Stability order GOC

[royal rock]

this is the increasing order of the stability
i think i got the right stuff
i cant differentiate between the stability of phenyl radical thing and the second one from left

[undone parcel]

<@&1227988359910260737>

Note for OP

+solved @user to close the thread when your doubt is solved. Mention the user who helped you solve the doubt. This will be added to their stats.

[tidal quest]

ywp

nice

[royal rock]

thats cool

[royal rock]

(just a suggestion ping after 10 minutes of inactivity in a forum would be better imo )

[silver lion]

Perhaps we can deal it with Inductive effect...

[royal rock]

not sure how will that work

[silver lion]

Same...still finding some way...

[silver lion]

Well I think...look at the hybridization Sp2 carbon ki electronegativity is greater than sp3 carbon to us par free radical(which for ease works similar to a positive charge) is more unstable and first Wale compound mein on addition to that -I of double bond is destabilizing it even more...

[royal rock]

in both first two casses in which i have confusion in has the free radical on a sp2 carbon

they both are unstable but why is the left most the most unstable

[silver lion]

Mean?
Actually Electronegativity goes hand in hand with -I here I think baaki hamare sir ne to Aisa karne ko bola ha, as EN increase, stability of neg charge increase and positive charges become more unstable, free radical ko positive charge manke kar lete

[royal rock]

yeah thats ok and cool but i cant seem to set it here (one sec attaching image , im on slowmode)

[silver lion]

Sure...

[royal rock]

like can you explain using your logic
while describing each of these one by one
im not able to recoup the info ig

[shell nacelle]

Less alpha hydrogens less hyperconjugation

[royal rock]

yeah noticed that for some reason i was thinking resonance in the first one
doubts solved now thanks!

[silver lion]

I don't think here we need to consider +H effect...can't go through-I?

[shell nacelle]

H is a stronger effect

[silver lion]

2 mein hyperconjucation ho hi nhi rha as free radical is present in sp2 orbital and is in the plane of the ring

So...the only way is inductive dude...I guess.
Perhaps m wrong...but this is what makes sense to me.

[mossy oriole]

yo @royal rock is this one done?

[royal rock]

oh yeah my bad sorry

[mossy oriole]

+solved @shell nacelle