Lab #3 Objectice Statement: The objective of this lab is to understand and depict the potential energy of simple alkanes in different conformations. The line angle drawing of long carbon chains does not properly depict the orientation of single bonds. Due to single bonds' ability to rotate, this lab will show the different shapes of conformational isomers (conformers for short) to understand the importance of varying potential energy in different shapes.
Embedded Files
Compounds of Study:
Propane
Propane
Butane
Butane
Pre-Lab Questions
1. After reading the introduction above, using your own words, describe a conformer.
A conformer is an isomer of a specific molecule that differs from other orientations by a rotation of single bonds within that molecule.
2. After reading the introduction above, open PowerPoint, select draw then draw an octane (8-carbons) zig-zag then draw two other bend structures, drawing a blue arrow to show what bond is rotating to give the bend. Group these drawings, right-click on the drawings save them as a picture then insert this picture into your Google site.
See drawing below
3. After reading the introduction above, When we talk about a bond angle, how many atoms define a bond angle? How many atoms define a “dihedral angle”?
Three atoms define a bond angle between two bonds. A "dihedral angle" is defined by four atoms across three bonds.
4. After reading the introduction above, what are all of the names given to each of these dihedral angles?
Conformation names
0 Degrees: Fully eclipsed
60 Degrees: Gauche
120 Degrees: Eclipsed
180 Degrees: Anti-guache
5. Explain the difference between the two kinds of “eclipsed” conformers.
The fully eclipsed conformation is where the dihedral angle is 0 degrees. Where the eclipsed conformation is 120 degrees of rotation.
6. In Wade 3.7 it is said that methane is “perfectly tetrahedral”. Every 3-atom bond angle (H-C-H) is exactly the same. What is that bond angle?
Bond angle is the angle between two bonds originating from the same atom in a covalent species.
7. In Wade 3.7 look at the picture of ethane but imagine the C-C bond as if it is the two ends of pencils, the eraser ends, meeting in the middle of the bond. You can spin one of the pencils but the two erasers will still stay touching, they will still stay bonded. This is how you can think of bond rotation. Instead of a pencil rotating to give a bent structure, what kind of bonding orbital is rotating?
The sigma bond connecting the two carbon atoms is rotating.
8. In Wade 3.7, when drawing the Newman projection for ethane,
A. What bond are we looking through? You can directly quote what is said in the text for your answer.
We are looking straight down the bond connecting two carbon atoms.
B. What shape do we draw for the “3 lines”?
A "Y" shape.
9. In Wade 3.7, for Sawhorse drawings, instead of looking through the C-C bond and creating a drawing, what are we looking at?
We are looking down at an angle toward the carbon-carbon bond.
10. After Viewing the solution for the problem 3-11 in Wade section 3.7, draw each of the following Newman projections 2-methyl propane (CH3CH(CH3)CH3) from C1-C2. You can copy the table into PowerPoint and add your drawings then upload them to your electronic notebook. Use the prompts below to help you with your drawings.
See drawing below
11. After reading 3-7, describe if you would expect the eclipsing of a C-H bond over the top of another C-H bond to be higher, lower, or the same torsional strain as the eclipsing of a C-C bond over the top of another C-C bond. Briefly explain.
An eclipsing C-H bond over the top of another C-H bond will have lower torsional strain than one of an eclipsing C-C bond over the top of another C-C bond. This is because torsional strain is dependent on the amount of electron density between the atoms coming in close proximity to each other. The more electron density, the higher the repulsion between the two bonds, the higher the potential energy, and therefore the higher the torsional strain. Where a C-C bond over the top of another C-C bond will have higher potential energy due to the increase of electron repulsion from the carbon atoms.
2.
10.
Experimental
NA for Lab #3
Results
Part I: Propane
Energy vs. Constraint
-117.613301 -180.00
-117.612045 -160.00
-117.609453 -140.00
-117.608096 -120.00
-117.609417 -100.00
-117.612010 -80.00
-117.613298 -60.00
-117.612052 -40.00
-117.609449 -20.00
-117.608099 0.00
Part II: Butane
Energy vs. Constraint
-156.432466 -180.00
-156.431045 -160.00
-156.428178 -140.00
-156.426702 -120.00
-156.428195 -100.00
-156.430685 -80.00
-156.431043 -60.00
-156.428632 -40.00
-156.424774 -20.00
-156.422833 0.00
Part III: Methyl ethyl ether
Energy vs. Constraint
-192.028675 0.00
-192.030009 20.00
-192.032938 40.00
-192.035347 60.00
-192.035899 80.00
-192.034719 100.00
-192.033830 120.00
-192.034755 140.00
-192.036618 160.00
-192.037548 180.00
Post Lab Questions:
In one paragraph, rationalize propane and butane's highest and lowest energy structures and your unique compound. Why does it make logical sense that certain conformers have high energy and some have low energy?
The highest energy states of all three of these compounds would have the lowest distance between the eclipsing atoms. This higher energy means that the compound is the least stable at the conformer. That aligns with the diagram of the conformers because when the two eclipsing atoms are closest to each other, their electron repulsion increases dramatically and causes the atoms to want to get away from each other and that is where the increase of energy comes from. The opposite of this is true when the eclipsing atoms are furthest from each other, their energy is the lowest, and that compound is at its most stable conformer structure.
In one paragraph, reflect on the difference in the plots of propane and butane. Why does butane have conformers intermediate in energy whereas propane conformers are all either high or low?
The main difference between the plots of propane and butane is that propane's plot is a bit more drastic at the higher and lower ends of energy than butane. Where butane seems to settle within a closer range of possible energy on the different ends of the conformer spectrum. I think that this is due to butane's extra C-C bond. The three carbon-to-carbon bonds in the dihedral structure of this lab are more stable than the two carbon-to-carbon bonds and one carbon-to-hydrogen bond in propane. The carbon-to-hydrogen bond is slightly polar and kind of tips the scale' a little compared to the three very stable bonds in butane. Without the extra stability in the propane molecules, the compound experiences higher levels of energy at different conformities. And that is where we see those higher spikes of energy when comparing the two graphs.
Draw the Newman Projection, about the indicated bond, of the most stable and least stable conformer of:
a. 3-methyl pentane (C2-C3) b. 3,3-dimethyl hexane (C3-C4)
CH3CH2CH(CH3)CH2CH3 CH3CH2C(CH3)2CH2CH2CH3
Page updated
Report abuse