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I am trying to understand Lewis bases (bear with me, I am a physicist). I was asking myself how diethyl ether and tetrahydrofuran compare with respect to their properties as solvents in electrophilic additions. Or if there is any difference at all.

So, for example, is it better to use a (boron trifluoride + diethyl ether) complex or a (boron trifluoride + tetrahydrofuran) complex? As to my understanding this is mainly a question about how strong the solvent acts as a Lewis base. If the solvent (ether or THF) is a stronger base, it might inhibit the catalysis of the actual electrophilic addtion reaction we are targeting, right? So under this assumption it is probably better to use the weaker base.

My guess is that diethyl ether is the stronger Lewis base. It has two hydrogen atoms more, so it is also contributing two more electrons, and moreover due to the slightly lower electronegativity of hydrogen, these electrons tend to be dragged towards the oxygen. On the other hand, THF misses the two protons and electrons, and moreover it must be electroneutral in the symmetry plane where the opposite C-C bond is, so it contributes a little less electron stuff to the nucleophilic oxygen.

Is this argumentation correct?

And what about this: the THF molecule is a little more compact than the diethyl ether and moreover has a tensed 5-ring. So shouldn't this push the electrons a little more to the periphery of the molecule, especially to the oxygen, and thereby increase its strength as a Lewis base again? Is this effect relevant at all?

PS: After writing this question I found a table at Wikipedia (https://en.wikipedia.org/wiki/Lewis_acids_and_bases) which states that the enthalpy of formation of the BF3 adducts is

  • Et2O: 78.8 kJ/mol
  • THF: 90.4 kJ/mol

This is about the opposite of what I would have expected and seems to indicate that the effect of the two additional protons/electrons of Et2O is not as big as it might seem at first.

But is it really support for the "compactness of THF in the 5-ring" theory? Or are there other effects responsible for THF being the stronger Lewis base?

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  • $\begingroup$ You call it compactness. Think of the lone pairs. In which case they are more exposed and thus readily form a complex? In THF. $\endgroup$ – Alchimista Mar 13 at 15:22

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