0
$\begingroup$

enter image description here According to this table, more heat is released in breaking a single double bond in 1,3-butadiene than in trans-2-butene. Does this mean that the resonance of 1,3 butadiene leads to a destabilization of the compound?

$\endgroup$
  • 2
    $\begingroup$ Your comparison is not correct. You are comparing two molecules with difference in internal bonding. You will not get trans-2-butene by removing 1 double bond from 1,3-butadiene. You should have compared 1-butene and 1,3-butadiene. Theoretically, addition of one double bond to 1-butene should give 1,3-butadiene, i.e. $\Delta H^0_{theo} = -60.6 $ kcal/mol. But practically it is lesser ($-57 $ kcal/mol) due to its higher stability due to conjugation. $\endgroup$ – Soumik Das Nov 11 '18 at 6:08
2
$\begingroup$

None of the comparisons you might make suggests any sort of deatabilization by conjugation.

Let's look at the four-carbon compounds. Depending on which source you believe for 1,3-butadiene, you could release up to 27.8 kcal/mol by hydrogenating one double bond (the figures given for this compound refer to hydrogenating both double bonds). But then you release 30.3 kcal/mol hydrogenating the remaining double bond in 1-butene. The combination of two double bonds in 1,3-butadiene has less energy to be released by hydrogenation than two independent bonds like the bond in 1-butene has.

Similarly for the five-carbon compounds. When you hydrogenated both double bonds in the pentadiene isomers, a pair of conjugated double bonds in 1,3-cyclopentadiene has less energy to release than the non-conjugated double bonds in 1,4-cyclopentadiene.

Finally, something a little more subtle is going on between 1-butene and 2-butene. There can no conjugation like that in 1,3-butadiene when only one double bond is present, yet 2-butene has clearly less energy to release by hydrogenation than 1-butene. Thinking of carbocation nomenclature, the former is a like a "secondary carbication by having an alkyl group on only one end of the double bonds, while the latter is "tertiary" with an alkyl group on both ends. Hyperconjugation, like that in carbocations, is at work here.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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