Selective Reduction of Ester and Carboxylic Acid Groups

To determine the relationship between the major products $\mathbf{P}$, $\mathbf{Q}$, $\mathbf{R}$, and $\mathbf{S}$, we analyze the chemoselectivity of the reducing agents and the resulting stereochemistry of the products.

1. Chemoselectivity of the Reagents

1. Lithium Borohydride ($\text{LiBH}_4$):

   • Chemoselectively reduces the ester group ($-\text{CO}_2\text{Et}$) to a primary alcohol group ($-\text{CH}_2\text{OH}$).
   • It leaves the carboxylic acid group ($-\text{COOH}$) unreacted under standard conditions: $-\text{CO}_2\text{Et} \xrightarrow{\text{1. }\text{LiBH}_4,\; \text{2. }\text{H}^+} -\text{CH}_2\text{OH}$

2. Borane ($\text{BH}_3$):

   • Chemoselectively reduces the carboxylic acid group ($-\text{COOH}$) to a primary alcohol group ($-\text{CH}_2\text{OH}$).
   • It reacts much faster with carboxylic acids than with esters, leaving the ester group ($-\text{CO}_2\text{Et}$) unreacted: $-\text{COOH} \xrightarrow{\text{1. }\text{BH}_3,\; \text{2. }\text{H}^+} -\text{CH}_2\text{OH}$

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2. Analysis of Products $\mathbf{P}$ and $\mathbf{Q}$

In the first starting material:

• The methyl group ($-\text{CH}_3$) is present on a wedge.

• At the substituted carbon position, the ester group ($-\text{CO}_2\text{Et}$) is on a wedge, and the carboxylic acid group ($-\text{COOH}$) is on a dash.

• Formation of $\mathbf{P}$ ($\text{LiBH}_4$):

  • The ester group ($-\text{CO}_2\text{Et}$) on the wedge is reduced to $-\text{CH}_2\text{OH}$ (wedge).
  • The acid group ($-\text{COOH}$) on the dash remains unchanged.
  • Thus, in $\mathbf{P}$, the $-\text{CH}_2\text{OH}$ group is cis to the $-\text{CH}_3$ group.

• Formation of $\mathbf{Q}$ ($\text{BH}_3$):

  • The carboxylic acid group ($-\text{COOH}$) on the dash is reduced to $-\text{CH}_2\text{OH}$ (dash).
  • The ester group ($-\text{CO}_2\text{Et}$) on the wedge remains unchanged.
  • Thus, in $\mathbf{Q}$, the $-\text{CH}_2\text{OH}$ group is trans to the $-\text{CH}_3$ group.

Since $\mathbf{P}$ and $\mathbf{Q}$ are non-superimposable stereoisomers that are not mirror images of each other (due to differing cis/trans spatial arrangements of functional groups relative to the fixed methyl group), $\mathbf{P}$ and $\mathbf{Q}$ are diastereomers.

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3. Analysis of Products $\mathbf{R}$ and $\mathbf{S}$

In the second starting material:

• The methyl group ($-\text{CH}_3$) is again on a wedge.

• The ester group ($-\text{CO}_2\text{Et}$) is on a wedge, and the carboxylic acid group ($-\text{COOH}$) is on a dash.

• Formation of $\mathbf{R}$ ($\text{LiBH}_4$):

  • The ester group ($-\text{CO}_2\text{Et}$) on the wedge is reduced to $-\text{CH}_2\text{OH}$ (wedge).
  • In product $\mathbf{R}$, the $-\text{CH}_2\text{OH}$ group is cis to the $-\text{CH}_3$ group.

• Formation of $\mathbf{S}$ ($\text{BH}_3$):

  • The carboxylic acid group ($-\text{COOH}$) on the dash is reduced to $-\text{CH}_2\text{OH}$ (dash).
  • In product $\mathbf{S}$, the $-\text{CH}_2\text{OH}$ group is trans to the $-\text{CH}_3$ group.

Similarly, $\mathbf{R}$ and $\mathbf{S}$ differ in their relative configuration (cis vs. trans relationship), making $\mathbf{R}$ and $\mathbf{S}$ diastereomers.

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Conclusion

• $\mathbf{P}$ and $\mathbf{Q}$ are diastereomers.
• $\mathbf{R}$ and $\mathbf{S}$ are diastereomers.

Hence, the correct option is C.