Why do radicals attack polyunsaturated fatty acids?

AI Thread Summary
The discussion centers on the reactivity of unsaturated fatty acids with radicals, particularly focusing on the role of double bonds. Unsaturated fatty acids are more susceptible to radical reactions due to the presence of double bonds, which have higher electron density and allow for easier overlap with radicals. This makes them prime targets for reactions, as radicals can effectively "steal" electrons from these double bonds. The mechanism of addition reactions is highlighted, explaining how radicals like bromine (Br2) dissociate into two Br atoms, which then react with the double bond. The bond enthalpy of the C=C double bond is lower than that of C-C single bonds, facilitating the reaction. The process results in the formation of a carbocation and the completion of the addition reaction when the second Br atom bonds with the remaining unpaired electron. A diagram illustrating this mechanism was also requested for clarity.
sameeralord
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Is it because they are unsaturated and can unfold. I understand how one radical would create another radical but I don't know exactly how radical could spread in a fatty acid chain. Any diagram would be useful. Thank you :smile:
 
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The double bonds in unsaturated fattly acids are especially vulnerable to reaction with radicals while radicals react much more slowly with aliphatic hydrocarbons (hydrocarbons with no double bonds).
 
Ygggdrasil said:
The double bonds in unsaturated fattly acids are especially vulnerable to reaction with radicals while radicals react much more slowly with aliphatic hydrocarbons (hydrocarbons with no double bonds).

Thank you for the reply but you are surely capable of telling me why that is so :smile: I read theory about hybridisation so you can use that if you want. Thanks :smile: Is this something to do with a pi bond?
 
I think this can be sufficiently explained through the mechanism of addition reactions.

Let's say Br2 dissociates to two electrically neutral Br atoms. Notice these are radicals because of the unpaired electron on each.

Since the double bond has a paticularily high electron density, it is a prime target for radicals, as orbitals can overlap and form covalent bonds. Because of electron-electron repulsion, "stealing" on of these electrons is much easier than in a C-C single bond. Simply put the bond enthalpy of the C-Br bond is lower than the bond enthalpy of the C=C double bond, so the Br radical has little trouble stealing one of the electrons from the C=C double bond. Then the Br- ion bonds to one of the carbons from the double bond through attraction on opposite charges (stealing the electron makes the acid a carbokation). The other karbon from the double bond now has an unpaired electron (see fig), and the other Br atom from the Br2 molecule can bond with it and complete the addition reaction.

http://img408.imageshack.us/img408/73/addisjon.png
 
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