center increase hyperconjugative stabilization. Each alkyl group brings additional sigma_mathrmC-H$\sigma_{\mathrm{C-H}}$ bonds adjacent to the empty p-orbital, increasing the total count of alpha$\alpha$-hydrogens and enhancing charge delocalization. Thus, Statement II is True. ### Step 2: Conclusion Therefore, Statement I is False but Statement II is True, matching Option (3). ### Pattern Recognition Permanent organic effects include: Inductive, Mesomeric (Resonance), and Hyperconjugation effects. Temporary electronic effects include: Electromeric and Inductomeric effects (which require an attacking reagent to manifest). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques