The molecules cis-dichloroethylene and transdichloroethylene shown on p. 324 can be interconverted by heating or irradiation. (a) Starting with cis-dichloroethylene, show that rotating the \(\mathrm{C}=\mathrm{C}\) bond by \(180^{\circ}\) will break only the pi bond but will leave the sigma bond intact. Explain the formation of trans- dichloroethylene from this process. (Treat the rotation as two, stepwise \(90^{\circ}\) rotations.) (b) Account for the difference in the bond enthalpies for the pi bond (about \(270 \mathrm{~kJ} / \mathrm{mol}\) ) and the sigma bond (about \(350 \mathrm{~kJ} / \mathrm{mol}\) ). (c) Calculate the longest wavelength of light needed to bring about this conversion.

Step by step solution

01

Understand and illustrate isomer conversion

Observe the structure of cis-dichloroethylene. Rotating the C=C bond by 90 degrees twice will only break the pi bond without disturbing the sigma bond as the sideways overlapping p orbitals forming the pi bond are weaker and can easily get disconnected due to rotation. After this π bond breakage, when we rotate it by another 90 degrees, we get trans-dichloroethylene.

02

Analyze the bond enthalpies

A σ bond (sigma bond) is stronger than a π bond (pi bond). This is because sigma bonds form from the end-to-end overlapping and hence the electron density lies between the nuclei of the bonding atoms, which ultimately decreases the potential energy and makes the bond stronger. Thus, σ bond (350 kJ/mol) has larger enthalpy than π bond (270 kJ/mol).

03

Calculate the wavelength of light

The energy required to break the pi bond and initiate the transformation can be found from the enthalpy of the pi bond, given by \(270 kJ/mol\). Convert this to J/mol using the conversion \(1 kJ = 1000 J\). The energy of a photon is given by the Plank's equation: \(E = h\nu\), where h is Plank's constant and \(\nu\) is the frequency. The frequency can be related to the wavelength as: \(\nu = c/\lambda\), where c is the speed of light and \(\lambda\) is the wavelength. Thus, substituting \(\nu = c/\lambda\) in \(E = h\nu\) gives: \(E = hc/\lambda\). Rearranging for wavelength, we get: \(\lambda = hc/E\). Substituting the known values of h, c, and E(from the enthalpy calculation) will give the required wavelength.

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