(+)-O-Desmethyltramadol and analogs

[clubcard]

Tramadol was originally patented in the 70s, tapentadol in 2008.

 

[Fruitofknowledge]

I think replacing one of the N-methyl groups at the nitrogen with a phenylethyl moiety would make it a selective mu opioid agonist and increase potency by a factor of 8-14X

 

[Fruitofknowledge]

 

[Fruitofknowledge]

 

[aced126]

This is what what piglet and ronald seid what do you think?


look up the structure of tramadol and look up heroin/morphine to get the general idea. This has been done with tramadol minus the methoxy group. However this particular compound does not appear in the literature anywhere but it should work all right.

the structure of tramadol is well researched. The patents explain that the most active of the series was O-desmethyl tramadol which the body converts tramadol into. It's about 3 times more potent. I suggested replacing the dimethyl moety with I piperidine or morpholine ring since these analogues in other opiates increase lipophilicy & so increase the speed of action (but reduce duration) for example methadone->diconal or dextromoramide.
I suggest that esterifying thst bare hydroxyl wil further increase the lipophilic character. The Quat hydroxyl will need protecting (with a triflate or something) during esterificatioh I guess, but you really SHOULD (knowing SAR of opiates pretty welll) produce something pretty damn nice.






get the structure of tramadol on your screen. Now imagine placing the hydroxyl (-OH) group with -O-C=O.CH3 by reaction with acetic anhydride and a catalytic amount of base (pyridine, triethylamine). In pharmacological terms compounds with hydroxyl groups are too polar to cross the blood brain barrier (BBB) easily. Hence compounds with hydroxyl groups have low potency. However once across the BBB if the molecule undergoes a transformation to make it more polar this will help with its retention leading to enhanced activity. [This is the sole reason why heroin is more potent than morphine] The acetyl tramadol should cross the BBB easily whereby the ester is hydrolyzed back to hydroxyl and thus retained (in the brain). Thus acetyl tramadol is behaving as a 'pro-drug'. Tramadol itself is still the active compound.


^^ Good explanation.
http://opioids.com/tramadol/structure.html
There is the structure. Most every opiate out there follows something called 'the morphine rule' which is:

1-Aromatic system
2-Quaternary Carbon
3-Two carbon chain
4-Tertiary amine

Test this with morphine, methadone, tramadol, demerol or whatever. Fentanyls, Tilidine, thiambutenes & etoniterzine are the only exceptions I know.

P-)

BTW Ronald, you don't need a 'catalyst' if you use anhydride, only if you use the acid chloride. It's not even really a catalyst since it salts to a quat amine to shift the balance to the right...

^^ No, the problem is that there are 2 -OH groups to acetylate. The aromatic one improves performance, the quat one decreases performance. That's why I said that the quat needed protection during acetylation. The patents point out the esterifying the quat reduces performance. If you just O-demethylate then add acetic anhydride you WILL get a less active compound. You need to go

1: Protect
2: Lewis acid (titanium tertachloride) to O-demethylate
3: esterify
4: deprotect.

You can, of course, swap the quat OH for a halide which is still active...

An acetyl group on the aromatic reduces activity; a free 3-phenolic group is an important part in the pharmacophore of traditional opioids. However in this case the tertiary alcohol might actually be playing a similar role to what the 14-hydroxy plays in oxymorphone or oxycodone. So acetylating that oxygen might result in a reduction in activity, through reduction in receptor affinity itself.

The problem with your synthesis is in the last step, by deprotecting with either acid or base catalyst, you inevitably also hydrolyse the ester, so the deprotection should be carried out in very concentrated acid or base so that there is no water present. This will result in a high equilibrium concentration of the ester. Alternatively to skip the protection steps, you can use the principle that the phenolic OH is less nucleophilic when it is not deprotonated (on the flipside, you can make it more nucleophilic by adding base to deprotonate it; phenolic protons come of easier (pKa ~ 10) than normal alcoholic protons (pKa ~ 15) because the phenolate ion is stabilised by the aromatic ring) and add 1 mol equiv of Ac2O so that only the alkyl alcohol is acetylated. This results in a more lipophilic compound which crosses past the BBB easily. However in this case the tertiary alcohol might actually be playing a similar role to what the 14-hydroxy plays in oxymorphone or oxycodone. So acetylating that oxygen might result in a reduction in activity, through reduction in receptor affinity itself. So it is essentially a competition of increased lipophilicity vs potential loss in receptor affinity.

 

[aced126]

some ideas might come from these...

(-)-(1S,2S)-O,N-Di-desmethyl Tramadol HCl

(+)-(1R,2R)-O,N-Di-desmethyl Tramadol HCl

N-Desmethyl Tramadol HCl

O,N-Di-Desmethyl Tramadol HCl

O-Desmethyl Tramadol-d6 HCl

O-Ethyl Tramadol

Tramadol-d6 HCl

sorry for all the posts

As you mentioned by yourself and others, N-demethylated opioids will be largely inactive. It's not such a bad idea, in terms of safety, if it results in a dosage range of 500-1000mg.

 

[aced126]

I think replacing one of the N-methyl groups at the nitrogen with a phenylethyl moiety would make it a selective mu opioid agonist and increase potency by a factor of 8-14X

Not necessarily and probably not in this case. This molecule has more possible conformations and the lowest energy conformation of this molecule is likely to conform to a slightly different shape than morphine (which is why, disregarding pharmacokinetic considerations, tramadol has quite a higher dosage than morphine). Thus the phenethyl moiety is in an unlikely position to fit in the hydrophobic pocket near the tertiary nitrogen of morphine in the mu opioid receptor. N-phenethyl substitutions only really work for opiates, and opioids of similar structure to morphine.

 

[Solipsis]

Does anyone remember what tramadol analogue it was that was tested by some people and found to produce seizures (don't think there were signs that it was from overdosing it)?

Was not the relatively common O-desmethyl...

 

[Fruitofknowledge]

Someone claimed that adding a n-phenylethylamine moiety would decrease potency if the opioid was an open chained opiois, which didn't make any sense to me. I found a research article that mentioned the tramadol analog I mention months ago, N-nor-methyl phenethyltramadol, the article stated it was equal-potent to o-desmethyltramadol, making it 10x more potent than the parent compound. I think another promising structural modification to produce an potent analog by modifying the cyclohexane ring, by adding a beta methyl group to the cyclohexane ring, just like 6,7-dihydromethyldesorphine, which in return increases the potency 50X of morphine of the latter compound mentioned, maybe this would make tramadol 4-5X the potency of morphine by weight and help making it a selective mu opioid agonist eliminating the norepinephrine/serotonin binding affinities. Any thoughts or comments? JUST A DREAM, MANY MORE IDEAS, I COULD WRITE A BOOK WITH MY IDEAS.

 

[clubcard]

2-[(Dimethylamino)methyl]-1-(3-hydroxyphenyl)fluorohexanol is the strongest of the series, just look at the patent.

 

[Brainbooster82]

I may have access to (-)-O-DESMETHYLTRAMADOL, CAS:144830-15-9
Does anyone here know if it's any good recreational wise?

I think when people talk about Desmethyltramadol they normally talk about (+)-O-Desmethyltramadol,right?

Is o-desmethyltramadol legal to purchase in the U.S.A.?