organic chemistry - Non-radiative transitions or what features of rhodamine result in it being so highly fluorescent?

Rhodamine is a fluorescent dye. What features of the molecule result in it being so highly fluorescent ?

I assume the extended conjugated system with a stabilized first excited state is necessary, to absorb light in the visible region of the spectrum, and I would also think a rigid, planar structure is preferable for good Franck-Condon factors for absorption. Are there other factors involving the energy/stability of the $S_{1}$ state I should be aware of ?

Answer

Rigidity is definitely an important factor. At $295\ \mathrm{K}$, indeno[2,1-a] has a fluorescence quantum yield $\Phi_F$ of 0.94, (DOI). In an organic glass at $77\ \mathrm{K}$, even (E)-stilbene, which hardly shows any fluorescence at room temperature, exhibits a $\Phi_F$ of 0.75.

Vibrational stretching of $\ce{C-C}$ bonds in substituents can significantly contribute to radiationless deactivation of an $S_1$ state and therefore reduce the fluorescence quantum yield $\Phi_F$. Admittedly, toluene isn't famous for its fluorescence ($\Phi_F = 0.15$), but the fluorescence of tert-butyl benzene is even lower ($\Phi_F = 0.03$).

Apart from radiationless deactivation through internal conversion (IC), an excited singlet state can be depopulated by intersystem crossing (ISC) to a triplet state. Under certain conditions, this is even possible for compounds that usually have very low triplet quantum yields.

Halobenzenes added to a solution of fluorescing anthracenes in dioxane (DOI) or ethyl iodide added to solutions of pyrene in alkanes (DOI) have shown to quench the fluorescence normally observed in the absence of the halogenated compounds. The latter exhibit a external heavy atom effect.

The lack of $S_1 \rightarrow T_1$ intersystem crossing is probably the most important factor that makes rigid compound show large $\Phi_F$ values.

It is striking that strongly fluorescent compounds typically have an extended $\pi$-system. Here, the $S_0\rightarrow S_1$ excitation is a $(\pi,\pi^*)$ transition.

On the other side of the fence, ketones, such as acetophenone or benzophenone, hardly show fluorescence. For these compounds, $S_0\rightarrow S_1$ excitation of the $\ce{C=O}$ group is a $(n,\pi^*)$ transition. From a $^1(n,\pi^*)$ state, spin-orbit coupling can both flip the spin and move one electron to a more distant and geometrically different orbital. $S_1 \rightarrow T_1$ thus is in fact a $^1(n,\pi^*) \rightarrow {}^3(\pi,\pi^*)$ transition.

For compounds like rhodamin B in their $^1(\pi,\pi^*)$ states, the corresponding transition to a $^3(\pi,\pi^*)$ state is largely spin-forbidden ans does not occur. Provided that no external effects come into play, these molecules exhibit strong fluorescence.