Why does the reduction of naphthalene using $\ce{H2/Ni}$ lead to trans-decalin instead of cis-decalin?

I learned in high school that catalytic hydrogenations are always syn. Why is this inconsistent with what I learned?

Reductions of naphthalene with different reducing agents

  • $\begingroup$ Because what you learned in high school is an oversimplification. This statement from Chem. Absts.: ”The reduced NiO-MoO3/Al2O3 catalyst at 350° and 35 bar produced the highest decalin yield, i.e., 40 wt.% trans- and 15 wt.% cis-decalin”. $\endgroup$
    – user55119
    Dec 15, 2021 at 21:43
  • $\begingroup$ A similar question about this diagram is on ChemSE. Look here $\endgroup$
    – user55119
    Dec 16, 2021 at 2:56
  • $\begingroup$ @user55119 Even with that simplification ('all' instead of 'some' trans-decalin) the question still remains why trans-decalin is generated at all. My initial assumption is that the bridge is not hydrogenated in a single step which leads to steric effects causing the hydrogens on either end to be installed in different directions but I haven't checked the literature. $\endgroup$
    – Jan
    Dec 16, 2021 at 15:45
  • $\begingroup$ This is because trans decalin is more stable than cis decalin. In cis decalin, 1,3-diaxial interactions (which are repulsive in case of decalin ) destabilize roughly by 8.4kJ/mol than trans decalin. $\endgroup$
    – Infinite
    Dec 21, 2021 at 16:12

1 Answer 1


I think to understand heterogenious hydrogenation you need to understand homogenious hydrogenation of metals by Wilkinson's cataylst and hydroformylation of alkenes by a mixture of hydrogen and carbon monoxide using a cobalt carbonyl cataylst. If we consider for a moment the reaction of an acetylene with a complex such as RuHCl(CO)(PPh3)2. Then what will happen is that the acetylene will bond to the vaccent site on the ruthenium to form a pi complex. The reaction of an acetylene with a ruthenium hydride complex Next a rearrangement known as an insertion reaction will occur which will convet the acetylene into a vinyl group. The metal and the hydrogen which was originally on the metal will be attached to the alkene such that they will be cis to each other. This is the basis of why the hydrogenation of an alkyne normally gives the cis isomer with things like partly poisoned palladium.

If we consider for a moment the reaction in the oxo process then we have a hydrogenation with a little bit added onto it. Here is the cycle for cobalt. Cobalt hydroformylation cycle

The hydrogenation cycle for Wilkinson's cataylst is very similar, you are likely to be thinking of the same sort of thing occuring on the nickel surface. Based on what we know about the chemistry of acetylenes with hydride complexes we can feel confident that an isolated alkene or acetylene will undergo cis hydrogenation. Please see the following picture. What will happen with your substrate is that the very unreactive benzene ring will hydrogenate once to form a very reactive 1,3-diene which will hydrogenate very quickly.

Two diagrams you need to see

A lot of the selectivity for hydroformylation (linear vs branched product) relates to the energies of the alkyl complexes which exist in part of the cycle.

There are three homogenous hydrogenation mechanisms which exist but the one for Wilkinson's cataylst is the most common and it is the closest to what happens on a nickel or palladium surface.

Now for the 1,3-diene I am sure that as it goes through the hydrogenation process that at some point it will pass through an alkyl complex. This is likely to be in the final stage of the hydrogenation.

The two alkyl complexes

The lower energy one will be the main one in the reaction mixture. We need to be careful as the platinum gave the other product. The problem is that when we change from a first row to third row transition metal is the bond energies change. It is possible for the rate determining step to be different even if the catatyltic cycle looks the same. Consider the Monsanto process for the carbonylation of methanol vs the iridium process for the same reaction (even the same substrates) which BP invented later. There the RDS is a different step in the cycle.

I hope that I am not too heavy going, you have just had the associate prof level of answer for the problem.


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