Trying to prevent conversion of naturally occurring organic acids in wood pulp to their derivative alcohols and aldehydes under temp and pressure conditions?

Let’s look at the major organic acids in a wet chemistry wood pulp and
see how the presence of sodium carbonate benefits us:

First, let’s examine the major wood acids in equilibrium:

Formic acid = CH2O2 = H+ + CHO2- [formate]

Acetic acid = C2H4O2 = H+ + C2H3O2- [acetate]

Propanoic acid = C3H6O2 = H+ + C3H5O2- [propanate]

Butanoic acid = C4H8O2 = H+ + C4H7O2- [butanoate]

Phenolic acid = C6H6O = H+ + C6H5O- [phenoate]

Benzoic acid = C7H6O2 = H+ + C7H5O2- [benzoate]



Now let’s add sodium carbonate to the equilibrium equation for each
acid:

2CH2O2 + Na2CO3 = 2NaCHO2 + CO2 + H2O

Sodium methanoate

2C2H4O2 + Na2CO3 = 2NaC2H3O2 + CO2 + H2O

Sodium ethanoate

2C3H6O2 + Na2CO3 = 2NaC3H5O2 + CO2 + H2O

Sodium propanate

2C4H8O2 + Na2CO3 = 2NaC4H7O2 + CO2 + H2O

Sodium butanoate

2C6H6O + Na2CO3 = 2NaC6H5O + CO2 + H2O

Sodium phenoate

2C7H6O2 + Na2CO3 = 2NaC7H5O2 + CO2 + H2O

Sodium benzoate

First....do you believe the organic acids above, in the presence of
sodium carbonate, would in fact react as I have shown in the equilibrium
equations...essentially forming sodium methanoate, ethanoate, propanate
etc. etc. compounds?

Second, how stable do you believe these sodium methanoate, ethanoate
etc. compounds to be under increased temperature and pressure....in
other words, if these “oate” compounds were raised to a temperature
of 400o F and 400 psi, do you believe that they would off-gas to
compounds like methanol, formaldehyde, acetaldehyde ----- my mission is
to try to prevent the generation of methanol, formaldehyde, phenol, and
acetaldehyde gaseous air emissions!

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