JWB Aromatic 2013.pptx PDF
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Aromatic and Heterocyclic Chemistry 1 Aromatic and Heterocyclic Chemistry 8 Lectures, Trinity 2013 [email protected] handout at: http://burton.chem.ox.ac.uk/teaching.html ◾ Advanced Organic Chemistry: Parts A and B, Francis A. Carey, Richard J. Sundberg ◾ Organic Chemistry, Jonathan Clayden, Nick Greeves, Stuart Warren, Peter Wothers ◾ Advanced Organic Chemistry: Reac7ons, Mechanisms and Structures; J. March ◾ Fron7er Orbitals and Organic Chemical Reac7ons; I. Fleming ◾ Heterocyclic Chemistry; J. Joule, K. Mills, G. Smith ◾ Aroma7c Heterocyclic Chemistry; D. Davies ◾ Reac7ve Intermediates; C. Moody and G. Whitham ◾ Aroma7c Chemistry; M. Sainsbury Aromatic and Heterocyclic Chemistry 2 Synopsis ◾ The Origin of AromaGcity and General CharacterisGcs of AromaGc Compounds ◾ Examples of AromaGcity ◾ Electrophilic AromaGc SubsGtuGon ◾ Nucleophilic AromaGc SubsGtuGon ◾ Arynes ◾ ReacGons with Metals: Ortho LithiaGon, Palladium Catalysed Couplings, Chromium Complexes and Birch ReducGons ◾ IntroducGon of FuncGonal Groups ◾ Pyridine: Synthesis and ReacGons ◾ Pyrrole, Thiophene and Furan: Synthesis and ReacGons ◾ Indole: Synthesis and ReacGons ◾ ReducGon of AromaGcs ◾ AromaGc Chemistry in AcGon Aromatic and Heterocyclic Chemistry 3 Origin of Aroma0city ◾ Typical reacGons of alkenes ◾ Typical reacGons of benzene ◾ retains aromaGc sextet of electrons in subsGtuGon reacGons ◾ does not behave like a “normal” polyene or alkene ◾ benzene is both kine7cally and thermodynamically very stable ◾ Heats of hydrogenaGon ΔHo = -‐120 kJmol-‐1 ΔHo = 3 x -‐120 = -‐360 kJmol-‐1 ΔHo = -‐210 kJmol-‐1 hydrog hydrog hydrog (hypotheGcal, 1,3,5-‐cyclohexatriene) ◾ benzene ≈150 kJmol-‐1 more stable than expected – (represents stability over hypotheGcal 1,3,5-‐ cyclohextriene) – termed the empirical resonance energy (values vary enormously) ◾ empirical reso nance energy is not the true resonance energy as this would be the difference between benzene and a symmetrical non-‐delocalised cyclohexatriene unit which does not exist ◾ we know that delocalisaGon is stabilising, but how much more stabilising is the delocalisaGon in benzene – should compare benzene with a real molecule – we will use 1,3,5-‐hexatriene ◾ require a theory which explains the stability of benzene Aromatic and Heterocyclic Chemistry 4 Understanding Aroma0city ◾ Hückel’s Rule: planar, monocyclic, completely conjugated hydrocarbons will be aroma%c when the ring contains (4n +2) π-‐electrons (n = 0, 1, 2….posi7ve integers) Corollary ◾ planar, monocyclic, completely conjugated hydrocarbons will be an%-‐aroma%c when the ring contains (4n) π-‐ electrons (n = 0, 1, 2….posi7ve integers) Hückel Molecular Orbital Theory (HMOT) ◾ applicable to conjugated planar cyclic and acyclic systems ◾ only the π-‐system is included; the σ-‐framework is ignored (in reality σ-‐framework affects π-‐system) ◾ used to calculate the wave funcGons (ψ) and hence rela7ve energies by the LCAO method i i.e. ψ = c φ + c φ + c φ + c φ + c φ ….. i 1 1 2 2 3 3 4 4 5 5 ◾ HMOT solves energy (E) and coefficients c i i ◾ there are now many more sophisGcated methods for calculaGng the stabilisaGon energy in conjugated systems; however, HMOT is adequate for our purposes. Aromatic and Heterocyclic Chemistry 5 Understanding Aroma0city HMOT in Ac0on ◾ For cyclic and acyclic systems: Molecular Orbital Energies = E = α + mβ i j ◾ α = coulomb integral -‐ energy associated with electron in an isolated 2p orbital (albeit in the molecular environment) – α is negaGve (stabilising) and is the same for any p-‐orbital in π-‐system ◾ β = resonance integral – energy associated with having electrons shared by atoms in the form of a covalent bond – β is negaGve (stabilising) and is set to zero for non-‐adjacent atoms. ◾ (all overlap integrals S assumed to be zero, electron correlaGon ignored) ◾ linear polyenes m = 2cos[jπ/(n+1)] j = 1, 2……n (n = number of carbon atoms in j conjugated system) ◾ cyclic polyenes m = 2cos(2jπ/n) j = 0, ±1, ±2……±[(n-‐1)/2] for odd n, ±n/2 for even n j ◾ 1,3,5-‐Hexatriene vs Benzene – linear polyene ◾ six 2p atomic orbitals give 6π molecular orbitals; n = 6, j = 1, 2, 3, 4, 5, 6 ◾ m = 2cos(π/7) = 1.80 2cos(2π/7) = 1.25 2cos(3π/7) = 0.45...... j and the corresponding negaGve values ◾ energy E = α + 1.80β α + 1.25β α + 0.45β α – 0.45β…..etc Aromatic and Heterocyclic Chemistry 6 ◾ 1,3,5-‐Hexatriene vs Benzene ◾ six 2p atomic orbitals give 6π molecular orbitals; n = 6, j = 1, 2, 3, 4, 5, 6 ◾ m = 2cos(π/7) = 1.80 2cos(2π/7) = 1.25 2cos(3π/7) = 0.45...... j and the corresponding negaGve values ◾ energy E = α + 1.80β α + 1.25β α + 0.45β α – 0.45β…..etc MO no. nodes energy HMOT MO (calculated) ψ 5 α -‐ 1.80β 6 ψ 4 α -‐ 1.25β 5 ψ 3 α -‐ 0.45β 4 α ψ 2 α + 0.45β 3 ψ 1 α + 1.25β 2 ψ 0 α + 1.80β 1 ◾ stabilisaGon energy = E = 2(3α + 3.5β) = 6α + 7β stab Aromatic and Heterocyclic Chemistry 7 ◾ Benzene m = 2cos(2jπ/n) j = 0 ±1, ±[(n-‐1)/2] for odd n; ±n/2 for even n j ◾ six 2p atomic orbitals give 6π molecular orbitals; n = 6, j = 0 ±1, ±2, ±3 ◾ m = 2cos(0) = 2, 2cos(±2π/6) = 1, 2cos(±4π/6) = -‐1, 2cos(±6π/6)= -‐2 j ◾ energy E = α + 2β, α + β, α -‐ β, α – 2β MO no. nodes energy HMOT MO (calculated) ψ 6 α -‐ 2β 6 ψ 4 α -‐ β 4,5 α ψ 2 α + β 2,3 ψ 0 α + 2β 1 ◾ stabilisaGon energy = E = 2(α + 2β) + 4(α + β) = 6α + 8β; stabilisaGon energy w.r.t. 1,3,5-‐hexatriene = β stab ◾ HMOT predicts benzene is more stable than 1,3,5-‐hexatriene ◾ aroma7c compounds are those with a π-‐system lower in energy than that of acyclic counterpart Aromatic and Heterocyclic Chemistry 8 ◾ Cyclobutadiene vs 1,3-‐butadiene ◾ four 2p atomic orbitals give 4π molecular orbitals; n = 4, j = 0 ±1, ±2 ◾ m = 2cos(0) = 2 2cos(±2π/4) = 0 2cos(±4π/4) = -‐2 j ◾ energy E = α + 2β α α – 2β cyclobutadiene 1,3-‐butadiene no. HMOT MO energy MO energy nodes (calculated) ψ 4 α -‐ 2β α -‐ 1.62β 4 α -‐ 0.62β ψ 2 α α 2,3 α + 0.62β ψ 0 α + 2β α + 1.62β 1 ◾ stabilisaGon energy = E = 4α + 4β ■ stabilisaGon energy stab E = 4α + 4.4β stab ◾HMOT predicts cyclobutadiene is less stable than 1,3-‐butadiene ◾ an7-‐aroma7c compounds are those with a π-‐system higher in energy than that of acyclic counterpart Aromatic and Heterocyclic Chemistry 9 Frost-‐Musulin Diagram – Frost Circle ◾ simple method to find the energies of the molecular orbitals for an aromaGc compound ◾ inscribe the regular polygon, with one vertex poinGng down, inside a circle of radius 2β, centred at energy α ◾ each intersecGon of the polygon with the circumference of the circle corresponds to the energy of a molecular orbital α0202β α0202β E α020β α α 2β α0+0β 2β α0+02β α0+02β cyclobutadiene benzene General CharacterisGcs of AromaGc Compounds ◾ planar fully conjugates cyclic polyenes ◾ more stable than acyclic analogues ◾ bonds of nearly equal length i.e. not alternaGng single and double bonds ◾ undergo subsGtuGon reacGons (rather than addiGon reacGons) ◾ support a diamagneGc ring current -‐ good test for aromaGc character of a compound H H δ "="56"ppm δ "="7.26"ppm H H aroma.cs"δ "="78"ppm H Aromatic and Heterocyclic Chemistry 10 Examples of AromaGc and AnG-‐AromaGc Compounds ◾Hückel’s rule [(4n +2) π-‐electrons for aromaGc compounds; 4n π-‐electrons for anG-‐aromaGc compounds] holds for anions, caGons and neutrals Cyclopropenium caGon ◾ (4n +2), n = 0, 2π electrons; stabilisaGon energy = 2α + 4β – stabilisaGon energy of allyl caGon = 2α + 2.8β α"%"β E α"+"2β ◾ insoluble in non-‐polar solvents; 1 signal in 1H NMR δ = 11.1 ppm -‐ aromaGc and a caGon H ◾ compare with cyclopropyl caGon which is subject to rearrangement to the allyl caGon Cyclopropenones
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