$\ce{LiAlH4}$ 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?


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).


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