Given the ¹H NMR spectrum and the molecular formula, suggest a...

Question

Given the ¹H NMR spectrum and the molecular formula, suggest a structure for each molecule. [The IR spectrum suggests the presence of two C=O bonds.]

(b)

Accepted Answer

Welcome back everyone to another video for the following proton NMR spectrum and the given molecular formula determine the plausible structure of the organic compound. Hint there is a presence of co double bond based on the ir spectrum of this compound that is not shown here. Now, essentially we want to begin this problem by calculating the hydro deficiency index because we're given the molecular formula, we're going to use one half multiplied by two multiplied by the number of carbon atoms plus T minus the number of hydrogen atoms plus the number of nitrogen atoms minus the number of Hagen atoms. Text. And in this case, we would have one half multiplied by two, multiplied by eight plus T minus 16. We don't have any nitrogen atoms or halogens. So we can essentially say that we have two divided by two, which is one and this is the CEO double bond that we are given in our hint, right? So essentially that means there are no other pi bonds or ranks and that would be it, we can now continue with our shifts starting with the first chemical shift at around 0.9 parts per million. We have an integration of six. So we're going to say six hydrogens and we also observe a, the blood, our next shift is at about one point 25 parts per million. We have six Hagens and a singlet. Now, what else do we have? Well, we have a peak at around 2.1 P PM. There's an integration of one and it looks like it's a pen set, right? 12345, we're not sure. However, because the area is so small, it's a low resolution. So let's skip this for now. And let's look at the next one, we have one more peak at about 2.2 P pm. And we're going to say that the integration is three. It's a singlet. Now, first of all, we will start with the word lost peak because it has a 2.2 B PM value is the most downfield peak. It's a singlet, meaning there are no neighboring hydrogens, right? So essentially, if we have our carbon group and because there are three hydrogens, this represents a methyl group, right? So we can say that this should be a methyl group with no neighboring hydrogens. And based on the shift, we understand that it must be connected to the carbonyl group itself. Now, the reason why this is true is basically because we find our methyl group at around 0.9 P pm and then there's an increase in the P PM value by about 1 to 1.5 if it's attached to an electron withdrawing group. So this is what we can conclude from the 2.2 B PM chemical shift, we have three Hagens, which is a metal groups glad no neighboring hydros high P PM value. This means that it must be connected to an electron withdrawing group. Now, let's consider our next peaks. So then we have a peak at around 1.25 P pm, six hygen and blood, right? So essentially what we can conclude from here is that six hydrogens generally represent two equivalent metal groups attached to a carbon atom. We can say that we have two methyl groups attached to the same carbon atom. And because it's a single la, that means these methyl groups have no neighboring hydrogens. So there should be possibly a questionary carbon involved. Now, moving on to 0.9 P PM, we have six Hagens do left. So once again, we have two methyl groups attached to a carbon atom, we one hydrogen. So we can essentially imagine this fragment as an isopropyl group, right? Because there's one hydrogen and the two metal groups would be split into the blood. And this is how we come to our final peak. Notice that we have one hydrogen and we said it looks like it's a pen sat. But in reality, it's actually a subset, assuming that there are no neighboring hydrogens and there are only two adjacent metal groups according to the M plus one rule, this would give us a subset. So let's try and combine our fragments into a structure. We can start with our carbon group. We're going to attach that methyl group assuming that it is a sad based on the two P PM value, we would have our C age group afterwards. That makes sense, right? Because there should be no neighbors and a downfield shift of about two P PM verifies that we will have our CH or it might be CH, right? So we are not yet sure. So essentially, let's assume that it's ach group. Now, if we attach two metal groups, there's nothing we can do later on to get our additional two metal groups at a quaternary carbon, right? So essentially, it implies that our quater carbon should be the next carbon. So we are going to state that let's introduce our questionary carbon. Instead, we will have that peak for six hydrogens with a singlet and this is where we can have our ch, right. So let's add that isopropyl group and let's check if we assigned everything correctly. So we have our metal group at 2.2 P PM, right? It's a slat because it's attached to carbonyl, then we have our questionary carbon with two methyl groups. So we are essentially stating six H slat, we have assigned it, the next carbon has one hydrogen and it would be a septet because there are six neighboring hydrogens from the two metal groups at 3.1 P pm. And finally, everything makes sense. We have six hydrogens that bled at around 0.9 p pm. So this tells us that this would be the correct structure for this problem. That would be it for today. And thank you for watching.

Given the ¹H NMR spectrum and the molecular formula, suggest a structure for each molecule. [The IR spectrum suggests the presence of two C=O bonds.]
(b) <IMAGE>

Key Concepts

¹H NMR Spectroscopy

Molecular Formula Analysis

IR Spectroscopy and Functional Groups

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