Author Topic: PP's General Chem Question Thread  (Read 15983 times)

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Rhodium

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diazonium
« Reply #40 on: May 07, 2002, 11:30:00 PM »
Both aromatic and aliphatic diazo compounds are made by treating the corresponding amine with aqueous HNO2 at 0-5°C...

PrimoPyro

  • Guest
Hm, That Simple?
« Reply #41 on: May 08, 2002, 04:50:00 AM »
Ah yes, this is what I had thought at first, but then I wondered if these "diazo" compounds were different than diazonium species I am already familiar with (along with their preparations, as they are wellknown here).

If they are the same, that really makes things a lot simpler, doesnt it?

This would mean that methylamine and nitrous acid would produce diazomethane, CH2N2, correct? Fear not, I have no intention of ever preparing this compound, it is only an example.

                                                    PrimoPyro

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Rhodium

  • Guest
diazomethane
« Reply #42 on: May 08, 2002, 05:18:00 AM »
As far as I know, the only viable ways for diazomethane is by degradation of MMNG or Diazald.

http://www.orgsyn.org/orgsyn/prep.asp?prep=cv4p0250


PrimoPyro

  • Guest
Ever Heard Of The Schlotterbeck reaction? ;-)
« Reply #43 on: May 08, 2002, 05:26:00 AM »

http://themerckindex.chemfinder.com/TheMerckIndex/NameReactions/ONR64.htm



If R = a phenyl group, and R' = a methyl group, we have reactants diazotoluene and acetaldehyde, and product phenylacetone and nitrogen. Prepare the diazotoluene from benzylamine.

8)

                                                   PrimoPyro

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Antoncho

  • Guest
Notes
« Reply #44 on: May 08, 2002, 08:03:00 AM »
Rhodium: is not the nitrosyl urea diazomethane prep. by Zealot good enough for you? Requires much simpler things and technique, may i say?


2 PrimoPyro: wow, this is VERY cool! Fuckin' cool! P2P from benzylamine and acetaldehyde, shit!!!


Care to find experimental details for that Schlotterbeck synth? ;)


Antoncho

halfapint

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Can't even catch a rabbit
« Reply #45 on: May 08, 2002, 08:20:00 AM »
Primo, you hound dog. If we didn't know you better, it might look like you sniff something lively out from time to time. No, I never heard of Schlotterbeck, and neither did you. I think you just hacked into Merck's site and stuck that stuff in to fool us. All right, now can you find a way to look any of that stuff up, so we can know what's going on?
Chempen sez:

Buchner-Curtius-Schlotterbeck Reaction.

E. Buchner and T. Curtius, Ber. 18 2371 (1885)
F. Schlotterbeck, Ber. 40, 40, 479 (1907); 42, 2559 (1909)
B. Eistert in Newer Methods of Preparative Organic Chemistry, English Ed. (New York, 1948), p521
C. D. Gutsche Organic Reactions, 8, 364 (1954)
J. B. Bastus, Tetrahedron Letters, 1963, 955.


but this adds only the Gutsche Organic Reactions ref to what Merck has.


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Rhodium

  • Guest
Buchner-Curtius-Schlotterbeck
« Reply #46 on: May 08, 2002, 09:07:00 AM »
Buchner-Curtius-Schlotterbeck

Methylation of aldehydes with diazomethane to produce methylketones, alpha-epoxides and aldehydes.
The reaction proceeds with aliphatic, aromatic and heterocyclic compounds.




halfapint

  • Guest
Interim Developments 001
« Reply #47 on: May 08, 2002, 06:28:00 PM »
Gotta reinstall my scanner in case I find neat portraits of Buchner and Curtius in any of my old chem books. 50/50 chance I'll find one or the other, 17.5% probability I'll have both. Schlotterbeck, snowballs' chance in hell.

While at the same time I'll locate the book that has the explanation of diazomethane derivation that made me say, "Ohhh...! So any n-methyl amino derivative of mumblety-jumble, on treatment with nitrous acid will give diazomethane...", and to see again what mumblety-jumble was. That old citation will be cast in plain language about aliphatic diazotization, so people can grasp it; later it can bee supplemented with more modern citations, stated more abstractly, which should allow us to generalize in a more nearly rigorous sense. "So it stands to reason that the n-benzylamino derivative of.." and start to check on good-better-best ways to get the critical intermediate. PrimoPyro won't gripe about a new diazotization thread, specially if it has those old pix decorating it. Now to work.

BTW funniest thing happened earlier. Rhodium and I both used Web search to find the Institute for Chemistry site in Skopje, Macedonia --- the only worthwhile site with this reaction cited that's easily found --- and both of us posted the same info simultaneously. Same words, Rhodium reworked the pic slightly, while I simply linked it in from Skopje. Looked funny to have the same stuff twice on the thread, so I zapped mine.

PrimoPyro

  • Guest
Grignard System Sensitivity
« Reply #48 on: May 15, 2002, 02:09:00 AM »
Hi.  :)

I know that grignard systems (organomagnesium halides stabilized in ether solvents) are very sensitive to the following things:

water
alcohols
carbonyl groups
imines
nitriles
halides
acids

But what I'm curious about is, are they sensitive toward alkalai metals and also are they sensitive toward bases, such as specifically sodium hydroxide?

Would a molecule RMgX react in some way with a species of MOH where M = alkalai? Would it react with metallic M? M is not lithium. Lithium is excluded from this entire post (unless you have something to add) because lithium has its own series of organometallic compounds, that I'm not interested in here.

Basically, the reason I am interested is because Grignard systems are sensitive toward water, and one application for such a system, unfortunately involves insitu generation of water, and it must be removed chemically. Kinetic drying agents are not satisfactory. The water must be destroyed immediately.

One such method would be to use an acid anhydride such as acetic anhydride, formic acetic acid. This is obviously 100% out of the question, becuse the grignard reagent will react with acetic anhydride, and the produced acid will also destroy the reagent. This is not satisfactory.

Another avenue is to use the opposite: an alkoxide. But this produces an alcohol, which, like the former, will destroy the grignard reagent and produce a useless alkane. This is also out of the question.

But I wonder about using sodium oxide, Na2O, as a water scavenger, as it would form 2NaOH as a product. Would this work, or will either entity, the oxide, or the hydroxide, interfere with the grignard adduct?

And also what of alkalai metals like Na metal? They will react with water to form the hydroxide as before, and evolve hydrogen gas. Are the free metals reactive toward the reagents?

I have never seen any examples before of these types of compounds being detrimental to such a system, but I have also never seen their employment, so I truly am in the dark.

Can anyone suggest a chemical reaction that involves combination with or destruction of water, but does not yield an acidic or basic compound, but rather a neutral one or perhaps something that reacts with water and becomes insoluble in a system, allowing for removal?

I know of no such methods, and I hope to get some ideas.

As for just how much water is needed to be removed, it is thought to be equimolar to the molar of the grignard reagent.

                                                  PrimoPyro

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Rhodium

  • Guest
hydroxides and grignards
« Reply #49 on: May 15, 2002, 11:19:00 AM »
Sodium hydroxide reacts with grignard reagents, forming NaOMgX and the alkane, as sodium hydroxide is more acidic than grignard reagents.

I believe you would have a problem removing in situ formed water, as I don't think anything would react faster with it than the grignard itself.

PrimoPyro

  • Guest
Using a Grignard As A Reducing Agent
« Reply #50 on: May 16, 2002, 06:42:00 AM »
Here is an idea (completely unrelated to my previous posts here on Grignard Reagents, I just thought of this) I know this must have been thought of elsewhere too, because honestly it's kind of obvious, I dont know why I havent thought of it before.

I always wished one could reduce chloroephedrine by forming its grignard reagent, then quenching with dilute acid, reducing the halide to a hydride, forming an alkane. This, unfortunately, is not a possibility because of the amine function in the ephedrine molecule. It prevents the initial formation of the grignard reagent.

But here is the obvious variation that just popped into my head 2 minutes ago: Form a simple grignard reagent of an irrelevant compound, whatever workable halide is available. The halide is not the important moeity.

Grignard reagents react with alcohols, and reduce them to alkanes. Ephedrine's benzyllic alcohol could be reduced by merely subjecting it to a preformed grignard reagent, such as butylmagnesium chloride. The amine hinders their formation, but I cant find any reason it would protect the alcohol in any way to prevent its reduction at the hands of the grignard adduct + aqueous acid. If it were an imine or a nitrile that would be a different story, but not an amine.

Im sure there is the possibility that some of the Gr. reagent might get interfered with by the amine, tied up so to speak, so maybe it would be wise to use a slight molar excess of the reagent to the alcohol, maybe 1.1-1.2 times the molar. I dont really know.

What I do know is that grignard reagents are not so impossible to form in reality as many people assume they are. I see them as a truly viable option, providing that they work.

                                                  PrimoPyro

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Osmium

  • Guest
Nope
« Reply #51 on: May 16, 2002, 06:58:00 AM »
R-OH + BrMg-R' -----> R-OMgBr + H-R'

The grignard reagent will be reduced to the alkane, not the alcohol! The alcohol (or amine or whatever C-H acidic compound) will merely provide the H+.

I'm not fat just horizontally disproportionate.

PrimoPyro

  • Guest
The alcohol is merely regenerated? What a spoof.
« Reply #52 on: May 16, 2002, 07:02:00 AM »
The alcohol is merely regenerated? What a spoof. Blah, did I read it wrong or something?

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PrimoPyro

  • Guest
Alkene Hydration
« Reply #53 on: May 22, 2002, 09:16:00 PM »
In an alkene hydration of an allylbenzene, the entity that actually adds to the double bond is H3O+, isn't it? The extremely acidic medium provided by ~80% H2SO4 must surely convert a vast majority of water to hydronium ions.

Could ammonium sulfate dissolved in sulfuric acid be used to add NH4+ to the alkene in the same fashion?

                                                  PrimoPyro

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Osmium

  • Guest
??? H3O(+) is just a way to say that the proton ...
« Reply #54 on: May 23, 2002, 03:09:00 AM »
???
H3O(+) is just a way to say that the proton is hydrated. I doubt that 'H3O(+)' really exists as a single molecule, in fact it's rather a shell of H2O molecules surrounding the proton somehow bonded to it.

NH4(+) is just another acid, just like H3O(+). The first step of alkene hydration is addition of H(+). Since there is no NH2(-) floating around in an aq. solution of ammonium sulfate in H2SO4, and since NH3 will not let go of one of its protons under these conditions you will not be able to get that NH3 onto the double bond.

I'm not fat just horizontally disproportionate.

Rhodium

  • Guest
hydroamination
« Reply #55 on: May 23, 2002, 09:00:00 AM »
PP: Read up on the hydroamination of alkenes (gave you loads of refs), where R2N- actually adds to alkenes.

PrimoPyro

  • Guest
Osmium: thanks, that is exactly the answer I was ...
« Reply #56 on: May 23, 2002, 10:00:00 AM »
Osmium: thanks, that is exactly the answer I was looking for. And I didnt think about there being more than one H2O per proton, I took the formula as a literal representation.

Rhodium: I will. I have a little bit, and I was somewhat confused about the reaction conditions is all.

                                                  PrimoPyro

PrimoPyro

  • Guest
I think I got it....
« Reply #57 on: May 23, 2002, 03:30:00 PM »
So the high proton concentration causes a proton to add to the alkene, reducing the double bond, to give the most stable carbocation, and then the nucleophile, which is actually neutral in charge, just polar, then adds to the carbocation and ejects a proton of it's own to sustain neutrality?

The nucleophile couldn't contain a negative ion, because the protons in solution would combine with it, wouldn't they? So the Nu in this case is just electronegative polar molecules/groups?

So RCH=CH2 + H+ --> RC(+)HCH3
RC(+)HCH3 + NuH --> RCH2(NuH+)CH3 --> RCH2(Nu)CH3 + H+

Right?

But if the nucleophile is attracted to positive charges, why doesn't it react with the readily available protons in the system, forming NuH2+, and this positive entity would exhibit electrostatic repulsion toward the alkene/proton complex, preventing it's coupling?

                                                   PrimoPyro

hypo

  • Guest
excess of nucleophile
« Reply #58 on: May 23, 2002, 11:46:00 PM »
because the nucleophile is in large excess compared to the H+ ions.
even at pH=1 you have a H2O:H+ ratio of about 55.5

further, i would believe the carbocation to be a strong acid
by itself, thus being allways in competition with the H+ for
nucleophiles. classical equilibrium reaction, where the nucleophile
is eliminated from the equilibrium.

hypo, as usual just guessing  ;)

foxy2

  • Guest
don't
« Reply #59 on: May 23, 2002, 11:57:00 PM »
I took the formula as a literal representation.

Most ways molecules are represented or drawn cannot bee taken as a literal representation of the molecule.  Take benzene rings and how their resonance is depicted, it definately not literal.

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