Triflic Acid
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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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DraconicAcid
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No. The terminal ends are the least favourable positions to brominate.
Please remember: "Filtrate" is not a verb.
Write up your lab reports the way your instructor wants them, not the way your ex-instructor wants them.
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Triflic Acid
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Any catalyst that would prevent a radical from forming on the secondary carbon, an inhibator?
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ArbuzToWoda
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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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Triflic Acid
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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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DraconicAcid
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Any free radical initiator is going to form the more stable secondary radical.
Please remember: "Filtrate" is not a verb.
Write up your lab reports the way your instructor wants them, not the way your ex-instructor wants them.
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Triflic Acid
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So FeBr3 won't work
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DraconicAcid
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FeBr3 will only catalyze electrophilic brominations, not free radical ones.
Please remember: "Filtrate" is not a verb.
Write up your lab reports the way your instructor wants them, not the way your ex-instructor wants them.
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Triflic Acid
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Ok I thought so
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Dr.Bob
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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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clearly_not_atara
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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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Cou
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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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Triflic Acid
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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
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j_sum1
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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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clearly_not_atara
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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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Triflic Acid
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@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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DraconicAcid
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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.
Please remember: "Filtrate" is not a verb.
Write up your lab reports the way your instructor wants them, not the way your ex-instructor wants them.
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Triflic Acid
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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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DraconicAcid
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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.
Please remember: "Filtrate" is not a verb.
Write up your lab reports the way your instructor wants them, not the way your ex-instructor wants them.
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unionised
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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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clearly_not_atara
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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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Texium
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