The IR spectrum of compound A with a molecular formula of C

Question

The IR spectrum of compound A with a molecular formula of C₅H₁₂O is shown below. Compound A is oxidized to give compound B, a ketone with a molecular formula of C₅H₁₀O. When compound A is heated with H2SO4, compounds C and D are obtained. Considerably more D is obtained than C. Reaction of compound C with O₃, followed by treatment with dimethyl sulfide, gives two products: formaldehyde and compound E, with a molecular formula of C₄H₈O. Reaction of compound D with O₃, followed by treatment with dimethyl sulfide, gives two products: compound F, with a molecular formula of C₃H₆O, and compound G, with a molecular formula of C₂H₄O. What are the structures of compounds A through G?
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Accepted Answer

All right. Hello everyone. So this question says that compound A has an IR spectrum shown below. It has a molecular formula of C 4h 80. When compound A is reduced compound B with a molecular formula of C 4h 10 0 is produced when compound B is dehydrated, compound CD and E are produced when compound D and compound E are subjected to O analysis separately, they only yield compound F which has the molecular formula C two H 40. When compound C undergoes oz aosis compounds G and H are produced. Both compounds are all the hides with their respective molecular formulas of C H2O and C three H 60 is the following structure. Identification of compounds A through H correct. So here we have all eight compounds listed and also the IR spectrum of compound A. Here options A or B simply state whether or not the identification is correct or incorrect. All right. So first and foremost, there's a lot to take in for this particular problem, right. So let's start with the first glaring question. The ira spectrum that's provided is supposed to be for compound A and compound A right now. Is proposed to be too. But so our first question to address for our answer here is does the IR spectrum provided match the structure of two? But no. So what is the most important structural component of two? But no. Well, that would be the car bonel, right. So if we want to verify whether or not the IR spectrum given is that of tube known, we should check for the presence of this car B. And recall first and foremost that in an IR spectrum, this car B tends to appear at around 1700 inverse centimeters. The exact number can change a little bit, but it should be in this overall range, this general area. And the Car Bono peak in an IR spectrum is characteristically very long and sharp. So going back now to the IR spectrum, we do see a very long and sharp peak towards the right side of the structure and considering the numbers, this peak itself is at around the 1700 range. Specifically, it's between 1725 and 1705. And so this particular peak does correspond to the car bal. So that's a step in the right direction. Now, when you consider this additional peak towards the left side of the spectrum that's pretty broad at around the 3000 inverse centimeter range. This band, if you recall represents both the SP two and SP three hybridized carbons within or component question within two. But so going back to the first question, does the IR spectrum given match the structure of two but known the answer would be yes. Right. So we can proceed with the second part of this question, knowing that compound A at the very least is correct. So let me scroll down to open up some space here. All right. So as mentioned before, compound A is correctly identified as two but no. So from here, we're going to follow the reactions described in the text of the question and see if those products match the ones that are given. All right. So now let's follow these reactions. So the first reaction described is when compound A is reduced and when tutone is reduced, Tutino is produced. So we're going from a ketone to a secondary alcohol and this would be compound B. So that's tiny now to butanol just so happens to have the molecular formula of C 4h 10 0. So B was also identified correctly. And so from here, our secondary alcohol to butanol undergoes dehydration and dehydration if you recall is the removal of water across two carbons to create a double bond to create an all key. So here, because this particular alcohol is inside of the carbon chain, we end up with two possibilities, right? We end up with a terminal ake where the double bond is in between carbons one and two. And we also end up with an internal ake where the double bond is in between carbons two and three. Now, in addition to the possibility of an internal alkene existing recall that we have to consider both the trends and cis isomers of that internal A key. And it just so happens that these three elimination products were also listed, right? They are compound CD and E respectively. So are terminal all keen is sea our trends. Internal Alkin was D and our cis internal Alkin was e you just scroll up one more time just to make sure, yes, everything is labeled correctly. So now the question is what happens when these two or excuse me, these three keens are subjected to conditions of Oz analysis. So let's discuss that right. Recall that o analysis is otherwise referred to as oxidative cleavage. So basically, just to go ahead and summarize what happens, the two sp two carbons of the double bond are separated from each other and created and rather create car bals. So they remain sp two hybridized. And so because the carbon chain is being split in half, essentially, you might end up with two carbon compounds depending on whether or not the starting material has a symmetry to it. So if we consider one butane here, that's our terminal akin. And we imagine the two sp two carbons of the alkene separating from each other, one product would have three carbons from the starting material. With the first being the carbo itself and the outermost carbon of the ALEN would end up being oxidized to form formaldehyde. So oz analysis of one beine gives us two products. Now, the same cannot be said for compounds D or E. And the reason for this is because if you imagine splitting or rather separating the two carbons of the double bond from each other, you actually have the same number of carbons on either side of the split. So really oz analysis just gives you the same product. So os analysis of both compounds D and E gives you a two carbon aldehyde. And fortunately, right, all three of these aldehydes were identified in the list of compounds given specifically compound F is the product of the ozono thesis of our internal akines. Whereas G and H were the ozono assis products of one beine of our terminal all K so G was formaldehyde. So there you have it. So bottom line is that our correct answer is going to be a because the structure identification of compounds A through H is correct. And so if you watch this video all the way through, thank you so very much for watching. And I hope you found this helpful.

Key Concepts

Infrared Spectroscopy (IR)

Oxidation and Reduction in Organic Chemistry

Ozonolysis