Question

Considering 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

• A. \(P\ \&\ Q\) are identical, and \(R\ \&\ S\) are diastereomers.
• B. \(P\ \&\ Q\) are diastereomers, and \(R\ \&\ S\) are identical.
• C. \(P\ \&\ Q\) are diastereomers, and \(R\ \&\ S\) are diastereomers.
• D. \(P\ \&\ Q\) are identical, and \(R\ \&\ S\) are identical.

Hint:

Remember the important chemoselectivity rules: \[ \mathrm{LiBH_4} \] reduces: \[ \text{Esters} \rightarrow \text{Alcohols} \] but generally does not reduce: \[ \text{Carboxylic acids} \] Whereas: \[ \mathrm{BH_3} \] selectively reduces: \[ \text{Carboxylic acids} \rightarrow \text{Alcohols} \] In stereochemistry problems, always compare:
• relative orientation of substituents
• wedge/dash configurations
• possibility of superimposition before deciding whether products are identical, enantiomers, or diastereomers.

Answer 1

The Correct Option is A

Concept: This problem is based on:
• Chemoselectivity of reducing agents
• Stereochemistry of substituted cyclobutane systems
• Relative spatial arrangement of substituents
• Comparison between products obtained from selective reductions The key chemical facts given in the question are:
• \( \mathrm{LiBH_4} \) reduces esters into alcohols
• \( \mathrm{LiBH_4} \) does not reduce carboxylic acids
• \( \mathrm{BH_3} \) selectively reduces carboxylic acids Therefore:
• Ester group \((\mathrm{CO_2Et})\) becomes alcohol with \( \mathrm{LiBH_4} \)
• Carboxylic acid group \((\mathrm{CO_2H})\) remains unchanged with \( \mathrm{LiBH_4} \)
• Carboxylic acid group becomes alcohol with \( \mathrm{BH_3} \)
• Ester group remains unchanged with \( \mathrm{BH_3} \) Thus, the problem mainly tests whether the products obtained after selective reduction become identical compounds or stereoisomers.

Step 1: Analyzing product \(P\). In the first substrate:
• The ester group \((\mathrm{CO_2Et})\) is shown as a solid wedge
• The carboxylic acid group \((\mathrm{CO_2H})\) is shown as a dashed bond Reaction conditions: \[ 1.\ \mathrm{LiBH_4} \qquad 2.\ \mathrm{H^+} \] Since \( \mathrm{LiBH_4} \) reduces only the ester group, \[ \mathrm{CO_2Et} \longrightarrow \mathrm{CH_2OH} \] while the carboxylic acid remains unchanged. Therefore, product \(P\) contains:
• one alcohol group replacing the ester
• one unchanged carboxylic acid group Importantly, the stereochemistry at the carbon center remains preserved during reduction because no bond to the stereogenic center is broken. Hence the relative stereochemical arrangement remains the same.

Step 2: Analyzing product \(Q\). Now the same substrate is treated with: \[ 1.\ \mathrm{BH_3} \qquad 2.\ \mathrm{H^+} \] We know: \[ \mathrm{BH_3} \] selectively reduces the carboxylic acid group. Thus, \[ \mathrm{CO_2H} \longrightarrow \mathrm{CH_2OH} \] while the ester group remains unchanged. After reduction, the molecule again contains:
• one alcohol substituent
• one ester substituent Now observe carefully: Originally the ester and acid groups occupied opposite stereochemical orientations (one wedge and one dash). After selective reduction, the identities of the groups interchange, but the final arrangement becomes superimposable on product \(P\). Hence both products represent the same compound. Therefore, \[ P \equiv Q \] Thus, \(P\) and \(Q\) are identical.

Step 3: Analyzing product \(R\). Now consider the second substrate. In this structure, the orientation of the methyl group and substituents differs from the first case. Again treatment with: \[ 1.\ \mathrm{LiBH_4} \qquad 2.\ \mathrm{H^+} \] reduces only the ester group. Thus: \[ \mathrm{CO_2Et} \longrightarrow \mathrm{CH_2OH} \] and the acid remains unchanged. The stereochemical arrangement present initially remains preserved.

Step 4: Analyzing product \(S\). Now treatment with: \[ 1.\ \mathrm{BH_3} \qquad 2.\ \mathrm{H^+} \] reduces only the acid group: \[ \mathrm{CO_2H} \longrightarrow \mathrm{CH_2OH} \] while the ester remains unchanged. After reduction, comparison of the three-dimensional arrangements shows that the relative configurations of substituents differ from product \(R\). They are not mirror images, but they are stereoisomers differing at one stereochemical relationship. Hence they are diastereomers.

Step 5: Final comparison of products. From the stereochemical analysis: \[ P \text{ and } Q \text{ are identical} \] while \[ R \text{ and } S \text{ are diastereomers} \] Therefore, the correct statement is: \[ \boxed{P\ \&\ Q\ \text{are identical, and}\ R\ \&\ S\ \text{are diastereomers}} \] Final Answer: \[ \boxed{\text{(A)}} \]