Question 4

$LiBH_4$ selectively reduces the ester group ($-CO_2Et$) to a primary alcohol ($-CH_2OH$), leaving the $-COOH$ group intact. Conversely, $BH_3$ selectively reduces the carboxylic acid group ($-COOH$) to a primary alcohol ($-CH_2OH$), leaving the ester intact.

In the first reaction sequence:

• In product P, the $-CO_2Et$ (trans to methyl) is converted to $-CH_2OH$. So, the $-CH_2OH$ is trans to methyl.
• In product Q, the $-COOH$ (cis to methyl) is converted to $-CH_2OH$. So, the $-CH_2OH$ is cis to methyl. Since the relative stereochemical arrangement of the resulting alcohol group and the methyl group differs, P and Q are diastereomers (assuming the workup leads to comparable functional groups or simply looking at the spatial configuration).

In the second reaction sequence:

• In product R, the $-CO_2Et$ (cis to methyl) is converted to $-CH_2OH$. So, the $-CH_2OH$ is cis to methyl.
• In product S, the $-COOH$ (trans to methyl) is converted to $-CH_2OH$. So, the $-CH_2OH$ is trans to methyl. Similarly, R and S are diastereomers because their relative spatial arrangements differ. Therefore, both pairs P & Q and R & S are diastereomers.