# Why does LiAlH4 reduce the double bond of cinnamaldehyde, whereas it does not reduce the double bond in crotonaldehyde?

$$\ce{LiAlH4}$$ is a strong reducing agent capable of breaking the double bonds. It does break the double bonds in cinnamaldehyde.

Shouldn't $$\ce{LiAlH4}$$ be able to break the double bond in crotonaldehyde as well? Why does the double bond in crotonaldehyde remain unaffected?

• Jul 28, 2021 at 3:03

## 1 Answer

α,β-Unsaturated carbonyl compounds, for the most stable intermediate with respect to $$\ce{H-}$$ ion prefer 1,4-addition. However, $$\ce{LiAlH4}$$ prefers 1,2-addition over 1,4-addition. This can be explained due to the following reasons.

• It is more reactive so it prefers a kinetically favorable pathway (1,2-addition is often kinetically favored while 1,4-addition is thermodynamically favorable).

• It is also plausibly explained by HSAB principle which also explains why $$\ce{NaBH4}$$ prefers 1,4-addition for α,β-unsaturated carbonyls.

Based on the above facts we should see that it is natural to assume that 1,2-addition is obeyed for both crotonaldehyde and cinnamaldehyde. However experiments contradict and instead show that for cinnamaldehyde 1,4-addition is more preferable even when $$\ce{LiAlH4}$$ is acting as the $$\ce{H-}$$ source. This can be easily explained by more stability of the carbanion by two-fold resonance, both by phenyl group and by the allylic (sort of) carbonyl group.

It would seem that stabilizing the β-carbanion can allow 1,4-addition to take place. The extent of stability required, however can only be predicted by experiments. It seems like phenyl group is sufficient to warrant 1,4-addition like it does for styrene (as it should since it distributes charge over larger set of nuclei).