Chem 118/218 second take-home W13
DUE: Friday February 22 nd
at 11:00 am in class.
Read the paper on Dual Fluorescence from N
6
,N
6
-Dimethyladenosine, by Bo Albinson JACS ,
119 , 6369-6375 (1997).
Answer the following questions concisely (maximum 1 paragraph each)
(1) How do you know that the quenching of the B state is to another excited state, rather than to the ground state (by internal conversion)?
The ground state would not fluoresce.
Also: as the B population decreases, the A population increases, as shown by the negative pre-exponential factor in the time fit.
Additional notes: The data show that the existence of the intramolecular transition to another excited state. Actually that does not exclude a possible pathway of internal conversion. The paper lists a number of arguments why the authors think this pathway does not occur.
Another question could be whether the A fluorescence is involves another ground state rather than from a different excited state. However there is no evidence for that in the absorption spectrum, which is the same for all fluorescence wavelengths.
(2) Why does the fluorescence quantum yield from the B state change with temperature?
Quantum yields vary with temperature when any of the competing deactivation processes of the excited state depend on temperature. In this case that is the conformational isomerization, k
(BA)
(T).
(3) Why does the fluorescence from the A state not depend on the temperature?
Actually, in fact, it does, but only weakly for several reasons:
Only a small part of the A state fluoresces, most of it decays non-radiatively. The fluorescence quantum yield from the A state is 5.3 x 10 -3 . The fluorescence rate K
FA
= 4.6 x 10
-6
s
-1
. The non-radiative rate K nrA
= 8.7 x 10
8 s
-1
.
By contrast, the B state has a radiative rate constant of the order of 10
8
s
-1
.
As a result, the B state fluorescence is much more sensitive to the B to A isomerization that the A state fluorescence.
Aditional notes: At about 130 K the isomerization rate K
BA
becomes about 10 8 s -1 . You can get this from the Arhenius plot in Figure 5. So at T > 130 K, Φ
BA
becomes practically equal to 1(the rate exceeds the fluorescence rate more and more) thus the A state fluorescence does not change much anymore above that temperature. At the same time the B state fluorescence becomes very small (Figures 4 and 5).
(4) Which are the deactivation channels from the excited state?
Fluorescence, isomerization, phosphorescence.

Additional note: Most nucleobases also undergo internal conversion and the paper presents a number of arguments why they the authors think that this channel does not play a role in this system (except for the A state).
(5) Why are there two fluorescence peaks in Figure 1 and why are they different in different solvents.
Fluorescence from the B state peaks at about 3,000 cm
-1
, fluorescence from the A state peaks at about 2,000 cm
-1
. A state fluorescence is quenched by protic solvents. B state fluorescence is the same in different solvents.
Additional notes: Which part of the photochemistry that leads to A fl
depends on the solvent? It is not the isomerization but the fluorescence itself. How can you tell? The lifetime tracks with the fluorescence intensity. So the excited state potentials are affected by the solvent polarity.
(6) In the lower panel of Figure 4 there is a large peak below 25,000 wavenumbers at 80 K.
Why?
Phosphorescence.
(7) How do you know that K
FB
is independent of temperature?
When T goes up both the excited state lifetime and fluorescence quantum yield go down in parallel (Figure 5). So whatever shortens the lifetime is the same process that reduces the intensity, namely depopulation of the excited state by k
(BA)
(T) .
(8) How can the fluorescence and phosphorescence quantum yields be equal at 80K while the respective lifetimes differ by eight orders of magnitude?
The intersystem crossing rate is about the same as the fluorescence rate. So within a few nanoseconds about half the excited molecules fluoresce from the B state and half wind up in the triplet state. From there that half may take about 5 seconds to phosphoresce.