Question

Why do 3° free radical is more stable than 1°and 2° free radical?​

Answer 1

Answer:

1. Stability Of Free Radicals Increases In The Order Methyl < Primary < Secondary < Tertiary

Let’s talk a bit about stability first, and then circle back to their structure. Being electron deficient, you might already have a hunch regarding factors that might stabilize free radicals. Waaaay back, we talked about how a considerable portion of organic chemistry can be explained simply by understanding that: 1) opposite charges attract (and like charges repel), and 2) the stability of charges increases if it can be spread out over a greater volume. These still apply here!

Electron poor species are stabilized by neighboring atoms that can donate electron density. [“if you’re poor, it helps to have rich neighbors”]. The most common way to interpret “rich neighbors” here is the observation that increasing the number of alkyl groups on the carbon bearing the free radical increases its stability. Radical stability increases in the order methyl < primary < secondary < tertiary.

2. Free Radicals Are Stabilized by Delocalization (“Resonance”)

Secondly, we have also learned that any factor which can lead to the electron deficient site being delocalized [spread out] over a larger area will also stabilize electron poor species. Previously, for example, we’ve seen that the positive charge of a carbocation is considerably stabilized when it is adjacent to a π bond.

That’s because the carbocation is sp2 hybridized and bears an empty p orbital, allowing for overlap with the adjacent p orbitals and therefore leading the positive charge to be delocalized over multiple carbon atoms, in a manner that is most easily grasped by drawing resonance structures.

3. The Geometry of Free Radicals Is That Of A “Shallow Pyramid”, Which Allows For Overlap Of The Half-Filled p-Orbital With Adjacent Pi Bonds

If we draw out the electrons in a typical alkyl free radical, we see that there are three bonding pairs and a single unpaired electron, for a total of four occupied orbitals. By analogy to, say, amines, we might expect that the hybridization of the molecule to be sp3 and geometry of a free radical would be trigonal pyramidal. That’s actually a good approximation, except that the “pyramid” is a little shallower than it is for molecules which have a full lone pair.