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Author: Subject: Chloroform Synthesis: Why does the common procedure uses Acetone instead of Ethanol?
asox
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[*] posted on 30-5-2019 at 12:00
Chloroform Synthesis: Why does the common procedure uses Acetone instead of Ethanol?


Synthesis of Chloroform is well documented by the reaction of Sodium Hypochlorite and Acetone. An exothermic reaction occurs with the formation of CHCl3 and from what I understood of the mechanism Acetic Acid as well.

It's stated in literature that the same reaction occurs if we mix Sodium Hypochlorite and Ethanol, however I haven't found a single description of the reaction being experimented with Ethanol for the synthesis of Chloroform. Considering obtaining Ethanol is much easier and possibly cheaper than Acetone, I'm thinking there might be some complication with running the reaction with Ethanol that I'm not aware of.

Has someone ever tried this reaction with Ethanol before? Were there any complications that made this synthesis harder than the synthesis with Acetone instead?
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[*] posted on 30-5-2019 at 12:36


I think it's because the reaction with ethanol uses more bleach for each gram of chloroform.
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[*] posted on 31-5-2019 at 05:36


Is it because the ethanol reaction is less favorable than the acetone reaction, requiring excess bleach to dislocate the equilibrium to the products side?

I'm thinking of giving it a go with ~90% v/v ethanol and see how it goes.
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[*] posted on 31-5-2019 at 22:32


No, it's not a question of equilibrium. You simply need more moles of NaClO per mole of chloroform produced. This is because the ethanol needs to be oxidized to acetaldehyde before the haloform reaction can take place.



As below, so above.
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[*] posted on 1-6-2019 at 01:57


The 2 reactions are:

3 CH3CH2OH + 12 NaClO = 10 NaOH + 4 CHCl3 + 2 HCOONa + H2O

3 NaClO + CH3COCH3 = CHCl3 + 2 NaOH + CH3COONa

As you can see, in both cases you get the same amount of chloroform per amount of hypochlorite used. I tried doing the reaction with isopropanol, since for me it was much more easily purchasable at that time, and it didn't work as expected. Apparently it needs heating for the oxidation to acetone step, but heating a potentially exothermic reaction mix seems dangerous to me. I know there was a thread on SM about this, but I can't seem to find it.


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[*] posted on 1-6-2019 at 17:15


@Metacelsus

I see, so I'd spend 1 extra mole of NaOCl for the oxidation step. Still think this extra consumption is compensated by the easier access to ethanol, but I can see now why when having either of the reactants the acetone reaction is preferred.

@CobaltChloride

I think you got the ethanol reaction wrong. I tried to balance the equations here and came up with the following:

CH3CH2OH + 4NaOCl -> CHCl3 + HCOONa + NaCl + 2NaOH + H2O

I tried to add the formulas for each step and then compensate for the ions used with the NaOCl aqueous solution equilibriums. It does match @Metacelsus answer that it consumes more NaOCl per mole of chloroform.
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Corrosive Joeseph
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[*] posted on 1-6-2019 at 17:26


Interesting topic.......... I was only discussing this with a friend not too long ago........... You might also like these -

'Haloform reaction' - https://en.wikipedia.org/wiki/Haloform_reaction

Oxidation of ethanol by hypochlorite - http://www.sciencemadness.org/talk/viewthread.php?tid=32346


/CJ




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[*] posted on 2-6-2019 at 01:49


If you use alcohol, in addition to the less favourable stoichiometry, you end up wasting bleach because it reacts with formate.

The problem isn't so much that you need more hypochlorite- it's cheap enough noty to matter greatly.
But, if you are using dilute solutions you add more water along with the bleach and that means larger losses of chloroform by dissolution.
It's not such an issue if you can distill the whole reaction mixture (rather than salting out the product)
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[*] posted on 2-6-2019 at 03:29


Are you sure that HCOONa would survive being in the same mixture with NaOCl?
Formiates are fair reducing agents and my gut feeling is that they would reduce hypochlorites, though it may be pH dependent.
But I may be mistaken.

Edit: I see unionised also mentioned this.

[Edited on 2-6-2019 by Pumukli]
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[*] posted on 2-6-2019 at 04:20


Quote: Originally posted by asox  
@Metacelsus

@CobaltChloride

I think you got the ethanol reaction wrong. I tried to balance the equations here and came up with the following:

CH3CH2OH + 4NaOCl -> CHCl3 + HCOONa + NaCl + 2NaOH + H2O

I tried to add the formulas for each step and then compensate for the ions used with the NaOCl aqueous solution equilibriums. It does match @Metacelsus answer that it consumes more NaOCl per mole of chloroform.

Yeah, you are right. Sorry, I wrote the reaction wrong.
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[*] posted on 5-6-2019 at 13:51


Quote: Originally posted by Corrosive Joeseph  
Interesting topic.......... I was only discussing this with a friend not too long ago........... You might also like these -

'Haloform reaction' - https://en.wikipedia.org/wiki/Haloform_reaction

Oxidation of ethanol by hypochlorite - http://www.sciencemadness.org/talk/viewthread.php?tid=32346


/CJ


I'll check the thread, it looks like the goal there is to synthesize acetaldehyde by using a catalyst but there's probably information that would be useful for the haloform reaction of ethanol as well. I'll read it through.

Quote: Originally posted by unionised  
If you use alcohol, in addition to the less favourable stoichiometry, you end up wasting bleach because it reacts with formate.

The problem isn't so much that you need more hypochlorite- it's cheap enough noty to matter greatly.
But, if you are using dilute solutions you add more water along with the bleach and that means larger losses of chloroform by dissolution.
It's not such an issue if you can distill the whole reaction mixture (rather than salting out the product)


Quote: Originally posted by Pumukli  
Are you sure that HCOONa would survive being in the same mixture with NaOCl?
Formiates are fair reducing agents and my gut feeling is that they would reduce hypochlorites, though it may be pH dependent.
But I may be mistaken.

Edit: I see unionised also mentioned this.

[Edited on 2-6-2019 by Pumukli]


I found an article describing this oxidation reaction and making a rate study: OCl- + HCOO- + OH- -> Cl- + CO3- + H2O
https://www.nrcresearchpress.com/doi/pdf/10.1139/v63-402

Maybe I could lower the dissolution losses by cooling the mixture before I took it to a separation funnel?
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