Presentazione sul tema: "Magnetochimica AA Marco Ruzzi Marina Brustolon"— Transcript della presentazione:
1Magnetochimica AA 2011-2012 Marco Ruzzi Marina Brustolon 5. Radicals with delocalized electron density
2Summary Molecules with delocalized electrons. MO of electrons (Huckel method).Radicals of even systems: Radical anions and cations.Proton hyperfine splitting and spin density. McConnell relationship.Uneven systems are radicals.
3Sistemi coniugati 1 butadiene OA 2p, che si combinano negli OM Overlap tra gli orbitali 2p: legami . L’overlap è tra ogni orbitale atomico 2p e i vicini, e quindi l’OM deve essere costruito usando la base di tutti gli OA 2p degli atomi di C del sistema coniugato.scheletro di legami di tipo
4Sistemi coniugati 21. Per trovare i coefficienti delle combinazioni lineari bisogna prima diagonalizzare il determinante secolare. Le radici sono gli autovalori dell’energia.2. Per costruire il determinate secolare si possono fare alcune approssimazioni, note come “metodo di Hückel”.3. Gli OM risultanti sono in numero eguale al numero di OA che contribuiscono al sistema . Se il numero è pari, metà OM sono di legame e metà di antilegame. Se il numero è dispari, un orbitale è di non-legame.
5Butadiene Alternante pari Sistemi coniugati 34. Un’utile serie di proprietà può essere anticipata se il sistema coniugato è alternante (è cioè possibile contrassegnare gli atomi alternando atomi contrassegnati e atomi non contrassegnati). Se il numero di atomi coniugati è pari o dispari, il sistema si dice rispettivamente alternante pari o alternante dispari.In questo caso i livelli energetici sono simmetricamente disposti attorno allo zero dell’energia; inoltre coppie di OM, uno di legame e uno di antilegame, corrispondenti a livelli energetici simmetrici, hanno i coefficienti della combinazione lineare di ciascun atomo eguali in valore assoluto, ma con segno opposto per gli atomi stellati.. Per gli alternanti dispari il numero di atomi contrassegnati (stellati) e non stellati è diverso. Si sceglie allora di stellare i C in numero maggiore.4312*Allile Alternante dispari.Butadiene Alternante pari**
6Sistemi coniugati 4. Esempio: butadiene Approssimazioni di Hückel1. Sii=1, Sij=02. Tutti gli integrali Hii si considerano eguali allo stesso valore .3. Tutti gli integrali Hij si considerano eguali a se gli atomi sono adiacenti, altrimenti = 043121 = 2 = 3 = 4 = Energie degli OM
7Le combinazioni lineari degli orbitali 2p OM di antilegame4 = p p p p43 = p p p p42 = p p p p4OM di legame1 = p p p p4
8bearing the unpaired e- Radicals 1Note that the MObearing the unpaired e-is called also"SOMO"Molecules with conjugated π electrons with an even number of atoms belonging to the delocalized MO can give radical anions when reduced (the unpaired electron goes to the LUMO), and radical cations when oxidized (an unpaired electron is left in the HOMO).-b-2b2bbLUMO-b-2b2bbHOMO
9 Radicals 2In a first approximation the spin density on each of the C atoms of a radical system is given by the square of the coefficient of its AO in the SOMO (probability of finding the unpaired electron on that C atom).The SOMO corresponds to the LUMO for radical anions, to the HOMO for radical cations, and to the NBMO for neutral radicals.
10 Radicals 3An experimental determination of the spin density distribution on the conjugated frame can be done by measuring the hyperfine coupling constants of H atoms linked to the C atoms*.As the unpaired electron is on an orbital which has a node on the H nucleus, we should expect a zero hyperfine coupling constant.+HCIt should be = 0 on H-*These protons are the so called -protons (in with respect to the C atom).
11π- spin polarization 1However, the spin density is transferred to the H nuclei by π- spin polarization .The two "spin up" and "spin down" electrons of the C–H bond, have both a repulsive Coulomb interaction with the unpaired electron. However, this repulsive interaction is slightly weaker when the electron of the C-H bond has the same spin state as the unpaired electron, than with the other one (Pauli principle). This unbalanced spatial distribution depending on the electron spin state is called spin polarization.
12π- spin polarization 2Consequently, the spatial distributions of the two electrons of the C-H bond become slightly distorted; the electron with the same spin as the unpaired electron “moves” towards the C atom, and the other one “moves” toward the H atom.In conclusion, on the H atom there will be a negative spin density (i.e. a spin density with opposite sign with respect to that of the unpaired electron).On the H nucleus there is therefore some spin density of opposite sign with respect to that on the 2p orbital.CHCH
13π- spin polarization 3 aH = Q · C The spin density transferred by spin polarization on the H is proportional to that on the corresponding C atom, and it has a sign opposite to that on the carbon atom.This simple relationship (McConnell equation), with the semiempirical constant Q < 0, allows to predict the value of the H h.c.c. in conjugated π systems.aH = Q · C
14MO and Spin density 1In the Hückel Molecular Orbital (HMO) model the spin density on the C atoms is equal to the square of the coefficient of the LCAO for the SOMO:the unpaired electron being in the i-th MO.aH = Q · ci2The measurement of hyperfine coupling constants in radicals has been a very important experimental benchmark for theoretical approaches to quantum calculations in the last fifty years. No other experimental parameter is so directly linked to electron distribution on a molecule!
15Benzene radical anionThe benzene radical anion is a π radical. The unpaired electron is delocalized between the 2pz orbitals of the benzene sp2 hybridized C atoms.For symmetry, the unpaired electron has a probability of 1/6 to be on each C atom. According to the model of π- spin polarization, a spin density is transferred to the adjacent hydrogen atoms (a-hydrogens).aH = GaH = (-22.5·G) x 1/6
16EPR spectrum of Naphtalene radical anion A quintet of quintets, each 1:4:6:4:1
18MO and Spin density 2 aH = Q = Q ci2 The elementary semiempirical McConnell relationshipaH = Q = Q ci2interpreted in the framework of the Hückel approach, relates directly the protons hyperfine coupling constants to the squared coefficients of the LCAO of SOMO (i.e. of HOMO for radical cations, of LUMO for radical anions and of non bonding orbital NBMO for neutral radicals).The Q “constant” varies approximately in the range |Q| = G ( mT).Today the h.c.c.’s are calculated by DFT methods with great accuracy. But anyway the McConnell relationship allows to foresee by a very simple approach the approximate pattern of the ESR spectrum of a radical.
19Radical rC CH3• -23,04 1 C5H5• -5,98 1/5 -29,9 C6H6–• -3,75 1/6 -22,5 aH/GrCCH3•-23,041C5H5•-5,981/5-29,9C6H6–•-3,751/6-22,5C6H6+•-4,28-25,7C7H7•C8H8–•-3,95-3,211/71/8-27,7Q/GMcConnell equationaH=QrCexpAlthough ESR spectra cannot give directly the sign of the hyperfine coupling constant (h.c.c.), there are methods to obtain this sign. It is found that all the h.c.c.’s of the Table are negative.
20MO and Spin density Odd alternant hydrocarbons 1 Of particular interest are the conjugated systems with an odd number of π centers.In alternant odd molecules we have a different number of starred and unstarred atoms. Let us choose to have the largest group starred.
21BenzylBenzyl has 7 C atoms in its conjugated system. It is an uneven alternant hydrocarbon.We star 4 atoms and leave 3 non starred. On these latter the spin density is zero.Spin density*zero spin densityNBMO**1.000 1.259 *2.101 Distribution of the unpaired electron in NBMO calculated with the Huckel approximation. Note that the spin density is zero on the non starred atoms.
22MO and Spin density Odd alternant hydrocarbons 2 On the basis of Hückel results, the SOMO in odd alternant hydrocarbons has nodes in correspondence of the unstarred atoms:The NBMO has a node on the unstarred atomNBMO
23MO and Spin density Odd alternant hydrocarbons 3 Therefore the h.c.c. of proton bound to C2 should be zero. However, the ESR spectrum of allyl radical shows the following aH h.c.c.’s:The signs of these h.c.c.’s are :positiveaH=QrC Q<0therefore aH<0 means rC>0 and viceversanegative
24MO and Spin density Odd alternant hydrocarbons 4 For benzyl :Huckel coefficients of the SOMOaH/mTExperimental h.c.c.’s
25MO and Spin density 6We have seen that for C > 0 we should have an aH< 0 (by π- spin polarization).A positive sign of aH therefore must correspond to C < 0. This is due to another type of spin polarization, i.e. the π- π spin polarization.Similarly to the π- spin polarization, it is due to spin dependent repulsive interaction between the unpaired electron and the electrons of the filled orbitals.
26Let’s indicate the spin orbital with alpha spin in red, and that with beta spin in blue. smaller repulsionlarger repulsionFollowing the monoelectron approach of the simple Huckel theory, the two alpha and beta spin orbitals should be identical and at the same energy. But we know that the repulsive electron-electron interaction will be smaller for alpha-alpha than for beta-alpha spin orbitals.
27Spin polarization on the central C atom (more “blu” spin than “red”= negative spin density) Following the monoelectron approach of the simple Huckel theory, the two alpha and beta spin orbitals should be identical and at the same energy. But we know that the repulsive electron-electron interaction will be smaller for alpha-alpha than for beta-alpha spin orbitals.
28Alternation of positive and negative spin density is like a bipartite magnetic lattice
29Non-kekulè moleculesA non-Kekulé molecule is a conjugated hydrocarbon that cannot be assigned classical Kekulé structures (all the π electrons in double bonds).Since non-Kekulé molecules have two or more formal radical centers, their spin-spin interactions can cause electrical conductivity or ferromagnetism (molecule-based magnets), and applications to functional materials are expected.However, as these molecules are quite reactive and most of them are easily decomposed or polymerized at room temperature, strategies for stabilization are needed for their practical use. Synthesis and observation of these reactive molecules are generally accomplished by matrix-isolation methods. The simplest non-Kekulé molecules are biradicals.
30Triplet ground statesAn interesting class of molecules with a triplet ground state is that of non-Kekulé hydrocarbons. The designation non-Kekulé implies that their conjugated system cannot be represented by any resonance structure containing n double bonds derived from their 2n electrons.
31Triplet ground statesAs a general rule, when the two interacting electrons in a neutral organic molecule belong to two half-filled molecular orthogonal orbitals, the triplet state is lower in energy with respect to the singlet.For example in carbenes and nitrenes the two unpaired electrons are accommodated in a orbital and in a orbital respectively on a single C or N atom. These species can be obtained by photolysis of a suitable precursor in a glassy matrix or in a crystal, since in solution they would be non persistent.NPhenylitrHPhenylcarb
32Non-kekulè hydrocarbons: a method for finding the number of unpaired electrons A benzenoid structure can be oriented in three different ways with some of its edges (approx. 1/3) in a vertical direction.A benzenoid structure so oriented has peaks (upward pointingvertices on the upper periphery) denoted by and valleys (downward pointing vertices on the lower periphery) denotedby V.Gordon and Davison have shown that whenever V then the corresponding benzenoid structure is a radical..V - = 2 for diradicals, etc.V - = 1 for monoradicals
33Phenalenyl and derivatives V - = 1 for monoradicalsV - = 2 for diradicals, etc.
34Synthetic organic spin chemistry for structurally well-defined open-shell graphene fragments - ppYasushi Morita, Shuichi Suzuki, Kazunobu Sato & Takeji Takui.Phenalenyl — a triangular neutral radical consisting of three adjacent benzene rings — and π-conjugated derivatives based on the same motif, can be viewed as 'open-shell graphene fragments'. This Perspective discusses their electronic-spin structures, the properties that arise from their unpaired electrons, and highlights their potential applications for molecular spin devices.
35Graphene can be viewed as a sheet of benzene rings fused together Graphene can be viewed as a sheet of benzene rings fused together. Three benzene rings can be combined in three different ways, to yield linear anthracene and angular phenanthrene, where the rings share two C–C bonds, and the phenalenyl structure where three C–C bonds are shared between the rings.This third structure contains an uneven number of carbon atoms and, hence, in its neutral state, an uneven number of electrons — that is, it is a radical.All three structures may be viewed as being sections of graphene. Extension of this concept leads to an entire family of phenalenyl derivatives — 'open-shell graphene fragments' — that are of substantial interest from the standpoint of fundamental science as well as in view of their potential applications in materials chemistry, in particular quantum electronic devices. Here we discuss current trends and challenges in this field..
37EPR spectrum of phenalenyl Negative spin densityPositive spin densityseptetquartet
38Why looking for organic ferromagnetism The development of new organic ferromagnetic materials is a challenge, and is being pursued vigorously owing to their useful and attractive properties, such as:being lightweight;their solubility in organic solvents giving rise to the possibilities of liquid magnets, colloidal dispersions and Langmuir- Blodgett films;their transparency in many spectral regions making them suitable for photomagnetic switches and optical data storage;and, the possibility of perpendicular magnetic ordering leading to higher density of data storage.