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How many dilferent monobrominated products can be obtained when the pllowing molecule treated with Brz and ultraviolet E light? Just Indicate the number No need dra...

Question

How many dilferent monobrominated products can be obtained when the pllowing molecule treated with Brz and ultraviolet E light? Just Indicate the number No need drawthem points)

How many dilferent monobrominated products can be obtained when the pllowing molecule treated with Brz and ultraviolet E light? Just Indicate the number No need drawthem points)



Answers


How many high-phosphoryl-transferpotential molecules are required to synthesize phosphatidylethanolamine from ethanolamine and diacylglycerol? Assume that the ethanolamine is the activated component.

In this problem, I'm writing the answer or the structure will be something like this. CS three Ch cl, CH two double born CH two. It will produce two product the can will CH two cl, CH two, ch double gone CH two. It will produce one product CS three CH double one, th ch two cl It will produce two product. And finally, I can write this structure. Age this CH two cl at CS three at as village CS Teresita Will one, see as CH two cl at so and said will be equal to fight When there will be equal to 55 will be the answer.

This question asked us to figure out the major product obtained from each of these reactions with feel to an ultraviolet light. So first thing we're going to do is remember that chlorine has a preference for tertiary hydrogen. So tertiary over secondary over primary. And that ratio is five 23.8 to 1. So we'll do some math year. So for this first structure right here we have six primary hydrogen, so six times one is six. For the relative probability primaries being replaced, We have 2468 secondary hydrogen. So eight times 3.8 is you see, eight times 3.8 is 30.4, and then we also have Teoh Tertiary hydro joins right here in here. So two times five is 10. So 30.4 is the biggest number we have here. So we're gonna replace a secondary hydrogen. Um, and there's a couple ways to do that on the structure so we can have this one of these ones being replaced. You could also have this one being replaced. What should be the same as this one right here. Um and then we can also replace this one over here with the chlorine. So those are the possibilities for a and then for B will do the exact same thing. Here we have the same number of each type of hydrogen we have, um, six primaries on our methyl groups here and here. And then we have eight secondaries on these four carbons and to tertiary on this carbon. And this carbon so again will replace a secondary carbon. And in this case, um, they are all the same because their symmetry here So our product will look like this. And then for this bottom one down here. Now we have nine primary hydrogen. So that's nine times one is nine, and then we have six secondary hydrogen, six times 3.8 is 22.8, and we have three tertiary as let's three times five is 15. So, once again, our secondary hydrophones have the biggest relative probability of being replaced. So will replace one of those. And that product will look like this. So there is our product for a seat

Treatment over 34 Where this is the issue off. Reporters off the formula C 68 10 great structures of those borders. And here's the H model. See Memar or used to process penalize them. Emulation produces integrated, which is more stable, and observers who would be over 200 million. Usually you would allow for election on a limb when double one here and here. The observed off such component will be the law. Tow it, the regulations over the walls from the different we owe the the and E the That's, um, uh has a free, different types of cover. So as Compound B, which is a resiliency, which is e Z. Six different purpose. So come on the scene, then there, defined by government, just doesn't. It will hear six signals and compounds and will have prison compounds, and we can be distinguished by each one of them are bythe green. Just 46 hydrogen. Those problem customs will be different. The Trends Council would be higher, and that could lead us distinguish indeed, by h one in the math

This is the answer to Chapter one problem Number 23 from the Smith Organic Chemistry textbook on this problem gives us two molecules that we're told are, uh, I guess components of common sunscreens. Um, and in these two molecules, there are several carbons indicated in each On were asked to identify how many's hydrogen atoms our present around each indicated carbon on. And we're also asked what the molecular formula is for each molecule. Um and so this is the first problem where we've been introduced Thio the line drawing type of notation that use the organic chemistry. In my opinion, this is by far the best way to write any molecule under any circumstance, because it's it's the most condensed way. And once you are used to looking at it, it's the easiest way to look at a molecule and to see what type of re activities that might have what functional groups are present. Um, I think it's the best of the possible ways on dhe, so we should try to become used to looking at these these line drawings or skeletal drawings. They're also called, um, and so remember every, uh, the end of a line is a carbon or like a corner or junction of two lines is a carbon, um, and remember, unless otherwise indicated, the carbon is making four bonds. And so if it appears to be drawn with only two bonds, uh, then we can assume that it has two hydrogen ZX. It appears to be drawn. Only one bond. It has three Hodgins, etcetera, etcetera. So, for example, even though we're not asked about this carbon that I'm indicating here, it looks like it only has one bond s So we can safely say that the three bonds that we don't see are 200 Jin's. Okay, Um and so that's that's all that this problem really is. So if we start at the left and go left to right um, I guess I could have used this 1st 1 is my example. So it looks like this carbon is only bound oxygen us. So there are three hydrogen sze that are not shown, but they are there. Um, so then moving to the right. This next Corbyn, that's indicated, actually has zero. Hydrogen is around it, so it has a double bond to another carbon, a single bond. Thio yet another carbon and then a single bond to 1/3 carbon for a total of four bonds. Ah, and so there are no Hodgins. All of its four bonds are two other carbons. So the next carbon carbon immediately to the right has a triple par Pardon me? A double bond shown and a single bond shown. So there's three of its four bonds accounted for, So that leaves one bond to be taken up by hydrogen on dhe. Then the last indicated carbon in this molecule has three single bonds showed s so we can assume that it has one hydrogen. Ah, and that's a and so will follow of the exact same reasoning for B. S. O. Again moving left to right. This 1st 1 at the top of the ring has a double bond and a single bond, so that leaves one bond to be occupied by a hydrogen. I'm moving to the right. This carbon, you'll carbon on, I guess if you guys air reading this book in order, you have no idea what a carbonell is. Yet Esso car video is a carbon double bound to oxygen. Eso This next indicated carbon is part of a carbon eel. Um, and it has to bonds to other Corbin's and to bonds the oxygen for a total of four, which leaves zero to be occupied by hydrogen Sze. Likewise, the next carbon to the right eyes bound to four other carbons. So it has zero bonds to be occupied by Hydra Jin's on the last carbon in this molecule appears to be bound only to one off carbon. So it has three bonds to be occupied by Hydra Jin's, so three ages. Okay, I mean, so the other thing that we're asked to do in this problem is to identify the molecular formula for each of these molecules. And really, To do that, all that you need to do is, um, Count carbons, Oxygen's and hydrogen Sze. Um, if it helps, you can go through and draw in every single hydrogen at least until you get good enough at line drawings to just know where they are. So a is gonna be C 18 c 18 each. 26 03 Um and then be is going to be C 20 h 22 03 c 20 each. 22 03 on again to get those formulas, you would just have to count. And that's the answer to Chapter one problem number 23.


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