Sciencemadness Discussion Board

Sodium and chloroform

libavius - 3-5-2009 at 13:21

(Sorry for my English), but about this reaction:

Na + CHCl3 -> NaCl + ? + ?

Who knows what are the stable products that are formed in addition to NaCl?
Those who actually come from the tube! (...dichlorocarbene...if... not stable!)
Thank you

Nicodem - 3-5-2009 at 23:48

This is not a definable reaction. It is just a cleavage of an organic compound with sodium. NaCl, some H2 and carbon should be the most characterizable products. I have no idea what form of carbon you get though.

Formatik - 4-5-2009 at 00:19

http://sciencemadness.org/talk/viewthread.php?fid=3&tid=... See the second attachment by Axt. See also COPAE book in the forum library, it discusses the reactions also.

libavius - 4-5-2009 at 04:41

OK Formatik!

Your link was perfect for me; exactly what I was looking for!
Thanks

sakshaug007 - 4-5-2009 at 17:14

Its funny you bring up this question libavius because I was wondering just this same thing when I was trying to hypothesize a mechanism for the formation of poly(hydridocarbyne) via the electrolysis of chloroform using NaCl as the electrolyte in acetonitrile solvent.

The mechanism I developed is as follows:

Na(+) + e- (cathode) => Na + CHCl3 => Na(+) + Cl(-) + *CHCl2 (radical)
4*CHCl2 => poly(hydridocarbyne) and Cl2 (anode)

Or at least something like this is going on.

DJF90 - 4-5-2009 at 18:04

If you want to form the dichlorocarbene then you can use NaOH and CHCl3, and this way you can form 1,1-dichlorocyclopropanes from alkenes, although unsure if the OH present will substitute for the chlorines... I dont think so seeing as the C-Cl sigma* orbital is pretty hindered... Also I believe this is one method of formylating phenyl rings, although I am not sure of the mechanism.

[Edited on 5-5-2009 by DJF90]

Nicodem - 4-5-2009 at 23:49

Quote: Originally posted by sakshaug007  
Its funny you bring up this question libavius because I was wondering just this same thing when I was trying to hypothesize a mechanism for the formation of poly(hydridocarbyne) via the electrolysis of chloroform using NaCl as the electrolyte in acetonitrile solvent.

The mechanism I developed is as follows:

Na(+) + e- (cathode) => Na + CHCl3 => Na(+) + Cl(-) + *CHCl2 (radical)
4*CHCl2 => poly(hydridocarbyne) and Cl2 (anode)

Or at least something like this is going on.

It seems to me you do not completely understand the basics of electrolysis. The electrons are not supposed to interact with the electrolyte at the electrode where the electrochemical reactions on the substrate goes on. This only leads to unwanted side reactions (=waste of electric current). The electrolyte is there only to provide electric conductivity and therefore must be chosen in such a way for its ions to have a greater overpotential at the working electrode as is the potential needed for the desired electrochemical reaction.
For example, if you want to electrolyse water you can chose any electrolyte that has a higher overpotential at the electrodes when compared to H2O. That is, you can not use NaCl as electrolyte because the chloride anion has a lower overpotential at the anode than H2O, which would lead to the formation of Cl2 instead of O2. You can also not use CuSO4 because Cu(2+) ions have a lower overpotential than H2O leading to the deposition of Cu instead of H2 formation. But you can use Na2SO4 because both, its anion and its cation have overpotentials higher than the potential required for the H2O electrolysis at each electrode.
Similarly, for the electroreduction of CHCl3 you need an electrolyte which at the cathode will not preferentially reduce the electrolyte (or solvent!) as opposed to the substrate. You need to favour the nCHCl3 + 3e(-) => (CH)n + 3Cl(-) reaction as opposed to M(+) + e(-) => M(0) side reaction (where M(+) is the cationic part of the electrolyte). Anyway, I think Na(+) has a higher overpotential than CHCl3 at the catode in just about any solvent, so essentially you don't have to worry about it. You have much more to worry about the solvent used.

UnintentionalChaos - 4-5-2009 at 23:54

Quote: Originally posted by DJF90  
If you want to form the dichlorocarbene then you can use NaOH and CHCl3, and this way you can form 1,1-dichlorocyclopropanes from alkenes, although unsure if the OH present will substitute for the chlorines... I dont think so seeing as the C-Cl sigma* orbital is pretty hindered... Also I believe this is one method of formylating phenyl rings, although I am not sure of the mechanism.

[Edited on 5-5-2009 by DJF90]



If the formylation reaction works, I would assume that you form a 1,3-cyclohexadiene with the attached gem-dichlorocyclopropane ring. Breaking of the cyclopropane ring would be promoted by the rearomatization of the benzene ring, giving a benzal chloride. This hydrolyzes pretty easily to a benzaldehyde.

[Edited on 5-5-09 by UnintentionalChaos]

DJF90 - 5-5-2009 at 00:00

Its called the Riemer-Tiemann reaction IIRC, and the mechanism you propose sounds very likely.

Nicodem - 5-5-2009 at 00:40

Not really, since it would then work with substrates other than phenoxides and similar. The reaction mechanism is currently believed to be based on the electrophilicity of the dichlorocarbene which C-attacks the phenoxide anion:
http://en.wikipedia.org/wiki/Reimer-Tiemann_reaction
(the insertion of any carbene into an aromatic Pi-system is too unlikely)

DJF90 - 5-5-2009 at 03:23

Ah right yes that makes much more sense. I dont know why I didnt just look up the mechanism in the first place. I guess I should really get more sleep :(

libavius - 5-5-2009 at 21:59

Quote: Originally posted by sakshaug007  
Its funny you bring up this question libavius because I was wondering just this same thing when I was trying to hypothesize a mechanism for the formation of poly(hydridocarbyne) via the electrolysis of chloroform using NaCl...

Sakshaug007, you wanted to do about my own reverse reaction... really funny!
I am not an expert in electrolysis, Nicodem certainly familiar with the topic.
Also interesting is the whole debate on dichlorocarbene ... very good!

But this reaction would also be discussed:

CCl4 + Na -> NaCl + C ?

Another explosion, black smoke, so simply?
Bye
[Edited on 6-5-2009 by libavius]

[Edited on 6-5-2009 by libavius]

sakshaug007 - 6-5-2009 at 16:35

Quote:

Sakshaug007, you wanted to do about my own reverse reaction... really funny!

But this reaction would also be discussed:

CCl4 + Na -> NaCl + C ?


Yeah I would like to get a hold of some CCl4 so I could attempt the electrolysis of it. Without the lone hydrogen it might produce a coating of tetrahedral diamond on the cathode. It may be more likely, however, that it would form graphite as this may be more favorable under these conditions.

libavius - 7-5-2009 at 13:19

Saks, I seem somewhat optimistic...
In any case, best wishes for your diamonds ...!

TheOrbit - 8-5-2009 at 04:41

i think H2 + CCl2 : chlorocarbene"

hissingnoise - 8-5-2009 at 05:28

Sodium reacts with chloroform to form NaCl, HCl and hexachlorobenzene---the flash of the explosion is visible in direct sunlight. . .

DJF90 - 8-5-2009 at 09:07

Do you have a reference for this? The stoichiometry seems right though.

hissingnoise - 9-5-2009 at 06:49

Can't remember where I saw it, but Davis has it on page 403 of COPAE. . .
It's curious that the reaction is so explosive.

libavius - 9-5-2009 at 14:06

Quote: Originally posted by hissingnoise  
...on page 403 of COPAE. . .
It's curious that the reaction is so explosive.

How can I find the COPAE book in the forums library?
I can not find it ...
(Yes, even for me this is very curious!)


hissingnoise - 10-5-2009 at 03:23

COPAE is "The Chemistry of Powder and Explosives" by Tenney L Davis and IMO, the most interesting book of an interesting collection. . .

Foss_Jeane - 6-6-2009 at 00:11

Quote: Originally posted by sakshaug007  
Yeah I would like to get a hold of some CCl4 so I could attempt the electrolysis of it. Without the lone hydrogen it might produce a coating of tetrahedral diamond on the cathode. It may be more likely, however, that it would form graphite as this may be more favorable under these conditions.


Well, that won't happen. Pour CCl4 into a partitioned cell, acidify with H2SO4 or HCl for good conductivity, and use a H2SO4 (3.0 -- 4.0M) anolyte and a copper cathode, and you reduce it to chloroform. Cathodes with higher overvoltages will reduce it to CH2Cl2 or CH3Cl.