Methylene blue has two absorbance peaks at 609 and 668 nm. Why does methylene blue have two peaks? What are the bonds that contribute to these peaks? How to analyze the peaks of a dye?
You'll find it here J. Phy. Chem. A 109, 6702 (2005)
Methylene Blue can aggregate and be protonated. You probably have two species. Check link below for specific references:
http://en.wikipedia.org/wiki/Methylene_blue#Light_absorption_properties
http://en.wikipedia.org/wiki/Methylene_blue#Light_absorption_properties
I agree with the good answers of Shajesh and Oksana BUT :
Prior to answer to your question, please give us more infos about the conditions of your investigation concerning the analytical aspect of the peaks of " a dye".
If you want to analyze the peaks of the MB, you must have at first pure MB for spectroscopical investigations. Commercial MB samples are impures and you must check the purity of you sample via TLC.The nature of the anion is important too!
Tetrafluoroborate is better than Chloride in some cases.
Please define your sentence: How can I analyze the peaks of " a dye " ? Do you mean another dye or MB dye?In this case write better the peaks of this dye.
I send another message about the bonds contribution the next time.
Good luck
JRG
Thanks Grezes,
Methylene blue's (MB) visible spectrum consist of two peaks in the visible region. I do not know which bonds contribute to these two peaks 605 and 669 nm peaks. I mean which of these bonds are responsible for these peak in MB -S-, -C=, N= or -C-CH3.
Hello,
Well the absorption spectrum of MB normally is characterized by an absorption band at high energy (π-π* of benzen ring) and a band at low energy around 660-670 nm (moving according to the pH of the solution) and corresponding to n-π* transitions (n is the free doublet on the nitrogen atom of C=N bond and free doublet of S atom on S=C bond). The peak at 605nm is not a band but a shoulder and it correspond to a vibronic transition 0-1 (level 0 of ground state to level 1 of the excited state).
Hope it was clear!
Adela
Dear Hariharan,
Thank you for your answer. According to the very good answer of Adela, I send you 2 links in order to refresh your background in M.O theory and theoretical UV-Vis spectroscopy .
For the sigma and N-R (Rydberg-Serie) transitions, another expensive and sophisticated spectrometers are needed.
- http://en.wikipedia.org/wiki/Molecular_orbital_theory
- http://www.plasticphotovoltaics.com/lc/characterization/lc-advanced-c/lc-uvvis.html.
Have you the possibility to draw 2 and 3 D molecules with software like ACD labs. It is free for university (see google).
After download of Chemsketch p.e., you have the possibility to draw the 3 resonances structures (two with N (CH3) and a positive charge and one with sulfur with a positive charge.
The delocalization of the positive charge is due to resonance and all the C, H, N atoms are in a plane. The C=N(CH3)2 cannot rotate .
M.B. Cation is planar and all these effects are responsible for the blue colour and fluorescence.
- Bindschedlers Green with a para-chinoide structure (see wikimedia for B.G. cation and M.B. Cation) has no fluorescence. However the introduction of a sulur bridge in this molecule gives M.B cation with a blue colour and a red fluorescence in solution.
- The concentration of a M:B. cation in solution (EtOH, MeOH, Water) give with
c= 10-7 Mol/l more monomer versus more dimer with c= 10-3 Mol/l.
see :
Paper:
Solvent isotope effect in the monomer–dimer equilibrium of methylene blue
Sandro L. Fornili, Giuseppe Sgroi and Vincenzo Izzo
J. Chem. Soc., Faraday Trans. 1, 1981,77, 3049-3053
I send you some files of interest per separate mailing.
Good luck and I hope I was and it was clear!!
JRG
I send you 2 files as agreed.
- Jablonski Energy Diagram
- page 1522 of the paper of David Creed with several M.B. cations bearing hydrophilic substituents (e.g. 3) or six-membered (cyclic) ring containing 2 nitrogen atoms at opposite positions in the ring.
Please read the value of the absorption maxima of these dyes (footnote) and compare with the value of methylene blue cation.
If you draw at first 2 D of the paper compounds and then 3 D, you can see the position of the cyclic amine at the positive charged nitrogen.
JRG
The peak centered at around 660 nm has an extinction coefficient between 78,000 M-1*cm-1 and 90,000 M-1cm-1, depending on the solvent. This is too high to be due to a n-pi* transition which is symmetry forbidden. This band is most likely associated with the resonance of the pi electrons from the sulfur resonating with those from the C's in thiazinic center (pi-pi* which is symmetry allowed).
- Badges
- Science method
- Similar topics
- Methodology
- Absorption Spectroscopy