We just learned an assumption that stated that all of the hydrogens on a single atom are going to share one peak on proton NMR. And while that assumption is very helpful for most cases, there are going to be some exceptions where these protons will actually have different relationships depending on the original chirality of the molecule. This is where the exceptions begin. And if you learn to accept the exceptions and just roll with them, this is going to go a lot smoother for all of us. So let's dig right into proton relationships and the q test. As I just mentioned, sometimes these hydrogens are going to have different relationships based on their original chirality. And the test that we use for that is called the q test. So what the heck is the q test? Well, the q test is simply this. What we do is let's just look at this example here. We have 3 hydrogens. We take 1 hydrogen. We cross it out and we replace it with a random letter. In this case, q. The only reason we're using q is just to be random so that we don't confuse it with an atom. We're going to replace H with the symbol q and we're going to analyze that carbon now, that atom that the q is attached to. I'm going to say, did I just achieve a new chiral center by adding that Q there? Imagining that this H just randomly turned into some atom, would this now be a new chiral center? So that's what the q test is. And now we're going to see how the q test can yield some different results based on the original molecule. Let's just take this example into consideration. Do you think that that is a new chiral center because the Q is there? No guys. Remember from Organic Chemistry 1 that chiral centers only exist when you have 4 completely different atoms all around the same carbon or all around the same central atom. And in this case, why is this not a new chiral center? Well, the q is different. This bromoethyl group is different. But I've got 2 hydrogens, 2 of the same exact atom on that carbon, This obviously cannot be a chiral center. What do we do? What do we conclude when your q test does not generate a new chiral center? Well this is going to be the relationships with your protons are going to be based on results like these from the q test. If your q test does not yield a new chiral center, then these protons are always going to have the same relationship that's called homotopic. These protons are always comotopic and because homotopic means the same, that means they're considered equivalent. That means that they actually do share a signal on proton NMR. Meaning that the assumption that I taught you about how you can just assume that all the hydrogens on a carbon have the same peak, It holds true in this case because we just said that they're homotopic so they're the same. Perfect. And in general guys, you don't have to use the q test on stuff like CH3, right? Because notice that CH3 is always going to have 2 hydrogens left over after you use the q test. So at this point, we don't need to be wasting time drawing q tests for methyl groups because we just know that it's going to yield no chiral centers. Perfect. Now the next two relationships that we're going to discuss are going to be when we actually do find a new chiral center. When it does create a new chiral center, what's that relationship now? It's not homotopic anymore. Let's check it out. This next category, let's just go ahead and look at the picture first and see what's different about this one. Notice that here instead of starting off with CH3, I'm starting off with a CH2. When I use the q test on one of the H's and get a q instead, do I get a new chiral center? Yes guys. This is a chiral center. This is a prime perfect example of a chiral center. Why? Because notice that this carbon now has 4 completely different groups on it. If we were labeling chiral centers, we could label them as 1 q, some kind of high priority atom. 2, there's a double bond there. That double bond is different from a single bond. Then lastly, my hydrogen would be in the back. That's number 4. Now we're not going to be naming R and S here. That's not important. All I care about is that you decide is it a chiral center. Yes, it is. Now is that enough to tell the relationship between those protons? No. We still need one more piece of information. Now that we know that it makes a chiral center, we need to analyze the chirality of the original molecule. We have to look at the original molecule and say, did that original molecule have a chiral center? What do you guys think? Now I'm going to inspect this one and you guys are going to tell me, is there a chiral center on that molecule? Remember that in order to have a chiral center, I need to have 4 different groups on an atom and absolutely not. There are no chiral centers. This has 2 hydrogens. This has 2 hydrogens. The other one has 2 hydrogens and then the double bond can't be a chiral center.
- 1. A Review of General Chemistry5h 5m
- Summary23m
- Intro to Organic Chemistry5m
- Atomic Structure16m
- Wave Function9m
- Molecular Orbitals17m
- Sigma and Pi Bonds9m
- Octet Rule12m
- Bonding Preferences12m
- Formal Charges6m
- Skeletal Structure14m
- Lewis Structure20m
- Condensed Structural Formula15m
- Degrees of Unsaturation15m
- Constitutional Isomers14m
- Resonance Structures46m
- Hybridization23m
- Molecular Geometry16m
- Electronegativity22m
- 2. Molecular Representations1h 14m
- 3. Acids and Bases2h 46m
- 4. Alkanes and Cycloalkanes4h 19m
- IUPAC Naming29m
- Alkyl Groups13m
- Naming Cycloalkanes10m
- Naming Bicyclic Compounds10m
- Naming Alkyl Halides7m
- Naming Alkenes3m
- Naming Alcohols8m
- Naming Amines15m
- Cis vs Trans21m
- Conformational Isomers13m
- Newman Projections14m
- Drawing Newman Projections16m
- Barrier To Rotation7m
- Ring Strain8m
- Axial vs Equatorial7m
- Cis vs Trans Conformations4m
- Equatorial Preference14m
- Chair Flip9m
- Calculating Energy Difference Between Chair Conformations17m
- A-Values17m
- Decalin7m
- 5. Chirality3h 39m
- Constitutional Isomers vs. Stereoisomers9m
- Chirality12m
- Test 1:Plane of Symmetry7m
- Test 2:Stereocenter Test17m
- R and S Configuration43m
- Enantiomers vs. Diastereomers13m
- Atropisomers9m
- Meso Compound12m
- Test 3:Disubstituted Cycloalkanes13m
- What is the Relationship Between Isomers?16m
- Fischer Projection10m
- R and S of Fischer Projections7m
- Optical Activity5m
- Enantiomeric Excess20m
- Calculations with Enantiomeric Percentages11m
- Non-Carbon Chiral Centers8m
- 6. Thermodynamics and Kinetics1h 22m
- 7. Substitution Reactions1h 48m
- 8. Elimination Reactions2h 30m
- 9. Alkenes and Alkynes2h 9m
- 10. Addition Reactions3h 18m
- Addition Reaction6m
- Markovnikov5m
- Hydrohalogenation6m
- Acid-Catalyzed Hydration17m
- Oxymercuration15m
- Hydroboration26m
- Hydrogenation6m
- Halogenation6m
- Halohydrin12m
- Carbene12m
- Epoxidation8m
- Epoxide Reactions9m
- Dihydroxylation8m
- Ozonolysis7m
- Ozonolysis Full Mechanism24m
- Oxidative Cleavage3m
- Alkyne Oxidative Cleavage6m
- Alkyne Hydrohalogenation3m
- Alkyne Halogenation2m
- Alkyne Hydration6m
- Alkyne Hydroboration2m
- 11. Radical Reactions1h 58m
- 12. Alcohols, Ethers, Epoxides and Thiols2h 42m
- Alcohol Nomenclature4m
- Naming Ethers6m
- Naming Epoxides18m
- Naming Thiols11m
- Alcohol Synthesis7m
- Leaving Group Conversions - Using HX11m
- Leaving Group Conversions - SOCl2 and PBr313m
- Leaving Group Conversions - Sulfonyl Chlorides7m
- Leaving Group Conversions Summary4m
- Williamson Ether Synthesis3m
- Making Ethers - Alkoxymercuration4m
- Making Ethers - Alcohol Condensation4m
- Making Ethers - Acid-Catalyzed Alkoxylation4m
- Making Ethers - Cumulative Practice10m
- Ether Cleavage8m
- Alcohol Protecting Groups3m
- t-Butyl Ether Protecting Groups5m
- Silyl Ether Protecting Groups10m
- Sharpless Epoxidation9m
- Thiol Reactions6m
- Sulfide Oxidation4m
- 13. Alcohols and Carbonyl Compounds2h 17m
- 14. Synthetic Techniques1h 26m
- 15. Analytical Techniques:IR, NMR, Mass Spect7h 3m
- Purpose of Analytical Techniques5m
- Infrared Spectroscopy16m
- Infrared Spectroscopy Table31m
- IR Spect:Drawing Spectra40m
- IR Spect:Extra Practice26m
- NMR Spectroscopy10m
- 1H NMR:Number of Signals26m
- 1H NMR:Q-Test26m
- 1H NMR:E/Z Diastereoisomerism8m
- H NMR Table24m
- 1H NMR:Spin-Splitting (N + 1) Rule22m
- 1H NMR:Spin-Splitting Simple Tree Diagrams11m
- 1H NMR:Spin-Splitting Complex Tree Diagrams12m
- 1H NMR:Spin-Splitting Patterns8m
- NMR Integration18m
- NMR Practice14m
- Carbon NMR4m
- Structure Determination without Mass Spect47m
- Mass Spectrometry12m
- Mass Spect:Fragmentation28m
- Mass Spect:Isotopes27m
- 16. Conjugated Systems6h 13m
- Conjugation Chemistry13m
- Stability of Conjugated Intermediates4m
- Allylic Halogenation12m
- Reactions at the Allylic Position39m
- Conjugated Hydrohalogenation (1,2 vs 1,4 addition)26m
- Diels-Alder Reaction9m
- Diels-Alder Forming Bridged Products11m
- Diels-Alder Retrosynthesis8m
- Molecular Orbital Theory9m
- Drawing Atomic Orbitals6m
- Drawing Molecular Orbitals17m
- HOMO LUMO4m
- Orbital Diagram:3-atoms- Allylic Ions13m
- Orbital Diagram:4-atoms- 1,3-butadiene11m
- Orbital Diagram:5-atoms- Allylic Ions10m
- Orbital Diagram:6-atoms- 1,3,5-hexatriene13m
- Orbital Diagram:Excited States4m
- Pericyclic Reaction10m
- Thermal Cycloaddition Reactions26m
- Photochemical Cycloaddition Reactions26m
- Thermal Electrocyclic Reactions14m
- Photochemical Electrocyclic Reactions10m
- Cumulative Electrocyclic Problems25m
- Sigmatropic Rearrangement17m
- Cope Rearrangement9m
- Claisen Rearrangement15m
- 17. Ultraviolet Spectroscopy51m
- 18. Aromaticity2h 34m
- 19. Reactions of Aromatics: EAS and Beyond5h 1m
- Electrophilic Aromatic Substitution9m
- Benzene Reactions11m
- EAS:Halogenation Mechanism6m
- EAS:Nitration Mechanism9m
- EAS:Friedel-Crafts Alkylation Mechanism6m
- EAS:Friedel-Crafts Acylation Mechanism5m
- EAS:Any Carbocation Mechanism7m
- Electron Withdrawing Groups22m
- EAS:Ortho vs. Para Positions4m
- Acylation of Aniline9m
- Limitations of Friedel-Crafts Alkyation19m
- Advantages of Friedel-Crafts Acylation6m
- Blocking Groups - Sulfonic Acid12m
- EAS:Synergistic and Competitive Groups13m
- Side-Chain Halogenation6m
- Side-Chain Oxidation4m
- Reactions at Benzylic Positions31m
- Birch Reduction10m
- EAS:Sequence Groups4m
- EAS:Retrosynthesis29m
- Diazo Replacement Reactions6m
- Diazo Sequence Groups5m
- Diazo Retrosynthesis13m
- Nucleophilic Aromatic Substitution28m
- Benzyne16m
- 20. Phenols55m
- 21. Aldehydes and Ketones: Nucleophilic Addition4h 56m
- Naming Aldehydes8m
- Naming Ketones7m
- Oxidizing and Reducing Agents9m
- Oxidation of Alcohols28m
- Ozonolysis7m
- DIBAL5m
- Alkyne Hydration9m
- Nucleophilic Addition8m
- Cyanohydrin11m
- Organometallics on Ketones19m
- Overview of Nucleophilic Addition of Solvents13m
- Hydrates6m
- Hemiacetal9m
- Acetal12m
- Acetal Protecting Group16m
- Thioacetal6m
- Imine vs Enamine15m
- Addition of Amine Derivatives5m
- Wolff Kishner Reduction7m
- Baeyer-Villiger Oxidation39m
- Acid Chloride to Ketone7m
- Nitrile to Ketone9m
- Wittig Reaction18m
- Ketone and Aldehyde Synthesis Reactions14m
- 22. Carboxylic Acid Derivatives: NAS2h 51m
- Carboxylic Acid Derivatives7m
- Naming Carboxylic Acids9m
- Diacid Nomenclature6m
- Naming Esters5m
- Naming Nitriles3m
- Acid Chloride Nomenclature5m
- Naming Anhydrides7m
- Naming Amides5m
- Nucleophilic Acyl Substitution18m
- Carboxylic Acid to Acid Chloride6m
- Fischer Esterification5m
- Acid-Catalyzed Ester Hydrolysis4m
- Saponification3m
- Transesterification5m
- Lactones, Lactams and Cyclization Reactions10m
- Carboxylation5m
- Decarboxylation Mechanism14m
- Review of Nitriles46m
- 23. The Chemistry of Thioesters, Phophate Ester and Phosphate Anhydrides1h 10m
- 24. Enolate Chemistry: Reactions at the Alpha-Carbon1h 53m
- Tautomerization9m
- Tautomers of Dicarbonyl Compounds6m
- Enolate4m
- Acid-Catalyzed Alpha-Halogentation4m
- Base-Catalyzed Alpha-Halogentation3m
- Haloform Reaction8m
- Hell-Volhard-Zelinski Reaction3m
- Overview of Alpha-Alkylations and Acylations5m
- Enolate Alkylation and Acylation12m
- Enamine Alkylation and Acylation16m
- Beta-Dicarbonyl Synthesis Pathway7m
- Acetoacetic Ester Synthesis13m
- Malonic Ester Synthesis15m
- 25. Condensation Chemistry2h 9m
- 26. Amines1h 43m
- 27. Heterocycles2h 0m
- Nomenclature of Heterocycles15m
- Acid-Base Properties of Nitrogen Heterocycles10m
- Reactions of Pyrrole, Furan, and Thiophene13m
- Directing Effects in Substituted Pyrroles, Furans, and Thiophenes16m
- Addition Reactions of Furan8m
- EAS Reactions of Pyridine17m
- SNAr Reactions of Pyridine18m
- Side-Chain Reactions of Substituted Pyridines20m
- 28. Carbohydrates5h 53m
- Monosaccharide20m
- Monosaccharides - D and L Isomerism9m
- Monosaccharides - Drawing Fischer Projections18m
- Monosaccharides - Common Structures6m
- Monosaccharides - Forming Cyclic Hemiacetals12m
- Monosaccharides - Cyclization18m
- Monosaccharides - Haworth Projections13m
- Mutarotation11m
- Epimerization9m
- Monosaccharides - Aldose-Ketose Rearrangement8m
- Monosaccharides - Alkylation10m
- Monosaccharides - Acylation7m
- Glycoside6m
- Monosaccharides - N-Glycosides18m
- Monosaccharides - Reduction (Alditols)12m
- Monosaccharides - Weak Oxidation (Aldonic Acid)7m
- Reducing Sugars23m
- Monosaccharides - Strong Oxidation (Aldaric Acid)11m
- Monosaccharides - Oxidative Cleavage27m
- Monosaccharides - Osazones10m
- Monosaccharides - Kiliani-Fischer23m
- Monosaccharides - Wohl Degradation12m
- Monosaccharides - Ruff Degradation12m
- Disaccharide30m
- Polysaccharide11m
- 29. Amino Acids3h 20m
- Proteins and Amino Acids19m
- L and D Amino Acids14m
- Polar Amino Acids14m
- Amino Acid Chart18m
- Acid-Base Properties of Amino Acids33m
- Isoelectric Point14m
- Amino Acid Synthesis: HVZ Method12m
- Synthesis of Amino Acids: Acetamidomalonic Ester Synthesis16m
- Synthesis of Amino Acids: N-Phthalimidomalonic Ester Synthesis13m
- Synthesis of Amino Acids: Strecker Synthesis13m
- Reactions of Amino Acids: Esterification7m
- Reactions of Amino Acids: Acylation3m
- Reactions of Amino Acids: Hydrogenolysis6m
- Reactions of Amino Acids: Ninhydrin Test11m
- 30. Peptides and Proteins2h 42m
- Peptides12m
- Primary Protein Structure4m
- Secondary Protein Structure17m
- Tertiary Protein Structure11m
- Disulfide Bonds17m
- Quaternary Protein Structure10m
- Summary of Protein Structure7m
- Intro to Peptide Sequencing2m
- Peptide Sequencing: Partial Hydrolysis25m
- Peptide Sequencing: Partial Hydrolysis with Cyanogen Bromide7m
- Peptide Sequencing: Edman Degradation28m
- Merrifield Solid-Phase Peptide Synthesis18m
- 32. Lipids 2h 50m
- 34. Nucleic Acids1h 32m
- 35. Transition Metals5h 33m
- Electron Configuration of Elements45m
- Coordination Complexes20m
- Ligands24m
- Electron Counting10m
- The 18 and 16 Electron Rule13m
- Cross-Coupling General Reactions40m
- Heck Reaction40m
- Stille Reaction13m
- Suzuki Reaction25m
- Sonogashira Coupling Reaction17m
- Fukuyama Coupling Reaction15m
- Kumada Coupling Reaction13m
- Negishi Coupling Reaction16m
- Buchwald-Hartwig Amination Reaction19m
- Eglinton Reaction17m
- 36. Synthetic Polymers1h 49m
- Introduction to Polymers6m
- Chain-Growth Polymers10m
- Radical Polymerization15m
- Cationic Polymerization8m
- Anionic Polymerization8m
- Polymer Stereochemistry3m
- Ziegler-Natta Polymerization4m
- Copolymers6m
- Step-Growth Polymers11m
- Step-Growth Polymers: Urethane6m
- Step-Growth Polymers: Polyurethane Mechanism10m
- Step-Growth Polymers: Epoxy Resin8m
- Polymers Structure and Properties8m
1H NMR:Q-Test - Online Tutor, Practice Problems & Exam Prep
In proton NMR, the q test helps determine the relationships between protons based on chirality. Protons can be homotopic (same signal) if no new chiral center is formed, or enantiotopic (same signal) if a new chiral center is created without an original one. If both a new and an original chiral center exist, protons are diastereotopic (different signals). Understanding these relationships is crucial for interpreting NMR spectra and predicting molecular behavior in organic chemistry.
Q-Test (Proton Relationships)
Video transcript
Identifying Proton Signals using Q-Test
Video transcript
Notice that this question is asking the same thing that we've already answered before. How many signals will each molecule possess in proton NMR? But now that we know about the Q test, the very first thing we actually need to answer for all these problems is going to be, do I use the Q test or not? Because you don't have to always use the Q test. So you might be wondering, well, how do I know when I have to use it? Okay. So here's the hint: We use the Q test when you already have 1 or more chiral centers. Now the logic behind that is this: If you don't already have a chiral center, then for sure it's going to be either homotopic or enantiotopic if you don't have any chiral centers present from the get-go. Homotopic or enantiotopic. And remember that homotopic and enantiotopic both result exactly the same in proton NMR. They share a signal. That means the only time I have to worry about using the Q test is if I already had 1 or more chiral centers, in which case it might change the answer.
My first question to you is, for question A, do we have to use the Q test? The answer is no. We don't because there are no chiral centers present. Notice that you might think this is a chiral center, but it has two of the same exact group on it. It's CH3 CH3. So that's not a chiral center. There are no chiral centers here. That means I don't have to use the Q test. I'm just going to use the old rule that said that every single atom gets its own signal and watch for symmetry. I would go ahead and I would say this is signal A. This is signal B. This is signal C. This is signal D. And notice that after atom D, we actually do have a plane of symmetry developing where both of these are going to be equivalent. E and E because after that, you get a plane of symmetry. So we have symmetry on one part of the molecule but not on the rest. That's still okay. It helps us to determine that these methyl groups are the same as each other. The answer here would be 5 signals. Notice that it's going to become really important when you try to answer these questions that first you ask yourself, do I need the Q test? On this next problem, problem B, that should be your first question. Do you need to use the Q test? If so, where do you use it? When do you use it? If not, then just go with the old rule we learned. Go ahead and solve question B.
Identifying Proton Signals using Q-Test
Video transcript
Did we have to use the q test on problem b? The answer is yes because I do have a chiral center present from the beginning. If you're wondering where that chiral center is, it's right in the middle guys because notice that I have a methyl. I have a hydrogen. I have an ethyl on one side and I have an isopropyl on the other. That is one chiral center. Now does that mean that I have to use the q test on every single atom? Actually, no. You only use the q test when you have one or more chiral centers and only on CH2s. Because the fact that I told you guys, CH3's are always homotopic no matter what because they don't make chiral centers. CHs, if you only have one H present, that kind of answers your question already because you only have one H. So, you don't have to worry about is it equivalent to another hydrogen. Let me just show you. For example, we can already conclude that this is going to get its own peak, A. We can conclude that these two are going to get their own peak. That's B because there's symmetry there. I'm trying to color code this for you guys. We can conclude that this is going to get its own peak C because it's a methyl group. On top of that, we've got this hydrogen which is obviously unique because it's the only one on that chiral center, so that must be hydrogen. We've also got a hydrogen here which, since it's the only hydrogen there, it must get its own peak because it's the only hydrogen that's in between two methyls like that. So far, we've been able to do all of this without the q test. Where does the q test really come into play? Only on any CH2s that we have. Do we have a CH2 present? Yes. We have a CH2 right here. And on that CH2, we need to use the q test. So, I'm going to take an H. I'm going to take an H and I'm going to replace one of the H's with a q. And now I'm going to ask myself, once I've done that, did I just make a new chiral center? What do you think? Is that a new chiral center now that I added a q? Yes, it is because I've got one group, two groups, a methyl and then four which is the rest of that junk. I can't even name it. It's a pretty big substituent. So that's definitely a new chiral center. So what does it mean when the Q test gives you a new chiral center and you already had one chiral center? What's the conclusion? It means that these hydrogens, hydrogen 1 and hydrogen 2 are diastereotopic. But most importantly, this question didn't ask me what the relationship was. It said, how many signals are we going to get. So, most importantly, what that means guys, and this is the important part, what that means is that this H gets its own letter F. It's an ugly own letter F. And this H gets its own letter G because remember that whenever you have diastereotopic protons, they each get their own signal. If you said 6 signals, that was the trick answer. The answer should actually be 7 signals because of the fact that those two diastereotopic protons get their own signal each. That was complicated. If you have any questions, let me know. But that's the way you got to approach these problems. That being said, what's the first thing you're going answer for C? You're going to tell me, do you use the q test or not? Go ahead and try to figure it out.
Identifying Proton Signals using Q-Test
Video transcript
So, did we have to use the q test on C? The answer is no. Guys, this molecule has no chiral centers. You might be thinking, but Johnny, isn't that a chiral center? No. Because I have two of the same exact group on both sides. So there's no chiral center here, meaning that there's no q-test. So that means that I'm literally just going to give every atom its own peak. Is there any symmetry on this molecule? That's the harder question. Unfortunately, yes. There's actually symmetry right down the middle. You might be wondering how in the world is that possible. One side obviously has the alcohol. One side obviously has the methyl. Johnny, you're nuts. What's wrong with you? No, because guys remember that tetrahedral molecules don't really look like that. What they look more like is that you've got two groups to the side and then you've got one group in the front and you've got one group in the back. Now we don't know which one is which. We don't know if the OH is in the front or the OH is in the back since wedge and dash wasn't given to us. But really when you split this molecule down the middle, you're splitting it down that front and back molecule. So when you're splitting it down the middle, you actually have you're splitting let's say the alcohol's in the front. You're splitting it right down the middle and you're also splitting the methyl right down the middle as well, meaning that this actually is symmetrical. So how many different bonds would I have? Well, obviously that H is unique. That's going to be type A. Obviously, this methyl is unique. That's going to be type B. Nothing else is like that. We have a carbon here that we're not going to count because it doesn't even have any H's. And then we've got molecules C. We've got proton C and we've got proton D, which are going to be mirrored on the other side because of symmetry. So this is also and this is also. So that means that this only had 4 signals. Got it? Cool. So let me know if that made sense. I'm sorry, it was a little bit tricky. Just got to get practice with this, alright? So let's move on to the next part.
Identify the indicated set of protons as unrelated, homotopic, enantiotopic, or diastereotopic.
Problem Transcript
Identify the indicated set of protons as unrelated, homotopic, enantiotopic, or diastereotopic.
Problem Transcript
Identify the indicated set of protons as unrelated, homotopic, enantiotopic, or diastereotopic.
Problem Transcript
Do you want more practice?
More setsHere’s what students ask on this topic:
What is the Q-test in 1H NMR and how is it used?
The Q-test in 1H NMR is a method used to determine the relationships between protons based on chirality. To perform the Q-test, you replace one hydrogen atom on a carbon with a hypothetical atom 'Q'. If this substitution creates a new chiral center, the protons are analyzed for their relationships. If no new chiral center is formed, the protons are homotopic and share the same signal. If a new chiral center is formed without an original one, the protons are enantiotopic and still share the same signal. If both a new and an original chiral center exist, the protons are diastereotopic and have different signals.
What are homotopic protons in 1H NMR?
Homotopic protons in 1H NMR are protons that are chemically equivalent and share the same signal in the NMR spectrum. This occurs when the Q-test does not generate a new chiral center, meaning the protons are in identical environments. For example, in a CH3 group, all three hydrogens are homotopic because replacing any one of them with a hypothetical 'Q' does not create a new chiral center. Therefore, they all produce the same NMR signal.
How do enantiotopic protons appear in 1H NMR spectra?
Enantiotopic protons in 1H NMR appear as a single signal because the NMR is not specific enough to distinguish between them. Enantiotopic protons occur when the Q-test creates a new chiral center, but the original molecule has no chiral centers. Although these protons are technically different, the NMR spectrum treats them as equivalent, resulting in a single peak. Advanced NMR techniques using chiral solvents can differentiate them, but this is beyond the scope of basic organic chemistry courses.
What are diastereotopic protons and how do they affect 1H NMR spectra?
Diastereotopic protons are protons that are in different chemical environments and thus produce different signals in the 1H NMR spectrum. This occurs when the Q-test creates a new chiral center and the original molecule already has one or more chiral centers. Because these protons are not equivalent, they result in separate peaks in the NMR spectrum. Although the signals may be very close together, they are distinct, reflecting the different environments of the diastereotopic protons.
When should the Q-test be used in 1H NMR analysis?
The Q-test should be used in 1H NMR analysis when you need to determine the relationship between protons on a carbon atom, especially if there is a possibility of chirality. Specifically, you use the Q-test if the molecule has an original chiral center. If there is no original chiral center, the protons will either be homotopic or enantiotopic, and the Q-test is not necessary. The Q-test helps identify whether the protons are homotopic, enantiotopic, or diastereotopic, which is crucial for interpreting NMR spectra accurately.
Your Organic Chemistry tutors
- If the imaginary replacement of either of two protons forms enantiomers, then those protons are said to be ena...
- Phenyl Grignard reagent adds to 2-methylpropanal to give the secondary alcohol shown. The proton NMR of 2-meth...
- a. For the following compounds, which pairs of hydrogens (Ha and Hb) are enantiotopic hydrogens? <IMAGE>
- a. For the following compounds, which pairs of hydrogens (Ha and Hb) are enantiotopic hydrogens? <IMAGE>
- If rotation is restricted, as in the case of the molecule shown, the hydrogens labeled a and b are nonequivale...
- Replace Hₐ, H₆, and H꜀ in methyl benzene with a deuterium. What is the relationship between the three product...
- Replace Hₐ and H₆ in acetone with a deuterium. What is the relationship between the two products you obtain? ...
- How can 1,2-, 1,3-, and 1,4-dinitrobenzene be distinguished by a. 1H NMR spectroscopy?