Question

(a) What is Hückel theory? Why is it used? What assumptions does it make? (b) Consider hexatriene, how many molecular orbitals would you expect there to be? (C) Write the secular determinant for the hexatriene system. (d) Using Hückel approximations, re-write the Hückel secular determinant using the appropriate variables, and explain what the variable represent. (e) Determine the energies of the molecular orbitals (1) Determine the total energy of the -electronic energy in hexatriene (9) Hückel theory is meant as a way to take into account delocalization of electrons. We can compare the energy of the delocalized system in hexatriene to the energies of localized bonds by look- ing at the difference between the hexatriene energy and the energy of three ethene molecules (each ethene molecule contains a single localized bond). The energy of the ethene system is 2a + 28. Thus determine the 'delocalization energy" in hexatriene. Is this a stabilizing or destabilizing energy? Comment (h) An isomer of hexatriene is fulvene, shown below. Write the Hückel secular determinant for this molecule (using the labelling as drawn). Compare your answer to your answer from (d). Are there any differ- ences? Why or why not?

Answer 1

(a) Nuckel Theory Profered by Esich Huckel in 1930, 9t io the |(LcAo-Mot) linear comblnotion ef atemie aulritalsMtthed molecular orbital methed to idutirmine enengy of Molecular Oubitale T-electrons in This waleculu delecaligid moleculu can lhe afflied ter cenfugatid gt is saed to detumine enrgy of Telectrens un -delocalised nalecule conjugatid ytan Assunbtions- hucksl made aeuumfeions albout three kinde of intighale Exchamge (Resenane) intigral, ouenlat intigual, to sinaflipy sedular equatien Coulenmb intigral (an (1) All Overlap integrals are Zero,i.es,. Stj=o D Coutemb intigual, Hi (which tefere to Jurergy f eleciem inp oraital on arlion atn) io deneted y o th li the exchange inligeal Hij= Hi=0 less amd hrelactals are on adiacmt thay carlonatems in uhich case arre as

(b) Jhere are 6 A-melecular orbitale hexatriene (c HI-ES Hi2-ES2 HIB-ES1 Hiv-ESy HISESIS Hie-ESi H2i-ES21 H22-ES22 H23-ES23 Hay-ESay Has-ES25 Hac-ES24 H31-ES31 Ha2-ES32 H33-ESas Hay-ESay Has-ESa5 Hac-ES3c H I-ESY HY-ES42 Hy3-ESvs Hyy-ESy Hys-ESus Huc-ESy Hs1-ESSI Hs2-E Ss2 Hs3-ESs Hsy-ESsy Hss-ESss Hsc-ESse H&I-ES Ha-ES2 He3-ESc3 Hey-ESy Hes-ESes Het-ESsc d-E O fIE =O d-E d-E d-E O d-E O whue, d= Hii Coulemb latigral 1 Hei if bend is preaent betureen Ct aud Ci CNOTE) 9 No bend bruaent between Ci and c, then C:- Carboni Hij -o = 1,if i-i Stj and Sij =O,i iti

eEnengy leelo for 1,3,5-henatriene- mi-2 cos -42,--n Edt mip Ey= o+1.902p) = 1. 802 = 2 cos IX 6+1 Ey=(at 1.247p) M2= 2 cos 2XT 1.247 641 = 0.445 E=(a+0446p M3=2 cos 3Xx Eyty=-0.445B my- 2 cos 4Xx 6+1 = -0.445 Eve =Ca-1.247) ms=2cos 5x 6+1 = -1. 2 47 me 2cos 6XT1.802 Ew (a- 1.802 B 6+1