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Write tho acid catalyzed reaction Haworth projections B-D- glucopyranose with methano include all productsName the reaction products problem #3 (1 pt)What E the new...

Question

Write tho acid catalyzed reaction Haworth projections B-D- glucopyranose with methano include all productsName the reaction products problem #3 (1 pt)What E the new functional group?Would the products be reducing sugars?Which carbohydrate(s) could have Ihe following tost results? points, total)(a} Produces reddish-brown solid wilh Benedicts reagent and red color wilh Seliwanoff $ reagent in minute. (2 points)(b) Gives color change with Benedict s test, light orange color with Seliwanoff $ reagen

Write tho acid catalyzed reaction Haworth projections B-D- glucopyranose with methano include all products Name the reaction products problem #3 (1 pt) What E the new functional group? Would the products be reducing sugars? Which carbohydrate(s) could have Ihe following tost results? points, total) (a} Produces reddish-brown solid wilh Benedicts reagent and red color wilh Seliwanoff $ reagent in minute. (2 points) (b) Gives color change with Benedict s test, light orange color with Seliwanoff $ reagent aler minutes, and produces no bubbles during fermentation. (2 points) Give no color change with Benedict's or Seliwanoffs reagents_ but tums blue-black color wth iodine reagent point) Page 3 of 10



Answers

Draw the products of the following reactions:
a. diethyl heptanedioate: (1) sodium ethoxide; (2) HCl
b. pentanoic acid $+\mathrm{PBr}_{3}+\mathrm{Br}_{2},$ followed by hydrolysis
c. acetone + LDA/THF: (1) slow addition of ethyl acetate; (2) HCl
d. diethyl 2-ethylhexanedioate: (1) sodium ethoxide; (2) HCl
e. diethyl malonate: (1) sodium ethoxide; (2) isobutyl bromide; $$\text { (3) } \mathrm{HCl}, \mathrm{H}_{2} \mathrm{O}+\Delta$$
f. acetophenone + LDA/THF: (1) slow addition of diethyl carbonate: (2) HCl

For everyone to hear. We're doing Chapter 18 Problem 36. And this problem asked us to draw the product form When Fino, which is a benzene with an alcohol, is reacted with each of these re agent. So there is quite a bit of re agent. So just bear with me while we work through this. But if you're if you understand why you form the products in each one of these regents, you guys will be in very good shape for any upcoming quizzes or assessments. So let's start with a band before we serve it, eh? So I struck a here before he started today. Let's identified. So a field we know that we have an auction in Reno. Auction has long prayer electrons. So we need a first identify where the substitute they're gonna add is going to be placed on our benzene ring. So we know that because auction has long pair electrons, they can go through some residents structures. That place is a full carbon on a full negative charge on the benzene ring, their force and increased electron density on the bench entering. Therefore, it is unless on donating group. Therefore, we know the lecture on Tony in groups direct or so and pere on our benzene rent. So every one of these creations will have it too. I summers or most of these products will have to ice summers one being orthe ally summer one being the Paradise mur. So for a we see that we're reacting this funeral with a metric oxide and sulfuric acids. We know that we're adding in nature group to this, so fortunately is promise and not ask us for the mechanism because then we'd be here forever doing his mechanism. But hopefully later in textbook, there will be some mechanistic questions that we can go through to fully confirm why we get and Eno to group when we have these regions in these type of problems, so one is going to be an NL too and nitro group on Ortho Bishan Post will get the Paradise Zimmer having that, you know to group at the Pera position now for B for adding s 03 with sulfuric acid. So we're adding the soul final group here, so be well once again citing the soulful group and now also three h but now adding it and the Ortho and proposition just like this nitro group. So all of these regions are in your text spoke so you can just refer to that to confirm the mechanism of these type of problems or look at the later problems that will cover in this textbook. But it's also good for organic. Commission does not have much memorization. But for simple things like these, these air not too hard to memorize. And once you understand the mechanism, it really helps you to clarify why a certain product will be formed. And that really takes the memorization completely out of it is because you can't just rationally work through the mechanism and obtain the product. So before you guys were doing that, or just the first time saving our system for some multiple layers of verification, memorizing the products for these reactions are helpful. So I just want to see, See, we have now this alcohol hale I group with Louis acids. So we know that this is going to be fearless craft calculation. So we're gonna add 12 carbons onto the ortho or onto the para position to Iceman is being formed one, the Ortho, and won the pair I smear once again, it's the Orthodox and have one to carbon And then the para is gonna have once again 12 carbon So in deal So we have now here we have field crafts I'll, uh, a solution now. So in D, I'll draw the easel group just so we can clarify. So we have one too. So we have one too. Then we have die. A thaw like this is writing this group onto the ortho and armed to the power position in separate I sinners. So plus that I simmer terrorism, getting this one. So now let's move on to e even now have halogen nation. But we have held a nation with some sort of Louis acid. So I'll do Ian on different color. Just help us clarify overhears a e writing bro Ming Group. But now, because we have this asset, we reform this Louis asset to every cycle off adding a bombing on this monkey. Also now consequence of that is that you will get Polly prom in ation so you'll get domination at all your available directing site. So you will get pronation old orphan and Paris site because you're reforming us Louis acid and you're making this bro me in group. That diatonic bro me into such a stronger electoral file Because, in essence, what you're forming is br b r plus down Teo E B R three minus. So you're now making this bro Ming group into such a stronger electoral fall now. So because you're adding it in such strong electoral, you're making to destroy Electra file on your reforming this Louis ass in every time you get a consequence of having this polychrome a nation occurring. So what you gonna be end up with is one product with every single directed sites of both Ortho in the one metre or so one para having domination a current now f that problem is resolved because now we don't have that Louis assets. So once again, we're gonna have only one bro Ming being added to either one of the ortho positions or the pair positions. So once again, because we have no lose acid in this time you will get the substitution products with having one. I simmer at the ortho position. And when I saw her at the Pera position Now, Gene, we're doing thie exact same thing as an E, but we're just changing the identity of the daylights out of the yard to we have sealed to instead of Ft. York Trivia Fncl three. So G is exactly like Eve announcing a blooming It's going to be including. And this is just because we're just changing the density of Hey, lives now on each. So in a JJ, What we're doing is we're taking the product of a and they were adding, reducing agent here. So what we gotta do is so let's draw the product of a CZ. We have two product, eh? So we'll draw the products of a first of its age on DH misses one ice mur. Then we have the other ice murder, and now we're adding tin on DH strong acid. So we know that tin with a strong acid using your textbook as a reference is the reducing agent. So because it's the reducing agent, we know that the only reducible groups here alcohol is already at this most reduced form. It was Aldo hide we could reduce that are herds of gastric introduced that but alcohol's already fully reduced. The aromatic ring here is too stable to be touched. So only thing that's left is the natural. Perhaps we can reduce natural to aiming so they can reduce the ortho and pear. I simmers into Ortho Pera ice summers of aiming so N H, too with him and a strong asset or any sort of reducing agent an inch, too. And then I'll switch colors against Let's do I now. So I was taking the products of Divan, adding another reducing agent here. So we're at in part of the D, which is the Fear Crafts isolation product now was I? And now we're adding zinc mercury with assess this, another reducing agent just like Tim Acid. So once again, alcohol's feli reduced. But now, instead of having the Nitro group, we have a key tone which could be reduced to just see it too. So we can fully reduce this basil drew into an alcohol group. This is actually method. How through feel cross isolation, we can get a straight Shane Carden. Yeah, So you get one item. It looks like this and the other I simmer. Dorcas, Dorcas, we get that also the parallelism don't lose that key tone to look like this. Now, J. J What are we doing in a chair? In generating the product of D, they were adding hydrazine and alcohol. So we add hydrazine and alcohol. This is actually another way to fully reduce your a social group. So this is actually very specialized way of reducing a little group. So you actually will get the same products as in I hear. So these three agents, so hydrazine in on hydroxide and I on with zinc mercury, um, and I said, strong acid. The Syrians alert different. They're both reducing agent. Specifically, they will reduce your carbon on group's heir key tones to stretching. Al came. So both I Andrzej, we'll give you the same products. So now what about in K and this hydrogen in the mechanism for this has also shown in the textbook just for reference. But this is one of the case where it's good to have a little bit of memorization, organic chemistry, just to be able to identify these things quickly. So you don't waste your time on a test so Jay will take a product and see, So let's take that product and see okay groups. We have to i summers and we're adding roaming and light, so each new is light. So this is a radical Bram in ation mechanism. And specifically this type of radical Dr Abomination occurs at the bins. Ilic carbons here been zilla Carbon is the first apartment off your Benzedrine. So these are your penzler carbons. Urine after brewing. Ah, hey, Light onto these positions. So you will have once again your toy summers one ad in the roaming off the ortho position and one IDing the brawny off the para position benzel of carbon. And now, last but definitely not least so. L were often the product of C And then we're using came and afford fully reduce that Oh, sorry, fully oxidized that. Sorry. So I was thinking the product of C and we're using came in a four toe fully oxidizes sow. What you will we know about came in before is that it won't touch this alcohol because if you were to put a double bond here than this, carbon here would get five bonds and we know that's not possible. It's only carbons that it can touch. Is this Ah, Ethel Group that we added here. But some people might say that well, we would have a carbon spacer and then oxidized that this terminal carbon. But what would they think that we often is terrible for mechanical force so strong that is gonna oxidize the first carbon off of benzene. So it's gonna happen in multiple steps. First will oxidize this into a key tone, then to an alga hide and into a car park Cilic acid. But overall product is going to be that car looks like acid product with on ly one carbon attached to it. So what I mean by that is that you'll have this carbon here being remains staying on. Then you oxidize it with the Taliban oxygen. But then we're gonna have your car back. So gas, it's so this carbon here falls off in a process of this oxidation because first we go through key tone that I had to hide them car back so gassid but came in before you. You cannot isolate the elder Hideaki Tone product. It's goingto oxides fully right away instantly to your fully oxidized Carmack silk acid form. And you're gonna have this with both of these ice immers, and that's it's for this problem

This is the answer to Chapter 21. Problem number 20 from the Smith Organic Chemistry textbook. Ah, and this problem asks us to draw the product of a series of reactions. Eso were given this starting material here, um, which have drawn which the problem identifies as molecule A. Um So there's molecule a Ah. And we're told that if we treat molecule a successively with first this vinegary agent that have drawn here, um, and then if we, uh, hydrogen eat that product that we get a product see on and were asked thio, identify product. See, um and so the best way to do this is to just take this a step at a time. So first will identify what product be is, and then we will hydrogen ate that product, and that will give us see. So the first thing to do is to look at the reaction of a with this vid agree agent. So, um, the six member aromatic ring portion of this molecule, of course, is not going to be affected. Bye. The vig reaction. So there we go. Six member of aromatic bring method occur, but the top is also unaffected. Um, And so remember the product of ah, vig reaction. Um, I like to think of it is just sort of like adding the double over, replacing the oxygen in the carbon deal with, um with the carbon that's double bound to the phosphorus and then drawing the rest of the chain and basically and so maybe that's not Maybe I didn't do a great job describing that, but so, um, if we were to redraw a right so we would have we would have this, um, and then, of course, in a we have an oxygen here, but because this is the product of, ah, big, we can just put our carbon there. So here's, um let me go ahead and label these in red. So this carbon that I've drawn here, uh, is this car in here? Um, and then we can just go ahead and draw in the rest of our chain. So, um um, so it's going to be ah, Theo to ch two ch three. So c 02 ch to see age three. Um, and of course, have drawn this double bond as a trans double bond. Um, you also might want to know that you would get a mixture, you'd get some cysts as well. However, it really doesn't. Doesn't matter at all for our purposes because, um, when we hydrogenated this in in the next step, um, and we're told to use, um, hydrogen gas and palladium on carbon so each to P D. C. Um, And when we treat this molecule Oh, by the way, this is Ah, this is be the problem. Problem references it as be on. So when we treat be with hydrogen gas in the presence of a palladium catalyst, we're going to lose that double bond anyway. So it doesn't matter. Um, and of course, our aromatic double bonds. This type of hydrogenation is not going to be strong enough to affect them s so they're gonna remain unaffected. Ah, and the rest of our molecule will be exactly the same. Just that that double bond, um, will have been reduced. So there we go. This is molecule. See, um, and so that's ah, the molecule that the problem is actually asking us to identify. So there it is. Ah, and that's the answer to Chapter 21. Problem number 20


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