$\ce{LiAlH4}$ 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?


1 Answer 1


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).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.