Which of the following is more stable ?

  1. $\ce{CH2=C(CH3)-CH=CH2}$

  2. $\ce{CH3-CH=CH-CH3}$

I was going for answer 2 because it has 6 hyper conjugative effects. I found the answer given to be 1, which has 3 hyper conjugations but more resonance structures. Will the R-effect meet the deficit of 3 hyper conjugations and make 1 more stable than 2?

  • 3
    $\begingroup$ I do not understand in which context should the one be more stable than the other. $\endgroup$ – Martin - マーチン Feb 25 '15 at 7:15
  • $\begingroup$ I dont get you, sorry can you elaborate ? $\endgroup$ – Dhruba Banerjee Feb 25 '15 at 7:17
  • 10
    $\begingroup$ Is it more stable towards hydration, oxidation, bromination, substitution, elimination, heating, pressure, etc, pp.? The context is missing. It is impossible to compare the stability of two molecules without knowing how to relate them. $\endgroup$ – Martin - マーチン Feb 25 '15 at 7:24
  • $\begingroup$ Overall stability (?) The question asks for stability only. And it is a verified question appeared in public exams in past years. $\endgroup$ – Dhruba Banerjee Feb 25 '15 at 7:33

We can hydrogenate unsaturated compounds and measure the heat given off, this is called the heat of hydrogenation. As the following figure illustrates, if we hydrogenate the various butene isomers they all wind up producing the same molecule, butane.

enter image description here

image source

By comparing their heats of hydrogenation we can see which butene was lowest in energy to begin with, that is, which butene is the most stable. The heats of hydrogenation reported in the figure tell us that trans-2-butene is more stable than 1-butene by about 2.3 kcal/mol. This extra stability can be attributed to the additional hyperconjugative resonance structures involving hydrogen present in trans-2-butene (6) compared to 1-butene (2).

Next let’s consider 1,3-butadiene. Here when we hydrogenate it, butane is again produced and we might estimate that it will have a heat of hydrogenation equal to twice that reported for 1-butene (1-butene is the best model because its double bond has the same substitution pattern as found in 1,3-butadiene) or 60.6 kcal/mol of heat given off. When we actually run the experiment we find that only 57.1 kcal/mol of heat is given off, 3.5 kcal/mol less than we estimated. In other words 1,3-butadiene is about 3.5 kcal/mol more stable than isolated double bonds and this is due to the resonance interaction between the two double bonds in 1,3-butadiene.

The resonance stabilization in 1,3-butadiene (3.5 kcal/mol) is larger than the hyperconjugative stabilization in trans-2-butene (2.3 kcal/mol).

Your question actually asked about 2-methyl-1,3-butadiene. In this case we have the resonance stabilization found in 1,3-butadiene plus some hyperconjugative stabilization from the additional methyl group. It will be even more stable than 1,3-butadiene. Further, 2-methyl-1,3-butadiene is more stable than any butene, so answer #1 is correct. Resonance stabilization is larger than hyperconjugative stabilization in this series of compounds.


Not the answer you're looking for? Browse other questions tagged or ask your own question.