JEE Advanced Chemistry: Organic Chemistry - Some Basic Principles and Techniques questions with solutions

Q1. Three carbon acids are listed with their pKa values: (I) Triphenylmethane (Ph3CH), pKa = 33.3 (II) Benzene (C6H6), pKa = 43 (III) Methane (CH4), pKa = 50 Arrange them in increasing order of acidity (least acidic first).

1. (I) < (II) < (III)
2. (III) < (II) < (I)
3. (II) < (I) < (III)
4. (III) < (I) < (II)

Answer: (III) < (II) < (I)

A lower pKa corresponds to a stronger acid. Since pKa(I) = 33.3 < pKa(II) = 43 < pKa(III) = 50, compound I is most acidic and compound III is least acidic. Increasing order of acidity: (III) < (II) < (I).

Q2. From the following compounds, how many will produce effervescence (release CO2 gas) when treated with aqueous NaHCO3 solution? (i) 2-methylbenzoic acid (o-toluic acid) (ii) cyclohexylsulfonic acid (iii) 4-aminobenzaldehyde (iv) 4-hydroxycyclohexanone (v) cyclobutane-1,2,3,4-tetraol

1. (A) 2
2. (B) 3
3. (C) 4
4. (D) 5

Answer: (A) 2

NaHCO3 reacts with compounds whose pKa < ~6.4 to release CO2. (i) o-methylbenzoic acid is a carboxylic acid (pKa ~3.9) - REACTS. (ii) Cyclohexylsulfonic acid is a strong acid (pKa < 1) - REACTS. (iii) 4-aminobenzaldehyde has no acidic proton with pKa below 6.4 (the amino group is basic, the aldehyde is not acidic) - NO REACTION. (iv) 4-hydroxycyclohexanone: the alcohol OH has pKa ~15, the alpha-H of ketone has pKa ~20 - NO REACTION. (v) Cyclobutane-1,2,3,4-tetraol: alcohol groups with pKa ~15 - NO REACTION. From the 5 named compounds: 2 react. Note: compound (vi) was image-only and is excluded from evaluation.

Q3. Which of the following correctly represents the order of acid strength for the given substituted benzoic acids?

1. o-hydroxybenzoic acid > m-hydroxybenzoic acid > p-hydroxybenzoic acid
2. m-methylbenzoic acid < m-trideuteromethylbenzoic acid < m-tert-butylbenzoic acid
3. Benzoic acid with -NH3+ substituent > benzoic acid with -NMe3+ substituent > benzoic acid with -CH3 substituent
4. o-chlorobenzoic acid > p-chlorobenzoic acid > m-chlorobenzoic acid

Answer: o-hydroxybenzoic acid > m-hydroxybenzoic acid > p-hydroxybenzoic acid

Option A: o-OH > m-OH > p-OH for hydroxybenzoic acids. The ortho isomer (salicylic acid) is most acidic due to intramolecular H-bonding in the anion. The para-OH is electron-donating via resonance (decreases acidity) while meta-OH only has inductive effect. So o > m > p is correct. Option B: -CD3 vs -CH3: deuterium has weaker hyperconjugation than H, so m-CD3 is slightly more electron-withdrawing than m-CH3 (less hyperconjugation donation). m-tert-butyl is more electron-donating than m-CH3 (more hyperconjugation). So acidity: m-tBu-benzoic acid < m-CH3-benzoic acid < m-CD3-benzoic acid. The stated order is reversed. Option C: -NH3+ and -NMe3+ are both electron-withdrawing (+I effect absent, strong -M absent but strong +charge EWG). -NH3+ is more acidic than -NMe3+ because NH3+ has additional H available for H-bonding/solvation. -CH3 is electron-donating (decreases acidity). Order should be -NMe3+ > -NH3+ > -CH3 for acid strength based on EWG power (NMe3+ has stronger inductive EWG than NH3+), or the stated order may be wrong. Option D: o-Cl > p-Cl > m-Cl for chlorobenzoic acids. Cl is EWG by inductive and EDG by resonance. At ortho: both inductive and field effects enhance acidity. At para: resonance donation decreases acidity. At meta: only inductive (no resonance at meta). So order: o > m > p. Option D states o > p > m which is incorrect. Option A is correct.

Q4. Match the following chemical tests in List-I with the functional groups or compounds they identify in List-II. List-I: (A) Baeyer test, (B) Ceric ammonium nitrate test, (C) Tollen's dye test (phthalein dye test), (D) Schiff test List-II: (I) Phenol, (II) Aldehyde, (III) Alcoholic -OH group, (IV) Unsaturation Choose the correct matching:

1. (A)-(III), (B)-(I), (C)-(IV), (D)-(II)
2. (A)-(II), (B)-(III), (C)-(IV), (D)-(I)
3. (A)-(IV), (B)-(I), (C)-(II), (D)-(III)
4. (A)-(IV), (B)-(III), (C)-(I), (D)-(II)

Answer: (A)-(IV), (B)-(III), (C)-(I), (D)-(II)

Baeyer test: alkaline KMnO4 is decolourised by unsaturated compounds (alkenes/alkynes) -> identifies unsaturation (IV). Ceric ammonium nitrate test: gives red colour with alcoholic -OH groups -> identifies alcoholic OH (III). Phthalein dye test (thalen dye test): used to identify phenols -> (I). Schiff test: Schiff's reagent (fuchsin-SO2) turns pink/magenta with aldehydes -> identifies aldehyde (II). Matching: A-IV, B-III, C-I, D-II.

Q5. How many carbon atoms are present in the first (simplest) member of the homologous series of esters?

1. 1
2. 2
3. 3
4. 4

Answer: 2

The first member of the ester homologous series is methyl formate (HCOOCH3), which contains 2 carbon atoms — one in the formyl group and one in the methyl group.

Q6. In the Dumas method for nitrogen estimation, 0.30 g of an organic compound liberated 82.1 mL of nitrogen gas collected over water at 300 K and 775 mm Hg total pressure. Given that the aqueous tension (vapour pressure of water) at 300 K is 15 mm Hg, calculate the percentage of nitrogen in the compound.

1. 31.11
2. 15.56
3. 28.0
4. 31.72

Answer: 31.11

After subtracting aqueous tension and reducing the volume to STP, the mass of N2 gives the percentage of nitrogen.

Q7. 0.8 g of an organic compound is analysed for nitrogen by Kjeldahl's method. If the nitrogen content is found to be 42%, how many mL of (M/2) H2SO4 are neutralized by the ammonia liberated? (N = 14)

1. 12 mL
2. 24 mL
3. 36 mL
4. 48 mL

Answer: 24 mL

Mass N = 0.336 g -> 0.024 mol NH3; H2SO4 needed = 0.012 mol; at 0.5 M that is 24 mL.