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:
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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
. 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