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CuReUS
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I have had this question for a long time. Could you reduce iodoform to DIM using an alkaline glucose solution ?
[Edited on 29-5-2016 by CuReUS]
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PHILOU Zrealone
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Maybe Silver malonate + I2 may give CH2I2 upon heating?
Silver carboxylate heating with halogen allows for decarboxylation and formation of alkyl halide and silver halide...
CH2(-CO2Ag)2 + 2 I2 -heat-> CH2I2 + 2 CO2(g) + 2 AgI(s)
PH Z (PHILOU Zrealone)
"Physic is all what never works; Chemistry is all what stinks and explodes!"-"Life that deadly disease, sexually transmitted."(W.Allen)
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Boffis
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@ CuReUS, I have looked at the reductions several times as I use methylene diiodide in mineralogy and its very expensive. It appears that there are
many reducing agents that can accomplish the reduction, the problem is stopping them at the removal of one iodine. An alkaline glucose solution looks
like a reasonable option provided there isn't a large excess of alkali.
The main issue I encountered seemed to be the protecting effect of the first formed CH2I2 on the remaining iodoform, the former dissolves the latter
taking it into the organic phase. I have always though that a organic-soluble reducing agent would be helpful. Any ideas?
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CuReUS
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Quote: Originally posted by Boffis  |
The main issue I encountered seemed to be the protecting effect of the first formed CH2I2 on the remaining iodoform, the former dissolves the latter
taking it into the organic phase. I have always though that a organic-soluble reducing agent would be helpful. Any ideas? |
Why use an organic solvent at all ? we could run the reaction similar to a benedict's test without the Cu ions.Just boil the iodoform in the
alkali-glucose solution.Since DIM is more soluble in water(1240 mg/dl) than iodoform(100 mg/dl),it will slowly dissolve in the aq solution whereas the
iodoform will stay behind,thus pushing the equilibrium forward.
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PHILOU Zrealone
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| Quote: Originally posted by CuReUS | | Quote: Originally posted by Boffis |
The main issue I encountered seemed to be the protecting effect of the first formed CH2I2 on the remaining iodoform, the former dissolves the latter
taking it into the organic phase. I have always though that a organic-soluble reducing agent would be helpful. Any ideas? |
Why use an organic solvent at all ? we could run the reaction similar to a benedict's test without the Cu ions.Just boil the iodoform in the
alkali-glucose solution.Since DIM is more soluble in water(1240 mg/dl) than iodoform(100 mg/dl),it will slowly dissolve in the aq solution whereas the
iodoform will stay behind,thus pushing the equilibrium forward. |
Why not simply cold distil/condense it away?
CH2I2 must be much more volatile than CHI3...thus shifting equilibrium...
PH Z (PHILOU Zrealone)
"Physic is all what never works; Chemistry is all what stinks and explodes!"-"Life that deadly disease, sexually transmitted."(W.Allen)
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CuReUS
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because the glucose-alkali reduction of iodoform will be done in an aqueous solution and the boiling point of DIM is
181'C
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clearly_not_atara
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Dichloroacetic acid or its esters should undergo the Finkelstein reaction far more rapidly than dichloromethane. From:
https://en.wikipedia.org/wiki/Finkelstein_reaction
you can see that chloroacetone reacts about 200x as fast as MeCl.
http://en.wikipedia.org/wiki/Dichloroacetic_acid
| Quote: | | DCA is prepared from chloral hydrate also by the reaction with calcium carbonate and sodium cyanide in water followed by acidifying with hydrochloric
acid. |
Might this be one of the odd reactions (cf. benzoin condensation) where thiamine is a useful stand-in for cyanide?
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Boffis
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Dichloroacetic acid is readily prepared from commercially available trichloroacetic acid by dissolving zinc in it. Instead of hydrogen being evolved
the acid looses a chlorine atom instead. I have a reference somewhere for this.
So now you can get DCA and lets assume the Finkelstein reaction works a treat on it, are you sure you can get it to decarboxylate readily or does this
happen spontaneously as with some other substituted acetic acids ie nitroacetic acids?
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PHILOU Zrealone
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| Quote: Originally posted by CuReUS |
because the glucose-alkali reduction of iodoform will be done in an aqueous solution and the boiling point of DIM is
181'C |
OK thanks!
Glucose solution with NaOH could boil very much higher than 100°C...but 181°C is maybe too high.
For the rest glycol might be used as a high boiling solvent maybe forming an azeotrope with CH2I2 (CH2Cl2 and CH2Br2 do form azeotrope with it)...
[Edited on 11-6-2016 by PHILOU Zrealone]
PH Z (PHILOU Zrealone)
"Physic is all what never works; Chemistry is all what stinks and explodes!"-"Life that deadly disease, sexually transmitted."(W.Allen)
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