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Chapter 4: Problem 2

The \((3 S, 4 S)\)-isomer of deuterated compound \(\mathbf{1}\) is a substrate for a dephosphorylase that gives an antielimination to the \(Z\)-isomer (2). The corresponding (3R,4S)-isomer is not a substrate. How do you explain that?

Short Answer

Expert verified
The difference in reactivity of the two isomers towards the dephosphorylase enzyme is due to their different spatial orientations (3D structures). The enzyme can recognize and bind with the (3 S, 4 S)-isomer due to its specific orientation, ensnaring it in its active site and facilitating the enzymatic process. The (3R,4S)-isomer, on the other hand, does not fit into the active site owing to its different 3D structure.

Step by step solution

01

Understanding the given isomers

The (3 S, 4 S) and the (3R,4S) are stereoisomers. This means they have the same molecular formula and sequence of bonded atoms, but a different spatial orientation of atoms. The (3 S, 4 S)-isomer is a substrate ripe for anti-elimination by a dephosphorylase to yield a Z-isomer (2). The (3R,4S)-isomer, however, doesn't serve the same role.
02

Analyzing the role of enzyme specificity

Enzymes, like the dephosphorylase in this case, are highly specific. Their activity depends on the ability to recognize a certain 3D structure. The spatial orientation of the (3 S, 4 S)-isomer allows it to fit into the active site of the enzyme and undergo the enzymatic process.
03

Explaining non-reactivity of the (3R,4S)-isomer

The (3R,4S)-isomer does not react because its 3D structure does not allow it to fit into the active site of the dephosphorylase. This is due to the specific orientation of atoms required by the enzyme to bind and react. Since the (3R,4S)-isomer does not possess this specific orientation, it remains virtually untouched by the enzyme.

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