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acetylation or diacetylation of certain opioids

butane said:
fastandbulbous: I'd heard the term "keto-enol tautomerization" before but never fully understood it. You're the man!

So, now that I've got that, this raises even more questions. I've done a fair amount of reading on opiate SAR as it interests me quite a bit, but I haven't really run across this one, so if these questions are already answered elsewhere forgive me.

Which does, for instance, the hydromorphone molecule favor, the keto or the enol, and by what percentage?

Actually, I'm not so sure that this molecule exhibits tautomerism, because then wouldn't a percentage of morphine be hydromorphone and vice versa? If they convert back and forth, then no matter which one you started with you would reach the same equilibrium point in the middle and they would be indistingiushable, but this is clearly not the case. In converting morphine to hydromorphone, you must use a palladium black catalyst to move the hydrogen from the hydroxy up to the double bonded carbon and hydrogenate it. I don't think that acetic anhydride would draw it back, and even if it did you'd be left with just diacetylmorphine anyway. Not that that's a bad thing, but it would have been a lot easier to just acetylate morphine in the first place.

I think with acetylating hydromorphone at least, all you're going to get is acetylmorphone (wikipedia link).

Edit: I failed to see that the enol of hydromorphone wouldn't have the double bond in the same position as morphine. I think that renders some of my earlier points moot...
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I'm not the expert on this, but sometimes you're looking at 99+% of the keto form, and miniscule amounts of the enol form (in most cases, the keto form is favored, due to bond energies/stability). But as you said, and as I asked above- the double bond is in a different place. I don't expect there to be much of the enol form present in say, water, but it could still acetylate (I won't say fully but quite a bit) if only 1% is the enol form, because a slight amount would interconvert and quickly be acetylated (trapped in that form, I think someone said that phrase above).
 
MattPsy said:
Codeineone is very easy to make from codeine, and the non-phenolic hydroxyl can easily be acetylated (the -6 position one, the ketone in this case) by acetic acid + catalyst. You only need acetic anhydride if you want to acetylate the -3 position hydroxyl, which is actually largely unnecessary for the potency increase.
However this isn't going to work well with 3-methylated opioids (like codeine, hydrocodone, etc) because the 3-methoxy must first be demethylated, and the acetyl group on the -6 position will get stripped off quite quickly...
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What does the 3-OH being methylated or not have to do with anything? Dihydrocodeinone enol acetate is on wikipedia, it's apparently slightly more potent:

http://en.wikipedia.org/wiki/Dihydrocodeinone_enol_acetate
 
I was under the impression you had to have a free hydroxyl at -3 for binding?
The activity of both codeine and hydrocodone are from their 3-demethylated metabolites aren't they?
Anyway, I thought that if this was the case, then you'd want to start off with -3 as a free hydroxyl, otherwise that acetyl enol at -6 is going to get stripped off during the process of demethylation.
If anything the above link supports this since not _all_ of the the molecules will have their 6-acetyl enols hydrolyzed during the demethylation, meaning you'd get a slightly higher potency than the parent compound, hydrocodone.
 
no?

synthesis discussion is verboten, you should know that by now.

But the answer is "no."

Does anyone know anything about 3-methyl-6-acetyl-morphine?

I assume that's what would result from the above 'synthesis"
 
My experience with opiates has shown overwhelmingly that, at least in my body, 3-methyl compounds suck compared to their plain and 3-acetyl equivalents.

Wikipedia says that Acetylmorphone is made with the reaction of acetyl chloride or acetic anhydride on hydromorphone, suggesting that acetic anhydride leaves the keto group alone, but that's all I've got to go on.
 
It's an equilibrium reaction, so as soon as you remove athe smalll amount of enol by acylating it, some of the ketone becomes the enol to maintain equilibrium (which is then acylated and the process conrtinues).

I don't want to see any details of the practical method of doing this or backsides will be kicked because I'm getting a bit pissed off having to deal with loads of synthesis posts recently
 
hard to say w/ hydrocodone. Jury still seems to be out on whether or not it's a prodrug.
 
rave23 said:
and then we have... something... ?

A post that is breaking the BLUA
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jesus christ 8) I know that, i wasn't actually serious. Did i ask for something? No... i was just trying to be funny :( Oh well... came back at me. That was my one-free-didn't-read-the-BLUA i guess :\ The point i was trying to make is that i got lost in all the weird drawings and diagrams of acetylated mollecules. Oh well, shit happens.
 
Of course, replacing the 6-OH with a halogen and then reducing it to a H seems to increase potency a lot. In fact, so dose making an ether of the 6-OH. Heterocodeine should increase potency.
 
Acetylation of opiates other than morphine

I thought about this some years ago and it came to my mind again when I read the post about acetylating tryptamines.

Now I found that acetylated hydromorphone even has it's own article on wikipedia as acetylmorphone. It says there it'd be slower acting because deacetylation is needed for it to become active.

But if I remember correctly, deacetylation from heroin to the monoacetyl happens very quickly, hence the quick onset of the drug. So it should be the same for deacetylation of hydromorphone.

And I believe it would be more potent not only by needed dose, but by recreational value than plain hydromorphone, because of the quicker onset.

If so, an acetylated oxymorphone would be even more potent.

But the reason for these substances being quite unknown probably is that while they are not hard to make the reagents are too hard to get in sufficient quantity.

Anyway, has anyone ever heard of these substances being tried by humans?
 
With hydromorphone or oxymorphone you'd only be able to acetylate the 3 position, and that won't do anything for you. Bulky substituents on that 3 position usually decrease activity, as 3-methyl morphine (a.k.a. codeine) is much less recreational. I think the only thing you'd do by acetylating either of these two would make the come-up longer as it would have to deacetylate and also make it less potent by weight.

I don't get where people say that heroin is simply a prodrug for morphine, as that's simply not true. The effects are clearly different, and heroin is more potent by weight. If it were a prodrug, you would need an equimolar amount of each to achieve the same effect, if not more, but you actually take less heroin to get the same effect. That 6-acetyl group remains attached to the molecule as it attaches to your receptor and increases its activity in a similar manner to the 6-keto group of hydromorphone and oxymorphone.
 
well, that's not entirely true butane. Even if it were simply a prodrug, it would be more potent. A drug with a faster onset will reach Cmax far more rapidly of course, and so it is more potent. Not based on ki or ic50, but pharmacologically it is more potent.

But yeah, it obviously isn't just a prodrug.
 
You couldn't acetylate the 3 position of codeine as you have a manky great methyl group attatched to the aromatic OH. Anyway if you do acetylate it, it hydrolyses off incredibly quickly to give 6-O-acetylmorphine (which is the compound heroin is technically the primary prodrug for) - this is eventually hydrolysed into morphine. It's the polarity of the phenolic 3-hydroxy group that slows BBB penetration, so reducing polarity (increasing lipid solubility) by esterifying it give a faster BBB penetration

I don't get where people say that heroin is simply a prodrug for morphine, as that's simply not true. The effects are clearly different, and heroin is more potent by weight. If it were a prodrug, you would need an equimolar amount of each to achieve the same effect, if not more, but you actually take less heroin to get the same effect.
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Not strictly true as the increased lipid solubility can result in the prodrug form creating a higher concentration of active metabolite inside the blood-braiin barrier (it's why heroin has a duration of action considerably less than that of morpjine - the increase in potency has to be paid for by not acting for as long. The fact that dipropionylmorphine is more potent than diacetylmorphine due to higher lipid solubility (technically it's called the water/1-octanol partition coefficient) showd that its nothing to do with the overall number of molecules but how fast they get into the brain

Of course you can still acetylate a keto group in position 6 by taking advantage of keto-enol tautomerism and acetylating the enol version.


I've never seen anyting about acetylation of the 14-hydroxy group of oxymorphone/oxycodone
 
fastandbulbous said:
Not strictly true as the increased lipid solubility can result in the prodrug form creating a higher concentration of active metabolite inside the blood-braiin barrier (it's why heroin has a duration of action considerably less than that of morpjine - the increase in potency has to be paid for by not acting for as long. The fact that dipropionylmorphine is more potent than diacetylmorphine due to higher lipid solubility (technically it's called the water/1-octanol partition coefficient) showd that its nothing to do with the overall number of molecules but how fast they get into the brain
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How much of the deacetylation at the third position is due to enzymes within the brain? If most of this happened outside of the brain wouldn't the metabolite have trouble getting back in due to its polarity?
Is the BBB a one-way or two-way filter? If it is a two-way filter then increasing the levels of active metabolites should increase the concentration of more polar metabolites, which would be somewhat trapped in the brain. If the enzymes responsible for this within the brain are saturable this should skew the equilibrium resulting in a longer half-life. This is a question I've been wondering about for years. Hopefully someone with more of a background in biology than I have will be able to answer this.
 
Iirc MPTP is converted in the brain by MAO-B to the polar and toxic NPP+, which can't make it out. That's the reason one dose could permanently give someone alzheimer symptoms; the drug could go on and on etc.

So I guess the BBB is a two way filter.
 
immad said:
Iirc MPTP is converted in the brain by MAO-B to the polar and toxic NPP+, which can't make it out. That's the reason one dose could permanently give someone alzheimer symptoms; the drug could go on and on etc.
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MPTP causes Parkinson's, not Alzheimer's disease.
 
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