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Predict the products when each of the following compounds is treated with NBS and irradiated with UV light:(a)...

Question

Predict the products when each of the following compounds is treated with NBS and irradiated with UV light:(a)

Predict the products when each of the following compounds is treated with NBS and irradiated with UV light: (a)



Answers

How can you use UV spectroscopy to distinguish between the compounds in each of the following pairs?

When we compare a couple of organic compounds and how they will behave in that UKIP's spectrum, in order to predict which will have the longer wavelength maximum absorbent. We look for several things. We look for the extent of conjugation. More conjugation means lower energy, which means longer weight length. If conjugation is the same, then we look at the stability of the trivial resonance structures that can be drawn and the stability is going to be affected by the polarity of the bonds or the electro negativity of the atoms involved. So, for example, in this first pair, the structure on the right has more conjugation. So that's going to have the longer wavelength maximum absorbent in the second pair. The congregation is the same. But the structure on the right has this double bond to an oxygen compared to double bond to a carbon in the first structure. So the uh double bond to the oxygen stabilizes um or create a trivial resonant structure that is more stable. So the longer wavelength will be for the key tone compared to the alkaline in this third pair, the key tone in the structure on the left is not conjugated with the ring, whereas the key tone in the structure on the right is conjugated, so greater conjugation leads to longer landau max, and then finally in the bottom pair, the resonant structure that can be drawn. The trivial resident structure that can be drawn has a positive formal charge on the oxygen atom, and that positive charge is going to be stabilised by a neighboring our group compared to a neighboring hydrogen. They are group has hyper conjugation and electron donating effects that will stabilised at positive charge. So the ring with them of foxy. The vessel should be have a longer wavelength of maximum absorbency than the final compound.

In this problem, the reaction will happen. Something like this. Cst. CH 20 CH two CS three plus or two in Potenza. Hp will produce the product, add this component, see http, see edge oh edge or CH two C s t D. And this compound C six at five ch CS three. CST in pageants of or two 95 degree 1:35° will produce the productive this compound C six H 5. See see http. CS three or or age. So according to the option in this problem. Obstinacy it, correct answer.

Have eyes that some problem. 81 28. You need to credit the product off the given direction. So the parcel is also more slices off. Cyclical came in the presence of zinc and the problem. We'll have both LD head and kitten groups in late, so the double won't pick him and Louisville Gator Kit on on and the hand functional group on the same lonely lives to drop the structure of the Maliki. So this is the kid on and another side. We have going to have this ever go at this Andy head and this is the problem for the first reaction Pontus of injection when the electrical king is directing with petition Parliament an idea in acid X elation these rebels want is going to break and we will get government problems a weak acid and again, in this compound there will be to provoke Cilic assets on the same molecule wanted to the beginning and one at the end. The bad reaction here again, the double bond is going to break. But this is has reparation oxidation reaction and it is going to give us, um and uncle so we know when we're doing head Operation Oxidation We get and go home with a high dosage group is going toe touch to the problem that we just having list substitution. So in this case, this company is having the least substitution. Therefore, um, Hendrickson group is going to end to this carbon and all these have revelation. Oxidation gives us end a product which mean always gives us sin products. In addition, Texas, which means that the have absolute group here and the hydrogen in this travel they're going to be on the same side off the molecule. Now let's draw by structure off, though. Strada, save these Me 10 group is on the back side, off the plane off the paper off the pain of the molecule And the hydroxy little group has been added to the least substituted Adam here and the hydrogen is added to the same sign. I don't want to kill and we have another hydrogen. So this is the product for the 4th 1 Distraction is oxen Operation Democratization reaction again, We're going to get alcohol and the problem problem double bond here will be broken and we will have only a single one. And when we do ox, inauguration, demarcation and absolute group added to the most substituted governmental. Which means that this government is going to have the hydro slip up and the hydrogen. And here and for oxen, decoration, demarcation we always get and a product which means the hydroxy group and the average in group, which is being added to these two car bombs. These are going to have opposite. They're going to add on from the opposite side, clean off the molecule. So let's draw by structure. The high looks ill group is added to the most substituted. Come on, Adam, and the hydrogen is going to be the same site on the opposite side. I hope I look simple and today's other hydrogen here and this is the product. So we get and day addiction Fonda and the sun is inundation.

This is the answer to Chapter 15. Problem number 45 Fromthe Smith Organic chemistry textbook. Ah, And in this chapter, we were asked to draw the products formed when each Al Keane is treated with NBS and light. Ah. And so remember that NBS and light are going to give us, um, broom a nation at the Lilic position oven. L keen. Uh, and so, um, for a, uh, since this molecule is ring, um, we're really only going to get the one product. Um, since any multiple products would look exactly the same. So that's gonna be a accident for B. Uh, we're going to get, um, rumination at the two possible positions here. Ah, and so they will look like this. So, um, we can get rumination here. Ah, and then the other potential Ah, that that we have is for, um, the double bond to re range. And so, uh, we can have the double bond here. Ah, and then we'll get rumination at this position because it's an electric system. So the double bond can, uh can move, and we can get a perm in ation, um, at the new Elik position. So those are our two options for B. So then, for sea, there are actually going to be four potential products here so we can get perma Nation here. We can get, um, Brahma Nation. Ah, here. Ah, and then Thea other potential products that we have are going to be due to a rearrangement of double bond so we can get Aroma Nation here. Uh, and we can get, um, broom in ation here as well. Okay. And so those are our four products for C. Move that up. So I have enough room for D here. Ah, and then in de, um, we're also gonna have four potential products on, and they're gonna be again to, um from a well, it positions in the starting material. Ah, and then two from a little egg positions that are sort of new alot of positions as the result of rearrangement. So here are the two that are a little like positions and starting material. And then if instead we get rearrangement, we will get thes two molecules as well. So to be clear, it's not. It's not like an either, or, uh in both cases, we we will get the rearrangement. And so we will get a mixture of these products. Okay. Ah, and so that's that's the answer. D um, and again, Thio, answer this problem. We just need to remember that NBS and light are gonna give us berman ation at a will. IQ positions. Um, but in these Loic systems of the double bond can rearrange, um, and so we also need to account for that rearrangement in our product molecules. And that's the answer to Chapter 15 problem number.


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