# Reactions on alpha tetralone to synthesize naphthalene

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.

• Well many reactions can go both ways... and that might help you solving this – SteffX Nov 24 '18 at 15:02
• Can you help by writing the exact steps – Moinak Banerjee Nov 24 '18 at 15:28
• The tetralin loses 2 molecules of hydrogen on the active Palladium surface. – Waylander Nov 24 '18 at 16:49