Structure and Reactivity: Acids and Bases, Electronic Effects, Stereochemistry, Conformers and Aromaticity
2. Electronic and steric effects
| Effect | How it acts | Examples | Typical consequence |
|---|---|---|---|
| Inductive (±I) | Polarisation along σ bonds; permanent; fades within about three bonds | −I: NO₂, CN, F > Cl > Br > I, OH, OR, C=O. +I: alkyl groups, O⁻ | Chloroacetic acid stronger than acetic acid; alkylamines more basic than NH₃ |
| Resonance or mesomeric (±M) | Delocalisation of π electrons or lone pairs through conjugation; felt at ortho and para positions | +M: NH₂, OH, OR, NHCOR, halogens. −M: NO₂, CN, CHO, COR, COOH, SO₃H | Phenols and carboxylic acids acidic; aniline weakly basic |
| Hyperconjugation | Overlap of a filled C–H σ bond on an α carbon with an adjacent empty p orbital, half-filled orbital or π bond | (CH₃)₃C⁺ has 9 α C–H bonds; CH₃CH₂⁺ has 3 | Carbocations and radicals 3° > 2° > 1° > CH₃; more substituted alkenes more stable |
| Steric | Crowding at the reacting site, or a group pushed out of plane | tert-Butyl, neopentyl, ortho substituents | Slow SN2 at a crowded carbon; bulky bases give Hofmann alkenes; steric inhibition of resonance |
When induction and resonance oppose, the outcome depends on the atom. For O and N (OH, OR, NH₂) the +M donation into the ring outweighs −I, so they are overall donors. For halogens −I outweighs +M, so they are overall withdrawers that still push electrons toward the ortho and para positions (the root of their odd behaviour in aromatic substitution). Hyperconjugation is also why heats of hydrogenation fall as an alkene gains alkyl groups: trans-but-2-ene < cis-but-2-ene < but-1-ene (the cis isomer also suffers a methyl–methyl clash).
Steric effects: crowding at a carbon blocks the backside approach that SN2 needs (Chapter 2), and a bulky base removes the most exposed hydrogen. Crowding can also twist a group out of the plane it needs for resonance: in 3,5-dimethyl-4-nitrophenol the two methyl groups flanking the nitro group turn it out of the ring plane, it can no longer accept the phenoxide charge, and the compound is a markedly weaker acid than 4-nitrophenol (steric inhibition of resonance). Similarly, most ortho-substituted benzoic acids are stronger than benzoic acid whatever the substituent (the ortho effect), because the crowded COOH twists out of the ring plane and loses the ring's electron donation.
3. Stereochemistry: chirality, R/S and optical activity
Stereoisomers share connectivity but differ in spatial arrangement. A molecule is chiral when it cannot be superimposed on its mirror image; a molecule with a mirror plane or a centre of symmetry is achiral. The usual source of chirality is a stereocentre, an sp³ carbon bearing four different groups: butan-2-ol is chiral, propan-2-ol (two CH₃ groups on C2) is not. The two mirror-image forms are enantiomers. Configuration is named by the CIP (Cahn–Ingold–Prelog) rules:
- Rank the four atoms attached to the stereocentre by atomic number (I > Br > Cl > O > N > C > H).
- Break ties by moving outward and comparing the sets of attached atoms at the first point of difference: –CH₂OH (O,H,H) outranks –CH(CH₃)₂ (C,C,H).
- Multiple bonds count as duplicated atoms: –CHO is treated as C(O,O,H), so it outranks –CH₂OH (O,H,H).
- View with the lowest-ranked group pointing away. If 1 → 2 → 3 runs clockwise the centre is R, anticlockwise S. In a Fischer projection the horizontal bonds point toward the viewer, so when the lowest group is horizontal, read the direction and reverse it.
Enantiomers share melting point, boiling point and solubility, and differ only in the direction they rotate plane-polarised light and in how they interact with chiral partners such as enzymes and receptors. Diastereomers are stereoisomers that are not mirror images; they differ in physical properties and can be separated by crystallisation or chromatography. A molecule with n stereocentres has at most 2ⁿ stereoisomers. Fewer exist when a meso form appears: a compound with stereocentres and an internal mirror plane, achiral and optically inactive. Tartaric acid, HOOC–CH(OH)–CH(OH)–COOH, has two stereocentres but only three stereoisomers: (2R,3R), (2S,3S) and meso (2R,3S).
Specific rotation [α] = α ÷ (l × c), with the observed rotation α in degrees, the path length l in decimetres and c in g/mL. A 50:50 racemic mixture shows no rotation. Enantiomeric excess ee = %major − %minor, and, when rotation is linear in composition, ee = (observed [α] ÷ [α] of the pure enantiomer) × 100. Worked example: 0.40 g of a chiral compound X in 10.0 mL (c = 0.040 g/mL) in a 1.0 dm tube rotates light by +2.60°, so [α] = 2.60 ÷ (1.0 × 0.040) = +65.0. If pure (+)-X has [α] = +130, ee = 50%, and the sample is 75% (+)-X and 25% (−)-X.
4. Geometrical isomerism and tautomerism
Geometrical (cis/trans) isomers need restricted rotation (a double bond or a ring) and two different groups on each carbon concerned: but-2-ene has cis and trans forms, but-1-ene and 2-methylbut-2-ene do not. The general E/Z system ranks the two groups on each carbon by the CIP rules: higher-ranked groups on the same side is Z, on opposite sides E. Cis is not always Z. In the isomer of 2-bromobut-2-ene, CH₃–CH=C(Br)–CH₃, whose two methyl groups are on the same side, Br outranks CH₃ on C3, so the higher-ranked groups (CH₃ on C2, Br on C3) are opposite and the isomer is (E).
Geometrical isomers are diastereomers and differ physically. cis-1,2-Dichloroethene has a dipole moment and the higher boiling point; the trans isomer has none, packs better and melts higher. Maleic acid (cis-butenedioic acid) loses water on heating to form a cyclic anhydride, which fumaric acid (trans) cannot do because its carboxyl groups are too far apart. Maleic acid is also the stronger acid in its first ionisation but the weaker in its second, because its monoanion is held by an internal hydrogen bond. Counting: n stereogenic double bonds give up to 2ⁿ isomers, fewer in symmetrical molecules (hexa-2,4-diene has three: EE, ZZ and EZ).
Tautomers are constitutional isomers that interconvert rapidly, usually by moving a proton and shifting a π bond. Both are real compounds in equilibrium, unlike resonance structures, which differ only in where electrons are drawn. In keto–enol tautomerism (catalysed by acid or base) simple aldehydes and ketones lie almost entirely on the keto side, because C=O is stronger than C=C. The enol gains ground when it is conjugated and held by an internal hydrogen bond, as in pentane-2,4-dione, where it is a large fraction, and wins outright when it is aromatic: phenol is the enol form of a cyclohexadienone. Other pairs are imine–enamine, nitro–aci-nitro and amide–imidic acid (lactam–lactim).
5. Conformers of ethane, butane and cyclohexane
Conformers are arrangements that interconvert by rotation about σ bonds; at room temperature they usually cannot be isolated. A Newman projection looks along the C–C bond: the front carbon is a point and the rear carbon a circle. In ethane the staggered form (dihedral angle 60°) is the minimum and the eclipsed form (0°) the maximum, about 12 kJ/mol higher. That barrier is torsional strain, about 4 kJ/mol for each eclipsed H/H pair.
| Conformer | CH₃–C–C–CH₃ dihedral angle | Energy, kJ/mol | Source of strain |
|---|---|---|---|
| Anti (staggered) | 180° | 0 | None; the most stable |
| Gauche (staggered) | 60° | about 3.8 | CH₃/CH₃ steric crowding |
| Eclipsed, CH₃ facing H | 120° | about 16 | Torsional plus CH₃/H strain |
| Fully eclipsed, CH₃ facing CH₃ | 0° | about 19 to 25 | Torsional plus CH₃/CH₃ strain; the maximum |
Cyclohexane adopts a chair in which every C–C bond is staggered and every angle is close to tetrahedral, so it is essentially strain-free. The boat (eclipsing plus a flagpole H/H clash) lies about 29 kJ/mol higher, the twist-boat about 23. Each chair carbon has one axial bond (parallel to the ring axis) and one equatorial bond (around the ring's equator). A ring flip (barrier about 45 kJ/mol, fast at room temperature) turns every axial bond equatorial and vice versa. A substituent prefers the equatorial position, because an axial group meets the two axial hydrogens on the same side of the ring (1,3-diaxial interactions, each like a gauche-butane clash).
Worked example. In methylcyclohexane the equatorial conformer is lower by about 7.3 kJ/mol. At 298 K, K = [equatorial]/[axial] = exp(ΔG/RT) = exp(7300 ÷ (8.314 × 298)) = e^2.95 ≈ 19, so about 95% of the molecules are equatorial. A tert-butyl group effectively locks itself equatorial. For disubstituted rings, mark each substituent up or down: cis-1,2, trans-1,3 and cis-1,4 are necessarily axial–equatorial, while trans-1,2, cis-1,3 and trans-1,4 can be diequatorial and are the more stable of each pair. In β-D-glucopyranose every ring substituent is equatorial in the usual chair, one reason the β anomer predominates in solution.
6. Aromaticity and Hückel's rule
A ring is aromatic, and unusually stable, when it is (1) cyclic, (2) planar and (3) fully conjugated, with a p orbital on every ring atom, and (4) holds 4n + 2 π electrons (2, 6, 10, 14…). Meeting 1 to 3 with 4n π electrons makes it antiaromatic and destabilised (cyclobutadiene). Failing any of 1 to 3 makes it non-aromatic: cyclooctatetraene puckers into a tub and reacts as a polyene; the sp³ CH₂ of cycloheptatriene breaks the loop. To count π electrons: each ring double bond gives 2, a carbocation's empty p orbital 0, a carbanion's lone pair 2, and a heteroatom lone pair 2 only if it sits in a p orbital perpendicular to the ring (at most one pair per atom).
| Species | π electrons | Verdict and reason |
|---|---|---|
| Benzene | 6 | Aromatic |
| Cyclopentadienyl anion, C₅H₅⁻ | 6 | Aromatic; hence the pKa of about 16 for cyclopentadiene |
| Tropylium (cycloheptatrienyl) cation, C₇H₇⁺ | 6 | Aromatic; tropylium bromide is an ionic salt |
| Cyclopropenyl cation, C₃H₃⁺ | 2 | Aromatic (n = 0) |
| Cyclobutadiene | 4 | Antiaromatic |
| Pyridine | 6 | Aromatic; the N lone pair lies in an sp² orbital in the ring plane, is not counted and stays basic |
| Pyrrole; furan | 6 | Aromatic; the N lone pair (one O lone pair in furan) completes the sextet, so pyrrole is barely basic |
| Imidazole | 6 | Aromatic; one pyrrole-like N in the sextet and one pyridine-like N, which is basic (pKaH near 7; about 6 in the histidine side chain) |
| Naphthalene; indole; purine | 10 | Aromatic fused bicyclic systems |
Aromaticity runs through biochemistry: the purine and pyrimidine bases are flat aromatic heterocycles that stack in DNA, tryptophan carries an indole, and histidine an imidazole whose basic nitrogen shuttles protons in enzyme active sites near pH 7. The large resonance stabilisation (benzene releases about 150 kJ/mol less on hydrogenation than three times cyclohexene) is why arenes prefer substitution, which keeps the ring intact, to addition (next chapter).
Key takeaways
- Acid strength follows conjugate-base stability: charge on a more electronegative or larger atom, more s character, nearby −I groups, resonance and aromaticity all lower pKa. Bases are compared by pKaH; a delocalised lone pair (aniline, amide, pyrrole) is a weak base. Proton transfer has K = 10^(ΔpKa) toward the weaker acid.
- Induction fades within about three bonds; resonance acts at ortho and para positions; O and N donors are overall +M, halogens overall −I. Hyperconjugation ranks cations, radicals and alkenes by α C–H count; bulk slows crowded reactions and can switch resonance off.
- R/S from CIP ranks (lowest group away; reverse if it is horizontal in a Fischer projection). n stereocentres give at most 2ⁿ stereoisomers, fewer with meso forms. [α] = α/(l·c), l in dm, c in g/mL; ee = observed/pure × 100; major enantiomer = (100 + ee)/2 per cent.
- Cis/trans needs restricted rotation and two different groups on each carbon; E/Z uses CIP ranks, and cis is not always Z. Tautomers (keto–enol, lactam–lactim) are real interconverting isomers, not resonance structures; enolisation racemises an α stereocentre.
- Butane: anti < gauche (about 3.8 kJ/mol) < eclipsed < fully eclipsed. Cyclohexane sits in a chair with substituents equatorial; the population ratio is K = exp(ΔG/RT), about 95% equatorial for methyl.
- Aromatic means cyclic, planar, fully conjugated and 4n + 2 π electrons; 4n is antiaromatic. A lone pair counts only if it occupies a p orbital perpendicular to the ring: pyrrole's does, pyridine's does not.
Practice questions (18)
Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.
Arrange phenol, 4-methylphenol, 3-nitrophenol and 4-nitrophenol in order of increasing acid strength (weakest acid first).
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Answer: C — 4-methylphenol < phenol < 3-nitrophenol < 4-nitrophenol
A methyl group donates electrons (+I and hyperconjugation) and destabilises phenoxide, so 4-methylphenol (pKa about 10.3) is slightly weaker than phenol (10.0). A nitro group withdraws: from the meta position only by induction (3-nitrophenol, about 8.4), from the para position also by resonance, which puts the negative charge onto the nitro oxygens (4-nitrophenol, about 7.2). The tempting order with 3-nitrophenol as the strongest forgets that resonance withdrawal reaches only the ortho and para positions.Which of the following is the weakest base in water?
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Answer: D — Pyrrole, C₄H₄NH
Pyrrole's nitrogen lone pair is part of its aromatic sextet; protonating that nitrogen would destroy the aromaticity, so pyrrole is an extremely weak base (its conjugate acid has a pKa far below zero). Aniline is the tempting choice: its lone pair is delocalised into the ring, yet it is still available enough to give a pKaH of 4.6, close to pyridine's 5.2. Cyclohexylamine, an ordinary alkylamine (pKaH about 10.6), is the strongest base of the four.Taking pKa = 4.76 for acetic acid and pKa = 7.15 for 4-nitrophenol, calculate the equilibrium constant K, in water, for CH₃COOH + 4-O₂N–C₆H₄–O⁻ ⇌ CH₃COO⁻ + 4-O₂N–C₆H₄–OH. Give K to the nearest integer.
Numerical answer — type the value.
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Answer: 245
K = Ka(acetic acid) ÷ Ka(4-nitrophenol) = 10^(pKa of the product acid − pKa of the reactant acid) = 10^(7.15 − 4.76) = 10^2.39 = 245.47, which is 245 to the nearest integer. The equilibrium lies toward the weaker acid, 4-nitrophenol. Reversing the subtraction gives 10^(−2.39) ≈ 0.0041, the usual slip.A substituent on a benzene ring withdraws electrons through its σ bond to the ring, yet is overall an electron donor to the ring because of resonance. Which substituent fits this description?
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Answer: A — Methoxy, –OCH₃
Oxygen is more electronegative than carbon (−I), but its lone pair delocalises into the ring (+M), and for oxygen +M dominates, so methoxy is an overall donor that activates the ring. Chloro is the tempting choice because it also combines −I with +M, but for halogens −I dominates and the ring is deactivated overall. –CF₃ withdraws by induction only, and –CH₃ donates through +I and hyperconjugation rather than withdrawing through its σ bond.Hyperconjugation in a carbocation uses the C–H bonds on the carbons directly attached to the positively charged carbon. How many such C–H bonds does the 2-methylbutan-2-yl cation, (CH₃)₂C⁺–CH₂CH₃, have?
Numerical answer — type the value.
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Answer: 8
The cationic carbon carries two CH₃ groups (3 + 3 = 6 C–H bonds) and the CH₂ of the ethyl group (2 C–H bonds): 6 + 2 = 8. The terminal CH₃ of the ethyl group is β to the cationic carbon, not α, so its three hydrogens do not count; including them gives the wrong answer 11.Which of the following statements about electronic and steric effects are correct?
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Answer: B — Moving the chlorine atom of chlorobutanoic acid from C2 to C4 raises the pKa, because the inductive effect weakens with distance; D — The tert-butyl cation is more stable than the ethyl cation partly because it has more α C–H bonds that hyperconjugate with the empty p orbital
2-Chlorobutanoic acid (pKa about 2.9) is far stronger than the 4-chloro acid (about 4.5) because induction fades with distance. The tert-butyl cation has nine α C–H bonds against three in the ethyl cation, and their hyperconjugation (with the +I effect of the methyls) stabilises it. Crowding slows SN2, which needs backside approach, so tertiary halides hardly react by SN2. Phenoxide is stabilised by resonance delocalisation of its charge into the ring, not by an inductive push from the ring.4-Nitrophenol (pKa 7.15) is a markedly stronger acid than 3,5-dimethyl-4-nitrophenol, although methyl groups at the 3 and 5 positions of phenol itself change its pKa by only about 0.2 units. Which explanation accounts for this?
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Answer: B — The methyls twist the nitro group out of the ring plane, so it can no longer accept the phenoxide charge by resonance
In 4-nitrophenoxide the charge reaches the nitro oxygens only if NO₂ is coplanar with the ring. Methyl groups at C3 and C5 flank the nitro group at C4 and twist it out of plane, leaving it only its inductive pull: steric inhibition of resonance. The inductive explanation is ruled out by the stem, since the methyls by themselves shift the pKa by only about 0.2. Methyl C–H bonds do not hydrogen-bond, the OH is meta to both methyls and far from them, and the aromatic ring itself stays planar.Which of the following alkanes is chiral?
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Answer: D — 3-Methylhexane
In 3-methylhexane C3 carries H, CH₃, an ethyl group and a propyl group, four different groups, so it is a stereocentre and the molecule is chiral. In 3-methylpentane C3 carries two ethyl groups, and in 4-methylheptane C4 carries two propyl groups, so neither has a stereocentre; 4-methylheptane is the tempting choice because its longer chain looks less symmetrical. In 2-methylpentane C2 carries two methyl groups.A Fischer projection of lactic acid, CH₃CH(OH)COOH, is drawn with COOH at the top, CH₃ at the bottom, OH on the left and H on the right. What are the CIP descriptor and the D/L label of this form?
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Answer: C — (S), and it is L-lactic acid
Priorities: OH > COOH > CH₃ > H. Going OH (left) → COOH (top) → CH₃ (bottom) is clockwise as drawn, which would read R, but H sits on a horizontal bond pointing toward the viewer, so the true descriptor is the reverse: S. With the OH on the left and the most oxidised carbon at the top, the form is L. (R) is the classic slip of forgetting to reverse when the lowest-ranked group is horizontal.Hexane-1,2,3,4,5,6-hexol, HOCH₂–CH(OH)–CH(OH)–CH(OH)–CH(OH)–CH₂OH, is the parent of sugar alcohols such as sorbitol and mannitol. How many stereoisomers does it have in total, counting meso forms?
Numerical answer — type the value.
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Answer: 10
C2 to C5 are four stereocentres, so 2⁴ = 16 configurations can be written. Because the two ends of the chain are identical (CH₂OH), a configuration read from C2 to C5 is the same molecule as the matching one read from C5 to C2, which pairs most of the 16. For a symmetrical chain with an even number n of stereocentres there are 2^(n−1) = 8 chiral forms (four enantiomeric pairs, D- and L-glucitol among them) and 2^(n/2 − 1) = 2 meso forms (allitol and galactitol, each with an internal mirror plane). Total 8 + 2 = 10. Answering 16 ignores the symmetry.A solution of 0.600 g of a chiral compound in 20.0 mL of ethanol, in a 1.00 dm polarimeter tube, rotates plane-polarised light by +1.092°. The pure (+)-enantiomer has a specific rotation of +52.0 under the same conditions. Assuming rotation is proportional to composition, what percentage of the sample is the (+)-enantiomer?
Numerical answer — type the value.
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Answer: 85
c = 0.600 ÷ 20.0 = 0.0300 g/mL. [α] = α ÷ (l × c) = 1.092 ÷ (1.00 × 0.0300) = +36.4. ee = 36.4 ÷ 52.0 × 100 = 70%. The (+)-enantiomer is (100 + 70) ÷ 2 = 85% and the (−)-enantiomer 15%. Reporting the ee, 70, as the percentage of the (+)-enantiomer is the commonest slip.Which of the following statements about stereoisomers are correct?
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Answer: A — Two enantiomers have the same melting point and the same solubility in an achiral solvent; C — Diastereomers generally differ in boiling point and solubility and can be separated by ordinary chromatography; D — meso-Tartaric acid has two stereocentres but is optically inactive because it has an internal mirror plane
Enantiomers differ only toward chiral influences, so their melting points and solubilities in achiral solvents are identical. Diastereomers are not mirror images and have different physical properties, which is why they can be separated by ordinary methods. meso-Tartaric acid, (2R,3S), has two stereocentres and a mirror plane between them, so it is achiral. The R/S descriptor is assigned by naming rules and has no fixed relation to the measured sign of rotation: (R)-glyceraldehyde is dextrorotatory, but (R)-lactic acid is laevorotatory.In one isomer of 3-methylpent-2-ene, CH₃–CH=C(CH₃)–CH₂CH₃, the C1 methyl group and the methyl group attached to C3 lie on the same side of the double bond. How is this isomer designated?
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Answer: B — (E), because the higher-ranked groups, C1 on C2 and the ethyl on C3, are opposite
On C2 the groups are CH₃ (C1) and H, so CH₃ ranks higher. On C3 they are CH₃ and CH₂CH₃: both attach through carbon, but the ethyl carbon carries (C,H,H) against (H,H,H), so ethyl ranks higher. The C1 methyl is on the same side as the C3 methyl and therefore opposite the ethyl group: the higher-ranked groups are opposite, so the isomer is (E). Calling it Z because the methyls are together compares the wrong pair of groups; the H on C2 is on the ethyl side, but it is the lower-ranked group on C2. Methyl and ethyl are different groups, so geometrical isomerism does exist.Optically pure (S)-3-phenylbutan-2-one, CH₃CO–CH(C₆H₅)–CH₃, is kept in NaOD/D₂O. Its optical rotation falls to zero and deuterium is found at C3. Which explanation accounts for both observations?
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Answer: A — Base removes the C3 proton, giving a planar enolate that is deuterated from either face to give both enantiomers
C3 bears a hydrogen α to the C=O. Base removes it to give a planar, achiral enolate (keto–enol tautomerism in base), and D₂O delivers deuterium to either face with equal probability, so the stereocentre is racemised and labelled at the same time. Phenyl is not a leaving group, so no SN2 occurs, and a full inversion would give the (R) enantiomer with full rotation, not zero. A hydrate at C2 carries two identical OD groups and is not a stereocentre, and it would not put deuterium on C3. Rotation about a single bond changes conformation, never configuration.Which dimethylcyclohexane can have both methyl groups equatorial in a chair conformation?
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Answer: C — cis-1,3-Dimethylcyclohexane
Mark each equatorial bond as pointing slightly up or down. On adjacent carbons (1,2) and on 1,4-related carbons the equatorial bonds point opposite ways, so two equatorial groups there are trans, and the cis isomers must be axial–equatorial. On 1,3-related carbons both equatorial bonds point the same way, so cis-1,3 can be diequatorial, while trans-1,3, the tempting choice, is forced to be axial–equatorial.In a monosubstituted cyclohexane the equatorial conformer is more stable than the axial one by 5.0 kJ/mol (ΔG°). Taking R = 8.314 J mol⁻¹ K⁻¹ and T = 298 K, what percentage of the molecules is equatorial at equilibrium? Give the answer to the nearest integer.
Numerical answer — type the value.
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Answer: 88
K = [equatorial]/[axial] = exp(ΔG°/RT) = exp(5000 ÷ (8.314 × 298)) = exp(2.018) = 7.52. Fraction equatorial = K/(1 + K) = 7.52/8.52 = 0.8827, i.e. 88%. Using kJ instead of J in the exponent gives K ≈ 1 and 50%, and reporting K/(1 + K) as K gives nonsense above 100%.Which of the following species are aromatic?
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Answer: A — Cyclopentadienyl anion, C₅H₅⁻; C — Tropylium (cycloheptatrienyl) cation, C₇H₇⁺
The cyclopentadienyl anion has two C=C bonds (4 π electrons) plus the carbanion lone pair (2) in a planar, fully conjugated ring: 6 π, aromatic. The tropylium cation has three C=C bonds (6 π) and an empty p orbital that completes the loop: aromatic. Tub-shaped cyclooctatetraene is not planar, so its 8 π electrons are not delocalised round the ring: non-aromatic. The cyclopropenyl anion would have one C=C (2) plus a lone pair (2) = 4 π electrons, a 4n count, so it is antiaromatic, not aromatic.Imidazole (pKa of its conjugate acid about 7) is far more basic than pyrrole, although both are five-membered aromatic rings containing an N–H group. Which statement explains the difference?
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Answer: D — Its other nitrogen keeps its lone pair in an in-plane sp² orbital, outside the sextet, so protonating it keeps aromaticity
Imidazole has one pyrrole-like nitrogen, whose lone pair is in the sextet, and one pyridine-like nitrogen, whose lone pair sits in an in-plane sp² orbital and is not part of the π system. Protonating that nitrogen leaves the six π electrons untouched, and the resulting cation is further stabilised because its positive charge is shared by both nitrogens. Pyrrole has only the sextet lone pair, so protonation costs its aromaticity. Imidazole has six π electrons, not eight, it is aromatic, and its N–H nitrogen is not the site of protonation.