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Topic: NaBH4 reduction  (Read 1190 times)

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Offline Jstub

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NaBH4 reduction
« on: November 12, 2019, 09:52:44 PM »
So the actual question I have to answer asks me to show the reaction of progesterone with excess sodium borohydride in methanol.
I know this reagent will reduce ketones to alcohols, but in this reaction, will both of the ketones be reduced? Progesterone has a ketone on either end, but will the both react with NaBH4 and methanol?
Also, if it only reacts with one of them, how do I know which one?
« Last Edit: November 12, 2019, 10:06:54 PM by Jstub »

Offline clarkstill

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Re: NaBH4 reduction
« Reply #1 on: November 13, 2019, 04:29:15 AM »
I think the keyword in the question is "excess" - under these conditions you'd expect both ketones to be reduced to the corresponding alcohols.

However, if you're really interested in the order of reactivity, generally speaking ketones are more reactive towards nucleophiles than enones (since the C=C pi-electrons partially donate into the C=O antibonding orbital, making the carbonyl less reactive). So, it is likely that the ketone will be reduced first. This has been studied here: https://pubs.rsc.org/en/content/articlepdf/1955/jr/jr9550003426.

If you wanted to favour reduction of the enone carbonyl, you would use the Luche Reduction (https://en.wikipedia.org/wiki/Luche_reduction), where addition of Lewis acidic CeCl3 enhances its reactivity relative to the ketone.

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