A reaction of A and B, gives C. the NMR of reaction mixture contain all peaks of A,B,C. I need to calculate yield or percentage conversion of B to C. A has no role . Yield of C need to calculate with respect to B. all have distinguish peaks. On integration, specific peak of C has 1.00 while B has 0.35. both contain 2 hydrogens each. Can I say yield of C is 1/1.35 *100= 74% ? Or is there any other methods to calculate the yield ?
Assuming that the spectrum was recorded under quantitative conditions (i.e. the relaxation delay is sufficiently long for full relaxation of both signals) and that neither of the 2 signals is exchangeable (NH, OH, SH etc), indeed your yield would be 74%.
Luc Alders thanks for your response. can you please explain more about quantitative conditions? is it related to instrument frequency, number of scans ?
NMR uses radiofrequency pulses to manipulate the nuclear spins of a sample. After a pulse, it takes some time for the spins to return to their equilibrium state. This process of returning to the equilibrium state is call 'relaxation'. For some protons it takes longer than for others, depending on their chemical environment. For quantitative NMR, or qNMR, the time between pulses needs to be long enough that the spins have fully relaxed because otherwise the response of subsequent pulses will decrease.
The instrument frequency does have a slight impact on the rate of relaxation. The number of scans does not, but the observed effect is exaggerated when the number of scans is further increased for spins that have not returned to their equilibrium state before the next pulse arrives.
In molecules with protons with a very different chemical environment, such as aromatic and aliphatic protons, this effect is easily observed when the relaxation delay (a.k.a. recycle delay) parameter is set to a very short time, e.g. 1 second. The aromatic protons will usually have a smaller integral than the aliphatic protons because they have a slower relaxation.
The rate of relaxation is something that can be experimentallly determined using a T1 experiment, but if the relaxation delay parameter is set to 1 minute, it will be quantitative under almost all conditions (although in most cases it will be overkill).
Dear A R Rayasingh .
Remember that it is not possible to calculate yield of the reaction only by NMR. Even using quantitative NMR you can only estimate B to C ratio changes in time. This information may be usefull to study reaction mechanism, kinetics etc. but it is not yield. Yield you can calculate only by knowing amount of the isolated product form the reaction. After optimization you should perform your reaction in preparative scale (e.g. 100g) to have reliable estimation of the reaction yield.
Even if you observe 95% conversion of B to C on NMR, you may end up with 35% yield. Why? Reasons may be numerous, i.e., decomposition of B, decomposition of C, formation of side-products, problems with isolation of C from the calalyst or these side-products etc.
Maciej Cieplak - this is not entirely true. It is possible to absolutely quantify components if you use a standard (internal, external or electronic). Many people use NMR this way routinely. Unless extremely accurate quantification is required, my preferred approach is PULCON-based, such as Bruker's ERETIC2 or QUANTAS.
dear Maciej Cieplak
If it says the ratio of B and C then, can we say there is 35% of B left and 65% of C is formed? or we need to include their molar masses.
isolate yield is not completely correct as there is a chance of loss in compounds during workup, CC, extraction etc. and also 100g scale is basically impossible.
Hi John Hollerton
can you please explain that how to use external and electronic standard? if possible please write the practical approach.
hello Luc Alders
Are you sure it's 74% yield? have you any reference kindly share. is there no role of amount and molar masses of B, C or mixture?
Carlos F. Marcos
- ERETIC2 and QUANTAS are both software approaches based on the PULCON method of Wider et al. (J Am Chem Soc. 2006 Mar 1;128(8):2571-6. doi: 10.1021/ja055336t). Note that the original ERETIC (Akoka et al.) is a hardware-based approach and may not be suitable. QUANTAS reference: Magn Reson Chem. 2010 Oct;48(10):753-62. doi: 10.1002/mrc.2647. PMID: 20803488.There is no reason why these approaches can't be used in a benchtop NMR.
Hi A R Rayasingh
What you see in the NMR spectrum is the molar ratio of the 2 molecules in solution. Weights, MW etc are only needed if the absolute yield needs to be calculated from the spectrum. That would also require some kind of internal of external standard.
The NMR yield (74%) would only be correct if you were 100% sure that nothing else happened with B in the reaction - and also be sure that you integrate the signals properly and you have high enough signal to noise ratio. If it decomposes and the decomposition product is insoluble in the solvent you use for NMR then the NMR yield is incorrect. Why don't you want to use an internal standard? I recommend 3,4,5-trichloropyridine, for example (depends on the nature of your compounds, of course).
And following your other question, you definitely don't need molar masses for calculating NMR yield, you are getting the ratio from NMR.
Eliska Matousova Thanking you for your explanation and suggestion, I was just curious about this type of yield calculation as it's used in GC yield determination. I am sure that B only give the product without any other consumption and my isolated yields are nearly matching . Yes I am using internal standard such as mesitylene during workup.
Good. Just be careful about mesitylene, from our experience it is quite volatile, so it cannot be put on the rotavap with the reaction mixture but must be added at the final stage of the sample preparation.
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