Considering LiBH4 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 [figure]
- P & Q are identical, and R & S are diastereomers.
- P & Q are diastereomers, and R & S are identical.
- P & Q are diastereomers, and R & S are diastereomers.
- P & Q are identical, and R & S are identical.
Answer
(3) P & Q are diastereomers, and R & S are diastereomers.
Solution
LiBH4 reduces only the ester (–CH₂CO₂Et → –CH₂CH₂OH); BH3 reduces only the acid (–CH₂CO₂H → –CH₂CH₂OH). On H+ treatment the new –CH₂CH₂OH closes a six-membered δ-lactone with the other arm, giving a spiro lactone.
P: [figure] LiBH4 [figure] [-H2O] H+ [figure]
The wedged (ester) arm becomes –CH₂CH₂OH, so the ring O of the lactone sits on the wedged side.
Q: [figure] BH3 [figure] [-EtOH] H+ [figure]
The hashed (acid) arm becomes –CH₂CH₂OH, so the ring O sits on the hashed side.
P and Q have the same configuration at the CH₃-bearing carbon but opposite configuration at the spiro carbon, so P & Q are diastereomers.
R: [figure] LiBH4 [figure] [-H2O] H+ [figure]
S: [figure] BH3 [figure] [-EtOH] H+ [figure]
In R the lactone O is cis to the 3-CH₃, in S it is trans: R & S are cis–trans (geometrical) isomers, i.e. diastereomers.
Answer: (C)