?Molecular orbital calculations may be used to predict trends in the standard potentials

Chapter 9, Problem I9.5

(choose chapter or problem)

Molecular orbital calculations may be used to predict trends in the standard potentials of conjugated molecules, such as the quinones and flavins, that are involved in biological electron transfer reactions. It is commonly assumed that decreasing the energy of the LUMO enhances the ability of a molecule to accept an electron into the LUMO, with an accompanying increase in the value of the molecule’s standard potential. Furthermore, a number of studies indicate that there is a linear correlation between the LUMO energy and the reduction potential of aromatic hydrocarbons.

(a) The standard potentials at pH = 7 for the one-electron reduction of methyl-substituted 1,4-benzoquinones (7) to their respective semiquinone radical anions are:

Use the computational method of your or your instructor’s choice (semi-empirical, ab initio, or density functional theory methods) to calculate \(E_{\text {Lмо }}\), the energy of the LUMO of each substituted 1,4-benzoquinone, and plot \(E_{\text {Lмо }}\) against \(E^{\ominus}\). Do your calculations support a linear relation between \(E_{\text {Lмо }}\) and \(E^{\ominus}\)? (b) The 1,4-benzoquinone for which \(R_{2}=R_{3}=\mathrm{CH}_{3}\) and \(R_{5}=R_{6}=\mathrm{OCH}_{3}\) is a suitable model of ubiquinone, a component of the respiratory electron transport chain. Determine \(E_{\text {Lмо }}\) of this quinone and then use your results from part (a) to estimate its standard potential. (c) The 1,4-benzoquinone for which \(R_{2}=R_{3}=R_{5}=\mathrm{CH}_{3}\) and \(R_{6}=\mathrm{H}\) is a suitable model of plastoquinone, an electron carrier in photosynthesis. Determine \(E_{\text {Lмо }}\) of this quinone and then use your results from part (a) to estimate its standard potential. Is plastoquinone expected to be a better or worse oxidizing agent than ubiquinone?

Text Transcription:

E_Lмо

E^ominus

R_2=R_3=CH_3

R_5=R_6=OCH_3

R_2=R_3=R_5=CH_3

R_6=H