Effect of catalytic NaOH on atom economy

I conducted a base catalyzed aldol condensation using NaOH. Now I'm supposed to calculate the [atom economy](https://en.wikipedia.org/wiki/Atom_economy), AE:


$$\text{AE} = \frac{{\text{MW of desired product}}} 
{\sum{\text{MW of reactants}}} \times 100\%$$


but I'm unsure on how to interpret the role of NaOH: It reacts with the reagent by deprotonation (base), but it also gets regenerated (catalyst) and is not reused (waste).


Since the AE is also a way of depicting waste generated during a reaction my approach would be to include it as a reactant although it has a catalytic role which as I learned is not considered in AE?
What would be the correct way?




You may consult e.g., the [Wikipedia article](https://en.wikipedia.org/wiki/Atom_economy) about the topic, and use the literature references it cites as an entry.


Though the very first example of Sheldon's paper is about an acid catalyzed reaction, you can draw a conclusion *by analogy*:


[![enter image description here](https://i.stack.imgur.com/RMjTs.png)](https://i.stack.imgur.com/RMjTs.png)


Both reactions intend to oxidize the alcohol to yield the ketone. In the first instance, the the acid is used in stoichiometric amounts and if you are not able to recycle / reuse the acid, it is waste; low atom efficiency. In the second case, you have a solid catalyst which after the reaction is separated, filtered-off and can be reused for the next batch of reaction; this improves atom efficiency.


So, how do you run your aldolization? Do you run it e.g., by phase transfer catalysis, can recover the base for the next batch of synthesis? Because you say no, so you count it fully into MW of reactants.


Sheldon, R. A. Atom efficiency and catalysis in organic synthesis. *Pure and Applied Chemistry* **72**, 2000, 1233–1246; <https://doi.org/10.1351/pac200072071233> (open access).


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