Show how to make these deuterium-labeled compounds, using CD 3 MgBr and D 2 O as your sources of deuterium, and any non-deuterated starting materials you wish. a. CH 3 CH(OD)CD 3

Hey everyone and welcome back in today's video, we are going to learn how to solve the following problem using tribe you theory of metal, magnesium, bromide and heavy water as your sources of deuterium and any non duty rated starting material show how to make this deuterium labeled compound. If you look at the given structure is CH three ch two ch O D. C D. Three branches here are really relevant because we want to understand that this part is bonded to that carbon as well as this part is bonded to that carbon. So if we draw our final structure, we understand that there is carbon oxygen bonds. So we are going to add od bonded carbon then see the three bonds that carbon there is that hydrogen bonded to that carbon. We will not worry about it. We are just going to draw that remaining ethyl group. Right? So this is our final product that we want to form Now here we want to propose a starting material and according to the problem, we are using the green yard region in the presence of ether that's important. Center stand right recall that cranial creations go with ethers that those are our solvents and number two addition of it might be due to aerated water riot or it might be simple hydro knee. Um It really depends on our further analysis. So let's perform that analysis. If we look at our structure in particular, this is the central carbon atom because we have formed an alcohol right? O H R O. D. It just the nicest of hydrogen. So this is the carbon atom that initially belonged to our carbon compounds. We can just draw it for analogy because the last step in the gradient reaction is the protein nation stuff. The elk oxide and iron gets protein ated and it got protein aided by the Ut area meaning we are going to use D. T. O. Just to suggest that for our reaction for that protein nation step, if it was O. H, we would have used H 30 plus here instead to protein it Alcock side. But now it's so these were using detail. Okay, So now this allows us to understand that the remaining part CD three comes from the green region and it's also understandable that there's only one equivalent of the granite regent use because we have only added one equivalent of CD three. And this is really important now for us if we are forming only actually listening about it, this is an alcohol. Right? And it has to alcohol submissions. So this is a secondary alcohol. One of the one of the remaining submissions to say hydrogen atom. So we want to propose the correct carbonell compound for us. First of all, we understand that okay, there's an ethyl group on the love. So we can just draw that ethyl group now, what other atoms should we draw here and because we're forming a secondary alcohol, it must be an alga hide, right because Aldo hides upon reacting with granite regions form secondary alcohols and in addition to that, we have that implicit hydrogen atom over here, we just verifies that. So we may use this al the height as our starting material. And we are basically done. We can just look at how this reaction takes place. So we're using our green york region D three C. Bonded to magnesium bromide. There's a nuclear attack that carbon ill. We're breaking the pi bond and forming the Alcock said an eye on So now we have oxygen negatively charged an ethyl group. There's still that hydrogen and there's an alcohol. Supposition that we have added that CT three and finally we need to protein ate it just as we said before the protein nation stop depends on what kind, what is the source of those products? In this case it's heavy water because our final product has D instead of age. So we're going to use heavy water D. O. D. So there will be an attack on the Syrian fold by the loss of the leaving group. And this is how we form our final alcohol. So that makes perfect sense. We looked at the mechanism the proposed structure works for us right now, essentially we will label this as our final structure. This will be our starting material. How many carbon atoms do we have? 123. That be Pro panel panel would be a suitable starting material for this reaction. If we used try the material method. Magnesium bromide in the presence in ether followed by hydra leases and heavy water. Now the correct answer to the small cultures question is C, However, if you have any questions, don't hesitate, leave your comment down below in the comment section and thank you for watching.

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1. A Review of General Chemistry5h 5m
2. Molecular Representations1h 14m
3. Acids and Bases2h 46m
4. Alkanes and Cycloalkanes4h 17m
5. Chirality3h 39m
6. Thermodynamics and Kinetics1h 22m
7. Substitution Reactions1h 48m
8. Elimination Reactions2h 30m
9. Alkenes and Alkynes2h 9m
10. Addition Reactions3h 19m
11. Radical Reactions1h 58m
12. Alcohols, Ethers, Epoxides and Thiols2h 42m
13. Alcohols and Carbonyl Compounds2h 17m
14. Synthetic Techniques1h 26m
15. Analytical Techniques:IR, NMR, Mass Spect7h 3m
16. Conjugated Systems6h 13m
17. Ultraviolet Spectroscopy51m
18. Aromaticity2h 34m
19. Reactions of Aromatics: EAS and Beyond5h 1m
20. Phenols55m
21. Aldehydes and Ketones: Nucleophilic Addition4h 56m
22. Carboxylic Acid Derivatives: NAS2h 51m
23. The Chemistry of Thioesters, Phophate Ester and Phosphate Anhydrides1h 10m
24. Enolate Chemistry: Reactions at the Alpha-Carbon1h 53m
25. Condensation Chemistry2h 9m
26. Amines1h 43m
27. Heterocycles2h 0m
28. Carbohydrates5h 53m
29. Amino Acids4h 20m
30. Peptides and Proteins2h 42m
31. Catalysis in Organic Reactions1h 30m
32. Lipids 2h 50m
33. The Organic Chemistry of Metabolic Pathways2h 52m
34. Nucleic Acids1h 32m
35. Transition Metals6h 14m
36. Synthetic Polymers1h 49m

13. Alcohols and Carbonyl Compounds

Grignard Reaction

Organic Chemistry

13. Alcohols and Carbonyl Compounds

Grignard Reaction

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5:16 minutes

Problem 55a

Textbook Question

Show how to make these deuterium-labeled compounds, using CD₃MgBr and D₂O as your sources of deuterium, and any non-deuterated starting materials you wish.
a. CH₃CH(OD)CD₃

Verified step by step guidance

Step 1: Begin with a non-deuterated starting material, such as acetone (CH3COCH3). Acetone is a good choice because it contains a ketone functional group that can undergo nucleophilic addition reactions.

Step 2: React acetone with CD3MgBr (deuterium-labeled methylmagnesium bromide). This is a Grignard reagent, which will attack the carbonyl carbon of acetone, forming a tertiary alcohol intermediate. The reaction mechanism involves nucleophilic addition of the CD3 group to the carbonyl carbon.

Step 3: After the Grignard reaction, the intermediate will have a tertiary alcohol group. To introduce the deuterium-labeled hydroxyl group (-OD), perform a proton exchange reaction by treating the intermediate with D2O (deuterium oxide). This will replace the hydrogen in the hydroxyl group with deuterium.

Step 4: Verify the structure of the final product, CH3CH(OD)CD3. The molecule should now have one deuterium atom in the hydroxyl group (-OD) and three deuterium atoms in the methyl group (-CD3).

Step 5: Ensure proper purification of the product using techniques such as distillation or chromatography to isolate CH3CH(OD)CD3 from any side products or unreacted starting materials.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Deuterium Labeling

Deuterium labeling involves substituting hydrogen atoms in organic compounds with deuterium, a stable isotope of hydrogen. This process is crucial in tracing reaction pathways and studying mechanisms in organic chemistry. In the context of the question, deuterium from CD3MgBr and D2O will be incorporated into the target compound to create specific isotopomers.

Grignard Reagents

Grignard reagents, such as CD3MgBr, are organomagnesium compounds that react with various electrophiles to form carbon-carbon bonds. They are highly reactive and can be used to introduce deuterium into organic molecules. Understanding how Grignard reagents function is essential for synthesizing the desired deuterium-labeled compound in the question.

Nucleophilic Substitution

Nucleophilic substitution is a fundamental reaction mechanism in organic chemistry where a nucleophile replaces a leaving group in a molecule. In this case, the deuterated Grignard reagent acts as a nucleophile, attacking a suitable electrophile to form the desired product. Recognizing the conditions and mechanisms of nucleophilic substitution is vital for successfully synthesizing the target compound.

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