Propose mechanisms and predict the major products of the follo...

Question

Propose mechanisms and predict the major products of the following reactions. Include stereochemistry where appropriate.

(c) (E)-dec-3-ene + Br₂ in CCl₄

(d) (Z)-dec-3-ene + Br₂ in CCl₄

Problem-Solving Hint: Models may be helpful whenever stereochemistry is involved. Write complete structures, including all bonds and charges, when writing mechanisms.

Accepted Answer

everyone and welcome back in today's video, We are going to tackle the following problem which says propose a mechanism and predict. The major products of the following reactions include stereo chemistry. In your answers were appropriate. We have two reactions of E pen to eat with bro mean in carbon tetrachloride and then the same reaction with Z pen to in because the two reactions are very similar in nature, we may just analyze them simultaneously. We may now start by drawing our starting materials. They are very similar structures. They just have different arrangements E N C. Starting with a double bond because we have two Elkins. The first structure has a double bond in position # two. So we have an additional carbon opposition number one and because it's spent essentially that implies the parent has five carbon atoms. So we need additional two carbon atoms to have five. And that would be our E arrangement which is essentially trans. As you recall for the second structure, it will look pretty much the same but now it's C. So that would be our sis I Zimmer. We may draw the method group on the same side as the ethyl group. These are our starting materials and we understand that the reaction conditions are pretty much the same. Right? We are using roaming and carbon tetrachloride. So we are going to indicate that roaming carbon tetrachloride is the solvent. So we are writing it below the arrow. Right bro me these two reactions will produce some products then we are interested in the major ones to predict a major products. We will think about the mechanism of each reaction if we start with the first reaction. Since we need to propose a mechanism, we made sure that al keen. Once again we have an electrolytic edition mechanism as your equal. Because roaming reacts with a double bond in particular, we have elk in hydrogen elimination. So the bromine molecule simply approaches that the taliban and gets polarized. There is a partial positive charge here because the electrons from the pi bond are repelling the bonding electrons so it gets partially positive disposition and then the double bond can attack booming while the lumpy and grooming will attack. The double bond, simultaneous will have the loss of the leaving group and in the stop we will form our first intermediate. The intermediate will not have any double bond at this position anymore. However, we will form a ceremony umbridge and then we have to notice that the two al kiel groups were on the opposite faces, meaning one of them will be pointing down. We made sure a method group pointing down so let's do that and the ethyl group pointing up Now this is our intermediate and we understand that we had a loss of the libyan group. We have formed the bromine and iron which will now perform a backside attack. Right, we have two different carbon atoms that these two positions because they are equally substituted. Even though this is a bit of a longer chain. Right? We have two possibilities to attack in general. A weak nuclear file such as bromide would attack a more substantive position but they're equally substituted. So we have two possibilities for the attack. One of them will result in the attack at the first carbon atom. We have a backside attack and then we would have the loss of the leaving group. So we would form our first product. And because we have a backside attack, we may simply indicate that central carbon carbon bond first because we have a ring opening the right carbon has exactly the same stereo chemistry. So roaming would be pointing up on a solid line and then we would have an ethyl group pointing up in regarding the first position due to the backside attack. This bro mean is now pointing down while in particular the methyl group will still be dashed since booming is already pointing down right, we're basically preserving that bond orientation. So we are good with that structure. But what if we attack this carbon atom and study? We may label this in a different color to understand that there's another pathway. So there will be an attack he backside attack a disposition ring opening in the formation of our next product. So now if we draw the central carbon carbon bomb and start with position number one, nothing really changes because that center did not participate in the reaction. The method group is pointing down still. But this bromide. Well essentially form a carbon bromine bonds pointing down while the ethyl group will preserve its original orientation. It will be pointing out now we have our two structures. If you perform a flip with respect to that horizontal line you will notice that they are indeed in and humors you will simply notice that this protein would be pointing up just as this one. But the configuration of an ethnic group that was originally pointing up will now be opposite will be pointing down and a similar thing happens with this carbon atom. The method group will be pointing up instead of pointing down in this direction. So we may essentially label them as two unanimous right they are and an anti repair. Great job. So we got our two products and we may just add them right, we may go back to reaction number one and we may draw our products that we got. So the first one that we got is this one. Right let's just carefully through it. So that's bro mean another bro mean over here. The first one plus it's an and humor. Great job. So we got our first two products for reaction number one and we are ready to look at reaction number two. What if you like a dizzy I zimmer participating in this reaction. If we re troy it just as with that for part two. My apologies for part one for part two we will have that see Hi Samara and if we recall the staff that we previously took, we will be able to draw a mechanism roaming approaches the double bond, it gets polarized. The level one attacks the lone pair of grooming attacks back to form that bridge. We have the loss of the leaving group. We are forming bro ammonium. Now it will look slightly differently because the two groups are pointing down. That means in particular in this case we are going to troll both of them on. Let's suppose wedges right, let's suppose that they are pointing up Now we may also change the perspective, we may just rotated and it will just look the same right to be the same structure as long as they are both registered ashes, that's perfectly fine. Now this is an ethyl group, this is the metal group. And now, similarly to the previous problem, we have two pathways, we will indicate the first pathway because these two carbon atoms are equally substituted, their cheap possibilities for a weak nuclear file to attack the pathway in blue and then another pathway in green. Now we are ready to draw the tea products, The 1st 1 in blue. If we start with the central carbon carbon bond, right, we may start with the second position roaming will be pointing up now this ethyl group still put services configuration because it didn't participate in the reaction, it was this carbon was just this moment which is the leaving group. So nothing changes in this position. And now here we have a backside attack. So grooming is pointing down and the ethyl group. My apologies. The metal group is pointing out on a watch where the second product we will now draw the central central central central central carbon carbon bond. My apologies. And in particular we have a backside of text. We may start with the first carbon atom. They're seeing muscle group pointing up roaming and now here after the attack. This means pointing down and the ethyl group is pointing up. If we flip this structure with respect to a horizontal line, we will notice again that the configuration has been reversed. An inversion of configuration results and the fact that both centers have inversion of configuration. So we may classify these two as in and tumors again. And now we have our mechanism let's label all of these steps. That's really important for us because one of the parts of this problem asked us to provide the mechanism. So it's pretty much everything right. Everything that we have shown including this part as well. So now the only thing that we want to we want to do our final products. Those will be our major products. They are formed in roughly equal quantities. That's why both of them are major. The first one would be the central carbon carbon bond and then we have a method group pointing up Hessel pointing up roaming on a solid line romain plus it's an and humor and it's an and humor is over here. So we may essentially draw it out that's booming, Essel muscle and grooming. So these are our products, that's our mechanism. You may now clearly see that these are our products to in and commerce for reaction number one and In an tumors for reaction # two. The correct answer to this multiple choice question is A and if you have any questions, don't hesitate to leave your comment down below in the comments section.

Key Concepts

Electrophilic Addition Reactions

Stereochemistry and Isomerism

Mechanism of Bromination

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