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OH CH35- This productcould be obtained from treatment of 1-Methyl cyclohexene withOHand6- Esters are reduced byNaBHA LiAIHA) to give(1" alcohol diol7 -In Grign...

Question

OH CH35- This productcould be obtained from treatment of 1-Methyl cyclohexene withOHand6- Esters are reduced byNaBHA LiAIHA) to give(1" alcohol diol7 -In Grignard reaction, the intermediate which formed during the mechanism is order to synthesis a alkyl chloride from 3" alcohol HCI, SOCIz reagent should 8 - In be used could be prepared from benzoic acid in a single step by adding 9 - Methyl benzoate and aldhydes_ If you want to prepare only 10- Primary alcohols are oxidized to carboxyl

OH CH3 5- This product could be obtained from treatment of 1-Methyl cyclohexene with OH and 6- Esters are reduced by NaBHA LiAIHA) to give (1" alcohol diol 7 -In Grignard reaction, the intermediate which formed during the mechanism is order to synthesis a alkyl chloride from 3" alcohol HCI, SOCIz reagent should 8 - In be used could be prepared from benzoic acid in a single step by adding 9 - Methyl benzoate and aldhydes_ If you want to prepare only 10- Primary alcohols are oxidized to carboxylic acids should be used aldyhdes from primary alcohols,



Answers

Preparation and reactions of alcohols Alcohols can by prepared from alkyl halides, epoxides, alkenes, etc. When treated with acid, $\mathrm{C}-\mathrm{OH}$ bond is cleaved heterolytically to form a carbocation which could then undergo substitution, elimination or rearrangement depending on the conditions and stability of the carbocation. When 1,2 -dimethyl cyclohexene is treated with cold dilute alkaline $\mathrm{KMnO}_{4}$, the product obtained is

Fortunately we have hell only containing compounds it is reacting with the X. Plus. Then what can we get? See edge to edge have longer electron. This is the thing with the X plus. So this long peregrine game that's blessed. Then it's two molecules will get eliminated. Yes or no. Then we will get to get strictness. No carbo is present on the Earth isn't carbon to the ring. Then bring expansion will takes place. So he revealed that ring expensive five members will convert into the six member by this week. No edge placement gets anymore and we will get a cycle of vaccine. So correct option will be it. Yes, my mom. Okay, thank you.

This question were given a series of alcohols and were asked how to prepare these alcohol's using, anchoring your free agent. And so we're going to start with a So for a we have two metal to pro panel. So if we draw the structure, we have pro pin all. So it's going to be three carbons chu Meaning are alcohols on the second carbon and two methyl meaning we have a metal substitue int also on the second carbons. Okay, so we see now that we have a tertiary alcohol. So to prepare, using a green yard re agent, we want to then start from a key tone. So if we draw a key tone, we see that we can then sever this linkage and add metal, magnesium, bromine. And we also know that if two or more alcohol groups are the same, that we can use an Esther because in Esther is going toe add twice. So we see that we do have more than one of the same alcohol group. And so we could also imagine starting with and ask her and then adding to quit once of the following Grenier. Okay, we're gonna draw a true arrow, Okay. And so these are the ways that we could prepare a green yeard, starting with alcohol in a So that's his on to be so in B. We have one metal cyclo, hexane all so cyclo hicks an means six members. Rain all means alcohol and one metal. So we have a metal substitue int in the one position. And so again, we see that we have a tertiary alcohol. And so then we want to start with a key tone, and in this case, we can add we could imagine it severing this bond and using metal magnesium bromide. Okay, so we can move on to part C now. So for see, we see, we have three metal three pen tunnel. So hence Han, a five member chain, 30 So the alcohol at the three position and free metal so mental at the three position as well. So again, just as imports A and B, we see we have a tertiary alcohol. So our first inclination then is going to be to start with the key tone. So we have our key tone weakened, then imagine breaking this bond and reacting with metal, magnesium remind and We can also think about this a severing this bond so we could also have a key tone. But this time it would be an asymmetrical key tone. That's Epple E needs Ambro mind this time. And we also see that we have two symmetrical groups to symmetrical out kill groups about the carbon where alcohol is located. So just is in part A. We could also imagine using a an Esther I've been adding to here and so those are our options for making the following alcohol from three metal freep internal. So now we can move on to part D. So in D we see we have to fennel to butin also butto tan that is a four member chain to all and to fennel So fennel here. So again Justus imparts either See, we have a tertiary alcohol. So what then? We know we want to begin by starting with a key tone. So we could imagine severing this bond and starting with the following key tone plus fennel, magnesium bromide and we could also imagine breaking bonds and other places so we could have three different variations of p tones because we have three different Al killed groups about our carbon containing the alcohol. So we could imagine this being our Brainard re agent, which would give us a little plus metal, magnesium bromide. Or we could imagine making this disconnection and starting with edible magnesium burn one. Okay, so those are our choices for making alcohols making the following alcohol. So now we're going to move on TV. So e, we have benzel alcohol. So this alcohol is quite different than the alcohol's that we've looked at in parts A through C because now we have a primary alcohol. So when we see a primary alcohol, that should be a clue that we need to use an alga hide as a starting material. So in this case, we could imagine breaking the following bond making this disconnection, in which case the alto hide that we would use would be formaldehyde. And then excuse me. Who would imagine making this disconnection? Who's wrong again? I was right the first time. Sorry, this disconnection, in which case we would use formaldehyde and then we could use fennel, magnesium bromide, And this is only going to add once into our Aldo hide and said, This is how we could achieve Benzel alcohol. So now finally, in part F, we have four metal, one pen tunnel. So pin tan, that's five 51 all. So I'm going to make this carbon one and four metal. So 1234 Okay, so again, Justus in part E we see we have a primary alcohol and that should be a clue for us that we need to start with an alga high. So then we can imagine making the following disconnection. So again we use formaldehyde, which is so one carbon LDA hide and using the following Grenier dri agent I'm just gonna make sure that I have all of my carbons. Correct. So 1234 1234 So this

This is the answer to Chapter 20. Problem number 23 from the Smith Organic Chemistry textbook. Ah, and this problem asks us to consider what Grenier, Drea, agent in Carbondale compound are needed to prepare each of the alcohol's that were given here. I saw a given four alcohols on, and then it also talks about a tertiary alcohol with three different our groups on the carbon that the alcohol is attached to. I could be prepared by three different Grenier three agents on DSO We encounter that in part D of this problem, okay. And so for part a Ah, there's really only one way that we make this s o it's gonna be with a metal Grenier three agents of ch three and g b r. Ah, And that is going to add to askyou Dalva hide so ah, like that. Okay. Uh, part B. Again, there's really only one way to make this. Um, and that is going to be, um, using a cyclone. Maxine Greene. You're free agent. Ah. And this time adding to formaldehyde. All right, now for a report. See, um, there are two ways that we could make this because we could make ah, either side of the, um, product molecule, the grin yeard. Ah, And you know, either side the carbon deal. So if we want to make the cyclo heck sane, Uh, the green yeard, uh, that would look like this. And so that would have to add to, uh, this Aldo hide. So pro Pianalto hide. Okay, um, war. We could make the other part of the molecule Grenier, so we would use an ethel magnesium bromide, Um, and use this Aldo hide and they would react and give us the exact same product. Uh, now, for part D. As I said in the beginning, since this is a tertiary alcohol with three different our groups, we can do this three different ways. And so the bicycling ring system is not really gonna participate in this reaction at all. So it's gonna be unchanged in all three of these re agents. Um and so basically, we could just make this the Grenier. And if that's gonna be our green yeard than the rest of our molecule, it's going to be to Butte known. So there we go. Um, we could also, um, make this ring system part of the starting carbon, Neil. So in that case, we could put, um, key tone here, and we could use a grin. You agree, Agent? We could use ethyl magnesium bromide. Okay, um or, ah, we could make it the, uh, the Ethel Key tone on and use metal, magnesium bromide. And so that would look like this. Get in. The ring system is not participating. So nothing's gonna change about that, Have our key tone here. Now we have an f o group to the right of the key tone. Ah, and we will use meth. Oh, magnesium bromide. Okay, All right. On. So that's the answer to this problem. The only way to get this answer ah, is to know how Grenier three agents work. Um, And to be able to look at an alcohol and work backwards and think, how can we split this alcohol into two pieces? Such that one piece is an Alfa Haider key tone. On the other piece is the granary agent, and that's the answer to Chapter 20. Problem number 23

Of course, in the cycle of painting, went to die on oxygen with the lead petra as it did gifts. Okay, so one is that correct product that we are getting in this reaction, psychlo panting went to die. Mhm. Is reacting with the nerd as you did. Mhm. Then what will happen then? We will get this region is behaving like a awful So we will break this single one and we will add hydroxy group at each this carbon. Then what can we get 123. See that's true. All right. See which Which here is the watch? Which? So from now he can eliminate that's two molecules from each carbon center. Then then what can be given CH two CH 2 feet. That so correct ups and build the city. Okay?


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