Draw the product or products that will be obtained from the re...

Question

Draw the product or products that will be obtained from the reaction of cis-2-butene and trans-2-butene with each of the following reagents. If a product can exist as stereoisomers, show which stereoisomers are formed.
1. HCl

1. HCl

Accepted Answer

Welcome back, everyone. Two of the products obtained in the reaction of Z 34 dimethyl hex three E and E 34 dimethyl hex three in with hydrochloric acid show the stereo chemistry. If stereo isomers are formed, first of all, we're going to throw X three E, the double bond starts at carbon. Number three, we have two more carbon atoms on the left and two more on the right because we must have a total of six bonnet to carbon's number we and number four, we have two metal groups. That would be our c isomer because eyl groups are higher in priority than methyl groups. And the highest priority groups must be on the same side of the double bond for the ze and how to draw an ee. The main difference is that we want to make sure that the sea ethyl groups are on opposite sides of the double bond. And that would be our ee due to the fact that the highest priority groups are on opposite sides of the double bond in the first step of the reaction. Since we are dealing with hydro hallucination of alkenes, we're using a proton to proton, one of our double bonded carbon atoms, the two structures are symmetrical and the two carbon atoms are equally substituted. So the more cognitive rule is not applicable and we only have one possibility to add hyen. Let's choose carbon number three. In each case, once again, according to the more cognitive rule, we're adding hydrogen to the less substituted carbon to form a more stable carbon ion. But the two carbon atoms are equally substituted and symmetrical. So we only have one possibility to form our carbon Canion. There is one problem. However, we're breaking our double bond and that we can add hydrogen from the front side or the back side, which implies that in one case, the metal group will start pointing up since hygiene is pointing down. And then our second possibility is to draw that metal group pointing down if hygiene is added from the front side. And then of course, we are forming our carbo ions. Now, here we have to be very careful and we have to understand that the chloride and ion that has been just formed can attack each carbo Canion from the front side or the back side one's game because we can form a choral center. So we are going to introduce a bit more space. And let's take a look at our possibilities. For the first carbo canon is the chloride and ion is added from the front side. Then the metal group will be pointing down and the chloral group will be pointing up. So we're going to add our chloral group pointing up and the muscle group winding down. So that is one of our products now for the other car will get on or actually let's be consistent. So this considering just the first Carbo Cannon, this is the first product that we can get. But then if we perform a backside attack, we will essentially swap chlorine with the muscle group in terms of configuration because chlorine can be added from the backside as well. Well done, we got two products and this is exactly what we expect when we consider the second carbo I, the second Carbo Canon has a metal group winding down and now the chloride can attack. The car will get on from the front side or the back side. So that's a draw. Both of them, we are going to eliminate our positive charges, right? Because now we are going to form our carbon chlorine bond. Once again, our first possibility is to perform a front side attack. So the chloral group will be pointing up and the metal group will be pointing down. Similarly, we can form a backside attack. So the metal group will be pointing up and the chloral group will be pointing down. This is how we get a total of four products. When stereo chemistry is considered for the Z isomer, we're going to label them and now we are ready to move on to the to the e isomer. Just as we stated before, our first step is the protonation step. And we already know that the two carbon atoms are equally substituted and symmetrical. So we're going to add a hydrogen forming a positive charge on the adjacent carbon atom. And we have to remember once again that there are two possibilities for us to add hydrogen from the front side or the back side. So in one case, the metal group will be pointing up if hydrogen is added from the back side. And considering the second possibility, the metal group will be pointing down if hydrogen is added from the front side. Now, instead of repeating the process, we have to be very sensible, there's free bond rotation. So we can actually redraw our carbo can ions, right, due to the free bond rotation, we can simply swap our ethyl group and the metal group so that the metal group will be pointing up and the group will be pointing down. So if we perform that bo bond rotation, because now we're not restricted since we don't have a pi bond, which restricts rotation, we can say that the carbo canon that we're forming will be the same now, right? Because we're just changing the orientation of the methyl group and the ethyl group. And the reason why we're doing this is because we now clearly see that if we also flip it specifically with respect to the horizontal line, we will get the previously drawn carbo can, right. So let's try to understand how this happens. We troy horizontal line and then we flip it so that the methyl group is pointing down. And then we have two methyl groups above the horizontal. And then of course, we have our remaining methyl group. So that's the same carbo Canion, meaning we can replace it with the previous drawing. And similarly, what about the other carbo? So once again, we can perform the bond rotation and swap the methyl group and the ethyl group. Now we will have an L group pointing up and the muscle group winding down. Now, we want to check if this is a different car, car. So once again, what if we simply flip it? But before we do, so let's just make sure that we have the two ey groups on the same side, right? So we performed a bond rotation and now we got it correct, right. The two groups are now on the same side. Now, if we flip it, we will get the first carbo canon that we drew previously for the C isomer, right? We're going to use a once again the horizontal line passing through our CC bond. And when we flip it with respect to that line or rotate 180 degrees, we will notice that the two ethyl groups will be pointing up or above the horizontal and the metal group will be pointing up on a wedge. So that means we once again get the same carol can iron. And now if we get exactly the same two carbo canons as before, we will get the same products. What we can conclude from this video is that whether we use an E or C isomers or hydro hallucination including stereo chemistry, such as in this case, we will get the same four products. Of course, it depends on how many chiral centers we have. But in this specific problem, we had a total of two chiral centers. So we got R stereo isomers well done. So we can actually label one of them as our final answer because they represent exactly same four products. We obtained our four products and we are ready to conclude the video. Both enc is imer in this case, produce the same four products. Thank you for watching.

Draw the product or products that will be obtained from the reaction of cis-2-butene and trans-2-butene with each of the following reagents. If a product can exist as stereoisomers, show which stereoisomers are formed.
1. HCl

Key Concepts

Alkenes and Their Reactions

Electrophilic Addition Mechanism

Stereochemistry and Stereoisomers

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