The reduction of the ketone shown below with sodium borohydnde results in the fo

Question

The reduction of the ketone shown below with sodium borohydnde results in the following product distribution. Explain why the first product is formed in such a larger amount than the second. Hint: remember that the carbon of the carbonyl group is sp2 hybridized and planar and that the reduction involves a hydride transfer from NaBH« to the carbon of the carbonyl group. In the above reaction a single starling material can form two stereoisomers, but yields one of them preferentially. Any reaction in which this occurs is defined as .

Explanation / Answer

(a) It's very simple. Reason is: Steric Hindrance

Here a Hydride (H-) from NaBH4 attacks the ketone group staying in the cyclohexyl plane

Now, the top side of the cyclohexyl plane contains two bulky -CH3 groups. So the approach of hydride from this face is resisted since that would cause steric congestion in the system. While, the bottom side of the cyclohexyl plane is free of any such steric hindrance, so hydride can easily approach from this side. That's why the isomer with H at the bottom is produced much more preferentially.

(b) Such reactions are called stereoselective reactions.

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