Frost rule is a expandable rule of polygon rule, to determine aromatic, antiaromatic and delocalized energy of ring system. There is a simple diagrammatic way of finding the HMO energy levels of any cyclic conjugated chain. Formerly, BM introduced with you about MO Huckel method (HMO), there is too a method to determine energy of each molecular orbital, after linear combination of atom orbital (LCAO). The most important is specification x in formula: E=ap:) - x.bt:), from there, we determine coefficient of wave equation of molecular orbital, to determine all properties of molecular, such as atomic charges, bond orders. With HMO, x is determined by secular determinant, and with Frost rule, it is determined by general triogonometric. Two important point of Frost rule:
- Inscribe regular polygon within a circle of radius 2 making sure that one apex is at the lowest point.
- Find the distances from the apices to diameter (co-ordinate), these are the roots x in fomular: E=ap:) – x. bt:) Example: Frost rule is illustrated for cycles containing N=3–>7 carbon atoms. Figure: Frost’s rule for the cyclopropenyl radical (C3H3 , N=3), cyclobutadiene (C4H4 , N=4), cyclopentadienyl radical (C5H5 , N=5), benzene (C6H6 , N=6), cycloheptatrienyl radical ( C7H7 , N=7), and cyclooctatetraene (C8H8 , N=8). Delocalized energy: -cyclopropenyl radical has three pi electrons. Two of these form an electron pair into the lowest MO energy level E0 = ap:) + 2.bt:) ; electron remain fill into one of the two degenerate levels E1 = E2 = ap:) – bt:) , say E1. The total pi electron energy of the cyclopropenyl radical is given by: E(C3H3) = 2E0 + E1 = 2(ap:) + 2bt:)) + (ap:) – bt:)) = 3ap:)+ 3bt:) It is not difficult to see the third electron in an antibonding orbital ( E1 = ap:) – bt:) > ap:)) increase the energy of the system and destabilizing effect. Compare the energy cyclopropenyl radical with cyclopropenyl cation -cyclopropenyl radical is lost one electron : E(C3H3+) = 2E0 = 2ap:) + 4bt:) Cyclopropenyl cation is lower energy than cyclopropenyl radical, so it is more stable than.
-Two of the four electrons in cyclobutadiene is filled into the lowest energy molecular orbital. According to Hund’s rule, two electron remain have to be filled so that maximum spins, the two orbitals corresponding to the next doubly-degenerate level are singly-occupied by electrons with parelle spins. E(C4H4) = 2E0 + E1 + E3 = 2(ap:) + 2.bt:)) + ap:) + ap:) = 4ap:) + 4bt:) Delocalized energy of C4H4: E(C4H4) – 2E(ethene) = 4ap:) + 4bt:) – 2(2ap:) + 2bt:)) = 0 (1) (We ignore the strain ring of cyclobutadiene) The zero delocalized energy is a further confirmation of the low stability and high reactivity of cyclobutadiene. 4ap:) + 4bt:), is higher than open chain butadiene 4ap:) + 4.472bt:) (2), therefore, cyclobutadiene is an antiaromatic system.
-See the scheme of cyclopentenyl radical, it is not difficult to consider stability of anion C5H5- (anion cyclopentadienium) , all of electron are filled bonding orbitals, delocalized energy: E(C5H5-) = 2E0 + 2E1 + 2E5 = 2(ap:) + 2bt:)) + 2[ap:) + 2cos(2pi/5)bt:)] + 2[ap:) + 2cos(2pi/5)bt:)] = 6ap:) + [4 + 8 cos(2pi/5)]bt:) It is very stable and appears in important organometallic compound such as ferrocene.
Benzene, all three bonding orbital with energy (ap:) + 2bt:)) and (ap:) + bt:)) (doubly-degenerate) are occupied by electron pairs making the system is very stable with lowest energy. Benzene involve three carbon-carbon double bonds, so we calculate delocalized energy system by subtracting 3E(ethene) E(C6H6) – 3E(ethene) = [2(ap:) + 2bt:)) + 4(ap:) + bt:))] – 3(2ap:) + 2bt:)) = 2bt:) Delocalized energy of benzene is much lower than delocalized energy of butadiene, (E(butadiene) – 2E(ethene) = 0.472bt:)). Even if we calculate thedelocalized energy per double bonds, in butadiene we get 0.472bt:)/2, while in benzene 2bt:)/2. the low energy of the pi-electron sextet in benzene explain the aromatic character of this molecular. :noel4 ( :chocwe ( :sacsua ( Note: (1), (2), you can search in “MO Huckel tự soạn” document which BM uploaded link: http://www.compchem.hcmuns.edu.vn/chemvn/showthread.php?t=57 Have fun!!! :noel4 ( :bole (