[Gregorian]

My professor is trying to make para-benzenedithiol, HSC6H4SH, starting from the diamine,
H2NC6H4NH2, by doing two diazonium reactions with NaSH.  It never worked,
always gave him the quinone instead.
 
Can you explain why?!

Thanks

[Gregorian]

I think that it is because you replace both -NH2 groups on the benzene ring with two SH- groups which are more nucleophilic than OH-. Then the thiol groups get protonated and they become good leaving groups, so now H2o attacks making a diphenol.
Hs on the phenol groups are acidic so they leave making it a benzene ring with two oxygens with a negative charge on each which is stabilized by reasonance into a quinone.

Does that sound right? what do you think?

[nj_bartel]

Well, you're saying that the thiol's are more nucleophilic than the hydroxyls, which is correct, but then you relate that to being a better Bronsted base.  Second, I'm pretty sure a phenyl LG won't actually leave.  Pretty sure phenyl substituents in general don't undergo that type of nucelophilic substitution.  Third, why is this reaction being run in water?  Do quinones even have resonance?  I'm tired, but I'm not seeing any immediately.

[Gregorian]

Well, you're saying that the thiol's are more nucleophilic than the hydroxyls, which is correct, but then you relate that to being a better Bronsted base.  Second, I'm pretty sure a phenyl LG won't actually leave.  Pretty sure phenyl substituents in general don't undergo that type of nucelophilic substitution.  Third, why is this reaction being run in water?  Do quinones even have resonance?  I'm tired, but I'm not seeing any immediately.

I never said that phenyls will leave, but the hydrogens on the phenyl groups are pretty acidic and when something is acidic it gives away its hydrogens that's why they leave so we end up with a benzene ring that has two oxygen with a negative charge. You use reasonance and this stabilize into a quinone.
Yes reaction is run in water
thank you   

[Gregorian]

my professor is saying the you replace the NH2 groups with N2Cl since it is a diazonium reaction, but I don't understand why this happens and where the Cl comes from and then when he tries to replace N2Cl with SH- groups, it doesn't work. It gives him the quinone

I really need *delete me*

[AWK]

my professor is saying the you replace the NH2 groups with N2Cl since it is a diazonium reaction, but I don't understand why this happens and where the Cl comes from and then when he tries to replace N2Cl with SH- groups, it doesn't work. It gives him the quinone

I really need *delete me*

Diazotization is usually performed in strong acid

http://en.wikipedia.org/wiki/Diazonium_salt

[Gregorian]

my professor is saying the you replace the NH2 groups with N2Cl since it is a diazonium reaction, but I don't understand why this happens and where the Cl comes from and then when he tries to replace N2Cl with SH- groups, it doesn't work. It gives him the quinone

I really need *delete me*

Diazotization is usually performed in strong acid

http://en.wikipedia.org/wiki/Diazonium_salt

I agree with you. Diazonium's conditions are NaNO2, HCl, 0 degree to generate N2Cl, but I don't know why he said he wanted to use the diazonium and then said NaSH

But based on the problem what do you guys think is going on?

[Gregorian]

Come on you guys, I am sure a lot of you have ideas.
let me correct my explanation on the process this reaction take because I made a mistake!

conditions of diazonium conditions are (NaNO2 HCl, 0 degrees)


it is a diazonium reaction so the NH2 is replaced with N2Cl which is a VERY good leaving group. My professor said than SH- attacks (because it is more nucleophilic than OH from water) and that's why we should end up with para-benzenedithiol. But in lab he got the quinone instead!

I think it is because the SH groups on the benzene rings got protonated so they became SH2+ and became leaving groups and then the H2O attacks forming a benzene ring with two phenyl groups which give up their acidic Hs and reasonate to form the quinone..

WHAT DO YOU THINK?

[shelanachium]

Quinones don't have benzene-type resonance, they behave like the reactive dienones they are. But they form easily by removal of two electrons from the conjugate base of hydroquinone (or by removing two H's from neutral hydroquinone):

-O-C6H4-O- -2e- ---> O=C6H4=O
also HOC6H4OH -2[H] ---> O=C6H4=O

That is, by oxidation. Nitrous acid, present in a diazotization mixture, and diazonium ions themselves have oxidising power, and if any hydroquinone is formed in the reaction it will be rapidly oxidised to quinone.

Because diazotisation mixtures are acidic, any HS- present will be lost as H2S gas, leaving H20 as the predominent nucleophile. This will indeed yield hydroquinone, which then oxidises to quinone.

Another possibility is that no diazonium salt was formed at all, and that the p-phenylenediamine was directly oxidised to the quinonimine HN=C6H4=NH which then hydrolysed to quinone and ammonia (at once converted to ammonium ion in the acid medium, rendering the reaction irreversible).

[Gregorian]

thank you, I think your first explanation is better.
Actually my professor just told me few days ago that he was able to get the ortho product... but when it came to the para product, it gave him the quinone instead of the diamine!!!! he can't figure out why is that? 

[ARGOS++]

Dear Farah Jarous;

Would it be allowed to have a “very” different Idea?

First of all for me it’s no surprise, that starting from p-Di-amine ends with the Quinone as the major product, because the Di-radical as the intermediate, what “must” end in the most stable product.
For that you may read on:

http://www.organic-chemistry.org/namedreactions/sandmeyer-reaction.shtm

I for myself would start from p-NitroAniline and do only the first “Sandmeyer”, but not with NaSH !!!
I would use Na-S-CS-O-Et instead, it’s the Sodium Ethylxanthate. This would end in a protected Thiol!
Now a mild reduction of the nitro and the whole “Sandmeyer” again.
Finally to release the free Di-thiol you use NaOH and hope that the end product is enough stable.

But I don’t know if such would be allowed.
Otherwise I would try to protect only one amine and continue as above.

Maybe it’s at least a base for future ideas.
Good Luck!
                    ARGOS⁺⁺

[ARGOS++]

Dear Farah Jarjous;

Sorry for my typing mistake in your name!

Good Luck!
                    ARGOS⁺⁺