Structure and Bonding:  Hindered Bond Rotation 

The arrangement of the 2-butene isomers shown below was examined in The Arrangement of Bonds.  Let's examine them a little closer this time.  The carbon atoms in the double bonds are trigonal planar in geometry.  Measure the C-C=C bond angle in each isomer.  Which isomer do you think is more stable (lower energy)?  Why?


cis-2-butene                                      trans-2-butene

C-C=C angle ______                            C-C=C angle ______

Here is some thermodynamic data at 25oC for the two isomers.

thermodynamic quantity cis-2-butene  trans-2-butene
DHof -7.7 kJ/mole -10.8 kJ/mole
So 220 J/K-mole 163 J/K-mole

Source: NIST Webbook

Which isomer is the more stable?  Explain.

Does your choice using the thermodynamic data match your choice based on electron repulsion?

For the conversion reaction given below, is it endothermic or exothermic?

trans-2-butene     =     cis-2-butene

View the two reaction boxes given below.  The box at 500oC started the same as the box at 25oC.  Both boxes are at equilibrium.  The same thing happens if you start with a box of cis-2-butene molecules too.  What happened in the 500oC box?

Rotation around a double bond is not suppose to occur; however, it seems to have happened above.  Since the two compounds have very similar heats of formation, the forward and the backward activation energies must be very similar.  Can we estimate the activation energy for this reaction?  We do not want to break the carbon-carbon double bond completely, but support we disrupted the bond so that it became a single bond, and then reformed the double bond.  Since a double bond contains a sigma bond and a pi bond, suppose we disrupted the pi bond only.  Click on the image of the pi (p) bond below to see what happens.

This would allow rotation to occur.  Look up the bond energies for the carbon-carbon single and double bonds in your textbook.

Bond   Bond Energy
C=C  double bond (s + p)  
C-C  single bond (s)  
p bond must be the difference  

The energy for the breaking of the pi bond must be the activation energy for the isomerization reaction.  Why does the pi (p) bond reform?

Draw and label a potential energy diagram for this reaction.  Draw the structure of the activated complex.

At room temperature the energy is not available for the reaction to occur; hence, the two isomers are stable.  The isomerization reaction is not possible at room temperature.  The activation energy is too high.  Click on diagram above for anwser.  Click here to see a Chime structure of the activated complex in a new window.  What would you estimate the value of the equilibrium constant to be for the reaction?  Explain.

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