Heteronuclear J-coupling
The communication of nuclear spins is not limited to nuclei of the same ‘type’. Atoms that have a nuclear spin (i.e., I ≠ 0) are affected by the same phenomena that cause protons to communicate and ultimately split each other’s NMR signal.
If you’re concerned that heteronuclear J-coupling is much more complicated than homonuclear J-coupling you can relax – well, you can mostly relax! Once you’ve worked out solving and assigning homonuclear splitting patterns – you have a pretty good handle on the situation!
The first step is to think about the types of coupling that are possible given a molecular structure (i.e., what nuclei are present, if they are NMR active, and what is the natural abundance of the NMR active isotope). There are three cases:
1) I = ½ and natural abundance of the NMR active isotope is ~100% e.g., 1H, 19F, 31P, 89Y, 103Rh).
2) I = ½ but the natural abundance of the NMR active isotope is < 100% ( e.g., 13C, 117/119Sn)
3) I ≠ 0 or ½.
For the sake of this discussion I’m going to omit the last case and focus on I = ½ nuclei.
Case 1: You already have homonuclear splitting down? EASY!
These spectra can be solved in the same manner as you would with homonuclear couplings. The ‘n + 1’ and Pascal’s triangle rules still apply.
Case 2: This is probably easiest to conceptualize by highlighting a specific example, so for sake of simplicity we look at a basic hydrocarbon: iso-propanol ((CH3)2CH(OH)). The proton NMR will have three signals: (i) –OH; (ii) –CH; and (iii) –CH3 (as the methyl groups are equivalent). This data would be reported as:
1H NMR (60.16 MHz, 10% (v/v), D2O): δ = 4.71 (bs, 1H, OH), 4.00 (septet,3JH-H = 6.2 Hz, 1H, CH), 1.15 (d, 3JH-H = 6.2 Hz, 6H, CH3) ppm.
In terms of homonuclear coupling this looks exactly like what we’re used to seeing. That is, the spectra can be explained with standard 3JH-H splitting:
1) A broad, uncoupled singlet for the alcohol;
2) A septet for the methine (as the methyl groups are equivalent ‘n + 1’ = ‘6 + 1’ = 7);
3) A large doublet at 1.15 ppm for the methyl groups (as ‘n + 1’ = ‘1 + 1’ = 2).
Okay – pretty standard – when do we learn about carbon-13 induced coupling?
Well, look closely at the base of this large doublet. See those equally spaced doublets? Well – there it is! Finally, some evidence of the elusive ‘Case 2’ coupling!
Due to the isotopic ratio of carbon, 98.9% of the time the protons in question will be bonded to a carbon-12, which, of course, is the NMR silent isotope. In this case, the nuclear spins will not communicate and no heteronuclear coupling will occur. Subsequently, 98.9% of the time we only observe the homonuclear coupling – that is, we ONLY see the 3JH-H splitting of the methine and that big central doublet. HOWEVER, 1.07% of the time, these methyl groups will ‘see’ a carbon-13 atom (NMR active, I = ½). This will result in that itty-bitty doublet of doublets that surround the central peak. This is due to both heteronuclear and homonuclear coupling (i.e., 1J13C-H = 125 Hz and 3JH-H = 6 Hz).
So, the cliff notes for understanding 13C satellites – for any given proton resonance – 98.9% of the peak area will not be split and 1.1% will be. This is why we call this type of coupling satellites – these heteronuclear multiplicities are small peaks that ‘hover’ around the big peak.
Practically carbon-13 satellites can be difficult to spot. They contain a total of only 1.1% of the area of any given peak and they are split so they contain a maximum of only 1.1%/2 = 0.55% of the intensity of the main peak. I selected the two-methyl group resonance in iPrOH to illustrate this coupling because: (i) the large peak makes these easier to spot; and (ii) the other signals are considerably downfield so the satellites are well resolved.
Typical 1J13C-1H coupling constants fall within the 115 – 270 Hz [1] range and are dependent on bond strength, bond angle, and amount of s character in the bond (where s character progresses as sp3 < sp 2 < sp). Long-range 13C-1H coupling constants are rarely observed. Although these satellites are rarely reported, they do have implications for (i) extracting structural information, and (ii) qNMR and generating very accurate integrals.
For another example of heteronuclear coupling you can check our BODIPY blog!
[1]Silverstein, R. M.; Webster, F. X.; Kiemle, D. J.; “Spectrometer Identification of Organic Compounds” 7th Ed. John Wiley & Sons Inc.: USA