Notes: Concepts in mass, charge, and isotopomers. Techniques: ESI mass spectrometry. Notes: Concepts in fragmentation, isotope patterns, and molecular ions. Techniques: EI mass spectrometry.
Analytical Chemistry – A Guide to 13-C Nuclear Magnetic Resonance (NMR)
Notes: Electron ionization EI mass spectra are shown for five transition metal acetylacetonate acac complexes. Techniques: 1 H NMR spectroscopy. Notes: This problem gets to the heart of coupling and diastereotopicity. It is one of my all-time favorites. Notes: This is a great little matching problem that gets to the heart of pattern recognition, coupling, and symmetry in 1 H NMR spectroscopy. Notes: This problem builds on some of my favorite concepts in sterochemical determination in a cyclohexane ring system 2-phenylcyclohexanol from the coupling pattern. Notes: A challenging matching problem that probes concepts of chemical equivalence and symmetry in 1 H NMR spectroscopy.
Notes: A matching problem that probes concepts of chemical equivalence and symmetry in 1 H NMR spectroscopy. Notes: A small but challenging molecule. Notes: This problem was designed to build on concepts of 1 H NMR non-first-order coupling pattern recogntion and symmetry.
Notes: This problem was designed to build on concepts of 1 H NMR coupling pattern recogntion and symmetry. Notes: We designed this molecule to illustrate principles of coupling patterns in the 1 H NMR spectrum and isotope patterns in the mass spectrum. Notes: A pretty spectrum with interesting coupling patterns. Notes: Concepts in pattern recognition, symmetry, and diastereotopicity. Notes: Concepts in pattern recognition and spin-spin coupling. So here's the carbon next door. Here we have a neighbor and here we have a neighbor. So this makes sense.
And then, we already talked about the blue protons, right? Having zero neighbors. So everything seems to make sense here. To sum everything up, make sure to count all of your atoms and you will get, of course, five Carbons, ten Hydrogens, and one Oxygen when you do that. So this is the Dot Structure that we were trying to find. Let's take a look at another one here.
So we have a molecular formula of C8 H Let's go ahead and calculate the Hydrogen Deficiency Index. So if we have eight Carbons, then we can have a maximum of two times eight plus two Hydrogens. So two times eight is 16, plus two is For eight Carbons, 18 Hydrogens is the max. Here we have 10 Hydrogens. So we're missing eight Hydrogens or missing four pairs of Hydrogens. The Hydrogen Deficiency Index is equal to four. I'm going to go ahead and draw a Benzene Ring in here. Let's look at the integration. All right, so sometimes you'll see the integration given like this.
This represents five protons for this very complex looking signal right here. This signal over here represents two protons and this signal represents three protons. Let's go back to these five protons with this complex signal, all right.
13C NMR Problem Set
This is in the Aromatic range, all right? So approximately 6. Those must be five Aromatic protons. We can go ahead and draw in five protons off of our Benzene Ring. Even though those protons are in slightly different environments, because this integration value is five here, we know a Benzene Ring is present, we're done, all right?
We don't have to worry about the slightly different environments.
We know that these five protons are giving us this complicated signal over here. All right, next, let's look at this signal. So two protons, so that must be a CH2. How many neighboring protons for the CH2? Well, there's one, two, three, four peaks. So if there's four peaks, four minus one is three.
So these CH2 protons have three neighbors.reiclimfastturtmong.tk
carbon NMR | MendelSet
All right, let's look at this next signal. So three protons. This must be a CH3. How many neighboring protons do we have for the CH3 protons? So three peaks. Three minus one is two, so two neighbors. Two neighboring protons, that must be these two neighboring protons. We'll go ahead and draw an Ethyl group, all right? So this is an Ethyl group pattern over here.
There's a CH2 and there's your CH3. So let's draw in all of our protons. So we have these protons and we have these protons. Let's color coordinate here just to make sure everything makes sense. So these red protons right here, they must be giving us this signal, so this Metal protons, all right?
We expect two neighbors because we have three peaks on our signal. And here are the two neighbors.
One, two, so two neighboring protons. Let's look at these two protons next. So these two protons are giving us this signal. We'd expect three neighbors because we have four peaks.