Show how you would accomplish the following synthetic conversi...

Question

Show how you would accomplish the following synthetic conversions.

a. 3-methylpent-2-ene → 2-chloro-3-methylpentan-3-ol

Problem-Solving Hint: The opening of a halonium ion is driven by its electrophilic nature. The weak nucleophile attacks the carbon bearing more positive charge.

Accepted Answer

Hey everyone and welcome back in today's video. We are looking at this problem which asks us to show how to accomplish the following synthetics transformation. We are going from team ethel beauty in chloral team ethel beats and Ceo were given a hint that the electra Felix at the electra Philip nature of a halo ni. Um Iron drives its opening carbon bearing more positive charge will be attacked by the week New club file. Let's begin this problem by simply observing our reaction. The reaction states that we have to Metal beauty in. So we are going to draw a bead chain chain or in other words, a four member chain because it's B two in position number two, you within has a double bond. So we're going to add that double bond Position. Number two also has a methyl group. So we are going to add a methyl group and we have our starting material. Now let's draw our product. The product is beautiful. So we are going to draw another four member chain to old Essential implies that we have an alcohol group at position number 22 method means that we have a method group at that same position and three chloral implies that there's chlorine at position number three. If we compare the starting material and the final product, we noticed that we started with an al king and this has chlorine which is a halogen and an alcohol group. So the double bond that we had originally over here undergoes addition. There's no more double bonds. Instead we added two different substations in a wage group and a chloral group and we have an additional reaction forming a halo hydrant. Now all that we have to do is just recall how halo hydrants are formed. So we take our original al Keen. The original al Keen will first will form a clay roni umbridge. In other words, a molecule of chlorine approaches the double bond and gets slightly polarized. If it gets slightly polarized, that means this chlorine will be slightly partially positive and the other one will be slightly partially negative. The double bond will attack the partially positive chlorine and chlorine with its lone pair will attack back forming the first intermediate. We may now draw that intermediate. If we look at our double bond and sort of horizontal arrangement, we will have the following substitutions bonded to chlorine. First of all, there will be chlorine forming a ring with a formal positive charge because now it has two bonds attached to it on the left side, we have two methyl groups. One of them will be pointing up. Another one will be pointing down. So we are indicating that by a wedge and a dash respectively on the right hand side there will be a methyl group pointing down and hydrogen pointing up great job. So we have one of the major parts of this mechanism and now we have to think about the next step if we think about the next step, we essentially have to take a look at the hand If we're not certain it says that the electra filic nature of the Helen yamin. That's exactly what we got. We can say that this is claire ammonium which is that polonium iron that is mentioned. And basically we are trying to open it. The more electro Philly carbon atom is the one that is more substituted. So this is the side that will participate in the reaction with the weak nuclear file. The weak nuclear violence we're trying to use is the one that would form a wage essentially we're trying to use water. We don't want to have any charge base. We're essentially trying to use a molecular substance. So recall that whenever we were trying to use a halo hydrant reaction, we're using a halogen and water in our next step. So we may indicate the presence of water. The lone pair on oxygen will attack a more substituted or a more electro Felix center. So in that case if we have this attack, we will simultaneously have the loss of the leaving group. Recall that this is a backside attack and therefore we will have inversion of configuration of this center. We will form our first product. We may essentially draw it starting with the right hand side, there is no change in configuration clearness and a solid line, hydrogen is pointing up a methyl group is pointing down and on the right hand side. After the deep protein Nation step, we will have O. H. And N. C. Arrangements to chlorine. They're on opposite sides with respect to that bond so that B. O. H. And then we have a muscle group pointing up and another missile group pointing down. This is just one of the two attacks. We also noticed that this compound has an anti mark. So we may basically flip it. So you also wish to recall that there will be another product form. If we essentially flip it with respect to the vertical line we will see another version of it which looks like does chlorine with a positive charge. We have a bridge. And now when you're flipping at this muscle group will be pointing up hydrogen will be pointing down. On the other hand when you flip a DCR equivalent metal group. So we are not that concerned. And now again we are taking a water molecule which will attack the carbon atom with a more with a greater electro filet character. So basically this carbon and simultaneously we will have the loss of chlorine. After the D. Protein nation step, we can draw a final product starting with the central carbon carbon bond on the left, nothing changes regarding configuration. We have chlorine that hydrogen that's pointing down. And that method group that is pointing up. Whereas here we have a backside attack and after the deeper the nation stopped at a wage has an anti arrangement to chlorine and there are two metal groups, one pointing up another pointing now. So these are our two. In an humors. We may essentially label them. These are our two products. They both correspond to the original product that we needed to get. It's just that here in particular were not shown any stereo chemistry. So in general, if we compare these, indeed, we got the same product as we were required to get. We have shown the mechanism for this halo high transformation. And we may conclude our answer to this problem is seeing the multiple choice because to form this hail hydrant, we are using chlorine and water. These standard conditions for any halo hydra information. If you have any questions, don't hesitate to leave down down below. Thank you for watching this video.

Show how you would accomplish the following synthetic conversions.
a. 3-methylpent-2-ene → 2-chloro-3-methylpentan-3-ol
Problem-Solving Hint: The opening of a halonium ion is driven by its electrophilic nature. The weak nucleophile attacks the carbon bearing more positive charge.

Key Concepts

Halonium Ion Mechanism

Nucleophilic Attack

Stereochemistry in Organic Reactions

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