Make models of the following compounds, and predict the produc...

Question

Make models of the following compounds, and predict the products formed when they react with the strong bases shown.

(a)

Accepted Answer

All right. Hello everyone. So this question is asking us to provide the product's form note, sin coplanar E two eliminations are rare, usually occurring when free rotation is not possible. Now, here we have 2 S3 R two chloral, three methyl hexane reacting with sodium hydroxide in or rather to produce both substitution and elimination products. Now, before we can go ahead and discuss the products for this particular reaction, let's first convert the fisher projection, we are given into a conventional line angle drawing. So the first thing I'm going to do is take my fissure projection and rotate the entire structure by 90 degrees in the counterclockwise direction. So as a result, the hydrogens on carbons two and three are going to be on the bottom. So now that we have rotated the entire Fisher projection, we're going to go ahead and change our perspective to look at the drawing itself from the bottom. So because of this, the vertical bonds on the bottom are going to be pointing towards us on a wedge and the top vertical bonds are going to be pointing away from us. On a dash. On the other hand, the internal and horizontal bonds are going to be remaining straight. So in this case, both hydrogens on carbons two and three are going to be on a wedge which they use chlorine and our methyl group on carbon three on a dash. So now I'm going to go ahead and rotate the bond in between carbons two and three to give that conventional zig zag structure. So after my rotation, carbon two is going to remain the same as before. But now carbon three has its substituents flipped. In other words, the methyl group is this time or is now going to be on a wedge and hydrogen is going to be on a dash. All right. So from here, we can go ahead and talk in a little bit more detail about the substitution and elimination reactions. In question, our other reagent is sodium hydroxide. And because hydroxide has a negative charge, it just so happens to be a strong Nile or base. Meaning here we're going to see a mixture of SN two and E two products. So I'm going to scroll down here once more and I'm going to redraw our starting material very briefly. All right. So first, let's talk about the E two reaction. Recall that the E two reaction is a bi molecular single step mechanism in which a beta hydrogen. In other words, a hydrogen attached to a beta carbon is eliminated from an al Halide, which subsequently displaces the halogen after a pi bond is formed between the alpha and beta carbons. Now to be a little bit more specific, the base is going to re remove a hydrogen from the beta carbon that is anti co planner to the leading group. And because hydroxide happens to be a relatively small base, the more substituted keen is going to be favored. So just to go ahead and clarify carbon two of this molecule contains chlorine, which is our leaving group. So the carbon atom that contains our leaving group is the alpha position and the beta carbon is directly next to the alpha carbon. So here it just so happens that we have two different beta carbons, each of which has at least one proton carbons, one and three respectively. However, a double bond between carbons, two and three is going to be more substituted and therefore, is going to be our major product. But we're going to have to rotate this molecule again because we have to showcase that the hydrogen of the beta carbon and chlorine are on opposite sides. This also gives us the correct positions of the substituents when the double bond is actually added. So we're going to rotate this molecule so that chlorine and hydrogen or on the plane and, and tied to each other. So now focusing our attention on carbon two, here we have hydrogen on a dash and a methyl group on a wedge which is really carbon one of the parent chain on the other hand, focusing our attention on carbon three, we have a methyl group on a wedge and we have the remaining carbon of our parent chain on a dash. Excuse me, it's the other way about the carbons of the parent chain are actually on a wedge and the methyl group is on a dash. So from here, we can go ahead and draw out the elimination product by removing the leaving group. In this case, chlorine and adding a double bond between the alpha and beta carbs. Now, it's worth mentioning that we also have to be mindful of the positioning of our substituents. So the substituents that are on a wedge here should be on the same side of the double bond and the substituents on a dash should be on the same side as well. This means that we end up with AC OK. There we go. I paused the recording just to make sure I had enough space because I realized that the screen cut off. But with that in mind, we can see that our elimination product is the re methyl hex two in. All right. So now let's talk about the SN two reaction recall that the S and two reaction is characterized by the concerted backside attack of the Nile. So the nuclei attacks the opposite side of the carbon of the leaving group which simultaneously breaks the bond between carbon and the leaving group. Now, if I scroll down here, I went ahead and redrew another copy of our starting material. And because of this backside attack, the SN two reaction is going to result in the inversion of stereo chemistry if a chiral center is created. So in this case, to draw out the SN two product, because hydroxide is the nucleo file, I'm going to go ahead and replace chlorine with a hydroxy group on carbon two. So I'm going to draw the same carbon chain and I'm going to redraw the substituents on carbon three since that is not going to change. However, on carbon two, because chlorine was originally on a dash, the hydroxy group is going to be on a wedge, which means that hydrogen is now on a dash and therefore our substitution or SN two product is two R three R three heth, excuse me, three methyl hexane 20 and there you have it. So here we have the elimination and the substitution products. So if you watch this video all the way through to the end, thank you so very much. And I hope you found this helpful.

Make models of the following compounds, and predict the products formed when they react with the strong bases shown.
(a)

Key Concepts

E2 Elimination Mechanism

Substitution Reactions

Steric Hindrance

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