Predict the major product(s) for each reaction. Include stereo...

Question

Predict the major product(s) for each reaction. Include stereochemistry where appropriate.

c. cis-but-2-ene + Cl₂/H₂O

d. trans-but-2-ene + Cl₂/H₂O

Accepted Answer

Hey everyone and welcome back in today's video, we're looking at the problem which reads for each of the following reactions, determine the major products includes stereo chemistry. In your answer. Reaction number one we have six hex three in reacting with chlorine and water. Reaction. Number two trans hex trian reacts with chlorine and water. Let's start this problem by drawing our substrates number one. Since X three in we have a six member chain and L keen sense arrangement. The double bond starts at position three. So we may draw a horizontal double one. Because we have this arrangement in particular, we will still have two more carbon atoms on the left side and on the right side because essentially that's vaccine notice 1234566 carbon atoms at positions. We have a double bond. This arrangement means that the two asshole substitutions are pointing in the same direction with respect to the double bond. The conditions that we're using are chlorine in water. And for number two we're using the trans version of that same substrate. So that means we're just going to change the direction of that ethyl group. So it will be pointing up. So the tooele kill groups will be pointing in opposite directions to form a trans eyes mur we are using exactly the same conditions. So that would be chlorine in water and we are done with our setup. We now simply wish to recall that because we're using al canes and its reaction and now we're adding a halogen and water. In each case we understand that such a reaction indicates the formation of a halo hydrant. So we're able to use a mechanistic approach to understand which products we form in these reactions. Let's start with number one. And we will try to draw the mechanism to understand what products are formed. We are going to draw our six version of that compound, the first stop, we're adding a chlorine molecule. We are going to expand the chlorine chlorine bond in order that when this molecule that is non polar in nature is approaching the double bond, it gets slightly polarized. The left chlorine is partially positive and the right chlorine is partially negative because the electrons are pushing the bonding electrons away from the left chlorine and we are forming a clay Rony umbridge the double bond at Saks chlorine attacks back, we're forming a bridge and simultaneously we're losing another chlorine as a chloride and I am now we mutual our intermediate first of all, there's no more double bond. Instead, we are forming a carbon chlorine bond but this chlorine now has a formal positive charge. And because this was a cis arrangement, the two ethyl groups could be indicated on wedges. They are on the same side while the two original hydrogen is that we're over here will be on dashes. Now we have formed our claire, ammonium and now this is where water comes in. We are going to use a water molecule and recall that in this mechanism, water will attack a more substituted carbon. But because they're equally substituted, we have two different possibilities for that nuclear attack. First of all, water might attack this carbon atom and simultaneously we will have the loss of the leaving group in this case chlorine. This is a backside attack, meaning we will have inversion of configuration. And therefore, if we look at the first pathway and the blue pathway in particular, we are forming a protein ated intermediate in which we may simply draw our middle bond chlorine has left. So we are just going to indicate chlorine on a solid line. Nothing changes regarding cereal chemistry. At this position we still have an ethyl group pointing up and high Chen pointing down. So that's the right side of the molecule regarding the left side of the molecule. There was a backside attack. So hydrogen will be on a solid line. It's an anti arrangement. My apologies is not hygiene. It's a water molecule. So it's an anti arrangement to chlorine now. So we may say that oh H two. It has a formal positive charge because it's protein ated. It forms three bonds and at this side essentially will have an ethyl group pointing up and hydrogen pointing down when this intermediate is then dip Roatan ated by another water molecule. We simply get rid of one of these protons. We may essentially understand it after the de protestation step, we lose that additional hydrogen to form our first product. Great job. So let's use a similar approach said. And for our next product in this reaction, another water molecule may also add tag the clay roni um intermediate at a different carbon because they are equally substituted as we already said. And we will have the loss of the leaving group chlorine after the D protein nation step. We will essentially not worry about the protein ated oxygen. We will draw the final product immediately and we may draw the central bond. Now, let's see what's happening here at this carbon, nothing changes. We will have chlorine on a solid line. There is no change in configuration hydrogen pointing down and sl pointing up. On the other hand over here, we have a backside attack. So a water molecule which is now deep protein ated will be pointing on a solid line down that B. O. H. After the D. Protein ation stop. And now if you look at the remaining two substitute parents, ethel will have a wedge basically pointing up and hydrogen will be on a dash pointing away from the viewer. These are our cheap products. We may notice that the relationship between them that this is a pair of finance humors and we are ready to go to our next problem for the next problem let's redraw the trans version of the substrate. We already know that it basically has a double bond with two ethyl substations pointing in opposite directions. In the first step of the hilo hi transformation. we again have another molecule of chlorine. The double bond attacks the partially positive chlorine chlorine attacks back and we have the loss of a chloride and iron. We're forming our bridge. Let's draw it out. Starting with the central bond, we will have a bridge chlorine with a formal positive charge. And I notice that the T ethnic groups where on opposite sides of the double bond, so we are going to draw them pointing in opposite directions. One of them will be pointing up and another one will be pointing down. This means the remaining hydrogen will correspondingly have one hydrogen here pointing away and one Haijun over here, pointing towards the viewer, this is our cloney um intermediate. And now we because these two carbon atoms are again equally substituted, hydrogen will attack both of them. We will have two different products recall that hydrogen will generally attack a more substituted carbon, but they are equally substituted. So we have two possibilities. One of them is to perform a backside attack at this carbon losing chlorine and we meet through our product, we will not worry about that additional proton bonded to oxygen which will then be removed by another water molecule. So we may indicate that we're losing hydrogen as well. So if we draw our central bond, if we start from the right side, that would be chlorine on a solid line. Nothing changes regarding configuration here, hydrogen pointing up sl pointing down and now on the left side this was a backside attack. So we will have an A wage after the deep rooted nation step in anti arrangement to chlorine. If we take a look at these two substitutions, we will have an ethyl group pointing up as well as a hydrogen pointing down. Great job we have our first product. Let's look at the second possibility in which another water molecule may attack that equally substituted carbon and we would have the loss off chlorine. We will not worry about the protein ated intermediate and we'll just try to draw the product. So if we draw our central carbon carbon bond and if we start from the left side, nothing changes there regarding configuration. That would be chlorine on a solid line. Then we have an ethyl group on a wedge Haijun on a dash regarding this carbon. We have a backside attack so and the wage group would be on a solid line pointing down due to the requirement of a wage being anti to chlorine. And now the remaining two atoms. Those are hydrogen pointing up and an ethyl group pointing down, We may again say that these two our our in and tumors. So we have our products for the first reaction legislative e what we have so far. These are our two products for the first reaction and on the bottom we have two additional products for the second reaction. Based on our analysis, we may now complete our reactions and we may drool our products. So going back over here, reaction number one, let's troll our products. Clearly, the first product that we get looks like this. There will be hydrogen here. Another Essel hydrogen. Oh, age. Well done. The second product in this reaction would be a wage chlorine. An anti position hydrogen point hang down, Essel pointing up and now for the second reaction, our products were chlorine, hydrogen pointing up Sl pointing down here, S. L. Pointing up hydrogen pointing down and then a wage anti to chlorine And the next one hydrogen pointing up S. O. Pointing down. Ohh on the solid line here, chlorine, anti to ohh! Another ethnic group here on a wedge and finally hydrogen pointing down so these are our products and as you can see the candies for products for the first reaction in the second reaction, the correct answer choice to this multiple choice question is the However, if you feel if you have any questions regarding this problem, don't hesitate to leave your comment down below. Thank you for watching

Predict the major product(s) for each reaction. Include stereochemistry where appropriate.
c. cis-but-2-ene + Cl₂/H₂O
d. trans-but-2-ene + Cl₂/H₂O

Key Concepts

Electrophilic Addition Reactions

Stereochemistry

Markovnikov's Rule

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