Draw the product of each of the following sigmatropic rearrang...

Question

Draw the product of each of the following sigmatropic rearrangements:

c. Chemical structure illustrating a [5,5] sigmatropic rearrangement with a phenyl group and a diene.

Accepted Answer

Welcome back everyone. To another video illustrate the product of the given sigma tropic rearrangement reaction. Our starting material undergoes the 55 stigma tropic rearrangement in the presence of heat. And what we want to do to begin with is just rotate our bonds so that the pipe bonds are closer to each other. We have to remember that in a sigma tropic rearrangement reaction, one sigma bond is broken and one sigma bond is formed. The bond that is broken is the carbon oxygen bond specifically for the 55 sigma tropic rearrangement. And the bond that is formed as the carbon carbon bond. What we're going to do is just locate our carbon oxygen bonds and we have two of them and we are actually going to break the one that is given in blue because it's a bond between an sp three hybridized carbon atom and oxygen instead of sp 21. And the reason is is simple because such a bond is weaker. So it's easier for us to break it. Now, what we want to do is just label our carbon atoms and we will actually all of our atoms and we are going to label oxygen as one and the adjacent carbon as one because those correspond to the atoms that are involved in the bond cleavage going down, we are going to label our carbon atoms. So that will be carbon number two, we four and five. And we can stop here and going to the left, we will have 234 and five. So now that we locate our positions five and five, we understand that there must be an attack from carbon number five to carbon number five. So we're going to use that pi bond and we are going to perform that attack. So let's see what we got. So, first of all, we are performing that attack by the pi bond on carbon number five. And then we have to change the position of the pi bond, then change the position of the other pi bond. And this is where we are breaking our carbon oxygen bond by simply turning this sigma bond into a pi bond between carbons one and two, which also implies that we have to shift the pi bond between carbons two and three and make it a pi bond between carbons three and four. Well done. So now we clearly see the arrows and that means we can predict our product correctly. So we are just going to redraw our starting material once again and break it down step by step. If we start with the benzene ring, we first of all, notice that we are changing the position of the pipe bond and we are breaking the other pipe bond. So now we get a pi bond between carbons three and four. So let's add that pi bond, we are also introducing our carbon oxygen double bond. So let's add a double bond, let's break the carbon oxygen single bond. And now let's add our pi bond between carbons one and two. Now that we have added our pi bond between carbons one and two, we essentially want to break the pi bond between carbons two and three. So let's go ahead and do that and introduce a pi bond between carbons three and four. Similarly, we want to break the pi bond between carbons four and five. So let's go ahead and break it and bond the resultant fragments to carbon number five within the ring. So what we want to do is just separate our fragments. Let's add that oxygen atom where it was. And now the best idea is to actually draw it horizontally. So what we are going to do is that we're just going to take our structure and rotate it slightly so we can get our horizontally looking product. In other words, we are just changing the angle of the perspective. So let's see what we get. If we simply rotate it, we now get it in a horizontal perspective. And now we want to bond our side chains. What we're going to do is just rotate our side chain as well. And we're going to rotate in a way such that we can actually bond that side chain to our structure. Now, let's introduce our result on carbon carbon bond. So let's just add that line to represent that bond. And finally, what we want to do from here is understand that this is not yet our final product because actually, we can get a more stable form since we can illustrate a Keto anal organization, which in this case will actually satisfy or favor the inal form. And we will soon understand why. So let's indicate the presence of the implicit hydrogen atom which will be captured in the totem organization. We're going to show that the pi bond at S captures the proton, right, we are going to form an oh group. And then we are going to use those electrons in the formation of a pi bond changing the position of the adjacent pi bond. We are shifting it and we are producing a more stable aromatic structure. And as we know, aromatic implies high stability. This is why we are performing the tote. And this is also the reason why toto is favored. In other words, our anal form is favored. So what we get is this, we're going to notice that there should be our benzene ring formed, which implies CMA and a higher stability. And now we get our old age group. So we essentially get a pol So that would be our final structure. We can label it as our final structure for this problem. Thank you for watching.

Draw the product of each of the following sigmatropic rearrangements:
c.

Key Concepts

Sigmatropic Rearrangements

Conservation of Orbital Symmetry

Regioselectivity

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