How does the number of hyperconjugates affect the $\ce{C=C}$ bond length in the compounds propene and 3,3-dimethylbut-1-ene?


In general, is bond length directly or inversely proportional to the number of hyperconjugates?

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    $\begingroup$ Have a look here for a detailed explanation on how hyperconjugation works on a frontier-orbital level. In propene and 3,3-dimethylbut-1-ene you have 3 bonds that can take part in hyperconjugation: in propene it's the 3 $\ce{C-H}$ bonds and in the other compound it's the 3 $\ce{C-C}$ bonds. So, the premise of your question seems to be wrong: The number of "hyperconjugates" is the same, only their types differ. Concentrating on the type, I would say the $\ce{C-C}$ bonds should be a bit more effective at hyperconjugation than the $\ce{C-H}$ bonds. $\endgroup$ – Philipp Nov 22 '15 at 13:40

The higher the number of alpha hydrogens present in a molecule, the more resonance there will be which will decrease the bond order which was earlier 2, thus decreasing its strength and increasing its length.

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    $\begingroup$ The answer is correct but could be explained a lot better. $\endgroup$ – Apoorv Potnis Aug 26 '16 at 17:15

It directly depends on the bond order. Because bond order=(number of covalent bonds in that molecule)/(number of resonance structures) Hence ,more the hyperconjugative structures less will be its bond order. As bond order is inversely proportional to bond length,less the bond order more the bond length. Therefore more hyperconjugative structures is directly proportional to bond length. Hence with greater no of alpha hydrogen they will have greater bond length. Hope this helps..


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