Animated Solution for Chemistry - Organic Chemistry: Considering ${\text{LiBH}}_4$ reduces an ester group to the corresponding alcohol and does not reduce a carboxylic acid group, the correct statement about the major products P, Q R and S is

Imagine you are an architect tasked with building a complex molecular structure. You have two different tools in your belt, and each tool only works on a specific type of building block. This problem is a beautiful demonstration of how chemoselectivity and 3D stereochemistry intertwine to create distinct molecular architectures.

Understanding the Reactants

We are given two starting materials, both of which are 1,3-disubstituted cyclobutanes.

In Reactant 1, the methyl group at C1 is on a wedge (pointing towards you). At C3, we have a carboxylic acid ($-{\text{CO}}_2\text{H}$) on a wedge and an ester ($-{\text{CO}}_2\text{Et}$) on a dash (pointing away from you).

In Reactant 2, the methyl group at C1 is still on a wedge. However, the positions at C3 are swapped: the ester is now on the wedge, and the carboxylic acid is on the dash.

Here is where the magic happens. We use two highly specific reducing agents:

1. Lithium Borohydride (${\text{LiBH}}_4$): This reagent is a specialist. It will reduce the ester group to a primary alcohol ($-{\text{CH}}_2\text{OH}$), but it completely ignores the carboxylic acid. 2. Borane (${\text{BH}}_3$): This reagent is the exact opposite. It selectively reduces the carboxylic acid to a primary alcohol, leaving the ester completely untouched.

Once the reduction occurs, the newly formed alcohol and the remaining acid/ester are perfectly positioned on the same carbon (C3) to undergo an intramolecular reaction, forming a stable 5-membered lactone ring. Because they share a single carbon with the cyclobutane ring, the resulting structure is a spiro compound.

Let's trace the stereochemistry for each product carefully.

Product P: We treat Reactant 1 with ${\text{LiBH}}_4$. The dashed ester is reduced to a dashed $-{\text{CH}}_2\text{OH}$. When it cyclizes with the wedged acid, the resulting lactone has its $-{\text{CH}}_2-$ group on the dash and its carbonyl ($-\text{C=O}$) on the wedge.

Product Q: We treat Reactant 1 with ${\text{BH}}_3$. The wedged acid is reduced to a wedged $-{\text{CH}}_2\text{OH}$. Upon cyclization with the dashed ester, the lactone forms with its $-{\text{CH}}_2-$ group on the wedge and its carbonyl on the dash.

Because P and Q have the same configuration at C1 but opposite configurations at C3, they are diastereomers.

Product R: We treat Reactant 2 with ${\text{LiBH}}_4$. The wedged ester is reduced to a wedged $-{\text{CH}}_2\text{OH}$. It cyclizes with the dashed acid. The resulting lactone has its $-{\text{CH}}_2-$ group on the wedge and its carbonyl on the dash. If you look closely, this is the exact same 3D structure as Product Q!

Product S: We treat Reactant 2 with ${\text{BH}}_3$. The dashed acid is reduced to a dashed $-{\text{CH}}_2\text{OH}$. It cyclizes with the wedged ester. The resulting lactone has its $-{\text{CH}}_2-$ group on the dash and its carbonyl on the wedge. This is structurally identical to Product P!

By carefully tracking the 3D orientation of the bonds during the chemoselective reduction, we have uncovered the relationships:

P and Q are diastereomers. R and S are diastereomers. P and S are identical molecules. Q and R are identical molecules.

Therefore, the correct statement is that P & Q are diastereomers, and R & S are diastereomers.