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Topic: NBS in CCl4  (Read 25362 times)

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Offline rleung

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NBS in CCl4
« on: August 14, 2005, 10:32:26 PM »
Hi,

I am confused about the product(s) of NBS/CCl4 reactions.  When you react trans-2-pentene with NBS/CCl4, it says the products should be trans-4-bromopent-2-ene and cis-4-bromopent-2-ene.  Why wouldn't you also get 1-bromopent-2ene and 3-bromopent-1-ene?  It seems to me that the resonance forms of the cations would account for those two structures as well.  Also, why would the structure invert to a cis formation?  The bromination occurs at an achiral carbon, doesn't it?  It does not involve the double bond at all.

Also, I do not understand why reacting methylbenzene with NBS/CCl4 would yield only (bromomethyl)-benzene.  Wouldn't the reonance-stabilized + charge delocalized across the benzene ring also account for the following other two structures below:
« Last Edit: August 14, 2005, 10:34:30 PM by rleung »

laotree

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Re:NBS in CCl4
« Reply #1 on: August 15, 2005, 07:04:07 AM »
Your reaction condition favors the free radical mechanism. Free radical, like carboncation, could be stable termodynamically. If you added AIBN into you reaction, and heated to 60oC. It could be a totally free radical reaction.  Movies will let you know the full answer.

Offline movies

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Re:NBS in CCl4
« Reply #2 on: August 15, 2005, 12:29:12 PM »
Yes, bromination with NBS goes through a radical mechanism.  Can you figure out what the steps are?  First think about what the weakest bond is.

The cis/trans change is due to one of the reaction intermediates.

For the methylbenzene case, think about what the lowest energy resonance structure will be and why it will be so low in energy.

The Good Doctor

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Re:NBS in CCl4
« Reply #3 on: August 16, 2005, 02:22:47 PM »
Hi!

Note first like movies allready told that NBS/CCl4 goes via a radicalar mechanism (i hope you see that the goal of N-bromo-succinimide is to keep the concentration of your bromine in your reaction low to supress addition to the dubble bound who is the bad dude side reaction)..anyway. First of all if u have to position your bromine first ask yourself wich allylic radical is the most stable, in this case it's 4-position due to hyperconjugative effect (the more branched, the more stable is the radical). So in the both possible positions the one intermediate radical is lower energy than the other resulting to position 4 as favorated and faster product. If u have more troubles about that check up Hammond's postulate...u'll see what i mean.
For the rotation like allready told by movies and Co. it's obvious that during radical intermediates the 2-3 double bound disconnect...and rotation occurs...giving the cis-isomer...
cheers. ;)

Offline movies

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Re:NBS in CCl4
« Reply #4 on: August 16, 2005, 04:18:19 PM »
Here, I drew out a detailed mechanism for after the answer had be posted.  I hope this helps.  Remember the three parts of every radical mechanism: initiation, propagation, and termination!

Offline carambar

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Re: NBS in CCl4
« Reply #5 on: November 19, 2006, 05:56:10 AM »
Hello,
Sorry I m not really confident with radical mechanism so it might be a stupid question :-[:
Is a reaction between two Br radical leading to Br2 is possible, and if not, why?

Offline Ψ×Ψ

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Re: NBS in CCl4
« Reply #6 on: November 19, 2006, 04:44:45 PM »
Hello,
Sorry I m not really confident with radical mechanism so it might be a stupid question :-[:
Is a reaction between two Br radical leading to Br2 is possible, and if not, why?

Recombination of radicals is definitely possible, but whether or not it actually occurs depends on the concentration of the radicals...this is related to the concentration of the initiator (if one is used).

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