Potency and a Methyl group

[Ham-milton]

Cross-Physical Dependence of Several Drugs in Methaqualone-Dependent Rats
Tsutomu SUZUKI1), Yoko KOIKE1), Yasuhiro CHIDA2) and Miwa MISAWA1)

1) Department of Applied Pharmacology, School of Pharmacy, Hoshi University
2) Gerontology Research Center, National Institute on Aging, National Institutes of Health
[Released: 2006/08/25]
Abstract:
We investigated the characteristics of physical dependence on methaqualone. Rats were made physically dependent on methaqualone by the use of the drug-admixed food (DAF) method for 33 days. Pentobarbital, barbital, ethanol and diazepam were cross-administered against methaqualone to evaluate the degree of suppression of methaqualone withdrawal signs as an index for the cross-physical dependence liability of these drugs to methaqualone. To evaluate the cross-physical dependence liability, we used AUC of body weight loss and withdrawal scores between the first cross-administration (9 hr after the withdrawal) and 27 hr after the withdrawal. AUC of weight loss was significantly suppressed by the four test drugs as compared to each control. Withdrawal scores were also significantly inhibited by the cross-administration of barbital, ethanol and diazepam. Considering that the rats given barbital or ethanol fell asleep after the cross-administration, diazepam seems to cause the strongest suppression of methaqualone withdrawal signs among the four test drugs. Thus, physical dependence on methaqualone may be similar by nature to that on benzodiazepines rather than barbiturates and alcohol.

Ihave the full paper if you're interested.

 

[immaturepoop]

That is not really correct. Yes, a methyl substituent got a +I-effect on the aromatic ring, but an amino group shows an -I-effect! That's the first difference.
While methyl exhibits no M-effects at all, an aniline does provide a +M-effect. Because that one weights more, the amino group exhibits in summary an electron-donating effect, as does methyl. But they for sure don't go "along", as 1. the similarity is just for pure coincidence (e.g. Ar-OH and Ar-NH2 have much more in common) and 2. the effects are nor comparable (+I by methyl is very weak, +M by NH2 is quite significant).


Methyl always decreases polarity.

Murphy

Hey thanks murph! I actually have learned a lot since this thread. Can you explain the (+/-) I and M effect more? From my understanding at this point, methyl (or ethyl, propyl, etc.) groups are simply weak ring activators, while groups, like N with a lone pair, or OH groups like phenol are strong ring activators because resonance always beats induction (except in the case of tri-halide-methyl groups and the like). We discussed this before I got into the topic of electrophilic aromatic substitution. remember I'm trying to learn all this without being in school at the moment! thanks!

edit: oh! and murphy, I forgot to clarify. when discussing electron donators I was heavily on the topic of carbocations and the stability relative to withdrawing groups (ie, halides, nitriles... etc.) and donating groups (ie alkyl, amino, alcohol... etc.) groups. I jumped ahead of myself, since I did not know how these affected aromatic rings!

 

[MurphyClox]

Ihave the full paper if you're interested.

Yep, I am! Post it @"Blacklite", plz!

I guess, the key to understand how methaqualone works (on a molecular level!) is within such research articles. I didn't find any reliable source of information about exact methaqualone targets or mechanisms of action...

 

[immaturepoop]

^ Oh and thanks for the PM! check back at blacklight when you get a chance!

 

[EN21]

Looking again at the particular structure of the mentioned methylated methaqualon derivative, I am pretty sure that the methyl group makes the phenyl ring stay twisted or orthogonally to the heterocycle, you know what I mean? The rings must stay 90° to each other.
When I remember correctly, this is also the active conformation of benzos.

 

[MurphyClox]

I made an energy minimization calculation, using the semiempirical AM1 algorithm. It yielded an dihedral angle between the heterocycle and the phenyl substituent of 108° resp. 110° (two different calculations). So, not exactly 90°, but almost.

I didn't know that this was actually the active conformation of benzos, too, but it goes hand and hand with other findings that suggest an mode of action for methaqualone at the GABA-receptors, similar to the one of benzodiazepines. Interesting indeed!

Peace! Murphy