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18. Treatment of the acyl chloride to the right with ispropylmagnesium bromide would yield which of the following products?...

Question

18. Treatment of the acyl chloride to the right with ispropylmagnesium bromide would yield which of the following products?

18. Treatment of the acyl chloride to the right with ispropylmagnesium bromide would yield which of the following products?



Answers

Identify the alkene that gives each of the following products upon ozonolysis, followed by treatment with dimethyl sulfide: a. b.

Okay, so for this problem you're reacting um you're reacting bugle bromide with some type of nuclear file to form each of these different products. Plus br minus as you're leaving group. So what this problem wants you to do is go through each of these choices and identify the nuclear file that you could use. So for these, remember that your product is going to have this general CH three CH two bonded to your nuclear file. So for let's look at the first one. So in this first one you have this beautiful group, right? And so everything not in this beautiful group is going to be your nuclear file. And because this is do this in green because this is a primary how locking we're going to preferentially talk about SN two reactions. So in SN two reactions, you want a nuclear file to have a negative charge because this is going to be a better nuclear file. So if we look at our first piece, our nuclear file is just going to be, oh each because of this sort of react with our beautiful bromide, It would form a new bond between this carbon and this oxygen. And kick off your leaving group. Okay, Let's look at the 2nd 1. So again we have beautiful group and then we have our new group. So in this case we're just we're just breaking this bond between this carbon and this oxygen. So our nuclear file is going to be oh CH three or meth oxy group. Again, we have our beautiful and then we have S. H. So we'll have S. H. Minus as our nuclear file. And I'm just gonna do this for a few problems now. So we can identify the bugle group in each of these. Uh Okay. Oh so we have in each of these cases the same beauty group now are substituted group. We're breaking this bond and so we're giving a negative charge to the sulfur. So s soups S CH two Ch three for this species here. This is a little bit different. I guess you could technically right N. H. C. H. Three. But because in nitrogen and such a strong base, You could also write an age to CH three for this one here, for our Siano group are negative charge. Because we're breaking this carbon carbon bond are negative charge is going to be on our carbon, going to be Tripoli bonded to our nitrogen. Here are negative charges on our oxygen. C. c. H. three for this car boxing group. And then lastly, see C. c. age three for our kill group. Now we have the negative charge on our carbon. And so this nuclear file, all the nuclear files here shown in blue would react with our beautiful bromide to give the given product.

This question asked us what the products of each of these e two reactions would be. So starting with a, um uh, in a typical e to reaction where we've got plenty of hydrogen is around and a leaving group, we are just going to make the most stable double bond. In that case, that would mean conjugating it right here with the existing l keen. So that would look like this for Part B. Again, we just want the most stable double bond. And in that case again, this is conjugating it with the existing, um, benzene ring. So that would look like this. Um, so I've taken one hydrogen from here and the Bruning off to make room for that double bond. And then for the this last problem again, we haven't existing, um, double bonds. And because of that, it's gonna be more stable to put the new one over here. Um, normally, you would put it in internal be more stable. L keen, but because this one is here, it actually really unstable to put it right here, um, and its various able to conjugated and put it over here. So that's what we'll dio So that will look like this. That'll be a product for part C.

Okay. So here's the drill mechanism for the following problems. And we're starting with the seed of chloride. So a stone is CH three C. O. C H. Three. And replacing one of the methods with the chloride. The first thing we want to create is um this Esther is keto ester. Is there a place in the chlorine with a master? So um these a circle around these are the most reactive of the car box of astor evidence because the chlorine is good leaving group. And if we react this with 73 Carbons, a three carbon alcohol, the oxygen is going to attack the carbon carbon. We'll get the L. Cox that I on and this is able to reform because if you could leave a group so that's going to leave and we're gonna get the structures intermediate. So if the kids come back, you have the addition of the astor. This is still bound to hydrogen. And the last step is for the chlorine atom. Those expelled two grab boots on and we get our products. Okay, so for B we have you want to create an amine and this is going to be similar to what we did in the first one. So you want to add, we want to create the structure and we can react this with um with uh so we need two carbons. We're gonna use ethyl immune. Its first going to it's one of the primary mean its first going to attack the carbon carbon and Similar Mechanism to 1st 1. You're the key to him back. We have presentation of the hydrogen. You have the products. Okay, so here's the second equivalent of method will mean because instead of chlorine reacting with or grabbing a parts off the action, it could also react with another method, I mean, or ethyl amine. And this would lead to this product. So we use twice the amount of methamphetamine. I mean, I thought I mean as the basic Lord so that we get the path we want. Okay, so for B we want to I mean for say we want to create um dimethyl amine. So you can start with the esa carod. Sure. Okay. So this is done with um two equivalents of metal. I mean as well for the same reason. So we're first going to react it with um dye my full of you. So we brought with us. Mm hmm. And I just don't wanna ask catalyst to get the reaction to go, okay, so it's going to attack the carbon carbon and reformed heaton, explode the chlorine. And again, when we have the structure we still have a hydrogen and mr of the intermediate. So instead of chlorine, this could also interact with another dimethyl amine. Do you get that product every day you want to create? Um We're gonna get a car box off acid so we can react this with sodium hydroxide and the oxygen is going to attack. So the same mechanism we get our products. Yes, sir. For e we want to create this anhydride and we're first gonna get alcohol and then deeper NATO with a strong base to the al cox that I am. And this can interact with another with another acid chloride. Sir. First going to get the alcohol and then treating the structure again with sodium hydroxide too depreciate the hydrogen. And they're reacting Alcock said ion with another asked florida. Good needs hydro Mhm. Then for press we have the structure so you can ask her again. Yes. Okay. So instead of um we're going to react with fennel as opposed to other methanol or ethanol. Because we need structure. They're reacting with the funeral to the same mechanism as the first one or the auction. On fetal. It's going to attack there 3 4 and expelled chlorine, But the chlorine is going to depart in eight. The hydrogen on funeral. Do you have the products this is bound to enter to as well. So in the power position, we have a nitro group that.

In this problem, the reaction will happen. Something like this. Just look at it carefully. C2 H five by plus or any T V. Will react to give C2 H five old food B. B plus four and I plus three BB. So according to the option in this problem, the correct and set age option B. Option B H, correct answer for this problem.


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