Chemistry 3730 Fall 2000 Quiz 4 Solutions

1.

This turned out to be a more interesting question than I thought it would be for a reason I should have anticipated: These molecules have several methyl rotamers (minima on the potential energy surface which differ by the rotation of CH₃ groups) which differ in energy by about 1kcal/mol each. The structure you get doing the obvious thing in HyperChem ( $\textrm{Build}\rightarrow\textrm{Optimize}$ without touching the methyl groups) is not normally the lowest energy state and thus not the main contributor to $\Delta E$.

I did not do an exhaustive search of the rotamers. However, for each isomer, I performed a directed search and found a relatively low energy structure:

$E = -97\,418.57\,\mathrm{kcal/mol}$	C-C: 1.51Å
CH3 C-H: 1.08Å	C=C: 1.32Å
=C-H: 1.08Å	C-C=C angle: $125^\circ$

2.

$E = -97\,417.05\,\mathrm{kcal/mol}$	C-C: 1.51Å
CH3 C-H: 1.08Å	C=C: 1.32Å
=C-H: 1.08Å	C-C=C angle: $128^\circ$

There are no significant quantitative differences in the bond lengths or angles. This is not altogether surprising given that the two compounds are almost completely identical.

3.

$\Delta E = E_\mathrm{trans} - E_\mathrm{cis} = -1.52\,\mathrm{kcal/mol}$. This is quite reasonable, given that the basis set used here is rather small. Moreover, we might have expected trouble with this calculation since we are attempting to measure a very small energy difference.

4.

The spectral lines correspond to transitions from energy level J to level J+1. Thus,

$$\begin{eqnarray*}\Delta E_J & = & E_{J+1}-E_J\\ & = & \frac{\hbar^2}{2\mu R^2}... ...+1)\left[(J+2)-J\right]\\ & = & \frac{\hbar^2}{\mu R^2}(J+1). \end{eqnarray*}$$

The spacing between two lines in the spectrum is therefore

$$\begin{eqnarray*}\Delta\Delta E & = & \Delta E_{J+1}-\Delta E_J\\ & = & \frac{... ...R^2}\left[(J+2)-(J+1)\right]\\ & = & \frac{\hbar^2}{\mu R^2}. \end{eqnarray*}$$

In this case, $\Delta\Delta E = 20.51\pm 0.19\,\mathrm{cm}^{-1} \equiv (4.07\pm 0.04)\times 10^{-22}\,\mathrm{J}$. Also,

$$\begin{eqnarray*}\mu & = & \left(\frac{1}{1.007\,825\,032\,1\,\mathrm{amu}} + \... ...hrm{amu}\\ & \equiv & 1.626\,651\times 10^{-27}\,\mathrm{kg}. \end{eqnarray*}$$

We can now solve for the bond length:

$$\begin{eqnarray*}% latex2html id marker 58 R^2 & = & \frac{\hbar^2}{\mu\Delta\De... ...\therefore R & = & (1.295\pm 0.006)\times 10^{-10}\,\mathrm{m}. \end{eqnarray*}$$