The spectral data below are presented in a manner similar to w...

Question

The spectral data below are presented in a manner similar to what you would find in a chemistry research journal. Identify the structure for each set of data.

(c) C₇H₁₄O₂ : ¹H NMR: δ 0.91 (3H, t, J = 7.0 Hz), 1.11 (6H, d, J = 7.0 Hz), 1.82 (2H, sextet), 2.40 (1H, sept, J = 7.0 Hz), 4.14 (2H, t, J = 7.1 Hz); IR (cm ⁻¹) : 1745, 1200

Accepted Answer

All right. Hi, everyone. So this question says that the spectrum information below is displayed as we see it in research journal draw the structure of the given dataset. So here the molecular formula is C six H 1202 proton NMR chemical shift values are 1.19 or a six proton doublet 1.21 for a three proton triplet 2.41482 proton quartet 4.934 a one proton septet and IR values in inverse centimeters are 1750 as well as 1210. So recall that the easiest step should always be to find the index of hydrogen deficiency or the IHD for short, based on the molecular formula. Now recall that the IHD is or is calculated by multiplying the number of carbons by two and then adding two to that value. And then the number of nitrogen is added while also subtracting the number of hydrogens and the number of halogens. So after everything has been taken into account or taken into account, you divide by two to find your final answer. So based on the molecular formula C six H 1202. The equation is as follows. It is two multiplied by six, add it to two zero nitrogens, subtracted by 12 hydrogens and zero halogens all of which gets divided by two. So after calculating everything, I get a total IHD of one, now recall that an IHD of one can correspond to two things, either I have one double bond or I have a ring. So to determine which of which scenario applies in this case, right, which possibility I go ahead and I consider the IR data, I have prominent peaks at 752,210 inverse centimeters. Now recall the signal at 1750 inverse centimeters is very characteristic of a car bal or ac double bonded to oxygen or a carbon double bonded to oxygen. So the presence of this car bundle confirms that we have a double bond present in this molecule which explains our ihd of one. Now addition, additionally, the peak of 1210 inverse centimeters indicates a bond between carbon and oxygen. So based on this data given right, the most likely functional group present in this molecule is an ester. So now we can go ahead and consider our proton and Mr data to find what our Esther is going to look like. So the question then becomes to find what fragments are missing and to determine whether or not they are connected to the carbon carbon or the oxy oxygen of the ester. Now of the four peaks listed in the NMR data, the one that sticks out to me is the one at a chemical shift of 4.93 because recall that this compound only has one double bond, which means that the remaining carbons that we're trying to piece together must be sp three hybridized or alk carbons. Now, a chemical shift value of 4.93 is relatively d shielded or an all kill carbon or an alk proton. Now, we call the D shielding is a consequence of that specific proton being adjacent to or close to a very electron group. So because oxygen is very electron, the protons here at a chemical shift of 4.93 must be from a carbon that is adjacent to the oxy oxygen in the ether. So from here, we can determine what that signal there actually looks like because we're given the integration, which is one as well as the multiplicity. Now recall that the integration is simply how many protons correspond to that one signal. But the multiplicity is a consequence of how many quote unquote neighbors can contribute to the splitting of that individual proton or that individual signal. Now, when it comes to finding the amount of neighbors, recall that we use the N plus one rule where N is the number of quote unquote neighbors. So if the multiplicity is equal to seven for a septet, then that means the number of neighbors is going to be equal to six. So we can conclude, then that N equals six or six neighbors means that there are two methyl groups, each with three protons connecting two, the proton giving a chemical shift of 4.93. And this makes sense when we consider the first signal here at a chemical shift of 1.19 parts per million because it has a chemical shift or excuse me, an integration of six as well as a splitting pattern or a multiplicity of two right or a doublet. This means that the protons contributing to the signal at 1.19 have to have one single neighbor, which would be the singular proton of the carbon connecting the two methyl groups together. So with this in mind, we do have the first fragment, it is going to be an isopropyl group on the ether or the all coy oxygen of the ether. So now the question is the find what the fragment next to the car bal has to look like. So to find that information, we draw our attention to the remaining two signals described in the data, we have one signal at a chemical shift of 1.21 parts per million and another at 2.41 parts per million. Now, what I want to point out here is that these two signals are a triplet and a quartet respectively. Now, if I scroll down for just a moment, recall that a triplet quartet duo is characteristic of an Ethel group, right? Because if we consider an Ethel group for a moment after I scroll down a bit more, right? An EO group has one CH two carbon which I will label A and one CH three, which our lab will be. Now, if we consider just the ethel group in isolation, right, hey, would have three neighbors considering the CH three adjacent to it, which means that when we consider the end plus three rule or N plus one rule, the multiplicity of the protons from carbon A here would be four or a quartet, right? Similarly B which is our CH three here has two neighbors from a right next to it, which means that its multiplicity would be three thus giving a triplet. And so if we consider the data provided here, the integration also checks out right, because the quartet with three neighbors should have two protons, which it does in the signal for at 2.41 and the triplet having two neighbors should have three protons. So what that means right is that the fragment next to the car bono must therefore be an ethyl group. And this further makes sense when we consider the fact that RCH two of the ethel group is directly next to the car Bal, which itself does not have any protons. So the only protons that should split that specific signal or the ones from our CH three at the very end. And there you have it here is our final answer. The structure of C six H 1202 is isopropyl propionic. And so with that being said, if you stuck around to the end of this video, thank you so very much for watching and I hope you found this helpful.

Key Concepts

NMR Spectroscopy

IR Spectroscopy

Molecular Formula and Degree of Unsaturation

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