The following questions illustrate how stereochemical...

The following questions illustrate how stereochemical considerations can be used to elucidate aspects of biological mechanisms and reactions. a. A mixture of ${ }^{3} \mathrm{H}$-labeled 14-A and 14-B was carried through the reaction sequence shown: D-amino acid oxidase will oxidize only serine having $R$ configuration at $\mathrm{C}(2)$. Glycolate oxidase will remove only the pro- $R$ hydrogen of glycolic acid. Does the product $\left(\mathrm{O}=\mathrm{CHCO}_{2} \mathrm{H}\right)$ contain tritium? Explain your reasoning. b. Enzymatic oxidation of naphthalene by bacteria proceeds by way of the intermediate cis-diol shown. Which prochiral face of $\mathrm{C}(1)$ and $\mathrm{C}(2)$ of naphthalene is hydroxylated in this process? <smiles>OC1C=Cc2ccccc2C1O</smiles>c. The biosynthesis of valine by bacteria involves the following sequence: <smiles>CC(O)C(O)C(=O)O</smiles> <smiles>CC(C)C(N)C(=O)O</smiles> The stereochemistry of the reaction has been examined using the starting diol in which each methyl group was separately replaced by $\mathrm{CD}_{3}$. The diol- $d_{3}$ of the $2 R, 3 R$ configuration produces $2 S, 3 S$-valine- $d_{3}$, whereas the $2 R, 3 S$ diol $-d_{3}$ produces $2 S, 3 R$-valine- $d_{3}$. From this information deduce whether the $\mathrm{C}(2)$ and $C(3)$ hydroxy are replaced with inversion or retention of configuration. Show the basis for your conclusion. d. A synthesis of the important biosynthetic intermediate mevalonic acid starts with the enzymatic hydrolysis of the diester 14-C by pig liver esterase. The pro- $R$ ester group is selectively hydrolyzed. Draw a three-dimensional structure of the product.

The following questions illustrate how stereochemical considerations can be used to elucidate aspects of biological mechanisms and reactions.
a. A mixture of ${ }^{3} \mathrm{H}$-labeled 14-A and 14-B was carried through the reaction sequence shown:

D-amino acid oxidase will oxidize only serine having $R$ configuration at $\mathrm{C}(2)$. Glycolate oxidase will remove only the pro- $R$ hydrogen of glycolic acid. Does the product $\left(\mathrm{O}=\mathrm{CHCO}_{2} \mathrm{H}\right)$ contain tritium? Explain your reasoning.
b. Enzymatic oxidation of naphthalene by bacteria proceeds by way of the intermediate cis-diol shown. Which prochiral face of $\mathrm{C}(1)$ and $\mathrm{C}(2)$ of naphthalene is hydroxylated in this process?
<smiles>OC1C=Cc2ccccc2C1O</smiles>c. The biosynthesis of valine by bacteria involves the following sequence:
<smiles>CC(O)C(O)C(=O)O</smiles>
<smiles>CC(C)C(N)C(=O)O</smiles>
The stereochemistry of the reaction has been examined using the starting diol in which each methyl group was separately replaced by $\mathrm{CD}_{3}$. The diol- $d_{3}$ of the $2 R, 3 R$ configuration produces $2 S, 3 S$-valine- $d_{3}$, whereas the $2 R, 3 S$ diol $-d_{3}$ produces $2 S, 3 R$-valine- $d_{3}$. From this information deduce whether the $\mathrm{C}(2)$ and $C(3)$ hydroxy are replaced with inversion or retention of configuration. Show the basis for your conclusion.
d. A synthesis of the important biosynthetic intermediate mevalonic acid starts with the enzymatic hydrolysis of the diester 14-C by pig liver esterase. The pro- $R$ ester group is selectively hydrolyzed. Draw a three-dimensional structure of the product.

Key Concepts

Isotopic Labeling in Stereochemical Studies

Isotopic labeling uses atoms such as tritium or deuterium as tracers to follow the fate of specific atoms through a chemical reaction. In stereochemical studies, labeling is especially useful to track the movement of atoms in reactions, identify the stereochemical course, and understand reaction mechanisms by determining which atoms are retained or lost during the transformation.

Inversion and Retention of Configuration

This concept involves understanding how the geometric configuration at a chiral center is either reversed (inversion) or maintained (retention) during a chemical reaction. Determining whether a reaction proceeds with inversion or retention of configuration helps to elucidate the reaction mechanism and the stereochemical pathway involved in the transformation.

Three-Dimensional Molecular Representations

Drawing three-dimensional structures is essential for visualizing stereochemistry because the spatial arrangements of atoms determine molecular behavior and reactivity. These representations help in analyzing how enzymes interact with substrates based on the orientation of functional groups, thereby explaining the stereospecificity observed in many biological reactions.

Stereospecific Enzymatic Reactions

This concept refers to the ability of enzymes to distinguish between different stereoisomers of a substrate and selectively catalyze reactions for one configuration over another. Enzymes recognize substrates through their three?dimensional active sites, which allows them to act only on specific enantiomers or diastereomers, leading to stereospecific outcomes in biochemical processes.

Prochirality and Prochiral Centers

Prochirality describes the potential of a molecule to become chiral when a single substituent is altered, such as replacing one of two identical groups attached to a prochiral center. This concept is critical in understanding how enzymes selectively remove or modify a specific hydrogen or group (designated as pro-R or pro-S) in a substrate, thereby introducing chirality into the molecule during catalysis.

Transcript

00:01 Hey guys, ladies to problem 50.

00:04 The first question in this problem is that which of these tsa cycle intermediates are chiral? we know that in a chiral compound the carbon atom is attached to four different atoms or groups and this type of carbon atoms are known as the stereo center.

00:20 Now in this dc cycle there are three intermediates that are having a stereo centers.

00:27 One is the isocytic acid.

00:31 And the stereocentors are this one and this one.

00:37 Another one is the oxalocinic acid.

00:41 The chiral carbon is this one with this star.

00:46 And the malic acid, it is also having a chiral center, which is here.

00:52 Then another additional question is, which intermediates has the greatest number of chiral center? from this structure we see that isome citric acid it is having two chiral center two chiral centers then the final question is which intermediate shows cis trans isomerism the cis trans isomerisms are isomers these are geometric isomers cis isomers of the same group same groups on the same side of double bond and trans isomer it is having same groups on the opposite side of the double bond.

02:12 Now in this case we need to find which isomers, which intermediates, they are having same groups and opposite groups on the sides of the double bond and then we can figure out which ones.

02:26 Which ones are the cis trans isomer in this case the aconetic acid here it is aconetic acid it is having it is having same groups on the same side of the double bond this is one cis trans isomer and the femuric acid this is another cis trans isomer because it is also having the it is a trans isomer it is having two hydrogens on the opposite side and carbonylic groups on the opposite side of the double bond this turned the cis trans isomers then the question b it says name the type of reaction that takes place in step one two and three what are the name of these reactions we need to name them so what we see that in the first step citric acid is converting into aconetic acid and in this conversion citric acid it loses water molecule it loses a water molecule and therefore this is a dehydration reaction since it is losing water molecule so it is reaction in the second step the aconetic acid it is converting into isocytic acid and in this conversion conversion there is addition of water molecule across the double bond of this aconetic acid so it is a hydration reaction hydration reaction because there is addition of water.

05:03 And the third step is that conversion of the isocytic acid into oxalus succanic acid.

05:10 In this case, alcohol it is converting into ketan.

05:16 The alcohol, it is converting into kitten.

05:19 Therefore, this type of reaction is known as oxidation reaction.

05:32 Okay, then question c it asks us notice that the hydration of aconetic acid to give isosacic acid does not follow marconikov's rule if it would follow marconikov's rule what product would be formed and why we are getting non -marconic of product but not the marconiakov product let us try to answer that therefore we are focusing on to this portion of this cycle.

06:19 We know marconikov's rule says that during an electrophilic addition reaction, unsymmetrical reagents into unsymmetrical reagents into unsymmetrical alkins and the negative parts of the reagents.

06:33 Always the negative part goes to the carbon of the double bond that has less hydrogen atom.

06:44 When an electrophilic addition happens between unsymmetrical reagents the negative part of reagent always goes to the carbon of the double bond that has less hydrogen atom that is the marconikov's rule now if the hydrogen of aconetic acid were followed marconycopsyl then the product should have been this one the product should have been this one the less hydrogen is having this metal carbon therefore this one should have got the hydroxyl group and the name of this compound is citric acid so if it would have followed marconycop's rule we should have got citric acid but this reaction it is not following marconypops shul and we are getting actually isocytic acid the reason behind this is in this cycle that is a cycle it is it proceeds due to due to the presence of enzyme and the enzyme positions the aconidic acid at the reactive site on the enzyme surface and the only one carbonation intermediate is form that is the isocytic acid and water is added through a reducelective mechanism due to the presence of enzyme water is added through regisolective mechanism therefore the addition of hydroxylic group is at the second carbon not the third carbon second carbon not the third carbon okay now the next question says what type of reaction takes place in a step 4 through 8 4 5 6 7 8 what are these reactions we need to answer here we see that the fourth step it consists of conversion of the oxalusaxonic acid to alpha ketogluteric acid in this reaction during this conversation a carbon dioxide molecule, it gets removed.

10:44 Carbon dioxide molecule gets remote, therefore it is a decarboxilation reaction, because carbon dioxide is remote...

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