Why is it that the major product of the reduction of chalcones the ketone and not the monoalcohol? In other words, Why isn't the major product a benzyl alcohol?
From what I understand, catalytic hydrogenation can be used to reduce carbonyls as well as alkenes.
My TA told me that nucleophilic hydrides are preferred for reducing carbonyls. Why wouldn't hydrogenation work as well? Does it have to do with resonance involving the carbonyl since in a chalcone, the carbonyl is adjacent to an aromatic ring and is also conjugated with the alkene? Could the reason that catalytic hydrogenation can't effectively touch the α,β-unsaturated carbonyl be the same reason catalytic hydrogenation can't effectively reduce carboxylic acids, esters, and amides — all of which are also resonance stabilized?
My first thought had to do with heats of hydrogenation and how resonance-stabilization found in carboxylic acids and its derivatives reduce the heat of hydrogenation. However, later I found a resource online implying that all pi bonds - even the delocalized, resonance-stabilized ones found in benzene — could be reduced through catalytic hydrogenation with enough time.
In addition, it was noted elsewhere that chalcones could be completely reduced to the benzylic alcohol, although not with ease.
This leads me to believe that there is an activation energy barrier that's impeding the hydrogenation of certain substrates within the confines of a 3 hour undergraduate lab period … am I on the right path? Sterics, perhaps? It was noted that the more highly substituted an alkene is, the slower it is reduced because of the difficulty in getting a highly substituted alkene to approach the catalyst's surface with the $\ce{M-H}$ bonds in an appropriate manner.
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