JEE Chemistry Organic Chemistry Some Basic Principles and Techniques MCQs Set 45

Answer: (a) I (azulene)

Explanation:
1. Aromaticity needs a planar, fully conjugated ring with (4n + 2) pi electrons.
2. Azulene (I) is planar with 10 pi electrons (n = 2), a fused cyclopentadienyl anion and tropylium cation, so it is aromatic.
3. Pentalene (II) has 8 pi electrons (4n), so it is antiaromatic; cyclooctatetraene (III) also has 8 pi electrons and avoids this by becoming tub-shaped (non-planar).
4. In IV the central sp3 carbon breaks the cyclic conjugation, so it is non-aromatic.

Teacher's Note:
1. Count the pi electrons only in a ring that is planar and unbroken by an sp3 atom.
2. Fused rings such as azulene follow the (4n + 2) rule over the whole periphery.

Answer: (a) I < III < II

Explanation:
1. Alkene stability increases with the number of alkyl groups on the double bond (hyperconjugation with the alpha C-H bonds).
2. I is a disubstituted alkene (two ring carbons), III is trisubstituted and II is tetrasubstituted.
3. Hence the order of stability is I < III < II.

Teacher's Note:
1. More substituted alkenes have more hyperconjugation structures and lower heat of hydrogenation.
2. Compare the number of alkyl groups on the two sp2 carbons.

Answer: (c) I

Explanation:
1. The more stable an anion, the stronger the acid it comes from.
2. Thiophenol (pKa about 6.6) is a stronger acid than phenol (about 10), aniline (about 30) and toluene (about 41).
3. Sulphur is large and polarisable, so the negative charge in I is best accommodated. The order of stability is I > II > III > IV.

Teacher's Note:
1. Down a group, larger atoms stabilise negative charge better.
2. Across a period, more electronegative atoms stabilise it better: O⁻ > N⁻ > C⁻.

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

Explanation:
1. A group that withdraws electrons stabilises a carbanion; a group that donates destabilises it.
2. NO₂ (III) withdraws by -M and -I, so it is the most stabilising. CH₃ (II) donates by +I and hyperconjugation; OCH₃ (IV) donates by a strong +M effect, so it destabilises most.
3. The order is III > I > II > IV.

Teacher's Note:
1. Think of the para group as pushing or pulling charge on the ring and benzylic carbon.
2. For carbanions, electron-withdrawing groups help and +M groups hurt most.

Answer: (d) II

Explanation:
1. II is the cyclopentadienyl anion: planar, cyclic, fully conjugated with 6 pi electrons, so it is aromatic and exceptionally stable.
2. IV (cycloheptatrienyl anion) would have 8 pi electrons and is antiaromatic; III has its lone pair in an sp2 orbital outside the pi system; I is only an ordinary conjugated carbanion.
3. So II is the most stable.

Teacher's Note:
1. Cyclopentadiene has pKa about 16, unusually acidic for a hydrocarbon, because its anion is aromatic.
2. Check planarity, continuous conjugation and 4n+2 for every cyclic ion.

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

Explanation:
1. Radical stability rises with delocalisation and with alkyl substitution.
2. IV (diphenylmethyl) is delocalised over two benzene rings (about 7 structures each side), so it is the most stable.
3. III is delocalised over three carbons of the ring; II is a primary radical stabilised by hyperconjugation; I (methyl) has none.
4. The order is IV > III > II > I.

Teacher's Note:
1. Resonance beats hyperconjugation, which beats no stabilisation.
2. More aryl groups on the radical carbon give more stability.

Answer: (d) II (cycloheptatrienyl cation)

Explanation:
1. The cycloheptatrienyl (tropylium) cation is planar and cyclic with 6 pi electrons, so it is aromatic and very stable.
2. The cyclopentadienyl cation has 4 pi electrons and is antiaromatic. The cyclopropyl cation has a strained ring and cannot delocalise the charge. The cyclohexyl cation is only a secondary carbocation.
3. So II is the most stable.

Teacher's Note:
1. Tropylium salts such as tropylium bromide can be isolated as stable solids.
2. Antiaromatic cations such as cyclopentadienyl are highly unstable.

Answer: (b)

Explanation:
1. The reactant is a secondary carbocation next to a quaternary carbon carrying three methyl groups.
2. A 1,2-methyl shift moves one CH₃ to the cationic carbon, so the positive charge moves to the carbon that lost the methyl group.
3. That carbon now carries two methyl groups and one isopropyl-type carbon, so the new ion is the tertiary 2,3-dimethylbutan-2-yl cation, which is more stable. The other options need a different skeleton or are less stable.

Teacher's Note:
1. Carbocations rearrange by 1,2-hydride or 1,2-alkyl shifts to become more stable.
2. A secondary to tertiary shift is strongly favoured.

Answer: (a) All C-C bonds are equal in length and all C-O bonds are equal in length.

Explanation:
1. The negative charge and the double bonds can be moved around the four-membered ring and the four oxygen atoms in equivalent resonance structures.
2. Each C-C bond is single in some structures and double in others in the same proportion, so all four C-C bonds are equal; likewise each C-O bond has the same average bond order.
3. Hence all C-C bonds are equal and all C-O bonds are equal.

Teacher's Note:
1. Equivalent resonance structures give equal bond lengths.
2. The squarate dianion is a stable, aromatic-like delocalised system.

Answer: (b) II > I > III

Explanation:
1. These are resonance contributors of the same protonated amide, so rank them by stability: structures with complete octets are best.
2. III has every atom with a full octet (N lone pair delocalised, positive charge on the more electropositive N), so it is the most stable.
3. I has a full octet but places the positive charge on the electronegative O; II has a carbon with only 6 electrons, so it is the least stable.
4. Stability is III > I > II, so potential energy decreases in the order II > I > III.

Teacher's Note:
1. The more stable the structure, the lower its potential energy.
2. A sextet carbon is worse than a positive oxygen with an octet.

Answer: (b) The C1-C2 bond is shorter than the C2-C3 bond.

Explanation:
1. Naphthalene has three main Kekule structures. The C1-C2 bond is double in two of them and single in one, while the C2-C3 bond is double in only one.
2. So C1-C2 has more double-bond character (about 1.37 Angstrom) than C2-C3 (about 1.42 Angstrom).
3. Therefore the C1-C2 bond is shorter than the C2-C3 bond.

Teacher's Note:
1. Count how often a bond is double in the resonance structures: more double-bond character means a shorter bond.
2. Naphthalene, unlike benzene, does not have equal C-C bonds.

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

Explanation:
1. The negative charge is stabilised by delocalisation onto carbonyl oxygens: more C=O groups give more stability.
2. IV has three CHO groups, III has two CHO groups, II has two acetyl groups and I has only one CHO group.
3. Methyl groups in II donate electrons (+I) and make it less stable than III, which has two unsubstituted CHO groups.
4. The order is IV > III > II > I.

Teacher's Note:
1. More electron-withdrawing conjugating groups mean a more stable carbanion.
2. Replace H of CHO by CH₃ and the carbonyl becomes a poorer acceptor.

Answer: (b) III > I > II

Explanation:
1. I is the allyl cation (charge delocalised over two carbons), II is a simple primary propyl cation, III is the benzyl cation (charge delocalised over four carbons).
2. Greater delocalisation means greater stability, so III > I, and the unstabilised primary cation II is the least stable.
3. The order is III > I > II.

Teacher's Note:
1. Rank by the number of resonance structures: benzyl 5, allyl 2, propyl 1.
2. Benzyl and allyl cations are stabilised by resonance.

Answer: (d) III

Explanation:
1. Benzene (IV), pyridine (II) and furan (I) are all aromatic and have large resonance energies.
2. III (4,4-dimethylcyclohexa-2,5-dienone) has a cross-conjugated dienone: the sp3 carbon breaks the ring, so there is only limited delocalisation through the carbonyl group.
3. So III has the least resonance stabilisation.

Teacher's Note:
1. Aromatic compounds have by far the highest resonance energy.
2. An sp3 atom in the ring stops cyclic delocalisation.

Answer: (b) III (borepin)

Explanation:
1. In borepin the boron is sp² with a vacant p orbital, which overlaps with the three C=C bonds: 6 pi electrons (4n + 2, n = 1) in a planar cyclic conjugated ring, so it is aromatic.
2. 1H-Azepine and the cycloheptatrienyl anion both have 8 pi electrons, so they are antiaromatic (or non-planar to avoid it). Cycloheptatriene has an sp3 carbon and is not fully conjugated.
3. So III is the aromatic compound.

Teacher's Note:
1. A vacant p orbital contributes zero electrons to the pi count.
2. A lone pair on N in azepine adds 2 electrons to the 6 of the three C=C, giving 8.

Answer: (d) II

Explanation:
1. I (cyclopropenyl cation, 2 pi electrons), III (cyclopentadienyl anion, 6 pi electrons) and IV (benzene, 6 pi electrons) all follow the (4n + 2) rule and are aromatic.
2. II is cyclobutadiene with 4 pi electrons (4n), so it is antiaromatic.
3. Hence II is the odd one out.

Teacher's Note:
1. Count the pi electrons: 2, 6, 10 are aromatic; 4, 8 are antiaromatic.
2. The cyclopropenyl cation is the smallest aromatic ion.

Answer: (c) I - 67, II - 88, III - 121

Explanation:
1. The heteroatom lone pair is delocalised into the ring; the more electronegative the heteroatom, the more tightly it holds its pair and the less aromatic stabilisation results.
2. Electronegativity: O > N > S, so aromatic character increases in the order furan < pyrrole < thiophene.
3. Hence furan 67, pyrrole 88, thiophene 121 kJ/mol.

Teacher's Note:
1. Furan is the most diene-like and thiophene the most benzene-like.
2. Remember: resonance energy increases as heteroatom electronegativity decreases.

Answer: (a) II

Explanation:
1. II is a tertiary carbocation (three alkyl groups on the positively charged carbon), stabilised by the most hyperconjugation and +I effects.
2. I and III are secondary and IV is primary.
3. So II is the most stable.

Teacher's Note:
1. Order of carbocation stability: tertiary > secondary > primary > methyl.
2. Count the alkyl groups on the C⁺.

Answer: (b) I only

Explanation:
1. I (pyrrole) is planar with 6 pi electrons (4 from two C=C and 2 from the N lone pair), so it is aromatic.
2. II (cyclopropene) has an sp3 carbon; III (cyclobutadiene) has 4 pi electrons and is antiaromatic; IV (cyclopentadienyl cation) has 4 pi electrons and is antiaromatic.
3. So only I is aromatic.

Teacher's Note:
1. The N lone pair in pyrrole is part of the pi system, giving the count 6.
2. A cation or anion changes the pi electron count by the charge.

Answer: (c) The carbonyl carbon and the beta carbon of the C=C

Explanation:
1. The C=O group is polarised: oxygen is delta minus and the carbonyl carbon is delta plus.
2. Through conjugation the C=C is polarised as in the resonance structure O⁻-C=C-C⁺ with the positive charge on the beta carbon (the C=C carbon farther from C=O), while the alpha carbon gains negative character.
3. So the atoms with delta plus are the carbonyl carbon and the beta carbon.

Teacher's Note:
1. In an alpha,beta-unsaturated carbonyl compound the beta carbon is electrophilic.
2. Draw the -M resonance structure to see where the positive charge goes.

Answer: (c) 3, 5, 9, 5

Explanation:
1. Hyperconjugable hydrogens are the alpha C-H hydrogens (on carbons attached to an sp2 carbon or a carbocation centre).
2. P: the cationic carbon bears one CH₃, so 3. Q: CH₃ (3) plus CH₂ (2) = 5.
3. R: two CH₃ on the C= end (6) plus the CH₃ on the other carbon (3) = 9. S: CH₃ (3) plus CH₂ (2) = 5.
4. So the numbers are 3, 5, 9, 5.

Teacher's Note:
1. Only alpha carbons count; hydrogens on the benzene ring or on the vinyl group are not hyperconjugable.
2. Count the H atoms on every carbon directly attached to the unsaturated or charged carbon.

Answer: (b) III

Explanation:
1. The NH₂ group is a strong +M donor, so the ring of aniline is much richer in electrons than that of toluene (which has only +I and hyperconjugation).
2. CHO withdraws electrons (-M, -I) and Cl withdraws more than it donates (-I outweighs +M), while benzene has no donor at all, so these rings are less electron rich than toluene.
3. So only III is richer than toluene.

Teacher's Note:
1. +M groups (NH₂, OH, OR) enrich the ring strongly.
2. Halogens deactivate the ring even though they have +M.

Answer: (d) IV

Explanation:
1. IV is a benzyl cation with a +M methoxy group at the para position. Its positive charge is delocalised into the ring and onto the oxygen lone pair, so it is the most stable.
2. I is an ordinary benzyl cation (less stabilised), II is a phenyl cation (the empty orbital is in the plane and cannot overlap with the ring), and III is an unstabilised primary alkyl cation.
3. So IV is the most stable.

Teacher's Note:
1. Delocalisation into oxygen's lone pair gives an extra resonance structure with full octets.
2. A phenyl cation is very unstable because its vacant orbital is perpendicular to the pi system.

Answer: (c) 5

Explanation:
1. Aromatic: pyrrole (I, 6 pi electrons), cyclopropenyl cation (II, 2), benzene (III, 6), pyrylium cation (IV, 6) and thiophene (V, 6).
2. Not aromatic: cyclobutadiene (VI, 4 pi, antiaromatic), cyclopentadienone (VII, 4 pi in the ring because of the C=O polarisation, antiaromatic) and 1,4-dioxine (VIII, 8 pi electrons including two O lone pairs, non-aromatic).
3. So five species are aromatic.

Teacher's Note:
1. Check planarity, full conjugation and the 4n+2 count for each ring.
2. A ring C=O pulls electron density out of the ring and can create an antiaromatic ring.

Answer: (c) 6

Explanation:
1. Resonance needs a pi bond (or charge, lone pair, radical centre) in conjugation with another pi bond or p orbital.
2. I has isolated double bonds (no resonance); VI (allyl chloride) has Cl on an sp3 carbon, so no conjugation.
3. The cyclohexadienyl anion, cation and radical (II, III, IV), vinyl chloride (V, lone pair on Cl conjugated with C=C), the phenoxide ion (VII) and N-methylpyrrole (VIII) all show resonance, i.e. six species.

Teacher's Note:
1. Isolated double bonds do not delocalise.
2. A lone pair on an atom next to a pi bond is in conjugation (+M effect).

Answer: (b) 5

Explanation:
1. A +M group has a lone pair on the atom attached to the ring that can be donated into the ring.
2. -NH₂, -SH, -F, -O⁻ and -OCH₃ each have such a lone pair, so these 5 show +M.
3. -CHO, -CONH₂, -CN and -NO₂ show -M; -NH₃⁺ has no lone pair; -CH₃ acts only by +I and hyperconjugation, not by a mesomeric effect.
4. So the number is 5.

Teacher's Note:
1. Ask: is there a lone pair on the first atom? If yes it is a +M group.
2. The -CONH₂ group withdraws through its C=O even though N has a lone pair.

Answer: (d) C

Explanation:
1. A is the neutral structure with all octets complete, so it is the most stable.
2. B and D carry charges, but every atom still has a complete octet (the negative charge is on oxygen).
3. C has a carbon atom with only six electrons together with separated charges, so it is the least stable and the least contributing structure.

Teacher's Note:
1. Rule: octet complete, then fewest charges, then negative charge on the more electronegative atom.
2. A carbon with a sextet is a very poor contributor.