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Mglethcr 6r Ln 1LLo h Pgn?Pcc ~oh~mgGn Xsynthesize 1,2,3- and any other necessary reagents to the only source of carbon, Use benzyl alcohol as triphenylpropan-2-ol:...

Question

Mglethcr 6r Ln 1LLo h Pgn?Pcc ~oh~mgGn Xsynthesize 1,2,3- and any other necessary reagents to the only source of carbon, Use benzyl alcohol as triphenylpropan-2-ol:OH2,3-triphenylpropan-2-ol

mglethcr 6r Ln 1 LLo h Pgn? Pcc ~oh ~mgGn X synthesize 1,2,3- and any other necessary reagents to the only source of carbon, Use benzyl alcohol as triphenylpropan-2-ol: OH 2,3-triphenylpropan-2-ol



Answers

Synthesize each compound from 3-pentanone $[(\mathrm{CH}_3\mathrm{CH}_2)_2\mathrm{C}=\mathrm{O}]$. You may also use benzene, organic alcohols having$ 3 C's, and any required inorganic reagents.

Problem, 9.32 were supposed to provide the reaction pathway. The convert one your time is that carbon source into the specified products. Let's begin in part A. We start with a four carbon chain and we will end with or carbon James. We can see that the only thing we need to do is add two carbons across the triple pot, so we will start with two equivalents of seal, too. Each of these will add in Trans and give us the four carbons a solvent and I can use for this is ch two C Out to now. Let's get part B reaction be No. If that there is still a four carbon chain in the final Montiel. Here's four carbons and we learned that we can make like a propane rings from double bonds and Chapter eight. Let's start by converting our triple bond into a double bond. We will do this using hydrogen and a little more catalyst, but other regions could be used for this step. This will give us for imbue, teen. Now we need a car bean make our single propane ring and we need to have two chlorine. So we will start by making our car being from C H. C l three and then by reacting this with a strong base we can attach or are being that will be created from the strong base into the cyclo propane ring. So by reacting C h c l three, we will get our cycle a propane ring with two coins on it and we'll also get assault. And this is how you work. Problem 9.32

This is Theo. Answer to Chapter twelve. Problem number seventy four from the Smith Organic Chemistry textbook. We're asked to construct three molecules the stipulations for this problem or that every carbon in the molecules has to come from a molecule of ethanol on DH. So for a we start with ethanol we treated with P B R three to get bromide, then potassium turkey Talk side Teo, get a scene. The Sal Cain treat that with bro mean gas on DH. We'LL get to die bromide on DH. Then we can use to equivalent of sodium, eh? My ID to get to a settling on. And then from a settling it's it's pretty much exactly what we've seen before. In other problems treated. Settling with sodium hydride on dh then sorry and alcohol. Hail ID out of form New carbon carbon bonds. Of course, because of the stipulations of this problem, we have to use s old bromide on. We already showed on beginning of this problem the very first step How how this ethyl bromide can be made from ethanol. So we are allowed to use it on DH. Then we'll de protein ate the other side of the al kind at another molecule of ethyl bromide on that gives us all the carbons that we need. So we just treat that al kind with sodium metal and liquid ammonia into a dissolving metal reduction on DH. We get our target molecule for a on now, for B, we can build on our final intermediate from, eh? So we would do everything to get get back to this final intermediate that we used on. Then this time, instead of using sodium metal and ammonia, we would use hydrogen. Yes. And when Lars Catalyst, that will give us the sis l keen instead of the trains all Kane on. And we can treat that with M c p b a to get to this park side. That is our target product for B. And then first see, we can used the Cecil Keen that we made in B on DH. Then, rather than treating it with M c p b a treated with oz me, um, tetroxide on, then sodium by soul fight and water. Ah, and that will give us this target. Uh, dialled. Sorry. I'm tryingto move it over so it's a little easier to see. Didn't go. So, um yes, that will give us this this target. Dia ll, uh, for si on DH. So that is the answer to number seventy four.

This is the answer to Chapter 11. Problem number 59 fromthe Smith Organic chemistry textbook. And this problem says, devise a synthesis of the key tone three hex unknown from ethyl bromide as three only organic starting material s. So all the carbons in three hacks unknown have to come from ethyl bromide on. We can use any other re agents that we need. Okay, Uh, and so to start, I'm gonna taken ethyl bromide. Um, and I am going to do an elimination reaction with potassium turkey talks ard, and that is gonna get me to the AL Keen here. Okay, then. So from that Al Keen, uh, we can know ruminate So elemental, bro. Mean, that's going to give us the dye bromide. Okay, um and so now we have the dye bromide. We can treat that with two equivalents of sodium a mine, and that is gonna get us to the al kind. Um, and now that we are to the AL kind now we can start forming new carbon carbon bonds so we can t protein ate this out kind with strong base. So sodium high, Dr. Um and that will get us this satellite eye on. Ah, And if we, uh, used the satellite, I on with effort bromide, Since this is where all the carbons have to come from, Ah will get a displacement on dso we end up with that. Okay. Ah, And so we now have a four carbon chain. Um, and we can just just repeat what we just did. So, sodium hydride again eyes gonna de protein eight Uh, the al kind like this now and so again, another equivalent of ethyl bromide is going to give us our six carbon chain, which is exactly what we need. So six carbon chain. So all of our carbons Aaron Place. Ah, And now, um, since we're trying to make, uh, three hex unknown, um, and our al kind is between carbons three and four. Um, and this molecule, it's metric, so we don't really need any further control than that. And so we can ah, just use water. So fear Gassid and mercury, so feet. Ah, and that is gonna get us to our key tone. 123 56 Key tone on carbon three. Okay. And so that's it. Um, yeah. And so since since we were limited Thio, starting only from ethyl bromide. We just had to get from ethyl bromide to ah, settling. And then from settling, we could d protein ate that to make the satellite eye on Ah, and then use that with two equivalents of ethyl bromide sequentially to build our six carbon chain on. Then just install Archy Tone, and that's the answer to.

On the white board. You can see that I have several different alcohols And these alcohols can be synthesized using an owl cane in a oxy Merck aeration Deemer creation reaction. And so first we have to figure out what l cane we need to do. So our 1st 1 the all Cane, will have the same number of carbon. So as you can see and our product, we have four. So you're gonna have four in a reactant, you're going to start off this and then since our alcohol is connected to this group, the double bond will be on one of these Alcoa groups. So we like this. And then a ox ox immigration dimmer creation reaction will start off museum H G print disease. Oh, a C to boy who dress, use, teach? Yes. And then secondly, you do in a BH for and these re agents will be used in all the other examples as well. So justice saved time writing it just for this next example I will have the same re agents don't here. And so for our next example, we have three carbon train because he won 23 and again we see that the alcohol group is on carbon too. So are all cane. We'll be on one of these alcohol groups as long as the double bond. It includes the carbon too. In them for a last example. We see we have one to three for carbon chain in 1/5 current chain. So I'm just gonna go ahead, draw that out, and we see that the alcohol or the age group is on carbon three. And so our double bond will be here. And then, as I said before, your re agents will be h dri parentheses o a c two c squared or just th f and then within a B H four.


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