(Original post by Yatayyat)
Okay so I have to find the value that is most negative as this would be the most exothermic enthalpy change.
Reading the above will greatly help you think on the right track
If a lot more of the energy used is done creating bonds then there is a lot less energy used breaking the bonds as this is the only way to get an exothermic reaction.
Does this mean that less energy is used up in breaking carbon double bonds? Why does it need to be C=C double bonds are not any other bond in the compound, can't it be single bonds instead?
Do I look for the compound that appears to be the least stable out of all 4 compounds then, as this need little energy to then re-stabilise?
I'm still confused to how resonance plays a big part in this?
Enthalpy change = Bonds broken - bonds made
When hydrogenating, you break:
1. pi bonds in C=C
2. H-H bonds
1. C-H bonds
You don't use energy to create bonds. When bonds form, energy is released. Energy is used to break bonds. It is as if stretching an elastic band. You have to input the force to break it, but it will release heat when contracting back (i.e heat is released when bonds are formed).
Look at A,B,C,D. They all have 2 pi bonds. So number of C-H bonds formed will be the same. You have to consider the strength of C=C bonds or C-C (bond is partial single and partial double (intermediate bond order)).
You do not need to complicate it.
Look at benzene again. Recall that the question asks you to consider bonding in benzene.
Benzene has a delocalised pi ring. Kekule structure is alternating single and double bonds. The pi electrons in p orbitals are delocalised, forming a pi ring. Reading the above will tell you that the bonds are stronger in the ring than simply static single and double bonds. This is due to the hydrogenation enthalpy being more endothermic than expected.
B, C, D have carbon atoms with alternating single and double bonds. The pi electrons in p orbitals are delocalised, strengthening the bond. Additional energy ("delocalisation energy") has to be provided in bond breaking. So the enthalpy change is more endothermic.
A does not have alternating single and double bonds.
Hence answer is A.