[21385]

Two proton NMR spectra of (CH3)3CF were obtained using neat (CH3)3CF and (CH3)3CF in SbF5, respectively. One spectrum, denoted as spectrum I, showed a singlet at g 4.35, and the other, spectrum II, revealed a doublet at g 1.30 with a coupling constant J=20 Hz. Which spectrum was obtained from (CH3)3CF in SbF5?

SbF5 is a superacid, so, it will react with (CH3)3CF to form (CH3)3C+ SbF6-

The correct answer is spectrum I but I don't know why the other spectrum, spectrum II, would have a doublet with a coupling constant J=20Hz? Is that because of a hydrogen on one methyl group is reacting with a hydrogen 2 carbons away? If so, does the coupling constant signify the distance between the two hydrogens (the more the coupling constant is, the further away the hydrogens are)?

Thanks a lot

[Rico]

Hey 21385

The reason why spectrum II is seen as a doublet, is that the hydrogens on the methyl groups couples to the fluorine atom, giving rise to this large coupling constant compared to H-H spin coupling. You must remember that the ¹⁹F is also an NMR-active nuclei.

Rico