June 05, 2020, 02:39:54 AM
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Organic Chemistry Forum / Re: Transformations
« Last post by Sarin chamod on Today at 01:37:18 AM »
@Rolnor,thank you sir..
Generic Discussion / Bellsouth email temporarily locked. What to do?
« Last post by annewilliam on Today at 01:35:26 AM »
If your Bellsouth email is locked or suspended for the temporary period, then it can because of the following reasons:

•   If you have entered the incorrect username and password multiple times.
•   When you are sending the email, which is beyond the SBCGlobal email size limit.
•   If a hacker has tried to login to your account.
•   Even when suspicious activities are reported to occur in your account.

Once you know the reason for the occurrence of the error the only, you can begin solving it. You can even contact Bellsouth customer service phone number to take the assistance of the experts in fixing the issue at hand.
Well that makes more sense. Thank you for your time  :)
Honestly it can be that the Cl- binds, but the major issue may come from the en ligand being protinated by the acid. If it makes NH3+, that would make it a much worse ligand so it would not be able to rebind to the Co3+
Organic Chemistry Forum / Re: Schotten-Baumann Benzoylation of Diamines
« Last post by jr620 on Yesterday at 09:57:53 PM »
So in this case is it just neutralizing the HCl, or is it soaking up the acidic hydrogen of the amine?
Organic Chemistry Forum / Re: Schotten-Baumann Benzoylation of Diamines
« Last post by wildfyr on Yesterday at 09:46:41 PM »
It's an acid scavenger. It's there so suck up any protons that would prevent reaction or cause something to be reversible.

So anything that can't react in protonated form, or if protonated could reverse react.

Thanks for the reply, makes a lot of sense. So when the en is exchanged for the solvent, I’m guessing Cl-, would that be enough to stop the oxidation of Co2+?

We also conducted Cyclic voltamettry with Coen but this time in pH6-7 and this gave a quasi reversible reaction. So at pH1 (0.1M HCl) it was irreversible and pH6-7 (KCl/H2O) it was reversible. does that have anything to with H+ concentration effect on the ethylenediamine ligand?

Again thank you in advance :)
Organic Chemistry Forum / Schotten-Baumann Benzoylation of Diamines
« Last post by jr620 on Yesterday at 08:57:48 PM »
I am an undergraduate student currently studying organic chemistry and i'm trying to come up with a full reaction mechanism (including electron movement) for benzoyl chloride + ethylene diamine. I have chosen Schotten-Baumann benzoylation as the method but i am unsure how NaOH works in this reaction, in some examples OH- reacts directly with the acidic hydrogen of the tetrahedral intermediate to form water and in other examples proton transfer occurs between the nitrogen and the carbonyl group and chloride ions remove the H from the resulting unstable C=O-H group. I have attached my progress so far, any advice or input would be appreciated.
Cool question Scissors99, the short answer is the chelate effect is more pronounced in the diNOsar ligand. I break it down a bit more in the post below, and have a video response at around 18 minutes in for my stream (midnightduck on twitch). It can really help to draw this out.

first lets go over why Co3+ does not lose its ligands before the reduction. Co3+ is "low spin", which means that all of the electrons are in the dxy, dxz, and dyz orbitals are much lower in energy than the dx2-y2 and dz2 Orbitals. This is what causes the Co3+ center to never give up its ligands. However when it gains just 1 electron the ligands become labile again and it can give and take ligands. When those ligands are labile, the en ligand is less likely to stay on (exchanging for solvent), because it has less bonds to the metal. The diNosar ligand has 6 bonds to the metal center whereas the en ligands have only 2.

Hope this helps!
When HCl adds to an alkene, it does not do so in a single step but instead the proton adds first and then chloride adds in a subsequent step.  In other words the reaction proceeds through a discrete intermediate.  Your class notes or textbook will help (this is common first-semester organic chemistry).  You have to be able to draw this intermediate to understand how C is formed in preference to B.  It is not closely related to the stability of the ring.

With respect to resonance, please show your drawing.  You may be suggesting a structure that violates the octet rule; if so, then you can be sure that it is not a correct answer.  What can we say is generally true about atoms that participate in resonance?

With respect to centers of chirality, let me make a general example.  Suppose there is a carbon that is bonded to H, OH, and to two other carbons.  Is this carbon a center of chirality?  It depends on the groups bonded to the two carbons. If both carbons are methyl groups then carbon-2 is not a center of chirality.  If one carbon is a methyl group and one carbon is an ethyl group, then carbon-2 is a center of chirality.
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