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Post by Mr Kwok on Feb 27, 2009 2:18:06 GMT -5
Hi everyone, You have learnt that free radical substitution mechanism occurs when a halogen is added to an alkane molecule in the presence of UV light. In that reaction, it is a replacement/substitution of a H from a C-H of alkane with a halogen. Curiously, do you think such reactions occur for alkenes and the aromatic rings (e.g. benzene)? If you do, why? If you don't, why? Thanks for sharing!  Regards Mr Kwok |
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Post by honcheong on Feb 27, 2009 8:26:24 GMT -5
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Post by belinda on Feb 28, 2009 0:53:47 GMT -5
For alkenes the double bond(s) present are electron rich. It will be attacked by halogens. Since the attraction is so strong, the halogen X2 will be added across the electron rich double bond instead of substituting just the H of the C-H bond.
I'm not sure if the electrons from the double bond of alkenes can be delocalised between all the carbons linked to it.. anyone knows?
For aromatic rings, the whole ring is electron rich. However, all the electrons are delocalised in the benzene ring. This will make the benzene less attracted to a free radical and it is less reactive. Therefore, I guess free radical substitution can take place in benzene rings.
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Post by tracetan on Mar 1, 2009 5:03:51 GMT -5
in free radical substitution, it is a replacement of a H from a C-H alkane with a halogen through the breaking of the C-H bond.
i agree that alkenes can undergo such reactions because the C-H bond in the alkene can still undergo substitution reactions despite the C is double bonded t another C. the C-H bond can still cleave homogeneously.
however, in benzene rings, because the C atoms all posses an overlapping p orbital that contains delocalised e, it is very stable. higher energy is needed to overcome the C-H bond in benzene rings. also, benzene undergoes electrophilic substitution, not free radical substitution.
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Post by seahyimin on Mar 1, 2009 5:27:57 GMT -5
honcheong: i think alkenes undergo electrophilic addition not free radical addition lol ;P belinda: i dont think electrons in the double bonds can be delocalised into the other carbons through the single bonds. i think the delocalisation of electrons from the halogen into the ring of alkyl halides is a special case @topic: i think free-rad substitution occurs at the sp3 carbons of the alkene but not at the sp2 carbons. As for aromatic rings, i think it occurs because Mr Kwok said that it in class(therefore cannot use free rad subst to test for phenyl but instead have to do nitration test). I guess it's because the reactive radical can still attack the benzene-H bond. hmmm correct me if im wrong, my memory isnt the best, for the best is yet to be...
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Post by Mr Kwok on Mar 1, 2009 6:33:14 GMT -5
Hallo everyone, now my turn to add some thoughts.
Well free radical additions do occur for alkenes, but that is something we do not study at the A level syllabus.
@tracie: If you suggested that free radical substitution reactions do occur for a C-H where the C is a sp2 carbon, then why is it that we do not see it happening?
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Post by davidlim on Mar 1, 2009 11:50:19 GMT -5
Free Radical Addition looks interesting. seahyimin Whether alkenes undergo electrophilic addition or free-radical addition depends on the conditions given I guess. belinda There's delocalisation by resonance but that's for a C* attached to one of the C in a C=C bond. Not the whole carbon chain gets their electrons delocalised here. My two cents: Free-Radical Substitution can occur for both alkenes and aromatic rings. Alkenes - For propene and above, there are saturated R-groups(no C=C bonds) so electrophilic addition reactions cannot take place there (C cannot expand it's octet anyway). On these R-groups, substitution reactions like free-radical substitution can take place as with alkanes. As for the C=C in the alkene molecules, free-radical substitution can take place but it requires a larger amount of activation energy than electrophilic addition so free-radical substitution is not favoured(though it can happen). Arenes - Free-radical substitution can happen at the R substituents as usual. It can also happen at the benzene ring itself because in the end it does not affect the ring of delocalised electrons. It also fits into the "Non-polar C-H bond broken, H substituted by halogen" classification of Free-radical substitution. Lim Boon Kiang David, 2SB2 (2CH203) |
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Post by Mr Kwok on Mar 2, 2009 0:48:27 GMT -5
Hi everyone,
I think I will wrap up this discussion. I must admit that I have done that because David has somewhat hit the jackpot.
In theory (I stress theory), free radical substitution can happens for all sort of C-H bonds, whether the C belongs to a sp3 carbon, arene's carbon, sp2 carbon or sp carbon.
At the A level syllabus, we learn that radicals are generated when UV light is injected to a halogen molecule, breaking them into radicals.
For the alkane molecules, the lack of reactivity since C-H bonds and C-C bonds are non-polar hence we get free radical substitution.
For alkene. The reactive C=C (it is exposed so it is reactive), hence electrophilic addition happens instead. Since this reaction has a lower Ea than free radical substitution, which makes the former happen more readily than the latter.
For arenes, the benzene pi electron ring is the one that promotes electrophilic substitution as this happens more readily than free radical substitution. - This reaction probably has a lower Ea than free radical substitution, hence the the observation that electrophilic substitution happens instead of free radical.
For the alkyl substituent on arenes, free radical substitution can happen because they contain only C-H and C-C bonds which are non-polar. There is no pi-electron cloud or polar bonds that will promote other sort of reaction.
Therefore, it is the appearance of the molecules - whether it has polar bonds, pi electron clouds, that determine what sort of reactions they have.
But for the convenience of teaching at the A levels, we usually confine it to specific reactions to make things less complicated. - Hence, for example free radical substitution do not occur for C-H where the C is the double bond's carbon.
Therefore, this goes to further reiterate that free radical substitution is a poor reaction, doesn't quite discriminate the different compounds and produces tons of unwanted side product. For synthesis questions, they are used only when we have NO choice.
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grace
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Post by grace on Oct 6, 2013 9:25:19 GMT -5
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grace
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grace
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grace
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Post by grace on Jan 10, 2014 10:28:20 GMT -5
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grace
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Post by grace on Jan 25, 2014 13:46:16 GMT -5
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