Author Topic: P2Ps from benzene, acetone and Mn(AcO)3  (Read 20727 times)

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PEYOTE

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Possible mechanism
« Reply #20 on: September 18, 2001, 12:12:00 PM »
I'm trying to make a possible (TOTALLY theoretical) mechanism for this process. But I've one question:

The reaction?

(CH3COO)3Mn(III) + CH3COOH + C6H6 + CH3COCH3 -------> C6H5CH2COCH3 + (CH3COO)2Mn(II) + WHAT?

I've lost a couple of protons....

Can you help me? Thanks

If Rhodium can help me, I'll get my response on his e-mail (or any anonymous else), cause I've to make some images but I don't have any http to get them.

PEYOTE

  • Guest
Re: Possible mechanism
« Reply #21 on: September 18, 2001, 01:36:00 PM »
Ehm.... another question:

There is a production of H2?


Rhodium

  • Guest
Re: Possible mechanism
« Reply #22 on: September 18, 2001, 02:28:00 PM »
I haven't got any email from you, try rhodium@privacyx.com - my hushmail is malfunctioning.

PEYOTE

  • Guest
Re: P2Ps from benzene, acetone and Mn(AcO)3
« Reply #23 on: September 19, 2001, 01:51:00 AM »
Ops, I'm stupid.... THIS IS the reaction involved...

Sorry!

:-[  :-[  :-[  :-[  :-[  :-[  :-[  :-[  :-[

Rhodium

  • Guest
Re: P2Ps from benzene, acetone and Mn(AcO)3
« Reply #24 on: September 19, 2001, 03:06:00 AM »
Taken from Fester's 5th Ed of his book:

To get Mn(III) acetate from Mn(II) acetate, we return to a recurring theme in industrial chemistry - the electric generation of Mn(III) from Mn(II). We saw one example of this kind of conversion back in Chapter Nine in the benzaldehyde recipe. For this next one see Acta Chemica Scandinavica B 33 (1979), pages 208-212. At a graphite or platinum anode in a simple, undivided cell, using a cathode much smaller than the anode to minimize reduction of the Mn(III) formed, the chemists produce Mn(III) acetate from Mn(II) in glacial acetic acid solvent.

One mole of Mn(II) acetate is dissolved in one liter of glacial acetic acid. A little bit of sodium lithium fluoroborate (a few grams) is added as current carrier to the solution. One could also try sodium or potassium acetate as current carrier; it may not interfere in this reaction. The fairly large graphite or platinum anode is placed in the solution, along with the smaller cathode. The mixture is warned, and then with stirring, DC current is made to flow through the cell. One should apply 4 milliamps of DC current for each square cm of anode surface facing the cathode. This is a one-electron oxidation, and one can count on getting around 66% efficiency in the oxidation. So one should pass about 1.5 faradays of current. One faraday is 96,500 amp seconds, so if for example one is passing one amp through the solution, the electrolysis to Mn(III) will take 40 hours. At four amps, it will take 10 hours, and so on.

At the end of the electrolysis, one has the Mn(III) acetate solution in acetic acid. Then to this solution, one can add the benzene and acetone with stirring, and react as usual. It's a lot of work to get 30 ml of phenylacetone, but those chemicals certainly are low profile, cheap, and easily available.

For another example of electric generation of Mn(III), see US Patent 4,560,775.

PEYOTE

  • Guest
Re: Possible mechanism
« Reply #25 on: September 19, 2001, 06:03:00 AM »
I've posted a possible mechanism to your email rhodium@privacyx.com, Chief.

Rhodium

  • Guest
Re: Possible mechanism
« Reply #26 on: September 20, 2001, 02:27:00 AM »
That mechanism is way over my head, but you can read it at

https://www.thevespiary.org/rhodium/Rhodium/archive/mechanism_beta.doc

(word file).

PEYOTE

  • Guest
Re: Possible mechanism
« Reply #27 on: September 20, 2001, 02:47:00 AM »
Thanks Chief!!!

Ehm, but I'm a silly man... I've forgotten to sign that .doc!!!!  :-[  :-[  :-[



DarkStarCrashesPouringItsLightIntoAshes

cilliersb

  • Guest
Re: Possible mechanism
« Reply #28 on: September 20, 2001, 11:30:00 AM »
The Mn(Ac)3 synth seems to work just fine, will post yields after crystals have finished falling out.

I didn't leave out the step of evapping the GAA first and then redisolving Mn(Ac)2 back in. It does take a 2 hour reflux to get it to dissolve.

The Mn(CO3)2 that I got was not white though, it was a very light pink color. (Which is probably why - in the original post - the solution after dissolving it in GAA is a "pleasant pink".

Who knows, I may have fucked up, but the rest of the reaction colors seemed exactly as described. Is it possible that I made something else? ::)

Rhodium

  • Guest
Re: Possible mechanism
« Reply #29 on: September 20, 2001, 01:58:00 PM »
I am interested in the preparation of Manganese(III)acetate.

We know from experimental evidence that the following reaction can occur:

KMnO4 + HCl + C6H12N4 => MnCl2 + KCl + ?

So the reaction could be something like that the hexamine is being hydrolyzed by hydrochloric acid to ammonium chloride and formaldehyde, the latter is then being oxidized by the potassium permanganate to formic acid like this:

2 KMnO4   5 HCHO   6 HCl => 2 MnCl2   2 KCl   5 HCOOH   3 H2O.

Or perhaps the formaldehyde gets oxidized all the way to carbon dioxide, like this:

4 KMnO4   5 HCHO   12 HCl => 4 MnCl2   4 KCl   5 CO2   16 H2O

Why is hexamine used, and not plain formaldehyde? Is the ammonium ions there to act as a pH buffer, or is is just convenient to use the less toxic hexamine? What is the reference for this reaction?

Anyway, the following reaction can occur, and could be another way of making MnCl2 (the MnO2 can be had from batteries):

MnO2   4 HCl => Cl2   MnCl2   2 H2O (this is how Scheele discovered chlorine in 1774)

The precipitation of manganese(II)carbonate is trivial:

MnCl2   2 NaHCO3 => 2 NaCl   MnCO3   CO2 or MnCl2   Na2CO3 => 2 NaCl   MnCO3

and its conversion to the manganese(II)acetate is too:

MnCO3   2 AcOH => Mn(OAc)2   CO2   H2O

But, how does the KMnO4 oxidation of manganese(II)acetate proceed? Have we references for this?

A very plausible reaction is 4 Mn(OAc)2   5 HOAc   KMnO4 => 4 Mn(OAc)3   KOAc   MnO2 according to experimental observations.

Rhodium

  • Guest
Re: Possible mechanism
« Reply #30 on: September 20, 2001, 03:09:00 PM »
What about the following mechanism for the coupling between acetone and benzene:



This explains the low yield (40%) based on the number of moles of Mn(III) used. One equivalent of Mn(III) is needed to first generate the free radical, and one equivalent is needed to terminate the radical reaction. No hydrogen are lost in this mechanism, and it explains the low molar yield, which is actually an 80% chemical yield, but it is halved because two equivalents are needed for the reaction.

Reference for the above:

http://www.york.ac.uk/depts/chem/staff/afpgrp/Work_Gregory.htm



These articles might also be of interest (haven't checked them out yet):

Linker, U.; Kersten, B.; Linker, T. "Potassium Permanganate-Mediated Radical Reactions: Chemoselective Addition of Acetone to Olefins" Tetrahedron 1995, 51, 9917-9926.
Linker, T. "Manganese(III)-acetate – A Versatile Reagent for Radical Generation and C–C Bond Formation" J. Prakt. Chem. 1997, 339, 488-492.

I find it amusing that a guy called "Linker" has written an article on "C-C bond formation"  :)

obituary

  • Guest
Re: Possible mechanism
« Reply #31 on: September 20, 2001, 04:26:00 PM »
nice mech proposal- both Rhodium and Peyote.  and Rhodium- you are too easily amused, you'd think that with your apparent intelligence you could find something a little more in depth to amuse yourself.  but hey, simply pleasures for simple people

Rhodium

  • Guest
Re: Possible mechanism
« Reply #32 on: September 20, 2001, 05:36:00 PM »
Life would be very boring if one wouldn't be able to find some level of amusement in even the dullest of places (like in a list of references).

obituary

  • Guest
Re: Possible mechanism
« Reply #33 on: September 20, 2001, 07:18:00 PM »
don't get obit wrong here- obit was also amused.  like obit said simple pleasures for simple people. obit just didn't state obit as being one before now.   :)

PEYOTE

  • Guest
Re: Possible mechanism
« Reply #34 on: September 21, 2001, 02:21:00 AM »
Referring to:

Is the ammonium ions there to act as a pH buffer, or is is just convenient to use the less toxic hexamine?



May the NH3 work as ligand?





Just A Touch Of Mojo Hand

PEYOTE

  • Guest
Re: Possible mechanism
« Reply #35 on: September 21, 2001, 02:25:00 AM »
Hey Chief, correct me If I wrong, but there is a 5 bonded carbon in your picture... is it an intermediate like a SN2 mechanism? I hope this.

However your mechanism is very very very interesting


Just A Touch Of Mojo Hand

PEYOTE

  • Guest
Re: Possible mechanism
« Reply #36 on: September 21, 2001, 02:29:00 AM »
Thanks obit, but I wasn't only amused, I was so curious, interested and in a complete creative ecstasy (not MDMA,  ;) ) that I had to research about this stuff.


Just A Touch Of Mojo Hand

Rhodium

  • Guest
Re: Possible mechanism
« Reply #37 on: September 21, 2001, 02:53:00 AM »
Yes, there is a 5-valence carbon drawn in the intermediate formed after the acetone radical has added to the benzene ring. The radical is in reality distributed throughout the aromatic ring though, thereby making it more stable.

I believe someone should try to perform the reduction of KMnO4 with only formaldehyde and see what the results are (like if 1.25 or 2.5 molar equivalents of formaldehyde is needed) and if more acid is required than indicated in the equations above, or if that is just enough.

halfapint

  • Guest
Re: Possible mechanism
« Reply #38 on: September 21, 2001, 05:51:00 AM »
Did I miss something? When Antoncho said "camping fuel (SWIM used hexamine cause he had it at hand - any ammonium salt will do just fine)" I was thinking ammonium sulfate (or ammonium nitrate, which I was very hesitant to try  :o ) might reduce KMnO4 appropriately. Ethylamine HCl? So I fail to understand the fuss about hexamine.

turning science fact into <<science fiction>>

Antoncho

  • Guest
Re: Possible mechanism
« Reply #39 on: September 21, 2001, 06:09:00 AM »
So do i ::)

You can reduce KMnO4 to Mn(II) w/just about anything, as long as pH is acidic.

Antoncho