Sorry for this silly doubt but why is CR2=CH2 alkene more stable than CHR=CHR alkene?

Both have same number of alpha hydrogens. I thought that the latter is more stable as in its trans isomer, the R groups are most seperated (least sterric hinderance) but in my book it is mentioned the other way.

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    $\begingroup$ I don't think this is universally true. $\endgroup$ Dec 17 '18 at 14:02
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    $\begingroup$ Perhaps because the sp2 CH2 in the first alkene is more electronegative than the CHR in the second one, so when electron density decreases more at the CR2 side of first alkene, it gets stabilized more by electronic releasing effect of the two R's as compared to CHR=CHR $\endgroup$ Dec 17 '18 at 14:13
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    $\begingroup$ I think that one reason is in the tags list you have chosen $\endgroup$
    – Alchimista
    Dec 17 '18 at 14:29
  • $\begingroup$ Do you mean to say that in the latter case, the flow of electrons of the two R groups interfere with each other during hyper conjugation? $\endgroup$
    – user600016
    Dec 17 '18 at 16:10
  • $\begingroup$ I think That's true only if the -R is an alkyl group without any other substituents and if -R is not a tert- group. $\endgroup$
    – Carrick
    Dec 27 '18 at 8:17

Unfortunatly, I can not comment yet, so maybe someone can move this to the comment section.

I don't think, this statement is generally true. It depends on the inductiv and mesomeric effects of the side groups (+-I/+-M).

You also have to choose, what you mean with stability. The energy that is needed to break the doublebond or the reactivity of the double bond. If a doublebond has four substituents (which are not Hydrogen of course) any reaction on the double-bound is stericly hindered. When there are still Hydrogens on it, the doublebond might be more likly to react. That doesn't mean, that it requires less energy to break the bond.


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