Avanine-Commuter
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Haloform Reaction help!
Hi, I'm having a problem figuring this formula out. I'll type up the reaction.
I added potassium iodide and iodine to a beaker of water. This should form potassium tri-iodine, KI3. Then, I added acetone and a little sodium
hydroxide. WHen I heated the mixture up to 60 degrees C, just a little above the acetone boiling point, the following reaction should occur:
3I2+C3H6O+3NaOH----->CI3COCH3+3NaI+3H2O
Then, I removed it from the heat and added about 200 drops of NaOH again, drop by drop. A precipate formed, and it should be:
CI3COCH3 + NaOH----> CHI3 + CH3COONa
Which ends up as Iodoform, the yellow precipitate, and sodium acetate.
I don't know and I need the following:
What is CI3COCH3? I looked on the internet, in Merck's Index, in my Chem textbook, etc. etc. Can't find it ANYWHERE!! 
What happened to the CH3COONa? Sodium Acetate. I know the iodoform is the yellow gunk left on the bottom of the beaker, but where did the Sodium
Acetate go? Did it dissolve in the water?
And Also, WHY does the 3I2+C3H6O+3NaOH-----> CI3COCH3+3NaI+3H2O reaction form? What kind is it? Is this the haloform reaction, or is it just a step
in the entire process?
CI3COCH3 + NaOH----> CHI3 + CH3COONa What kind of reaction is this? I don't know this one as well.
If someone can help me with these questions or guide me in the right way, I'd gladly appreciate it. Thanks in advance.
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guy
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How did you get the formula to be C2H3OCl3? It should be C3H3OCl3 which is trichloroacetone. And you are using iodine so it should be C3H3OI3,
triiodoacetone!
For the mechanism of a haloform reaction:
http://www.organic-chemistry.org/frames.htm?http://www.organ...
[Edited on 5/5/2006 by guy]
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chemoleo
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CI3COCH3 is triiodo acetone. Its product is unstable in base because the OH- from NaOH reacts with it, according to
RCOCI3 -->OH- --> RC[O-][OH]CI3 --> RCOOH + [CI3]-
CI3- is very reactive (makes an excellent leaving group), and produces with H2O I3CH and OH-.
So the OH- is regenerated, while the Na+ reacts with the free acid. So not strictly a catalysis.
Why does the reaction form the triply substituted RCOCI3? It is because mono-iodinated (halogenated) alky displays an increasing acidity of the
remaining alpha hydrogen, making the respective enolate ion RCOCH2I <=> [-RC(OH)=CIH]- + H3O+ more susceptible to further halogenation.
Yes, the acetic acid is soluble, so you won't ever notice it.
Never Stop to Begin, and Never Begin to Stop...
Tolerance is good. But not with the intolerant! (Wilhelm Busch)
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BromicAcid
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Guy, he has it as CI3COCH3, that's not a l, that's an I (I had to change the font to make that I look like
an I because the default font has capital I's and lowercase L's look the same). So he had the triiodoacetone although there is likely someway to
distinguish that all of the iodine are on the same carbon. Maybe 1,1,1-triiodoacetone? Or because it is a carbonyl compound (alpha), (alpha),
(alpha)-triiodoacetone (I forget how to make Greek letters).
The haloform reaction is pretty spontaneous and I've never had the need to heat it, that seems like it would just drive off acetone. Your sodium
acetate, being water soluble, stayed in the water.
The inital formation of the triiodo compound is due to multiple S<sub>N</sub>2 rxns and this compound is more or less stable under acid
conditions, its only when basified that the driving force, formation of the iodoform becomes prevelent (though I've never tried to do this stepwise
with replacement under acid followed by basification). The reason you end up with a triiodo substitued compound (providing you have enough iodine in
the solution) is because each subsequent S<sub>N</sub>2 reaction becomes more favored then the previous as the iodides stabilize the
enolate intermediate.
Your questions kind of ramble as does my answer so feel free to post a revised version of your question in response to any answers you might get.
[Edited on 5/5/2006 by BromicAcid]
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Avanine-Commuter
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Thanks for all the help!
A few more questions or just for clarification:
If the heating of the solution is not necessary, why was it part of the procedure? I THINK it has something to do with the acetone, because it had to
boil, but I'm not sure. Does acetone need to be at boiling point in order to react? Or does it react immediately with the other substances on contact?
Also, What would be some chemical principles involved in this? I don't know what distinguishes 'principles' from "general group of experiments" that
this belongs to. So far, for my general group of experiments I put:
Physical changes
Ions in solutions
elements in Group IA
hydrolysis (acid/base) and equilibrium
oxygen and sulfur
organic chemistry
halogens
analytical
These are the ones I chose out of the categories given. Is there a difference between the involved chemistry principles, and if so, what are they?
And last of all, What kind of reaction is 3I2 + C3H6O + NaOH?
[Edited on 5-5-2006 by Avanine-Commuter]
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solo
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Reference Information
Carbonyl Compounds - Iodoform Reaction
The iodoform test is a test for the existence of the CH3-CO- group in a molecule. This could be part of an alcohol (C-O single bond) or part of a
carbonyl compound (C=O double bond).
The hydrogen atoms on the methyl group are slightly acidic and can be removed with sodium hydroxide (stage 1 below).
The carbanion formed then react with iodine molecules to give an iodide ion and an organic iodo compound (stage 2 below).
This substitution continues until a triiodo group has been formed (stage 3 below) by repeated use of sodium hydroxide and iodine.
Another hydroxide ion can then attack the carbonyl carbon atom, giving a carboxylic acid and releasing the CI3 group which abstracts a proton from a
water molecule to give CHI3 (triiodomethane or iodoform) (stage 4 below).
Triiodomethane is a straw yellow solid, insoluble in water. This test works for ethanal and all methyl ketones,
...........source,
http://www.rjclarkson.demon.co.uk/candrands/carbonyls.htm#io...
It's better to die on your feet, than live on your knees....Emiliano Zapata.
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Dr.3vil
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reaction conditions - chloroform from Cl2
rather then start a new thread....
Reaction:
C3H6O + Cl2 in base => chloroform + carboxcylic acid
what would be the best method of introducing Cl2 into the reaction flask?
1) bubble chlorine by means of capillary tube or fish tank aerator
2) allow the reaction flask to fill with chlorine and use magnetic stirring to promote a reaction at the phase boundary.
Dr.3vil
--- Dig hard, dig deep.....deserve what you reap.... don't you know the truth will set you free ---
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Sauron
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A fritted glass piece of medium porosity on the end of a tube as sold by all the major glassware companies such as Ace, Kimble/Kontes, Corning etc
etc.
That gets you many small bubbles at same time.
Capillary gets you a single stream of relatively larger bubbles.
Important condition is temperature, the haloform reaction is done in the cold. There's won't be any free carboxylic acid, only the sodium salt.
This reaction will proceed with any compound containing the MeC(=O)- group. MEK gets you same chloroform. Ethanol first oxidizes to acetaldehyde then
gives chloroform. Acetophencone yields chloroform. And so on.
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