JEE Chemistry Organic Chemistry Some Basic Principles and Techniques MCQs Set 46

Answer: (c) (CH₃)₃N > (CH₃)₂NH > CH₃NH₂ > NH₃

Explanation:
1. In the gas phase there is no solvation, so only the +I effect of the alkyl groups decides.
2. More alkyl groups push more electron density on N, so the order is 3° > 2° > 1° > NH₃.

Teacher's Note:
1. In water the order changes because the ammonium ion is stabilised by hydrogen bonding with the solvent.
2. Remember: gas phase uses +I only; aqueous solution uses +I, solvation and steric effect.

Answer: (b) I > III > IV > II

Explanation:
1. R₄N⁺OH⁻ is a fully ionised quaternary hydroxide (like NaOH), so it is the strongest base.
2. For the amines in water the order is R₂NH (pKb 3.27) > RNH₂ (3.38) > R₃N (4.22), because of the balance of +I effect, solvation and steric hindrance.
3. So the order is I > III > IV > II.

Teacher's Note:
1. The tertiary methylamine loses out in water because its ammonium ion has only one N-H bond to hydrogen bond with water.
2. A quaternary ammonium hydroxide is a strong base.

Answer: (a) I > III > II > IV

Explanation:
1. Piperidine has a free sp³ nitrogen lone pair on an amine (pKa of conjugate acid about 11).
2. Morpholine has the same type of N, but the oxygen in the ring withdraws electrons (-I), so it is weaker (pKa about 8.4).
3. Pyridine's lone pair is in an sp² orbital (pKa about 5.2). In pyrrole the lone pair is part of the aromatic sextet, so it is practically non-basic (about -4).
4. The order is I > III > II > IV.

Teacher's Note:
1. sp³ N > sp² N in basic strength; delocalised lone pairs are the weakest.
2. A ring oxygen two atoms away lowers the basicity of the nitrogen.

Answer: (a) II < I < III

Explanation:
1. pKa of the conjugate acid measures the basicity of the amine: the stronger the base, the larger the pKa.
2. In diphenylamine the lone pair is delocalised over two rings (very weak base), in aniline over one ring, whereas cyclohexylamine has a localised lone pair and also +I effect of the ring.
3. So basicity is III > I > II and the pKa increases in the order II < I < III.

Teacher's Note:
1. Each phenyl group on N reduces basicity because it delocalises the lone pair.
2. Aliphatic amines are about 10^5 times more basic than aniline.

Answer: (c) II > I > IV > III

Explanation:
1. The protonated amidine is stabilised by resonance; electron-donating substituents on the ring help and electron-withdrawing ones reduce this stabilisation.
2. NH₂ (II) is a strong +M donor and increases basicity most. Cl (IV) withdraws by -I and NO₂ (III) by -M and -I, with NO₂ stronger.
3. Order: II > I > IV > III.

Teacher's Note:
1. Rank the para substituents by electron donation: NH₂ > H > Cl > NO₂.
2. The amidinium ion has two equivalent resonance structures, which makes amidines strong bases.

Answer: (c) Acetamidine > pyrrolidine > CH₃NH₂ > aniline

Explanation:
1. In each step the added base takes the proton from the weaker base. So CH₃NH₂ > aniline, pyrrolidine > CH₃NH₂ and acetamidine > pyrrolidine.
2. The order is acetamidine (conjugate acid stabilised by resonance, pKa about 12) > pyrrolidine (11.3) > CH₃NH₂ (10.6) > aniline (4.6).

Teacher's Note:
1. A stronger base removes H⁺ from the conjugate acid of a weaker base.
2. Cyclic secondary amines are stronger than primary amines of similar type.

Answer: (d) CH₃OH

Explanation:
1. The pKa values are approximately: CH₃OH 15.5, HC≡CH 25, C₆H₆ 43 and C₂H₆ 50.
2. Oxygen is much more electronegative than carbon, so the alkoxide ion is better stabilised than any of the carbanions, even the sp-hybridised acetylide ion. Methanol is the strongest acid.

Teacher's Note:
1. Compare the atom that carries the negative charge first (O > C), then its hybridisation.
2. Among hydrocarbons alone the order is ethyne > benzene > ethane.

Answer: (a) C

Explanation:
1. Acidity increases with the -I effect of the substituent, and the effect falls off with distance from COOH.
2. In A the strongly withdrawing NO₂ is on the alpha carbon, so A is the strongest acid. In B, C and D the substituent is on the beta carbon.
3. Among these, -I order is NO₂ > CN > Cl, so Cl (C) gives the weakest acid and the smallest Ka.

Teacher's Note:
1. Position and strength of the withdrawing group both matter.
2. -I order of common groups: NO₂ > CN > F > Cl > Br > I.

Answer: (b) PhSO₃H > PhCOOH > PhOH > PhCH₂OH

Explanation:
1. Sulphonic acids are strong acids (pKa about -2.5) because the sulphonate ion has three equivalent S-O bonds sharing the charge.
2. Benzoic acid has pKa 4.2, phenol 10 and benzyl alcohol about 15.4 (its anion is not resonance stabilised).
3. So PhSO₃H > PhCOOH > PhOH > PhCH₂OH.

Teacher's Note:
1. Rank by the stability of the conjugate base: sulphonate > carboxylate > phenoxide > alkoxide.
2. The CH₂ in benzyl alcohol insulates O⁻ from the ring.

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

Explanation:
1. The nearer the second COOH group, the larger its -I effect on the first ionisation. Oxalic acid has the two groups directly attached (pKa1 1.25), malonic acid 2.85, succinic acid 4.2, adipic acid 4.4.
2. So acidity increases in the order adipic < succinic < malonic < oxalic, i.e. IV < III < II < I.

Teacher's Note:
1. The inductive effect falls off by a factor of about 2 for each CH₂ group.
2. Compare the number of carbon atoms between the two carboxyl groups.

Answer: (a) III < II < I

Explanation:
1. NO₂ acts by -M and -I at the ortho and para positions, but only by -I at the meta position, so the meta isomer is the weakest acid (pKa 8.3).
2. The para isomer (pKa 7.1) is slightly stronger than the ortho isomer (7.2) because in the ortho isomer intramolecular hydrogen bonding holds the H of the OH group.
3. A stronger acid has a lower pKa, so pKa increases in the order III < II < I.

Teacher's Note:
1. Ortho nitrophenol is steam volatile because of its intramolecular hydrogen bond.
2. Do not confuse the order of acidity with the order of pKa: it is reversed.

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

Explanation:
1. Phenol (pKa 10) is a much weaker acid than any benzoic acid because the carboxylate ion shares its charge between two equivalent oxygens.
2. Among the benzoic acids, the electron-releasing CH₃ in p-toluic acid weakens the acid (pKa 4.4), Cl withdraws by -I and strengthens it (3.98), benzoic acid is 4.2.
3. So the increasing order is IV < III < I < II.

Teacher's Note:
1. Rank: acceptor-substituted benzoic > benzoic > donor-substituted benzoic > phenol.
2. Chlorine at para is a net acceptor (-I stronger than +M).

Answer: (b) The sodium salt of the sulphonic acid group (the -COOH is untouched) and acetic acid

Explanation:
1. A stronger acid reacts with the conjugate base of a weaker acid, giving the weaker acid. Acetic acid (pKa 4.76) is weaker than both -SO₃H and -COOH.
2. With only one mole of CH₃COONa, the proton is taken from the strongest acid group first. The sulphonic acid group (pKa about -2.5) is far stronger than the carboxylic acid group.
3. So the -SO₃Na compound with free -COOH forms, together with CH₃COOH.

Teacher's Note:
1. With limited base, the strongest acid group reacts first.
2. Reaction is feasible only if the product acid is weaker than the reacting acid.

Answer: (a) X, because its NO₂ group stays in the plane of the ring and exerts a -M effect

Explanation:
1. In Y the two methyl groups flank the NO₂ group and twist it out of the plane of the ring (steric inhibition of resonance), so it cannot show its -M effect on the phenoxide ion.
2. In X the methyl groups are next to OH, not NO₂, so the nitro group remains coplanar and strongly delocalises the negative charge.
3. So X is the stronger acid.

Teacher's Note:
1. Steric inhibition of resonance switches off a -M or +M effect.
2. The OH-flanking methyl groups only add +I and a little steric effect.

Answer: (d) I > II > III > IV

Explanation:
1. The ortho isomer is the strongest of all (ortho effect, pKa 3.91), stronger than benzoic acid (4.20), even though CH₃ is electron releasing.
2. The meta isomer (4.27) is weakened only by +I, and the para isomer (4.38) is weakened by +I and hyperconjugation as well.
3. So I > II > III > IV.

Teacher's Note:
1. The ortho effect overrides the electronic effect of the group.
2. Para donors reduce acidity more than meta donors.

Answer: (b) I > II > III > IV

Explanation:
1. Salicylic acid (I, pKa 2.98) is by far the strongest: its anion is stabilised by an intramolecular hydrogen bond (plus the ortho effect).
2. At the meta position OH works only by -I (pKa 4.08, slightly stronger than benzoic 4.20); at the para position its +M effect dominates and weakens the acid (4.58).
3. So I > II > III > IV.

Teacher's Note:
1. Ortho-OH gives chelation with the carboxylate.
2. Meta groups only show -I; para groups show both I and M.

Answer: (d) K₁ of maleic acid > K₁ of fumaric acid, but K₂ of maleic acid < K₂ of fumaric acid

Explanation:
1. In the maleate monoanion the carboxylate and the remaining COOH are close, so an intramolecular hydrogen bond stabilises the anion and K₁(maleic) > K₁(fumaric).
2. In the second step the maleate dianion has two negative groups close together, so there is strong repulsion and the second proton is hard to remove: K₂(maleic) < K₂(fumaric).
3. So the first statement is correct.

Teacher's Note:
1. Intramolecular H-bonding raises K₁; charge repulsion lowers K₂.
2. The same reasoning explains why maleic acid is much stronger than fumaric in the first step but weaker in the second.

Answer: (c) CH₃COCH₂COCH₃ (acetylacetone)

Explanation:
1. The enol is favoured by conjugation, by intramolecular hydrogen bonding and by an acidic active methylene group.
2. Acetylacetone has a CH₂ flanked by two ketone groups: its enol is conjugated and chelated by an O-H...O bond (about 75 % enol). In ethyl acetoacetate the ester withdraws less strongly, so only 7 to 8 % is enol; simple aldehydes and ketones have negligible enol content.

Teacher's Note:
1. Active methylene compounds are the ones with high enol content.
2. Enolisation is favoured in non-polar solvents.

Answer: (d) The oxime CH₃CH₂CH=NOH

Explanation:
1. A primary nitroso compound with a hydrogen on the carbon next to N=O behaves like a carbonyl compound: the alpha hydrogen shifts to oxygen.
2. This gives the double bond between C and N, i.e. the oxime CH₃CH₂CH=N-OH, which is the more stable tautomer.

Teacher's Note:
1. Nitroso-oxime tautomerism is analogous to keto-enol tautomerism.
2. The oxime is far more stable than the nitroso form.

Answer: (d) I > II > III > IV

Explanation:
1. pKa1 values: oxalic 1.25, malonic 2.85, succinic 4.2, glutaric 4.34.
2. The -I effect of the second carboxyl group weakens with each CH₂ group between them, so the strongest acid is the one with the groups closest.
3. So I > II > III > IV.

Teacher's Note:
1. A shorter chain between the COOH groups means higher acidity.
2. The second dissociation is always weaker than the first.

Answer: (c) B > A > D > C

Explanation:
1. Three fluorine atoms give a stronger -I effect than three chlorine atoms, so CF₃COOH (pKa 0.23) > CCl₃COOH (0.65).
2. Formic acid (3.75) has no alkyl group, so it is stronger than acetic acid (4.76).
3. So B > A > D > C.

Teacher's Note:
1. -I effect: F > Cl, and three halogens beat one.
2. Alkyl groups reduce acidity; H does not.

Answer: (a) F⁻

Explanation:
1. The conjugate acids are HF (pKa 3.2), H₂O (15.7), NH₃ (38) and CH₄ (about 50).
2. The strongest acid has the weakest conjugate base, so F⁻ is the weakest base.

Teacher's Note:
1. Basicity decreases from left to right across a period.
2. Fluoride is a very weak base but a good hydrogen-bond acceptor.

Answer: (d) 2-Pyridone (in equilibrium with 2-hydroxypyridine)

Explanation:
1. Lactam-lactim tautomerism is the amide analogue of keto-enol tautomerism: the N-H hydrogen moves to the carbonyl oxygen, giving an N=C-OH (lactim) form.
2. 2-Pyridone has the N-H next to the C=O group and is in equilibrium with 2-hydroxypyridine. Pyridine, benzene and cyclohexane have no such group.

Teacher's Note:
1. Amide N-H next to C=O is needed.
2. In the solid state the lactam (pyridone) form is the more stable tautomer.