[Lucek]

Hi, does anyone know what will happen in this reaction :


what if I have sodium naphthoxide with excess of NaOH and heated to 200*C and 300*C?
What If I introduce NaCl and NaOH simultaneously?

Does anyone know?

[pgk]

1). What will happen if the water is acidic, in the second step?
2). What is the necessity of NaCl and when must it be added?
Hint: This reaction occurs  in absence of solvent and above the melting point of sodium naphthalenesulfonate.

[Lucek]

Hi, thanks,

if water is slight acidic it will be a naphthol, but in this case it not.

I changed a little bit the process, do you know that process? is it known? Or maybe you know something like this form benzenesulfonic acids reactions? Can you please give me a hint what could happen or where should I look for an answer?



I am not sure about the NaCl influence, does it push the reaction towards a sodium naphthoxide? without, would It be just a naphthoxide?

[pgk]

Yes, a naphthol and this is the main industrial method of preparing phenols and naphthols by further neutalization with aqueous HCl, followed by recrystallization.
There is no need to add water therein. This reaction is solventless and occurs in liquid state, above the melting point of the reactants. Thus, NaCl behaves as an impurity that decreases the melting point of the mixture and leads to significant energy savings in industrial scale, together with salting out the recrystallization procedure.
Mechanistically, this reaction occurs via aromatic nucleophilic substitution SnAr2.
But how an aromatic nucleophilic substitution occurs herein, in the absence of conjugative R- groups (electron attractive groups)?

[Lucek]

1) but will the same reaction take place in a solvent? I understand that you don't need to use it, but I am curious
2) NaCl - decreases the melting point! right, thanks
3) what will happen if you not acidify it, will it going to stay just as a naphthoxide / sodium naphthoxide ?

is it the mechanism? :


isn't it easy to desulfonate the ring?

[pgk]

1). Yes, the reaction can take place in a solvent with high boiling point, but it will significantly delay due to the dilution, as well due to the decrease of the overall polarity (polar reactions are favored in polar medium).
2). You are welcome.
3). Yes, it will remain as sodium naphthoxide, which is quite alkaline (salt of a strong base and a weak acid) and highly hygroscopic (inconvenient storage).
4). No, the mechanism is not correct:
a). In the transition state, the leaving arrow does not point to the leaving group, neither to a conjugative aromatic carbon.
b). There is a trivalent carbon, which is uncharged.
c). There is a negatively charged hydrogen that is already bonded with an oxygen.
d). There is a monocharged sulfurous anion, in the final step.
5). Aromatic nucleophilic substitutions have high activation energy and take place either under drastic conditions (e.g. high temperatures, benzyne formation, etc.) or in presence of R- groups that decrease the activation energy by extending the conjugation of the transition state.

[Lucek]

Thank you again,

I am not sure if I am able to fully understand what is happening

b) where is the carbonation. How, then, can it be drawn? like that?:


c) do you mean OH- connected to C?
d) SO3 instead of SO3-?

[pgk]

Take also into account, the aromatic conjugation forms of the transition state.

[Lucek]

Do you mean possible resonance structures?



how can it influence the mechanism? does it make it more stable?

and regarding this:


Form where do you have your OH- groups, on the left site of reaction?

then on the right side will you end with Ar-ONa    and SO3 and H2O?

[pgk]

1). In the transition state, either an hydroxyl group or an oxanion but not a negatively charged hydrogen that is already bonded with an oxygen because the oxidation state of hydrogen is +1 (except in metal hydrides).
2). Nucleophilic attack always creates a carbanion and not a carbocation.

[Lucek]

Thank you PGK,
thanks for your time


so lets say that your source of OH is NaOH, than you end up with that:


or with sodium naphthoxide?

if in the system there is no free oxygen, no acid, just naphthalenesulfonate NaOH and heat 200 - 300*C, can I obtain naphthol as you presented?

[pgk]

1). ArOH +  HSO3-  +  2NaOH  →  ArONa +  Na2SO3  + 2H2O
Sodium naphthoxide plus sodium sulfite plus water
2). Theoretically, yes.
Because, sulfurous acid (H2SO3) has pka1 = 1.81 and pka2 = 6.91, while 1-napthol has pka = 9.34. Thus and according the reactions mechanism, the generated bisulfite anion can further acidify the so formed sodium naphthoxide.
But in practice, this doesn’t happen due to the excess of NaOH that is added because 1-naphthalenesulfonic acid has pka = 0.17, which means a strong acid that will consume an additional equivalent of NaOH.
1). If you prefer, the representation below, is more pontual:
ArSO3H + NaOH  → ArSO3Na + H2O
ArSO3Na + NaOH (heat) → ArOH + Na2SO3
ArOH + NaOH  → ArONa + H2O

[Lucek]

Thank you!

Do you know any publication, where I can read about this particular situation??

cheers,
Lucjan

[pgk]

Process for the preparation of phenols, US Patent 1988156, (1935)
Process for the preparation of phenol, US Patent 2007327, (1935)
Preparation of phenols, US Patent 2578823, (1951)