Sciencemadness Discussion Board

Dimethocaine - using Panthenol?

unome - 31-1-2010 at 20:39

Panthenol has some structural similarity to Dimethocaine, especially since it cleaves, on basic hydrolysis, to the a-hydroxyacid and B-alaninol.

If one formed the ester from the acid, then the amide from the ester, a Hoffman would give an aminoalcohol which VERY closely resembles that of the Dimethocaine side-chain (in fact, apart from that a-hydroxy, it would be identical.

Has anyone got any ideas how (or maybe, when) to remove that secondary alcohol?

PS The drawing doesn't take into account what would happen to the alcohols, primary or secondary, during the Hoffman Degradation. I think I would check to see if that secondary alcohol can be selectively oxidized then run a modified Clemmensen reduction on it (both Zn & HCl are cheap & readily available). Form an ester with the primary alcohol after the Clemmenson and prior to the Hoffman. I'll fix the drawing later, I'm out of time now.

Panthenol.SM.bmp - 457kB

[Edited on 1-2-2010 by unome]

JohnWW - 31-1-2010 at 21:09

Catalytic hydrogenation or reaction with something like LiAlH4 would reduce both -OHs to Hs, offering no differentiation. Dehydration of the -OHs would be possible to form alkene double bonds only if there was also some energetically-favored structural rearrangement, but that looks unlikely, and in any case would not be wanted. But oxidation with (di)chromate or permanganate, under certain conditions, would oxidize the primary alcohol -OH to a carboxylic acid and the secondary alcohol to a ketone which in fact would be part of an amide group, which differentiation should provide some sort of means of subsequently reducing the ketone; although there is possibly some danger that the primary amine group might also be oxidized.

The carboxylic acid group could be esterified with glacial acetic acid plus a small amount of H2SO4, or with acetic anhydride, to reduce its reactivity. The problem would then become one of how to reduce the ketone part of the amide group to Hs while leaving the ester untouched. However, esters also undergo catalytic reductive hydrogenation and with LiAlH4, as do ketones, so some other more selective method would have to be found.

[Edited on 1-2-10 by JohnWW]

Arrhenius - 31-1-2010 at 21:24

Hydrogenation is not a reasonable option. Stoichiometric lithium aluminum hydride, or cocktails of NaBH4 (say with H2SO4 or I2 to give borane) will reduce the hemiaminal (last structure you have drawn. hydrolytically unstable functional group) to the amine, which it seems you're after. It will not, however, reduce the primary alcohol to the alkane. I have no idea if that Hofmann rearrangement will work - my gut feeling is no, because the reaction is run under aqueous conditions wherein the product is not stable. Thus, if you can even obtain the aldehyde product (without oxidizing it during the reaction) you might consider reductive amination.

Dehydration of any of the alcohols drawn will not occur, as none possess alpha-hydrogens to affect the beta-elimination.

Look up the Lindgren oxidation for the hypochlorite oxidation, but this is under acidic conditions.

[Edited on 1-2-2010 by Arrhenius]

Sedit - 31-1-2010 at 22:21

Wouldn't acid hydrolysis of the Panthenol leave you with the amide cutting out the need for the ammonialysis of the ester. Theres also the option of using hydrozoic acid on the carboxylic acid to yeild the amine also IIRC.

I recall a thread here sometime back that discussed the conversion of the amide into a halogen which could be aminated in a number of fashions.

Arrhenius - 1-2-2010 at 00:20

Acid or base hydrolysis of amide or esters will give the carboxylic acid. Base hydrolysis is faster.

Yikes on the hydrazoic acid idea. Azides are (IMHO) not usually worth working with due to their energetic nature. I think aminolysis is reasonable, but unlikely to be high yielding. It is possible to do a coupling reaction using a carbodiimide (EDCI, HBTU, DCC, etc.), but you'd need to use a moderate excess of ammonium acetate (or diethylamine or whatever) to avoid formation of dimeric esters of panthenol.

unome - 1-2-2010 at 01:00

Quote: Originally posted by JohnWW  
Catalytic hydrogenation or reaction with something like LiAlH4 would reduce both -OHs to Hs; while dehydration would in both cases form alkene double bonds, offering no differentiation. But oxidation with (di)chromate or permanganate, under certain conditions, would oxidize the primary alcohol -OH to a carboxylic acid and the secondary alcohol to a ketone, which differentiation should provide some sort of means of subsequently reducing the ketone; although there is possibly some danger that the primary amine group could also be oxidized.

The carboxylic acid group could be esterified with glacial acetic acid plus a small amount of H2SO4, or with acetic anhydride, to reduce its reactivity. The problem would then become one of how to reduce the ketone group to Hs while leaving the ester untouched. However, esters also undergo catalytic reductive hydrogenation and with LiAlH4, as do ketones, so some other more selective method would have to be found.


Hydrogenation "might" remove the secondary hydroxyl, it ain't gonna scratch the primary alcohol. I'm seriously considering looking into ways to selectively oxidize that secondary alcohol to a ketone, then remove that with Zn/HCl.

Sedit - all the references I can find on the subject, and there are lots, all say that the product of hydrolysis is the B-alanol - which says the acid salt would be the other product.

In terms of doing this with the proper equipment, anyone with access to hydrides could do this a whole lot easier

(a) hydrolysis of the amide

(b) form the ester

(c) Form the amide via the Ester

(d) oxidize both OH's to the ketone and the carboxylic acid respectively

(e) modified Clemmensen Reduction of the ketone

(f) Hoffman degradation of the amide to the amine

(g) Benzylation of the amine

(h) Ethylation of the amine to the N,N-diethylamine

(i) Reduce the terminal, primary carboxylic acid to the alcohol + debenzylation

(j) Form the ester with p-aminobenzoic acid.

Whole lot of steps, requiring lots of reagents, but starting from either panthenol or pantothenic acid (aka Vitamin B5).

Some of these steps can be skipped, if and only if, we can work out a way to avoid having to oxidize that primary alcohol in order to remove the secondary one.


[Edited on 1-2-2010 by unome]

Someone will probably notice - I actually left the Hoffman degradation step out - instead of doing it I've changed the amide to the amine prior to the Clemmensen Reduction, which would be daft. I'll change the pic later. Still, the point is, we need a shorter route to this.

Panthenol.SM2.bmp - 1.1MB

[Edited on 1-2-2010 by unome]

Bolt - 1-2-2010 at 15:04

Aren't you missing a carbon after step D?

Klute - 2-2-2010 at 11:50

I see a much quicker way to this: dimethylate diethyl malonate, monosaponification followed by amidation with diethylamine (forming the acyl chloride or using DCC), reduction with LiAlH4, esterification with p-aminobenzoic acid. Pretty staright forward.

unome - 3-2-2010 at 01:31

Yeah, quicker by far - not so easy to source:P

Seriously though, I was looking through some references on hypochlorite oxidation of alcohols, secondary alcohols are oxidized selectively with hypochlorite/GAA in the presence of primary alcohols...* Seems nice, easy, relatively simple - modified Clemmensen (I don't even want to go near ANY FUCKING MORE Hg - I actually used to play a game with it (when I was much younger & even dumber - imagine that:o), I found some old tube-lights with a small ball of it in the bottom (obviously Hg vapor Neon type lights) and was breaking them in the air and playing with the shit:() from that form the amide, hypochlorite again to the amine... Fuck all in terms of chemicals, all OTC, cheap and easy to source.

The next step, forming the ethylation of the amine, that will be trickier...

* Sooner or later someone will try forming the amide from the diolacid and then try hypochlorite on that - it would probably give a flask full of crud, might just give the imine, or even the aminoketone, thus saving some fucking around, how do you think the tertiary amine would handle Zn/HCl?