Chemistry multiple choice question HELP!

What would I need to do to answer this?

https://imgur.com/vkJd7XC

I know that for an enthalpy to be most exothermic, it must release the most amount of energy from all 4 given options.

But I can't see how I would figure out which one to be the most exothermic?

Any help would be great. Thanks!

Use your understanding of the bonding in benzene. Alternating single and double bonds would form a resonance structure whereby the pi electrons are delocalised. The enthalpy of hydrogenation would be less exothermic as the bonds broken are stronger.

Why do you exactly mean when you mention 'resonance structure' I have never heard of that term before in chemistry?

The answer would be B because the -1,3- double bonds are closer than the -1,4- double bonds in A. This means the p orbitals would overlap, hence more energy is required to break the bonds.

Use your understanding of the bonding in benzene. Alternating single and double bonds would form a resonance structure whereby the pi electrons are delocalised. The enthalpy of hydrogenation would be less exothermic as the bonds broken are stronger.

But it is asking for the most exothermic enthalpy change

The answer would be A because the -1,3- double bonds in B are closer than the -1,4- double bonds in A. This means the p orbitals would overlap, hence more energy is required to break the bonds for B which is more stable. A is less stable so it releases more energy??

Is it not that when bonds are broken it is an endothermic process, because you have to supply some energy to overcome the attractive forces between the atoms. So if the bonds are stronger then, wouldn't it be more exothermic since stronger bonds release more energy than weaker bonds where they break down? Sorry, but I just can't see how a resonance structure will help identify if the enthalpy of hydrogenation is more exothermic or not?

Why would it be A in terms of the resonance structure?

too far apart to show as much delocalisation so it is less stable and less energy is required to break the bonds. The same amount of energy is required to form the C-H bonds in all the structures. This means A is more exothermic.

Energy is not required to form C-H bonds. Energy is released when C-H bonds form.

A does not form a resonance structure

B,C,D contain alternating single and double bonds and there will be delocalisation of the pi electrons. The bonds would be stronger and require more energy to break.

Okay so I have to find the value that is most negative as this would be the most exothermic enthalpy change.

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?

https://www.chemguide.co.uk/basicorg/bonding/benzene1.html

Reading 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.