[jamesrb]

Am I correct in assume that in ¹³CNMR spectroscopy proton decoupled spectrums still show ¹³C-¹⁹F coupling? And that a proton coupled spectrum would show both ¹³C-¹⁹F and ¹³C-¹H coupling? So in CH₂F₂ for the proton decoupled spectrum we would see 3 lines (1 triplet). And in the proton coupled spectrum we would see 6 lines, 2 triplets (as the coupling constant for C-H and C-F are different).

[Babcock_Hall]

That looks mostly correct, except for your last sentence.  Make a coupling tree and see what happens.

[jamesrb]

1 pentet?

H₁=α
H₂=α
F₁=α
F₂=α

H₁=α
H₂=β
F₁=α
F₂=α

etc. etc. etc.

[sjb]

Would you expect ²J_CH and ²J_CF couplings to be the same?

[Babcock_Hall]

@OP, I am not following what you wrote.

[jamesrb]

Would you expect ²J_CH and ²J_CF couplings to be the same?

I would not expect ²J_CH and ²J_CF to be the same.

[jamesrb]

@OP, I am not following what you wrote.

Instead of scanning my drawing of the coupling tree I just summarized what I thinking regarding the spins of the four atoms attached to the carbon. There are two hydrogen atoms: H₁ and _H2 and two fluorine atoms: F₁ and F₂. When I drew the tree I came up with five lines. Now I am thinking there should be 8. I have never drawn a coupling tree before and this may not be the best molecule to start with but none the less I tried:

[Babcock_Hall]

The H-1 decoupled spectrum is a 1:2:1 triplet, due to coupling from fluorine.  If one mentally turns off the H-1 decoupler, each triplet should split into a triplet.  I believe that there will be 9 lines.

[Irlanur]

It should be a triplet of a triplet (=9 lines). the number of lines will depend on the magnitude of the couplings and the resolution.