bucket reductive animation

[greeter]

Swim ran a 4x MM in a 6 gal bucket with a 3" stirbar (3" round x 1/2" thick).  Swim has a decent hotplate/stirrer.  110g foil were added to 2.5L methanol and stirring started.  The stir bar did an ok job but was not able to move the foil at the very edges of the bucket.  .5L methanol w/ 1g HgCl2 was added, stirring was fine until the reaction got going and about 30g of ketone added then swim could not see any indication that the stir bar was still turning (no vortex, and for that matter, no movement).  The bucket was swirled gently by hand, but swim could still see no activity from the stir bar.

Now the reaction mix was pretty hot, and it was clear that that Al was amalgamating, so swim said WTF and kept going, basically adding ketone to a mix that wasn't visibly stirring.  Total addition of 100g ketone took about 22 minutes (it appeared to drip on TOP of the sludge, run across it and get really hot and turn white).  Methanol bubbled out of the condenser from time to time when it got really hot and more was added down the condenser.  Once all the ketone was in swim added another 500mL of methanol to the mix.  This helped quite a bit with regards to stirring.  Once the foil had broken down substantially then the stirbar was able to get going and the vortex was back.  Swim doesn't know if the bar spun the entire time or if it got stuck in the mud halfway through.

Swim also doesn't yet know what the yield is on this batch, since the product is sitting in DCM waiting to be distilled, but it seems like it was ok.  He will likely recover the DCM, boil off any water and the weigh the crude oil he has just to check.  Swim would like to know if he should expect a significant loss in yield from the mix not being stirred efficiently the entire time?

Swim may try another batch with an extra liter of MeOH in from the start, or just build an overhead stirrer.  Nevertheless, this is really pushing it with a 3" stirbar.
A watched pot will indeed boil.

[greeter]

According to the MM writeup you should of used 1.6g HgCl2

It has been mentioned in this thread and others not to use so much HgCl2.  1g was perfect.  Any more would have torn the aluminum up too fast.  As it was *all* of the aluminum was consumed after about 1hr 45min, no little flecks or anything.  Swim waited 20 minutes before starting to drop ketone.  The timing was good.

Total addition of 100g ketone took about 22 minutes

    A bit of a hurry you were in?? Swim would of stretched that out a lot more. at least 45min.

Again, other threads have mentioned that the faster you add it, the better your yields.  A 6gal bucket has so much overhead that it probably could have been dumped in wholesale, 22 min was no problem in a container this large.

Swim thinks the only potential problem was the lack of adequate stirring.

A watched pot will indeed boil.

[Chromic]

[RoundBottom]

maybe i missed something, but isn't the HgCl2 only responsible for preparing the Al foil for reduction by stripping the aluminum oxide off the surface?
i learned a thing or two from charlie dontcha know.

[RoundBottom]

ok, here's the next logical question then.  what reaction/mechanism reduces the Al foil?  SWIM just did a 2x MM AlHg Nitro, and got most of the Al foil left in the flask after the usual amount of time.  this has been a reoccuring theme lately, and is causing a bit of consternation.

the reagents have all been used before.  the reaction was nowhere near as active as usual.  the temp did make it to 64°C as per usual. 

let's say someone ran a MM AlHg Nitro without ketone, but used everything else.  would the Al foil still get reduced?  thx.
i learned a thing or two from charlie dontcha know.

[goiterjoe]

but it is hydrolyzed by the water.  If you add nitro to an Al/Hg mixture, it will reduce the nitromethane to methylamine, and therefore eat up some of the aluminum.  Without adding the ketone, not as much aluminum will be used up because there is no imine to be reduced.

[RoundBottom]

> B. Your nitro isn't as pure as you thought. Did you distill fuel to get your nitro?

as i stated, nothing is different about the reagents.  the nitro is the same ACS grade that has been in use for a year already.  never had a problem with it.  ketone, MeOH, and HgCl2 are all literally the same bottles/batches.

correction, the Al foil is different in this single reaction, but it is a brand that has been suggested by canadian bees as wholey acceptable (alcan heavy duty).  it is thinner than american reynolds heavy duty, which theoretically means FASTER reduction.  though this doesn't explain the previous 2 "failures" (greatly reduced yields) which used the same imported american foil as all previous succeses.

> A. You didn't let the AL amalgamate long enough, and the nitro fell on Al203 and didn't reduce

this intrigues me somewhat, as it did appear to amalgamate slightly slower than normal.  previously have started addition at between T+10 and T+15 minutes.  this time addition was started at T+15.

the foil nuggets were indeed shiny when the liquid was decanted off at T+290m.  when a small amount of 25% NaOH was poured on the nuggets, they started fizzing, and kept on fizzing for hours.

gassing will be done tonight.  update on yield, if any, will be posted.
i learned a thing or two from charlie dontcha know.

[SuperStar]

When you guys are done with these scaled up AL/Nitro how much Toulene are you using? Are you scaling this up equal as well. Seems like large sep funnel will become a problem?
Pardon my friend officer, He's a little slow.

[greeter]

Swim scaled the solvent up 4x as well.  He thinks this was overkill (see swim's post in the Newbee Forum asking for some help on this very thing, maybe someone can...).  Swim did the mixing with a stirbar in the bucket and separated with multiple pours into a 2L sep funnel.  It was a pain in the ass but even at 4x, the workup didn't take more than a day.
 
A watched pot will indeed boil.

[SuperStar]

Thanks,

These guys have made some of this shit so easy it seem ridiculous to complain about anything, but hey constant improvement is what we are about.
Pardon my friend officer, He's a little slow.

[El_Zorro]

I'm confused now.  I didn't think that the HgCl₂ was to strip the Al₂O₃ off the surface, I thought it was there to dissolve with the Aluminum metal, which would release three electrons, which allows the MeNO₂ and imine to steal some hyrogen from water.  Can some of you chem experts help me out?
Do not go gentle into that good night.  Rage, Rage, against the dying of the light.  --Dylan Thomas

[SuperStar]

NOTE: SuperStar is not a chem expert.  Just relaying what SWIM read.

SWIM's understanding is that the HgCl2 or HgI2 makes the AL surface ready to donate those electrons.  Something about AL in air/oxygen forms a thin surface film that blocks access to the real AL underneath.  The Hg salts are sort of like transporters of the electrons.


Pardon my friend officer, He's a little slow.

[El_Zorro]

Al does form an Al₂O₃ film in contact with oxygen, but I thought the definition of an amalgum was two metals dissolving in each other.  Kind of like an alloy, right?  Where's Rhodium at?
Do not go gentle into that good night.  Rage, Rage, against the dying of the light.  --Dylan Thomas

[Rhodium]

The Al reduces the Hg²⁺ to elemental Hg, which deposits on the surface of the Al (forming an Al/Hg amalgam), preventing it from forming the protective oxide coating.

Then there is an electrochemical effect due to the electrical overvoltage of Hg which makes the amalgam more prone to reduce the organic substrates than creating hydrogen gas which would just bubble away (like when you throw aluminum into aqueous acid). UTFSE for "overvoltage" and you'll probably find a more precise explanation.

[El_Zorro]

then why does the Al completely dissolve before the organic substrates are added?
Do not go gentle into that good night.  Rage, Rage, against the dying of the light.  --Dylan Thomas

[Rhodium]

I said that the Al becomes "less prone" to give off hydrogen gas, not that it is eliminated.

If your Al dissolved before you have had any time to add your reactants, then your Al is too thin (too large surface area), you have amalgamated the Al for too long, or you are simply too slow.

Cooling the solution also slows the reaction.

[El_Zorro]

yeah, I got what you said, I'm just confused.  I thought I knew the way the amalgamation worked, but I guess I don't.
Do not go gentle into that good night.  Rage, Rage, against the dying of the light.  --Dylan Thomas

[goiterjoe]

I remember a thread about this being done a long time ago, although I don't remember who said what or even how to begin searching for it.  You will probably end up with 1000 hits of static if you search for nitro, Al/Hg, and methylamine.  I'm not sure how one could go about extracting this mess but I imagine it would go something like this. 

After the reaction has cooled back down, you would add sufficient hydrochloric acid to make the mix slightly acidic, filter the mix through filter paper to remove any excess aluminum, and then evaporate off the water.  You could probably then set up an apparatus for distilling methylamine gas and reclaiming it in water or methanol.  You would take the dried up mass from the evaporation and put it in the distilling flask, and then slowly titrate in concentrated NaOH to release the methylamine gas.  The MeAM gas would be dispersed into a tall beaker of cold water, where it could be reclaimed or possibly used as is.

I'm not recommending anyone try this;  if anyone sees any problems with it, please shoot it down.

[El_Zorro]

I don't get this:

Re:
aluminum is never reduced in the reaction but it is hydrolyzed by the water.  If you add nitro to an Al/Hg mixture, it will reduce the nitromethane to methylamine, and therefore eat up some of the aluminum.

I don't think hydrolized would be the correct term, would it?  Somebody help me out.
Do not go gentle into that good night.  Rage, Rage, against the dying of the light.  --Dylan Thomas

Edit: I really don't think the Al is hydrolized by the water now, I think it's oxidized to Al₂O₃, right?  So I guess the Hg doesn't have to dissolve with the Al, it just prepares the Al by cleaning the surface, and when the ketone/nitro is dipped in, the reaction proceeds as follows, right?

1)2Al+3H₂O-->Al₂O₃+6H+6e^-
2)MeNO₂+6H+6e^--->MeNH₂+H
₂O
3)ketone+MeNH₂-->imine+H₂O
4)imine+H+e^--->amine

Is this correct?  Is it balanced?

[lugh]

Post 106388 (missing)

(scwam: "Re: Aluminum Sludge", Chemistry Discourse) &

Post 271975

(Osmium: "Re: Factors of Al sludge consistancy", Chemistry Discourse) might bee helpful