Conformations and Cycloalkanes
Conformational Isomers of Propane
Last updated: March 27th, 2020 |
The Conformational Isomers of Propane (Are Awesome)
In this post we’ll explore the different conformations of propane and see that it has a slightly higher barrier to rotation than ethane due to a new steric interaction. We’ll draw out the Newman projection of propane from multiple perspectives, show the different Newman projections as we rotate through 360° , and map out the energy diagram vs dihedral angle!
Table of Contents
- Recap: Staggered, Eclipsed, Newman Projections, Dihedral Angle.
- What About Propane?
- Converting Propane to Newman Projections
- The Role of Perspective: Analyze One Bond At A Time
- Eclipsing Interactions in Propane
- Visualizing The Rotational Barrier In Propane
- Quiz Yourself!
- (Advanced) References and Further Reading
In the previous post [see: Staggered vs Eclipsed Conformations of Ethane] we saw that ethane has two important conformational isomers, depending on the orientation of the two methyl groups along the C–C bond.
- In the eclipsed conformation, the hydrogens on the front carbon block (or “eclipse”) the hydrogens on the back carbon when one looks directly along the C-C bond, like two Mercedes-Benz symbols superimposed on top of each other.
- In the staggered conformation, the hydrogens on the front carbon bisect the hydrogens on the back carbon, so that if one looks along the C–C bond, the hydrogens are all spaced out by 60 degrees.
- We saw that the Newman projection is a useful visual aid for visualizing the eclipsed and staggered conformations, as well as bond rotations.
- We also defined the dihedral angle as the angle between an arbitrarily chosen individual substituent on the front carbon and an individual substituent on the back carbon when looking along a carbon-carbon bond. A useful analogy for dihedral angle is the angle between the hour hand and minute hand on a clock face, where the dihedral angle is 0° at 12:00, 60° at 2:00, 120° at 4:00, and so on.
- When two substituents have a dihedral angle of –90° ≤ 0° ≤+90 they are syn. When they have a dihedral angle of +90° ≤ 180° ≤ –90° they are anti
- Conformational isomers can differ in energy; in ethane the staggered conformation is more stable than the eclipsed conformation by 3.0 kcal/mol which works out to about 1.0 kcal/mol per eclipsed H-H interaction. One way to think about this is repulsions between the hydrogens (or their electron clouds, to be more specific). There are other ways to think about it too.
Now that we’ve looked at ethane, let’s go to the next higher alkane: propane.
Propane is simple to draw as a line diagram. Just like stick figures of people omit granular details like fingers and toes, line diagrams of molecules omit the placement of hydrogens.
Most of the time that’s OK, but if we’re going to discuss conformations, we need to expand out our line diagram of propane to show where the hydrogens are.
Obviously, the first order of business is to make sure that the hydrogens are arranged in a properly tetrahedral fashion around each carbon. (Yes, it’s possible to mess this up, as Khan Academy sometimes does [see: Common Mistakes: Drawing Tetrahedral Carbons] )
Keep two substituents in the plane of the page, with one wedge and one dash, and an acute angle (<90°) between the wedge and dash.
Here’s a drawing of propane showing all the hydrogens. I’ve arbitrarily numbered the carbon on the left as C-1.
In expanding out the hydrogens on each carbon we now encounter the pesky details of conformational isomerism!
In the figure above, when I expanded out propane, I chose to draw an eclipsed conformation along both the C1-C2 and C2-C3 bonds. That’s one way to do it, and it’s not wrong (even if it is higher in energy).
However, there are other ways to draw out propane as well. I just as well could have drawn the C1-C2 carbon as staggered and the C2-C3 as eclipsed (below) or C1-C2 and C2-C3 as both staggered.
(There are of course other conformations that could be drawn out – an infinite number, but since they aren’t energy minima or maxima they generally aren’t as interesting to us. There are 3600 seconds in an hour, but when was the last time someone asked to meet you at 7:46:23 instead of just “8 o’clock” ?)
It doesn’t really matter too much what conformation we depict propane in. Since there is free rotation about the C1-C2 and C2-C3 bonds, all the conformations are interconvertible.
In propane, the lowest-energy conformation is the one where all the carbons are staggered relative to each other (top right).
That said: is there an easier, more intuitive way to analyze these conformations? Yes!
The best way to analyze the orientation along a carbon-carbon bond is by using a Newman projection.
The first step in drawing a Newman is to choose a perspective.
In our first drawing of propane, let’s choose to look along the C1-C2 bond, like this:
How did we get this Newman projection? By looking along the C1-C2 bond, like this:
Ultimately, a Newman projection is just a perspective drawing of a three-dimensional object; a molecule.
While artists and photographers have to put a lot of thought into choosing the right perspective for drawing people or buildings, our task is much easier because for the Newman projection, we must look directly along a bond.
Below, I’ve drawn propane such that there is an eclipsed conformation along the C1-C2 bond and a staggered conformation along the C2-C3 bond.
When drawing a Newman, we need to specify which bond we are looking along, and in what direction.
That’s because in even a simple molecule like propane, there are 4 different ways we could look along C-C bonds, and therefore 4 different Newman projections just for this one conformation!
None of these perspectives are “wrong”, just like there is no “wrong” perspective for drawing a house. But each perspective reveals different facets of the molecule’s conformation.
The important thing to note is that we analyze one carbon-carbon bond at a time and ignore the conformations along the other bonds.
So in the Newman projection for looking along C1-C2, we just draw C3 as “CH3” regardless of whether it’s eclipsed or staggered with respect to C2.
When we draw the Newman for looking along C2-C3, ,we draw C1 as “CH3“, no matter what. One bond at a time.
[Breaking down a big problem into a lot of tiny problems, solving them in isolation, and then building them back up to get the big picture (“reductionism”) is a powerful problem-solving technique in organic chemistry. By analyzing the conformations in a tiny molecule like propane, we can then apply what we’ve learned to much bigger molecules. [Note] ]
In the previous post, we analyzed the conformations of ethane and saw that the eclipsed conformation of ethane was about 3.0 kcal/mol higher in energy than the staggered conformation.
That breaks down to an H-H eclipsing interaction as being worth about 1.0 kcal/mol.
The rotational barrier for propane has also been measured. [Note] It is 3.4 kcal/mol.
Why might the rotational barrier for propane be higher? This is where looking at the Newman really helps.
In propane there is a CH3-H repulsive interaction that is not present in ethane.
What’s happening here is that the CH3 group is in relatively close proximity to the C-H bond, and the electron clouds around each group repel each other. This is known as Van der Waals repulsion.
Since CH3 is bigger than H, we would expect this repulsive interaction to be much larger.
But how much larger?
If we assume that each of the two H-H eclipsing interactions in propane is worth 1.0 kcal/mol, this gives the value of the CH3-H eclipsing interaction as “costing” 1.4 kcal/mol in strain energy.
(this is the value of reductionism, by the way: we can use the value of 1.0 kcal/mol per H-H eclipsing interaction we got from ethane and apply it to bigger molecules with H-H eclipsing interactions).
It turns out that this value of 1.4 kcal/mol for eclipsing CH3-H is pretty consistent, as seen in some other molecules (for an application, see Quiz Yourself! below).
Next, let’s have a closer look at the energy profile of the rotational barrier in propane. We’ll start by looking along the C2-C3 bond of propane in the eclipsed conformation with the C1 methyl group pointing straight up and choose our dihedral angle as the angle between the methyl group and the H directly eclipsed by it. We start at a dihedral angle of 0°. In the video below we mark each of them as red.
Now, we’ll rotate the back carbon (C3) in 60° increments, noting the energies as we go:
Note that the conformations repeat themselves every 120° !
If we’re really keen, we can even graph these energies, which would look something like this.
Note that the shape of this graph is exactly the same as what we saw for ethane, except that the rotational barrier is higher.
(It won’t always be so simple, as we’ll see in the next post with butane).
The analysis for the C2-C3 bond is exactly the same as that for the C1-C2 bond.
Therefore, we now have enough information to calculate the relative stability of the three conformations of propane drawn at the top of the post.
The (staggered, staggered) is lower in energy than the (eclipsed, eclipsed) by about 6.8 kcal/mol.
- The rotational barrier is 3.4 kcal/mol in propane versus 3.0 kcal/mol in ethane. That’s due to the 1.4 kcal/mol CH3-H eclipsing interaction. Since CH3 is bigger in size than H, we should expect to see greater van Der Waals repulsion when the groups are held closely together, such as when they are syn.
- There are multiple perspectives one can choose when drawing a Newman projection.
- When drawing a Newman projection, we focus on the conformation of one C-C bond at a time, and ignore all the other conformations.
- The rotational energy diagram of propane was a sine wave with three maxima, just like ethane.
Pretty straightforward, hopefully. But this is just a warm-up for the main act.
In the next post, we’ll look at the conformations of butane, which has two new types of steric interactions we haven’t seen before!
Conformations are incredibly important in nature – the three-dimensional shape of proteins, to give one example, is affected by the rotation of hundreds or thousands of bonds, and figuring out the energies of these conformations helps us design better therapeutics. Important advances on how to calculate these energies was the basis for the 2013 Nobel Prize in Chemistry.
Thanks to Jeremy Tran for assistance with the rotational energy diagram.
- Barriers To Internal Rotation About Single Bonds
John P. Lowe
Progress in Physical Organic Chemistry (eds A. Streitwieser and R.W. Taft).1969, vol. 6.
This book chapter compiles studies from many sources on barriers to internal rotation in small molecules such as propane.
- Low‐Frequency Modes in Molecular Crystals. IX. Methyl Torsions and Barriers to Internal Rotation of Some Three‐Top Molecules
J. R. Durig, The Journal of Chemical Physics 53:1, 38-50 1970
Very technical, but has a calculation of the rotational barrier to 2-methylpropane at 3.94 kcal/mol.