[shubhamrawal]

Oxygen has lone pairs so it can stabilise the adjacent carbo cation but it can not stabilize adjacent carbon free radical because it has only four orbitals. Then why are we justifying by saying oxygen stabilises alpha carbon free radical in this case

[Babcock_Hall]

What stabilizes radicals?

[shubhamrawal]

radicals are stabilised by hyperconjugation,+I effect or if it is adjacent to a carbon with double bond(allylic and benzylic free radicals).

[Babcock_Hall]

What about primary, secondary, and tertiary radicals?  How does the degree of substitution affect radical stability?

[pgk]

It’s a question of the radical’s life time. To detail, hydroperoxide formation occurs via initiation, propagation and termination step, as being a radical reaction. β-Ether radical has a short life time and favors propagation. Contrary, α-ether radical has a longer life time due to the conjugation with the oxygen and therefore, it favors immediate termination of the radical reaction.
¨O=O¨ +  hv (light)  →   ∙O∙=∙O∙                   Initiation
C-H  +  ∙O∙=∙O∙  →  C∙  +  ∙O-OH                   Propagation
C∙  +  H-C   →    C-H  +   C∙                          Propagation
C∙  +  ∙OOH  → C-OOH                                 Termination
Example: di-(n-propyl) ether that both α- and β- radicals are secondary.
CH3CH∙CH2¨OCH2CH2CH3  (propagation) →  CH3CH2C∙H¨OCH2CH2CH3 ← → CH3CH2CH=O(+∙)CH2CH2CH3 (termination) → CH3CH2CH(OOH)OCH2CH2CH3
However, it must be noted that tertiary radicals have a longer life time compared with secondary and primary ones and that secondary radicals have a longer life time compared with primary ones.
The same question, as above: How does the degree of substitution affect radical stability?

[shubhamrawal]

i am only a 12th standard student, so i don't know anything about radical's lifetime. but i know that +I effect stabilises the free radicals. so, more the alkyl groups attached to the free radicals , more the stability. so, tertiary carbon free radical should be more stable than primary and secondary.
according to my tender understanding of the subject, ethereal oxygen cannot stabilise the alpha free radical because oxygen has got only 4 orbitals . if it makes a bond with the alpha carbon in the resonating structures it has no orbital to keep its one odd electron.
please clear the confusion. waiting for reply

[pgk]

A stable radical means a higher life time.
The stabilization of the α-radical by the etheral oxygen occurs by conjugation, as follows:
The carbon free radical electron and an oxygen electron form a C=O(∙+) double bond and formation of an oxygen cation radical, as shown above.
Try to draw the mechanism (For the moment, don't worry if you cannot).

[shubhamrawal]

if there is something else that i don't know then please let me know.

[pgk]

What is missing, is:
One electron of oxygen and the electron of the methyne free radical, are donated and form a double bond which is conjugated with the so formed oxonium free radical (oxonium = oxygen with positive charge, because of one electron deficiency). The said oxonium radical has 4 electrons in four bonding orbitals and one electron (the radical one) in an andibonding p orbital. The overall is stable because the bonding electrons outnumber the antibonding electrons (please, see the attached file).
Furthermore, the hydroperoxide radical that is formed during propagation step, binds with the double bond carbon and terminates the radical reaction by forming the ether peroxide in the α-position.

[shubhamrawal]

thanks for the reply.
so, in short this can be explained only by molecular orbital theory and not by valence bond theory.
thanks

[pgk]

Molecular orbital theory and valence bond theory are not competitive.
Bond valence between oxonium radical and the methyne radical are equilibrated, if the positive charge is taken into account.