Question

1. Butane. Obtain the difference in energy $ (\Delta \mathrm{E}, \mathrm{kJ} / \mathrm{mol}) $ between gauche and anti forms of butane, and the fraction butane existing in the gauche form as follows: a) Draw butane, adjusting the angle around the central C-C bond so that it is roughly in the anti conformation (methyls at $ 180^{\circ} $ ), then minimise - the MMFF energy is displayed in the box below the minimize button. Rotate by hand to the gauche (methyls at $ 60^{\circ} $ ), minimise again, then calculate $ \Delta \mathrm{E}=\mathrm{E}_{\text {gauche }}-\mathrm{E}_{\text {anti. }} $ [1] \[ \Delta \mathrm{E}=-17.9651-21.2379=-39.203 \mathrm{KJ} / \mathrm{mol} \] b) Use the Boltzmann distribution $ \left(N_{i} / N_{T O T}=\exp \left(-\Delta G_{i} / R T\right)\right) $ to calculate what fraction of butane molecules exist in the gauche form at room temperature (use $ \mathrm{R}=8.314 $, $ \mathrm{T}=300 \mathrm{~K} $ ). Assume that the calculated energy difference $ \Delta \mathrm{E} $ can be used directly in the Boltzmann factor, i.e. $ \Delta \mathrm{E}=\Delta \mathrm{G} $. [1]

          1. Butane. Obtain the difference in energy \( (\Delta \mathrm{E}, \mathrm{kJ} / \mathrm{mol}) \) between gauche and anti forms of butane, and the fraction butane existing in the gauche form as follows:
a) Draw butane, adjusting the angle around the central C-C bond so that it is roughly in the anti conformation (methyls at \( 180^{\circ} \) ), then minimise - the MMFF energy is displayed in the box below the minimize button. Rotate by hand to the gauche (methyls at \( 60^{\circ} \) ), minimise again, then calculate \( \Delta \mathrm{E}=\mathrm{E}_{\text {gauche }}-\mathrm{E}_{\text {anti. }} \) [1]
\[
\Delta \mathrm{E}=-17.9651-21.2379=-39.203 \mathrm{KJ} / \mathrm{mol}
\]
b) Use the Boltzmann distribution \( \left(N_{i} / N_{T O T}=\exp \left(-\Delta G_{i} / R T\right)\right) \) to calculate what fraction of butane molecules exist in the gauche form at room temperature (use \( \mathrm{R}=8.314 \), \( \mathrm{T}=300 \mathrm{~K} \) ). Assume that the calculated energy difference \( \Delta \mathrm{E} \) can be used directly in the Boltzmann factor, i.e. \( \Delta \mathrm{E}=\Delta \mathrm{G} \). [1]

$1. Butane. Obtain the difference in energy (ΔE, kJ / mol) between gauche and anti forms of butane, and the fraction butane existing in the gauche form as follows: a) Draw butane, adjusting the angle around the central C-C bond so that it is roughly in the anti conformation (methyls at 180^∘ ), then minimise - the MMFF energy is displayed in the box below the minimize button. Rotate by hand to the gauche (methyls at 60^∘ ), minimise again, then calculate ΔE=Egauche -Eanti. [1] ΔE=-17.9651-21.2379=-39.203 KJ / mol b) Use the Boltzmann distribution (Ni / NT O T=(-Δ Gi / R T)) to calculate what fraction of butane molecules exist in the gauche form at room temperature (use R=8.314, T=300 K ). Assume that the calculated energy difference ΔE can be used directly in the Boltzmann factor, i.e. ΔE=ΔG. [1]$

Added by Rebecca F.

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