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### Course: Organic chemistry>Unit 3

Lesson 3: Conformations of alkanes

# Newman projections 2

Newman Projections 2. Created by Sal Khan.

## Want to join the conversation?

• what is a dihedral angle?
• In the context of the video, a dihedral angle is the angle between two points on two different parallel circles or planes. For example, imagine a clock. The angle between the hands is a dihedral angle because the minute hand has to pass over the hour hand (indicating that they are not on the same plane, if they were they would've collided).
• Why wouldn't there be another confirmation where the back CH3 group is eclipsed by a hydrogen and the front CH3 group is eclipsing a hydrogen?
• There is. He just didn't show them all in the video.
Just as there are three staggered conformations, there are three eclipsed conformations.
One has the two CH₃ groups eclipsed. The other two have CH₃ eclipsed with H.
• The angles between the hydrogen atoms and the carbon atom in ethane are 109.25 degrees since it is sp3 hybridised .then dihedral angles should have been 109.25 divided by 2 i.e 54.6 degrees .then how is the dihedral angle 60 degrees???
• In the case of the sp3 you are looking at 4 atoms in a tetrahedral (3D) configuration, which it turns out the farthest apart they can all be in 109.25 (think a triangular pyramid) On the Newman projection you are projecting the object into a 2D plane which will give you a 360 degree circle with 3 atoms in view, which will put them 120 degrees apart, add 3 more atoms and they are now 60 degrees apart (the 4th atom in the tetrahedrons, C, is behind the other C so it is not in the "circle" of the projection)
• I was hoping this video would touch on this, but it didn't:

If you have an electronegative atom, like Cl or F, would the eclipsed conformation of two Cl or two F have a higher potential energy or a lower potential energy then the eclipsed conformation of two methyl groups? And if so why?
Thank you!
• At , Sal mentions the Gauche conformation to be 2nd most stable. I was wondering which conformation would be more stable: 1. Gauche where there is a staggered state, but the methyl groups are close to each other and hence more electronic repulsions, or 2. Dihedral angle between the methyl groups is 120 degrees, there is interaction b/w methyl and hydrogen twice but hydrogen has very less electron density as compared to methyl. Thanks!
• The energies of interaction are approximately
• 3.3 kJ/mol for each CH₃-CH₃ gauche
• 4.2 kJ/mol for each H-H eclipsed
• 5.0 kJ/mol for each CH₃-H eclipsed
• 23 kJ/mol for each CH₃-CH₃ eclipsed
The major interactions are those between CH₃ and CH₃.
The CH₃-CH₃ eclipsed conformer is the least stable.
That is the case as long as the CH₃ groups are within about 30 ° of each other.
Then comes the CH₃-H eclipsed conformer (CH₃ groups at 120 °). It is still high-energy but more stable than the CH₃-CH eclipsed conformer.
• At , when draws the 3d structure of butane, how does he decide that H will go up and CH3 would go down?
(1 vote)
• It's an arbitrary choice.
But the CH₃ looked "down" in the original structure, so he put it down in his saw-horse projection.
It is convenient to draw the biggest groups either up or down in saw-horse and Newman projections, so most chemists do that automatically without even thinking.
• If we rotated the back methyl group by 60 degrees as Sal was initially saying, wouldn't that be more stable than the Gauche conformation ?? And would that be a Gauche conformation ?? Thanks !! :)
• In that conformation, it is still eclipsed. However, the back methyl group (in this context) is only eclipsed by a single hydrogen, which has a much smaller electron cloud than a methyl group.
• When he first draws the butane molecule at around the 1 minute mark, does the drawing have to be exactly like that? For example, when he draws the hydrogens for the middle 2 carbons, do they have to be both at the top or both at the bottom? If so, why? I was just thinking that they would be as far away as possible because the electrons want to repel each other.
(1 vote)
• They don't have to be at any place, no. Think of them as in constant rotation but spending the majority of the rotational time in the more stable positions.
• If you initially rotated it only 60 degrees, would it still be considered eclipsed, even though one of the hydrogen and the ch3 group wouldn't be sitting behind their counterparts? It would be (H/H), (H/CH3), (CH3/H).
Hypothetically, would this formation have only a slightly raised potential energy?
• I want to point out first that it's difficult to know exactly which conformation you're describing if you talk about an initial rotation of 60°. From what starting point are we rotating this 60°? The totally eclipsed? The anti conformation? It's unclear. We usually describe conformations by a dihedral angle of some kind. In this case looking at the C2 and C3 carbons of butane we would use the dihedral angle between the methyl groups.

But if we did have a conformation like the one you've described at dihedral angles of 120° and 240° then butane would be in an eclipsed conformation. It wouldn't be a totally eclipsed conformation, but a slightly lower in energy one instead. A totally eclipsed conformation of butane would have a potential energy of ~21 kJ while the more stable eclipsed conformation would have a potential energy of 15 kJ. Compared to the staggered conformations this is quite a lot larger. Even the least stable staggered conformations, the gauche ones at dihedral angles of 60° and 300°, would have potential energies of only 3.8 kJ. The most stable eclipsed conformations always have more potential energy than even the most unstable staggered ones.

Hope that helps.
(1 vote)
• in butane, why is the second and third carbon used to draw the newman projection, why not the first and second?
(1 vote)
• That's just the bond which Sal chose to discuss, but yes we could have drawn a Newman projection focused on the first and second carbons as opposed to the second and third carbons in butane too. Again the whole point of this exercise is to show the different conformations which arise from the internal rotation of the C-C single bonds. So carbons 2 and 3 have a single bond joining them and can rotate creating multiple conformations which can be shown in a Newman project, but the same can also be said for carbons 1 and 2.

Hope that helps.