Berkeley Review (TBR) Q&A

[ImDiene0412]

I wanted to start a thread specifically for questions on TBR. I hope it's helpful .

So I'll start,

I came across a practice questions that said "In both the E and Z isomers of an alkene, the atoms directly bonded to the alkene carbons are all coplanar"

The explanation says "Because the two p-orbitals of the pi-bond are coplanar with an orientation perpendicular to the substituents on the alkene carbons, the four atoms bonded to the two carbons of the alkene must be coplanar"

Can anyone elaborate? I don't quite understand why the orbitals being perpendicular makes the substituents coplanar?

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[TheLongGame]

In a sp2 hydrid carbon atom, the unhybridized p orbitals are perpendicular to the 3 hybrid sp orbitals. Since we know that the electrons want to be in the lowest energy state possible, this translates into the orbitals maximizing their distance from one another forming the trigonal bipyramidal arrangement we see in the picture. (Imagine (2) three sided pyramids that point in opposite directions and share a common base)

Since the primary bonds between the two double bonded carbon atoms utilize the p orbitals and one sp orbital, the substituents must bond to one of the remaining sp orbitals. Since all 3 sp bonds are coplanar, the substituents atoms directly attached to the double bonded carbon atoms end up being coplanar as well.



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[ImDiene0412]

View attachment 209062

In a sp2 hydrid carbon atom, the unhybridized p orbitals are perpendicular to the 3 hybrid sp orbitals. Since we know that the electrons want to be in the lowest energy state possible, this translates into the orbitals maximizing their distance from one another forming the trigonal bipyramidal arrangement we see in the picture. (Imagine (2) three sided pyramids that point in opposite directions and share a common base)

Since the primary bonds between the two double bonded carbon atoms utilize the p orbitals and one sp orbital, the substituents must bond to one of the remaining sp orbitals. Since all 3 sp bonds are coplanar, the substituents atoms directly attached to the double bonded carbon atoms end up being coplanar as well.



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Okay got it! Thank you! It is also safe to say that the planar nature of a pi bond is why there is no rotation about the bond aka it has trigonal planar geometry?

 

[TheLongGame]

Okay got it! Thank you! It is also safe to say that the planar nature of a pi bond is why there is no rotation about the bond aka it has trigonal planar geometry?

I'm not 100% sure what you're saying, but I will explain and hopefully answer your question in the process.

Starting off, we know that molecules are seeking the lowest energy state possible. In order to break bonds, we must add energy and conversely when we create a bond we release energy.

Now let's imagine what it would look like to rotate the C=C double bond. So let's hold one sp2 carbon fixed and rotate the other.

As you can see, the single pi bond actually consists of two orbitals that exist "above" and "below" the sigma bond. In order for us to rotate the carbon on the right, we would have to break the both pi bonds, rotate the carbon about the sigma bond, and then reform the pi bonds.

Given the diagram and the fact that molecules are constantly seeking their lowest energy state, we can see that the C=C atoms will not rotate because it is unfavorable energetically.

Conceptually, if you take two marshmallows and stick them on either end of a toothpick. It is easy to spin them freely about the axis of the bond. Now add a second toothpick parallel to the first and attempt to rotate either marshmallow. You'll find the act impossible without destroying your delicious marshmallows or by first removing (breaking bonds) a toothpick.

We'll refrain from speaking about bonding and anti-bonding orbitals, but this material only strengthens the argument above.

I hope this makes it clear!

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