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. b. CH 3 C(OH)(CD 3 ) 2

Hey everyone and welcome back in today's video. We are going to solve the following problem using ride deuterium metal, magnesium bromide and heavy water is your sources of the deuterium and any non do iterated starting material, show how to make this deuterium labeled compound. When we look at the structure of the government compound, we may essentially indicate that central carbon atom over here because if you look at frenzies, it essentially implies that there are two submissions bonds that carbon atom. So if we draw that central carbon atom first of all, we're going to draw oh age, there's an alcohol group there to try to tear a metal group. So let's draw them out as well. Those will be our C. D three groups. And in addition to that, we also have an ethyl group on it said. So that's our final structure that we want to synthesize. We want to understand how to perform that synthesis. What sort of material do we want to use now in any granite reaction? The first step is to use the granite region itself. So it says the granite region Cd three MGB R. So that's try the zero method magnesium bromide. We also want to make sure that we're adding ether as our solvent and number two would be addition of generally hydro ni. Um Right because we went to perform ballistic work up but we still have to determine whether or not this is right if you look at the alcohol group because chronic reactions form alcohols. Indeed, the protein nation stop involves addition of hydrogen. So yes, we are good to add hydro ni um instead of using the heavy water. Right? If we use heavy water, we are essentially adding deuterium here and bonding it to auction. But because there's hydrogen present, we are adding hydrogen. We are adding hydro knee. Um So now the last thing is to understand which green yard substitutions have been added to the substrate. If we look at these C. D. Three parts over here, we understand that they both come from the same green region, meaning we had to add two equivalents of the green yard region. And because we're adding two equivalents, it essentially implies that we are ready to draw our Carbonell compound. But we want to make sure that there is a good leaving group. And before we look at that leaving group, let's just draw the remaining submissions. So this is our central carbon atom which initially form that carbonell functional groups. We have drawn our Carbonell on the left, we have an ethyl group. So let's draw an ethyl group and recall that two equivalents of the green region can be added whenever we have a good leaving group over here and generally that can be an ester and acid chloride, an acid bromide or any group. That is a good leaving group. Right, so for the purposes of this problem, we can just go with an ester and we control a meth oxy group and with oxy group whichever group you prefer, that's perfectly fine. And we can essentially throw one of them which would be an ether oxy group. So that would be oh ch two ch three. That's a perfect leaving group for us within an ester. So we can propose this is our starting material and we can label that this structure is our material for this problem. Now, how does this reaction take place? If we look at our starting material essentially we're using one or based on the first equivalent of the korean yard regions, we can say D. Three C. Bonnets, magnesium bromide. They were saying nuclear feel like a tag the cleavage of the pi bon. Then the restoration of the pi bond and then the loss of the leaving group. So in the first step we are forming our intermediate which looks like this, the carbonell group is now bonded to C. D. Three. We still have an ethyl group. And because again this is a carbon group, we can use another equivalent. So another equivalent of the cranial region is used in this nuclear attack. We're forming an elk oxide and iron and there is no more good leaving group. So that would be our elk oxide with two. See the three groups and ethyl group and now we're protein eating it right as we said before, we are going to add an alcohol group. Oh H. Instead of O. D. That's why we didn't even use detail. We're saying that the negative charge would just attack that acidic proton of hydro ni. Um And that's how we get our found product. Now, the correct answer to this multiple choice question is a however, if you have any questions, don't have States, leave your comment down below in the comments 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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5:10 minutes

Problem 55b

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.
b. CH₃C(OH)(CD₃)₂

Verified step by step guidance

Start with a non-deuterated ketone, such as acetone (CH3C(O)CH3), as your starting material. Acetone has a carbonyl group that can undergo nucleophilic addition reactions.

React acetone with CD3MgBr (a Grignard reagent containing deuterium). The nucleophilic CD3⁻ group will attack the electrophilic carbonyl carbon of acetone, forming a tetrahedral intermediate.

Protonate the tetrahedral intermediate by adding D2O (deuterium oxide). This step will replace the oxygen atom of the carbonyl group with an OD group, resulting in the formation of CH3C(OD)(CD3)2.

Recognize that the OD group can be converted to an OH group by exchanging the deuterium atom with a hydrogen atom. This can be achieved by treating the compound with a mild acidic or neutral aqueous solution.

The final product is CH3C(OH)(CD3)2, where the two CD3 groups are deuterium-labeled, and the hydroxyl group is non-deuterated.

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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 the pathways of chemical reactions and understanding molecular structures. In this context, deuterium-labeled compounds can provide insights into reaction mechanisms and kinetics.

Grignard Reagents

Grignard reagents, such as CD3MgBr, are organomagnesium compounds used in organic synthesis to form carbon-carbon bonds. They react with electrophiles, including carbonyl compounds, to create new carbon centers. Understanding how to utilize Grignard reagents is essential for constructing complex organic molecules, including those with deuterium labels.

Alcohol Formation

The formation of alcohols from carbonyl compounds is a fundamental reaction in organic chemistry. In this case, the reaction of a Grignard reagent with a carbonyl compound leads to the formation of a tertiary alcohol. Recognizing the mechanisms of alcohol formation, including the role of deuterium in the final product, is vital for synthesizing the desired deuterium-labeled compound.

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