# Proton NMR splitting in 2-substituted furan

I have recently run proton NMR on 1-(furan-2-yl)ethan-1-ol I synthesised in the lab.

My 400 MHz spectrum consists of the following peaks:

• 7.36 ppm (dd, J = 1.9, 0.9 Hz, 1H)
• 6.31 ppm (dd, J = 3.3, 1.9 Hz, 1H)
• 6.22 ppm (dt, J = 3.2, 0.9 Hz, 1H)
• 4.87 ppm (q, J = 6.6 Hz, 1H)
• 1.53 ppm (d, J = 6.6 Hz, 4H).

For the peak at 6.22 ppm, I am wondering why a doublet of triplets is observed rather than a doublet of doublets. I understand the long-range splitting that results in the dd, but am just a little surprised by the dt. I attach a picture of the doublet of triplets.

• Maybe you have a ddd with two similar Js. The additional doublet might be due to coupling to the proton at 4. 87 ppm. Does that one show an additional small splitting? Oct 24 '20 at 12:14
• Can you run a 2D NMR experiment? ¹H-¹H COSY would clear this right up. Oct 24 '20 at 13:32
• The compound is commercial, available from Merck. Unfortunately they don't show NMR spectra for free anymore. You can try a simulation here. See also ChemSpider. Oct 24 '20 at 14:17
• I agree with @BuckThorn. I suspect the coupling in the quartet is just not fully resolved. You may be able to play around with window functions (or apodisation as it's referred to in Mestrenova) to try and tease out this small coupling. Suggestion: try selecting gaussian and exponential, set the exponential value to -2 Hz and the gaussian value to 0.2 GF, or something similar. See if the quartet is still a quartet. Oct 24 '20 at 22:54
• @orthocresol you are correct! Thanks for the advice on the window functions. Once adjusted, the quartet of doublets was clear! What exactly do the different window functions do? Oct 25 '20 at 19:07

The exponential one basically multiplies the FID by another decaying exponential. This is largest at the start and decreases as time passes, so effectively causes peaks to look as if they're decaying even faster. This leads to broader lines, but because the decaying exponential is largest at small $$t$$ where the signal is large, it "emphasises" the parts of the FID that have more signal. Effectively, you lose spectral resolution but gain some signal-to-noise.