[lutesium]

Dear Sir/Ma'am

How can alkyl potassium sulphates be synthesised???

R-OH + K ---> R-OK + H° + e-
R-OK + H2SO4 ---> R-KSO4 + H2O

Is this equation true??? Please help me!!!

[ARGOS++]

Dear Lutesium;

Sorry that will not work!  At least not this way.
Both equations are incorrect, because the first is not balanced and “free Electrons” exist only in Ammonia (aq) at ~ -78°C (A very deep blue colour.)
In your second equation H₂SO₄ would have to spend an OH^- group/ion to build water, but it is a strong Acid instead.

Better would be:                                                + 2KOH
  HO-SO₂-Cl + H-O-R   –––>  HO-SO₂-O-R  + HCl –––––––––-–>    KO-SO₂-O-R + KCl + 2H₂O

The first Educt is called: “Clorosulfonic Acid” and is normal available.

Good Luck!
                   ARGOS⁺⁺

[lutesium]

Thanks ARGOS but this sentence puzzled my mind:

"In your second equation H2SO4 would have to spend an OH- group/ion to build water, but it is a strong Acid instead."

In my second equation it has already been mentioned that the OH- group of the acid has been spent to build water. So why not it can't be true???

And doesn't by simply adding H2SO4 into an alcohol produce the intermediate alkyl hydrogen sulphate???

H2SO4 + R-OH ---> R-HSO4 + H2O

Thank you very much!!!

[ARGOS++]

Dear Lutesium;

With this my second equation sentence I tried to imply you that a compound which “spends” an OH_- group to build water must be a Base and that’s “not” possible for a strong Acid.
Quintessence of that: 
For a real extraction of –OH from a strong acid you need a very more drastic reaction as the one you postulated. And for such an extraction I know in the first moment only two reaction schemes:

Building of Acid Anhydride or building the Acid Chloride.

For Sulfuric acid you are nearly not able to start therefore from the free Acid, you have to go quite different ways for building them.

Of course that you can build the Esther also with you last postulated reactions und adequate Conditions, but there is a small difference in your reaction scheme:

The surviving Oxygen is originating from the other Educt, the Alcohol.

(Allowedly: Not many times important!)

Good Luck!
                   ARGOS⁺⁺

[lutesium]

Thank you very much ARGOS++!!!

I'm not telling that my reaction scheme is true but if you pay attention also in my reaction scheme H2 was taken from the acid and O from the potassiumalcoholate...

R-OK + H2SO4 ---> R-KSO4 + H2O

And you told that the surviving O would be from the alcohol but at your equation also the surviving O is from the alcohol...

So we can say that simply by adding H2SO4 to an alcohol then adding the KOH can produce the desired Alkyl Potassium Sulphate... Right???

Thank you very very very much!!!

[ARGOS++]

Dear Lutesium;

SORRY!  –  for my Libertinism!

The surviving Oxygen is originating from the other Educt, the Alcohol.

Of course!  that in reality I meant the opposite: From the Acid.

To both reaction schemes I can’t tell about the real yield you may be able to realise and also about the yield of sideproducts, so I cannot absolutely favour the one or the other.
(It’s more the “organic feeling” that favours the Acid Chloride.)

Sorry once again.
Good Luck!
                   ARGOS⁺⁺