The relative configurations of the stereoisomers of tartaric a...

Question

The relative configurations of the stereoisomers of tartaric acid were established by the following syntheses:

(1) D-(+)-glyceraldehyde $\overset{H C N}{\to}$ diastereomers A and B (separated)

(2) Hydrolysis of A and B using aqueous Ba(OH)₂ gave C and D, respectively.

(3) HNO₃ oxidation of C and D gave (-)-tartaric acid and meso-tartaric acid, respectively.

(a) You know the absolute configuration of D-(+)-glyceraldehyde. Use Fischer projections to show the absolute configurations of products A, B, C, and D.

(b) Show the absolute configurations of the three stereoisomers of tartaric acid: (+)-tartaric acid, (-)-tartaric acid, and meso-tartaric acid.

Accepted Answer

All right. Hi, everyone. So this question says that the following syntheses are used to produce the stereo isomers of 234 tri hydroxy pentane dios acid. Here, we've got three steps which I will narrate as we go along. But the last part says to draw the fisher projections of ABC D and the two isomers of 234 tri hydroxy pentane OIC acid formed in the end. So up on the screen here, I ha went ahead and I actually drew the fisher projection of the THOS. So we can go ahead and proceed with the first step right now, if I scroll up briefly, right, the first step involves reacting D theros with HCN to produce die stereo mirror A and B. Now, the reason for this, if you recall is because when you go ahead and you take a simple sugar here like D thos and you react it with HCN, then what happens here is that cyanide attaches itself to the aldehyde carbon on position two. And that makes the aldehyde carbon atom asymmetric, which produces three different, excuse me, two different cyano hydrants. Now, the reason why it's asymmetric is because now that the carbon of the night trial is the first carbon carbon two, which was formula formerly from the aldehyde can have two different configurations right. The hydroxy group can either be on the left at carbon two for one die stereo mirror or it can be on the right in the other. So first I am drawing the first is steri air with the hydroxy group of carbon too on the left side, but the remaining ones are going to remain the same. So this is dictum ee and the stereo air B is going to be the same thing. But with my second or rather the hydroxy group on carbon too on the right instead. So when I go ahead and compete that structure, I end up with dyster repeat. So if I go ahead and I scroll upwards to examine the second step, the second step involves the hydrolysis of A and B using an aqueous solution of barium hydroxide to produce C and D respectively. Now, here when I go ahead and I take either of my cyano hydrants or both of them actually, and I react with barium hydroxide in the presence of water. Recall that this is going to be the hydrolysis step of my two night trial groups, right? So by subjecting both night trials to um hy hydrolysis, excuse me with aqueous barium hydroxide. Then the night trial is broken down and I end up with a carboxylic acid. So you're going to end up with your carboxylic acid, but the rest of the structure is going to stay the same. So here, I'm just making sure that I'm keeping my hydroxy groups in the same positions and my hydrogens as well. Now, a compound A produces compound C after hydrolysis, which means that compound B is going to produce compound D after hydrolysis, tear all my hydrogens. Now and like I mentioned previously, compound B is going to produce compound D. So now let's go ahead and examine the third step. But before I do so, I just want to make sure that I have space for everything, which is why I'm moving everything off to the side before I scroll up. But part three is nitric acid oxidation of C and D to give two S 4 s3, 4 tri hydroxy pentane OIC acid and meso 234 tri hydroxy pentane OIC acid respectively. So in the very last step, right, recall that nitric acid is a very powerful oxidizing agent. So by subjecting compound C and D to oxidation with nitric acid, then your terminal C H2O H group or your terminal metox group is going to be converted into a carboxylic acid as well. So you end up with a dic carboxylic acid at the very end. So instead of drawing ac H2O, right, I end up with another carboxylic acid. So according to the description, right, compound C produces two, four S try hydroxy panting di OIC acid, just make sure I'm spacing out my words correctly. And now we can go ahead and draw the product after compound D is subsequently oxidized with nitric acid. So for this step, you're keeping all of your previous carbons the same. But the only difference is at the very last one, the very last carbon is instead going to be a carboxylic acid, not a meth boxy group. So here is my final carboxylic acid. And according to the description of the problem, right, compound D produces maso three comma four try hydroxy panting di OIC acid. And you can see that right? Because this final carboxylic acid here, this dic carboxylic acid has an element of symmetry to it. And now we have all of our products. So compounds A and B where are cyano hydrants which were then hydrolyzed into carboxylic acid C and D and later oxidized to yield the final two dic carboxylic acids. So with that being said, thank you so very much for watching if you stuck around to the end. And I hope we found this helpful.

Key Concepts

Stereochemistry

Fischer Projections

Oxidation and Hydrolysis Reactions

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