https://www.chemguide.co.uk/basicorg/bonding/benzene1.htmlReading the above will greatly help you think on the right track
Enthalpy change = Bonds broken - bonds made
When hydrogenating, you break:
1. pi bonds in C=C
2. H-H bonds
You form:
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.
Similarly,
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.