I have been going through reduction of aldehydes using $\ce{LiAlH4}$ and $\ce{NaBH4}$. If there is a double bond conjugated with the carbonyl group, $\ce{LiAlH4}$ doesn't reduce it, leading to an allylic alcohol. However, using $\ce{NaBH4}$, some of the fully reduced alcohol will also be formed. Why is this so?
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asked Nov 18, 2013 at 13:01
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$\begingroup$ Better look at this: chemistry.stackexchange.com/questions/87138/… $\endgroup$– user55119Commented Aug 21, 2020 at 14:35
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1 Answer
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Neither $\ce{LiAlH_4}$ nor $\ce{NaBH4}$ are able to reduce an isolated $\ce{C=C}$ bond. But if you have an enal (a conjugated aldehyde) it can react (as an electrophile) either at the β-carbon or at the carbonyle group's carbon. According to the HSAB Principle the β-carbon is a “soft” center and would react preferably with “soft” nucleophiles while the carbonyle carbon is a “hard” center and prefers to react with “hard” nucleophiles.
Now, $\ce{NaBH4}$ is a rather soft nucleophile and thus it reacts with an enal at the β-carbon. After this reaction the $\ce{C=C}$ bond is gone. But what is left behind is still a very reactive simple aldehyde that gets reduced to the alcohol by $\ce{NaBH4}$ in a second step.
$\ce{LiAlH4}$ on the other hand is a rather hard nucleophile and thus reacts with an enal at the carbonyl carbon. After this reaction there is just an isolated double bond left that can't be reduced by $\ce{LiAlH4}$ in a second step.
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5$\begingroup$ Cinnamaldehyde is reduced to Hydrocinnamyl alcohol by $\ce{LiAlH4}$. Both the double bond and carbonyl group are reduced $\endgroup$ Commented Nov 26, 2015 at 11:48
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2$\begingroup$ For cinnamaldehyde, both reductants first effect direct reduction of the carbonyl group; however, LiAlH4 is capable of further reducing the double bond. This has been recently explained here. $\endgroup$ Commented Dec 10, 2017 at 23:01
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2$\begingroup$ Isn't HSAB a lie? onlinelibrary.wiley.com/doi/full/10.1002/anie.201007100 $\endgroup$– PCKCommented Jul 18, 2019 at 7:26
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$\begingroup$ Reduction by LAH: What would be the product in case the double bond is conjugated to both carbonyl group and benzene ring? $\endgroup$– ApurviumCommented Jun 10, 2020 at 6:59
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$\begingroup$ @orthocresol Can I conclude that NaBH4 can reduce the conjugated π-bonds? $\endgroup$ Commented Aug 20, 2022 at 6:55