Sciencemadness Discussion Board

Why is bromochlorofluoroiodomethane so difficult to synthesise?

deltaH - 1-11-2013 at 04:34

I tried to find synthetic references for this compound and failed. Wiki states about it that it is a hypothetical compound often cited as the prototypical chiral compound.

Why is it so difficult to prepare? Any insights would be greatly appreciated!

[Edited on 1-11-2013 by deltaH]

bfesser - 1-11-2013 at 06:17

I'm bored, so I'll transcribe this by hand:<a href="" target="_blank">
<strong>;&nbsp;&nbsp;&nbsp;&nbsp;Bromochlorofluoroiodomethane, CBrClFI</strong><br /><br />&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;Bromochlorofluoroiodomethane, although often mentioned as <em>the</em> prototype of a chiral compound, has not been prepared yet. MO calculations indicate that it is the second least stable chiral halomethane judged from the <a href="" target="_blank">enthalpies of formation</a> <img src="../scipics/_wiki.png" /> <90BCJ1278>. The only chiral halomethane which has been prepared is CHBrClF, and it has been shown that this derivative hydrolyzes much faster than other mixed halomethanes <56JA479>; this may also be the case with bromochlorofuluoroiodomethane.<br />&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;It would be interesting to try the <a href="" target="_blank">Hunsdiecker reaction</a> <img src="../scipics/_wiki.png" /> on silver bromochlorofluoroacetate which reacts with both chlorine and bromine as mentioned above. It may, however, turn out to be a very low yielding reaction, as was the case with the treatment of silver bromodifluoroacetate with iodine (Table 3) <52JCS4259>.<br />&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;Theoretical discussions of the physical properties of so far unsynthesized tetrahalomethanes are on record <58MI 607-01, 59MI 607-01, 66ZOB1355, 72MI 6070-1, 76ZC377, 77MI 607-01, 79JMR599, 83MI 607-01, 91MI 607-01>.
</a>You really should put more effort into your topics; this was cited by Wikipedia and is on the second page of search results on Google. Frankly, it should be apparent to anyone with even a rudimentary understanding of O.Chem. why this compound would be difficult to synthesize. I'm a little disturbed that you're asking such questions while carrying on more complex arguments in other topics.

I think the Hunsdiecker reaction may be my new favorite&mdash;such a beautiful and elegant mechanism.
<a href="" target="_blank"><img src="" /></a><img src="../scipics/_wiki.png" valign="top" />

[still editing! don't quote me!]

[Edited on 1.11.13 by bfesser]

woelen - 1-11-2013 at 06:40

Quote: Originally posted by bfesser  
Frankly, it should be apparent to anyone with even a rudimentary understanding of O.Chem. why this compound would be difficult to synthesize.

Please give your explanation.
I have some ideas why this is difficult to synthesize, but I first want to see yours.

[Edited on 1-11-13 by woelen]

bfesser - 1-11-2013 at 06:50

Heh, I was afraid someone would say that&mdash;now I will demonstrate how clueless yet judgmental I can be. I'm still editing my post above, but here are a few of my immediate thoughts:<ul><li>suspected high enthalpy of formation</li><li>difficulty of controlling halogenations for single substitution</li><li>steric & <a href="" target="_blanK">electronic effects</a> <img src="../scipics/_wiki.png" /> (<em>intuit a lot of repulsion between halides</em>;)</li><li>sharing anything with fluorine :P</li><li><em>LOOK AT THAT THING! It's a beast! (call it 'chemical intuition')</em></li></ul><img src=" g" />

[edit] Also, I was unaware of the Hunsdiecker reaction before making my above post. <em>Thanks, partial public university education!</em>

[Edited on 1.11.13 by bfesser]

deltaH - 1-11-2013 at 06:54

bfesser, you get disturbed too easily :) I understand that enthalpies of formation suggest that this compound is strongly unfavoured, but I don't understand WHY this enthalpy of formation is so large.

Why is it that a C-F, C-Cl, C-Br, C-I bonds are all strong and fine and all manner of multiple version of it so long as it's of the same atoms, so C-F, whether F3C-F, or HF2C-F or H2FC-F or H3C-F are all fine and the same for all other permutations, but when you start cross halogenating on the same carbon, things are suddenly very unhappy. Even CI4 is fine inspite of the crazy steric hindrance around the carbon, but a mix of all four is suddenly particularly not freundlich? This is very odd behaviour to me.

I don't think the explanation is so simple, at least not to me (just being honest here) and I was hoping someone could shed some light on this chemical riddle.

The enthalpy of formation answer is not a reason, it's more a symptom if you understand my meaning (so are all the other symptoms mentioned in your article... hydrolyse tendencies and the likes, these are just properties because of poor stability).

Anyhow, I hope that makes sense...

[Edited on 1-11-2013 by deltaH]

bfesser - 1-11-2013 at 06:59

That does make sense. I agree that the explanation is not so simple, and look forward to reading replies from some of our Organic Chemists in Residence (Nicodem et al.). Keep in mind, I'm a repeat university dropout, which is why I often prefer to read and learn rather than share my uninformed opinions.

[edit] Lazy link of what I'm looking into at the moment:
Gotta go! Back later for more editing.

[Edited on 1.11.13 by bfesser]

deltaH - 1-11-2013 at 07:04

Please don't do that bfesser, I mean not share your opinion. No opinion is unworthy and I will never judge or ridicule. If I disagree with something you say, I will try to explain my thoughts as to why and in the discourse we may both learn something. This is my belief and the way I behave myself on this forum.

There are no masters or novices, we are all students of science forever.

...and besides, in science, nothing is 100% certain, we can merely come closer to the 'truth'.

Shit... sounds very preachy and religious lol Sorry not meant like that!

[Edited on 1-11-2013 by deltaH]

woelen - 1-11-2013 at 07:15

My line of thought is not about the stability (or lack of stability) of the compound, but simply that there is no known mechanistic path towards the compound. I can imagine that on the way of making this compound (e.g. sticking I to C, Br to C and so on, that adding one atom removes the other and undoes the effort put in attaching that atom).

A totally different example has been the perbromate ion (BrO4(-)), which was not made until the 1960's. Perbromate ion is perfectly stable, actually it is of all oxo-anions of the halogens the second stablest one (only ClO4(-) is more stable) and quite inert, just like ClO4(-). Its salt KBrO4 is more reactive than KClO4, but much less reactive than KIO4, KIO3, KBrO3, KClO3, NaClO2, or NaClO. The problem was the path towards this ion. It was synthesized first in microgram quantities from the SeO4(2-) ion, using a special isotope of Se, which is converted to Br by means of beta-decay. Still, no easy chemical synthesis is known for perbromate, the best known methods require the use of elemental fluorine under precisely controlled conditions and these methods have a yield of at most 2%.

This example demonstrates that there may be totally different reasons for non-existence of chemical compounds than stability or extreme reactivity.

[Edited on 1-11-13 by woelen]

deltaH - 1-11-2013 at 07:17

I think what I was trying to say is that in the curious case of CIBrClF (see I even remembered to order them in reverse electronegativity--pats self on back) it's almost as if the carbon suffers from xenophobia lol

I suspect that electronic effects is the problem, but again WHY? It can't merely be a push pull problem, because CF2I2 is stable!

It really only fails when you have all flavours of halogens!

kristofvagyok - 1-11-2013 at 07:19

The problem is not how, the problem is why. This molecule is good for theoretical chemists, but since they usually calculate things, they do not need to make it.

I would try it on this route:
(just an idea)

All easily performed reactions, usually work well, the only thing is to separate the products, so after the Swarts reaction separate the dichloro diethyl malonate, the chloro fluoro diethyl malonate and the difluoro malonate, same at the last part, at the Finkelstein reaction, distillation and distillation.

Would it be worth? If someone would pay for the product or it would be good for something sure, but doing it for fun? No.

deltaH - 1-11-2013 at 07:25

Hey kristofvagyok Nice! It would be worthwhile simple for kudos factor :) Not for me, I'm broke though :(

EDIT: decided to change coolness to kudos because I know how people react to anything k3wl on this forum :P

Having all four stable halogens dangling on one carbon... it's like molecular work of art!

Ok I will take a very crude stab at an explanation myself for fun... but this is totally over my head!

Could it be that the poor stability is due to the worst possible mismatch of atomic valence orbitals (over the widest possible range in energy) to a singular carbon who has to try to distort it's four valence electrons (from energy orbitals of the same energy) over the broadest range in order to form molecular orbitals simultaneously with each different atom in turn, thus resulting in the smallest gains from being atoms in the first place and not a molecule, hence by definition having a poor delta H formation.

What I'm trying to say is that its four carbon electrons have to hybridise over the broadest possible range of energies to create the molecular orbitals of the four sigma bonds making this maximally unfavoured?

Hmm... perhaps some orbital diagrams can illustrate this...

I know it's not a very scientific argument and feels distinctly fluey... but I think you get the gist of what I am trying to say, perhaps somebody can put it in better scientific terms if it's not total nonsense.


I agree that mechanism looks very elegant, but maybe also a little wishful think/overly simplistic, for example, if R* and Br* radicals formed as an intermediate, wouldn't you also expect a whole bunch of other products like R-R and Br2(excess) by other termination reactions? The way the mechanism is written on wiki suggests that R* and Br* neatly and selectively 'find' each other... well I think that would only happen half of the time. What am I missing here?

[Edited on 1-11-2013 by deltaH]

watson.fawkes - 1-11-2013 at 12:00

It's almost always worth actually reading before speculating. All it takes to find the reference 90BCJ1278 is reading the end note.

&nbsp;&nbsp;Electronic Structure of Chiral Halomethanes
&nbsp;&nbsp;Igor Novak
&nbsp;&nbsp;Bulletin of the Chemical Society of Japan Vol. 63 (1990) No. 4 P 1278-1280

Full text PDF available from the site. It takes some familiarity in understanding molecular orbital (MO) calculations to know what's going on, but there are clues in there. Here's one that struck me immediately:

I 5s and Br 4s orbitals lost their distinct atomic characters altogether forming linear combinations with each other and other bonding orbitals.

I'd guess that I-Br selectivity is rather low for any attempted synthetic route. The cited paper doesn't have them, but MO calculations for CI2ClF and CBr2ClF would be relevant here.

deltaH - 1-11-2013 at 12:15

Thanks for that watson.fawkes, this looks promising, I will have to have a hard read through it though to wrap my head around it all, but I see molecular orbital diagrams and I think that the answer must lie there, so let me see if I can get any of this to make sense to me.

bfesser - 1-11-2013 at 12:17

deltaH, I didn't mean to imply that the reaction had any significance here. Simply that I saw it on Wikipedia, liked it, and thought I'd share.
watson.fawkes, I concur re: speculation. Sadly, these days, I've been accumulating more papers than I can read, and I was probed for a speculative opinion.

deltaH - 1-11-2013 at 12:52


No not at all, I understand completely, was just pointing out something odd in that mechanism as seen on wiki, was hoping the org. chemists here could comment on whether what I thought was odd was really so, that these radicals were expected to neatly recombine like that.

Just had a read through the Novak paper. These guys used semi-empirical methods for their calculations, so one has to be wary of the accuracy of the results, however, I would imagine that it's probably ok for comparative purposes and to look for trends.

Sadly however, they are not comparing what I would like to see compared (ideally), they are comparing different varieties of apples with each other where the answer I would like is in the comparison of an apple with a pear and orange :) Their compounds are all highly cross halogenated, granted 3 halogens for most and then the CIBrClF specie, but still, ideally you would like to see their fig. 2 with many other compounds. But you've got to give them credit, it was 1989 and in those days computers weren't exactly the beasts that they are today.

I've stared long and hard at fig. 2 in that paper because if the answer is here, it would lie in those results. From those results I conclude that the valence electrons from the halogens do not change their energy very much between the various compounds, but those of carbon appear to have a slight trend upwards in energy from left to right. I think that "I 5s and Br 4s orbitals lost their distinct atomic characters altogether forming linear combinations with each other and other bonding orbitals." is more of a peculiarity that doesn't have direct bearing on the WHY here.

The take home message here in my opinion is that there is a marginal trend going from left to right (more extreme cross halogenated with more greatly differing atom sizes) on that figure for carbon's orbitals only, so clearly carbon's not happy and the halogens are unperturbed more or less, so that lends some support that it's carbon struggling to hybridise over the increasingly broader energy ranges, but doesn't exactly confirm it either.

Anyhow, that's what I get out if, don't know if I've interpreted it correctly!

There's still plenty of scope here for 'the answer' lol

[Edited on 1-11-2013 by deltaH]

arsphenamine - 2-11-2013 at 08:41

1st and 2nd row elements in the same column are more extreme in their chemical properties, and the 3rd,4th row ones are more alike than they are to their lower row brethren. 3rd row is where it grows computationally long at the ab initio level.

The Novak paper is from 1989. The semi-empirical methods gave "qualitatively" encouraging results in a reasonable time, unlike ab initio methods, which take forever and may not converge if you don't set them up without a lot of tweakage. Their mention of Clementi (first plausible ab init estimate of NH3 inversion barrier) may have been political.

The semi-empirical methods available then (AM1, FH,REX,EMT) are either abandoned or obsolete today. AM1 is still around to confirm antique calculations but is supplanted by PM6 and PM7. These latter ones are so fast on current hardware that the largest effort is drawing the molecules -- the PM7 runs take less than 10 seconds each.

Excuse me while I STFU and run the numbers.
CFClBrI, CHClBrI, CHFBrI, CHFClBr are the species of interest.

[Edited on 2-11-2013 by arsphenamine]

watson.fawkes - 2-11-2013 at 09:23

Quote: Originally posted by bfesser  
I was probed for a speculative opinion.
You can always say "I would prefer not to".

bfesser - 2-11-2013 at 09:29

But woelen asked so nicely; and making assertions without at least attempting to back them up is foolish.

arsphenamine, thank you. I understood a little of that. ;) I posted that link because I was in a hurry and the title sounded interesting and possibly relevant. :P

arsphenamine - 2-11-2013 at 10:18

Quote: Originally posted by bfesser  
arsphenamine, thank you. I understood a little of that. ;) I posted that link because I was in a hurry and the title sounded interesting and possibly relevant. :P
Thanks right back for kicking off an interesting discussion. The modern PM7 method gives results similar to the paper of Watson-Fawkes mention, although the older methods overestimate the C-H stretch energy at 3000 cm-1 (PM7 puts it at ~2600).

deltaH - 2-11-2013 at 12:29

arsphenamine, what was the goal of your modelling, to verify/improve upon the accuracy of the previous results or did you have something else in mind?

[Edited on 2-11-2013 by deltaH]

arsphenamine - 2-11-2013 at 17:55

Quote: Originally posted by deltaH  
arsphenamine, what was the goal of your modelling, to verify/improve upon the accuracy of the previous results or did you have something else in mind?[Edited on 2-11-2013 by deltaH]
Verification was one goal, but figuring out Novak's methodology was more important, answering questions about how they assign eigenvalues to specific orbitals.

I wanted to sharpen my general skills at using MOPAC since calculation of equilibrium geometry, vibration modes, and thermodynamic properties can't be chained the way they can in GAMESS-US.

OpenBabel has a few code defects when extracting equilibrium geometry from a MOPAC output file, and MOPAC has a code defect that emits defective Z-matrix coordinate outputs; obviously, some hand rolling in data extraction is essential here.

[Nota Bean: I do not mean to savage either OpenBabel or MOPAC; both are outstanding applications that have a very obscure set of bugs when used together. The code maintainers have reproduced the defects and the repairs will be in the next releases.]