The beauty of this multiplet lies in the 18 peaks arranged in an almost perfect mirror symmetry. Even more, there are a lot of spin I = ½ nuclei present in this molecule, namely, 13C, 1H, 19F, and 31P (in this blog post I will focus on the latter three ones). Clearly love at first sight but trying to rationalize this signal might require some effort.
Let’s start it off easy.
1. Let’s say each of the two aromatic hydrogen atoms in 3J (HA and HA’) and 4J (HB and HB’) distance would have the same coupling constant
2. The hydrogen atoms in closer proximity to the fluorine atom (HA & HA’) give the higher coupling constant (see Figure 2)