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Author: Subject: Free radical bromination on only the terminal ends of hexane
Triflic Acid
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[*] posted on 13-11-2020 at 13:04
Free radical bromination on only the terminal ends of hexane


I am trying to synthesize some 1,6 dibromohexane. I was planning on doing a UV free radical halogenation with bromine, but this will give me a mix of isomers. Is there any way to prevent multiple isomers from forming, only brominating the terminal ends of the alkane?
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[*] posted on 13-11-2020 at 13:12


No. The terminal ends are the least favourable positions to brominate.



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[*] posted on 13-11-2020 at 14:29


Any catalyst that would prevent a radical from forming on the secondary carbon, an inhibator?
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[*] posted on 13-11-2020 at 14:37


I don't think it's that simple. According to the Markovnikov's rule the secondary carbons would love to get brominated, you'd essentialy need to turn the reaction upside down. It may be possible, but I have no knowledge of such inhibitors that would be easier to acquire than just 1,6-hexanediol or 1,6-hexanediamine that can be turned into your desired compound with good yields. The diol can be prepared by hydrogenating adipic acid, unluckily a strong reducing agent (like LAH) is needed. I haven't dug very deep into the borohydride modifications, maybe the nickel boride route could be of use? That remains to be researched.
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[*] posted on 13-11-2020 at 14:38


Is there any other way that I could selectivly brominate the terminal ends of the alkane, FeBr3 comes to mind, but I think that only works for aromatic compounds. A bulky brominating agent might favor the ends over the middle?
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[*] posted on 13-11-2020 at 15:13


Any free radical initiator is going to form the more stable secondary radical.



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[*] posted on 13-11-2020 at 19:02


So FeBr3 won't work
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[*] posted on 13-11-2020 at 19:10


FeBr3 will only catalyze electrophilic brominations, not free radical ones.



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[*] posted on 14-11-2020 at 12:07


Ok I thought so
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[*] posted on 14-11-2020 at 17:13


You could get adipic acid and reduce it to the hexane-1,6-diol, then convert that to the 1,6-dibromohexane. I have done something like that with butane-1,4-diol before. That would be possible, although it might have some other isomers form from dehydrohalogenation and then reaction with bromine, but that should be manageable. Adipic acid and the hexane diol are both readily available, so those would be practical to find.

But halogenating hexane will create a mixture of many things, I suspect.
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[*] posted on 14-11-2020 at 18:38


Hm... Wurtz coupling of allyl bromide and anti-Markovnikov dihydrobromination of the 1,5-hexadiene with HBr in benzene?



[Edited on 04-20-1969 by clearly_not_atara]
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[*] posted on 14-11-2020 at 21:45


it would be cool if there was a way to selectively brominate only the terminal primary ends of an alkane. I wish I knew how to discover new reactions, but I just don't know how. i don't know how to be the next grignard, or the next suzuki, or the next sharpless.

But it's not really necessary, because several dicarboxylic acids of differing chain lengths are readily available.




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[*] posted on 14-11-2020 at 22:11


I amn't looking at this from a home-chemist point of view, I need to get this directly from hexane for a separate research. So, while helpful, suggestions using chemicals other than hexane are not going to help me. Also, just a thought, but out of the isomers that I looked at, the 1,6 has the highest boiling point. So, maybe by keeping the temperature just below the boiling point of the 1,6 isomer, I could have all the other isomers boil out, exchange bromines with other molecules until they happened to become 1,6, and then drip back down into the flask and stay as a 1,6. Then again, this is probably complete madness:P
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[*] posted on 14-11-2020 at 23:47


I am out of my depth with this question, but I thought it worth asking...
Is there a route beginning with cyclohexane and cracking open the ring?
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[*] posted on 15-11-2020 at 08:55


TA: if you brominate at random the chance of getting 1,6 is one in 15, a 6.7% yield before you even consider over and under bromination. It's statistically impossible. Free radicals are going to form where they're stable not where you asked them nicely to.

There is really only one way to go from hexane to 1,6-dibromohexane: you dibrominate at mostly 2/3/4/5, dehydrohalogenate to 2,4-hexadiene (Zaitsev favored), free radical allylic brominate to 1,6-dibromo-2,4-hexadiene, and selectively hydrogenate the alkenes. This is not feasible for a makeshift lab and not practical for a professional.

So if you want to be stubborn, here's what you get: you won't complete this project. Approaches from suberic acid or allyl alcohol or cyclohexene might actually work.




[Edited on 04-20-1969 by clearly_not_atara]
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[*] posted on 15-11-2020 at 10:12


@clearly_not_atara Is there any way to create free radicals on carbons without using a halogen in the gas phase. Something that would break C-Br bonds indiscriminately to create a bunch of free radicals. Messy, but then with a bunch of radical chains floating around in the gas phase, a Br* and a long carbon radical might eventually form in just the right way to create the 1,6 dibromohexane. Since that boils at a higher temperature than the other isomers, it would liquify and drip back down into the flask. Would it work?
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[*] posted on 15-11-2020 at 10:21


If there was any way you could get the bromoalkanes to rearrange, they would rearrange to secondary bromoalkanes, because of the stability of the secondary radicals. You may as well be asking what kind of wand would be best to wave over the flask to get the reaction to work.



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[*] posted on 15-11-2020 at 10:32


No, but by causing the 1,6 to go back into the wouldn't I be removing the 1,6 isomer and stopping it from rearranging further. So I would be enriching that 6.7% of 1,6 to get near complete 1,6 since the other isomers would constantly be getting rearagned until they formed 1,6, then they would go back into the flask and remain unchanged
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[*] posted on 15-11-2020 at 11:07


No, because the boiling points are probably not that different, and you would get ideal solutions in the liquid phase (meaning you'd have the other isomers contaminating the liquid, even if you held the temperature below the bp of the isomer you want and above the bp of everything else).

Also, that 6.7% is calculated based on random halogenation. Bromination isn't random, it strongly favours secondary bromination.




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[*] posted on 15-11-2020 at 11:48


Theoretically, imagine you mixed 2 moles of HBr and a mole of hexane and heated it until the HBr started to decompose.
You would get a tiny yield of 1,6 dibromohexane, and a lot of other stuff.
If you careful;y ran that through a prep scale chromatography system you could isolate the stuff you wanted.

And then you could capture the by-products and heat them up again.
Once more you would get a tiny yield of the stuff you wanted and you could separate that.
In principle, with a repeated sequence like this you might gat a yield that was merely "very bad" as opposed to tiny. (and you would also get some short chain stuff and some tar.)
So, it's hypothetically possible, but it would be insanely expensive.
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[*] posted on 15-11-2020 at 19:30


Quote: Originally posted by Triflic Acid  
@clearly_not_atara Is there any way to create free radicals on carbons without using a halogen in the gas phase. Something that would break C-Br bonds indiscriminately to create a bunch of free radicals. Messy, but then with a bunch of radical chains floating around in the gas phase, a Br* and a long carbon radical might eventually form in just the right way to create the 1,6 dibromohexane. Since that boils at a higher temperature than the other isomers, it would liquify and drip back down into the flask. Would it work?

Radical initiators only work for stable radicals. Thermal processes will probably give ethylene. Radicals can recombine to give oligomers. Photochemistry is theoretically possible (C-Br is the weakest bond in the molecule) but practically unlikely (the cross-section is negligible). And there is absolutely no rule anywhere that says each chain will end up with exactly two bromine atoms on it, as I pointed out before.

In short, no.

[Edited on 16-11-2020 by clearly_not_atara]




[Edited on 04-20-1969 by clearly_not_atara]
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