Chemical Forums
Chemistry Forums for Students => Physical Chemistry Forum => Topic started by: natalie5933 on January 22, 2022, 01:46:29 PM
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Question-
A standardization experiment was completed for the following reaction: S and P were mixed and the H NMR was taken. For the aromatic singlet closest to the ester, the integration were S=1.133 and P = 0.923. This same mixture was injected onto an HPLC and a ration of S:P = 36.7:63.3 was seen (based on peak integrations).
By what factor (response factor) do I need to multiply the starting material to get correct molar ratios for S:P?
Doubts-
what does nmr integration have to do anything with molarity?
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What do you know about NMR integration?
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That it tells you the number of protons
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So I have edited the post to show you the pic. Now correct me if I am wrong. acc to the structures and considering I am just the using the singlet signal close to the ester. so the number of protons of A:B will be 1: 2 right? and then integral A:B after setting A to 1 will be 1:0.815 . So finally my ratio of compounds will be A:B = 1: 0.4075? is this correct??
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So finally my ratio of compounds will be A:B = 1: 0.4075? is this correct??
Correct.. but why does the HPLC experiment give different result (P is the major product) ?
I think that H-NMR is not the best method for that type of quantitative measure, because both compounds are very similiar.
I mean - the instrument must have very high resolution.
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Correct.. but why does the HPLC experiment give different result (P is the major product) ?
That's what even I can't understand. The problem said this is a standardization experiment. But what are they trying to standardize? NMR or HPLC??
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As of H-NMR it is clear that peak integration is proportional to quantity of "S" and "P" (if instruments resolution is high enough), but HPLC - detectors dont measure quantity but rather properties of solution at given time - absorbance at given wave-length or change of conductivity or something else.
I mean - peak integration in HPLC is not the same as its quantity as it strongly depends on the type of the detector.
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I mean - peak integration in HPLC is not the same as its quantity as it strongly depends on the type of the detector
Can u explain again? I didn't get it. I am using UV-Vis for analysis for its measuring absorbance.
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For example - Cu(NH3)4Cl2 will absorb more than CuCl2. Higher peak at the same concentration.
If detector is ill-suited some compounds can even leave the column undetected. Detector measures only some properties, not quantity (concentration).
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Okay so how do I go about calculating response factor? By what factor should I multiple my starting material to get correct molar ratios then?
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Ratio of compounds S to P is 1: 0.4075 - this is the molar ratio already.
Standarization of HPLC....
36.7 x a = 1
63.3 x b = 0,4075
"a" and "b" are factors for HPLC quantitative standarization.
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okay. Ya even I did that. Thanks a lot for your help.
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THANKS A LOT FOR YOUR HELP !! One last thing, they have given retention time for S & P - S = 7.9 min, P(ent 1) = 11.0 min, P(ent 2) = 27.2 min. How do I calculate % conversion of product assuming no other by product is present?
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Final ratio of compounds S to P is 1 : 0,4075. This means initial quantity of S = 1,815.
1,815 S means that the final P concentration will be 0 for 0 % and 0,9075 for the 100 % conversion.
I think that retention time does not have anything with it.