Predict the major product(s) of the following elimination reac...

Question

Predict the major product(s) of the following elimination reactions, paying close attention to the stereochemical outcome of the reactions.

(f)

Accepted Answer

Welcome back everyone. Our next problem says, give the correct major product of the elimination reaction. Given below note, pay attention to stereo chemistry and we have a molecule here that has a pentane ring with two double bonds attached to a hexane ring. The two double bonds are on the carbons that are shared between the two rings. Going from those shared carbons out to the next carbons in the pentane ring, our hexane ring has no double bonds on the top carbon. So adjacent to the pentane ring, there's a dashed wedge indicating a methyl group. Then the next carbon to its right has a solid wedge with a chlorine and a dashed wedge indicating a methyl group. This reaction is taking place with eton a so sodium oxide and is taking place in an etoh solvent. So ethanol solvent. So we know we have an elimination reaction, we're going to be removing a hydrogen losing a leaving group and forming a new double bond. So we just need to determine will this be an E one or an E two reaction? So let's think about the factors that favor one or the other. First, let's look at our leaving group. So we know it will be this chlorine and this chlorine has bonds to three different carbons. So it's a tertiary Halide. And when we think about elimination reactions, that's going to favor the E two reaction. So an inaccessible leaving group favors the E two reaction. That can be a little confusing because we often are comparing the E one and E two with SN one and SN two and an inaccessible leaving group favors the SN one reaction, but it favors the E two reaction. So important to keep that in mind. So we know that favors the E two reaction. Now we look next at our Nile. Is it a strong or weak nuc file? Well, we have sodium oxide that sodium is going to be our key that this will dissociate into water, dissociate in water or in ethanol and become charged. So that sodium will be a sodium of ion with a positive charge or oxide will have a negative charge. So since we have a charged nuclei file, we automatically have a strong Nile, any of the oxides are strong nucleo files. So strong Niles favor E two also because they're strong enough to pull off the beta hydrogen even before the leaving group leaves. Finally, we look at our solvent, ethanol, this is a polar pro solvent that favors E one. But the solvent is sort of the least important of these factors. And since the shape of our structure with a leaving group and our strong nucleo file both favor E two. Those are more important and we're going to favor E two overall. So we'll say least important by our solvent. So we're going to decide overall on E two will be favored. We remember that this is a concerted reaction where the beta hydrogen will be removed as the leaving group leaves all in one step. And then we also need to recall that as we think about, we have to label our major product. We need to keep in mind the Zaitsev rule, which means that the most substituted alkene will be the favorite or major product here because we're asked to identify the major product. So let's take a look at our molecule and identify the possible beta hydrogens that could be removed in this reaction. So our alpha carbon will be the carbon with the chlorine on it and of beta carbons, we have three because we have a tertiary hill light here. The first beta carbon will be to the left of our alpha carbon. It has that methyl group on the dashed wedge, the other beta carbon will be to the right, which has no substituents. And then finally, we have another beta carbon on the end of that methyl group. However, since we're looking for the most substituted alkene, the two beta carbons in the ring are going to lead to double bonds within the ring. That is automatically, we know be more substituted than if we use the beta hydrogen from that methyl group is any beta hydrogen removed, there will lead to a double bond coming out from the ring that will automatically have no substituents on the right side. So we know that will be less substituted than our internal ones. So this will be lead to a less favoured a king. So important to be aware of all our possible beta carbons, but we'll not waste a lot of time thinking about that mechanism or product because we know we're only looking for the major product. We know that won't be it. So let's think about our possible alkenes. I'll draw molecule here that's reacting, put my pen ring, those double bonds. And then we have, there's our methyl group, we have a hydrogen on a solid wedge, a chlorine on a solid wedge here and another methyl group on a dashed wedge. So first in blue, we'll think about our first possible beta hydrogen removal up here on the top. So I'm drawing my ep oxide with all those lone pairs around that negatively charged oxygen. And one of these lone pairs will grab this beta hydrogen in the form of a proton, the electrons from that ch bond will come down and make a double bond as the chlorine leaves. So three electron pushing arrows there that will lead to let's try our product here. Six carbon ring, five carbon ring with double bonds. Wow this is getting messy here. Try and read all that ring a little better there. I don't know if that was any better, but that's not the part we're focusing on. Now, we have a double bond is formed between the top carbon and that six carbon ring. And the next one to the right, we have a methyl group on the top carbon. A note in terms of the stereo chemistry, it was on a dashed wedge indicating it was going back behind the plane of the ring. We now have a trigonal planar geometry because of the sp two hybridization of that carbon in the double bond. So now we just draw it as a solid line straight up. It's no longer a dashed wedge that also goes for the methyl group on the carbon where the chlorine left. So the other carbon in the double bond, just a solid line due to the trigonal planar geometry. So note, the salkin has four substituents, the two carbons in the ring and then the two methyl groups. Now our other product I'll show happening in red, we have two hydrogens, two beta hydrogens, we have our oxide coming in all its own pears. So two hydrogens on that. Now, we're looking at the carbon down to the right of where our chlorine is going to leave or down below one of the lo those long pairs grabs a proton, the electrons from the ch bond go down and form the new double bond and the electrons in the CCL bond go to the chlorine. So this mechanism is shown in red and that will lead to this red molecule. So I drew them on the same reactant but one pathway in red and one pathway in blue. So draw our molecule. Now, in this case, the methyl group on the top, we keep it as a dashed wedge because that is still in a single bond and therefore still sp three hybridized. So we keep that sti chemistry. In this case, our chlorine has left, there's a methyl group still. Now the double bond is going between the carbon that had the chlorine and the carbon below it. Since that's now a carbon in a double bond, the methyl group on that carbon is a straight line and then the carbon below it has no substituents. So notice this is the less substituted Akeem because one of the carbons that double bond has hydrogen on it. Whereas in the other alchem, all the carbons in the both carbons in the double bond have bonds to other or to substituents. So our blue product up here will be the more substituted alkene, am therefore our major product. So again, you probably wouldn't even have to draw that second product if you recognized right away that with two substituents on the carbons that were going to participate in the double bond, that would be the more substituted alkene. But there is our major product. We'll see you in the next video.

Predict the major product(s) of the following elimination reactions, paying close attention to the stereochemical outcome of the reactions.
(f)

Key Concepts

Elimination Reactions

Stereochemistry

Zaitsev's Rule

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