Question

Using figure 1, answer the following questions:

a.) Based on the data in figure 1, estimate the heat of hydrogenation for benzene (heat released when benzene is reduced to cyclohexane).

b.) The heat of hydrogenation for benzene is -49Kcal/mole. Add a solid line to figure 1 indicating this value for benzene. Is benzene lower in P.E. than you expexted?

Model 1: Amazing Benzene! Figure 1: Heats of Hydrogenation 아 Pt" -54 kcal/mole PY Hal 1-28 kcal/mole cyclo- hexane -- - -

Answer 1

Stabilizato r by Kcal/mole 284 pt Ikcal H2 -54 mole H2 pt pt 1-49 keamole kcal/mole H2o -28 Kcal/mole cyclohexane Heat of hydrogenation of one double bond = –28 kcal/mole ; two double bonds – – 54 kcal smole almost equals to (-28x2)=-567 recastrove (a) d be a (-28x3 = - 84 kcal/node There fore for three double bonds (benzene), the heat ot hydrogenation should be a (-28x3) = 8. But the actual value is – 49 kcal/mole. Therefore Therone Therefore benzene benzene is is lower loween in P.E. than the expected value by (84-49) = - 35 kcal/mote, (6)

The lowered P.E. is due to the aromaticity of benzene. So, from the above calculations, Benzene's aromatic stabilization energy is -35 kcal/mole.(Actually it's -36 kcal/mole; which is pretty close to our calculated value.)