5
$\begingroup$

According to UV-Vis spectroscopy, mesitylene (1,3,5-trimethylbenzene) absorbs at $\lambda_\text{max} = 210\ \mathrm{nm}$ while benzene absorbs at the slightly shorter wavelength of $\lambda_\text{max} = 204\ \mathrm{nm}$.

Why is that? One would think that $\pi$-ring's hyper-conjugation with the $\sigma^*$ orbitals of the methyl hydrogens would help stabilize the system thereby increasing the gap between the HOMO and LUMO (in turn decreasing absorption wavelength).

An opposite effect is observed when nitrobenzene ($\lambda_\text{max} = 252\ \mathrm{nm}$) is compared to 2-nitrotoluene ($\lambda_\text{max} = 250\ \mathrm{nm}$) and 2-isopropylnitrobenzene ($\lambda_\text{max} = 247\ \mathrm{nm}$). In these cases, the addition of larger and larger alkyl groups seem to be decreasing the absorbed wavelength (therefore increasing gap between HOMO and LUMO).

Is that because of sterics? Or because the the strongly electron withdrawing nitro group (conjugation) is being "counteracted" by the alkyl electron donating groups and therefore electrons are more concentrated in the ring rather than being spread though the molecule?

EDIT:

My question in more general terms: I understand HOW more conjugation results in a decrease in absorption frequency (due to molecular orbitals overlapping constructively vs destructively therefore decreasing gap between HOMO and LUMO). What I don't understand is HOW pumping electron density into a conjugated system (or withdrawing) effects the HOMO and LUMO. An explanation from a MO perspective would be greatly appreciated.

$\endgroup$
  • $\begingroup$ The beginning of an answer might be found here. Electron donating/withdrawing groups have different effects whether they are located on an odd/even carbon from the chromophore. $\endgroup$ – SteffX Aug 11 '16 at 18:52
  • $\begingroup$ @SteffX Right, but why is that? Is there a review article or textbook section that explores these concepts in more detail? $\endgroup$ – Nova Aug 12 '16 at 4:25
1
$\begingroup$

The general observation is that alkyl substitution intensifies and shifts the benzene spectrum to longer wavelengths. The effect is similar to that of groups with lone pairs such as $\ce{NH2}$ and O-alkyl. So I don't think that opposite effects are observed as you suggest. Some data are given below.

The 0-0 band transition, from v=0 vibrational level to v' =0 in the first excited state of benzene is at 262 nm (due to symmetry this is a forbidden transition so its position has to be calculated. The maximum in the S0-S1 transition in benzene spectrum is at 245nm, and S0-S2 at 203 nm). Toluene does not have a forbidden 0-0 transition, and has a 0-0 band at approx 266 nm (with maxima at 261 nm and S0-S2 at 207 nm), 0-zylene 0-0 is at approx 273 nm, so the substituents do stabilise the excited state very slightly. This has been attributed to 'hyperconjugation', i.e. some charge being donated by the methyl group to the ring, or the alkyl groups behaves as it it had a lone pair. (The idea of hyperconjugation is still hotly contested).
Toluene shows a slight increase in intensity as the symmetry is broken $\ce{D_{6h} to C_{2v}}$ but otherwise the effect on the electronic spectrum is slight.
Aniline has a relatively broad and intense absorption maximum (not 0-0) at 280 nm, nitrobenzene max very broad spectrum with a maximum at approx 268nm, but with a long shoulder in absorption stretching out to almost 400 nm, which is thought to cover the n-$\pi^*$ transition.

With very broad and structureless spectra (almost all the molecules you mention except benzene and toluene) the maximum intensity does usually always correspond to the 0-0 energy of the transition, due to Franck-Condon factors (overlap of wavefunctions which control intensity of transitions) and the fact that extra groups add considerable vibrational intensity to transitions and so broaden them. Thus comparing absorption peaks is not a reliable way of comparing state energies. (Experimentally the only ways are to measure vapour phase absorption spectra or measure in a solid matrix at 4k if possible.) Secondly spectra are stabilised differently in different solvents, even relatively non-polar ones and this can lead to spectral shifts.

The general sort of handwaving explanation, that is nevertheless useful, is to suppose that the molecule has energy levels that can be described like those by a particle on a ring. The bigger the ring the lower the energy levels become and so the lower the energy of the electronic transition. Thus incorporating a substituent that involves conjugating the electron over more atoms, increases the ring size very slightly and so produces a longer wavelength transition.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.