Show how to make these deuterium-labeled compounds, using CD

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.

d. Ph(CD₃)₂COD

Accepted Answer

Hey everyone and welcome back in today's video, we are going to solve the following problem using tribal theory method. Magnesium, bromide and heavy water is your sources of deuterium and any non duty rated starting material, show how to make this deuterium labeled compound and were given the formula of it. Right? So we're going to show how the structure looks like C three C S two, C O D C D 32 times. Now we have a central carbon atom. This is our carbon atom mon itty O. D. Right? This is a very similar group to the alcohol group that we generally C O D O H. Right? We're just replacing hydrogen with deuterium. We have to see the three groups. So we're going to indicate the presence of these groups. Now, this would be C with actually let's just write Cds twins. That's so that's one of them. This is the second one. And then on the left we have an ethyl group. So that's our ethyl group. We went to form that compound using the Green Yard reaction and we understand that number one is the addition of the Green Yard region. C D three MGB R. Number two, that would be addition of Water D 20. That's basically because we already recall that number one is the addition of the growing out of region and the presence of ether number two is the proton transfer step for the protein nation step of the elk oxide. So when we look at our material, we want to understand what kind of substrate we can use for this reaction. This is the central carbon atom. Because if you notice we are forming alcohols, right? And we understand that this carbon atoms bonded C. O. D. Which looks similarly to the alcohol group that we're seeing. We're using that deuterium. From details, we can essentially say that this deuterium comes from. I need to oh, after the protein nation stop of the elk oxide and ion. So definitely we understand that this deuterium has already been used now. Thinking about the two alcohol substations notice that they contain deuterium and we can say that they are identical. We understand that these two have deuterium meaning we are adding two equivalents of the green yard region. And that makes perfect sense because cd three respondents magnesium bromide. So these two L kill groups have been added in a green yard reaction. So there are two equivalents of the caribbean region. That was used. Now, this essentially implies that because there are two equivalents added, we must have a good leaving group. So the carbon compound that we're starting with, the carbon atom over here must have a good leaving group. So it must be an ester or an acid chloride for example. But for the purposes of this problem, we can say that a good leaving group that we're using here. Let's suppose that it's roaming even though actually you may use any Esther you want because Esther's have good leaving groups which are Alcock side and are informed but we may also use acid chloride or acid bromide. That's perfectly fine. Now if you think about the remaining part of the chain it just has an ethnic group. So that's our starting material. And let's see why this reaction works. It's basically because in the first step of the reaction where we have c. d. three or D. Three C. Bonnets MGB R. We're performing that nuclear feel like attacks stop breaking the pi bond which then restores itself. Followed by the loss of the leaving group. We're adding our first equivalent. So that would be our carbonell compound. But now instead of roaming we are adding C. D. Three and now we are ready to add our second equivalent D three C. M G B R. There was a nuclear attack. There's no more good leaving groups that were just breaking the pi bond and forming the elk oxide and I am oxygen negatively charged. See the three CD three once again and that ethyl group and finally we are pro tin ating it using the D. Iterated water. D. O. D. Now deuterium is abstracted for by the loss of O. D. Negative and we're forming our final product. This is how it's done. So that definitely works for us. Now let's just think about the strategy of the whole problem. Okay so once again when we looked at our final structure we understood that the alcohol group that we're seeing is the major point that we want to start off with OD essentially implies that this proton comes from Dita rated water because it's a source of protons for the protein nation stop. So this allowed us to eliminate do iterated water and understand that the granite region cannot be analyzed. We see that the two remaining groups are equivalent to share their cD three and CD three is present and the granite region, meaning there must be two equivalents of the granite region used in this reaction. There were many part of the chain corresponds to the alcohol chain within the car menial groups. We are ready to draw that alcohol chain that ethel substitution bonded to carbon ill. And we understand that jesus, the fact there that there are two equivalents, not one used, We need to make sure that there's a good living group bonded to carbon ill, so it might be an ester group, it might be chlorine or bromine, that's perfectly fine. And now we have shown the mechanism and we understand that the mechanism is consistent with our product. So we may simply say that the starting material for this problem would be this one. The correct answer choice for the multiple choice question is a However, if you have any questions don't have to leave your comment down below in the comment section and thank you for watching

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.
d. Ph(CD₃)₂COD

Key Concepts

Deuterium Labeling

Grignard Reagents

Deuterated Alcohols

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