Question

Which of the following two molecules is more likely to fluoresce? Explain what properties of that molecule encourage fluorescence and why.

Answer 1

Fluorescence happens when an atom or molecules relaxes via vibrational relaxation to its ground state after being excited electrically. The availability of conjugation allows the delocalisation of double bonds over the complete molecular system making them to effectively absorb the indident radiation energy. The characteristic frequencies of excitation and emission are dependent on the respective molecule or atom.

Molecule 1 (structure in the left hand side)

You can observe the availability of alternative presence of double bond and single bond in both the molecules given. This is called conjugation and it happens by the presence of continuous SP2 hybridised carbons. The careful look into the molecule 1 shows, this conjugation is not continous and there is a breakage at the carbon (numbered 5) bearing two H. The double bonds in either of the phenyl ring (at carbon 1 or carbon 4) can not undergo delocalisation through this carbon 5. This carbon 5 is SP3 hybridized and block the chain of delocalization through other SP2 carbons.

Therefore, molecule 1 is less likely to flouresce.

VH

Molecule 2 (structure in the right hand side)

The molecule 2 in the right hand side of the given question is more likely to fluoresce than the molecule in the left hand side.

Unlike the molecule 1 discussed in the earlier case, this biphenyl moelcule 2 has no any intermediate SP3 hybridised carbon. All the Carbons are SP2 hybridized. There can be effective delocalisation of electrons between, 1,2,3 and 4.

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