1
$\begingroup$

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.

Improve this question
$\endgroup$
7
  • 1
    $\begingroup$ I’d just use $s_\ce{A}$ for signal. $\endgroup$ – orthocresol♦ Jan 28 '19 at 14:33
  • $\begingroup$ And what for Conversion? Or just leave Conv? $\endgroup$ – Inselino Jan 28 '19 at 16:21
  • $\begingroup$ 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. $\endgroup$ – orthocresol♦ Jan 28 '19 at 16:59
  • 1
    $\begingroup$ 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." $\endgroup$ – MaxW Jan 28 '19 at 18:27
  • 1
    $\begingroup$ 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.... $\endgroup$ – Inselino Jan 28 '19 at 20:33
1
$\begingroup$

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.

Therefore, you could write

$ 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.

Improve this answer
$\endgroup$
3
  • 1
    $\begingroup$ Thanks for the answer how to write it. I did not really wanted to go into discussion about what I'm doing there. Maybe my expression was wrong. All three signals are in one spectrum of the crude polymer. So I'm not comparing 2 spectra with different concentration etc. This method is very common in polymer history to determine the conversion in a crude solution as it is fast and cheap and when it is accurate around +-2-5% which is in range of the NMR Error its totaly fine. $\endgroup$ – Inselino Jan 28 '19 at 20:32
  • $\begingroup$ @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. $\endgroup$ – Buck Thorn♦ Jan 28 '19 at 21:21
  • $\begingroup$ OP I think that you got the way to handle this. As far dynamics is concerned and just for non specialist readers: indeed the dynamics of the polymer can be so different that a signal from a polymer might be difficult to be even attained using the same conditions that readily give the monomer spectrum. When this is not the case than this method can be used, limitations taken into account as usual. $\endgroup$ – Alchimista Jan 29 '19 at 8:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.