Butane provides a relatively simple demonstration of the energetic
issues we face in evaluating rotational isomerism. Our goal is to
understand several things:
- What will be the most stable conformer of
an alkane (or
other molecule?
- What general principles can we find in
order to make a
prediction for a molecule we've never seen?
- What are the underlying physical laws that
control energy
vs. conformation?
- How should we think about what a collection
of molecules
will "look like"?
The analysis requires is to define a measure of
physical
structure. We will use the dihedral angle, the angular measure of
the "twist" of groups around a bond. In this analysis, we are
looking at the dihedral angle between the two terminal methyl groups in
butane.
A very useful drawing convention is the Newman projection, named after
the chemist who invented it in the 1960s. To draw a Newman
projection, use the following process:
- Draw a circle. This represents the
bond down which we
look.
- If the front carbon is sp3-hybridized,
draw three lines
that meet at the center of the circle, 120° apart. (An sp2
carbon
will use 2 lines 180° apart)
- If the back carbon is sp3-hybridized,
draw three lines
outside the circle that stop at the circle's edge. These three
will be 180° apart; the angular separation between a front-atom bond
and a rear-atom bond is the dihedral angle.
The rotation about the center bond in butane is shown in the chart
below using Chime structures. You should build a model.
Compare to Figure 3.11 in the text.
butane_0.pdb
|
butane_60.pdb
|
butane_120.pdb
|
butane_180.pdb
|
butane_240.pdb
|
butane_300.pdb
|
butane_360.pdb
|
Eclipsed,
syn, 0°
|
Staggered,
gauche, 60°
|
Eclipsed,
120°
|
Staggered,
anti, 180°
|
Eclipsed,
240°
|
Staggered,
gauche, 300°
|
Eclipsed,
syn, 360°
|
Energy:
+5 kcal/mol
+21 kJ/mol
|
Energy:
+0.9 kcal/mol
+3.8 kJ/mol
|
Energy:
+3.6 kcal/mol
+15 kJ/mol
|
Energy: 0
|
Energy:
+3.6 kcal/mol
+15 kJ/mol
|
Energy:
+0.9 kcal/mol
+3.8 kJ/mol
|
Energy: +5
kcal/mol
+21 kJ/mol
|
You will want to rotate each structure to see the perspective looking
down C2-C3. In each structure, you can have Jmol report the
dihedral:
- Right-click to get the menu;
- Select "Measurements", then "Click for
torsion(dihedral) measurement"
.
- In the structure, double left-click
on C1, then single click on
C2, C3, C4.
- The dihedral angle will appear in diagram.
You have to go through the process for each
structure--the actions of each structure are independent.
An understanding of the source of the higher energies can be seen with
space-filling models. (right-click, delect "Style", "Scheme" and
"CPK Spacefill".) The space-filling model attempts to
convey the location of the electron cloud around each atom. We
see that the high energies come from moving electron clouds close
together. The closer they get, the higher the energy.
Generally:
- Staggered conformations are more stable
than eclipsed
- Anti conformations are most stable; each
gauche interaction costs about 0.9 kcal/mol.
From examination of ethane, we know that a CH-CH
eclipsing interaction costs 0.9 kcal/mol. We can deduce, then,
that a CH-CC eclipsing interaction costs (3.6 kcal/mol - 2x0.9
kcal/mol)=1.8 kcal/mol. A CC-CC eclipsing interaction costs (5.0
kcal/mol -2x0.9 kcal/mol)=3.2 kcal/mol.
The energies tell us how much of any form will be present in a
collection of the molecules. Remember the equation for
equilibrium:
ΔG° = -RTlnKeq
The equilibrium between anti and gauche is a function of
the energy difference (0.9 kcal/mol) and comes out to be Keq = 4.57 at
room temperature. This works out to a mixture that is 82% anti,
18% gauche. (Work out the K for the eclipsed conformations, and
you will find there will be <0.1% of either.)
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