# Rank the following radicals in order of decreasing stability

Question

Rank the following radicals in order of decreasing stability

Aromaticity makes a cyclic compund more stable. Here, 1 and 3 are aromatic since they follow Huckels rule(Or am I wrong here?). So, I thought maybe 3 is most stable followed by 1, then followed by 2 and 4(I dont know which is more stable out of 2 and 4). But, that's not the case. 3 is most stable, but the next stable one is 2 and not 1. I don't get how.

Here, 1 and 3 are aromatic since they follow Huckels rule(Or am I wrong here?

Yes you're wrong over here, for applying the Huckel's rule the compound must be a closed circuit of delocalized electrons. In both the cases the circuits are not closed

If you're not aware of Conjugation then give that a read .
Now The most stable compound is 3 since the extent of conjugation is the greatest i.e 2 pi bonds.

3 is more stable than 4. Even though 4 is aromatic it is an aryl free radical the p orbital containing the radical is out of the plane and can't be stabilized through resonance/Hyperconjugation.
In fact I would say that option 4 is the least stable compound .

The second most stable compound is is 2 since it is stabilized via resonance which is more effective than Hyperconjugation; 1 is stabilized by Hyperconjugation.

3 is more stable than 2 since the extent of conjugation is greater.

The order of stability will be
$$\mathrm{3 > 2 > 1 > 4}$$

• Sorry, but I didn't get what closed circuit means. Would you please explain? I would be really grateful! – PolarBear Mar 17 '18 at 13:30
• I meant that the loop should be closed see this PDF for more details – Avnish Kabaj Mar 17 '18 at 13:33
• So they are open. How? I mean, what is exactly being closed. I am really sorry for bothering, but since I am still at basics, I am confused at petty things as well. – PolarBear Mar 17 '18 at 13:34
• For the flow of delocalized $\pi$ electrons to take place a complete loop must be the compounds 2 and 4 do not have complete loops you'll notice that the carbont atoms are sp3 hybridized, they are all $\sigma$ bonded . No $\pi$ orbitals are present for the flow of electrons – Avnish Kabaj Mar 17 '18 at 13:37
• Oh, now I get it. I knew about that fact, but just didn't get what loop you meant. Thanks! – PolarBear Mar 17 '18 at 13:39