Organic Chem From Hell

[Teleologist]

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Find the strongest Bronsted-Lowry acid which can exist at high concentration in HOCH2NHCHNCN(l).

Okay, I know that the strongest acid that can exist in a solvent at high concentration is the conjugate acid of the solvent. The CA has gained one hydrogen proton. So I just gotta figure out where I can stick an extra hydrogen to this badass solvent. I first gotta draw it out ... and I need practice on drawing out condensed formulas. This is what I get - does this look right?

I tried following these rules:

1) Hydrogen only forms one bond. So it's not going to be connecting any other atoms. It's just going to be hanging off something else
2) Formal charge for each atom should come out to 0 since that's Columbically ideal, and since the question didn't indicate that there was any pre-existing charge present on any part of the molecule. In following this rule I even made some nitrogens radicals by giving them lone electrons.
3) Carbon forms four bonds (duh).

The structure I drew seems to make sense since there's exactly one lone pair on a nitrogen and that's where the hydrogen would attach to form the conjugate acid.

So, what do you think? Did I draw it right? Any other rules to keep in mind while drawing out condensed formulas?

 

[Czarcasm]

Yeah, my drawing is wrong as NextStep pointed out. I assumed you can rearrange the atoms in any order to obtain the most acidic structure.

 

[Blueprint MCAT Tutor]

Czarcasm, you're missing the CH2

I think the OP is basically right.

What Czarcasm drew was HOC(CN)(NH2)CHNH

 

[Czarcasm]

Czarcasm, you're missing the CH2

I think the OP is basically right.

What Czarcasm drew was HOC(CN)(NH2)CHNH

Yeah, I see what I did wrong now. I'd say OP is correct as well. It satisfies degrees of unsaturation, has formal charges of zero, and the O-H is reasonably acidic.

 

[Schenker]

I would prefer the nitrile resonance form of your diradical compound. It makes it easier to see that protonation of the nitrogen on the end is resonance stabilized.

 

[Teleologist]

I would prefer the nitrile resonance form of your diradical compound. It makes it easier to see that protonation of the nitrogen on the end is resonance stabilized.
View attachment 178853

Nah, that's not right. The H would attach to the lone pair on the sp3 hybridized nitrogen. This mirrors that of ammonia becoming ammonium ion. We want the site with the strongest nucleophilic tendencies. That would be the sp3 nitrogen because in the sp3 nitrogen the electrons spend their time furthest from the nucleus (as opposed to sp2 or sp nitrogen - both of which are present on this molecule).

Here's a correct picture of the molecule. I accidentally made nitrogen bond five times (!!) in the original diagram.

 

[syoung]

I would hazard that Schenker's is better (for one, he has no radicals, while OP does), and two, resonance stabilization plays a stronger role at stabilizing the charge!
The reason I don't like OP's drawing is that the Nitrogens' have 9 electrons... breaking the octet rule.

 

[Chrisz]

I believe the OH is the strongest acid in this case. Once the lone pair on nitrogen is moved to the other side through resonance, there will be a positive charge on the nitrogen, through inductive effect this would definitely make more acidic than regular alcohol. Gonna upload a pic later

 

[Chrisz]

The alcohol group is more acidic after the inductive effect. Since the alcohol group is part of the solvent molecule, it also fits the definition of high concentration. Nitrogen on the right hand side now has a pair of extra lone pair which can be used to pick up the H+. This could be a typical exam question in orgo class. Do not misunderstand, OH is still the strongest acid