Reading papers about organic synthesis, I see many cases like setting one material as 1.2 eq. I presume that reagents used more than 1 equiv are relatively inexpensive or less reactive than other reagents. (Am I right?) But relative percentage of catalysts is still a mystery. For example, when I searched for Buchwald–Hartwig amination, I saw various mol% of catalysts, ranging from 0.5 mol% to 10 mol%. What criterion or rule of thumb is used? Is this just a result from experimental trials varying mol% of catalysts?

  • 4
    $\begingroup$ There can be no "rules of thumb" that work across the gargantuan variety of possible chemical reactions. The best you can do is to look for what worked in similar reactions in the literature. If the reaction is completely novel, then the only solution is to experiment and test what works. A lot of chemistry is, after all, an experimental not a theoretical science. $\endgroup$
    – matt_black
    Aug 3 at 11:16
  • 1
    $\begingroup$ A minor sidenote: Setting one material at 1.2 equivalent is not always done just because it is cheap, but it is usually done to make sure the material it reacts with completely reacts away. $\endgroup$ Aug 4 at 11:57

The most precious component of the reaction is the starting material you made yourself so the materials used in excess are usually the bought ones, unless you are at the first step of a sequence.

In some cases the % catalyst will be the result of running a series of experiments to establish how much was needed - these studies would normally form part of the published paper or would be referenced if published separately. In other cases it may be "this is what we did and it worked so we carried on using it".

There can be considerable variation between batches of catalyst depending age, supplier etc which are out of your control unless you are making your own (and your own experimental technique may introduce variability). Even then you may fail to achieve the reported yields (and this opens a whole other debate).

I would suggest looking at the published procedure that is closest to the reaction you are trying to run and using their conditions as your starting point but be prepared to alter them as each reaction is different.


The group that initially publishes a new method will usually run a series of test reactions using different loadings of catalyst, different relative equivalents and different reagents to deterine which combination gives the most optimal results. For example, if n-BuLi is a reagent in a new method, it could be that the optimal amount of n-BuLi is $\pu{1.2eq}$ to ensure complete deprotonation of the starting material; however, it could also be that $\pu{0.95eq}$ of n-BuLi are ideal to prevent overreaction and side reactions. The same testing is typically performed for most reagents and thus you will end up with reactions that work best with, say, $\pu{3eq}$ of amine base although only catalytic amounts of base would be needed according to the proposed mechanism.

The original publication will often contain one or more optimisation tables that detail which combinations and variations were tried (including leaving out the catalyst to prove the reaction is indeed catalysed).

Once a later group uses the previously published results by Previous Group to apply a reaction to their own synthesis, they will typically simply use the best conditions they can find. Sometimes, there may be a publication that uses $\pu{5mol\text{-}\%}$ while another uses $\pu{10mol\text{-}\%}$ for essentially the same transformation. A synthetic group may try whichever they found first, whichever gave the higher yield, whichever they trust more or both.

Then, a reaction may not work as intended in the later synthetic works; the product will be there but the yield will not be sufficient or a side reaction may occur. In those cases, synthetic groups will often just guess which reagent is insufficient in the published method and up that reagent. For example, if the published reaction uses $\pu{3 mol\text{-}\%}$ catalyst but the later group only achieves $\pu{30\%}$ yield, they may just re-run the reaction with $\pu{9 mol\text{-}\%}$ catalyst and hope that pushes the yield to $\pu{80\%}$ or more. If it does: they score. If not: ‘well, maybe if we triple the amount of base …’


Not mentioned yet by the other answers is this: Many (not all) organic reactions lead to a chemical equilibrium, and only substantially increasing the relative amount of one / of multiple reagent(s) compared to an other, or / and removal of one / of multiple reaction products renders these reaction synthetically useful.

Recall e.g., esterification reactions where a distillation (Dean-Stark apparatus) constantly removes water from the reaction mixture. Or, the reversible introduction of $\ce{-OH}$ / $\ce{-NH2}$ groups in the course of the Bucherer reaction. Depending on the concentration of ammonia, either the formation of the naphthol 1 or the if naphthylamine 6 is favourable:

enter image description here


A bit related to the answer provided by @Waylander is the observation that many catalysts are expensive. This may be by their elemental composition, or their cost of fabrication. This is true for many transition metal-based catalysts (you count by unit of money per mole), e.g. Grubbs' catalysts for olefin metathesis. Or, see e.g., Friedel-Crafts acylations, where the stoichiometric use of Lewis acids like $\ce{AlCl3}$ in may be ok at lab scale. For a manufacturer, they however may represent a heavy cost factor after the reaction for their safe hydrolysis and the subsequent neutralization of waste water. Thus, you seek for (reasonable) short, yet reliable efficient syntheses with conversions requiring low stoichiometric equivalents of reaction partners. Process chemistry (one representative journal) is a whole branch within chemical research to scale bench chemistry to large scale, often using different solvents / reagents / reactions to perform eventually same transformations safely (a relevant question on chemistry.se).


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.