The naphthalene excimer is at a different wavelength. I don't know about the xantone, but if you are worried about intermolecular dimers as the source of your signal, then you can try the same experiment at lower concentrations, which should affect the relative intensity of the signals. If its itramolecular, then the shape of the spectrum should be the same regardless of the concentration.

It looks to me that your model is correct

Lowering the concentration is a good idea, unless you have the very unlikely scenario that the molecule forms a strong dimer complex in urea. Another general technique is to split the pair.Prepare two single labeled types of molecule. Put xanthone at the 10th position on one molecule and Naphtaline on the 12 position the other. A dimer spectrum in this situation would be proof of an intermolecular dimer. However, this is probably not necessary.

I would be surprised if the Kd of a hypothetical dimer would hold it at low nM, but it's true, if the Kd is tight enough he wouldn't see any changes in the relative intensities.

I'm not sure about synthesizing new peptides, because then its a new system altogether, and it could associate in the first place, but not in the new control version.

Thanks for the answers!

@eugenio: i have measured at different concentrations (10µM and 100µM), the spectra look the same, so it's probably an intramolecular dimer formed by naphthalene and xanthone. but how do i know whether it only forms after excitation or whether it's always present? a normal dimer shifts absorbance bands, right?

I think you would probably have to do time-resolved measurements to know that.

You're probably right, that might be the best way. I was hoping i could use my absorbance data which i already have. Because i'm running quite low on peptide for the other measurements. Thanks again!

I don't know what is the focus of your paper, but I'd say that is a relatively minor mechanistic issue.

A normal dimer must absorb, an excited dimer not. Did you record absorption spectra in the same experimental conditions at various concentrations ?

In principle, the Lambert-Beer law holds only if the type of species/molecules in a solution is the same at any concentration. thus, you have a monomer-dimer equilibrium, which depends necessarily on the concentration, your extinction or emission intensity will be concentration-dependent and the Lambert-Beer law will not be obeid.

I agree with M. Vasquez comment about lowering the concentration.

If your fluorescence spectrometer is equipped with a peltier cell, you can also do a thermal scan, where you heat up your sample to e.g. 90 degree. If your fluorophores change environment (e.g. dimer-> monomer) you will see a change in the fluorescence intensity and probably also in the emission band location.

Another suggestion is to use anisotropy. This may be less sensitive, but it could still be informative. The basic idea is that fluorescence lifetimes are of the same order as molecular reorientation. Thus a fluorophore attached to a large molecule will tumble slowly in solution, whereas if it is attached to a small more rapid tumbling molecule, the latter will show less anisotropy.

Clearly you mix the two : thermal scan and anisotropy as well.

In order to detect the ground-state dimer formation one should compare the shapes of absorption spectra of the two solutions. No shape difference would indicate no ground-state dimerization. One can also perform time-resolved fluorescence measurements at 400 nm and 550 nm, to verify whether the flurescence decay is multiexponential and bears the signature of an excited-state process (see my 30 years old papers with Lakowicz...).

You should measure excitation spectra of both emissions (best would be a full excitation emission spectrum, EEM). The spectral shape of the excitation spectrum resembles the absorbance spectrum, because the emission intensity is proportional to the light intensity absorbed by the ground state.

If the excitation spectra of both emissions are the same and identical to the absorbance spectrum, it is a clear indication of an excited-state process (Excimer formation in your case). If the excitation spectrum of the long-wavelength emission is different, the emitting entity already exists in ground state.

The excitation spectrum of the short-wavelength emission should resemble the absorbance spectrum in any case...

Be aware, there are more processes that can lead to dual emission behaviour (ESIPT, TICT, phosphorescence, impurities)

And, xanthone is already special by itself, you might have to do some literature search, there is much of it out there... Just as a starting point:

http://www.ncbi.nlm.nih.gov/pubmed/16855722

http://infoscience.epfl.ch/record/181783

http://harker.chem.buffalo.edu/group/publication/Xanthone.pdf

Because you have the possibility of an intermolecular diner that will not respond to concentration changes, you need to think of another approach. Because it is a peptide, you can denature it to change the folding. I would suggest temperature or pH or an organic solvent. The ground state diner should have a shift in absorbance (but be careful - pH and solvent can change the spectrum as well). There is a lot of literature from Birks on - we have a paper from journal of applied polymer science 2007 104 p360 where we used fluorescence excitation spectroscopy with excimer detection to determine the phase diagram of polystyrene solutions.

Looks like H-aggregate emission. This may not show up in absorption, since the transition is weak. It shows up in fluorescence because the state is populated by non radiative decay. I find it difficult to believe that the dimer is formed after excitation. But, as Menges says, if the excitation spectrum of the longer wavelength emission is different then it's certain that the dimer exists before excitation. In this case, if it is an H aggregate, then the absorption should be blue shifted.

Time resolved fluorescence with fs resolution would be interesting.

If you do a temperature scan, you first have to check how both fluorophores behave with the increase in themperature, otherwise you might interpret as interaction what is just an effect of the temperature on the individual fluorophores.

I also agree with M. Vasquez comment of lowering the concentration which should result in spectral bandshape change, if you are dealing with intermolecular dimer, if no change in bandshape, then the second longwavelength emission may be due to either excimer, TICT, ESIPT, etc. as suggested by Friedrich Menges or from isomers. BTW, by running excitation wavelength dependent emission measurements you may be able to differentiate whether or not the origin the second emission band is due to effects after excitation (excimer, TICT, ESIPT, etc.)

Thanks to all of you for all the interesting answers! Some aspects were mentioned, that I would have never thought of!

Emission fluorescence spectra at different concentrations (10,50,100µM) all show the same shape.

Excitation spectra of both emissions are very similar to each other and to the absorbance spectrum. So it seems like we're dealing with an excimer here.

And it's also being planed to do time-resolved fluorescence.

Dear Dominik,

"excited dimers" are mostly called excimers (if between identical molecules) or exciplexes (if between molecules differing in electron affinity). Excimner is the usual name and there is a rich lterature, especially in the 1970s and 80s. John Birks and Theodor Foerster were the pioneers, lots of articles annd books. Excimers and exciplexes in DNA and similar molecules was the topic of my lecture before the faculty (habilitation procedure at uni Mainz) in 1975.

I am now more concerned about environmental problems and do not follow up the literature on excimers. It was a fascinating field of research. Experimentally I studied aromatic polymers, especially Poly- N- vinyl carbazole. If you go to my website www.kloepffer.de you can find a complete list of publications.

kind regards

Walter Kloepffer

It may be helpful to look at our small molecule intramolecular excimer work from the late 1970s where we looked at temperature and solvent viscosity. Kramers' reaction-rate theory is often used in modeling assuming you are in the kinetically controlled region---this is quite dependent on the chromophore. As others have noted, there is a large body of work in this field including both small and large molecules.

Regards,

Merrill Goldenberg

The time resolved spectroscopy should clearly show whether the dimer is present in the ground-state or is formed in the excited state by monitoring the formation and decay of the excimer (dimer) emission.

Have you recorded an emission spectrum while pumping the long wavelength band?

An excimer flourescence or phosphorescence is in general broad, structureless and red-shifted with regard to the momomer emission. An excited dimer - without the strong inteaction needed for excimer formation - may be slightly red shifted and still have vibrational structure.

An interesting question is whether or not "excimer-forming sites" act as traps for mobile S- or T- excitons (exciton hopping). I did not follow this Research Topic in recent years, but some fundamental questions should still be the same: if you work in solid Solutions, the excimer-forming sites have to be preformed and the Emission has to be due to to some kind of energy Transfer - either by singlet- or triplet excitons, or by both. I published a lot about the possible mechanisms (see my web site). rthere are also excellent books on exciton migration, excimer formation etc.

Communication via the Internet cannot replace the study of the original work published!

http://www.kloepffer.de.

Kind regards

Walter Kloepffer

How to distinguish the excimer and aggregation using TCSPC technique?

both have the red shifted phenomina.

Do they have different lifetime scale?

Dear Dominik,

definitely, you should analyze time-resolved fluorescence at about 400 nm and 550 nm. If the 550 nm emission originates from an excited-state reaction product (excimer), it should exhibit a biexponential behaviour of the form:

I(t) = a1*exp(-t/T1) - a2*exp(-t/T2).

Please, pay attention to the signs of preexponential factors (amplitudes a1 and a2). The existence of a negative preexponential factor constitutes a definite proof that the emitting species is an excited-state reaction product.

Regards,

Aleksander Balter