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Chapter 4: Problem 3

Compound 3 is a substrate for an enzyme that catalyzes the removal of the \(\mathrm{H}_{\mathrm{S}}\) proton exclusively. If compound 4 is used as the substrate, the deprotonation occurs mostly at \(\mathrm{H}_{\mathrm{S}}\), but a small amount of products are observed from removal of \(\mathrm{H}_{\mathrm{R}}\). This second reaction was shown to be catalyzed by the same enzyme as well. How could this be rationalized?

Short Answer

Expert verified
The same enzyme catalyzes the removal of both \( \mathrm{H}_{\mathrm{S}} \) and \( \mathrm{H}_{\mathrm{R}} \) in compound 4 because it has slight binding affinity for \( \mathrm{H}_{\mathrm{R}} \). The structure or properties of compound 4 may allow enzyme to reach \( \mathrm{H}_{\mathrm{R}} \). Flexibility or spacious active site of the enzyme could accommodate for multiple reactions.

Step by step solution

01

Understanding Enzyme Specificity

Enzymes are known for their specificity, generally speaking, they will only bind to certain substrates and catalyze certain reactions. This specificity is based on the shape and structure of the enzyme's active site, where the reaction takes place.
02

Examine Compound 3 Reaction

In the case of Compound 3, the enzyme cleaves the \(\mathrm{H}_{\mathrm{S}}\) proton due its binding affinity and specificity. The enzyme is designed in such a way that it recognizes and reacts with the \(\mathrm{H}_{\mathrm{S}}\) site, allowing for the hydrogen to be removed.
03

Examine Compound 4 Reaction

In the case of Compound 4, the same enzyme mostly reacts with the \(\mathrm{H}_{\mathrm{S}}\) proton. However, some \(\mathrm{H}_{\mathrm{R}}\) protons are also removed. This suggests that while the enzyme has a high affinity for the \(\mathrm{H}_{\mathrm{S}}\) proton, it is also able to weakly bind and remove the \(\mathrm{H}_{\mathrm{R}}\) proton.
04

Rationalization of the Reactions

The enzyme targeted both \(\mathrm{H}_{\mathrm{S}}\) and \(\mathrm{H}_{\mathrm{R}}\) in Compound 4 because it has a mild binding affinity for \( \mathrm{H}_{\mathrm{R}} \). It could be that the structure or chemical properties of Compound 4 allowed for the enzyme to reach and interact with \( \mathrm{H}_{\mathrm{R}} \). It's also possible that the enzyme has a certain flexibility or spacious active site that could accommodate for multiple simultaneous reactions.

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