# Why is conjugation not possible if there is not present alternate $\pi$ molecules according to MO theory?

I was reading this page at Chemwiki; there they were discussing conjugated-$\pi$ bond.

In order to show that conjugation requires alternate double bonding, they cited the example of 1,3,6-heptatriene:

To be considered conjugated, two or more π bonds must be separated by only one single bond – in other words, there cannot be an intervening sp$^3$-hybridized carbon, because this would break up the overlapping system of parallel 2p$_z$ orbitals. In the compound below, for example, the C1-C2 and C3-C4 double bonds are conjugated, while the C6-C7 double bond is considered to be isolated, due to the effect of the sp$^3$-hybridized C5.

I've a problem with their explanation; firstly are they following MO theory or VB theory? While describing molecules with MO, they, suddenly from nowhere, used the concept of hybridisation. Why?

Can anyone explain only with he help of MO theory why, there needs to be alternate $\pi$ bonding in the molecule for conjugation to happen without use of hybeidisation?

• I have a feeling that you don't exactly understand what MO theory is so you may don't know what your asking for - there was no MO stuff in what you mention. – Mithoron Jan 2 '16 at 19:21
• I'll try and answer when I have access to my Mac. But in the meantime, your exaxt question is answered in organic chemistry by Warren, Clayden and Greeves... – NotEvans. Jan 2 '16 at 19:57
• I think some help will be to read "Orbitals are not real". Orbitals are mathematical constructs. What's more important is that MO and VB theory are not competing, but complimentary, approaches to modeling multi-electron wavefuntions. They emphasize different aspects of electron behavior and depending on what you want to know, the maths may be simpler for one over the other (or whatever approximations were needed). – Ben Norris Jan 2 '16 at 19:58
• And to second NotNicolaou, just about any organic chemistry textbook will answer this question using MO theory only. It has to do with accessible orbitals having appropriate symmetry. – Ben Norris Jan 2 '16 at 19:59
• @Mithoron: Are you saying MO is not required for conjugation? – user5764 Jan 3 '16 at 2:03