destroyer reaction question #1
Started by 2PacClone23
just look at the picture and help me figure this bad boy out:
The first rxn is a radical mech on an alkane second one is a benzene so u would need to do EAS to get it on the benzene but Br2 and hv is still a radical mech rxn so it would attack the alkane (CH2CH3) on the benzene....if u want to add to the benzene u need either FeBr3 or FeCl3 (lewis acid catalyst) to get it on the benzene ring.
hope this helps 😀
What he said..except that the first reaction involves the benzene ring, while the second is a cycloalkane.
Benzene rings are relatively robust structures that don't react with just UV. You need something a little bit stronger. That is why the radical occurs on the ethyl branch, at the secondary carbon. The second reaction is pretty standard, with a tertiary carbon radical, like you mentioned.
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But why, in the first reaction, does it react with the secondary carbon and not the primary?
just look at the picture and help me figure this bad boy out:
It's trend is the same as a carbocation so it will attack the most stable radical and that's a tertiary carbon. If there is none then secondary. I believe that you can't form a radical on a benzene ring. I'm not sure but I have never seen a reaction with benzene ring with a radical before. Also, this reaction is for alkanes (only?)?
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First one attacks the BENZYLIC position. This is a favorite spot for radical substitution reactions!!! Remember that!
Second one just has to do with making the most stable radical intermediate (tertiary)
Second one just has to do with making the most stable radical intermediate (tertiary)
Okay so for benzene, it always attacks the benzylic position and not the tertiary carbon?
You also have to take into consideration the highly selective nature of bromine as compared to least selective nature of more electronegative halogens (Cl,F). Since we are dealing with a selective bromine radical you will see the most stable product(one that originates from the most stable intermediate) in this case a secondary benzylic radical. If instead of Bromine radical you gave me a chlorine radical you would most likely get a mixture of products as a result of chlorines high electronegativity and low selectively (beggars cant be choosers get it)just look at the picture and help me figure this bad boy out:
B. In this case you are dealing with a simple methylcyclopentane, since gain we are using a free-radical halogenation mechanism initiated with light with bromine as our radical we expect to yield the most stable product (again one that originates from the most stable free-radical intermediate) in this case a tertiary radical. Therefore the product is what we expect it to be a tertiary alkyl bromide.
Feel free to ask any questions 🙂
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