Ch 8 Exercise 77

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Ch 8 Exercise 77

Postby Ariana_Kermani_2G » Thu Jan 19, 2017 4:10 pm

8.77) Benzene is more stable and less reactive than would be predicted from its Kekule structures. Use the mean bond enthalpies in Table 8.7 to calculate the lowering in molar energy when resonance is allowed between the Kekule structures of benzene.

What is a Kekule structure and what effect does resonance have on Kekule structures?

Theresa Dinh 3F
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Re: Ch 8 Exercise 77

Postby Theresa Dinh 3F » Tue Jan 24, 2017 12:57 pm

According to an online source ( it is a Lewis structure in which bonded electron pairs in covalent bonds are shown as lines and avoids showing lone pairs. Resonance just makes the kekule structure show two different versions of benzene, which explains how resonance works.

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Re: Ch 8 Exercise 77

Postby Jenna_Hakel_2A » Tue Jan 24, 2017 1:05 pm

So, following up on that, in terms of question 77, when you use the bond enthalpies that show that 3 of the C-C bonds are double bonds and 3 are single bonds, you get a total energy of 2880 kJ; however, when you use the given values for bond energies given that all 6 of the bonds share the delocalized electrons (which is the C-C bond in the table connected with one line and one series of dots), you find that the total energy of the molecule is 3108 kJ. Therefore, it requires more energy to break the bonds when the electrons are delocalized than in each individual resonance structure, so it is more stable.

Alyssa Ishimoto 1A
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Re: Ch 8 Exercise 77

Postby Alyssa Ishimoto 1A » Tue Jan 24, 2017 1:09 pm

A kekule structure of benzene shows that benzene is a ring with alternating double and single bonds between the carbon atoms and single bonds between the hydrogen atom and carbon atom. Resonance causes the bonds between the carbon atoms to be single bonds 50% of the time and double bonds the other 50% of the time.

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