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.

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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.

  • $\begingroup$ In your question, do you mean reduce down to the allylic alcohol or the saturated one? $\endgroup$
    – Beerhunter
    Commented Nov 3, 2015 at 19:04
  • $\begingroup$ @Beerhunter - the allylic alcohol. $\endgroup$
    – Dissenter
    Commented Nov 3, 2015 at 22:35
  • $\begingroup$ you are considering the chalcones to break down into two parts making allylic alcohol as one product or something else? enlighten me $\endgroup$
    – slhulk
    Commented Nov 12, 2015 at 17:53
  • $\begingroup$ I am considering the "complete" reduction of the chalcone. In other words why isn't the product an allyl alcohol with no double bond? $\endgroup$
    – Dissenter
    Commented Nov 12, 2015 at 17:55
  • 1
    $\begingroup$ It is the vacation now and I do not have access to my university library - however, March's Organic Chemistry 6th ed has a table listing the same reactivity order you have given above towards catalytic hydrogenation (except the aldehyde and alkene are swapped). The reference given is House, Modern Synthetic Reactions, 2nd ed - so there might be info in there - and if nobody gets around to doing it, I will try to do some research when I am back in the UK. Hudlicky, M., J. Chem. Educ., 1977, 54, 100 also contains many literature links. $\endgroup$ Commented Jan 2, 2016 at 9:34

2 Answers 2


You hit it right on the nose. The real key piece of information is that given enough time, all the unsaturated bonds will be reduced. This tells you that though the reduction is thermodynamically favorable, it is the difference in the energy barriers ($\ce{\Delta \mathrm{G^{‡}}}$) that prevents the carbonyl reduction from occurring at the same rate as the alkene reduction. This means to produce the alcohol faster, we must manipulate the kinetics of the reaction.

To understand why the carbonyl-reduction transition state is higher in energy, we should consider the differences between the carbonyl and alkene bonds. The bonds are much more polarized, and as such the carbonyl $\pi$ bond is considerably stronger than alkene ($93\ \mathrm{kcal\ mol^{-1}}$ vs. $63\ \mathrm{kcal\ mol^{-1}}$)$^{\mathrm{[1]}}$. This means that at low temperatures, only a small fraction of the molecules have enough energy for the carbonyls to associate onto the $\ce{Pd/C}$ surface. Catalytic hydrogenation of carbonyl compounds (aldehydes, ketones, and especially esters) requires high temperatures and pressures to increase the presence of both the substrate and hydrogen on the catalyst.

$^{\mathrm{[1]}}$ Fox, M. A.; Whitesell, J. K. Organic chemistry; Jones and Bartlett: Sudbury, MA, 1997.

  • $\begingroup$ Implication being, adsorption requires the breaking of the pi bond? $\endgroup$ Commented Oct 1, 2016 at 19:16
  • 1
    $\begingroup$ @orthocresol I believe so. At any rate, the bond must be broken at some point in the reaction. $\endgroup$
    – ringo
    Commented Oct 1, 2016 at 19:19

This issue is discussed in the following references:

Hydrogenation of chalcones using hydrogen permeating through a Pd and palladized Pd electrodes Electrochimica Acta vol. 55, pages 5831–5839.


On the role of promoters in hydrogenations on metals; α,β-unsaturated aldehydes and ketones Applied Catalysis A: General 149 (1997) 27-48

the latter of which says:

Palladium is a very good catalyst for the C=C or $\ce{C#C}$ bond hydrogenations, but a very bad catalyst for hydrogenation of carbonyl groups. It is known that this is due to a too weak adsorption of carbonyls under reaction conditions [reference 28]. The mentioned weakness of adsorption through the carbonyl group could be in its turn caused by the change in the electronic structure of the Pd surface atoms ($4d^{9.7}$ $5s^{0.3}$ $\ce{->}$ $4d^{10}$ $5s^0$) induced by the hydrogen atoms in the interstitial positions [reference 29].

Reference 28 is: Concentration Dependence of Ketone Hydrogenation Catalyzed by Ru, Pd, and Pt. Evidence for Weak Ketone Adsorption on Pd Surface Bull. Chem. Soc. Japan, 55 (1982) 2275.

Reference 29 is: V. Ponec and G.C. Bond, Catalysis by Metals and Alloys, Series: Studies in Surface Science and Catalysis, Vol. 95, Elsevier, Ansterdam, 1995. (google books link)


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