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2", 3" alcohol with mol lecular: formula C H:O 0f 19 pts) stfuctute Draw theof enol with molecular foriula CHs0 b) Draw the structuresymmetrical and unsym...

Question

2", 3" alcohol with mol lecular: formula C H:O 0f 19 pts) stfuctute Draw theof enol with molecular foriula CHs0 b) Draw the structuresymmetrical and unsymmetrical ethers: the structure of Drawof unsaturation present in the following compounds: 3 pts) How many degree CsHs; CsHus, C,HaO, C HnBr, CsHsNRing opening of Epoxides pts}HBr1) HzSO4 2) HzoDraw the all possible productsPBra SN; Rerction with alcohols: (5 pts)Draw lhe pogeible machanls with arrow pushingHBr1} TaiPy 2) NaerPBig

2", 3" alcohol with mol lecular: formula C H:O 0f 19 pts) stfuctute Draw the of enol with molecular foriula CHs0 b) Draw the structure symmetrical and unsymmetrical ethers: the structure of Draw of unsaturation present in the following compounds: 3 pts) How many degree CsHs; CsHus, C,HaO, C HnBr, CsHsN Ring opening of Epoxides pts} HBr 1) HzSO4 2) Hzo Draw the all possible products PBra SN; Rerction with alcohols: (5 pts) Draw lhe pogeible machanls with arrow pushing HBr 1} TaiPy 2) Naer PBig



Answers

Draw the condensed structural formula for each of the following: a. ethanol b. 3 -methyl- 2 -iodo- 1 -butanol c. 2 -propyn-1-ol (propargyl alcohol) d. 2 -fluorophenol

So for this problem, you're given the name and you want to draw the condensed formula that corresponds to this name. So the easiest way that I found to do these is to start with your own coaching. So for part A, you have provable alcohol. So Pro Bowl is going to be ch three ch two, ch two because purple is three carbons one and because it doesn't give a number, you're going to assume the alcohol is going to be on carbon one. So in this case, this would be curry one carbon to carbon three. So this should be Pro Bowl alcohol For the second problem, I could just scroll down here for a minute. You have three pensionable. So you want to start from the longest carbon champ, which, in this case, because it's pensionable, let's change this to green. Hence, General, it would be five carbons so we can draw all C c c c c. And if we know that three is going to be the site of our alcohol, So you have o. H here. And so then you just feel in all of your hydrogen is and you can do this either by drawing bonds to each of these carbons. Or you could just do kind of convinced formula like this. So ch three, ch two by two ch with this o age group to C H two to ch three. So now just double check. You have 12345 carbons from pentacle and your alcohol group is on card in three right here. Okay, so let's go to letter C. You have you been all which tells you that you have four carbons, So let's start by drawing it out. 1234 with your alcohol group on carbon too. So if we number, this is 1234 or alcohol group is going to be right here. Now what's different about this problem as opposed to part A and part B, as we also also have a substitute to metal group. And so this tells you that on carbon to you also have a metal group so we can draw that in a swell. So now all we have to go do is fill in our carbon. It's so we have ch three bonded to carbon this carbon to is buying to an alcohol and a metal group and so it doesn't have any hydrogen is on it. And then you have ch two bonded to see each three. And then again, you just wanted a kind of double check. So you have to mess along with metal group on carbon to two beautiful, which is a four carbon chine with your alcohol on carbon too. So for you, the last problem. You want to kind of break this up again. So you have a final ring which, if you remember, is a six member of Ring six member benzene ring with an O age group that carbon one because it doesn't specify what number you're alcohol design. It's always going to be on card and one in these cases. Now, you notice again you have substituted on carbon for so few number these carbons. 1234 then bro mean would be on carbon for and this would be promo fennel for promotional

Now work on problems. Seventy two from chapter twenty one here are asked to draw the structure forfour alcohols from the names that were given. So let's start with party, eh? Were given the name too beautiful. So we know from the butte prefix here we're going to have four carbons so we can write just the skeleton here and then from the ol Suffolk's. We had know that there's going to be an alcohol at position too, so we can put it on the second carbon. Either Carbon would be correct in the middle here, this one or this one. So now we can write the whole structure Ch three ch ohh ch too ch three. So in part B were given a name of two metal one protein off. So we know that from probe that we're going to have three C's, C, C and C. And we know from the the numbers here, we're going to have a method, that position, too, and we're going to have an alcohol at position one on either side is correct. So we can write our full structure. Ch three c h c h three ch, too. Oh, h. So that's the structure for part B We'LL go ahead and open a new page moved to Park Si but we were given three Ethel one Hexen Hexen All means we have six carbons. We can have an ethnic group of position three and a alcohol at position one. So let's go ahead and draw in our structure here. We know that we're going to need to put an ethel at position three Instance, We're putting the alcohol on this side here. This is one, two, three. So we put in our Ethel route here. So that's their section for part, See, and finally for party were given to meth or three Penton all so we know from the pent here that we need five C's. We have a method opposition to and an alcohol at position three. So we know that because we have five carbons or alcohol is going directly in the middle, we can put our method on either side and be correct. Well, go ahead and number from the right to the left so that it goes on the right of the alcohol and that's our structure for her d

Thank you. Let's try drawing the structure of alcohols based on the names provided two. Beautiful. So we know that we have four carbons and on carbon to is an alcohol. So we have ch three. This would be ch two, but there's an alcohol group C H two C h three and that is too beautiful. Two metal, one pro panel. So we have C H to O H C H two c h three And this carbon has a methyl group bonded to it currently two. And that will be our molecule. Three Ethel, one hexagonal. We will have Ohh! C h two c h two c h because there is a method group on carbon three c h 21234 five and six Carbons three Metal earth three Ethel lips to be a ch two ch three group, not a CHD group. That would be three metal instead of three. Ethel. So that is going to be three Ethel. One Hexen all and finally to metal three Penton All so we will have ch three c h ch three. Okay. C h Ohh! C h two c h three And that will be our final molecular

In this problem. We're drawing the structure for each alcohol. And so we begin with two. Beautiful. So we have four carbons since it is a ah Butte tunnel, and then our alcohol group is on the 2nd 1 For letter B. We have two metal one probe in all, and so will draw our three carbon probe chain. And then the, uh, alcohol is on the one carbon, and there is a method group on the two. For letters C, we have three Ethel, one Hexen all So we'll draw our six carbon hex chain. We'll put our alcohol on the one, and then our three will have an Ethel. And for letter D, we have two metal three Penton all And so because of the pence, will draw five carbon chain. We have our alcohol group on the third Carbon and aim Ethel on the second


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