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Draw structural formula for the major organic product of the reaction shown belowBr[(CH3JCJzCuLiConsider ElZ stereochemistry of alkenes Do not show stereochemistry ...

Question

Draw structural formula for the major organic product of the reaction shown belowBr[(CH3JCJzCuLiConsider ElZ stereochemistry of alkenes Do not show stereochemistry in other cases. You do not have to explicitly draw H atoms. Do not include organocopper Or inorganic ion by-products in your answeropy AeleChemDoodle'

Draw structural formula for the major organic product of the reaction shown below Br [(CH3JCJzCuLi Consider ElZ stereochemistry of alkenes Do not show stereochemistry in other cases. You do not have to explicitly draw H atoms. Do not include organocopper Or inorganic ion by-products in your answer opy Aele ChemDoodle'



Answers

Draw a structural formula for an alkene with the indicated molecular formula that gives the compound shown as the major product. Note that more than one alkene may give the same compound as the major product.

Okay, So for this problem, you're doing the dehydration reaction and in dehydration reactions. All right, this here for dehydration, basically removing water, and you're going to be forming an AL Keen, which is a carbon carbon double bond. So your water is going to come from your alcohol and this water H 20 is going to come from your alcohol, which is going to be on one carbon, and you're going to grab ah, hydrogen, one of these hydrogen ins from an adjacent carbon. So if you kind of draw it out, call it this way so you can kind of see it easier. So you have moved to call. This are for the rest of your effort cheese. And on this terminal carbon, which is the one in blue, you have a no h and you have your hydrogen. So just to kind of keep it in the back of your mind So again for a dehydration reaction. So ch three ch two ch. So now you loose this hydrogen shown here in green. You lose this alcohol, so you're going to form a new double bond between this carbon and green and the carbon in blue each que plus we'll call this each. Ohh. So this would be your out keen plus water. Okay, let's go to the next one. So again, you're losing h 20 so you're losing an alcohol or Euro age group from this carbon? And because this cyclo plantain is symmetric, it doesn't matter if you grab a hydrogen from either of these two indicated carbons as long as you're grabbing a hydrogen from the adjacent carbon. So, for example, for this problem, let's grab the one on the right. So now you would have your cyclo pen teen plus your water because now you lose this, this form your water and you're left with a double bond on this carbon. No, the last one. So now instead of your alcohol being on a ring at the end of the chain now it's in the middle of a chain. Um, for these for this problem, you can either grab this hydrogen just to make it easier to see where you can grab this hydrogen. And again, just like the cyclo plantain, it's symmetric. So no matter which hydrogen, you remove your going to form the same product. So just for the sake of argument, Let's remove the one on the left in green. So now we have ch three ch. And now this bond right here is going to be come a double bond. So ch plus for a rather ch two ch three plus now your work. So that's all you need. Just so just remember for dehydration, you were losing water and forming an LP.

Okay, so we're doing dehydration reactions. What you want to remember? Is that for dehydration? No, just kind of Write this down here. You're losing water, and you're going to be forming in l cane. We're sorry. You're going to be forming out keen, which is a carbon carbon double barked. So which just kind of separate? This are kind of keep it in the back of your mind. So in terms of this problem, you're going to lose water by. I guess your water is going to come from an O. H and from ah, hydrogen. And these were going to be on a Jason carbons. So if we look at this first example, we can grab this alcohol here. So this is on this carbon. So now we just need so this carbon is attached to this carbon. So we just need to grab one of thes hydrogen ins. And so let's So we grabbed one of these hydrogen will be left with ch two, and now this is going to form a double bond between this green carbon and this blue carbon. So ch two plus h o h. So now this is our green carbon with our hydrogen. This are blue carbon where he we lost very alcohol And okay, so let's do a different let's do the next problem. So again, we have our alcohol here, and this is attached to this carbon here. So this is our carbon next door. And in order to former water, this carbon needs to give up one of its hydrogen. It's so if we have ch three see with our methyl group and now these two carbons are going to form a double bond plus each O h. And again, if we kind of go back to our original problem, you'll see that this green hydrogen was originally bonded to this green carbon in this green alcohol. We're sorry. This blue alcohol was originally bondage to this blue carbon. Okay. And so for our last problem. So you have an alcohol here attached to this carbon here. Okay. And what you want, remember, is you also have hydrogen ins on each of these skeletal carbon. And because this ring is symmetric, meaning there's no other substitue INTs. It doesn't matter which hydrogen you remove, because it's going to be the same regardless. So let's move this green hydrogen here from this carbon. So if we remove our water now, we're forming a new double bond between these two carbons, and we're left with each owe each were. Now, this is our hydrogen, and this is from our alcohol.

Strong structural formula that meets the given criteria for a richard draw and al caine That has six carbons all secondary carbons. So cyclo hexane mean meets this criteria where all the carbons, our secondary carbon atoms. Mhm. Mhm. For B. Where to draw an al caine. Mhm. That is eight carbons. Mhm. Where there are all primary hydrogen. Yeah. And 2233 tetra methyl beauty mean? Yeah. Right fits this criteria. Okay. Mhm. Yeah. Yeah. Should be yeah. Mhm stretcher here. Yeah. And for C we are to draw an al came That is seven carbons With two Isopropyl groups. Yeah. This is It's going to be 24 diamond for painting before we Yeah. Mhm. Yes, you're here, Yeah. Mhm. Me Which C? Mhm. So you weeks? Yeah. Mm three and CH three.

So here we are, just looking at some chemical transformations where we have the final product as well as the re agents, and we are needing to work backwards in order to deduce the formula on the structure off are starting materials. So our formula being C five h 10 was listed up up over the top of the screen, as you can see there. And so for each chemical transformation 123 I've drawn out my starting material on this is the starting material that reacts with R B R. Two. Now the site of all reactivity is at that double bond where were simply adding a bro mean to each end of our double bond, and this trend is present in all three of our chemical transformations.


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