Sciencemadness Discussion Board

Indoles forming salts?

madmaxman - 12-4-2018 at 16:28

Can anyone help me understand indole salts?
If you look at tryptophan.hcl it is a salt but due to the primary amine.
If you look at tryptophol, there is no mention of the salt forming.

Is it possible to form a 'double salt' by adding HCL to the indole and amine group?
Can any indole group with a minimum pka form HCL salts? Would that require anhydrous hydrogen chloride or just regular HCL?

I just can't find much information about indole salts, if there was a good textbook or reference on the subject that would be great.

AvBaeyer - 12-4-2018 at 18:39

The indole ring does not form any useful salts. The nitrogen is simply not basic. You should consult a heterocyclic chemistry text to expand your knowledge.

AvB

PhenethylamineMachine - 13-4-2018 at 09:21

Hmmm...

I think I understand what you are saying.

You are asking why the NH nitrogen of the indole ring moiety of tryptamine compounds can't also be used to accept a hydrogen ion (H+), as the amine nitrogen of the compounds does, right?

There is good reason for this, indole is not basic. Because of the aromatic ring the electron pair on that nitrogen are not available for protonation. I mean, you can protonate indole with certain strong acids, but you get what I am saying.



LearnedAmateur - 13-4-2018 at 09:30

Of course there are strong enough acids, I mean fluoroantimonic acid can create salts from hydrocarbons but common acids just aren’t enough due to resonance stability.

PhenethylamineMachine - 13-4-2018 at 09:40

Quote: Originally posted by LearnedAmateur  
Of course there are strong enough acids, I mean fluoroantimonic acid can create salts from hydrocarbons but common acids just aren’t enough due to resonance stability.


Yup, that's what I told the poster, that you can protonate indole with certain strong acids.

However, by context it seemed to me that this was not what the initial poster was asking about.

LearnedAmateur - 13-4-2018 at 10:29

Specifically about HCl, but that’s pretty much the strongest acid one could find in a domestic setting, although I wonder if hydrobromic or hydroiodic acids could be strong enough as laboratory reagents; doubtful, but there has to be some pKa value which enables acidic protonation at the nitrogen atom.

madmaxman - 14-4-2018 at 04:07

Thank you very much, this clears up a lot of confusion for me.

Other compounds similar to this, such as Lysergic Acid Diethylamide, form salts which I thought was at the indole position.

In the case of LSD, is the salt formed at the methylated N position instead?

PhenethylamineMachine - 14-4-2018 at 06:51

Quote: Originally posted by madmaxman  
Thank you very much, this clears up a lot of confusion for me.

Other compounds similar to this, such as Lysergic Acid Diethylamide, form salts which I thought was at the indole position.

In the case of LSD, is the salt formed at the methylated N position instead?


Yes, the methylated nitrogen 6 position is the protonation site...primarily. ...it's a actually pretty complicated, but basically N6 is the protonation site, most of the time.
Quote:

From the pKb , it is certain that about 75% of the LSD is protonated at the N6 site. Whether this proton is abstrated by the asparate residue is not known.
http://www.pitt.edu/~soriano/models/lsd/lsd.htm


Below is some review, so you can skip past if needed.

Basics:

·An acid readily gives up a proton (H+)
·A base readily accepts a proton

The conjugate acid is the protonated base, and the conjugate base is what's left of the acid after its donated it's hydrogen ion (proton)

Its pretty cool, if you have the non-protonated from of an amine, the free-base, (called as such because it's "free" from protonation) you know that an acid/base neutralization will produce a salt, also that the acid used determines the salt produced, so hydrochloric acid gives the hydrochloride salt, fumeric acid gives the fumarate, acetic acid gives the acetate salt and so on. Each salt form causes the molecule to differ in size, weight, solubility, and so on. I'm sure this is all review, actually I know it's review based in the conversation so far, it's cool stuff though.

PhenethylamineMachine - 16-4-2018 at 09:19

Wow, I remember there being a related question on a Dartmouth chemistry exam:
Quote:
1. Lysergic acid N,N-diethylamide ("LSD") :

mt1-a1a.gif - 1kB

(a) Q: The condensed formula of LSD is:
A: C20H25N3O.

Q: (b) The hybridization of the three nitrogen atoms are N1: sp2; N2: sp3; N3: sp2.

Q: (c) The most basic nitrogen atom in the molecule is:
A: N2.

https://www.chemistry.mcmaster.ca/~chem2o6/problems/96ansmt1...

https://www.google.com/url?sa=t&source=web&rct=j&...



In the attached depiction we see that on the test "N2" is representing the methylated nitrogen at position 6 on the LSD molecule.

Ok, here is how this relates to this thread: As the methylated nitrogen at position 6 is the most basic nitrogen atom it would be the most obvious site for protonation, which again, is a little more complicated, but the information and the link from my last post should have that covered.

Really fascinating stuff. I almost forget that I know some of these things until someone asks about them, then the memories all come rushing back, coupled with the enthusiasm that was present while first learning these topics. Its been fun responding to this one.

LearnedAmateur - 16-4-2018 at 10:39

Interesting; slightly off topic but does the same hybridisation of the amide group (and amides in general like APAP and acetamide) mean that it is very difficult to protonate them hence a general inability to form salts? I’m only really familiar with amines, even then just the linear ones as opposed to heterocyclic compounds, plus I’ve never really gone too far into hybridisation back when I did OC in school - really just covered sp2 with alkenes.

aga - 16-4-2018 at 10:43

"The hybridization of the three nitrogen atoms are N1: sp2; N2: sp3; N3: sp2:

It would be fascinating just to know how to work that out.



[Edited on 16-4-2018 by aga]

Sigmatropic - 16-4-2018 at 11:05

N1 is an amide, the other main resonance structure of an amide has a N=C bond. The nitrogen lone pair is thus said to be delocalized and has full sp2 hydrbridization in that resonance structure. Real world consequences are that this nitrogen is non basic, rotation about C-N bond is restricted and the N is trigonal planar.
N2 is a normal amine, it has a sp3 hybridization typical of 2nd row elements. This N is basic and trigonal pyrimidal.
N3 is part of a hetereoaromatic ring (see Huckel's rule for aromaticity). The nitrogen lone pair is needed to fufill Huckel's rule and can be considered to be delocalied into the ring. This nitrogen is non basic and trigonal planar.

Now for a bonus, what are the most acidic protons in LSD and their relative order?

aga - 16-4-2018 at 11:21

Cool ! Thanks for that explanation.
Quote: Originally posted by Sigmatropic  
Now for a bonus, what are the most acidic protons in LSD and their relative order?

Eeek! That's a hard one.

Umm, 'acidic' meaning more H-plussy isn't it ?

If so, electronegativity should come into it, so my guess:

The H on N3
The C next to N1

No idea really.
It'd be great to know how to work that out too.

Sigmatropic - 16-4-2018 at 11:57

It is indeed the H on N3 but this indolic NH can only be deprotonated by strong bases. Although I did indeed mean this proton when I said protonS, I doubt it can actually be deprotonated in ergolines (which is, I think the result of the acidic protons elsewhere on the molecule. What I mean is I don't think there will be a double anion).

It is not the C next to the N1 but very close by alpha protons of the amide. Deprotonation at that position gives quite a large conjugated system and is actually the basis of the isomerization between iso-LSD and LSD.

aga - 16-4-2018 at 12:06

Holy crap !

I guessed right !

So electronegativity is part of how you work that kind of stuff out ?

OC is still Dark Magic to me.

Sigmatropic - 16-4-2018 at 12:46

It is all guided by pKa values which are inherently connected with the stability of the conjugate bases. This stability is indeed a function of electronegativity but also other factors such as size (1), inductance (2) and resonance (3).

(1) Take for example the difference between the pKa of an alcohol and a thiol. On first inspection Oxygen is more electronegative and hence should be more 'willing' to accept electrons. This would lead to the wrong conclusion that alcohols are more acidic than thiols. Sulfur is a third row element and it's valence electrons are in a larger orbital, this results in a more spread out charge and is thus more stable.
(2) effect created by the presence of nearby, but not connected electronegative groups. For example compare trifluoroacetic acid, trichloroacetic acid and acetic acid.
(3) caused by the presence of connected electron withdrawing groups (note the difference between electronegative and electron withdrawing). For example compare the acidities of p-nitrophenol and m-nitrophenol.

This get's complicated in organic molecules, although it is really more of the same, as one can have multiple acidic protons. Deprotonation at one site affects the other site (as can clearly be seen the simple case of oxalic acid). A general knowledge of pKa values for various functional groups is very helpfull at this point. I would say the way to figure this out is by taking into account all these factors and a bit of knowledge. That is noting that 'This proton is next to an electron withdrawing carbonyl' or 'this proton in on a hetereo atom of a certain type'. Then knowing that amide bases (LDA, KHMDS, LiHMDS, NaNH2, ect) can be used to deprotonate carbonyl compounds (aldehydes, ketones, esters, amides, nitriles, ect) and the relative order follows: amide bases are stronger bases than enolates. At best it is performed on intuition and will not necessarily be correct.

aga - 16-4-2018 at 12:55

blogfast25 did teach something of what pKa is, yet my addled brain does not recall right now.

pKa is a measure of how acidic a certain thing is, i think.

Great teaching so far, keep it up !

Edit :

PLEASE

[Edited on 16-4-2018 by aga]

LearnedAmateur - 16-4-2018 at 13:55

It’s basically the measure of how an acid dissolves in water - the lower the pKa (it’s a negative logarithm) the more protons are released into solution. Hence, sulphuric acid releases more protons than acetic acid, and hydrochloric acid releases more than both, as examples. Each decrease by -1 means that 10 times as many H+ ions are released.

Edit: I may be a bit inebriated.

[Edited on 16-4-2018 by LearnedAmateur]

zed - 16-4-2018 at 16:26

Form a salt at position 1? Not impossible, but in doing so, Indole would cease to be Indole.

Indoles can fussily self-destruct under some conditions.

It is possible by various means, to hydrolyse the enamine area of the ring.... opening it. But, then of course, it isn't an Indole anymore.

Kyanuric acid comes to mind. I'll check.

Well, I checked. Might have the wrong spelling. I searched, but in vain.

[Edited on 17-4-2018 by zed]

Sigmatropic - 17-4-2018 at 00:10

Salts the one position are intermediates in some reactions (methylation, tosylation) and the products are still very much indoles. I don't understand what you mean with that it would cease to be indole.

LearnedAmateur - 17-4-2018 at 01:24

Wouldn’t it remove the electron delocalisation if you withdrew electrons from the ring during protonation, hence why it would destroy the indole structure?

PhenethylamineMachine - 17-4-2018 at 08:26

Quote: Originally posted by LearnedAmateur  
Wouldn’t it remove the electron delocalisation if you withdrew electrons from the ring during protonation, hence why it would destroy the indole structure?


Electrons are not withdrawn during protonation.



Indole_anion_reactions (1)_kindlephoto-23188205.png - 11kB

LearnedAmateur - 17-4-2018 at 08:39

Alright, thanks for that, wasn’t sure if the lone pair was donated to the ring or not. Quite a helpful diagram.

byko3y - 17-4-2018 at 14:38

https://books.google.com/books?id=srxzzUskq4wC&pg=PA51
pKb of unsubstituted protonated indole is -4 (pKb = 18), thus this acid is stronger than sulfuric acid. It will be formed under special conditions, but that stuff is very unstable.
2-Methylindole hydrochloride is sold commercially though, just like dimethylformamide hydrochloride and other stuff with pKb above 14.

PhenethylamineMachine - 23-4-2018 at 06:20

Quote: Originally posted by LearnedAmateur  
Alright, thanks for that, wasn’t sure if the lone pair was donated to the ring or not. Quite a helpful diagram.


I actually had some long highly detailed post discussing protonation, deprotonation, and discussion revolving around that non-bonding pair, but during editing a glitch in my device caused me to loose the post, so I quickly typed out a single sentence and posted. ...however, the picture I uploaded was still there.

I am glad it helped, but apologize that the actual post was destroyed, but if you got what you needed from it I guess it all worked out.