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Author: Subject: preparing 2-phenylethan-1-amine?
does_it_work?
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[*] posted on 7-8-2010 at 05:31
preparing 2-phenylethan-1-amine?


Hello!

in the synthesis of 2-phenylethan-1-amine, which method should be more effective?

a) benzene + n-chloroethylamine +iron (III) chloride

or

b) NaBr + (bromomethyl)benzene ---> 2-phenylacetonitrile

2-phenylacetonitrile +lithium aluminum hydride + h2o

-thank you
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entropy51
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[*] posted on 7-8-2010 at 06:34


Quote: Originally posted by does_it_work?  
Hello!

in the synthesis of 2-phenylethan-1-amine, which method should be more effective?

a) benzene + n-chloroethylamine +iron (III) chloride

or

b) NaBr + (bromomethyl)benzene ---> 2-phenylacetonitrile

2-phenylacetonitrile +lithium aluminum hydride + h2o

-thank you
If you think the first reaction under (b) is possible, it seems unlikely that you could pull off either one.

Maybe you meant NaCN instead of NaBr?

LAH isn't required to reduce the nitrile to the amine. And you sure don't need H2O in a LAH reduction:o
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[*] posted on 7-8-2010 at 07:03


ah, thank you- that was a typo. sodium cyanide, not sodium bromide.

LAH reduction was written improperly. i seem to recall that being at least a two step procedure.

if LAH is not needed for the reduction, what would work?
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[*] posted on 7-8-2010 at 07:46


Real chemists call it beta-phenethylamine and bromotoluene.

a) sounds like it would give a mess.
b) is certainly feasible; for reaction conditions, catalysts and reducing agents check the literature. Learn how to use a library.

But usually this is done at lab scale by condensation of benzaldehyde and nitromethane in acidic medium followed by one of various reduction. The literature is full of examples.

Of course LAH reactions need a source of H+, e.g. water. So your total reaction equation wasn't entirely wrong.
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[*] posted on 7-8-2010 at 09:15


Quote: Originally posted by turd  
Real chemists call it beta-phenethylamine and bromotoluene.

I would call it benzyl bromide or a-bromotoluene, but bromomethylbenzene is IUPAC.

To A) Does not work like this... you would have to protect the amine very well... Also anhydrous aluminium chloride would be a better catalyst than FeCl3.

Instead of n-chloroethylamine and lewis acid you can work with benzyl-N-vinylcarbamate (can be made from acrylic acid chloride, sodium azide, pyridine, hydrochinone and benzylalcohol) and 9-BBN⋅THF... than you would have to remove the Cbz-group with HAc/HBr to get your wanted amine... but that would be a pretty hard way. It may be only interesting if very sensitiv groups are attached on the aryl ring.

To B) I would prefer this method.

LiAlH4, LiAlH4+H2SO4, LiBH4 or NaBH4+H2SO4 in THF should work very nice for reduction of the nitril.

[Edited on 7-8-2010 by Methansaeuretier]

[Edited on 7-8-2010 by Methansaeuretier]
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[*] posted on 7-8-2010 at 09:25


Quote: Originally posted by Methansaeuretier  

I would call it benzyl bromide or a-bromotoluene, but bromomethylbenzene is IUPAC.

D'oh. Yes, benzyl bromide of course (my brain...). Real chemists don't use IUPAC. :)

[Edited on 7-8-2010 by turd]
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[*] posted on 7-8-2010 at 10:01


Quote:
a) benzene + n-chloroethylamine +iron (III) chloride


Of course 2-chloroethylamine not at all stable. It reacts with itself forming a polymer. However...2-bromoethylamine hydrochloride is commercially available. It is soluble in CHCl3, and possibly under the right conditions you might get a friedel-crafts working. The alternative would be to take the 2-bromoethylamine and add 2eq of an anhydrous tertiary amine (such as TEA). This would form aziridine in situ which is known to give phenylethylamine with benzene in the presence of AlCl3.

Anyways..I dont quite understand the question as phenylethylamine is readily available online to everyone. In my opinion the poster wants to make substituted phenylethylamines but instead of asking us his intended question he is hiding behind the synthesis of a benign compound. This is annoying because it wastes our time, there are many strategies to synthesize phenylethylamine that will not work for the synthesis of its derivatives.

[Edited on 8-7-2010 by smuv]




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[*] posted on 7-8-2010 at 10:01


Click on the search button. When you get to the search page.....Type in Phenethylamine.

For what end, pray tell, might you be wanting phenethylamine?

Perhaps there is an alternative to making it.

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[*] posted on 7-8-2010 at 11:04


Quote: Originally posted by smuv  
Quote:
a) benzene + n-chloroethylamine +iron (III) chloride


Of course 2-chloroethylamine not at all stable. It reacts with itself forming a polymer. However...2-bromoethylamine hydrochloride is commercially available. It is soluble in CHCl3, and possibly under the right conditions you might get a friedel-crafts working. The alternative would be to take the 2-bromoethylamine and add 2eq of an anhydrous tertiary amine (such as TEA). This would form aziridine in situ which is known to give phenylethylamine with benzene in the presence of AlCl3.

And there is also an alternative using a more benign reagent. The Friedel-Crafts alkylation of benzene with oxazolidin-2-one also gives 2-phenylethylamine (but not giving the reference because I see no references in this thread).

@does_it_work?: Open referenceless and/or beginner's threads only in the Beginnings section where I'm moving this.

[Edited on 7/8/2010 by Nicodem]




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[*] posted on 7-8-2010 at 13:01


We could argue all day about nomenclature, but as my first year lecturer said: The important thing is that what you call it means nothing other than what you mean.

That said, if it doesn't exist it doesn't help :P though as zed sed - there are lots of references and articles all over SM and the rest of the internet about such compounds. :)




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[*] posted on 7-8-2010 at 17:33


Quote: Originally posted by turd  
Of course LAH reactions need a source of H+, e.g. water. So your total reaction equation wasn't entirely wrong.
That's odd considering that everyone I know runs LAH reductions in anhydrous media. Benzyl bromide, bromotoluene, whatever.
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[*] posted on 7-8-2010 at 17:39


Quote: Originally posted by entropy51  
Quote: Originally posted by turd  
Of course LAH reactions need a source of H+, e.g. water. So your total reaction equation wasn't entirely wrong.
That's odd considering that everyone I know runs LAH reductions in anhydrous media. Benzyl bromide, bromotoluene, whatever.


He must be referring to an acid work-up, because LAH reactions are typically done in ether. LAH ignites on contact with water.




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[*] posted on 7-8-2010 at 17:51


Quote: Originally posted by DDTea  
He must be referring to an acid work-up, because LAH reactions are typically done in ether. LAH ignites on contact with water.
Well, he said reaction, not workup. None of the labs where I worked ever used a workup that didn't involve NaOH, not acid.

Quote:
LAH ignites on contact with water.
Thanks for clearing that up for me.

[Edited on 8-8-2010 by entropy51]
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[*] posted on 7-8-2010 at 19:19


You could get all techical and say that the workup in this case is part of the reaction but thats hassle. I'm aware of the NaOH method (I believe thats the one Feiser teaches correct?) but personally prefer the nicer sounding workup utilising aqueous ammonium chloride and rochelle salt. Technicalities aside, the reaction is of course perfomed under anhydrous and aprotic conditions.
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[*] posted on 8-8-2010 at 06:55


Thank you for the depth of you responses!

a few words on nomenclature:

understanding that IUPAC is seldomly used in chemistry circles, i don't have the advantage of knowing the appropriate common names. IUPAC allows what could otherwise be a very muddled communication to have some clarity.

The gross areas of knowledge that are missing on my part reflect the fact that my organic chem lab experience is next to nil. The unfortunate thing about many chem texts, as well, is that the lab conditions and procedures are frequently excluded for "ideal" reactions. That being said, a student with limited field experience may be hard pressed to understand the nuances in chemical procedure that make the critical differences in outcome.

Now, on to the chemistry.

This inquiry started as a result of my trying to understand the process and theory behind poly-substituted benzene chemistry. On paper, it nearly always seems simple. Bromo, nitro, hydroxy - groups, no problem. Activating, deactivating, wonderful. So the addition of a more complicated alkyl group is a progression from this beginning. My difficulty comes from understanding how to prioritize reactivity of a more complex reaction from simple textbook rules.

@turd: is the benzaldehyde and nitromethane reaction similar to the following: http://www.orgsyn.org/orgsyn/prep.asp?prep=cv1p0413 ?

@smuv: could you elaborate on the 2-bromoethylamine method?

@Nicodem: Thank you for moving the thread. I don't understand the mechanism in the Friedel-Crafts alkylation of benzene with oxazolidin-2-one. is there a particular resource in the SM library that outlines this general procedure?

-To address concerns over intent:

Is this a suspicious synthesis? Certainly. There's no point in pretending that the compound doesn't justifiably raise some alarms. Is it entirely beyond speculation that someone would want to learn the ropes of such a reaction for educational, benevolent purposes?

I had originally come across this compound while reading about MAO in neurotransmitter chemistry. Since I was learning about substituted benzene compound concurrently, it does not seem such a wild leap that I began to wonder about the synthesis for a (visually) simple substance.

I'm only trying to expand my knowledge of chemistry beyond a fickle, textbook-based amateur standpoint. I'm trying to actively consume and understand basic organic chem via textbooks and internet resources, but without the guide of experience the process is both daunting and dangerous (if attempted in a lab).

Thank you very much for your time and your patience.







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[*] posted on 8-8-2010 at 09:58


In that case, pick the safest, easiest way. All things being equal, you have picked a very strange route to Beta-Phenethylamine.

Common routes are:

Phenylalanine+Acetophenone+Heat------------->Phenethylamine+Acetophenone+CO2

Or, Benzaldehyde+Nitromethane------>Beta-Nitrostyrene, then H2/Pt H+----->Phenethylamine.

Or,BenzylChloride+NaCN----->Phenylacetonitrile, then H2/Pt------>Phenyethylamine

For a range of FREE, well illustrated, quasi-practical methods of producing Phenethylamines....And other materials. I would refer to "The Rhodium Archive" hosted by The Vaults of Erowid. Whatever, its original intention was, it is a pretty good organic chemistry text. Just click on the link, and scroll down. It's a giant chemistry text. http://www.erowid.org/archive/rhodium/chemistry/index.html

For tried and true preparations, consult Vogel "Textbook of Organic Chemistry" in our own library. Or, Organic Synthesis, generously available, free online. http://www.orgsyn.org/

If you just want some Phenethylamine, you can buy it outright, very inexpensively. They generally sell it at your local healthfood store, or online. It has become a common supplement.

When you ask the Genii to grant you a wish, you have to phrase your request very carefully. Otherwise, they may become grumpy, and a "Monkey's Paw" type of situation can develop. The directions are technically correct, but they only lead you deeper into the woods.

Working with LiAlH4, as some have suggested, is do-able. But, it isn't the most wholesome of approaches. It works OK, but it's fairly dangerous to handle.

[Edited on 8-8-2010 by zed]

[Edited on 8-8-2010 by zed]
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[*] posted on 8-8-2010 at 13:00


Quote: Originally posted by does_it_work?  

This inquiry started as a result of my trying to understand the process and theory behind poly-substituted benzene chemistry. On paper, it nearly always seems simple. Bromo, nitro, hydroxy - groups, no problem. Activating, deactivating, wonderful. So the addition of a more complicated alkyl group is a progression from this beginning. My difficulty comes from understanding how to prioritize reactivity of a more complex reaction from simple textbook rules.

Do you seriously think members here are so dumb to take that seriously? What does 2-phenylethylamine has to do with that? How is that supposed to quell the suspicions you are (hypothetically) just some greedy person interested in making fentanyl? So either avoid making up excuses (because here you don't need any as long as you conform to the rules) or be honest and cope with what's to come. The worst you can do is lying. That pisses me of more than all the kewls asking for "recipes".

Quote:
I had originally come across this compound while reading about MAO in neurotransmitter chemistry. Since I was learning about substituted benzene compound concurrently, it does not seem such a wild leap that I began to wonder about the synthesis for a (visually) simple substance.

And you chose this as your first hand on experience in organic synthesis? OK, fine, but with LiAlH4? That makes you look like someone with suicidal tendencies. Also a bit insane given that you would want to use a reagent like LiAlH4 that costs ten times more than the target compound. And where would you buy it? On the other hand you can buy the target compound without any paper signing or hazardous material shipping costs. Seriously, start first with reading books about practical chemistry, like Vogel's or Organikum or some older ones, like the ones in our library. Then start with simple experiments on small scale, possibly consulting the members here first. You don't start building a house from the roof. Then, when you will have some experience, you can use the beginner's friendly two step route to 2-phenylethylamine with benzaldehyde, nitromethane and aluminium amalgam.

Quote:
I'm only trying to expand my knowledge of chemistry beyond a fickle, textbook-based amateur standpoint. I'm trying to actively consume and understand basic organic chem via textbooks and internet resources, but without the guide of experience the process is both daunting and dangerous (if attempted in a lab).

Then perhaps it is time to start using primary literature as well.

Quote:
@Nicodem: Thank you for moving the thread. I don't understand the mechanism in the Friedel-Crafts alkylation of benzene with oxazolidin-2-one. is there a particular resource in the SM library that outlines this general procedure?

And also learn how to UTFSE. This skill comes very useful in these chaotic times.

Welcome to our forum. ;)




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[*] posted on 9-8-2010 at 13:43


Quote: Originally posted by entropy51  
Well, he said reaction, not workup.

Sheeesh. I've written "total reaction equation" (which I probably should have more appropriately worded "overall reaction equation") as in the sum of multiple partial reactions. That should be understandable by anyone who is not purposefully obtuse. It's perfectly fine to write the reduction with LAH as
R-CO-R' + 0.25LiAlH4 + H2O --> R-CHOH-R' + 0.25LiOH + 0.25Al(OH)3

That allows every chemist with a minimum of common sense to use partial equations that do not even take place. E.g. how many equivalents of LAH do I need at least to reduce a nitrostyrene?
Well:
Ar-CH=CH-NO2 + 4H2 --> Ar-CH2-CH2-NH2 + 2H2O
LiAlH4 + 4H2O --> 4H2 + LiOH + Al(OH)3
gives:
Ar-CH=CH-NO2 + LiAlH4 + 2H2O --> Ar-CH2-CH2-NH2 + LiOH + Al(OH)3
Dead. Simple. (And in this case not applicable.) Every chemist should have those trivialities down long before even knowing what a nitrostyrene is.

Off-topic, but well-meant advice: get a few friends and get out more. Apparently I'm not the only one to perceive lots of negativity and unwarranted arrogance in your posts. You don't seem to enjoy your life.

Quote:
understanding that IUPAC is seldomly used in chemistry circles, i don't have the advantage of knowing the appropriate common names. IUPAC allows what could otherwise be a very muddled communication to have some clarity.

Well, you should learn the trivial names. Because if you order "(bromomethyl)benzene" the clerks will think you are "probably a nice guy, but doesn't know anything about chemistry". Bad way to start a business relation.

IUPAC nomenclature was important in a time when you had to lookup compounds in paper based databases. Nowadays you search compounds using graphical user interfaces and if a journal wants IUPAC names, you let a program generate it. :P
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[*] posted on 9-8-2010 at 15:48


Quote: Originally posted by turd  
Ar-CH=CH-NO2 + 4H2 --> Ar-CH2-CH2-NH2 + 2H2O
LiAlH4 + 4H2O --> 4H2 + LiOH + Al(OH)3
gives:
Ar-CH=CH-NO2 + LiAlH4 + 2H2O --> Ar-CH2-CH2-NH2 + LiOH + Al(OH)3
Dead. Simple. (And in this case not applicable.) Every chemist should have those trivialities down long before even knowing what a nitrostyrene is.
And so the H2O on the left hand side comes from, well, where, when the reaction is conducted in anhydrous ether? Although I make no claim to being a chemist, I find your explanation somewhat trivial. Perhaps a real chemist like Nicodem could clarify this turdid, I mean turbid, situation for me. Does the reaction form a complex, which is only hydrolyzed to the products during an aqueous workup?

I think I smell some unwanted arrogance myself.

Edit: Sorry, Mr Turd, I did mean left hand side. I think that some people do indeed interpret reactions as recipes, and so writing a formalism that mixes LAH and H2O seems a little dangerous.

[Edited on 10-8-2010 by entropy51]
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[*] posted on 10-8-2010 at 01:35


Quote: Originally posted by entropy51  
And so the H2O on the right hand side comes from, well, where, when the reaction is conducted in anhydrous ether?

???
Your (rhetorical?) question doesn't make any sense unless you mean left hand side. Anyway I hoped I was clear that it's a formalism. When calculating how much LAH you need, you typically think about H2 equivalents and how much H2 equivalents one LAH molecule can provide in the overall reaction. And it can only provide H2 together with a H+ ion, which we canonically say comes from a H2O molecule. Could also come from an alcohol or whatever. Doesn't matter. As well as it doesn't matter if the H+ is introduced before, during or after the main reaction. Total reaction equations are not recipes, but abstract representations of what happens. And not really useful in the example I gave, since you need distinctly more than 1 eq. of LAH in this reaction.

BTW: I'm not here to be right (as in I'm right, you're wrong, bwahaha). Being right is totally uninteresting and a waste of time. Being shown wrong with convincing arguments is much more interesting since it means you learn something new and progress mentally. So what you perceive as arrogance from my part is just a sincere feeling that some people know hundreds of name reactions and mechanisms by heart before having thought about the basics.
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[*] posted on 14-8-2010 at 23:46


Which one do you want? Phenacylamine (aka 2-aminoacetophenone) which "could" be reduced to the B-phenethylamine(which could be made in other ways). Alternatively, do you want the a-phenethylamine?

Yes, naming conventions are kind of hard, but try naming the right chemical(s)...




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[*] posted on 15-8-2010 at 13:29


Seems that this thread looses focus, while the big, antisocial egos say what they just have to say. Big mouths. But really, as one poster suggested, doing a Knov. w/ nitromethane is probably your best way. No need for LAH. Just plain activated Al in isopropanol will reduce to the amine. Pretty much everything is OTC - the only way to go. If it's not OTC, it's not fun any more. Best luck, and don't let the immature and insecure huge egos get to you. Just let them show their true colors to everyone else;)!
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