How many dibrominated products could each of the compounds for...

Question

How many dibrominated products could each of the compounds form if stereoisomers are included?

Accepted Answer

Hey, everyone. And welcome back in today's video, we are going to predict how many di brominated products each of the following compounds can produce including stereo is mars, let's start with structure. A, this is a dying. We have two double bonds in our structure so we can through its bond line structure. And what we're going to do here, we're just going to indicate our double bond. And when we have our double bond, we noticed that then we have our cc single bond and we have our next double bond. Okay. So this is our dying and we essentially want to think about our di brominated products to draw our di brominated products. We have to keep in mind that we are similar replacing our hydrogen atoms with bromine atoms. First of all, the very simplest possibility is to add to blooming atoms by replacing hydrogen is these two implicit Hae Jin's at position one. So this is our first structure that we can get. Now, we immediately think can we get the stereo is more? The answer is no, because we have to exactly the same roaming substitue ints bonded to that carbon, which is sp two hybridized and we have to hide Jen's bonded to the other end of the second double bond. So we cannot have any stereo isthmus. That's the only structure. Now, for the next one, we can just change the position of one of our roaming atoms, right? We can essentially state that what if we keep our top broom in fixed? And for the bottom one, we are just going to add it at position two. So this gives us the following structure which is essentially CIS or Z an arrangement, right? Because the two bromo substitutions are pointing up with respect to the double bond, meaning we can actually get a stereo Esma or basically it's E or trans eyes more because we can make, we can essentially I have two BROMO substitutes pointing up and down with respect to that double bond. So let's see that and let's simply draw its E form where one of the bromo substitutions will be pointing down. So let's suppose that we essentially take the first Roman atom and make sure that it's pointing down. So this would be our E isa okay. So we got it. Now moving on, we can also state that if we change the position of the bromo substation at carbon number two to carbon number three, we get our next structure. So now we are essentially going to add it at carbon # three. Okay. And now here we have to be extra careful. This is our E eyes Mar because we have one double bond where stereo is mars are possible, bromo substitution is pointing up and our L Q group is pointing down their highest priority. So essentially that sorry eyes more or simply speaking, you can imagine that you're expanding these two implicit hydrogen atoms, they're on opposite sides. So that's basically our trans structure, right? The two hydrogen atoms are on opposite sides. So now what you can do, you can actually get a Z I summer if you get two hydrants pointing up, simply swapping this bro mean with the adjacent hydrogen and making sure that Brahman is pointing down. So this will give you your c structure. So let's add that Ci Zimmer by simply adding bromo substitution, which is now pointing down. So the two hydrants are pointing up and therefore, this must be Z okay. So we got it. Now, what if we change the position of our second Roman atom to carbon number four, right? We can do that as well and we get the following. So now we got roaming bonded to carbon number four, number one and number four, right? So we got it. And now we have to notice that for this one, both double bonds can produce theory of Eisenman's, right. So we, we will have four total from here, right? Because we have at position number one, we have one e because the two hydrants are pointing in opposite directions. And now thinking about position number three, right? Because this is where the double bond starts. That would be we e as well because once again, your hydrants are pointing in opposite directions with respect to the double bond. So we can have one E three, E, one E three, Z one C three E and one C three C, right? Two to the power of two, that's four for the total here. So let's throw them out for complete clarity and that will be all the items that we can get. So we got our one E three E, if we think about our next structure, we can simply make sure that the first Roman Adam is pointing down keeping the other one fixed, right. So two hydrogen atoms are now pointing up. So instead of one E, we actually get one Z okay, we got one E 31 Z three. Now we want to do the forming, we want to get our third one. So let's see what we can see. We can essentially start with the first structure that we got for our die in with two configurations. So first of all, we're going to keep that Romo supposition at carbon number one fixed. And we are just going to make sure that the bromo substation at carbon number four is actually pointing up, right? We can do that for sure. So if this promo is pointing up, then we will have one E and then three easy because the two hydrogen atoms are pointing down. So that's our Z I Zimmer, That would be three Z. And let's see what else we need. We want to get one Z three Z. So essentially, we're going to take the second structure from this set, this one. And we are simply going to state that we can also swab hygiene and roaming at position number four. And basically bombing will be pointing up. Hye Jin will be implicit. And we noticed that this structure would be one Z right to hydrants pointing up Sweezy to hydrants pointing down. And this would correspond to our final possibility for this dying from the final set where we said we would have four total here and we got them because we have either E or C possibilities for two positions two to the power of two that gives us four. So we got all of them right? Let's count how many we have. So 123456789. And let's see there can there actually be more notice that we only took our booming fixed at position number one. But what if we change our BMO substitute into position number two, we can get the following which would be our final visibility, right? Because the structure is symmetrical with respect to that carbon carbon bond, meaning we can still change the position of the first of the booming atom at carbon number one to position number two, and this gives us our next possibility in which we will perform the following. We are, we are just going to start with our original structure, our dying And we're going to replace the following two hydrogen atoms. One at carbon number two and one at carbon number three. And this structure doesn't have any stereo wiseman's notice how each double bond has to exactly same hydrogen substitutions on each end. So there will be no sterilize mars because we need to have two different substitutions right on each end of the double bond. So this is our additional one and that would be it, right? Essentially, if we had this problem into position number four, then that would give us the previously analyzed structure, right? Basically one comma two di substituted. So now we have all of them and let's count them again. So that's 123456789, 10. So we have a total of 10 possible structures. Now, moving on to be we are given an alkaline, right, we have CC triple bond and then we have two metal groups on each end. It's really important to understand that this molecule is symmetrical. So we can just think about one part of it. On the left side, we have a metal group. So we can actually replace two hydrogen atoms with bromine atoms immediately. And we get our first possibility. So we're going to draw C H two Bromo substitutions valid. Now, there's not much to do here because the other possibility is just to replace another hydrogen from the other metal group. Because these two metal groups are the only groups that are a source of hydrogen. So we can essentially have another structure where we are replacing hydrogen from the other metal group with roaming. So get cease to be our group on each end. Now, if we move this room into the other carbon and have two bromo substations bonded to that carbon, that would be exactly the same structure structure number one, due to the fact that it's symmetrical just as we analyzed. So here, it's much simpler because we only have two possibilities, either replace two hydrogen atoms from the first carbon Or one hydrogen from the first carbon and the 4th 1. So we only have two possibilities. There are no stereo is mars because Elkins do not correspond to stereo is mars. So we only have two here. Okay. So we got our answers. For the first part, we got some possible di brominated structures. And for B we got only two possible di brominated structures based on our answer. The correct answer to the multiple choice question is B Thank you for watching.

How many dibrominated products could each of the compounds form if stereoisomers are included?

Key Concepts

Dibromination

Stereoisomers

Product Distribution

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