When 1 mole of buta-1,3-diene reacts with 1 mole of HBr, both ...

Question

When 1 mole of buta-1,3-diene reacts with 1 mole of HBr, both 3-bromobut-1-ene and 1-bromobut-2-ene are formed. Propose a mechanism to account for this mixture of products.

Accepted Answer

Hey everyone and welcome back in today's video, we are going to tackle the following problem, the reaction of one mole of four metal panta. Once we die in with one mole of hydrochloric acid produces a mixture of the following three products. Four bromo T metal pan two in E. For Romo for metal pants in and Z. Four bromo four metal pencil. You in writing mechanism that accounts for the formation of these products. We may begin our problem by drawing the original structure, which is for metal danta once we die in. So by drawing a five member carbon chain because that's Penta dane, We are going to add double bonds at position number one and 3. And in addition to that, we noticed that carbon number four has a metal substitution, that's our starting material and we're using one mole of hydrochloric acid to produce three different products starting with the first one, which is panting The position of the double bond. Is that carbon number 2? Carbon # two also has a metal substitution And carbon # four has a bromo substitution. The next product that we obtain is E. So that would be trans for bromo for metal pencil you in. So let's draw pen to in again. The double bond starts at position number two and that's the correct structure because recall that he means trends. So the two lkoh suspicions are on opposite sides with respect to the double bond. And now carbon # four has a bromo substitute print and another methyl group. And similarly the third product is just the c. Is more of it. So we are going to change the position of the method group and we will make it point upwards. So if we do that we will essentially get the following the double bond and now there's a bromo and a metal substitution. So these are E. N. C. This is just the this is version of the previous structure. Now we see our products, we are now going to think about a possible mechanism and how can we get them? Because the system is conjugated. We might just choose the first double bond in the mechanism and see if that works for us because if it's conjugated, that means the carbon Karen will be resonance stabilized and we will most likely get all of our products by choosing any of the two double bonds. So let's go with the first one and let's try and to our mechanism if we redraw the original structure. The first step in the electra filic edition mechanism as you recall, is the protein nation step where we are capturing the protein and we are forming a carbo Karen. So the double bond would attack a molecule of hydro bro, Marcus said, then we would have the loss of the leaving group that would form us the bro might and I am now it's very important to remember that in this step, we are forming the more comical edition meaning hydrogen will be added to a less substituted position to form a more stable car Bull Karen. So now we are forming our secondary carbo can iron With a positive charge at position # two. And now we have to be careful. Previously we discussed the concept of conjugation. So it's a resonance stabilized car Bull Karen, meaning what we're going to do now, we will simply Draw its resonance forms. Let's start with the 1st 1. If it undergoes resonance, the pi electrons would shift here, forming a more stable tertiary carb. Okay, Ryan, so we would have a double bond and a positive charge at a different position. But also before we look at residents, it's often important to think about bond rotation. So why don't we just perform bond rotation and then see how that structure forms another tertiary carbon Karen, that will be very important for our mechanism. So if we rotate that bond in particular, that's not residence. So let's just indicate an arrow and say that we are just rotating one of the bonds, we can choose that ethyl group. And if we perform that rotation, we will essentially have this structure with a positive charge still at position number two over here and now we can think about the residence form that it shows. If we let's take another color for complete clarity. The double bond, we'll go here to form a tertiary carbo car Ryan and that's it. We have our possible carbon Karen's. Now let's think about each carbon Karen separately, starting with the blue one. And now let's remember that the next step is the nuclear attack step. and starting with Carbo Karen one, which in each case is exactly the same. These blue ones are exactly the same representations of the same carbon Karen because the bond is constantly rotating. So let's just take the first one to indicate our next step. If we start with the blue one, the kabul Karen undergoes a nuclear attack step, in which bromide acts as a nuclear file and it basically attacks that car broke Orion. That's how we will be forming our first product. So let's see what we get, romain gets attached and everything else stays the same. This is indeed one of our products, that's our product, number one. Great job. So now let's look at the second cardboard cat iron in green, we have the second representation of the carbo Karen, this is a tertiary carbo cara. Now let's try it more clearly similarly to the previous stuff, we just have a nuclear attack. So the bromine and iron attacks the carbo cast iron to form our next product which now has blooming attached tip. Well done. That's indeed one of our next products which was the e version or the trans version of the al King. And finally, if we look at the red one, which will explain us why it was very important to consider the bond rotation. We will get the following redrawing our kabul cast iron. We understand that again, the bromide and ion with attack it and that's how we get. I will see I Zimmer or desist I Zimmer. We now noticed that there will be two method groups and roaming etched, so that's indeed our third product and therefore we have successfully shown this mechanism. We have shown how to form these three products step by step. If we go up a bit, we may now see a complete mechanism and we may just label it as our final mechanism. That explains a complete formation of these three products. That said, the correct answer to this multiple choice question is B However, if you have any questions, don't have states to leave your comment down below in the comment section and thank you for watching.

When 1 mole of buta-1,3-diene reacts with 1 mole of HBr, both 3-bromobut-1-ene and 1-bromobut-2-ene are formed. Propose a mechanism to account for this mixture of products.

Key Concepts

Electrophilic Addition

Carbocation Stability

Markovnikov's Rule

Watch next