Sciencemadness Discussion Board

Amines from alcohols??

dan_teod - 17-2-2008 at 13:44

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.

solo - 17-2-2008 at 13:50

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]

Klute - 17-2-2008 at 14:16

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]

The_Davster - 17-2-2008 at 14:46

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?

dan_teod - 17-2-2008 at 15:20

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

Klute - 17-2-2008 at 16:11

So is the Zn reduction higher-yielding than the LiAlH4?

leu - 17-2-2008 at 16:59

It seems that the wrong question is being asked, but this paper describes the use of stannous chloride in methanol and yields 98% :D If one is paying that much for the precursor one should be looking for ways to synthesize it from less expensive materials :P

Attachment: Tet_Lett_27_Azides_to_amines_with_SnCl2.pdf (128kB)
This file has been downloaded 1169 times

leu - 17-2-2008 at 17:36

This paper describes methods for obtaining azides in a lot less than 72 hours :cool:

Attachment: Synthesis_1976_PTC_preparation_of_alkyl_azides.pdf (81kB)
This file has been downloaded 1317 times

chemrox - 17-2-2008 at 18:08

two nice refs-thanks!

Nicodem - 18-2-2008 at 01:43

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!).

leu - 18-2-2008 at 10:46

Rearrangements are common during the Ritter reaction, the attached diagram illustrates the phenomenon well :cool: 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 :P

Attachment: ritterrearrngements.djvu (7kB)
This file has been downloaded 796 times

dan_teod - 18-2-2008 at 12:31

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).

shemati - 18-2-2008 at 13:34

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.

Nicodem - 19-2-2008 at 01:01

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).

Methyl.Magic - 19-2-2008 at 09:35

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.


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-

DJF90 - 19-2-2008 at 15:00

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...

leu - 19-2-2008 at 18:37

HI has been used to reduce azides to amines:

Attachment: Tet_Lett_43_Azides_to_amines_with_HI.pdf (58kB)
This file has been downloaded 1395 times

leu - 19-2-2008 at 18:40

Zn/NiCl2 is used to reduce azides to amines:

Attachment: Synlett_1997_Azides_to_amines_with_zinc_and_NiCl2_6H20.pdf (134kB)
This file has been downloaded 1546 times

leu - 19-2-2008 at 18:45

Fe/NH4Cl can reduce azides to amines:

Attachment: Tet_Lett_37_Azides_to_amines_with_Fe_and_ammonium_chloride.pdf (136kB)
This file has been downloaded 885 times

leu - 19-2-2008 at 18:49

NaBH4/CuSO4 also reduces azides to amines:

Attachment: Syn_Comm_24_Reduction_of_azides_with_NaBH4_and_CuSO4.pdf (401kB)
This file has been downloaded 704 times

leu - 19-2-2008 at 18:52

Ammonium formate reduces azides to amines via CTH:

Attachment: Tet_Lett_24_Azides_to_amines_via_ammonium_formate_CTH.pdf (148kB)
This file has been downloaded 747 times

leu - 19-2-2008 at 18:54

NaBH4/CoCl2•6H2O reduces azides to amines:

Attachment: Synthesis_2000_CoCl2_catalysed_NaBH4_reduction_of_azides.pdf (67kB)
This file has been downloaded 2269 times

Nicodem - 19-2-2008 at 22:53

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.

DJF90 - 20-2-2008 at 06:10

Damn... thats something they're not teaching us at school...

Klute - 20-2-2008 at 06:17

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.

DJF90 - 22-2-2008 at 14:45

We did nucleophillic reactions last year but we didn't really cover why halogenoalkanes react with alcoholic alkali to give an alkene whereas a reaction with aqueous alkali gives an alcohol. And I'm doing a proper chemistry course, wouldnt touch biology with a 10 ft pole lol. But as I remember a tertiary alcohol would react with conc. HCl to give the respective chloroalkane? I'm pretty sure we did the reaction in class, using 2-methylpropan-2-ol

PHILOU Zrealone - 25-2-2008 at 04:36

Quote:
Originally posted by DJF90
We did nucleophillic reactions last year but we didn't really cover why halogenoalkanes react with alcoholic alkali to give an alkene whereas a reaction with aqueous alkali gives an alcohol. And I'm doing a proper chemistry course, wouldnt touch biology with a 10 ft pole lol. But as I remember a tertiary alcohol would react with conc. HCl to give the respective chloroalkane? I'm pretty sure we did the reaction in class, using 2-methylpropan-2-ol

True with tri-aliphatic alcohol, but much less evident once an unsaturated C=C link is in direct viccinity of the C holding the OH group...benzylic/allylic position will favourise elimination
(C6H5)2C(OH)-CH3 --> (C6H5)2C=CH2 + H2O

DJF90 - 25-2-2008 at 11:19

Ah I thought the phenyl rings would mess things up a bit. Is there any way to make the reaction favour the substitution route? Possibly by controlling the equilibrium of the reaction (temp/conc/pressure?)? I'm not really sure about how substitution/elimination is favoured, guess its something I should read up on :P

[Edited on 25-2-2008 by DJF90]