rushang

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Jan 6, 2008
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Just want to clarify, isn't Br2 and Light give you anti-markinikov addition than why in this question, destroyer tells us that it prefers the 'c' attached to benzene rather than the last C in the chain ????

I thought, Br2 with UV LIGHT is a Free Radical Bromination, right ???
 

PreDent2009

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Dec 20, 2007
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Br2 and hv is a free-radical reaction that prefers the benzylic site> allyic> tertiary>secondary>primary>methane. It does not work according to the markovnikov addition. Only Br2 with peroxide, such as HOOH will provide you with the anti-markovnikov addition product. Since reactant is n-butylbenzene, and Br2 with light is a dot reaction, this is going to be a benzene side-chain reaction. It will not under any circumstances attack the ring. Only electrophiles do so, with the exception of some nucleophiles which can attack benzene if the benzene is stabalized with enough withdrawing groups. Hope this helps.
 

Zerconia2921

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Mar 24, 2008
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Just want to clarify, isn't Br2 and Light give you anti-markinikov addition than why in this question, destroyer tells us that it prefers the 'c' attached to benzene rather than the last C in the chain ????

I thought, Br2 with UV LIGHT is a Free Radical Bromination, right ???
The last C in the chain will form a 1 deg radical which is not as stable as the position on C1.