Are the C-C bond lengths different in 2-methyl propene When we were first taught resonance it was with the example of benzene and I myself justified it by thinking that the double bond oscillates because no two carbon atoms are special and hence there is no reason there will be a double bond between any two and not any other two. I now know that's not true. But just to avoid any latent confusion, can you verify my reasoning is wrong by answering this question? By my method, all three bonds will have same length and bond order 4/3
The cases of benzene and 2-methyl propene are different.

In benzene, all carbon atoms are equivalent. One can also imagine the electron resonance.

In case of $\ce{CH3-C(CH3)=CH2}$ , the electrons have no place to move to create a delocalization cloud.

Additionally,  it would have to switch a hydrogen atoms between 3 atoms,
or create highly unstable $\ce{C=CH3}$

Such molecule would be highly reactive.