I read an answer here about why $\ce{LiAlH_4}$ does not reduce conjugated double bonds, and that $\ce{NaBH_4}$ does. But in the case of cinnamaldehyde, $\ce{LiAlH_4}$ reduces the conjugated double bond and $\ce{NaBH_4}$ does not. Why is it so?
Citation for $\ce{LiAlH_4}$: Quora
Citation for $\ce{NaBH_4}$:
Cinnamaldehyde (1) does not undergo 1,4-reduction but rather 1,2-reduction with LiAlH4. The product of the reduction is 3-phenylpropan-1-ol (2). If conjugate addition were to occur, enolate 3 would be produced (stable to the reductant; enolates can be prepared to protect ketones from this reductant) and aldehyde 4 would be expected, as long as the aqueous workup were to destroy the reductant before any enolate 3 could be protonated.
All of this aside, the answer was provided experimentally in 1954.1 Reduction occurs 1,2 to afford the aluminate salt of cinnamyl alcohol 5. This species undergoes hydroalumination of the double bond by a presumed intramolecular process to give alkyl aluminate 6. This cyclic aluminate was deuterated (D2O) at the benzylic position to provide 9. In addition, trapping of 6 with benzophenone afforded diol 7 in >80% yield! Reduction of cinnamaldehyde with LiAlD4 and aqueous workup would be expected to give dideuterio 3-phenylpropan-1-ol (8).
Sodium borohydride reduces cinnamaldehyde via 1,2-reduction in hydroxylic solvent to cinnamyl alcohol in 97% yield.2 The intermediate alkoxyborane (Cf.; structure 5) either is incapable of reacting with the double as the aluminate does or the alkoxyborane rapidly exchanges with solvent.
1) Dornow, A., Winter, G., Vissering, W. Chemische Berichte, 1954, 87, 629.
2) Chaikin, S.W., Brown, W.G., J. Am. Chem. Soc., 1949, 71, 122.
