Orgo: Br2, hv ==> most stable radical?

[joonkimdds]

I learned that Br2, hv will swap Br with H from a carbon that will form the most stable radical.

In the first reaction, the one that I marked with green is tertiary carbon that will form the most stable radical, so that's where Br attacks.

In the 2nd reaction, the ones that I marked with pink are 2ndary carbons so they are more stable than the CH3 which is primary carbon so why doesn't Br attack one of the 5 2ndary carbons rather than CH3?

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[tranv117]

I learned that Br2, hv will swap Br with H from a carbon that will form the most stable radical.

In the first reaction, the one that I marked with green is tertiary carbon that will form the most stable radical, so that's where Br attacks.

In the 2nd reaction, the ones that I marked with pink are 2ndary carbons so they are more stable than the CH3 which is primary carbon so why doesn't Br attack one of the 5 2ndary carbons rather than CH3?

its a benzene ring, must be activated. ie. Lewis acid. So Br2/hv would not react with the ring. My test is tomorrow, so off to bed. Good luck and hope that answered your question.

 

[xaznxcountx]

Yeah the only way you are going to get that type of attack is if you use BR2/FEBR3 Catalyst. Hope that helps.

 

[chessxwizard]

Also, bromine addition to benzene rings is an aromatic electrophillic substitution reaction, while its addition via hv to alkanes is done by a radical mechanism. It would add to ortho/para positions if reacted with the benzene ring shown above with Lewis acid catalyst.

 

[joonkimdds]

Ah~ right right.

But then how about this one?
CH3-CH=CH2 + Cl2 with heat

the answer is Cl-CH2-CH=CH2

why can't it be CH3-CCl=CH2?
This one will have more stable radical because the one in the middle is secondary carbon.

 

[chessxwizard]

The first one is a primary allylic carbon, which is stabilized by resonance. You can't add it directly to a carbon that's part of a double bond. Might want to review your class notes... vinyl radical is a no-no.

 

[joonkimdds]

You can't add it directly to a carbon that's part of a double bond. Might want to review your class notes... vinyl radical is a no-no.

could that be the reason why I couldn't add Br to the aromatic structure because all the carbons in the ring are part of double bond?


If you don't mind, I found a similar question from my old note and don't know if it's right or wrong.

if Cl2 + UV light is added,

and the answer choices are
a)1-chloro-1-phenylpropane
b)2-chloro-1-phenylpropane
c)1-chloro-4-propyllbenzene
d)1-chloro-2-propylbenzene
e)5,6-dichloro-1-propyl-1,3-cyclohexadiene

since Cl cannot attack the carbon that is part of double bond, and it prefers to attack the one that forms the most stable carbon radical, I am guessing that it's gonna be attacking either pink or black carbon from the picture.
so I think answer is B.
am I right?

 

[chessxwizard]

It would attack the pink carbon, since the radical electron can be distributed about the ring (resonance). The reason why you can't add it to the ring is because it has the unusual stability characteristic of aromaticity. Only Br2 with a lewis acid catalyst will do it. The reason why you can't add it to that position in the alkene is because of the terrible radical stability. I think the answer would be A, 1-phenyl-1-chloropropane.

 

[joonkimdds]

Thank you very much~

 

[Sea of ASH]

also cl radicals attacks primary carbons

 

[joonkimdds]

also cl radicals attacks primary carbons

But then doesn't that mean

+ Cl2 with UV light
the Cl should attack the primary, the last carbon which I marked with blue color?

 

[klutzy1987]

The allylic and benzylic positions are very stable for NBS/br2/hv. The reactivity order is tertiary>allyl/benzyl>secondary>primary>methyl>vinyl>aryl.

 

[yakuza]

But then doesn't that mean

+ Cl2 with UV light
the Cl should attack the primary, the last carbon which I marked with blue color?

light/peroxide doesn't affect Cl2, so the Cl2 will attack the pink..I'm PRETTY SURE

Also never deprotonate or kill a double bond from benzene, that'll ruin its aromaticity.

 

[Sea of ASH]

UV light does affect cl2 but peroxide does not.

the reason why free radical bromination is tertiary is becuz the reaction is very slow. however free radical chlorination is very very fast, so you will get mostly primary!

 

[joonkimdds]

UV light does affect cl2 but peroxide does not.

the reason why free radical bromination is tertiary is becuz the reaction is very slow. however free radical chlorination is very very fast, so you will get mostly primary!

I read the same thing from the textbook.
It says Br is very selective and Cl is not.

But then shouldn't Cl attack the blue carbon?

 

[chessxwizard]

It depends on the reactivity as well as availability of hydrogens. In a compound that has alot 1* carbons, Cl would preferentially attack those because it is not as selective and there is an abundance of sites. However, when considering hydrogens that are present in relatively the same amount (1 hydrogen difference between the benzylic and 1* position), I would say that the # of hydrogens doesn't make that much of a difference, and it would still attack the benzylic carbon due to a much more stable mechanism pathway.