When HBr adds across the double bond of 1,2-dimethylcyclopente...

Question

When HBr adds across the double bond of 1,2-dimethylcyclopentene, the product is a mixture of the cis and trans isomers. Show why this addition is not stereospecific.

Accepted Answer

Hey everyone and welcome back in today's video we are going to solve another problem. The hydro chlorination of one to the method cycle hex one in results in the formation of four eyes America products writing mechanism. That explains why this addition reaction is not stereo specific. We may begin the problem by drawing our substrate and as the name suggests we have cycles vaccine. We are going to add the double bond at position number one. It doesn't matter which one is your first carbon atom because we're only adding a double bond so far. So the double bond defines position number one and extends to position number two and these two carbon atoms have methyl groups just as the name suggests. One committee dimethyl. The problem also tells us that this is a hydro chlorination reaction. So we are going to add hydrogen chloride in our first step and we know that this is an addition reaction. We call that an electrical condition is observed whenever we start with an elk in and we are adding a hydrogen. Hey light to our system. Now in the first step, the pi bond attacks the protein and we have the protein nation step followed by the loss of the leaving group. And this step is reversible because the carbon cara informed can be D protein ated to form an L. Keen in an elimination reaction easily. Right so now the question is where do we add that hydrogen based on the mechanical school? We know that hydrogen is added to a less substituted carbon atom and because they are equally substituted and also symmetrical notice that there is a line of symmetry which is horizontal passing through our compound. It doesn't really matter which one we choose. Let's suppose that we add hydrants, the bottom carbon atom. So the structure that we get here looks as follows. We are going to draw our six member ring than because we're adding hygiene, a disposition it's now fully saturated and at the adjacent position position number one we have our car Bokeria informed because it's lacking electrons. Now when this car will get informed, it's really important to understand that there's a free rotation at this position. Notice first of all that it's Cairo because this carbon has four different substitutions and therefore we may either observe a wedge indicating that the muscle group is pointing up or a dash indicating that the muscle group is pointing down during the formation of the car broke Orion. The free bond rotation allows us to achieve two different configurations at this position. Right? Because as Scarlett has four different substations, we may have another version of this car, both Caroline and green. So we may simply draw a method group pointing down in our next structure and this one is plainer, right? It doesn't have any stereo chemistry. So we just draw a solid line. So we have formed our T. Boss, we'll carbo carry ons and then we are going to think about our next step. First of all, let's start with the blue pathway and let's think about the first car bokhari informed, so we are just going to retry it and now the chlorate and are informed in the first stop can now perform a nuclear attack. So the chlorate and iron would act as a nuclear file. It would attack that carb. Okay Ryan. And we understand that during this attack we can attack it from both phases from the top phase and the bottom phase which would give us a pair of finance commerce. Right? In this case it wouldn't really be correct to say an anti immerse because this position will not be affected. So it would be more accurate to say a pair of di S theory emerges right? Because we will only have an R. As configurations at this position but this position will retain as configuration. So we can now draw The cheap possible products for this one. For the first one we have an attack from the same side. So we can say same side attack. And the product that we are observing here would essentially correspond to the glory in chloral substitution pointing up because we have the same side right. We are noticing that the original methyl group is pointing up if the chlorine, the client and and is attacking the carbon Karen from the same side that would be pointing up as well. So that would be our first structure. We can essentially indicate that chlorine is pointing up and that means the remaining method group must be pointing down while the adjacent position retains as configuration. And similarly, if we think about the attack from the opposite side, so that would be opposite side attack. We would then see that chlorine is added from the bottom side and then chlorine would be pointing down. So that would be our second structure. So that would be Our six member dring. Our method group is pointing up and chlorine is pointing down now while the adjacent position retains its configuration. And since that retention we have a pair of the S tremors. These two structures are by di astronomers. Okay, so we got our two products for the blue pathway and this starts to make sense. We need to have more products meaning we will observe something really similar for the green pathway. If we take a look at the green carbon cara informed, we can now simply redraw it. We understand that there's a positive charge here and the adjacent methyl group is pointing down. We also know that there would be the chloride and I and acting as a nuclear file and we have our attack right from the same side or the opposite side. So first of all we can say the attack from the same side, same side attack right? So because the method group is pointing down chlorine will be pointing down as well, it will be attacking from the bottom and this would correspond to our first structure chlorine would be pointing down and the other muscle group will be pointing up as a result and now the adjacent position retains its configuration. We just want to make sure that we have chlorine here And for the other one we have an opposite side attack in this case chlorate and I would attack from the top phase. And then chloral substitution would be pointing up. So this is how we get our next structure which is the fourth product meaning we are nearly done. The chloral substitution is pointing up meaning the metal substitution is pointing down and nothing changes regarding this position. It's not participating in the reaction. This method group is pointing down and we may notice that because the inversion of configuration only takes place at one position. These two are again diocese farmers. Okay, so we got our four products and this explains why this reaction is not stereo specific. In other words, it produces even four possibilities for our configurations. You may notice that we have two pairs of the STM Rs. If we think about the first structure over here and the third structure, they are in and commerce each other and we have four possibilities for our configurations. This essentially implies that there is no preferred stereo chemistry and we have multiple possibilities due to the fact that we are forming a car broken iron that has a caramel center. So the two methyl groups can have different configurations and then the car will carry on itself will become Cairo after the nuclear filic attack, giving rise to two more products, right? We will have four possible products total, and now, based on that, the correct answer choice, This multiple choice question is C. However, if you have any questions, feel free to leave your comment down below in the comments section and thank you for watching.

When HBr adds across the double bond of 1,2-dimethylcyclopentene, the product is a mixture of the cis and trans isomers. Show why this addition is not stereospecific.

Key Concepts

Electrophilic Addition

Carbocation Stability and Rearrangement

Stereochemistry and Isomerism

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