How do we arrange the following in increasing order of stability? The answer given is: $5<3<1<2<4$ (numbering according to their order in the image).

My Approach:

I can figure that in the second and the fourth structure, resonance stabilization will be applicable and the fourth structure will be more stable than the second.

But I am not able to figure out how do I arrange the remaining structures.

• Well, 1 isn't stabilised and 3 and 5 are less table because of stronger s character of orbitals with unpaired electron. – Mithoron Nov 16 '16 at 1:20
• Did this question just get asked again today? chemistry.stackexchange.com/questions/83360/… @Jan – Zhe Sep 29 '17 at 11:54
• @Zhe Yes it did, so I decided to give the dupe target a proper answer. – Jan Sep 29 '17 at 11:55

Radicals are electron-deficient compounds. Therefore, according to Bent’s rule a radical should always populate an atomic orbital with a maximised $\mathrm p$ contribution. Ignoring 2 and 4 for a second, the radical occupies an $\mathrm{sp^2}$ orbital in 3 and 5 while it occupies an $\mathrm{sp^3}$ orbital in 1. This already gives us $3\approx5<1$.

In a next step, let’s consider 2 and 4. These, as you noticed, are adjacent to a π bond, so the radical can occupy a $\mathrm p$ type orbital and participate in π resonance. This is the main way to differentiate 1, 2 and 4, since all three could essentially adopt a planar configuration with the radical confined to a $\mathrm p$ orbital — yet in 2 it can also resonance with the allyl system and in 4 it can resonate across two double bonds. Thus, we now intermediately have $3\approx5<1<2<4$.

All that remains is to determine the difference between 3 and 5, which is not trivial. In 3, you have a single double bond. This structure can distort a little bit, increasing the bond angle on the radical-bearing carbon towards something closer to $\mathrm{sp}$. The freely rotatable single bonds making up most of the ring allow this distortion. The phenyl radical can distort much less, since its geometry is basically already defined from the six-membered aromatic ring. Thus, we may assume 3 to be more stable than 5. Therefore: $5<3<1<2<4$.

Note:

: In fact, rehybridisation is also possible in 1, giving the radical a pure $\mathrm p$ orbital and making the arrangement of substituents planar around the radical centre.

• I hope with rehybridisation you mean distorting towards a trigonal planar coordination. – Martin - マーチン Sep 29 '17 at 10:45
• @Martin-マーチン Yes. – Jan Sep 29 '17 at 11:54
• How do you explain radicals are electron-deficient? And apply Bent's rule to them in this way? The formal charge of a radical is 0. Could you cite some sources to justify your claims? – Tan Yong Boon Jan 27 '18 at 9:51
• @TanYongBoon Very simple. Atoms desire to achieve an octet. A radical is a septet. Thus, an electron is missing; therefore it is electron-deficient. – Jan Jan 27 '18 at 17:46
• Electron deficiency of the radical clearly does not suggest that the lone electron should be held in orbitals with higher p character, by Bent's rule. Bent's rule looks at the electropositivity/electronegativity of the substituents, not the overall electron deficiency of the central atom (in this case, carbon). – Tan Yong Boon Jan 28 '18 at 2:25

While checking stability of Carbon free radical it is treated as positive charge but we don't see resonance or mesomeric effect because for these effects we have to break bonds Homolatically, which is possible only at high temperature. So we see only H effect or I effect. Now 5 and 3 are sp2 which are least stable and + I of sp3 carbon chain is more so 3 is more stable than 5.
In 4,2,1 we see H effect so 6,5,4 conjugated hydrogen atoms are there respectively in 4,2,1 so stability order is 4 > 2 > 1 > 3 > 5

• We should definitely not treat a radical like a positive charge — even though I will admit that they are both electron-deficient compounds and similar rules apply to both of them. The main difference is that positive charges should always be confined to p-type orbitals while radicals can also occupy $\mathrm{sp}^n$ hybrid orbitals. Also, radicals do benefit from mesomeric stabilisation where applicable; we don’t have to ‘break bonds’. – Jan Nov 16 '16 at 19:40
• @Jan Call me kleinlich, but you cannot put a positive charge into an orbital, you cannot confine a positive charge either. However, if an orbital is empty, it will be the one with the least s character. – Martin - マーチン Sep 29 '17 at 9:13