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| | From:  al-Astair (Original Message) | Sent: 8/22/2005 10:28 AM |
In a Lab, for my Chemistry paper, we had to prepare a racemic mixture of [Co(en)3]Cl3. We then had to resolve the enantiomers by forming diastereoisomers with tartaric acid. The thing is, for the lab report we have to write the series of equations for the reactions that took place. I have no problem understanding the resolution and the isomerism, it's the actual reactions to form the [Co(en)3]Cl3.
We dissolved CoCl2.6H2O and en.2HCl (en=ethylenediamine) in water. To this we added NaOH and then H2O2, followed by more water. We precipitated out some crystals with the help of an ice bath, then dried them with a Büchner funnel. These crystals are rac-[Co(en)3]Cl3.0.5NaCl.3H2O.
As far as I can tell: the NaOH removes the HCl from the en.2HCl and the H2O2 oxidises the Co(II) to Co(III), thus making the formation of the complex possible. I would really appreciate some help with this.
Cheers, Al.
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Oh and sorry, but I wasn't able to make the subscripts um, subscripted.
So I apologise for it being difficult to read.
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| | | From: ·Steve· | Sent: 8/22/2005 8:33 PM |
Hi Al, yes, I think you identified the reactions correctly. The NaOH converts the en dihydrochloride to the "free base" form of the diamine which has the nitrogen lone pairs now free to donate to the Co3+ ion: [H3NCH2CH2NH3]2+, 2Cl- (aq) + 2NaOH (aq) ---> H2NCH2CH2NH2 (aq) + 2NaCl (aq) + 2H2O (l) The H2O2 oxidation of Co2+ ion to Co3+ ion is: oxidation Co2+ (aq) ---> Co3+ (aq) half-rxn reduction H2O2 (aq) ---> H2O (l) half-rxn from which we get 2Co2+ (aq) + H2O2 (aq) + 2H+ ---> 2Co3+ (aq) + 2H2O (l) (acidic solution) 2Co2+ (aq) + H2O2 (aq) ---> 2Co3+ (aq) + 2OH- (aq) (basic solution) BTW no problem with sub & superscripts... the subscripts are not so bad, just select the text and choose font size 1, not a true subscript but it looks OK. For superscripts, click on the "Use HTML to create your page" link at the bottom of the editing window and enter <sup> and </sup> tags. Co2+ would be Co<sup>2+</sup> in the HTML view. You can do subscripts this was also, using the <sub> tag. H<sub>2</sub>O gives H2O. Steve |
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| | | From: ·Steve· | Sent: 8/22/2005 9:40 PM |
From the order the reactants are added, the formation of the [Co(en)3]3+ complex should be a) Reaction of the hexaaquocobalt(II) ion with ethylenediamine: [Co(H2O)6]2+ (aq) + 3en ---> [Co(en)3]2+ (aq) + 6H2O (l) b) Oxidation of the tris(ethylenediamine)cobalt(II) ion with H2O2 in basic solution: 2[Co(en)3]2+ (aq) + H2O2 (aq) ---> 2[Co(en)3]3+ (aq) + 2OH- (aq) A helpful site is at http://www.chemguide.co.uk/inorganic/transition/cobalt.html which mentions that the hexaamminecobalt(II) ion can be oxidized by air or H2O2 also. Although waters of hydration are common in ionic compounds, I'm not sure why your crystalline product contains NaCl, or if the NaCl would be removed by recrystallization. That is interesting in itself! Steve |
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Awesome. Thanks, that's a great help. I'm not sure either what all that extra stuff of crystisation is.
Bout the subscripts though, I don't have those option things at the top of the page, or the "Use HTML to create your page" option either. But I use Mozilla Firefox not Internet Explorer, so If I really wanted them...
Anyways, the crystals that I resolved yesterday approached a disappointing 50% purity (with the (+)enantiomer being even less) but I got the biggest yield in the class and was able to recrystallise the excess tartaric acid, that went pink & candyfloss-like. Quite cool.
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| | | From: ·Steve· | Sent: 8/24/2005 6:43 AM |
Hi Al, glad your yields were at least reasonable! Any idea what the impurity was? >> But I use Mozilla Firefox not Internet Explorer << I didn't think of that... but I've noticed that Netscape also does not give the editing tools like IE does.  Steve |
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We are told to assume that the only other impurity, for our calculations, is the other diastereoisomer. This, not only making the calculations doable, seems very reasonable as we only 'fractionaly' crystalised it once.
Although, given by the fact the excess Tartaric acid crystalised out so easily, it could also be a majorly contributing factor to my lack of purity. |
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Actually, that's how we checked our purity.
That was part of the second part of the lab, along with the crystalising out of the enantiomers, that I didn't write out ( up there ^ ).
That lab seems much more complicated than ours though. |
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