Author Topic: PP's General Chem Question Thread  (Read 15986 times)

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PrimoPyro

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PP's General Chem Question Thread
« on: October 30, 2001, 08:00:00 AM »
Hi. I am starting this thread to ask several general chemistry questions relevant to the topic at hand here. I don't want to start a myriad of threads however for such shallow discussions, consisting of Q --> A, and thats it. I don't think the Hive would appreciate a bunch of 3 post threads starting with questions. So I will consolidate them here. If you know the answers, by all means, please answer. If you have a similar question spawned off one currently operating, go ahead and ask. If you do not have anything at all to contribute, please dont shit on my simple thread here. However, if a search query would happen to yield the necessary info, by all means cite utfse. I will use tfse b4 ever asking here, keep that in mind ppl. So just because the answer may seem simple to those in the know of advanced organic chem, keep in mind that some simply havent run across that information yet, and are trying to do so now.

Thank you for even reading this thread and considering my curiosity, I highly appreciate it.


The first question that comes to mind is:

Are there any known reactions that dyhydroxylate an aromatic ring in a single reaction? In either one step, or in a one-pot multiple step reaction. The ideal desired reaction is one that dihydroxylates the ring with the position of the OH's ortho respective to one another, but para/meta to another group already substituted on the ring. For example to illustrate my point: benzaldehyde --> pyrocatechualdehyde would be the most relevant reaction I can think of at the moment.

However, it need not be perfectly this case. At the moment I am interested in all reactions similar to this, but have no clue if there are any, and if so, what they are called, and thus have nothing to search for further than dihydroxylation and derivatives of the word.

Thank you for any info that comes to light in this thread, I really appreciate everyone's interest and/or tolerance.  ;)

                                                  PrimoPyro

How does one win a race when the finish line runs faster than you?  :(

foxy2

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Re: PP's General Chem Question Thread
« Reply #1 on: October 30, 2001, 11:32:00 AM »
I don't think an easy selective one step is possible.  The only easy additions I can think of are halogens and nitro groups

Do Your Part To Win The War

PrimoPyro

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Re: PP's General Chem Question Thread
« Reply #2 on: November 07, 2001, 11:47:00 PM »
Thanx foxy.

I have another series of interesting questions. (well I find them interesting  :P )

PrimoPyro is now interested in the hypothetical synthesis of ephedrine-like compounds. Specifically 1-phenyl-1-hydroxy-2-amino-propanes. This should be no big issue, until:

Im trying to find a synthesis route from the propenyl-benzene. On top of that, the product has to be the correct chirality for an oxazoline cyclization. Now to be honest, PP never put too much interest in learning chirality. I get the concept, but some of the terms evade me. I know what the difference between different isomers of ephedrine are, but I cant name which is which, the names escape me. So, that said, bear with my alternative language for them in this post. Thank you.

Now, I assume that the needed configuration is between the hydroxy and the amine for cyclization. I assume they need to be near each other, to be in proximity. this would denote a sort of cis isomer right? I dont think you name optical isomers this way, but I am here anyway. So, assuming what I have said thus far is ok, I would like to propose this:

The propenylbenzene is epoxidated, then reacted with hydrobromic acid to open up the epoxide to the alpha-hydroxy-beta-bromo-propane. This is then aminated with ammonia to form the 1-OH-2-NH2-propane.

I would think that the epoxidation would occur, having the correct bonding, as stated as having been needed above. (cis) as trans bonding would have a longer stretch on the oxygen bond and be less stable, or it would simply cause torsion in the C-C bond to revert to the stated "cis". Forgive me its hard to explain myself without a lot of pictures. Let me know if this is totally not understandable and I will draw some up.

Since this epoxide should be the correct configuration, so would be the products upon opening it up.

But, I wonder, could the 1-bromo-2-hydroxy-propane be a byproduct? I remember reading old ADC posts awhile ago stating that HBr only adds to the beta carbon in safrole, but I wonder if this is only applicable to alkenes.

If this step worked as well, when you aminated with ammonia, the hydroxy and amine groups would be in the assumed correct positions for cyclization.

I'd like to hear any questions on this, as well as many bees comments on anything posted herein. Thank you.

P.S. buying the correct isomer is not an option. This topic is about the synthesis of such a compound, so while all pointers on where to buy or extract this from are a nice thought, they arent the desired information.

Also, I have only presented here what I thought of in my head, id like to know if it will work etc., but I also want to know of any known syntheses of such compounds, or any known similar reactions that come to mind. It doesnt HAVE to be regarding this synthesis. Thanx again.

                                                   PrimoPyro

Vivent Longtemps la Ruche! STRIKE For President!

Osmium

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Re: PP's General Chem Question Thread
« Reply #3 on: November 08, 2001, 01:58:00 AM »
Ring opening of an epoxide by H(+) will not produce the cis diol. As far as I remeber it always produces the trans.

PrimoPyro

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Re: PP's General Chem Question Thread
« Reply #4 on: November 20, 2001, 03:29:00 AM »
Ok, bummer, but thank you. Now I know not to proceed further assuming that part to be correct, and waste more time.

Next, we have to move to good old electrochemistry, very simple electrochemistry, I hope.

Has anyone ever thought about the two very useful species locked up in Epsoms before? The o so nice element Magnesium, and the better half of the Sulfuric Acid structure. Wouldn't it be nifty to seperate em into two good clandestine tools?

Im wondering about doing this. Solvate Anhydrous magnesium sulfate (if the solvent is anhydrous. If its aqueous, who gives a shit) and electrolyze it to seperate the ions.

I assume magnesium could be absorbed into a mercury cathode to form an amalgam, like the method used for sodium production? As for the sulfate ion {SO4-2} It would be changed to SO4: What the hell would you do to procure sulfuric acid out of this? Would it decompose to sulfur trioxide? If so, this could be led into water to produce the H2SO4. But thats a random guess.

Is this even a possibility? Id think it should be. I just dont know how to do it. I did utfse btw, learned a decent bit, too, browsing the not-what-I-was-looking-for results.

So people, tell me what you think and/or know.

                                                 PrimoPyro

Vivent Longtemps la Ruche! STRIKE For President!

Rhodium

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Re: PP's General Chem Question Thread
« Reply #5 on: November 20, 2001, 03:57:00 AM »
It is theoreticaly possible, but not practical at all. Magnesium is abundant OTC (see

https://www.thevespiary.org/rhodium/Rhodium/chemistry/hydrogen.html

) and there are much better sources for sulfuric acid - even if you cannot get the concentrated acid, you can always resort to concentration of 30-40% battery acid, using a lot less money and equipment than if you were to do an electrolysis (in which I don't know if sulfuric acid would be formed at all).

The only instance where electrolysis of a salt is warranted at all would be in the preparation of sodium metal, and even that is a big hassle (just look in the search engine).

PrimoPyro

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Re: PP's General Chem Question Thread
« Reply #6 on: November 20, 2001, 04:19:00 AM »
Thanx Rhodium. I read that page, but Im reminded of the thought that Mg forms a metal oxide coating in air. Would this be suitable for grignard reactions? I know the surface is oxidized or an alloy of some sort, because you can add water to these parts "sources" and Mg(OH)2 is never formed.

Vivent Longtemps la Ruche! STRIKE For President!

Rhodium

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Re: PP's General Chem Question Thread
« Reply #7 on: November 20, 2001, 04:31:00 AM »
Magnesium alloys are not suitable, it must be pure magnesium. In preparing magnesium (preferably in the form of turnings or shavings, but a coarse powder may work satisfactorily) for use in a grignard reaction, it is first washed with dilute (about 1M) hydrochloric acid to remove any oxide coating (which always forms on the Mg surface upon exposure to air), followed by washing away the acid/water with alcohol, and finally washing away the alcohol with anhydrous ether. The clean magnesium is then directly transferred to the reaction flask and covered with anhydrous ether, and the reaction is begun as usual.

There is a way around the trouble of dirty magnesium, and that is the use of an ultrasonic bath to "shake" away the oxide coating, and this procedure also allows the chemist to use normal ether, and it does not have to be rigorously dried (usually over sodium wire), ether dried over MgSO4 is sufficiently dry.

Ultrasonicated grignard reactions:

https://www.thevespiary.org/rhodium/Rhodium/chemistry/grignard.wet.html






Osmium

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Re: PP's General Chem Question Thread
« Reply #8 on: November 20, 2001, 04:42:00 AM »
Never cleaned Mg for grignards at all. The oxide coating of Mg isn't that hard to overcome, stirring alone will expose some of the metal and allow the reaction to start. Whenever it didn't start a crystal of iodine took care of it.

But if you have access to an ultrasonic bath then of course use it.

PrimoPyro

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Re: PP's General Chem Question Thread
« Reply #9 on: December 10, 2001, 03:12:00 PM »
Is there a reaction that changes from an ester to a ketone by eliminating the terminal acid-alcohol oxygen, leaving the acid carbonyl, and thus creating a ketone?

I searched for esters, ketones, and rearrangements here and abroad, and didn't find what I was looking for  :(

I did find the reverse: the rearrangement of a ketone to an ester. Sigh, this is the opposite of what I want to do.

Anyone know?

                                                PrimoPyro

Vivent Longtemps la Ruche! STRIKE For President!

Rhodium

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Re: PP's General Chem Question Thread
« Reply #10 on: December 10, 2001, 06:41:00 PM »
You can hydrolyze the ester, and react the carboxylic acid with 2 equivalents of methyllithium, see the two last steps in:

https://www.thevespiary.org/rhodium/Rhodium/chemistry/p2p.azlactone.html


PrimoPyro

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Re: PP's General Chem Question Thread
« Reply #11 on: December 10, 2001, 07:01:00 PM »
Thank you, Rhodium. It is more knowledge, but sadly, doesn't apply to what I desire to do.

The example I wanted it for was to take either ethyl benzoate and make acetophenone, or even more desireable, to take benzyl acetate and make phenylacetone.

For the process you mentioned, I'd need phenylacetic acid, or a phenylacetic ester. I was hoping to find a method to do away with these reagents.

Benzyl alcohol + acetic acid + sulfuric acid --> benzyl acetate

Very OTC. If I could find a way to eliminate that oxygen to produce phenylacetone, or the acetophenone from ethyl benzoate, I'd be very very pleased.

Thanks for your reply.

                                                PrimoPyro

Vivent Longtemps la Ruche! STRIKE For President!

PrimoPyro

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Re: PP's General Chem Question Thread
« Reply #12 on: December 16, 2001, 07:56:00 PM »
Is it possible to perform ANY type of condensation between any of these:

an ether, and either ammonia, or a primary amine?

The ether plus ammonia would make a secondary amine by replacing the oxygen in the ether, with the nitrogen, forming the secondary amine and water.

If it were a primary amine, a tertiary amine would form, plus water.

Is this possible at all? I have never sen this talked about ever, so I wonder if it can even be done? If so, I have some new ideas to write out. Thanks for any help.

                                                 PrimoPyro

Vivent Longtemps la Ruche!

Rhodium

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Re: PP's General Chem Question Thread
« Reply #13 on: December 16, 2001, 09:12:00 PM »
Not with ordinary reaction conditions. It might happen something at 450°C and 300 atm pressure over some zeolite catalyst, but under such conditions it is usually possible to turn donald duck into dihydro-mickey mouse, so I guess that doesn't count.

PrimoPyro

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Re: PP's General Chem Question Thread
« Reply #14 on: December 16, 2001, 09:17:00 PM »
LOL!

Thanx Rhodium. I didnt put too much hope in it, but I was really curious about that possible conversion and had to ask. Thanx again.

                                                   PrimoPyro

Vivent Longtemps la Ruche!

PrimoPyro

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Re: PP's General Chem Question Thread
« Reply #15 on: December 23, 2001, 07:34:00 PM »
Back to a previous topic outlayed above: The formation of ephedrine derivatives from propenyl benzene. This time I wonder about PPA instead of the methylated ephedrine version.

https://www.thevespiary.org/rhodium/Rhodium/chemistry/ppa.synthesis.html



This shows how NBS and H2O can be used to change propenyl benzene into the amino-alcohol. It also says that it will work for substituted benzenes as well. Good.

My question is: Will this procedure produce the correct chiral aminoalcohol for clyclization to 4-MAR with HNCO, as outlayed in the thread "4-MAR w/o CNBr" ?

I would think it would depend on the chirality of the propenyl benzene right? Like I said before, I don't really know nomentclature for chirality, but speaking from a picture point of view, in the standard picture scheme for alkylbenzenes, with the phenyl on the left, and the alkyl coming out to the right from the 1 o clock position (god I sound so stupid, all because of a lack of vocabulary...)

Well, if the propenyl alkene part curves down like shown in that url, then the halogen is on the same side as the hydroxy, but if the alkene curves to the right, the hydroxy will be "up" and the amino "down" and I would think this isomer to be the incorrect one for cyclization.

Am I on the right track here, or am I speaking utter nonsense? I am taking a wild stab in the dark here trying to figure it out.

So specifically: Will this alkene-opening reaction produce the chirally correct compound for cyclyzing to 4-MAR with HNCO?

Don't forget to explain your answer.  ;)

Thank you everyone. You are most helpful, and it is very appreciated.

                                                   PrimoPyro

Vivent Longtemps la Ruche!

terbium

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Re: PP's General Chem Question Thread
« Reply #16 on: December 23, 2001, 08:51:00 PM »
then the halogen is on the same side as the hydroxy, but if the alkene curves to the right, the hydroxy will be "up" and the amino "down"
This seems to be cis/trans isomers that you are refering to in the above, not optical isomers. It is only when you are speaking of optical isomers that you use the term chirality.

Rhodium

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Re: PP's General Chem Question Thread
« Reply #17 on: December 25, 2001, 03:33:00 PM »
Primopyro: Halohydrin formation is trans-regiospecific (regardless if the starting alkene is cis or trans, as the intermediate is a flat bromonium ion), and produces the incorrect isomer (the same as pseudoephedrine) for use with the cyanate route to 4-MAR.

PrimoPyro

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Re: PP's General Chem Question Thread
« Reply #18 on: January 04, 2002, 03:33:00 AM »
Powers that bee, and bees buzy a buzzin' right now:

Here's a simple yes/no question  :)

If para-Dichlorobenzene is exposed to NaCN or KCN in solution, or w/o solution (two different scenarios) will a reaction commence to swap the chlorines on benzene with the nitriles of the alkalais?

Essentially:

PhCl2 + 2KCN ___> Ph(CN)2 + 2KCl

That is what I am curious about. I know that such reactions readily take place on alkyl halides, and on arylalkyl halides, but does this apply to aryl halides as well? Please say yes.  ::)

I think you know why I am curious about this.  ;)

                                                        8) PrimoPyro 8)

Vivent Longtemps la Ruche!

Rhodium

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Re: PP's General Chem Question Thread
« Reply #19 on: January 04, 2002, 04:55:00 AM »
It generally does not apply to aryl halides unless extreme rxn conditions are used. See any org chem textbook for the conversion of chlorobenzene to phenol (350°C and a lot of pressure).

PrimoPyro

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Re: PP's General Chem Question Thread
« Reply #20 on: January 05, 2002, 05:51:00 AM »
Could a pressure cooker supply enough pressure for a small scale reaction of this sort? I can heat it to 350C, but the only way I know of increasing pressure is in a pressure cooker, an autoclave, or a bomb.

Bomb is out of the question. NO WAY. I of all bees know what it's like when bombs go off unexpectedly. It's not fun.

I don't own an autoclave, nor do I have the funds to buy one for such a reaction.

So what about a pressure cooker?

                                                   PrimoPyro

Vivent Longtemps la Ruche!

Rhodium

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Re: PP's General Chem Question Thread
« Reply #21 on: January 05, 2002, 10:03:00 AM »
I would not reccommend heating cyanides to those temperatures under pressure, regardless of the quality of the container, there is a too large risk of poisoning yourself and others.

PrimoPyro

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Re: PP's General Chem Question Thread
« Reply #22 on: January 05, 2002, 06:51:00 PM »
I was kinda worried about that, too. Acids liberate HCN from NaCN/KCN/Ca(CN)2 therefore you cannot have any acid present.

This case would be extra special. Bases such as NaHCO3 or Na2CO3 may not be strong enough to react with the weak HCN, so they would be worthless, but bases such as NaOH, KOH, etc will. Well that sounds fine, it would regenerate NaCN for the reaction, BUT: Under these conditions of pressure and heat, it would also react with the dichlorobenzene to produce sodium chloride and hydroquinone, a dihydroxybenzene. Well, as we all know, dihydroxybenzenes are not basic, they are acidic. It would possibly react with the NaCN present to form the hydroquinone salt and HCN.

So, in this situation, the pH must be neutral, or disasters could result either way. A thin line to tread to say the least.

                                                  PrimoPyro

Vivent Longtemps la Ruche!

PrimoPyro

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Re: PP's General Chem Question Thread
« Reply #23 on: March 02, 2002, 10:58:00 AM »
Can someone please help me out with a basic chemistry question, I just outright don't know the answer... ::)

If you were to halogenate 2-butanone, where would the halogen go? Would it go to the lone methyl on one side of the carbonyl, or to the ethyl on the other side?

Please say ethyl. If it is the ethyl, which carbon, the terminal carbon, or the one next to the carbonyl? Please say the one next to the carbonyl....

Thanks.  :)

                                                    PrimoPyro

Wir pumpen..... (Klatschen) Sie oben!

Rhodium

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Re: PP's General Chem Question Thread
« Reply #24 on: March 02, 2002, 11:23:00 AM »
It will end up on the most substituted carbon (the most stable enol form of the ketone involves that carbon, and the halogen only reacts with the enol form), and in this case that is in the 3-position, next to the carbonyl on the ethyl side chain.

PrimoPyro

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Re: PP's General Chem Question Thread
« Reply #25 on: March 02, 2002, 11:28:00 AM »
Thanks Rhodium!  :)

Does this mean that since only the enol tautomer reacts, the halogenation with say bromine will be very slow?

                                                     PrimoPyro

Wir pumpen..... (Klatschen) Sie oben!

Rhodium

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Re: PP's General Chem Question Thread
« Reply #26 on: March 02, 2002, 12:06:00 PM »
It depends on the stability of the enol form. Acetone, for example, is in its enol form 1% of the time, while acetaldehyde is there 25% of the time. Say the bromination will then take 100 times longer than a regular halogenation of a double bond - it is still over in a few minutes.

PrimoPyro

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Re: PP's General Chem Question Thread
« Reply #27 on: March 02, 2002, 01:09:00 PM »
Thanks again.  :)

Can you use alkyl chlorides and bromides for the Ullmann reaction with copper, or only iodides? Anyone who knows can answer these questions by the way.

And anyone who has other hypothetical questions can post them as well, as long as they involve chemistry, not a recipe.  :)

                                                    PrimoPyro

Wir pumpen..... (Klatschen) Sie oben!

Rhodium

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Re: PP's General Chem Question Thread
« Reply #28 on: March 02, 2002, 02:16:00 PM »
I may rain on your parade somewhat here, but I think only aryl iodides are suitable for the ullmann reaction, not aliphatic iodides. You may have to look into stille/suzuki coupling reactions instead, alternatively Cu(I)-catalyzed grignard couplings.

PrimoPyro

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Re: PP's General Chem Question Thread
« Reply #29 on: March 02, 2002, 02:20:00 PM »
Rain all you need to Rhodium. I already figured that was going to come out.  ;)

I was just seeing if anyone knew for certain that aryl iodides were the requirement. I figured if I asked, everyone would jump in and say "probabaly" but if they were absolutely sure, they'd catch me in my "error."

No grignards allowed Im afraid lol. But I will look into the other one. Thanks!  :)

                                                   PrimoPyro

Wir pumpen..... (Klatschen) Sie oben!

psychokitty

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Re: PP's General Chem Question Thread
« Reply #30 on: March 02, 2002, 04:23:00 PM »
First off, PrimoPyro, you have got to get yourself the 2001 edition of Advanced Organic Chemistry:  Reactions, Mechanisms, and Structure (Fifth Edition) by Jerry March.  This sucker is like Beilstein in one volume.  I am not kidding.  I only have the fourth edition (1992) but it too is fantastic!  Bees are always saying that I have a knack for finding obscure references (right under Foxy2) but I'll let you in on a little secret:  Practically all of my sources are actually from one source--namely the aforementioned textbook!

Anyway, to answer some of your questions (some from this thread and some from that other one of yours):

1.)  Benzyl iodide could be had from benzyl alcohol by way of NaI or KI and phosphoric acid or by AlI3 (formed in situ through reaction of Al and I2).  Al is otc; I2 can be had through reaction of NaI or KI and oxone).  But toluene and calcium hypochlorite are more OTC, I guess.

Regarding pool chemicals, what about that "brominating" shit that they carry.  I imagine that if it brominates, then it could possibly be used for brominating toluene or something.

2.)  The amino alcohol that you desire could be had by reacting any epoxide with hexamine and then hydrolyzing the intermediate complex.  Yields from styrene oxide and hexamine are, according to a reference that I've posted before, 100% of the desired amino alcohol (and by desired I mean that both the hydroxyl and amino groups are in the right place).  To get the halohydrin, one could react the epoxide with AlI3 made in situ.  Iodides are easier to swap with amines so this might be preferred over reacting the epoxide with HBr (which isn't all that OTC, BTW) and going from there.

3.) As for your question about whether or not aryl chlorides or bromides could switch for aryl iodides in the Ullman reaction, I refer you to "3-16 The Ullman Reaction (De-halogen-coupling)" on p 665 of the aforementioned fourth edition textbook:

". . . The best leaving group is iodo, and the reaction is most often done on aryl iodides, but bromides, chlorides, and even thiocyanates have been used."

Hope this has been helpful.  However, the best advice that I have given you is to buy that fucking book.  Enough said.


PrimoPyro

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Re: PP's General Chem Question Thread
« Reply #31 on: March 02, 2002, 04:30:00 PM »
Thank you!  :)

I will purchase it the minute I locate a copy. I will begin searching for an online source right now. Thanks!  :)

I bought the last book recommended to me by foxy2, and it was worth the money. Operational Organic Chemistry, by John W. Lehman. It is full of procedures, especially for analysis, I love it.  :)  :)  :)

                                                     :) PrimoPyro :)

Wir pumpen..... (Klatschen) Sie oben!

PrimoPyro

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Re: PP's General Chem Question Thread
« Reply #32 on: March 02, 2002, 05:19:00 PM »
Guess what I just bought?  ;)

It's on it's way, hopefully my copy of the 5th edition will be here within a week. I look forward to its arrival. Found it really fast actually.

You can never have too many books, and you can never "read too much" no matter what anyone says. Money spent on information is money well spent. That's my story and I'm sticking to it.

Thanks again!  :)

                                                   PrimoPyro

Wir pumpen..... (Klatschen) Sie oben!

psychokitty

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Re: PP's General Chem Question Thread
« Reply #33 on: March 02, 2002, 05:25:00 PM »
Let me know when you get it.  I'll help you with the navigation.  I'm NOT saying you're a dumb-ass (in other words, that you can't figure it out for yourself), but actually I'm interested in a few things in there and helping you zero in on it quicker will help me as well.

cheeseboy

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Re: PP's General Chem Question Thread
« Reply #34 on: March 02, 2002, 07:40:00 PM »
So much great fun isn't it?So far cheese has invested in The Organic Chem Lab Survival Manual by Zubrick and Total Synthesis II by Strike.Thanks for the two other books, Pshychokitty and PrimoPyro  ...So toluene and calcium hypochlorite makes I2? Coolness.Is that synth on Rhodium's too?

Cheeseboy-a whiteboy with Soul Like a black guy without soul
May The Source Bee With You Always.

psychokitty

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Re: PP's General Chem Question Thread
« Reply #35 on: March 02, 2002, 08:29:00 PM »
No, toluene and calcium hypochlorite yields benzyl chloride; NaI or KI and oxone yields I2.

PrimoPyro

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Re: PP's General Chem Question Thread
« Reply #36 on: March 03, 2002, 06:09:00 AM »
I am wondering, is Pd+2 the only oxygenation catalyst known, or are there others for catalytic oxygenation?

I know very little about catalysis for hydrogenation and oxygenation (still learning), but I seem to recall the reactions being an equilibrium that one influences by flooding the system with excess reactant to shift the equilibrium in the desired direction, such as adding shitloads of hydrogen to the catalyst to hydrogenate, and oxygen to oxygenate of course.

Would this imply that hydrogenation catalysts such as Nickels would also be applicable to catalytic oxegenation? If so, can you elaborate on any specific known hydrogenation catalysts that can double as oxygenation catalysts that are not palladium compounds?

If the answer to that is just a dead end, then I ask: Are there any other catalysts other than bipositive palladium, that can be used in Wacker-like oxygenation of olefinic bonds to ketones?

Thank you for your help. My interest is shifting from specialty reagents to catalysis lately. Very interesting topic. But palladium is expensive, and recycling it is a joke.

Thanks again.  :)

                                                    PrimoPyro

Wir pumpen..... (Klatschen) Sie oben!

PrimoPyro

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Acid-Base Reaction Variations
« Reply #37 on: March 26, 2002, 01:44:00 AM »
What would happen if one performed various acid base reactions, with using more powerful, non-hydroxide alkalai bases?

The instance Im thinking of, for example, is the splitting of acetamide with sodium hydroxide.

AcNH2 + NaOH --> AcONa + NH3

In these type reactions, the hydrogen and oxygen of the basic hydroxy are torn apart, something that is not "normallly seen" in reactions. The oxygen stays with the sodium in this case to form the acetate salt, while the hydrogen forms ammonia. This is a relative perspective, achieved by comparing reactants to products in various simple reactions. This one stands apart.

What if one were to use sodium methoxide? Would the sodium acetate still form, and the alkyl group attach to the amide, like the hydrogen did from the hydroxy? Using sodium methoxide would of course then procure methylamine, but this is but a small venture into this possibility.

If this would happen, I dont think it wise to prepare traditional alkylamines such as methylamine, nor to aminate alcoholates like P2Pol alcoholate; this is too unrealistic.

What I wonder about, is possibly using the Leuckart reaction with only formamide, not N-methyl formamide, to change P2P/MDP2P into the formyl-MDA, and react the formamide with sodium methoxide to form sodium formate and the N-methylated amine: MA/MDMA.

                                                    PrimoPyro

Vivent Longtemps La Ruche!

Rhodium

  • Guest
Amide hydrolysis with methoxide
« Reply #38 on: March 26, 2002, 02:07:00 AM »
Nope, doesn't work that way - if you check the reaction mechanism of basic amide hydrolysis (reaction mechanisms is something you should look up every time you are unsure of reactants/products), you'd see that the basic anion (in my example methoxide) adds nucleophilically to the carbonyl, forming a tetrahedral intermediate, which then kicks out an amide anion, forming methyl acetate (if hydroxide was used, acetic acid would be the intermediate product). As the amide anion is so incredibly basic, it will not last for many microseconds, but will directly add to anything acidic, like the solvent protons (in my example methanol) re-forming sodium methoxide and free ammonia. If hydroxide was used as the base, the amide anion would abstract its hydrogen from there, forming ammonia and acetate).


PrimoPyro

  • Guest
Aliphatic Diazo Compound Preparations
« Reply #39 on: May 07, 2002, 07:29:00 PM »
How to?

If someone would be so kind (Rhodium, you are setting a trend in my thread here, would you be so kind as to grace my thread once again? Osmium, you too?)

I'm wondering what the common methods (if any) are for the preparation of aliphatic diazo compounds (RN2) from whatever other compounds you know of. I don't think I mean diazonium salts, as the formula is only RN2. I really don't know a thing yet haha, and Im trying to look stuff up, when I figured it couldnt' hurt to ask here in this old thread.  :)

Thanks in advance for providing very informative answers to all my questions, including this one.  ;)  :P

Yes, Im Using FSEs right now.

                                                   PrimoPyro

Vivent Longtemps La Ruche!

Rhodium

  • Guest
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

Vivent Longtemps La Ruche!

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

Vivent Longtemps La Ruche!

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

  • Guest
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.


a half a pints a half a pound a half a world a half a round
Sidearm n. Flask neck tube.

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

Vivent Longtemps La Ruche!

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

Vivent Longtemps La Ruche!

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?

Vivent Longtemps La Ruche!

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

Vivent Longtemps La Ruche! How's my posting? Call 1-800-EAT-SHIT

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.

Those who give up essential liberties for temporary safety deserve neither liberty nor safety

b159510

  • Guest
and then the nucleophile, which is actually ...
« Reply #60 on: May 24, 2002, 05:00:00 AM »
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?
this would be the case for the acid catalyzed addition of
water to an alkene, but likely a water molecule removes the proton.

The nucleophile couldn't contain a negative ion, because the protons in solution would combine with it, wouldn't they?
a nucleophile can have a negative charge even in a strongly acidic solution, but not in this particular reaction.
But if the nucleophile is attracted to positive charges, why doesn't it react with the readily available protons in the system, forming NuH2+,
maybe it does. maybe it gets protonated and deprotonated
a thousand times a second. but when the deprotonated form
comes in contact with the carbocation and sticks, the next
step of the reaction can take place (this actually happens
very fast compared to the formation of the carbocation).
The rates of some acid catalyzed reactions are very pH
dependent, like an amine with a ketone.
As hypo says, there is an equilibrium factor. As foxy says,
reaction diagrams are usually not literal depictions.
They also almost never represent all the reactions that
are taking place. 

Back to the Primitive

PrimoPyro

  • Guest
Hmm
« Reply #61 on: May 24, 2002, 05:18:00 AM »
Is this how acids of the species HX add to alkenes? The acid produces a carbocation and the halide anion attaches to it? Makes perfect sense.

Thank you for elaborating so much on this question, all of you.  :)  I've started reading a book on Functional Group Chemistry, and it relies very heavily on electronegativities of functional groups, and how they play a crucial role in reaction mechanisms. It is highly interesting, I think I'll put the organometallics book aside for the moment and read this one first.

                                                   PrimoPyro

Rhodium

  • Guest
Is this how acids of the species HX add to ...
« Reply #62 on: May 24, 2002, 06:48:00 AM »
Is this how acids of the species HX add to alkenes? The acid produces a carbocation and the halide anion attaches to it?

Yes, definitely:


PrimoPyro

  • Guest
But the more stable carbocation is the secondary ...
« Reply #63 on: May 24, 2002, 10:40:00 AM »
But the more stable carbocation is the secondary one, so wouldnt ISOpropyl halide be the chief product, chief?

hermanroempp

  • Guest
But the more stable carbocation is the secondary..
« Reply #64 on: May 24, 2002, 01:48:00 PM »
Correct, according to Markovnikov's rule, the 2-chloro-propane will form.

Quidquid agis, prudenter agas et respice finem!

Rhodium

  • Guest
Sorry, my bad
« Reply #65 on: May 24, 2002, 02:13:00 PM »
You're right. I found it at

http://www.chemguide.co.uk/mechanisms/eladd/unsymprob.html

by googling for "HX addition alkene", and copied the first addition mechanism I saw, which was an example of what doesn't happen  :-[

Here is the correct image:


PrimoPyro

  • Guest
Nickel Amalgam
« Reply #66 on: May 25, 2002, 09:15:00 AM »
Does precipitated nickel form good reducing amalgams/reducing catalyst with mercuric chloride? Is the power of the reducing metal in an amalgam determined by the metal's electronegativity? Nickel is slightly more electronegative than aluminum, so perhaps aluminum is a better reducing metal, but nickel has hydrogenation catalysis properties.

Im wondering if these properties could be enhanced by formation of a reducing amalgam with the precipitated nickel, and used in conjunction with a hydrogen donor. Would this be a better reducing system than precipitated nickel alone?

                                                   PrimoPyro

Rhodium

  • Guest
Does precipitated nickel form good reducing ...
« Reply #67 on: May 25, 2002, 10:07:00 AM »
Does precipitated nickel form good reducing amalgams/reducing catalyst with mercuric chloride? Is the power of the reducing metal in an amalgam determined by the metal's electronegativity?

Yes, the power is determined (at least in part) by the metal's electronegativity. As Nickel is not especially electronegative, it would give an amalgam with very bad reducing properties. Very few organic compounds would oxidize Ni to Ni2+ (as in the Ni reducing the organic compound).
 
I'm wondering if these properties could be enhanced by formation of a reducing amalgam with the precipitated nickel, and used in conjunction with a hydrogen donor. Would this be a better reducing system than precipitated nickel alone?

I don't understand what you are trying to achieve here. Read up on Urushibara Nickel - the combination of NiCl2 with either Al or Zn metal makes an excellent reducing system (again, the Ni is not reducing anything, it is the Al or Zn that does, the Ni is just acting as a catalyst for transferring electrons to the organic, just like the Hg does in Zn/Hg or Al/Hg amalgams.

PrimoPyro

  • Guest
I have read up on Urushibara Nickel, that is why ...
« Reply #68 on: May 25, 2002, 03:40:00 PM »
I have read up on Urushibara Nickel, that is why I am asking. I am trying to see if added HgCl2 would enhance the power of urushibara by acting as an "electron conduit."

Thanks.

Rhodium

  • Guest
No carbons? Is it then still chemistry?
« Reply #69 on: May 25, 2002, 05:56:00 PM »
The electronegativity of Ni is just slightly lower that that if Hg - I wonder if Hg would deposit properly on Ni just like that. Let's ask one of the inorganikers here.

PrimoPyro

  • Guest
Nickel is only slightly more electronegative (0.
« Reply #70 on: May 25, 2002, 06:03:00 PM »
Nickel is only slightly more electronegative (0.3 more) than aluminum.

Ny the way, in a completely unrelated question, do you think I would be successful in an attempt to reduce an organomercury acetate to an alkane with something other than sodium borohydride/hydroxide mix used on your website?

Im particularly interested in using sodium formate/hydroxide instead. Maybe sodium formate/Hydroxide/hydrazine if more power is needed.

                                                   PrimoPyro

Rhodium

  • Guest
Search at www.infotrieve.
« Reply #71 on: May 25, 2002, 07:23:00 PM »
Search at www.infotrieve.com and see if someone has done it before. NaBH4 it the perfect reagent for the reaction, so people has probably not bothered to look any further.

Osmium

  • Guest
> Nickel is only slightly more electronegative ...
« Reply #72 on: May 26, 2002, 01:08:00 PM »
> Nickel is only slightly more electronegative (0.3 more)
> than aluminum.

I don't think so!

Ni/Ni(2+) = -0.25V
Al/Al(3+) = -1.662V

Adding Hg to Ni will probably kill the catalytic acivity of the Ni.

I'm not fat just horizontally disproportionate.

PrimoPyro

  • Guest
I didnt just pull that out of my ass you know.
« Reply #73 on: May 26, 2002, 05:11:00 PM »
Ni: 1.9
Al: 1.6

Ch. Elschenbroich, A. Salzer: Organometallics, A Concise Introduction, 2nd Ed. pg. 8, Table 2-1. VCH Publishing, New York, ISBN: 0-89573-983-6

Ref cited in book for table source:

L. Pauling, The Nature of the Chemical Bond, 3rd Ed., Ithaca (1960); A.L. Allred, J. Inorg. Nucl. Chem. 17 (1961) 215.

Would you like a picture of the page?

                                                 PrimoPyro

hermanroempp

  • Guest
Electro-negative
« Reply #74 on: May 26, 2002, 06:27:00 PM »
PP, you're right with the electronegativities for Ni and Hg, I've just looked them up in the Elschenbroich/Salzer myself. But these are of little interest here, important are the standard potentials E0 in volts and here is Os right with his statements. The Al/Hg reaction works by the principles described in this thread before, it's the reductive power of the aluminium that makes the reaction go.
But I think I understand what you mean, you want to use the reductive power of the hydrogen that is also produced in this reaction, your goal is to boost the yield. This could be done by addition of Raney-Nickel to the reaction mixture. Amalgamation of the nickel is unnecessary, it's even bad practice because the mercury will "clog up" the highly porous and active surface of the nickel catalyst, thus drastically reducing its performance.

Quidquid agis, prudenter agas et respice finem!

PrimoPyro

  • Guest
Chavicol as a Safrole Precursor
« Reply #75 on: May 27, 2002, 03:10:00 PM »
Has anyone ever tried to use Chavicol, a.k.a. 4-hydroxy-allylbenzene, as a safrole precursor?

If one used sodium hydroxide and chloroform to perform the Reimer-Tiemann formylation, the only available position for the formyl group is ortho to the hydroxy group.

Then if this reaction were followed up by reaction with sodium hydroxide and hydrogen persoxide in a Dakin Reaction, the ortho aldehyde is oxidized to a phenol in the same ring position, giving 3,4-dihydroxy-allylbenzene.

Methylenation of this compound gives safrole.

                                                   PrimoPyro

Rhodium

  • Guest
My opinion
« Reply #76 on: May 27, 2002, 03:34:00 PM »
Hydrogen peroxide on a hydroxy-allylbenzene is asking for polymerization in my opinion.

PrimoPyro

  • Guest
Hm, and doing it at the acetone stage would form ...
« Reply #77 on: May 27, 2002, 03:46:00 PM »
Hm, and doing it at the acetone stage would form peroxides...

I see why it hasnt been done.  ;D

PrimoPyro

  • Guest
Formic Dicarboxylic Half Esters
« Reply #78 on: May 29, 2002, 07:55:00 PM »
Hi.  :)

Would a viable method of forming dicarboxylic aicd half esters of the formula HOOC-(CH2)nCOOR be the ring opening of the corresponding cyclic dicarboxylic anhydride with alkalai alkoxide, resulting in the half alkyl ester half alkalai salt of the dicarboxylic acid?

Illustratory example: Succinic anhydride would react with sodium ethoxide, forming NaOCO-CH2-CH2-COOEt.

The nature of the alkyl group is not important, it serves here only as a protecting group for the carboxyl. The half ester is desired for a Kolbe-like reaction where the product is to have a preserved terminal carboxyl group. This half ester faciliatates a protecting group for a single carboxyl while admitting the second carboxyl to facilitate the coupling reaction.

                                                   PrimoPyro

lugh

  • Guest
Amalgams
« Reply #79 on: May 30, 2002, 05:41:00 PM »

The electronegativity of Ni is just slightly lower that that if Hg - I wonder if Hg would deposit properly on Ni just like that.




Nickel amalgam cannot be formed by the direct union of nickel and mercury, according to J Nicklès in Comptes Rendus 104 154 (1853) While it's possible nickel amalgam could be formed under the described conditions, it's rather unlikely. Mercuric chloride is reduced by most metals, which could result in nickel chloride, and possibly form nickel amalgam at that point, though normally current or sodium are necessary to perform the task. Thus what's most likely to occur is what has been already stated, nickel's catalytic properties are dependent on being finely divided into small particles to adsorb hydrogen onto it's surface; which would be impaired by mercury on the surface  :(  On the other hand, when nickel amalgam is formed, it does produce hydrogen, nascent hydrogen is a well known reducing agent, and thus may bee useful  :)


Osmium

  • Guest
I seem to remember that very fine metal powders ...
« Reply #80 on: May 31, 2002, 01:41:00 AM »
I seem to remember that very fine metal powders (including nickel) can be formed by distilling away Hg from their amalgams. But I doubt that they can be used as catalysts.

I'm not fat just horizontally disproportionate.

Bwiti

  • Guest
If making methyllithium, phenyl-Mg-bromide, or ...
« Reply #81 on: May 31, 2002, 06:39:00 AM »
If making methyllithium, phenyl-Mg-bromide, or some other metal complex, is it generally OK to use potassium as a substitute? Patent GB861350 mentions Na and K in producing phenylcyclohexylamines, and I was just wondering if this can be expanded to amphetamine chemistry. Peace! 8)

Love my country, fear my government.

PrimoPyro

  • Guest
Bwiti, I dont really know if potassium is an ok ...
« Reply #82 on: June 02, 2002, 07:36:00 PM »
Bwiti, I dont really know if potassium is an ok substitute, but it would make sense. It may be more reactive though.

Ok, I need someone to throw me a bone here. What would be a good/best way (in your opinion) to prepare any salt or ester derivative of 2-formyl acrylic acid? I don't mean the process has to work for a wide range of salts and esters, I mean if you know a method/reaction to produce any salt or ester, of 2-formyl acrylic acid, please indulge me, and tell me how.

Don't say by Retro-Diels-Alder reaction of the corresponding 2-norbornene, because that is what this is being used to produce!  ;D

I can think of about 6 or 7 different crazy methods, but they all have a flaw in them that will prevent their success. I'm truly stumped, everywhere I turn, I find a reason why I cant use such and such method.

Any ideas? Thanks.... ::)

P.S. ten extra points goes to the person who guesses where the interest in this stems from, i.e. why would someone want this compound?  ;)

                                                   PrimoPyro

PrimoPyro

  • Guest
Idea
« Reply #83 on: June 02, 2002, 08:48:00 PM »
I just had an idea. Would this work?

Acrolein could be halogenated at the second position (how would you substitute the H for an X without saturating the olefin?) and then reacted with sodium cyanide, forming the 2-formyl acrylonitrile. This is subjected to alcoholysis with methanol and aqueous HCl, forming the methyl ester of 2-formyl acrylic acid, which is precisely the compound desired.

Would that work well? How would one halogenate the acrolein at the 2 position without reducing the double bond? Is there a better starting material for this?

P.S. hint for 10 point bonus: Oppolzer is a very elegant man. He has my utmost respects.  :)

                                                  PrimoPyro

Rhodium

  • Guest
Anhydride to monoester
« Reply #84 on: June 04, 2002, 11:22:00 AM »
Regarding

Post 315835

(PrimoPyro: "Formic Dicarboxylic Half Esters", Chemistry Discourse)
- Yes, that would work, but you don't need to use an alkoxide, just react the cyclic anhydride with the alcohol of choice.

PrimoPyro

  • Guest
Thank you
« Reply #85 on: June 04, 2002, 01:17:00 PM »
Thanks Rhodium, for the answer here and at the tryptamine forum as well.  :)

Regarding

Post 317041

(PrimoPyro: "Idea", Chemistry Discourse)
Thanks goes to hypo for helping me out there, providing a Chem Berichte reference and also physical data on (and the preparation of) 2-bromoacrolein.

If anyone is interested in the article, one of us can copy it here.

Then the 2-bromoacrolein is reacted with sodium cyanide to give 2-formylacrylonitrile, which is alcoholyzed with methanol and HCl to 2-formylacrylic acid, methyl ester, ready for reaction with cyclopentadiene to form that funky bicyclic.  :)

                                                  PrimoPyro

PrimoPyro

  • Guest
Revival
« Reply #86 on: October 08, 2002, 04:10:00 AM »
Its time to revive my thread once again with another basic question. I almost made a new thread asking this, until I remembered I had made this thread for these type purposes.

Question: What is the mechanism of the safrole + KOH ---> isosafrole reaction?

I am uncertain how it proceeds, and am curious why this occurs.

I imagine that the base -OH removes a proton from the alkene in the first step, forming an intermediate carbanion. I am *guessing* that the formed carbanion would be in the isopropyl position, because it would be the most stable, as it has the least protons/is the most substituted?

If this happens, yay, "go me" but what happens next is beyond me. Logically some sort of electronic rearrangement would take place, but I am unsure just what happens. Carbanions are not prone to rearrangement like carbocations are, so I don't know why or where any formed carbanion would move to, and what effect this would have on the double bond.

Formation of an isopropyl carbanion wouldn't even affect the alkene, it would just turn R-CH2-CH=CH2 into R-CH2-C(-)=CH2, right? Hmm, I wonder:

This intermediate, if *somehow* could get an electron onto that terminal carbon (I have no idea how, so don't ask) then the double bond would be reduced to a terminal carbanion, and a radical at the isopropyl position, along with the isopropyl carbanion, and that radical might rearrange to a benzyl radical. The isopropyl carbanion could give up its electron, which would also form an intermediate radical, and of course two radicals means bond strengthening, restoring the alkenyl bond in a different location: The benzyllic position that corresponds to isosafrole.

Then the terminal carbanion would react with the water produced in the first step to steal back a proton and regenerate the hydroxide catalyst.

Am I close? Not even? What happens? If I am close, where does this mystery electron come from? It can't come from any sodium atoms because they are all already existing as Na+ and won't yield any more electrons.....

How does this isomerization work? Help please? I'm not asking for pictures in ISIS or anything, just a word description like I have given, and maybe an explanation using the "R-CH2-CH=CH2" diagrams I use.

Any help would be appreciated. Thanks.

PrimoPyro

Firm supporter of the "Purge The Couch!" movement. Vote for the purge today.

Osmium

  • Guest
Please be patient with me and don't expect too ...
« Reply #87 on: October 08, 2002, 08:14:00 AM »
Please be patient with me and don't expect too much, I'm tired and exhausted, but I'll try to answr your question:

The KOH removes an H+ from the BENZYLIC position. That is the most stable carbanion that can form, since there is an electron deficient aromatic ring next to it, and it is also the benzylic position of the alkene. Check pKa values of benzylic and allylic hydrogens. This also means that electron can nicely delocalize all over the molecule.
I'm not sure if the H+ really fully separates and produces H2O with the OH- (which would then be absorbed by the KOH or removed by heat/vacuum) or not. I suspect the former, but it doesn't really matter.

So how is the isoalkene produced?

The (delocalized) anion now grabs itself another H+ from the next molecule of safrole. It ends up as predominantly isosafrole since that is the thermodynamically preferred product. It's kinda like a chain reaction proceeding through the whole reaction, with protons jumping from one molecule to the next one. If water mets such a carbanion it will probably give away its H+ and end up as OH- again. Of course all these steps are equilibriums, and the mixture slowly moves energetically downhill, to the preferred trans-isosafrole. The longer you perform the reaction the more trans you end up with, since the cis-isosafrole lies somewhere in between the safrole and the isosafrole.

Hope that was somewhat understandable.
I dunno if that's the real mechanism, if someone has a better explanation please correct me.

I'm not fat just horizontally disproportionate.

Protium

  • Guest
Hyperconjugation Stabilization
« Reply #88 on: October 10, 2002, 03:43:00 AM »
To supplement Osmiums explanation, and please correct the details of this if part is incorrect because i'm not very knowledgable with orbitals and such, this is just an educated guess based on memory and now some checking back.  I would say that the KOH dissolves and the hydroxyl abstracts a protium at the most stable position once the temp. allows for the C-H bond dissociation energy requirement, which will be the lowest for the most stable (benzylic) position, to form a singly occupied orbital, which would then bend over and try to grab onto another protium by overlapping of the p-orbital at the radical center with it's neighboring allylic C-H bond, forming into a more planar-type geometry, and this delocalization would continue almost instantaneously down the line until the delocalized radical finds the most stable position that it can on the alkyl chain, naturally the greater number of substituents the greater the strength of the hyperconjugation (stability), and at that point the hyperconjugation stops, and the double bond has shifted to this position of stability.

And so this reaction would be a free radical reaction by the homolysis of an H+ , the formation of a carbon centered radical at the benzylic position, and a hyperconjugation stabilization (as opposed to a resonance delocalization) of the substituent chain containing the pi bond.

Hold on i'll have a picture in a minute too. :)

Who say's MS Paint is useless?



I can't pull out too great of an analogy right now as Primo can do so well but basically the 4(benzylic) position of the carbon chain loses an H+ and so it reaches over and grabs one from a more stable neighbor on the chain, and in turn the newly deficient molecule does the same thing to it's neighbor, until you get the guy on the end of the rope (alpha carbon) who has even less hydrogen due to it's p-bond, and so the second to last carbon gets screwed for a hydrogen, and being more stable than the end carbon, is forced to take on the pi-bond. 

I've been awake for a long time now, and there's probably a bunch of mistakes, so if you see a correction please point it out. I'm tired as hell but couldn't sleep if I wanted to. The real answer Primo's question, however, is probably a lot more complex than this, as these types of reactions often have different series' of propagation cycles, different initiation and termination and all that as well, so like I said this is my best guess and I share Primo's curiousity in learning more about this reaction.

And if Primo would be so kind as to allow me to impose a follow up question to his own in his thread, does anyone know the mechanism of the equalibrium which determines the cis/trans isomerism ratio?  For most cooks I guess that it's a non-issue but I am rather curious as to the science of why there are differing isomers, as clearly this could not be simply based on chaotic mathematics and complete randomness if there is an equalibrium.


Pr(+)tium