Chromic:

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if only we could find an OTC way to reduce the 2,5-dimethoxynitrostyrene...

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What about

Post 217679

(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!]

True, sunlight's work is amazing... I guess it just seems to me that every step in the process from hydroquinone to 2cb is a huge uphill battle...  

Yeah..I know exactly what you mean.
 
Btw sometime last year I had a bash at Cheapskate's CHCl3 synth and also noticed something cranky with the stoichiometry. I think I tried to correct for it but at the same time do some clever modification of the reaction (can't remember what) which resulted in the decomposition of all the CHCl3 and so I couldn't conclude anything. Anyhow, the point is that I'm glad you've sorted the synth out cos it will make my life easier later (if I can just get the p-meo phenol prep to work properely   ).

Good work Chromic.

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

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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.

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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

You're absolutely right!    The two properly (hehe... am I really sure this time?) balanced equations are:

CH3COCH3 + 3 NaOCl -> CHCl3 + NaCH3COO + 2 NaOH

2 CH3COCH3 + 3 Ca(OCl)2 -> 2 CHCl3 + (CH3COO)2Ca + 2 Ca(OH)2

NOT:

2 CH3COCH3 + 6 Ca(OCl)2 -> 2 CHCl3 + (CH3COO)2Ca + 2 Ca(OH)2 + 3 CaCl2

CH3COCH3 + 6 NaOCl -> CHCl3 + NaCH3COO + 2 NaOH + 3 NaCl

I should be sent back to grade 10 chemistry.

Does this mean that I should remove your 'correction' from my page or that I should upload a third?

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

chloroform is slightly soluable in water. if you discard the aqueous layer you could be losing up to a third of your product. you cannot heat chloroform in basic water above 60c or it will decompose, but if you add hcl to above ph7 you can distill the aqueous layer. the amount you recover may vary but you can tell by the odor of the aqueous layer there will be some.

is there a way to cause this reaction to continue to CCl4 ... even in poor yeilds?

No.

No, you cannot use this reaction (haloform) to produce CCl4. But it is possible to drive further chlorination.

The reaction of CH4 + 4 Cl2 -> CCl4 + 4 HCl is a favorite introductory free-radical reaction in organic chemistry textbooks. Chlorination of course occurs with intermediate chlorine-containing compounds as well methane. CHCl3 + Cl2 + direct sunlight should do the trick.