Free Radical Substitution vs Addition

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jdmsamurai

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• May 25, 2007

• #2

hmmm from my understanding

X2/UV will cause an addition reaction causing the alkene to become an alkane radical with the most stable radical formation (3 > 2 > 1 > meth) which leaks to anti-markinovs.

the only allylic H leaving I can think of is polymerization?

 

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bkim

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• May 25, 2007

• #3

jdmsamurai said:

hmmm from my understanding

X2/UV will cause an addition reaction causing the alkene to become an alkane radical with the most stable radical formation (3 > 2 > 1 > meth) which leaks to anti-markinovs.

the only allylic H leaving I can think of is polymerization?

Click to expand...

But doesn't the Br2/hv also give free radical substitution? Especially when there is a loose H such as the allylic one

 

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K

Kingora

• May 25, 2007

• #4

bkim said:

But doesn't the Br2/hv also give free radical substitution? Especially when there is a loose H such as the allylic one

Click to expand...

i think its purely an addition reaction. Substitution would come in when alkane reacting with highly rective group 8 element

 

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bkim

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• May 25, 2007

• #5

In destroyer Ohcem section # 57, the question asks what will happen when a heptene is reacted with NBS/ROOR Heat. The solution says that NBS will remove the allylic H and replace it with Br. Isn't NBS/ROOR heat the same effect as Br2 radical?

 

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laundry

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• May 25, 2007

• #6

bkim said:

In destroyer Ohcem section # 57, the question asks what will happen when a heptene is reacted with NBS/ROOR Heat. The solution says that NBS will remove the allylic H and replace it with Br. Isn't NBS/ROOR heat the same effect as Br2 radical?

Click to expand...

Here's what happens when NBS/ROOR reacts with ethene.

Cheers

 

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bkim

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• May 25, 2007

• #7

laundry said:

Here's what happens when NBS/ROOR reacts with ethene.

Cheers

Click to expand...

Hey laundry thanks for the pic 🙂
So I guess NBS radical is not the same is Br2 radical, since NBS is more willing to take away the allylic (or benzylic) H than breaking the double bond.

Will an NBS radical ever break a double bond? For example, if I have ethene, will an NBS break the double bond?

Also, will a Br2 radical give both substitution and addition when reacted with an alkene?

 

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laundry

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• May 25, 2007

• #8

[email protected] wrote you a great reply and the stupid SDN lost it....

Here's the sunday school version... NBS can't break a double bond because it has no Hydrogen to donate...In fact, pure Br2/hv won't do it either, unless it's followed by HBr, which will give the H and form a free radical Br.

I don't think there is such a thing as free radical substitution because the radical cleaves off an H...always...as far as my little brain knows....

cheers...still mad...😡

 

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bkim

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• May 25, 2007

• #9

ah...bad SDN!!! Sorry that they lost your previous reply 🙁

Thanks again laundry.

As for fre radical substitution, I was referring to the free radical halogenation - taking a H out and putting Br radical in, just like an NBS would do.

So if I react Br2 radical with an alkene, would it take out the allylic hydrogen as well and perform free radical halogenation? I see that it will break the double bond, but wouldn't it also want to take out an H (preferably the allylic H)and subtitute with Br radical?
Thanks!

 

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bkim

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• May 28, 2007

• #10

Or maybe there is a difference b/w Br2 radical and HBr radical?
Is it only HBr radical that will give break the double bond and not Br2(which will take out H)?

 

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bkim

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• May 28, 2007

• #11

ok here is my temporary conclusion:

Br2/hv will not break the double bond of alkene, but would take out the most loose H and substitute with Br (Free radical halogenation). The same applies to NBS/ROOR

HBr/ROOR(initiator) will cleave the double bond and add the Br in an anti-malkovnokov fashion.



Correct me if I am wrong

 

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