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Is $\ce{NaBH4}$ a suitable reducing agent for the conversion of 2-cyclohexenone to cyclohexanol?

In Clayden (page 506, second edition), I have found the reduction of cyclopentenone to cyclopentanol using $\ce{NaBH4}$ in $\ce{Et-OH}$. (99% yield!). First NaBH4 performs 1,4 addition and then 1,2 addition to give the product.

However, I am unable to find a reference for reduction of cyclohexenone to cyclohexanol.

I found this in Smith: enter image description here

I don't understand why it can't reduce the double bond through Michael addition in case of cyclohex-2-enone as well. What is the reason behind it?

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  • $\begingroup$ Whether NaBH4 reduces the C=C double bond seems to be substrate dependent (and more often than not, it actually doesn't reduce it). And I am not sure if there is any pattern to it. $\endgroup$ – orthocresol Jul 14 '18 at 22:10
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According to Johnson, M. R.; Rickborn B. Sodium borohydride reduction of conjugated aldehydes and ketones J. Org. Chem., 1970, 35, 1041. DOI: 10.1021/jo00829a039:

2-Cyclohexenone can certainly be reduced fully. It has very similar reactivity to cyclopentenone. In this report, mixtures of products were observed. The experimental indicates the researchers used 0.5 eq of $\ce{NaBH4}$ per carbonyl.

enter image description here

Furthermore, the use of Luche conditions can again help you favor only 1,2 reduction:

enter image description here

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