With information from the ¹H NMR and the ¹³C DEPT spectra, str...

Question

With information from the ¹H NMR and the ¹³C DEPT spectra, structure elucidation becomes even easier. Provide the structure that corresponds to the following data. [The identity of the carbons comes from the DEPT experiment.]

C₆H₁₁BrO₂

IR: 1745 cm ⁻¹

¹H NMR: δ 1.25 (t, 3H), 2.18 (quint, 2H), 2.58 (t, 2H), 3.46 (t, 2H), 4.15 (q, 2H)

¹³C NMR: δ 14.2 (CH₃) , 27.8 (CH₂) , 32.5 (CH₂) , 32.6 (CH₂) , 60.5 (CH₂) , 172.4 (C)

Accepted Answer

All right. Hi everyone. So this structure, sorry. This question says that structure elucidation is made even simpler with the help of the proton and Mr and the carbon 13 DEPT spectrum give the structure corresponding to the data below. And the de the identity of the carbons was determined using the DEPT experiment. So recall that when it comes to structure determination, right, our first step is to calculate the index of hydrogen deficiency or IHD for short. So the formula for IHD is equal to two multiplied by the number of carbons plus two plus the number of nitrogens and then subtracted by the number of hydrogens and subtracted by it, the number of halogens as well, all of which you're going to go ahead and divide by two. So in this case, right, the molecular formula is C six H 11 bro two. So from there, you would plug in each respective number or value or each type of atom that you have. So I'm substituting my C for six and in this case, there's no nitrogen. So I'm not going to write anything there, but I do have one halogen that's bromine and 11 hydrogens. All of which is divided by two. So this comes out to two divided by two, which gives me a total IHD of one. Now recall that when the IHD or degree of saturation comes out to be one, we have a few different options, right? We either have one double wand or we have a ring structure. So in addition to the molecular formula that was just given, we do have one significant piece of information obtained from the Ir spectrum, which if you recall is crucial for identifying certain functional groups in organic compounds. So based on the one peak highlighted at approximately or at 1735 inverse centimeters, recall that a peak at approximately 1700 inver centimeters identifies a car bottle. So because our car bundle is a double bond right between carbon and oxygen, our one degree of saturation comes from the car bono. So this narrows down the types of functional groups that we may see in this compound, right? But because we have two oxygens, the most likely functional group is an Esther, right? Because we have two oxygens and there's no data in the IR spectrum suggesting that a hydroxy group may be present if it was a carboxylic acid. So our most likely candidate is an. So from here, right, we can go ahead and start to analyze the proton and more data and recall that when it comes to the proton NMR spectrum, the chemical shift range of approximately 3 to 5 parts per million corresponds to protons attached to carbons that are next to electron atoms. And in this case, we have two right, we have the oxygen from our ester and our bromine as well. So that brings me to the first two peaks that are listed in the actual proton. And Mr data, I've got a peak at 4.09 as well as a peak at 3.27 parts per million. Now both of them correspond to two protons of that type and they're both triplets. Now recall that the the multiplicity of a given proton NMR spectrum is dictated by the N plus one rule in which N is the number of neighbors, so to speak, or the number of neighboring non equivalent protons. So in this case, the signals at both 4.09 and 3.27 correspond to ch twos that have two adjacent protons considering it is a triplet. So what I want to bring to your attention here, right, is that 4.09 as opposed to 3.27 is more d shielded, right? It is more downfield in the sense that it appears more to the left of the spectrum. Now that has to do with the relative electronegativity of the electron withdrawing substance or atom next to it. Right. So between oxygen and bromine recall that oxygen is actually more electron negative than bromine So the triplet at 4.09 parts per million that integrates the two protons reflex A CH two next to an oxygen that has two adjacent protons on the other side. Whereas the 3.27 it is also a CH two with two neighboring protons, but that one is going to be next to the Bromy. So with that in mind, right, let's go ahead and focus our attention on the next signal at 2.85 parts per million in the proton and a more spectrum because recall that the chemical shift range of 2 to 3 parts per million represents protons attached to carbons that are next to car bottles. So there's only one such signal within the range, which is the one at 2.85 parts from really. So based on the fact that our compound is most likely an Esther, right? And the fact that the signal at 2.85 integrates to two protons and is a triplet, we're once again going to have a ch two next to the car bal that has two protons next to it in order to make it a triplet. So lastly, right, our last two signals in the proton and Mr spectrum, one spanning between 1.81 and 1.70 parts per million and the other at 0.96 parts per million, those correspond two you all keen protons, right? Because alkene protons tend to fall between approximately 0 to 2 parts per million. So because the signal spanning between 1.81 to 1.70 parts per million is a multiple as indicated by the lowercase letter M. It most likely is an Alcaine carbon in between two other al can carbons to account for the number of neighbors that it has, right. So if it has or if it inte integrates to two protons, right, that signal has to be a CH two and it can most probably be between another CH two and ACH three at the end of the chain to account for its number of neighbors. And so the 1.0 point the one at 0.964 million, excuse me, because it integrates to three hydrogens, it's going to be ac three with two neighbors considering it is a triple. So now that we have all these fragments, right, let's go ahead and see if we can put them together using the carbon 13 NMR data coupled with the dept experiments. The first of which is this carbon here that has zero hydrogens at approximately 171.56 parts from really, I should say now because of the fact that it is incredibly de shielded, considering the high chemical shift that signal must correspond to the car bono. So our car bono here is going to explain our signal at 171.56 parts per million. So the signal right next to it, right, or CH two at 66.13 parts per million, it has to be an Alcaine carbon considering the fact that it has been identified as a CH two. However, it is relatively downfield most likely due to its proximity to the carbon. Or actually let me correct myself really quick because a signal for an all can carbon appearing that downfield like at 66.13 it would more so be due to the fact that that carbon is connected directly to an electron atom. Therefore, having that signal be more d shielded. So therefore, right between 66.13 and 37.21 which are the two ch two signals that are relatively d shielded, right? Because they fall within the range of 30 to 80 parts per million, they indicate carbons attached to electron atoms. So because oxygen is more electron than bromine, the ch two at 66.13 correspond to the carbon right next to the oxygen. Whereas the one at 37.21 parts per million is going to correspond to the carbon next to the bromine. So now peaks that fall in between the range of 5 to 45 parts per million usually indicate all cane carbons. Now of the three peaks that fall within that range, the one with the highest chemical shift at 26.14 indicates an electron withdrawing substituent in close proximity to that particular carbon, which in that case must indicate the carbon next to the car bottle. And then last but not least, right, we've got one more secondary, all cane carbon and only one primary, all cane carbon. So when it comes to putting together the pieces that we discussed previously, right, if I draw the ether in the center here, I have to have two CH twos on either end right next to the car bal and next to the sp three oxygen. However, the bromine has to be at the very end of one of the chains. In this case, the one attached to the car bal because that ensures that I only have one CH three signal according to the det data. So my final answer is Propo three promo pro Panno and there you have it. So I know it was a lot, right? But if you stuck around to the end, thank you so very much for watching. And I hope we found this helpful.

Key Concepts

¹H NMR Spectroscopy

¹³C DEPT NMR Spectroscopy

Infrared (IR) Spectroscopy

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