Hopefully this is not a stupid question, but I have problems finding information about this. This question is especially targeted for adhesives. A relatively large amount of NaOH is needed to catalyze the polymerization reaction between phenol and formaldehyde. As it only functions as a catalyst, it is not consumed. When making plywood, what happens to the NaOH? Is it simply trapped within the polymer matrix, or is is pushed out of the curing polymer as its solubility decreases? Is it not problematic?
If enough sodium still appears in the analysis of your samples, it may be due to excess NaOH, but if it does not appear, the synthesis may have been successful and it is not interfering with the result and if you want to eliminate the excess NaOH you can do several washes per example with water or absolute ethanol. I hope it is helpful and not confuse more :)
Dear Alexander, thank you for posting this interesting technical question. This is absolutely not a stupid question. In fact, I never really thought about this, and many references about phenol-formaldehyde adhesives / resins simply do not mention what the fate of the added NaOH catalyst is. I assume that part of it just remains in the polymer. In this context I found two relevant articles which could provide some useful hints. This first paper is entitled:
Effect of synthesis parameters on thermal behavior of phenol–formaldehyde resol resin
Article Effect of synthesis parameters on thermal behavior of phenol...
Unfortunately this article has not been posted as public full text on RG, but perhaps you can access it through your institution. It is mentioned in this study that NaOH can react with formaldehyde to give methanol and sodium formiate, Na+HCOO–. This shows that the added NaOH does not necessarily remain unchanged in the polymerization process.
The second article is entitled:
NaOH and Ba(OH)2 Compound Catalyzed Phenol-Resorcinol-Formaldehyde Copolycondensation Resin Adhesive for Recombined Bamboo
Article NaOH and Ba(OH)2 Compound Catalyzed Phenol-Resorcinol-Formal...
In this study NaOH and Ba(OH)2 were used alternatively as catalysts. The authors point out that the experimental results showed that the NaOH and Ba(OH)2 catalyst shortened the curing time of the phenol-formaldehyde adhesive and also ensured suitable water solubility of the adhesive. Thus apparently it has certain advantages to have remaining NaOH in the polymer.
I hope this helps. Good luck with your work and best wishes, Frank Edelmann
Dear Alexander Henn , In the polymerization reaction:
A fraction of OH- attack to the carbonyl centers of the formaldehydes and double bond turns to one and negative charge comes on the head of Oxygen atom. Then H pushed out and shift on O atom next phenols attack to the carbon centers afterwards OH- pushed out and H shift back to it's first position and double bond forms again.
NaOH + C6H5OH --> C6H5O- + H2O
H2CO + C6H5O- + H2O --> HOCH2-C6H5OH + OH-
And a fraction of OH- goes on the side equilibrium reaction:( This is the reason for need of excess NaOH )
2 HCHO + NaOH --> HCOONa + CH3OH
or
H2CO + H2O H2C(OH)2
Article Research status, industrial application demand and prospects...
Dear Sina Safavi thank you for providing this interesting answer. It is just not clear to me what the intermediate compound "HC(OH)2" could be? As such, it would contain formally trivalent carbon and could not exist as a neutral molecule. It would make more sense if it was a hydrate or some sort of ortho-form of the formiate anion. In that case it should be formulated as HC(OH)2–. Could you please clarify? Thank you and best wishes!
Dear Frank T. Edelmann thank you for your delicate look to the answer and mention that to me. It was just a typo. The compound is methanediol with this formula H2C(OH)2. I corrected that in my answer. respectfully
Dear Carlos F. Marcos
the complete and correct form of equation is :2 HCHO + NaOH --> HCOONa + CH3OH (cannizaro reaction)
I wanted to show substitution of H with OH. But you are absolutely right about charge balance in equation. Thank you very much for reminding it to me so I have corrected it.
Thank you all for the discussion! Although my question regarded mostly the practical aspects of NaOH after the polymerization has already happened, the sources that Sina Safavi provided will be useful, so thank you for the discussion on chemistry! I especially thank Frank T. Edelmann for your very kind, precise, and helpful answer. It is encouraging for a young researcher to see that it is worth asking also things that may feel simple, as they in the end may actually be somewhat complicated.
I think some sodium hydroxide can end up as phenolate salt (water formed, Na+ ion remains close to the negatively charged aryl ring, probably coordinated/liganded with residual water (or perhaps like a metallocene like ferrocene or uranocene, liganded by aromatic ring?). The negative change attached to phenol ring by deprotonation of phenol would activate otho and para substituent sites of Phenol to polymerize with methanal (formaldehyde); Or perhaps some unreacted Formaldehyde with Sodium hydroxide may undergo Cannizzaro disproportionation reaction to form sodium methanoate and methanol.
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