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AQA A2 Chem: Aromatic Compounds help!

In my GCP textbook it says "more energy must have been put in to break the bonds in benzene than would be needed to break the bonds in the Kekule structure".

Why is this?


The Kekule structure has a lower enthalpy change (-208 kJmol-1) than would be expected (-360kJmol-1).

Why would it be lower though? Is that because less energy is needed to break the bonds? Yet the enthalpy change is bond formation and breaking. Though less energy can't be needed because the benzene structure is more stable than the kekule structure, due to its delocalised ring of electrons.

Basically I've confused myself, so could someone just break this down for me and explain it? Thanks! :smile:
we talked about this at school and i think we concluded that the book was wrong ( teacher agreed) if that helps :smile:
Reply 2
Original post by Amber May
we talked about this at school and i think we concluded that the book was wrong ( teacher agreed) if that helps :smile:


Okay thanks, I'll look try and find another textbook and have a look in there as well then.
Benzene is less reactive than we would predict from the kekule structure. Its bonding is therefore stronger and requires more energy to break.

Are we talking about hydrogenation? If so, the same amount of energy is given out by forming 6 x C-H bonds in both cases to give the same energy:

Benzene (kekule) + 3H2 -----> Cyclohexane: Enthalpy change = 3(C=C)+3(H-H) broken + 6(C-H) formed = X - [6 x 413] = -360

Benzene (actual) + 3H2 ------> Cyclohexane: Enthalpy change = aromatic ring broken+3(H-H) + 6(C-H) formed = Y - [6 x 413] = -208 = less energy given out to make the same thing

This means that Y > X i.e. it takes more energy to break the aromatic ring than the 3 double bonds
(edited 9 years ago)
Original post by EierVonSatan
Benzene is less reactive than we would predict from the kekule structure. Its bonding is therefore stronger and requires more energy to break.

Are we talking about hydrogenation? If so, the same amount of energy is given out by forming 6 x C-H bonds in both cases to give the same energy:

Benzene (kekule) + 3H2 -----> Cyclohexane: Enthalpy change = 3(C=C)+3(H-H) broken + 6(C-H) formed = X - [6 x 413] = -360

Benzene (actual) + 3H2 ------> Cyclohexane: Enthalpy change = aromatic ring broken+3(H-H) + 6(C-H) formed = Y - [6 x 413] = -208 = less energy given out to make the same thing

This means that Y > X i.e. it takes more energy to break the aromatic ring than the 3 double bonds


To add to the above ...

... here's an animation on the hydrogenation of benzene showing how it works. Just hover around to get more info!

What's with the signature EVS?

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