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Química Orgânica Física · Orbitals, conformation and selectivity

Efeitos estereoeletrônicos e controle conformacional

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O que você vai aprender

  • Explain hyperconjugation, the gauche effect and the anomeric effect as donation from a filled orbital into an aligned, anti-periplanar empty orbital, and predict the conformations and spectroscopic signatures that follow.
  • Use A-values, 1,3-diaxial and allylic (A1,3) strain to calculate conformational equilibria and to predict which chair or rotamer a molecule adopts.
  • Apply the anti-periplanar requirement to E2 eliminations, epoxide formation and opening, and 1,2-shifts, and recognise reactions that instead need syn geometry.
  • Predict the face of attack on a carbonyl group from the Bürgi–Dunitz trajectory, axial versus equatorial approach, the Felkin–Anh model and chelation control.
  • Classify ring closures with Baldwin's rules and use them, with ring strain, to predict the ring size formed.

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Hyperconjugation: filled orbitals feed empty ones

A filled bonding orbital (σC–H, σC–C, σC–Si) or a lone pair can donate into a nearby empty orbital: an empty p orbital, a π* orbital or a σ* orbital. The interaction is strongest when the two orbitals are parallel and in the same plane, which for two σ bonds on neighbouring atoms means anti-periplanar (dihedral angle 180°). Hyperconjugation explains why the tert-butyl cation (nine C–H bonds next to the empty p orbital) is far more stable than the methyl cation (none). It also explains the gauche effect: 1,2-difluoroethane prefers the gauche conformer, because only there is each good donor (σC–H) anti to a good acceptor (σ*C–F). Silicon makes the best common σ donor: a C–Si bond anti-periplanar to a leaving group speeds ionisation by many orders of magnitude (the β-silicon effect).

The anomeric effect

In a tetrahydropyran (oxane), an electronegative group on C2, the carbon next to the ring oxygen, often prefers the axial position even though sterics favour equatorial. When the C2–X bond is axial, a lone pair on the ring oxygen is anti-periplanar to it and donates into σ*C–X (n → σ*). The axial preference grows with the acceptor quality of C–X: 2-chlorooxane is almost entirely axial, 2-methoxyoxane is modestly axial in nonpolar solvents, and 2-methyloxane (a poor acceptor) is equatorial like any alkyl group. In water, glucose is still 64% β (anomeric OH equatorial) because hydrogen bonding shrinks the effect, but the 36% α is far more than the steric preference of an OH alone would allow. The same donation weakens an anti-periplanar C–H bond: an axial anomeric C–H shows a smaller one-bond ¹³C–¹H coupling (about 160 Hz) than an equatorial one (about 170 Hz). Spiroketals adopt the conformation in which each C–O bond is axial on the other ring, gaining two anomeric interactions.

Conformational energies: A-values and allylic strain

The A-value of a substituent is ΔG° for moving it from the equatorial to the axial position of a cyclohexane; an axial group suffers two gauche-butane-like 1,3-diaxial interactions with the ring. Typical A-values (kJ/mol): F 1.0, OH about 3.6, CH₃ 7.3, CH₂CH₃ 7.5, CH(CH₃)₂ 9.2, C₆H₅ 11.7 and C(CH₃)₃ about 20. Isopropyl can turn its C–H toward the ring, so it costs little more than ethyl; tert-butyl must point a methyl inward and effectively locks the ring with the tert-butyl equatorial. The equilibrium constant follows from ΔG° = −RT ln K: 7.3 kJ/mol at 298 K gives K ≈ 19, about 95% equatorial methylcyclohexane. A-values are roughly additive when groups are 1,3 or 1,4 apart. Allylic 1,3-strain (A1,3) is the clash between a substituent on the far alkene carbon (cis) and a group on the allylic carbon when both lie in the alkene plane; the allylic carbon therefore rotates its smallest group, usually H, into that plane. Amides behave like alkenes here, so an N-acyl-2-alkylpiperidine puts its 2-alkyl group axial.

Anti-periplanar requirements

Reactions that break one σ bond while a neighbouring pair moves in need the breaking σ* orbital aligned with the incoming electrons. In an E2 reaction the C–H and C–leaving-group bonds must be anti-periplanar; in a cyclohexane that means both axial (trans-diaxial). The geometry fixes the product: meso-1,2-dibromo-1,2-diphenylethane eliminates HBr to (E)-1-bromo-1,2-diphenylethene, the racemic diastereomer to the Z alkene. The same rule governs intramolecular SN2 and 1,2-shifts: trans-2-chlorocyclohexanol closes to cyclohexene oxide because its alkoxide can sit anti to C–Cl, whereas the cis isomer cannot, and instead the axial C–H anti to the axial C–Cl migrates as a hydride to give cyclohexanone. Epoxides on cyclohexane rings open to trans-diaxial products (the Fürst–Plattner rule). Bridgehead atoms of small bicycles cannot align the required orbitals (Bredt's rule). Thermal Ei eliminations (Cope, selenoxide) are the exception: they pass through a cyclic transition state and need syn-periplanar geometry.

Which face of a carbonyl?

Nucleophiles approach a carbonyl carbon at about 107° to the C=O bond (the Bürgi–Dunitz angle), not at 90°, because they must overlap with π* while avoiding the filled π orbital and oxygen. On a cyclohexanone, small nucleophiles such as LiAlH₄ attack axially (giving the equatorial alcohol, trans-4-tert-butylcyclohexanol, about 9 : 1), because the axial path avoids torsional strain with the neighbouring axial C–H bonds; bulky L-Selectride attacks equatorially to avoid the 3,5-axial hydrogens and gives the axial alcohol. For an acyclic carbonyl next to a stereocentre, the Felkin–Anh model places the largest (or most electronegative) α-substituent perpendicular to the carbonyl, anti to the incoming nucleophile, which then passes the smallest group. When the α-substituent is an alkoxy group and the metal can bind both oxygens (Mg²⁺, Zn²⁺, Ti⁴⁺), a five-membered chelate locks C=O and C–OR syn, and the nucleophile attacks from the face of the small group, giving the opposite (anti-Felkin) diastereomer. Bulky silyl ethers and monodentate Lewis acids such as BF₃ prevent chelation.

Baldwin's rules for ring closure

Name a ring closure by three labels: the ring size formed, exo or endo (whether the bond that breaks lies outside or inside the new ring) and the geometry of the attacked atom (tet for sp³, trig for sp², dig for sp). Closures are favoured when the nucleophile can reach the electrophile along the trajectory the orbitals demand: backside (180°) for tet, the Bürgi–Dunitz angle for trig, and about 120° for dig. Favoured: 3- to 7-exo-tet, 3- to 7-exo-trig, 6- and 7-endo-trig, 5- to 7-exo-dig and 3- to 7-endo-dig. Disfavoured: 5- and 6-endo-tet, 3- to 5-endo-trig and 3- and 4-exo-dig. 5-Endo-trig fails because the chain is too short to bring the nucleophile over the π face, whereas 5-endo-dig works because an alkyne offers a second π* in the plane of the ring. Within the favoured closures, five-membered rings usually form fastest, so pent-4-enoic acid with iodine gives a γ-lactone (5-exo-trig) rather than a δ-lactone.

What to carry forward

Most stereoelectronic effects are one idea: a filled orbital donates best into an empty orbital that is parallel to it, which for neighbouring σ bonds means anti-periplanar. That idea explains conformational preferences (gauche and anomeric effects), the geometric demands of E2, epoxide chemistry and 1,2-shifts, the angle and face of nucleophilic attack on carbonyls, and Baldwin's ring-closure rules. Steric costs (A-values, 1,3-diaxial and allylic strain) then decide which reactive conformer is available.

Problemas de prática (40)

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Desenhe o mecanismo 8

Attack on carbonyl faces

  • Chelation-controlled addition of methylmagnesium bromide to an α-methoxy ketoneUse the Cram chelate model to predict the diastereomer formed when methylmagnesium bromide adds to (S)-2-methoxy-1-phenylpropan-1-one, and draw the addition and the ammonium workup.
  • Felkin–Anh addition of methylmagnesium bromide to (S)-2-phenylpropanalUse the Felkin–Anh model to choose the face of (S)-2-phenylpropanal that methylmagnesium bromide attacks, then draw the addition and workup that give (2S,3S)-3-phenylbutan-2-ol as the major diastereomer.
  • Axial hydride attack: LiAlH₄ reduction of 4-tert-butylcyclohexanoneDraw hydride delivery to 4-tert-butylcyclohexanone from the axial face and the workup that gives the equatorial alcohol, trans-4-tert-butylcyclohexan-1-ol.

Anti-periplanar requirements

  • cis-2-Chlorocyclohexan-1-ol gives cyclohexanone by an anti-periplanar hydride shiftExplain why cis-2-chlorocyclohexan-1-ol cannot close to cyclohexene oxide in base, and draw the anti-periplanar 1,2-hydride shift that converts its alkoxide into cyclohexanone.
  • Anti E2 of meso-stilbene dibromide gives (E)-1-bromo-1,2-diphenyletheneUse the anti-periplanar requirement of E2 to draw the single elimination of HBr from meso-stilbene dibromide and to predict the geometry of the bromoalkene it gives.

Baldwin's rules for ring closure

  • Iodolactonisation of pent-4-enoic acid: a 5-exo-trig ring closureDraw carboxylate formation, iodonium ion formation and the 5-exo ring closure that gives 5-(iodomethyl)oxolan-2-one, and classify the closure with Baldwin's rules.
  • Intramolecular oxa-Michael addition: a 5-exo-trig ring closureDraw the base-catalysed conjugate addition of a tethered alkoxide to an α,β-unsaturated ester that closes a five-membered ring, and classify the closure with Baldwin's rules.

Hyperconjugation and the anomeric effect

  • Acid-catalysed spiroketal formation from 1,9-dihydroxynonan-5-oneDraw the acid-catalysed formation of 1,7-dioxaspiro[5.5]undecane through a hemiketal and an oxocarbenium ion, and explain with the anomeric effect which conformation the spiroketal adopts.

Preveja o produto 7

Conformational energies and allylic strain

  • Allylic strain picks the rotamerPredict the preferred rotamer about an allylic bond from allylic 1,3-strain with a cis alkene substituent.
  • Which isomer holds its OH axial?Use a conformationally locking tert-butyl group and cis/trans relationships to decide whether a ring hydroxyl is axial or equatorial.

Anti-periplanar requirements

  • Why more (E)- than (Z)-but-2-ene?Predict the major alkene stereoisomer from E2 of a secondary bromide by comparing the anti-periplanar conformers that lead to each.
  • E2 of the racemic stilbene dibromidePredict the geometry of the alkene from anti-periplanar E2 of an acyclic diastereomer with two stereocentres.
  • E2 of trans-1-bromo-2-methylcyclohexanePredict the E2 product of a trans-1,2-disubstituted cyclohexyl bromide by finding which β-hydrogen can be trans-diaxial to bromine.

Attack on carbonyl faces

  • A bulky hydride on a locked cyclohexanonePredict the alcohol diastereomer formed when a bulky hydride reagent attacks a locked cyclohexanone equatorially.

Hyperconjugation and the anomeric effect

  • Which of these prefers an axial substituent?Recognise the structural requirement for the anomeric effect: an electronegative substituent on the carbon next to a ring oxygen.

Classificar 5

Hyperconjugation and the anomeric effect

  • What stabilises an axial 2-chloro group?Identify the orbital interaction responsible for the anomeric effect in a 2-substituted oxane.

Baldwin's rules for ring closure

  • Classify an intramolecular Williamson ring closureAssign ring size, exo/endo and tet/trig/dig to an intramolecular SN2 ring closure.
  • Which ring closure is disfavoured?Identify a disfavoured ring closure among several classified closures using Baldwin's rules.

Attack on carbonyl faces

  • The Bürgi–Dunitz trajectoryIdentify the angle at which a nucleophile approaches a carbonyl carbon and the orbital reason for it.

Anti-periplanar requirements

  • Which elimination must be syn?Classify eliminations by their geometric requirement and identify the thermal Ei reaction that needs syn-periplanar geometry.

Comparar 2

Conformational energies and allylic strain

  • Rank substituents by A-valueRank common substituents by their equatorial preference on a cyclohexane ring.

Hyperconjugation and the anomeric effect

  • Rank 2-substituted oxanes by axial preferenceRank 2-substituted oxanes by the strength of their anomeric preference from the acceptor quality of the C2–X bond.

Calcular 5

Conformational energies and allylic strain

  • Chair preference of cis-1-methyl-4-phenylcyclohexaneUse additive A-values to calculate the free-energy difference between the two chairs of a cis-1,4-disubstituted cyclohexane.
  • Estimate the methyl A-value from gauche interactionsEstimate the A-value of a methyl group by counting gauche-butane interactions in the axial conformer.
  • Equatorial : axial ratio for methylcyclohexaneConvert an A-value into a conformational equilibrium constant with ΔG° = −RT ln K.

Attack on carbonyl faces

  • From ΔΔG‡ to diastereoselectivity at −78 °CConvert a difference in transition-state free energies into the percentage of the major diastereomer at low temperature.

Hyperconjugation and the anomeric effect

  • How much is β-glucose favoured in water?Calculate the free-energy difference between the glucopyranose anomers from their equilibrium ratio and relate it to the anomeric effect.

Encontre o erro 1

  • A bridgehead elimination that cannot happenDiagnose an impossible bridgehead E2 using the requirement for parallel p orbitals in the product alkene (Bredt's rule).

Interprete as evidências 1

  • Reading the anomer from a C–H couplingInterpret one-bond C–H coupling constants at an anomeric carbon as evidence for n → σ* donation into an axial C–H bond.

Trabalhe de trás para frente 1

  • Which alkenol gives 2-(iodomethyl)oxolane?Work backwards from an iodoether to the alkenol whose 5-exo closure forms it.

Planeje uma síntese 1

  • Choosing conditions for chelation controlPlan the protecting group and stoichiometry that allow chelation-controlled Grignard addition to an α-oxygenated aldehyde.

Explique por quê 9

Conformational energies and allylic strain

  • An axial methyl next to an amide nitrogenExplain why an N-acyl-2-methylpiperidine prefers the methyl axial, using allylic 1,3-strain across the amide C–N bond.
  • Why does tert-butyl jump so far?Explain the jump in A-value from isopropyl to tert-butyl by the rotamers available to an axial alkyl group.

Hyperconjugation and the anomeric effect

  • A silicon atom that speeds up ionisationExplain the β-silicon effect as σC–Si hyperconjugation that requires anti-periplanar alignment with the breaking C–O bond.
  • The gauche effect in 1,2-difluoroethaneExplain the gauche preference of 1,2-difluoroethane by pairing the best σ donors anti-periplanar to the best σ* acceptors.
  • Why is the tert-butyl cation so much more stable than CH₃⁺?Explain carbocation stabilisation by alkyl groups as hyperconjugation from aligned C–H σ bonds into the empty p orbital.

Baldwin's rules for ring closure

  • Why is 5-endo-dig allowed when 5-endo-trig is not?Explain, from the orbitals of alkenes and alkynes, why 5-endo-dig closures are favoured while 5-endo-trig closures are disfavoured.

Anti-periplanar requirements

  • Why does a methyl group refuse to move intramolecularly?Explain the endocyclic restriction test result from the collinear geometry an SN2 transition state requires.
  • The trans-diaxial rule for epoxide openingExplain the regioselectivity of nucleophilic opening of a conformationally locked cyclohexene oxide with the Fürst–Plattner rule.

Attack on carbonyl faces

  • Which group goes anti in an α-chloro aldehyde?Apply the polar Felkin–Anh model, in which an electronegative α-substituent is placed anti to the incoming nucleophile.

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