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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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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 $\beta$-carbon or at the carbonyle group's carbon. According to the HSAB Principle the $\beta$-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 $\beta$ carbon. After this reaction the $\ce{C=C}$ bond is gone. But, if what is left behind is still a very reactive simple aldehyde that gets reduced to the alcohol by $\ce{NaBH4}$ in a second step. $\ce{LiAlH_4}$ on the other hand is a rather hard nucleophile and thus reacts with an enal at the carbonyle carbon. After this reaction there is just an isolated double bond left that can't be reduced by $\ce{LiAlH_4}$ in a second step.

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    $\begingroup$ Cinnamaldehyde is reduced to Hydrocinnamyl alcohol by $\ce{LiAlH4}$. Both the double bond and carbonyl group are reduced $\endgroup$ – yasir Nov 26 '15 at 11:48
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    $\begingroup$ 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. $\endgroup$ – orthocresol Dec 10 '17 at 23:01
  • $\begingroup$ Isn't HSAB a lie? onlinelibrary.wiley.com/doi/full/10.1002/anie.201007100 $\endgroup$ – PCK Jul 18 at 7:26

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