Question

Compare bond energy of c-f and c-h in hexafluoro benzene and benzene

Answer 1

Answer:

C=C (0.134 nm)

That would mean that the hexagon would be irregular if it had the Kekulé structure, with alternating shorter and longer sides. In real benzene all the bonds are exactly the same - intermediate in length between C-C and C=C at 0.139 nm. Real benzene is a perfectly regular hexagon.

The Kekulé structure has problems with the stability of benzene. Real benzene is a lot more stable than the Kekulé structure would give it credit for. Every time you do a thermochemistry calculation based on the Kekulé structure, you get an answer which is wrong by about 150 kJ mol-1. This is most easily shown using enthalpy changes of hydrogenation. Hydrogenation is the addition of hydrogen to something. If, for example, you hydrogenate ethene you get ethane:

CH2=CH2+H2→CH3CH3(1)

In order to do a fair comparison with benzene (a ring structure) we're going to compare it with cyclohexene. Cyclohexene, C6H10, is a ring of six carbon atoms containing just one C=C.

When hydrogen is added to this, cyclohexane, C6H12, is formed. The "CH" groups become CH2 and the double bond is replaced by a single one. The structures of cyclohexene and cyclohexane are usually simplified in the same way that the Kekulé structure for benzene is simplified - by leaving out all the carbons and hydrogens.

In the cyclohexane case, for example, there is a carbon atom at each corner, and enough hydrogens to make the total bonds on each carbon atom up to four. In this case, then, each corner represents CH2. The hydrogenation equation could be written:

The enthalpy change during this reaction is -120 kJ mol-1. In other words, when 1 mole of cyclohexene reacts, 120 kJ of heat energy is evolved. Where does this heat energy come from? When the reaction happens, bonds are broken (C=C and H-H) and this costs energy. Other bonds have to be made, and this releases energy. Because the bonds made are stronger than those broken, more energy is released than was used to break the original bonds and so there is a net evolution of heat energy.

If the ring had two double bonds in it initially (cyclohexa-1,3-diene), exactly twice as many bonds would have to be broken and exactly twice as many made. In other words, you would expect the enthalpy change of hydrogenation of cyclohexa-1,3-diene to be exactly twice that of cyclohexene - that is, -240 kJ mol-1.

In fact, the enthalpy change is -232 kJ mol-1 - which isn't far off what we are predicting. Applying the same argument to the Kekulé structure for benzene (what might be called cyclohexa-1,3,5-triene), you would expect an enthalpy change of -360 kJ mol-1, because there are exactly three times as many bonds being broken and made as in the cyclohexene case.

In fact what you get is -208 kJ mol-1 - not even within distance of the predicted value! This is very much easier to see on an enthalpy diagram. Notice that in each case heat energy is released, and in each case the product is the same (cyclohexane). That means that all the reactions "fall down" to the same end point.

Heavy lines, solid arrows and bold numbers represent real changes. Predicted changes are shown by dotted lines and italics. The most important point to notice is that real benzene is much lower down the diagram than the Kekulé form predicts. The lower down a substance is, the more energetically stable it is. This means that real benzene is about 150 kJ mol-1 more stable than the Kekulé structure gives it credit for. This increase in stability of benzene is known as the delocalization energy or resonance energy of benzene.

Contributors

Jim Clark (Chemguide.co.uk)

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