Eiteiou Posted November 3, 2012 Posted November 3, 2012 Going to be performing this lab experiment next week and I'm trying to figure out the mechanism just for a better understanding. The textbook does not offer a detailed enough mechanism for my liking. Let me explain the procedure and I'll mention my gaps of understanding as I go...First we are mixing acetic acid with sodium dichromate dihydrate. I'm assuming that this activates the dichromate to make it an oxidizing agent, but the transfer of atoms confuses me... Does the acetic acid attract the sodium ions to make the poly atomic ion dichromate cr2o7^2-? In order to do that it must lose a hydrogen, does it lose it to water molecules to form hydronium ions, or just dissociate and form free protons in the mix? Either way I know that dichromate ^2- must form. The next step in the reaction mechanism would be water plus the dichromate ion to form hcro4^-. Again, here I don't understand how that step works, where do the three oxygens go?After this we add the dichromate mixture to cyclohexanol to form a chromate ester and water. This reaction will then involve a water molecule acting as a base, taking a hydrogen off the carbonic carbon, so then the electrons from that bond make the carbonyl bond to the oxygen, while the chromate ester is eliminated. This step must be the actual oxidation, because here is where the ketone forms. However, my next issue is that the products, which are the cyclohexanone, hydronium ion, and hcro3^- confuse me. Specifically the hcro3^-, because I know we started with a cr6+ chromium, which appears. Orange, but the lab book says the reaction will finish green because of a cr3+ chromium, but the oxidation number on the chromium atom in hcro3^- is cr4+ is it not?
weiming1998 Posted November 3, 2012 Posted November 3, 2012 Going to be performing this lab experiment next week and I'm trying to figure out the mechanism just for a better understanding. The textbook does not offer a detailed enough mechanism for my liking. Let me explain the procedure and I'll mention my gaps of understanding as I go...First we are mixing acetic acid with sodium dichromate dihydrate. I'm assuming that this activates the dichromate to make it an oxidizing agent, but the transfer of atoms confuses me... Does the acetic acid attract the sodium ions to make the poly atomic ion dichromate cr2o7^2-? In order to do that it must lose a hydrogen, does it lose it to water molecules to form hydronium ions, or just dissociate and form free protons in the mix? Either way I know that dichromate ^2- must form. The next step in the reaction mechanism would be water plus the dichromate ion to form hcro4^-. Again, here I don't understand how that step works, where do the three oxygens go?After this we add the dichromate mixture to cyclohexanol to form a chromate ester and water. This reaction will then involve a water molecule acting as a base, taking a hydrogen off the carbonic carbon, so then the electrons from that bond make the carbonyl bond to the oxygen, while the chromate ester is eliminated. This step must be the actual oxidation, because here is where the ketone forms. However, my next issue is that the products, which are the cyclohexanone, hydronium ion, and hcro3^- confuse me. Specifically the hcro3^-, because I know we started with a cr6+ chromium, which appears. Orange, but the lab book says the reaction will finish green because of a cr3+ chromium, but the oxidation number on the chromium atom in hcro3^- is cr4+ is it not? Ok, I'll explain: First, the acetic acid (a stronger acid is typically used) protonates the CrO4(2-) ion formed from dissolving sodium chromate in water). The formation of the dichromate can be summed up by two reactions: 1, Protonation (CrO4(2-)+ H+--->HCrO4-) 2, Dehydration (2HCrO4- <-----> Cr2O7(-2)+H2O) Note that Cr2O7(-2) can be rehydrated back into HCrO4-. The proper oxidation reaction is catalysed by H+. The balanced equation is as: Cr2O7(2-)+3C6H11OH(cyclohexanol)+8H+---->2Cr3+ +3C6H10O(cyclohexanone)+7H2O. Cr(IV), if it exists and is formed, would be a short-lived product of the reduction of dichromate, and decomposes to Cr(III), which is green-coloured. This site (http://www.organic-chemistry.org/namedreactions/jones-oxidation.shtm) can explain more clearly of the exact mechanism of the oxidation.
Eiteiou Posted November 3, 2012 Author Posted November 3, 2012 Wonderful, thank you for your reply and for the resource. So now i understand the formation of the dichromate ion. One more question, if you have a minute...the H+ that catalyzes the oxidation reaction, is this from the acetic acid as well? Also, just to confirm this, the cr4+ containing molecule would be HCrO3- in the final product, but is unstable, so would breakdown into cr3+ and water, correct? Thanks again!
weiming1998 Posted November 4, 2012 Posted November 4, 2012 (edited) Wonderful, thank you for your reply and for the resource. So now i understand the formation of the dichromate ion. One more question, if you have a minute...the H+ that catalyzes the oxidation reaction, is this from the acetic acid as well? Also, just to confirm this, the cr4+ containing molecule would be HCrO3- in the final product, but is unstable, so would breakdown into cr3+ and water, correct? Thanks again! To be exactly correct, the H+ ion doesn't really catalyse the reaction, as it is a reactant itself. It's just that the reaction proceeds a lot slower without the acid, because the H+ concentration in a plain solution of K dichromate and water is too low for the reaction to proceed quickly, but it will still proceed due to the fact that when H+ is consumed, more dichromate/water hydrolyses to form H+ and CrO4(2-) or H+ and OH- to "balance out" the loss of H+. But yes, acetic acid is added to provide the H+. I'm not sure why acetic acid was used though, HCl or H2SO4 is a much more common acid for this purpose. And yes, at the end, HCrO3- is formed. This then disproportionates into Cr3+ and more HCrO4- in the presence of an acid. Edited November 4, 2012 by weiming1998
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