I Like to Draw Pictures of Random Molecules

[Soulfake]

Some Taurine analogues of Gabapentinoids. (R= Methyl, Fluoro, Methoxy etc.). Replace the NH2 units with HO and you get the corresponding GHB analogues.

The ones at the bottom are just-for-fun "Nitrophenibut" and "Ethoxy-Nicotinoyl-Nitropheni-GHB" analogues. But who knows (well, probably those who understand chemistry better than me), maybe they would be the ultimate gaba drugs ^^

 

[Hodor]

These are the ones I had in mind. The double carbon bond should be at the sulfur, makes a more stable compound.

No. No, it wouldn't.
Putting three of its electrons into covalent bonds (and 2 in a lone pair) leaves the last electron on the sulfur unpaired, making it a radical, i.e. pretty much the exact opposite of "stable".

Where are you getting the idea that this would be more stable?

 

[S.J.B.]

No. No, it wouldn't.
Putting three of its electrons into covalent bonds (and 2 in a lone pair) leaves the last electron on the sulfur unpaired, making it a radical, i.e. pretty much the exact opposite of "stable".

Where are you getting the idea that this would be more stable?

The sulfur should at least be given a positive charge, but even so, it would immediately lose a proton at the 3 position of the bicycle to quench the positive charge and bring the five-membered ring into aromaticity.

 

[Pomzazed]

Shift the double bond to a 2,3-position then it will be a stable benzothiophene!
Gaffy’s 1,2- double bond version cannot be made (sulfur with 3 bonds huh?)

 

[Gaffy]

Sorry about that. I only have a very basic (if not plain bad) chemistry level.

 

[Hodor]

Sorry about that. I only have a very basic (if not plain bad) chemistry level.

No reason to apologize. I was genuinely curious as to why you would assume that three bonds on the sulfur atom would make it more stable, so I could better explain it to you.

Maybe this will make it easier to understand:
As I've said, sulfur has 6 electrons in its outer ("valence") shell. Usually, atoms in compounds will enter a certain number of bonds with other atoms in order to achieve a shell with 8 electrons in it.

Take, for example, hydrogen sulfide (H2S): The sulfur enters two of its electrons into single bonds with hydrogens, so it now has 2 x 2 electrons in the S-H bonds, in addition to 2 pairs of electrons not involved in bonds (so-called "lone pairs").
Since (2x2) + (2x2) = 8, it has now achieved the desired 8-electron ("octet") configuration.

Compare this to, for example, carbon or nitrogen:
Carbon has 4 electrons, so it going to try and enter bonds to gain 4 more electrons; nitrogen has 5, so nitrogen needs to form 3 bonds to complete its octet:
Thus the formula for methane is CH4, and that of ammonia is NH3.
In benzene, each carbon is connected to one carbon atom via a single bond, and to another via a double bond. Its fourth bond is to hydrogen, but carbon-hydrogen-bonds are typically not drawn in a skeletal formula so as not to make it overly complicated.

By replacing one of the carbon atoms in benzene with nitrogen, we get pyridine. Note that here, you'd have to draw a hydrogen atom if there was one connected to the nitrogen - but there isn't, because the nitrogen is quite happy with the 3 bonds (a single bond and a double bond) it has formed with the adjacent carbons, giving it the desired 8-electron-configuration.

Now, it is possible to have a compound where the sulfur is connected to the adjacent carbon atoms via 3 bonds, namely by removing one of its electrons. With 5 electrons in its valence shell, it is now "isoelectronic" with nitrogen, and will thus bond to the adjacent carbons just like the nitrogen in pyridine. Removing the electron leaves it with a positive charge; however, it can be stabilized by forming a salt with a negatively charged counterion (e.g. chloride). The image below shows a pyrylium cation; replace the oxygen with sulfur, and you'd have the thiopyrylium cation.

 

[Gaffy]

I got it . Now;
Up and rolling; Dark and horsing, I present to you:
The world's strongest DRIs (money back if you find better elsewhere)​

The world's most cyclised R-30490s, and one of cebranopadol's relatives, un invite de marque:

​

And to top it off:

BTW Please kill me; CHARMM costs 600? (if anyone has a torrent, or an invite for uni codes granting access, I'd be honored)

 

[S.J.B.]

Analogously to the situation with the sulfur-containing molecules you had posted previously, the two structures on the right will quickly rearrange to give enamines, thereby forming indoles.

 

[Gaffy]

That's alright

 

[Gaffy]

Gaffy's MORs

 

[Hodor]

That's alright

This is basically your introduction to a concept called "aromaticity". The term "aromatic" describes ring systems in which electrons can be freely shared between atoms across conjugated double bonds. However, for this to happen the molecule needs to be flat ("planar"), and be able to contribute a specific number of electrons into a pi-system, i.e. a halo of electrons distributed above and below the ring.

Most of these electrons come from double bonds, with each double bond contributing 2 electrons. Benzene and pyridine both have three double bonds, meaning there are 3x2=6 pi electrons. 6 happens to be the exact number of electrons needed for a single ring to achieve aromaticity.

In naphthalene and quinoline, there are 5 such double bonds. 5 times 2 equals 10, and ten happens to be the exact number of electrons needed to achieve aromaticity in a system with two rings.

But what about pyrrole? It only has two double bonds, yet it is still aromatic. How so?
Well, look at the way the molecule is drawn on the left.

As we've established earlier, nitrogen has 5 electrons in its outer shell, 3 of which it uses to form bonds. Notice those two dots above the "N", though? That's the remaining "lone pair" (of electrons), which can be contributed to the pi system, just like the electrons from a double bond. Adding up the 2x2 pi electrons from the double bonds and the two pi-electrons from the nitrogen's lone pair we get 2x2+2=6 electrons, just the right amount to make the whole thing aromatic.

If we fuse such a pyrrole ring with a benzene ring, we get a system with 4 double bonds and, again, a lone pair on the nitrogen, giving it 10 pi electrons. The compound (called an "indole") is thus aromatic, just like the aforementioned quinoline, despite not involving the nitrogen in a double bond:

And as I mentioned in the beginning, an aromatic molecule needs to be planar. A carbon with a double bond forms a flat triangle with the atoms it is bonded to, which is what we want. Without the double bond, the 4 bonds would form a 3-dimensional tetrahedron, and it would no longer be planar, and thus not aromatic. This is another reason we need the double bond to be on the carbon instead of the nitrogen.

 

[Gaffy]

Happy New Year Everybody, may Chems be with you.

 

[Gaffy]

Disso cannabinoid? Some new CBs:

A little recap+ some new ones

Novel opioids

 

[polymath]

Is there any information about the cyclic product that forms in the reaction of phenylalanine with 2-chloroethanol? The chlorine end would alkylate the amino group and the hydroxyl end would form an ester with the -COOH.

 

[Gaffy]

2-Br-EtoH would be more suited I guess. I think you hydrolysis between COH-HOC=COC+H2O would take waaaayy longer than the actual reaction between halogen/amine; but anyway here's your product;​

 

[S.J.B.]

This compound is a hemiaminal, which would be very prone to hydrolysis, in which case an equivalent of formaldehyde would be released.

NOTE: this is not in fact the compound that Polymath was talking about; in that case, a hemiaminal would not be formed.

 

[Gaffy]

Could you explain how and why?

 

[S.J.B.]

Could you explain how and why?

Basic nitrogens have a propensity to donate a lone pair of electrons. If there is an oxygen connected to a basic nitrogen through a one carbon bridge, the nitrogen can then donate its lone pair to form a nitrogen carbon-double bond and kick out the oxygen. If there is no water around, the oxygen will just add back and the compound will likely exist mostly as the hemiaminal (especially since it is a five-membered ring, which are particularly stable and are the fastest to form of any ring size). In an aqueous medium such as the body, however, a molecule of water could add instead, taking the place of the oxygen. This would lead to a sequence of steps that eventually liberates phenylalanine and formaldehyde. The entire sequence (what we would call the "mechanism" of the reaction) is shown below:

 

[Pomzazed]

polymath said 2-chloroethanol!
so the product would be a 6-membered ring and that hydrolysis of hemiaminal will not occur.

 

[S.J.B.]

polymath said 2-chloroethanol!
so the product would be a 6-membered ring and that hydrolysis of hemiaminal will not occur.

Ah sorry, I was looking at Gaffy's drawing! That is correct - I will edit my post. Thanks!