I am attempting to interpret an experimental section on a scientific paper. The paper calls for 0.51 mmol of 3,5-dimethoxy phenol, 1.0 mmol of 3-methyl-3-butenal, 5 mol% of phenylboronic acid, 20 mol% of benzoic acid, and 2.5 mL of heptane. I am trying to figure out what is meant by mole% in this case. There are many different interpretations of this I have seen online, including mole percent being based on a fraction of the limiting reagent (the phenol in this case), the mole percent being based on the fraction of all of the total moles of reactant and catalyst (most fitting of the definition of mole percent), and the mole percent being based on all of the moles of reactant, catalyst, and solvent. It makes most sense that the mole percent for this protocol would be based on the limiting reagent, but I included the calculations based on the total moles of reactants and catalysts for reference. Does anyone know what the term "mole percent" would appropriately based on in this case?

0.51 mmol (phenol der.) + 1.0 mmol (aldehyde) + x mol PhB(OH)2 + y mol PhCOOH = z total mol

0.2 z (total mol) = y mol PhCOOH 0.05 z (total mol) = x mol PhB(OH)2

0.51 mmol (phenol der.) + 1.0 mmol (aldehyde) + 0.2z + 0.05z = z total mol 0.75z = 1.51 z = 2.013

0.2 (2.013) = 0.40 mol benzoic acid 0.05 (2.013) = 0.10 mol phenyl boric acid

If the mole percent was based on the moles of the LR, the moles of catalyst would be as follows: 0.10 mol benzoic acid and 0.026 mol of phenyl boric acid

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    $\begingroup$ In the absence of any other information, I would always assume that it refers to the limiting reagent. This is by far the most common usage (in fact, I don't think I've ever seen your other definitions before). But you know what would be really helpful - if you provided a link to this paper... $\endgroup$ Dec 26, 2019 at 19:42
  • $\begingroup$ The information I provided is about all the paper provides in terms of the experimental section, however, here is a link the the paper in case anyone is interested: DOI: 10.1039/C6OB01026A $\endgroup$
    – Eli Jones
    Dec 26, 2019 at 19:43

1 Answer 1


Eli, you should have read the paper very carefully before asking questions. I think you are very intelligent person, so don't choose easy way out. Now, to answer your question, about mol%. It is right in the paper. As you already figured out and also Orthocresol pointed out in his comment, The amount equals to given mol% is $LR \times \frac{given \ mol\%}{100}$. It is in the General procedure A of given reference:

Condensations of phenols and naphthols with $\alpha,\beta$-unsaturated ketones. An oven-dried $\pu{25 mL}$ Schlenk tube equipped with a teflon-coated magnetic stir bar was evacuated and purged with argon. The phenol ($\pu{0.2 mmol}$), pentafluorophenylboronic acid ($\pu{8.5 mg}, \ \pu{0.04 mmol}, \ \pu{20 mol\%}$), diphenylphoshinic acid ($\pu{8.7 mg}, \ \pu{0.04 mmol}, \ \pu{20 mol\%}$), $\alpha,\beta$-unsaturated ketone ($\pu{0.4 mmol}$) and heptane ($\pu{1 mL}$) were added to the tube under argon.....

Accordingly, The phenol used: $\pu{0.2 mmol}$ ($LR$), pentafluorophenylboronic acid used: ($\pu{0.04 mmol}$ meaning $0.2 \times \pu{20 mol\%}$), and diphenylphoshinic acid ($\pu{0.04 mmol}$ meaning $0.2 \times \pu{20 mol\%}$).

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    $\begingroup$ Thank you for pointing this out. I did not think to read general procedure A since it was not referred to in the sub-procedure I was following; however, I see how valuable information can be held within every aspect of the paper. $\endgroup$
    – Eli Jones
    Dec 27, 2019 at 16:26

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