Alpha-Tetralone is reacted by $1.\ce{Zn(Hg),HCl}\ \ 2.\ce{ Pd/C}$ What is the product and mechanism that is followed for the 2nd reaction?

A book by SN Sanyal says that Naphthalene will be produced . I know that in the 1st reaction, clemenson reduction is followed thereby reducing the Carbonyl group to $\ce{CH2}$ group .Then how the reaction proceeds is not clear.

I may be wrong of the 1st step also. Please correct me.

  • $\begingroup$ Well many reactions can go both ways... and that might help you solving this $\endgroup$
    – SteffX
    Nov 24, 2018 at 15:02
  • $\begingroup$ Can you help by writing the exact steps $\endgroup$ Nov 24, 2018 at 15:28
  • $\begingroup$ The tetralin loses 2 molecules of hydrogen on the active Palladium surface. $\endgroup$
    – Waylander
    Nov 24, 2018 at 16:49

1 Answer 1


Moinak Banerjee: Your concern about the mechanism of the conversion of tetrahydronaphthalene (tetralin) 1 to naphthalene 5 has been addressed in a comment by Waylander. It is important to recognize that the process is a 4-electron oxidation of tetralin 1 and a four electron reduction of the hydrogen that is liberated as 2 equivalents of hydrogen gas. Ideally, the reaction is catalytic in palladium (Pdo). Mechanistically, the reaction likely proceeds by oxidative addition of Pdo into the labile benzylic C-H bond of tetralin 1 to afford Pd(II) species 2, which, in turn, isomerizes to π-complex 3. These steps are shown as reversible because complex 3 is the likely intermediate in the hydrogenation of 1,2-dihydronaphthalene 4 to tetralin 1. Collapse of complex 3 regenerates the catalyst Pdo and an equivalent of hydrogen gas. A repetition of these steps at either the benzylic or allylic C-H bonds of 1,2-dihydronaphthalene 4 leads to naphthalene 5 and a second equivalent of hydrogen gas. These reactions are driven to the right by the loss of hydrogen gas in an open vessel. --continued--

enter image description here

This reaction played a critical role in early attempts to synthesize estrone since it served to reduce the number of asymmetric carbons, thereby facilitating stereochemical analysis. An example of this process is illustrated in the sequence 6 --> 8 from the work of Bachmann and Chemerda1. The double bond in tetracycle 6 is isomerized by Pdo to the tetrasubstituted position in isomer 72. The conversion to the naphthalene 8 follows the same steps as in the case of 1,2-dihydronaphthalene 4.3

1) W. E. Bachmann and J. M. Chemerda, J. Am. Chem. Soc., 1948, 70, 1468.
2) The double bond migration involves a similar C-H insertion and subsequent π-allyl complex. But that is another story.
3) See also: J. Heer and K. Miescher, Helv. Chim. Acta, 1949, 32, 1572.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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