# How to write correct NMR integrals in a formula?

for my masterthesis I want to include the Equation how I calculate my conversion in a polymer reaction. It is pretty easy, but I still want to write it down. Basicly the equation is: (Integral A usally 4-5 and Integral B always 1 beacuse its one Proton, Integral B and C are equal) $$Conv=\frac{Integral_A-Integral_C}{Integral_A+Integral_B}=\frac{Integral_A-1}{Integral_A+1}$$ But the formular with "Integral" in it looks not very nice while I thin the $$\int{}{}$$ sign seems to be a bit over the top.

If someone wants to know how I end up with the equation, here is the explanation:

It comes from the point, that my Monomer has 2 protons on each side of the double bond. When the polymer is formed they shift forming one broad signal. In the polymersolution I can identify one proton of the monomer which gets the Integral 1 while the other big peak contains the polymer as well as one proton from the monomer. Because both monomer protons have the same integral, both get a 1. My "clean" polymersignal is therefore the broad Integral -1. This has to be devided by the overall Integral which is the broad signal (which already includes one monomer proton)+ the other proton.

• I’d just use $s_\ce{A}$ for signal. Jan 28 '19 at 14:33
• And what for Conversion? Or just leave Conv? Jan 28 '19 at 16:21
• I don't think there's a standard symbol. For what it's worth, in the context of kinetic resolutions, conversion is usually denoted with $c$ (e.g. doi.org/10.1002/…), which I suppose is intuitive. The bottom line is that you can choose whatever you want, as long as you define them in the text. Jan 28 '19 at 16:59
• To me this opens a can of worms. Using $s_A$ is fine, but how is background handled? How are overlapping peaks resolved? The signal/noise ratio, and hence relative error, are critical in instrumental analysis. // Remember student in Chem 101 course who got 121% iron for a sample. me - "Didn't having more than 100% Fe seem strange?" student - "That's what the calculator said."
– MaxW
Jan 28 '19 at 18:27
• Thanks but I think you are a bit over the top. This is not chemistry 101 this is a common method around polymer chemistry for a fast and cheap calculation of the conversion. Maybe my explanation was not good, all signals are in one spectrum. Therefore the S/N Ratio should be the same unless it is not changing significantly with higher ppm.... Jan 28 '19 at 20:33

You assume the total number of protons summed over reactants and products is constant during the course of the reaction, which is perfectly reasonable, and that all monomer protons are incorporated into the polymer (perfectly reasonably), but also that the total signal is a constant. Note that because polymer signal will relax more rapidly by $$T_2$$ relaxation, usually, signal ratios are not expected to be strictly proportional to concentration, but we'll ignore that point, assuming this is a solution experiment and the polymer is highly dynamic.
$$f_{conv}= (s_A-s_C)/(s_A+s_C) = (s_{A,norm}-1)/(s_{A,norm}+1)$$
where $$s_C$$ is the isolated signal from one type of monomer proton and $$s_B=s_C$$, the other monomer signal, overlaps with the polymer signal to generate $$s_A$$. Then you define units for the area $$s_B=s_C=1$$ according to the expression on the right. This is explicit in the way I have written the equation. You should explain the meaning of the symbols in the text, of course.
• @Inselino No problem, I understood how you intend to do the analysis from one spectrum. You can also write $f=(I_A−I_C)/(I_A+I_C)=(I_A−1)/(I_A+1)$, but then should mention that the areas are normalized to the area of one monomer peak. Jan 28 '19 at 21:21