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dan_teod
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Amines from alcohols??
I am interested to obtain amines from a tertiary (diphenyl-alkyl) alcohol. For the moment I use a method: I make chloride with SOCl2/CH2Cl2+Et3N, I
make azide with NaN3/DMF and for the amine a reduce it with Zn+NH4Cl/EtOH-H2O. I want a better method because this has 2 problems: first - the
reaction with sodium azide tkaes 72 hours and the second is that de reduction doesn't work very well.
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solo
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Why not use HI it will reduce your alcohol to the alkane and give you the desired amine.......or catalytic hydrogenation to remove the Cl with
Pd/C........or ZnCl +HCl to reduce the chlorine........solo
Edit: In my reading of the post I thought I read it was a tertiary amine with an alcohol group.....so my suggestion is not valid......solo
[Edited on 17-2-2008 by solo]
It's better to die on your feet, than live on your knees....Emiliano Zapata.
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Klute
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Just an idea, not sure if everyhting would be terribly efficient yeild-wise, but maybe you try the following sequence:
-Turn the alcohol to an alkylhalide
- Monoalkylate diethylmalonate and hydrolysis/decarboxylation to the corresponding acid
-Formation of the corresponding acyl chloride
-Formation of the amide with NH3
-Hofmann degradation to the amine
There's quite a few steps, adding a carbon to pull it off after, but as you can't use a Gabriel synthesis (correct me if i'm wrong) or form a
nitrile, the azide pathway might be your best shot after all.
Anyone else got any ideas? It's a pretty challenge
EDIT: We posted at the same time Solo 
I don't undertand, the reductions ytou propose would just give the alkane, how would he form the amine after that?
PS: if the reduction doesn't work well enought, you could try LiALH4 or CTH/catalytic hydrogenation instead, no? If the azide swap works well, 72h
reaction at room temp can't be such a problem, can it?
[Edited on 17-2-2008 by Klute]
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The_Davster
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Why such chlorination conditions on a tertiary centre? Shouldent a tertiary alcohol go to chloride well by just aqueous HCl by SN1 mechanism?
Triphenylphosphine followed by aqueous workup should convert the azide to amine with elimination of nitrogen. Assuming you can easily separate TPPO
from your product.
The reduction can also be done by PtO2 and 1atm hydrogen using polyethylene glycol as solvent(MW=400)
Whats wrong with the gabriel synthesis here? Tertiary chloride too big?
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dan_teod
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I can't use Gabriel because there is a big steric hinderance
The original recipe was with LiAlH4 but I changed to increase the yield.
I can't use to many steps or low yield reactions because my starting material cost around 100 € per gram and I'll have to work a lot with the amine.
Thanks for the interest
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Klute
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So is the Zn reduction higher-yielding than the LiAlH4?
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leu
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It seems that the wrong question is being asked, but this paper describes the use of stannous chloride in methanol and yields 98%  If one is paying that much for the precursor one should be looking for ways to
synthesize it from less expensive materials 
Attachment: Tet_Lett_27_Azides_to_amines_with_SnCl2.pdf (128kB)
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leu
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This paper describes methods for obtaining azides in a lot less than 72 hours 
Attachment: Synthesis_1976_PTC_preparation_of_alkyl_azides.pdf (81kB)
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chemrox
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two nice refs-thanks!
"When you let the dumbasses vote you end up with populism followed by autocracy and getting back is a bitch." Plato (sort of)
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Nicodem
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| Quote: | Originally posted by dan_teod
I am interested to obtain amines from a tertiary (diphenyl-alkyl) alcohol. For the moment I use a method: I make chloride with SOCl2/CH2Cl2+Et3N, I
make azide with NaN3/DMF and for the amine a reduce it with Zn+NH4Cl/EtOH-H2O. I want a better method because this has 2 problems: first - the
reaction with sodium azide tkaes 72 hours and the second is that de reduction doesn't work very well. |
You should provide more information for a reliable answer. Does your diphenylalkyl alcohol contain any other functionality? If not, or at least no
acid labile ones, you could try a Ritter reaction with HCN or acetonitrile on it. However, keep in mind that the resulting t-alkyl
formamide/acetamide will be a bitch to hydrolyze. Maybe you could get it off with hydrazinolysis or KOH/ethylene glycol.
For preparation of the diphenylalkyl azide try another system, like NaN3/DMSO. Maybe it could give you better results in shorter time. Did you noticed
any elimination products and how much is the yield?
For the azide reduction there are really so many methods. Ph3P as The_Davster suggested might work, but perhaps it would be even less efficient than
Zn/NH4Cl since the intermediate would be a quite sterically crowded. Hydrogenation could also be applied if you find the correct conditions as to stop
it before the amine falls off (the benzhydryl group is even more easily removed by hydrogenation than the benzyl!).
…there is a human touch of the cultist “believer” in every theorist that he must struggle against as being
unworthy of the scientist. Some of the greatest men of science have publicly repudiated a theory which earlier they hotly defended. In this lies their
scientific temper, not in the scientific defense of the theory. - Weston La Barre (Ghost Dance, 1972)
Read the The ScienceMadness Guidelines!
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leu
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Rearrangements are common during the Ritter reaction, the attached diagram illustrates the phenomenon well Without more information it's impossible to give a good answer to the question, but one thing is clear and that's
since the diphenylalkyl alcohol in question is so expensive there are undoubtedly ways to synthesize it from far less costly reagents
Attachment: ritterrearrngements.djvu (7kB)
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dan_teod
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I'm not paying that price for the alcohol, but I mentioned it to underline his commercial value. The yield of reduction with ZnCl2 is comparable with
the one with LiAlH4, but it's more easy to work I'll try the reduction with SnCl2, hope will work. Thanks for help ( multumesc Leu).
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shemati
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A short method for converting of an alcohol to amine is accomplished by Mitsunabu. In this method by reaction of alcohol with PPh3 (tripheny
phosphine) in the presence of NaN3 and Diethylazodicarboxylate and then addition of water you can perparing of related amine.
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Nicodem
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Shemati, do you have any references for any kind of Mitsunobo nucleophilic substitution on such a sterically hindered substrate that usually does not
even want to react through SN2 pathways?
The paper about azides reduction with SnCl2 that Leu uploaded gives a nice review also of other systems used for reducing azides. I would check those
as well if the azide reduction is problematic.
Leu, carbocations do tend to rearrange whenever there are possibilities to form a more stabile carbocation. However, rearrangements are not to be
expected on this particular substrate since it already forms an excellently stabilized carbocation that has no way to stabilize further (except by
reacting with the nucleophile or by eliminating a proton - which is reversible anyway).
…there is a human touch of the cultist “believer” in every theorist that he must struggle against as being
unworthy of the scientist. Some of the greatest men of science have publicly repudiated a theory which earlier they hotly defended. In this lies their
scientific temper, not in the scientific defense of the theory. - Weston La Barre (Ghost Dance, 1972)
Read the The ScienceMadness Guidelines!
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Methyl.Magic
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| Quote: | Message original : Nicodem
You should provide more information for a reliable answer. Does your diphenylalkyl alcohol contain any other functionality? If not, or at least no
acid labile ones, you could try a Ritter reaction with HCN or acetonitrile on it. However, keep in mind that the resulting t-alkyl
formamide/acetamide will be a bitch to hydrolyze. Maybe you could get it off with hydrazinolysis or KOH/ethylene glycol.
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yeah this methode is easy, the yield is not very high but the reagent are very cheap.
Ph2Me-COH + H2SO4 > Ph2Me-C(+)
Ph2Me-C(+) + MeCN > MeCN(+)-C(Ph2Me)
hydrolyse to Ac-N-C(Ph2Me) and cleave your amide with KOH to make your corresponding (Ph2Me)-C-NH2 .
I've done a trial with t-butanol on acetonitrile to form t-butylamine : it's woked great ¨!
You can probably use urea instead of Acetonitrile. Personnally I'll avoid using hydrogene cyanide because of its high toxicity-
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DJF90
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ok so you want to replace the hydroxy group with an amine group? Heres how I would tackle it:
React the alcohol with conc. Hydrochoric acid to produce chloride
Reflux with alcoholic ammonia to yield amine
Is it really that simple? Surely I'm missing something...
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leu
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HI has been used to reduce azides to amines:
Attachment: Tet_Lett_43_Azides_to_amines_with_HI.pdf (58kB)
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leu
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Zn/NiCl2 is used to reduce azides to amines:
Attachment: Synlett_1997_Azides_to_amines_with_zinc_and_NiCl2_6H20.pdf (134kB)
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leu
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Fe/NH4Cl can reduce azides to amines:
Attachment: Tet_Lett_37_Azides_to_amines_with_Fe_and_ammonium_chloride.pdf (136kB)
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leu
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NaBH4/CuSO4 also reduces azides to amines:
Attachment: Syn_Comm_24_Reduction_of_azides_with_NaBH4_and_CuSO4.pdf (401kB)
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leu
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Ammonium formate reduces azides to amines via CTH:
Attachment: Tet_Lett_24_Azides_to_amines_via_ammonium_formate_CTH.pdf (148kB)
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leu
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NaBH4/CoCl2•6H2O reduces azides to amines:
Attachment: Synthesis_2000_CoCl2_catalysed_NaBH4_reduction_of_azides.pdf (67kB)
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Nicodem
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Variants of the Ritter reaction exist where urea, thiocyanate or cyanate salts were used (forming HSCN or HOCN in situ). These only work with
tertiary alcohols (which is fine here). With urea one gets the corresponding N-alkylureas. The cyanate one only work with substrates that can
form a very stabilized tertiary benzylic carbocation (a benzhydryl carbocation should do). The thiocyanate one gives a mixture of R-NCS and R-SCN with
the ratio highly dependant on the intermediate carbocation stability (benzylic carbocations give practically exclusively R-NCS). The hydrolysis of all
these species yields the corresponding amines. Let me know if references to any of these methods are wanted.
| Quote: | Originally posted by DJF90
ok so you want to replace the hydroxy group with an amine group? Heres how I would tackle it:
React the alcohol with conc. Hydrochoric acid to produce chloride
Reflux with alcoholic ammonia to yield amine
Is it really that simple? Surely I'm missing something... |
Yes, you missed that the substrate is is an alpha-alkyl benzhydryl alcohol and therefore the reaction with HCl would give a mixture of the
corresponding elimination and substitution product in a ratio dependent on the HCl concentration.
For similar reasons the reaction of an alpha-alkyl benzhydrylchloride with NH3 does not yield the corresponding amines but the elimination product
instead. NH3 is nucleophilic, but it is also somewhat basic thus promoting the elimination pathway. For similar reasons, the use of phthalimide as
nucleophile, or other variants of the Gabriel amine synthesis, is also not possible since phthalimide is also quite basic (pKa 8.3). That is why the
azide anion is used as the nucleophile – it is a good nucleophile and a very weak base at the same time.
…there is a human touch of the cultist “believer” in every theorist that he must struggle against as being
unworthy of the scientist. Some of the greatest men of science have publicly repudiated a theory which earlier they hotly defended. In this lies their
scientific temper, not in the scientific defense of the theory. - Weston La Barre (Ghost Dance, 1972)
Read the The ScienceMadness Guidelines!
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DJF90
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Damn... thats something they're not teaching us at school...
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Klute
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Really? It a very well know fact that tertiary alcohol undergo easy elimination reactions in such conditions. Never heard of Substitution/Elimination
competition? That would be in the first chapter dealing with nucleophilic substitutions!
Of course if you are not in a chemistry-specific course, such as biology or simialr, they might have overgone this, but it's pretty surprising
anyway.
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