Cheapskate messed up in his calculations in https://www.thevespiary.org/rhodium/Rhodium/chemistry/chloroform.html (https://www.thevespiary.org/rhodium/Rhodium/chemistry/chloroform.html)
3 mol bleach is required for 1 mol acetone. The overall reaction is:
2 CH3COCH3 + 6 Ca(OCl)2 -> 2 CHCl3 + (CH3COO)2Ca + 2 Ca(OH)2 + 3 CaCl2
340g of acetone is 431mL or 5.85mol. 5.85mol of acetone requires 3x the equiv of NaOCl, or 17.56mol of NaOCl. That is 1.306kg of pure crystalline NaOCl, or 12.098kg of 10.8% NaOCl. 10% NaOCl has a density of 1.16g/mL. So that is a volume of 10.4L of NaOCl. Or 3.2 U.S. gallons (1gal = 3.78L). Cheapskate is using 2 gallons, ie he is 1.2 gallons short on bleach. Not enough bleach to oxidize all of the acetone...
Where I live, bleach is sold in 4L containers... so all this makes no sense to me anyways! :)
So I'll redo the calculations.
The final word is:
For every 1g of acetone (0.788mL) you want to oxidize, use 35.5g (30.6mL) of 10.8% NaOCl.
Also, if you're having trouble oxidizing anything with NaOCl, use 1 mol of HCl to 1 mol of NaOCl required, and drip it in with what you're oxidizing. Ie, if you want to make sure the NaOCl oxidizes the acetone easily, add 1 mol eq's of 10% HCl relative to the NaOCl mixed together with the acetone, and drip the acetone/HCl into the mixture slowly (with the whole thing on an ice-bath and ice-water running through the condenser). Otherwise what can happen is you'll have nothing happen, then all of a sudden the reaction picks up, and with the heat, it picks up even more... until it starts to boil over... 100s of mLs of chloroform evaporating into your working area within seconds, is NOT fun.
Data I used for the calculations:
10.8% NaOCl (w/w)
31.25% HCl(aq) (w/w)
0.03646 g/mmol HCl
0.0744 g/mmol NaOCl
0.05808 g/mmol acetone
0.788 g/mL acetone
1.16 g/mL 10.8% bleach
Chromic:
if only we could find an OTC way to reduce the 2,5-dimethoxynitrostyrene...
What about Post 217679 (https://www.thevespiary.org/talk/index.php?topic=11591.msg21767900#msg21767900)
(sunlight: "Russian patents works!!! Zn-HCl nitrostryrene rdxn", Novel Discourse) ?
Or am I missing something ?
[Hope not cos otherwise I've been wasting a lot of fucking time, effort and money on the hydroquinone route!]
CH3COCH3 + 6 NaOCl -> CHCl3 + NaCH3COO + 2 NaOH + 3 NaCl
this equatation is unbalanced.
Cl2 + 2NaOH => NaOCl + NaCl + H2O
this equatation is balanced
No, I am not eager to find mistakes in the work of others.
I found it here:
http://www.sciencemadness.org/talk/viewthread.php?tid=2050 (http://www.sciencemadness.org/talk/viewthread.php?tid=2050)
The mistake has probably been made because when chlorine is used half goes to NaCl, so the author based the equation on half of the chlorine from hypochlorite going to chloride. NaOCl doesnt do this being effectivly the half of the Cl2 molecule that disproportionated high, ie its in oxidation state +1, chlorine is oxidation state 0.
The person who posted it there is known to know his chemistry well.
This affects also:
https://www.thevespiary.org/rhodium/Rhodium/chemistry/chloroform.html (https://www.thevespiary.org/rhodium/Rhodium/chemistry/chloroform.html)
Actually, just replace the correction. I will ass-u-me I'm right... could someone check these calculations? I don't want to make another mistake (it's rather embarassing when a mistake gets published)..
and the correction should read:
-----------
Chromic: Corrections to Cheapskates chloroform writeup (update May 24, 2004)
Cheapskate messed up in his calculations in https://www.thevespiary.org/rhodium/Rhodium/chemistry/chloroform.html (https://www.thevespiary.org/rhodium/Rhodium/chemistry/chloroform.html)
. This is certainly not an uncommon mistake, as I too messed up the corrections before this update (twice!)
3 mol OCl- is required for 1 mol acetone. Depending on the source of hypochlorite, the overall reaction is:
* CH3COCH3 + 3 NaOCl -> CHCl3 + NaCH3COO + 2 NaOH
* 2 CH3COCH3 + 3 Ca(OCl)2 -> 2 CHCl3 + (CH3COO)2Ca + 2 Ca(OH)2
To repeat the calculations with Cheapskate's original scale of 340g of acetone, that is 431.5mL or 5.85mol of acetone. 5.85mol of acetone requires 3x the equiv of NaOCl, or 17.56mol of NaOCl. That is 1.307kg of pure crystalline NaOCl, or 12.0985kg of 10.8% NaOCl (26.132kg of 5% NaOCl).
10% NaOCl has a density of 1.16g/mL. So that is a volume of 10.5L of NaOCl. Or 2.76 U.S. gallons (1gal = 3.78L). Cheapskate is using 2 gallons, ie he is only using 72% of the amount he should be using (the chloroform would be contaminated with a fair qty of acetone).
However, where I live, bleach is sold in 4L containers... so all this makes no sense to me anyways! So I'll redo the calculations.
The final word is: For every 1g of acetone (1.27mL) you want to oxidize, use 35.6g (30.7mL) of 10.8% NaOCl (71.2g of 5% NaOCl).
Also, if you're having trouble oxidizing anything with NaOCl, use 1-2 mol of HCl to 1 mol of acetone required, and drip it in with what you're oxidizing. Ie, if you want to make sure the NaOCl oxidizes the acetone easily, add 1-2 mol eq's of 31% HCl relative to the acetone, and drip the acetone/HCl into the mixture slowly. This works by consuming the hydroxide and pushing the reaction to the right. If you're not using acid, the reaction hardly comes to reflux (a condenser or some cooling is needed though).
Beware, if you use too little bleach, this is a fast exothermic reaction as Cheapskate wrote about. Distillation of chloroform is not anything to be worried about, no more so than methylene chloride or any other chlorinated solvent. There is no need for a fractionating column when distilling all of the acetone is gone, but do add some ethanol as Cheapskate indicates! Good luck with the Reimer-Tiemann formylation!
Data I used for the calculations:
* 10.8% NaOCl (w/w)
* 31.25% HCl(aq) (w/w)
* 0.03646 g/mmol HCl
* 0.0744 g/mmol NaOCl
* 0.05808 g/mmol acetone
* 0.788 g/mL acetone
* 1.16 g/mL 10.8% bleach
* 1.08 g/mL 5% bleach