Chapter 25, Problem 25.35AP

Interpretation Introduction

(a)

Interpretation:

A curved-arrow mechanism for the interconversion of $3-\mathrm{p}\mathrm{h}\mathrm{o}\mathrm{s}\mathrm{p}\mathrm{h}\mathrm{o}\mathrm{g}\mathrm{l}\mathrm{y}\mathrm{c}\mathrm{e}\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{e}$ to $2-\mathrm{p}\mathrm{h}\mathrm{o}\mathrm{s}\mathrm{p}\mathrm{h}\mathrm{o}\mathrm{g}\mathrm{l}\mathrm{y}\mathrm{c}\mathrm{e}\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{e}$ is to be drawn.

Concept introduction:

The interconversion of $3-\mathrm{p}\mathrm{h}\mathrm{o}\mathrm{s}\mathrm{p}\mathrm{h}\mathrm{o}\mathrm{g}\mathrm{l}\mathrm{y}\mathrm{c}\mathrm{e}\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{e}$ to $2-\mathrm{p}\mathrm{h}\mathrm{o}\mathrm{s}\mathrm{p}\mathrm{h}\mathrm{o}\mathrm{g}\mathrm{l}\mathrm{y}\mathrm{c}\mathrm{e}\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{e}$ is carried by enzyme catalysed nucleophilic substitution. The nitrogen of histidine residue in a protein acts as a nucleophile whereas the phosphorus of phosphate ester acts as an electrophile.

Answer to Problem 25.35AP

Explanation of Solution

The interconversion equation of $3-\mathrm{p}\mathrm{h}\mathrm{o}\mathrm{s}\mathrm{p}\mathrm{h}\mathrm{o}\mathrm{g}\mathrm{l}\mathrm{y}\mathrm{c}\mathrm{e}\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{e}$ to $2-\mathrm{p}\mathrm{h}\mathrm{o}\mathrm{s}\mathrm{p}\mathrm{h}\mathrm{o}\mathrm{g}\mathrm{l}\mathrm{y}\mathrm{c}\mathrm{e}\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{e}$ and the structures of histidine and phosphohistidine residue is shown below:

A curved-arrow mechanism for the interconversion of $3-\mathrm{p}\mathrm{h}\mathrm{o}\mathrm{s}\mathrm{p}\mathrm{h}\mathrm{o}\mathrm{g}\mathrm{l}\mathrm{y}\mathrm{c}\mathrm{e}\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{e}$ to $2-\mathrm{p}\mathrm{h}\mathrm{o}\mathrm{s}\mathrm{p}\mathrm{h}\mathrm{o}\mathrm{g}\mathrm{l}\mathrm{y}\mathrm{c}\mathrm{e}\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{e}$ is shown below:

The nitrogen of histidine residue acts as a nucleophile after loss of proton by a base $A^{-}$ and forms substitution product with inversion of configuration at phosphorous. In second step the nucleophilic attack of oxygen from acetate group gives $2-\mathrm{p}\mathrm{h}\mathrm{o}\mathrm{s}\mathrm{p}\mathrm{h}\mathrm{o}\mathrm{g}\mathrm{l}\mathrm{y}\mathrm{c}\mathrm{e}\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{e}$ where the configuration at the phosphorous reverts back to original.

Conclusion

The two step curved arrow mechanism for the interconversion of $3-\mathrm{p}\mathrm{h}\mathrm{o}\mathrm{s}\mathrm{p}\mathrm{h}\mathrm{o}\mathrm{g}\mathrm{l}\mathrm{y}\mathrm{c}\mathrm{e}\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{e}$ to $2-\mathrm{p}\mathrm{h}\mathrm{o}\mathrm{s}\mathrm{p}\mathrm{h}\mathrm{o}\mathrm{g}\mathrm{l}\mathrm{y}\mathrm{c}\mathrm{e}\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{e}$ is drawn.

Interpretation Introduction

(b)

Interpretation:

If the isotopically chiral $3-\mathrm{p}\mathrm{h}\mathrm{o}\mathrm{s}\mathrm{p}\mathrm{h}\mathrm{o}\mathrm{g}\mathrm{l}\mathrm{y}\mathrm{c}\mathrm{e}\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{e}$ formed would have the same configuration, the opposite configuration, or equal amounts of two configurations at phosphorus when an enantiomerically pure $2-\mathrm{p}\mathrm{h}\mathrm{o}\mathrm{s}\mathrm{p}\mathrm{h}\mathrm{o}\mathrm{g}\mathrm{l}\mathrm{y}\mathrm{c}\mathrm{e}\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{e}$ derivative containing isoptopically chiral phosphate group is converted to $3-\mathrm{p}\mathrm{h}\mathrm{o}\mathrm{s}\mathrm{p}\mathrm{h}\mathrm{o}\mathrm{g}\mathrm{l}\mathrm{y}\mathrm{c}\mathrm{e}\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{e}$ is to be determined.

Concept introduction:

The enzyme catalyzed substitution reaction of phosphate ester derivate at phosphate proceeds through $S_{N}2$ reaction with inversion of stereochemistry. If substitution occurs twice that is with double inversion results in retention of stereochemistry.

Answer to Problem 25.35AP

If enantiomerically pure $2-$ phosphoglycerate is subjected to this reaction, the isotopically chiral $3-$ phosphoglycerate formed would have the same configuration at phosphorous.

Explanation of Solution

The enzyme catalyzed substitution reaction of phosphate ester derivate at phosphate proceeds through inversion of stereochemistry. The substitution occurs twice with double inversion of the product formed with retention of configuration. Thus, if enantiomerically pure $2-\mathrm{p}\mathrm{h}\mathrm{o}\mathrm{s}\mathrm{p}\mathrm{h}\mathrm{o}\mathrm{g}\mathrm{l}\mathrm{y}\mathrm{c}\mathrm{e}\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{e}$ is subjected to this reaction, the isotopically chiral $3-\mathrm{p}\mathrm{h}\mathrm{o}\mathrm{s}\mathrm{p}\mathrm{h}\mathrm{o}\mathrm{g}\mathrm{l}\mathrm{y}\mathrm{c}\mathrm{e}\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{e}$ formed will have the same configuration at phosphorous.

Conclusion

The interconversion of $2-\mathrm{p}\mathrm{h}\mathrm{o}\mathrm{s}\mathrm{p}\mathrm{h}\mathrm{o}\mathrm{g}\mathrm{l}\mathrm{y}\mathrm{c}\mathrm{e}\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{e}$ to $3-\mathrm{p}\mathrm{h}\mathrm{o}\mathrm{s}\mathrm{p}\mathrm{h}\mathrm{o}\mathrm{g}\mathrm{l}\mathrm{y}\mathrm{c}\mathrm{e}\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{e}$ by enzyme catalysed substitution occurs with retention of stereochemistry at phosphorus.