### Related Formula The relative rate of nucleophilic acyl substitution follows the leaving group ability: textRate of Hydrolysis propto textLeaving Group Ability propto frac1textBasic Strength of Leaving Group$\text{Rate of Hydrolysis} \propto \text{Leaving Group Ability} \propto \frac{1}{\text{Basic Strength of Leaving Group}}$

### Core Logic Let's analyze the leaving groups across all choices layout-by-row: * For **(p)**, the leaving group is mathrmCl^-$\mathrm{Cl^-}$ (Very weak base, excellent leaving group). * For **(q)**, the leaving group is mathrmRCOO^-$\mathrm{RCOO^-}$ (Resonance stabilized carboxylate, good leaving group). * For **(r)**, the leaving group is mathrmRO^-$\mathrm{RO^-}$ (Alkoxide, strong base, poor leaving group). * For **(s)**, the leaving group is mathrmNH_2^-$\mathrm{NH_2^-}$ (Extremely strong base, exceptionally poor leaving group due to nitrogen lone pair resonance into the carbonyl). This structural comparison yields the final sequence: mathrmp > q > r > s$\mathrm{p > q > r > s}$. ### Pattern Recognition Acyl chlorides (p) are always the most reactive acid derivatives, while amides (s) are consistently the least reactive due to strong amide resonance stabilizing the carbonyl group. ### 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 Elemental content calculation system equations: textMoles of C = fractextMass of mathrmCO_244$\text{Moles of C} = \frac{\text{Mass of } \mathrm{CO_2}}{44}$ textMoles of H = 2 times fractextMass of mathrmH_2O18$\text{Moles of H} = 2 \times \frac{\text{Mass of } \mathrm{H_2O}}{18}$ ### Core Logic Let's perform the stoichiometry layout step-by-step: * Moles of mathrmC$\mathrm{C}$ inside sample system: mathrmn_C = frac1.4644 = 0.033mathrm~mol$\mathrm{n_C} = \frac{1.46}{44} = 0.033\mathrm{~mol}$ textMass of C = 0.033 times 12 = 0.396mathrm~g$\text{Mass of C} = 0.033 \times 12 = 0.396\mathrm{~g}$ * Moles of mathrmH$\mathrm{H}$ inside sample system: mathrmn_H = 2 times frac0.56718 = 0.063mathrm~mol$\mathrm{n_H} = 2 \times \frac{0.567}{18} = 0.063\mathrm{~mol}$ textMass of H = 0.063 times 1 = 0.063mathrm~g$\text{Mass of H} = 0.063 \times 1 = 0.063\mathrm{~g}$ * Determine Oxygen mass by subtracting values from total starting mass: textMass of O = 1.0 - (0.396 + 0.063) = 0.541mathrm~g$\text{Mass of O} = 1.0 - (0.396 + 0.063) = 0.541\mathrm{~g}$ mathrmn_O = frac0.54116 = 0.033mathrm~mol$\mathrm{n_O} = \frac{0.541}{16} = 0.033\mathrm{~mol}$ * Find atomic whole-number ratio profile: mathrmC : H : O = 0.033 : 0.063 : 0.033 approx 1 : 2 : 1$\mathrm{C : H : O} = 0.033 : 0.063 : 0.033 \approx 1 : 2 : 1$. * This gives an empirical configuration of mathrmCH_2O$\mathrm{CH_2O}$. ### Step 1: Evaluation Calculating formula mass: textEmpirical Mass = 12 + (2 times 1) + 16 = 30mathrm~g$\text{Empirical Mass} = 12 + (2 \times 1) + 16 = 30\mathrm{~g}$ ### Pattern Recognition When calculated mole properties output identical numbers for two elements (0.033$0.033$ for both C and O), their structural subscript ratio is exactly 1:1$1:1$. This pattern significantly speeds up empirical calculations. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques

### 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