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14_ What is the final product of the following reaction sequence? HzCrO4 Mg organic product Br ether 2) H3o" Msk COzH ~CHO B) OHFinal ProductCOzHD)...

Question

14_ What is the final product of the following reaction sequence? HzCrO4 Mg organic product Br ether 2) H3o" Msk COzH ~CHO B) OHFinal ProductCOzHD)

14_ What is the final product of the following reaction sequence? HzCrO4 Mg organic product Br ether 2) H3o" Msk COzH ~CHO B) OH Final Product COzH D)



Answers

Identify the final product in the following sequence of reactions. (a) $\mathrm{CH}_{3} \mathrm{CH}\left(\mathrm{CH}_{3}\right) \mathrm{COOH}$ (b) $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}$ (c) $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{COOH}$ (d) $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}$

Portion is Predict eight and the c compound right for me are going to discuss back compound will get between a industry In the past tops and we have 3 3 at seven CH two CH 2 Ch the uh CS to be ill. This is your thing with magnesium. Yes or no. Yeah. So what will we can get then? We will get C three S seven CH two the ETS fierce group. MdbR MG we are here MGB are very now just compound is reacting with the six. That's five minutes to then we know that. And it's to have the acidic hydrogen. So due to present acidic hydrogen As the base reaction takes place and then we will get C6 H12. That's forbidden normal accent because These both MDB will get eliminated and one experts will be on this carbon and on this carbon right in the B option B part we have see they were born O. c. s. three. See it's true four cl reacting with the MG proprietor then this MDV react with chlorine late then what will happen then we will get the development oh we'll see yesterday. Three X food all four I. M. D. C. Now in that this compound is that did not result. So Carbon have negative charge and this 500 transfer here. Well the charge will attack on this carbon and then we will get cycle depending on uh forget amazing of the O. C. S. T. Group I think. And in the C option we have B. R. C. H. Two All four We are and drinking with any which that's George. Then what can we get? They will get You can even go to you because all of the urban gets eliminated and high interested. I will be added so we will get the edge to All four. Yes or no. This compound we are getting okay. Thank you.

In this problem, I can write the reaction edge. Just look at it carefully watched CS three shell will react with K. C. N. To give the component see http see and and this compound in pageants of At plus or H two will change into this compound see HTTP C. O edge. So according to the given option, this is compound. We therefore, we can say that according to the given option, option, C option C. Each but act. And for this problem, I hope you understand the reaction of this problem.

So this question we're talking about predicting major products from your action off HBR with different Al Keane's. So soon with eight, we're gonna start with this structure, right? So is god something that looks like this. I'm reading HBO to this. Um, the product is actually gonna be a little different than what would seem obvious you're actually going, Teoh produce. This is a product. And the reason for that is because as an intermediate obs draw this over here and read, I'm gonna form this car broke, and I now this has a spare hydrogen off here which can then to draw in blue kind of shift over causing, uh, the car behind on to move the tertiary carbon in a one to hydride shift, making our my major products be on that tertiary carbon. Right? So moving on to ah, to be eso this time we have a slightly different, uh, thing Looks like this HBR this time were a little too far away to do any of those fancy 1 200 shifts. Oh, or even a metal shift. So in this case, you'd expect just a normal reaction. All right, part C, we're going to start with, get this product and there's no need to do any sort of shifting for this one. I simply due to the fact that they're already conform a church here. Carver Cod, eso weaken. Just I expect this to be in a stable product. Uh, D is gonna be a very similar structure where we have the double bond one single way and again you conform a tertiary cardboard cut on here. So there's no need to do any sort of shifting or changing its stability. Uh, so you're actually going to form the same product? Uh, which is kind of exciting, So to move onto Parky eso. Now we've got something that, um he's gonna have another four carbons this time to muffles and then a guy right there. So we're actually going to form a metal shift on this one with addition of HPR eso. What that means is when I add HBR, my final product is actually going to look like come this. And the reason for that is I have Ah, this is an intermedia. And what can happen is since there's no available hydrants to move to a tertiary car back on this whole methyl group can actually shift over, causing the car behind on to be on a metal, but also armature sharing site, but also moving the methyl group completely, which causes are major product to be for him to the way it is. All right, then. Lastly, moving on to f We have something that looks like this. Ah, And again, there's not really a good way to shift this or change this, so we can just expected to add kind of on the edge like that.

In this problem, the reaction will happen. Something like this. Just look at it carefully. Siesta, we feel in presence of cation will react to give CS three CN and this on further reaction with AT plus as to will give CS three c. o. h. This is big. So according to the option in this problem, the correct answer? Age, option C, option C. It's correct answer for this problem. Option C. H. Correct answer for this problem.


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