JEE Chemistry Organic Chemistry Some Basic Principles and Techniques MCQs Set 69

Answer: (c) Internal compensation

Explanation:
1. Meso-tartaric acid has two chiral centres of opposite configuration.
2. The rotation produced by one half is cancelled by the other half inside the same molecule.
3. This is internal compensation.
4. Answer: (c)

Teacher's Note:
1. First draw the possible structures systematically (chain, position, functional group), then count each distinct one only once.

Answer: (d) Dicyclohexyl ketone

Explanation:
1. Syn-anti isomerism of an oxime C=N–OH needs two different groups on the carbon.
2. Acetophenone, benzaldehyde and methyl cyclohexyl ketone all have different groups (or H) on the carbonyl carbon.
3. Dicyclohexyl ketone has two identical cyclohexyl groups, so only one oxime is possible.
4. Answer: (d)

Teacher's Note:
1. First draw the possible structures systematically (chain, position, functional group), then count each distinct one only once.

Answer: (c) –CN

Explanation:
1. Ka rises when the substituent withdraws electrons and stabilises the benzoate ion.
2. OCH₃, CH₃ and NH₂ are electron-releasing at para and decrease Ka.
3. CN is strongly electron-withdrawing (–I, –M), so it increases Ka.
4. Answer: (c)

Teacher's Note:
1. Decide first whether the effect is inductive, resonance or hyperconjugation, then compare the stabilisation of the ion or the strength of the effect.

Answer: (c) ii > iii > i

Explanation:
1. A –I group on the alpha carbon stabilises the carboxylate ion and raises acidity.
2. CHO has the strongest –I effect, phenyl has a weaker –I effect, and the ethyl group is electron-releasing.
3. Order of Ka: ii > iii > i.
4. Answer: (c)

Teacher's Note:
1. Decide first whether the effect is inductive, resonance or hyperconjugation, then compare the stabilisation of the ion or the strength of the effect.

Answer: (d) 3 > 2 > 1

Explanation:
1. Basicity of a carbanion depends on how stable it is: the greater the s-character of the carbanionic carbon, the more stable and the less basic it is.
2. (1) is sp (50% s), (2) is sp² (33% s) and (3) is sp³ (25% s).
3. Basic strength: 3 > 2 > 1.
4. Answer: (d)

Teacher's Note:
1. Decide first whether the effect is inductive, resonance or hyperconjugation, then compare the stabilisation of the ion or the strength of the effect.

Answer: (c) iii > iv > ii > i

Explanation:
1. A larger pKb means a weaker base.
2. Diformylamine (iii) is an imide-type compound whose lone pair is delocalised over two C=O groups: it is the weakest base.
3. Aniline (iv) has the lone pair in resonance with the ring. Benzylamine (ii) is weaker than ethylamine (i) because of the –I effect of phenyl.
4. Order of pKb: iii > iv > ii > i.
5. Answer: (c)

Teacher's Note:
1. Decide first whether the effect is inductive, resonance or hyperconjugation, then compare the stabilisation of the ion or the strength of the effect.

Answer: (c) Acid

Explanation:
1. H₂SO₄ is the stronger acid and protonates HNO₃.
2. HNO₃ + H₂SO₄ → H₂NO₃⁺ + HSO₄⁻, then H₂NO₃⁺ loses water to give the nitronium ion NO₂⁺.
3. Here H₂SO₄ donates a proton, so it acts as an acid (and HNO₃ acts as a base).
4. Answer: (c)

Teacher's Note:
1. Decide first whether the effect is inductive, resonance or hyperconjugation, then compare the stabilisation of the ion or the strength of the effect.

Answer: (c) Benzaldehyde

Explanation:
1. Friedel-Crafts reactions work on benzene and activated or weakly deactivated rings.
2. Benzaldehyde has a strongly deactivating CHO group (and complexes with AlCl₃), so it does not react.
3. Chlorobenzene is only weakly deactivated and still reacts.
4. Answer: (c)

Teacher's Note:
1. Decide first whether the effect is inductive, resonance or hyperconjugation, then compare the stabilisation of the ion or the strength of the effect.

Answer: (c) Phenyl acetate

Explanation:
1. The O of the acetate group donates its lone pair to the ring (+M), activating it relative to benzene.
2. Chlorobenzene (–I > +M) and nitrobenzene (–M, –I) are deactivated.
3. Phenyl acetate is the most reactive.
4. Answer: (c)

Teacher's Note:
1. Decide first whether the effect is inductive, resonance or hyperconjugation, then compare the stabilisation of the ion or the strength of the effect.

Answer: (b)

Explanation:
1. Compare the carbocations formed.
2. (a) gives diphenylmethyl, stabilised by two phenyl rings. (b) gives triphenylmethyl (trityl), stabilised by three rings. (c) gives a cycloheptadienyl cation, which is only allylic. (d) gives benzyl, stabilised by one ring.
3. The trityl cation has the greatest resonance delocalisation, so (b) is the most stable.
4. Answer: (b)

Teacher's Note:
1. Decide first whether the effect is inductive, resonance or hyperconjugation, then compare the stabilisation of the ion or the strength of the effect.

Answer: (a) CH₃–NH₂

Explanation:
1. In methylamine the lone pair is localised on nitrogen and the methyl group pushes electrons towards it.
2. In the aniline derivatives the lone pair is delocalised into the ring, so all three are much weaker bases.
3. Methylamine is the most basic.
4. Answer: (a)

Teacher's Note:
1. Decide first whether the effect is inductive, resonance or hyperconjugation, then compare the stabilisation of the ion or the strength of the effect.

Answer: (c) o-Nitrophenol

Explanation:
1. NaHCO₃ dissolves only compounds more acidic than carbonic acid (pK_a about 6.4), releasing CO₂.
2. Benzoic acid, benzenesulphonic acid and picric acid are all stronger acids.
3. o-Nitrophenol has pK_a about 7.2, weaker than carbonic acid, so it does not react.
4. Answer: (c)

Teacher's Note:
1. Decide first whether the effect is inductive, resonance or hyperconjugation, then compare the stabilisation of the ion or the strength of the effect.

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

Explanation:
1. In the gas phase there is no solvation, so only the +I effect of the alkyl groups matters.
2. More alkyl groups mean more electron donation to nitrogen.
3. Basic strength: (CH₃)₃N > (CH₃)₂NH > CH₃NH₂ > NH₃, i.e. III > II > I > IV.
4. Answer: (c)

Teacher's Note:
1. Decide first whether the effect is inductive, resonance or hyperconjugation, then compare the stabilisation of the ion or the strength of the effect.

Answer: (a) II > I > III

Explanation:
1. (II) is stabilised by the lone pair on oxygen directly attached to the cationic carbon (+M effect): it is the most stable.
2. (I) is a secondary cation stabilised by hyperconjugation.
3. In (III) the –OCH₃ group is one carbon away and withdraws electrons (–I), which destabilises the cation.
4. Order: II > I > III.
5. Answer: (a)

Teacher's Note:
1. Decide first whether the effect is inductive, resonance or hyperconjugation, then compare the stabilisation of the ion or the strength of the effect.

Answer: (b) NaHCO₃

Explanation:
1. Both phenol and carboxylic acids react with Na to give H₂, and both can turn blue litmus red or show acidity in tests.
2. Carboxylic acids are strong enough to release CO₂ from NaHCO₃; phenol is weaker than carbonic acid and does not.
3. NaHCO₃ therefore separates them.
4. Answer: (b)

Teacher's Note:
1. Decide first whether the effect is inductive, resonance or hyperconjugation, then compare the stabilisation of the ion or the strength of the effect.

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

Explanation:
1. In (I) OCH₃ at the meta position reinforces the activating OH group.
2. Phenol (IV) is the reference.
3. Cl (II) is deactivating and NO₂ (III) strongly so.
4. Order: I > IV > II > III.
5. Answer: (a)

Teacher's Note:
1. Decide first whether the effect is inductive, resonance or hyperconjugation, then compare the stabilisation of the ion or the strength of the effect.

Answer: (b) I ≈ II ≈ III

Explanation:
1. In nitration, the slow step is the attack of NO₂⁺ on the ring; breaking of the C–H (or C–D, C–T) bond is a fast step.
2. There is therefore no primary kinetic isotope effect.
3. The three rates are about equal.
4. Answer: (b)

Teacher's Note:
1. Decide first whether the effect is inductive, resonance or hyperconjugation, then compare the stabilisation of the ion or the strength of the effect.

Answer: (c) II, II, I

Explanation:
1. Polarity of a bond increases with the electronegativity difference.
2. (a) C–Br is more polar than C–H: II.
3. (b) O is more electronegative than N, so C–OH is more polar: II.
4. (c) O is more electronegative than S, so C–OH is more polar: I.
5. Answer: II, II, I.
6. Answer: (c)

Teacher's Note:
1. Decide first whether the effect is inductive, resonance or hyperconjugation, then compare the stabilisation of the ion or the strength of the effect.

Answer: (c) SO₃, H₂O, CH₃⁻

Explanation:
1. A nucleophile is an electron-rich species with a lone pair or negative charge.
2. SO₃ is electron-poor (sulphur is bonded to three oxygen atoms) and acts as an electrophile.
3. So the set containing SO₃ is not a set of nucleophiles.
4. Answer: (c)

Teacher's Note:
1. Decide first whether the effect is inductive, resonance or hyperconjugation, then compare the stabilisation of the ion or the strength of the effect.

Answer: (b) CH₃–Cu → CH₃⁻ + Cu⁺

Explanation:
1. In heterolysis the shared pair goes to the more electronegative atom.
2. In CH₃–Cu carbon is more electronegative than copper, so CH₃⁻ and Cu⁺ form.
3. In CH₃–Cl chlorine is more electronegative, so CH₃⁺ and Cl⁻ form. In CH₃–S–CH₃ sulphur takes the electrons, giving CH₃⁺ and CH₃S⁻.
4. Answer: (b)

Teacher's Note:
1. Decide first whether the effect is inductive, resonance or hyperconjugation, then compare the stabilisation of the ion or the strength of the effect.

Answer: (a) a = b

Explanation:
1. The nitro group is a resonance hybrid of two equivalent structures, in which the double bond and the single bond exchange places.
2. Hence both N–O bonds have the same length (partial double bonds).
3. Answer: (a)

Teacher's Note:
1. Decide first whether the effect is inductive, resonance or hyperconjugation, then compare the stabilisation of the ion or the strength of the effect.

Answer: (b) –X

Explanation:
1. A +R group donates a lone pair to the ring or to a π-system.
2. Halogens have lone pairs and donate by resonance (although they withdraw by –I).
3. C=O, COOH and CONHR groups withdraw electrons by resonance (–R).
4. Answer: (b)

Teacher's Note:
1. Decide first whether the effect is inductive, resonance or hyperconjugation, then compare the stabilisation of the ion or the strength of the effect.

Answer: (c) Formation of acetaldehyde cyanohydrin

Explanation:
1. In the +E effect the π-electrons move towards the atom to which the attacking (electrophilic) reagent attaches.
2. Reactions of alkenes and alkynes with electrophiles (a, b, d) are examples.
3. In cyanohydrin formation the nucleophile CN⁻ attacks carbonyl carbon and the π-electrons move to oxygen: this is the –E effect.
4. Answer: (c)

Teacher's Note:
1. Decide first whether the effect is inductive, resonance or hyperconjugation, then compare the stabilisation of the ion or the strength of the effect.

Answer: (b) It does not have (4n + 2)π electrons and moreover is non-planar

Explanation:
1. Aromaticity needs a planar, cyclic, fully conjugated system with (4n + 2) π electrons.
2. Cyclooctatetraene has 8 π electrons and adopts a tub shape (non-planar) to avoid antiaromaticity.
3. Answer: (b)

Teacher's Note:
1. Decide first whether the effect is inductive, resonance or hyperconjugation, then compare the stabilisation of the ion or the strength of the effect.

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

Explanation:
1. Piperidine (I) is a saturated amine with a free lone pair on sp³ N: the most basic.
2. Pyridine (II) has the lone pair on an sp² nitrogen and is less basic.
3. Aniline (III) has its lone pair delocalised into the ring.
4. Phenol (IV) is acidic, not basic, so it is the weakest base.
5. Answer: (a)

Teacher's Note:
1. Decide first whether the effect is inductive, resonance or hyperconjugation, then compare the stabilisation of the ion or the strength of the effect.

Answer: (b) I, II and III

Explanation:
1. Meta directors are electron-withdrawing groups.
2. CCl₃ (–I from three Cl), NO₂ and CHO deactivate the ring and direct meta.
3. The phenoxide ion O⁻ is a strong activator and directs ortho/para.
4. Meta in I, II and III.
5. Answer: (b)

Teacher's Note:
1. Decide first whether the effect is inductive, resonance or hyperconjugation, then compare the stabilisation of the ion or the strength of the effect.

Answer: (c) C₆H₅CH₂⁺ (benzyl cation)

Explanation:
1. The benzyl cation is delocalised over the ring (resonance).
2. The isopropyl cation is stabilised only by hyperconjugation, and the ethyl cation even less.
3. The phenyl cation has an empty sp² orbital in the plane of the ring that cannot overlap with the π system, so it is very unstable.
4. Most stable: benzyl cation.
5. Answer: (c)

Teacher's Note:
1. Decide first whether the effect is inductive, resonance or hyperconjugation, then compare the stabilisation of the ion or the strength of the effect.