General Concepts and Stereochemistry: Acidity, Aromaticity, Chirality, Topicity and Conformational Analysis
1. Structure, acidity and basicity; aromatic stability
Acid strength is decided by the stability of the conjugate base. Electronegativity and hybridisation: C–H acidity rises with s-character, ethane (pKa ≈ 50) < ethylene (44) < acetylene (25). Induction falls off with distance: CF₃COOH (0.2) < ClCH₂COOH (2.9) < CH₃COOH (4.8). Resonance delocalises the charge: phenol (10) is far more acidic than cyclohexanol (16), and a p-NO₂ group lowers phenol’s pKa to 7.1 because the nitro group accepts the phenoxide charge directly; acetylacetone’s central CH₂ (9) is acidified by two carbonyls. Aromaticity can be decisive: cyclopentadiene (pKa ≈ 16) is as acidic as water because its anion is a 6π aromatic ring, while cycloheptatriene (≈ 36) gives an antiaromatic-leaning 8π anion. For substituted benzoic acids electron-withdrawing groups raise acidity (p-nitrobenzoic acid 3.4 < benzoic 4.2 < p-methoxybenzoic 4.5), and almost any ortho substituent raises it (the ortho effect: the carboxyl is twisted out of conjugation with the ring).
Basicity is quoted as the pKa of the conjugate acid (pKaH). A lone pair in an sp orbital (nitriles) is much less basic than one in sp² (pyridine, pKaH 5.2) or sp³ (piperidine, 11.1). Delocalisation lowers basicity: aniline (4.6) is a far weaker base than cyclohexylamine (10.6), and in pyrrole the nitrogen lone pair is part of the aromatic sextet, so pyrrole is barely basic (pKaH ≈ −3.8, protonated on carbon). Delocalisation of the protonated form raises it: guanidine (13.6) and amidines such as DBU are strong neutral bases because their conjugate acids share the charge over three or two nitrogens. In water, methylamines are ordered Me₂NH > MeNH₂ > Me₃N > NH₃ because solvation of the ammonium ion competes with the inductive effect; in the gas phase the order is simply Me₃N > Me₂NH > MeNH₂ > NH₃.
Aromatic stability in physical properties. Benzene’s heat of hydrogenation (−208 kJ/mol) is about 152 kJ/mol less than three times that of cyclohexene (−120 each), the classic measure of its resonance energy. The aromatic ring current deshields protons outside the ring (benzene δ 7.3) and strongly shields those inside, as in [18]annulene (inner H near δ −3). Charge separation that creates aromatic rings produces large dipole moments: azulene (1.0 D, blue) behaves as a cyclopentadienide fused to a tropylium, and tropone and cyclopropenone are unusually polar and basic at oxygen because their protonated forms are aromatic cations.
2. Chirality, configuration, optical purity and relative stereochemistry
A molecule is chiral when it is not superimposable on its mirror image, which is the same as having no improper axis Sₙ. The commonest cause is a stereocentre (a carbon with four different groups), labelled R or S by the Cahn–Ingold–Prelog rules: rank the substituents by atomic number (at the first point of difference, duplicating atoms for multiple bonds), view with the lowest-ranked group pointing away, and read 1 → 2 → 3 clockwise as R. Chirality without a stereocentre is examined too: axial chirality in allenes abC=C=Cab and in biaryls with bulky ortho groups whose rotation is hindered — atropisomers such as 6,6′-dinitrobiphenyl-2,2′-dicarboxylic acid and BINOL, separable when the rotation barrier exceeds about 90 kJ/mol; planar chirality in trans-cyclooctene and substituted ferrocenes; and helical chirality in hexahelicene.
Enantiomers rotate plane-polarised light equally and oppositely. The specific rotation [α]ᵀ_λ = α/(l·c), with the observed rotation α in degrees, the path length l in decimetres and c in g/mL. The composition of a mixture is expressed as enantiomeric excess, ee = ([major] − [minor])/([major] + [minor]) × 100%, which for an ideal sample equals the optical purity [α]_obs/[α]_pure × 100%. So 80% ee is a 90:10 mixture, and a racemate (0% ee) is optically inactive. With several stereocentres, n centres allow at most 2ⁿ stereoisomers, fewer when meso forms (achiral, with an internal mirror plane) exist: 2,3-dibromobutane has three (a pair of enantiomers and a meso form), tartaric acid three. Diastereomers — stereoisomers that are not mirror images — have different physical properties; relative configuration is described as erythro/threo or syn/anti on a zig-zag chain, and E/Z (CIP ranking) for alkenes.
3. Homotopic, enantiotopic and diastereotopic groups and faces
Test two apparently identical groups by replacing each in turn with a new group. If the two products are identical, the groups are homotopic (the protons of CH₂Cl₂, related by a C₂ axis); if they are enantiomers, the groups are enantiotopic (the CH₂ protons of ethanol, related only by a mirror plane); if they are diastereomers, the groups are diastereotopic (the CH₂ protons of 2-butanol, or the two methyls of an isopropyl group attached to a stereocentre). Homotopic and enantiotopic nuclei are equivalent in NMR in an achiral solvent; diastereotopic nuclei are chemically inequivalent and can have different shifts and couple to each other. Enzymes, being chiral, distinguish enantiotopic groups: in the oxidation of ethanol by alcohol dehydrogenase only one CH₂ hydrogen is removed.
The same classification applies to the two faces of a trigonal centre. The faces of acetone’s C=O are homotopic; those of acetaldehyde are enantiotopic, labelled Re and Si by the CIP order of the three substituents seen from each face (clockwise = Re); addition of a nucleophile to one face gives one enantiomer. A carbonyl next to an existing stereocentre has diastereotopic faces, and attack on them gives diastereomers in unequal amounts — the basis of the Cram and Felkin–Anh models of the synthesis chapter.
4. Conformational analysis of acyclic and cyclic compounds
Rotation about C–C single bonds interconverts conformations. In ethane the eclipsed form lies about 12 kJ/mol above the staggered, mostly from torsional strain. In butane the anti conformer is lowest, the gauche about 3.8 kJ/mol higher (a steric CH₃/CH₃ interaction), and the fully eclipsed form about 19 kJ/mol above anti. Heteroatoms change the preferences: in 1,2-difluoroethane and in ethylene glycol the gauche form is favoured (the gauche effect and intramolecular hydrogen bonding).
Cyclohexane adopts a strain-free chair in which every C–C bond is staggered; the boat is about 29 kJ/mol higher and the twist-boat about 23. Each carbon has one axial and one equatorial position, and ring flipping exchanges them. A substituent prefers the equatorial position, avoiding 1,3-diaxial interactions; the free-energy preference is its A-value: about 7.3 kJ/mol for CH₃ (95% equatorial at 298 K), 9 for i-Pr and more than 20 for t-Bu, which effectively locks the ring. For disubstituted rings: trans-1,2 and cis-1,3 and trans-1,4 can put both groups equatorial, whereas cis-1,2, trans-1,3 and cis-1,4 must have one axial. trans-Decalin is rigid and cannot flip; cis-decalin flips. In pyranoses an electronegative group at the anomeric carbon prefers the axial position — the anomeric effect (n_O → σ*C–X).
5. Configuration, conformation and neighbouring groups in reactivity; stereospecific and stereoselective reactions
Conformation controls reactivity when a reaction has strict geometric demands. E2 elimination needs the H and the leaving group anti-periplanar, which in cyclohexanes means both axial. Neomenthyl chloride, whose Cl is axial in the favoured chair, eliminates fast and gives mainly the more substituted 3-menthene (Zaitsev); menthyl chloride must first flip to put Cl axial, reacts about 200 times more slowly, and then has only one anti-periplanar hydrogen, so it gives only 2-menthene. In rigid systems, axial and equatorial isomers differ in S_N2, oxidation and ester-hydrolysis rates.
Neighbouring-group participation (anchimeric assistance) is internal nucleophilic attack by a group positioned anti to the leaving group. It shows as a rate enhancement and as retention of configuration, the result of two successive inversions. trans-2-Acetoxycyclohexyl tosylate solvolyses several hundred times faster than the cis isomer and gives the trans product through a bridging acetoxonium ion; a β-aryl group forms a phenonium ion; and a suitably placed C=C can accelerate ionisation enormously — anti-7-norbornenyl tosylate reacts about 10¹¹ times faster than the saturated 7-norbornyl ester.
A reaction is stereospecific when stereoisomeric starting materials give stereoisomerically different products because of the mechanism: anti addition of Br₂ converts cis-2-butene into racemic (2R,3R)/(2S,3S)-2,3-dibromobutane and trans-2-butene into the meso compound; S_N2 inverts; syn-dihydroxylation with OsO₄ turns a cis-alkene into a meso or erythro diol. A reaction is stereoselective when one stereoisomer forms in preference to others that could form: reduction of 4-tert-butylcyclohexanone by LiAlH₄ gives about 90% of the trans (equatorial) alcohol by axial attack, while the bulky L-selectride attacks equatorially and gives mainly the cis (axial) alcohol.
Key takeaways
- Acidity follows the stability of the conjugate base: s-character, induction, resonance and aromaticity (cyclopentadiene pKa ≈ 16); pyrrole is barely basic because its lone pair is aromatic, guanidine very basic because its cation is delocalised.
- Chirality means no Sₙ; allenes, hindered biaryls (atropisomers) and helicenes are chiral without stereocentres. [α] = α/(l c) with l in dm; ee = optical purity; 80% ee = 90:10.
- Replacement test: identical products → homotopic, enantiomers → enantiotopic, diastereomers → diastereotopic; only diastereotopic nuclei differ in NMR in an achiral solvent; Re/Si name the faces.
- Butane gauche is 3.8 kJ/mol above anti; cyclohexane chair with A-values (CH₃ 7.3 kJ/mol → 95% equatorial, t-Bu locks); trans-decalin cannot flip.
- E2 needs anti-periplanar H and leaving group (menthyl vs neomenthyl chloride); NGP gives rate enhancement with retention; Br₂ + cis-2-butene → racemate, + trans → meso (stereospecific).
Practice questions (16)
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.
A solution of sucrose containing 0.050 g/mL in a 2.0 dm tube shows an observed rotation of +6.65°. What is the specific rotation, in degrees, to one decimal place?
Numerical answer — type the value.
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Answer: 66.5
[α] = α/(l·c) = 6.65/(2.0 × 0.050) = 66.5. The path length must be in decimetres and c in g/mL; using l = 20 cm gives 6.65, and c in g/100 mL gives 0.665.A sample of a chiral compound has [α] = +39.9°, while the pure (+) enantiomer has [α] = +66.5°. What is its enantiomeric excess, in per cent?
Numerical answer — type the value.
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Answer: 60
Optical purity = 39.9/66.5 × 100 = 60%, which equals the ee if rotation is linear in composition. The mixture is then 80:20 (+):(−); answering 80 gives the percentage of the major enantiomer, not the excess.A product is obtained in 80% ee. What percentage of the mixture is the major enantiomer?
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Answer: 90
ee = major − minor with major + minor = 100: major − minor = 80 and major + minor = 100 give major = 90%, minor = 10%. Reading 80% ee as 80% major is the classic slip.How many stereoisomers does 2,3-dibromobutane have?
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Answer: 3
Two stereocentres allow at most 2² = 4, but (2R,3S) and (2S,3R) are the same meso compound with an internal mirror plane, leaving (2R,3R), (2S,3S) and meso: 3. Answering 4 ignores the meso form.How many stereoisomers does 2,3,4-trihydroxypentanedioic acid, HOOC–CH(OH)–CH(OH)–CH(OH)–COOH, have?
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Answer: 4
C2 and C4 are stereocentres and C3 is pseudoasymmetric. When C2 and C4 have the same configuration (R,R or S,S) C3 is not stereogenic and the two forms are a pair of enantiomers; when they differ, C3 can be r or s, giving two meso forms. Total 2 + 2 = 4, not 2³ = 8.Which of the following is chiral although it has no stereogenic carbon atom?
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Answer: A — 6,6′-Dinitrobiphenyl-2,2′-dicarboxylic acid
Four bulky ortho groups prevent rotation about the aryl–aryl bond, and the twisted biaryl has no plane or centre of symmetry: separable atropisomers. Biphenyl rotates freely; an allene needs different groups on each end (Cl₂C= has two identical ones); meso-tartaric acid has an internal mirror plane.Which pairs of protons are diastereotopic?
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Answer: A — The two CH₂ protons of 2-butanol; D — The two CH₂ protons of (S)-2-chlorobutane
A CH₂ next to a stereocentre (2-butanol, 2-chlorobutane) has diastereotopic protons: replacing each gives diastereomers, and they can show different NMR shifts. Ethanol’s CH₂ protons are enantiotopic (mirror-related, NMR-equivalent in an achiral solvent) and CH₂Cl₂’s are homotopic.The two faces of the carbonyl group of acetaldehyde, CH₃CHO, are
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Answer: A — enantiotopic (Re and Si)
The carbonyl carbon carries O, CH₃ and H, three different groups, so the two faces are mirror images: nucleophilic addition to Re or Si gives the two enantiomers of a secondary alcohol. Acetone’s faces are homotopic (two identical CH₃), and diastereotopic faces need a stereocentre already present.The A-value of a methyl group on cyclohexane is 7.3 kJ/mol. What percentage of methylcyclohexane has the methyl group equatorial at 298 K, to the nearest whole number? (R = 8.314 J mol⁻¹ K⁻¹)
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Answer: 95
K = [eq]/[ax] = exp(7300/(8.314 × 298)) = exp(2.946) = 19.0, so the equatorial fraction is 19.0/20.0 = 0.95, i.e. 95%. Taking K itself as the percentage, or forgetting to convert kJ to J, are the usual errors.Which disubstituted cyclohexane can have both substituents equatorial in one chair conformation?
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Answer: A — trans-1,4-Dimethylcyclohexane
On a chair, trans-1,2, cis-1,3 and trans-1,4 arrangements can be diequatorial, while cis-1,2, trans-1,3 and cis-1,4 always have one axial and one equatorial group. So trans-1,4 is the diequatorial isomer and the more stable of its pair.The heat of hydrogenation of cyclohexene is −120 kJ/mol and that of benzene −208 kJ/mol. Using these, estimate the resonance (aromatic stabilisation) energy of benzene, in kJ/mol.
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Answer: 152
Three isolated double bonds would release 3 × 120 = 360 kJ/mol; benzene releases only 208, so it is 360 − 208 = 152 kJ/mol more stable than a hypothetical cyclohexatriene. Using one double bond (120 − 208) gives a meaningless negative value.Which compound is the strongest acid?
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Answer: A — Cyclopentadiene
Deprotonating cyclopentadiene gives the aromatic 6π cyclopentadienide, so its pKa is about 16. Cycloheptatriene’s anion has 8π electrons and gains no aromatic stabilisation (pKa ≈ 36); propene’s allylic anion is only resonance-stabilised (≈ 43); cyclopentane has none (≈ 50).Arrange in order of increasing basicity (pKaH) in water: aniline, pyridine, pyrrole, piperidine.
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Answer: A — pyrrole < aniline < pyridine < piperidine
Pyrrole’s lone pair is in the aromatic sextet (pKaH ≈ −3.8); aniline’s is delocalised into the ring (4.6); pyridine’s sp² lone pair is free but s-rich (5.2); piperidine’s sp³ lone pair is fully available (11.1). The trap is placing pyridine below aniline because both are aromatic amines.Menthyl chloride and neomenthyl chloride differ only in the configuration at the C–Cl carbon. On E2 elimination with ethoxide, menthyl chloride gives only 2-menthene, slowly. Why?
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Answer: A — Its Cl is equatorial in the stable chair; after flipping to make Cl axial, only one β-hydrogen is anti-periplanar
With all three substituents equatorial, Cl in menthyl chloride cannot be anti-periplanar to any H; the ring must flip to a strained all-axial conformer, which slows the reaction, and in it only the C2 hydrogen is axial and anti. Neomenthyl chloride already has Cl axial with two anti H, giving mainly Zaitsev 3-menthene quickly.Addition of Br₂ to trans-2-butene gives
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Answer: A — meso-2,3-dibromobutane
Bromination goes through a bromonium ion opened anti, so the reaction is stereospecific: trans-2-butene gives the meso dibromide and cis-2-butene the racemate. An achiral alkene and reagent cannot give a single enantiomer.trans-2-Acetoxycyclohexyl tosylate undergoes acetolysis much faster than its cis isomer and gives trans-1,2-diacetoxycyclohexane. This shows
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Answer: A — neighbouring-group participation by the anti acetoxy group through an acetoxonium ion, with overall retention
Only in the trans isomer can the acetoxy oxygen reach the back of the C–OTs bond; it displaces tosylate (first inversion), and acetate opens the bridged acetoxonium ion from the back (second inversion), so configuration is retained and the rate enhanced. An S_N2 by solvent would give cis product; a free cation would give mixtures.