# Rearrangement of double bonds

Consider:

I found these rearrangements, but I could not understand what the mechanism is behind this.

When I searched on the Internet I found the Wikipedia page Di-pi-methane rearrangement.

This could be due to hydride shift.

In the first image, the pi-bond shifts on the carbon from which $\ce{H}$ is eliminated. As the pi-bond leaves the first carbon, it becomes positively charged due to lose of electron, and the $\ce{H}$ gets converted to $\ce{H^-}$(hydride ion) as it accepts electron. Therefore, it is called hydride shift. It also stabilises the molecule as the resulting molecule is conjugated. The same can be applied for second and third images.

This is for the second image. The third one you can do it yourself. All the resulting molecules are stabilised due to conjugation.

• This makes sense as far as the stability of the final product (except maybe the third example), but there still has to be something to overcome the initial activation energy. Does it need to happen at reflux? Is a catalyst needed? – NH. May 8 '17 at 16:58
• @NH. I think it mostly happens in presence of light or when heated. – user237650 May 8 '17 at 17:03
• No, no, no... Hydride shift does happen here but you need strong acid to reversibly protonate alkene - it's simple carbocation rearrangement. – Mithoron May 8 '17 at 21:29
• @Mithoron As no reagent was given, so this can also be a mechanism for it. But I don't understand your talk about carbocation. – user237650 May 9 '17 at 1:58

@Mesentery has already proposed a possible mechanism, and @CupC_56 has tried to explain why it might happen. I found @Cup_C56's points to be correct, but the explanation to be quite less than ideal, so to speak, so I'll try to provide an alternative and I hope somewhat more sensible explanation as to why this happens.

It is indeed due to the resonance-stabilization of the conjugated products formed due to the rearrangements you quote. Resonance-stabilized compounds have lower energy than non-resonance stabilized ones, but why is that? One of the possible explanations is that resonance "averages out" energy minima and maxima between the resonating parts of the compound. This "smoothed-out" compound usually has more symmetry - and in most cases, symmetrical configurations are more stable than asymmetrical ones if both are under comparable amounts of steric stress.

The rearrangement is to a conjugated diene because conjugation facilitates resonance - if the $\pi$-bonds are too far apart, the $\pi$-clouds cannot overlap, which is resonance (or electron delocalization over $\pi$-bonds). "Delocalization" quite literally means that the $\pi$-electrons, instead of being associated with a particular bond between 2 atoms of the molecule, are part of an extended $\pi$-cloud over the molecule (an extended molecular orbital formed by the overlap of partially- and fully-occupied molecular orbitals and unoccupied atomic orbitals (near the $\sigma$-bonds) - the premise of Molecular Orbital Theory - please do find the time to read about it!). The overlap probability and extent, which determines the overall stabilization, falls off very quickly with the separation between the overlapping orbitals - usually beyond one bond it is very weak, and beyond 3 it is negligible.

So now you realize why it is favorable for an organic diene system to undergo rearrangement to a conjugated system - the latter is stabilized by $\pi$-cloud overlap, or "resonance".

Are you familiar with delocalization of pi-bonds? Well, if not, i suggest you read about that as it will help you in understanding such mechanisms, and also various other reactions related to Benzene and its homologues.

All the structures which you have depicted in your diagram are conjugated compounds (alternate double and single bonds), and hence the delocalization of pi-bonds becomes fairly easy, as compared to normal unsaturated compounds. Delocalization is basically the breakage of pi-bond at one site, and shift of the electrons to another site (forming a double bond in the process at the same site) in the same molecule.

WHEN conjugated compounds exist, the delocalization is easier because the intermediate complexes formed are quite stable, and hence lower energy is required for the breaking of such pi-bonds. (I strongly suggest you read more about stability of conjugated compounds as it is too broad a topic to be mentioned here).

NOTE: Not all compounds in your diagram are conjugated ones. To be more specific, the product of each reaction is.

This is one more clue as to why conjugated compounds are more stable. These reactions obviously proceed to form such products which are more stable than the reactants.

• Delocalization is basically the breakage of pi-bond at one site, and shift of the electrons to another site - no it isn't. See here for details on this common misconception. – bon May 8 '17 at 17:12
• Well, im pretty sure that my explanation holds good for the said question. What i have defined for delocalization is accurate, and i strongly disagree with the answer provided in your link. Please do consult Morrison Boyd's book for Organic Chemistry to clear your doubts. – CupC_56 May 8 '17 at 17:22
• I don't think that anyone can claim that one is more wrong than the other, but I agree with @bon, as MOT provides a more sensible (and more in line with experimental observations) description of what we call "resonance". Also, I take strong offence to the implication that bon had any doubts - from his answer, it does not seem so. On the contrary, your answer is somewhat suspect - note intermediate complexes formed are quite stable, and hence lower energy is required for the breaking of such pi-bonds - should it not be that more energy is required for breaking stronger bonds, not less? – Tamoghna Chowdhury May 8 '17 at 19:45
• And I very much disagree with your definition of delocalization - it seems quite like the "lies-for-children" definitions that teachers are likely to give to younger students. See Wikipedia: In chemistry, delocalized electrons are electrons in a molecule, ion or solid metal that are not associated with a single atom or a covalent bond. Most definitely not what you say it is. – Tamoghna Chowdhury May 8 '17 at 19:49
• @TamoghnaChowdhury "intermediate complexes formed are quite stable, and hence lower energy is required for the breaking of such pi-bonds - should it not be that more energy is required for breaking stronger bonds, not less? –" Obviously, if it is going to a lower energy state, then the energy required for the breakage of pi-bonds is compensated because of the greater energy released during product formation. Is that that hard to comprehend? – CupC_56 May 9 '17 at 5:39