$\ce{[BrF6]+}$ is an octahedral species, and should have only one fluorine environment due to the symmetry. $\ce{Br}$ has nuclear spin $I=3/2$ and therefore is unlikely to cause splitting due to rapid interconversion between its spin up and spin down states averaging to a zero magnetic effect. Therefore I would expect one singlet in the $\ce{^19F}$ NMR spectrum.
However... the $\ce{^19F}$ NMR is actually two $1:1:1:1$ quartets (at low temperatures). Can someone help me find the flaws in my reasoning?
Answer
$\ce{BrF6^+}$ does have an octahedral structure consequently all of the fluorines are equivalent. As you point out, bromine is a spin = 3/2 nucleus, therefore, it is also a quadrupolar nucleus (all nuclei with spin > 1/2 are also quadrupolar). A spin 3/2 nucleus that is coupled to another nucleus (fluorine in this case) will split the nmr signal for the other nucleus into a 1:1:1:1 quartet. In the case of bromine we don't normally see this coupling because bromine typically undergoes rapid quadrupolar relaxation (some quadrupolar nuclei do, some don't) which washes the coupling out (e.g. rapid quadrupolar relaxation effectively decouples the nuclei, that's why the protons in $\ce{CH3Br}$ are a singlet, for example). However, when the quadrupolar nucleus exists in a highly symmetric environment (octahedral, tetrahedral geometry) the electric filed gradient at the nucleus is near zero and quadrupolar relaxation becomes slow (on the nmr timescale) and coupling can often be observed in such cases.
Finally, bromine occurs naturally as a mixture of two isotopes, $\ce{^{79}Br}$ (50.7%) and $\ce{^{81}Br}$ (49.3%). Both isotopes are quadrupolar with spin = 3/2. Because of the slight difference in electronegativity of the two bromine isotopes, they will produce a slight difference in chemical shift for the fluorines bonded to them. Hence, this mixture of $\ce{^{79}BrF6^+}$ and $\ce{^{81}BrF6^+}$ will produce two 1:1:1:1 quartets in the $\ce{^{19}F}$ nmr spectrum.
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