Propose a structure consistent with each set of data. a. Compound J: molecular ion at 72; IR pea

Propose a structure consistent with each set of data. a....

Key Concepts

Mass Spectrometry

Mass spectrometry is used to determine the molecular weight of a compound by detecting the molecular ion peak. This information is essential as it provides the base molecular formula that all subsequent spectral analyses must be consistent with, guiding the overall structure elucidation process.

Infrared Spectroscopy

Infrared spectroscopy identifies functional groups within a molecule by measuring vibrational transitions at specific wavelengths. Recognizable absorption bands, such as those associated with carbonyl groups or hydroxyl groups, are key in confirming the presence of these functionalities and narrowing down possible structural frameworks.

Nuclear Magnetic Resonance (1H NMR) Spectroscopy

1H NMR spectroscopy provides detailed insights into the hydrogen environment in a compound. The chemical shifts, which indicate the magnetic environment of protons, along with splitting patterns and integration, help determine the connectivity of atoms, the proximity to electronegative groups, and the overall molecular structure.

Splitting Patterns and Integration

The splitting patterns observed in NMR, such as triplets, quartets, and singlets, offer information about the number of neighboring protons through spin-spin coupling. Integration of these signals quantifies the number of hydrogens that give rise to each signal, playing a critical role in deducing how various fragments of a molecule are arranged.

Structure Elucidation

Structure elucidation is the process of piecing together the molecular architecture by integrating data from different spectroscopic techniques, such as mass spectrometry for molecular weight, IR spectroscopy for functional groups, and NMR spectroscopy for detailed proton environments. This holistic approach allows chemists to propose structures that are consistent with all provided spectral evidence.

Transcript

00:03 This is the answer to chapter 14, problem number 65 from the smith organic chemistry textbook.

00:11 And this problem asks us to propose two structures that are consistent with two sets of data that we've been given.

00:22 So for each structure that we're supposed to propose, we've been given a molecular ion, an ir spectrum, and an nmr spectrum.

00:33 Okay, and so starting with jay, we're told that jay has its molecular ion at 72.

00:42 And so from there we can start to think about some formulas.

00:47 So six carbons equals 72.

00:50 So five carbons and 12 hydrogens could get us to 72.

00:56 But i'm going to jump around a little bit.

00:59 And in light of the fact that jay has an ir peak at 70, 1710 wave number, which is consistent with a carbon oxygen double bond, i'm going to say that there is an oxygen in j.

01:15 And so when i see the m to z at 72, i think that this is going to be c4h80.

01:27 And i think that because, again, this ir tells me that there has to be a carbon oxygen double bond in here.

01:36 And so if the formula is c4h80, the degrees of unsaturation for that formula are going to be 2 times 4 plus 2 minus 8, all of that over 2.

01:53 And so that's going to be one degree of unsaturation.

01:57 And so i think this is the correct formula.

02:00 And i think that we have a carbon -oxygen double bond in this molecule.

02:05 So then we need to look at our nmr signals.

02:08 So 1 .0 ppm triplet that integrates to 3.

02:13 That's going to be a methyl group.

02:17 And i'm going to skip the next signal and look at this 2 .4 ppm quartet that integrates to 2.

02:24 And that's going to be a methylene group.

02:28 And so these two groups are splitting each other, right? because we have a quartet on the, on the methylene group.

02:35 And so that means it has three neighboring protons.

02:40 And the signal for the methyl group is split into a triplet, which means that it has two neighboring protons.

02:46 And so these two are next to each other.

02:48 They're splitting each other.

02:50 And then the last signal, the second signal here, the 2 .1 ppm singlet that integrates to three, that's going to be a methyl group with no neighboring protons because it's a singlet.

03:02 And so if we put all of this information together, what we get looks like this.

03:12 So here's our carbonyl, and then there's another methyl group.

03:18 And just to sort of spell this out, so this signal is going to be this methyl group, this signal is going to be this methyl group, and this signal is going to be this methylene group.

03:37 Okay, so there's there's j, and now we just need to do the exact same thing.

03:42 Thing for k.

03:45 So looking at k, we have a molecular ion at 88.

03:50 And since looking at the ir for j proved so helpful, let's start with the ir here.

03:55 So this time, the ir is a peak between 3 ,200 and 3 ,200 wave number, so it's a broad peak.

04:03 And so this is consistent with an oh.

04:07 So we have an alcohol in this molecule.

04:11 And so again, we need to account for the presence of an oxygen.

04:15 And so c5, okay, so that's 60, and then an oxygen is going to be 16.

04:25 So that's 76.

04:27 So let's go c5h12...