The following reduction is best accomplished with:

(a) Zn/Hg, conc. HCl
(b) N2H4, C2H5ONa
(c) H2, Pd
(d) All of the above

My thoughts: By looking at the question it is evident that only the C=O group should be affected. H2/Pd is used for hydrogenation which might affect the C=C double bond as well. Option (a) gives the reagents which are used for the Clemmensen reduction which is used to convert carbonyl groups to alkanes. So, I think (a) is the correct answer. Is that right?


1 Answer 1


The Clemmensen reduction is really only effective at reducing ketones to alkenes when one group is aromatic. Additionally, the Clemmensen reduction uses concentrated HCl, which might cause problems. Another functional group in your molecule, the tertiary alcohol, will be removed by HCl via an elimination reaction.

Choice (b) is the Wolff–Kishner reaction, which also converts ketones into alkenes. It requires strong base (not acid) and works on almost all ketones.

Choice (c) will react with the alkene also, and so cannot be correct.

  • $\begingroup$ The double bond will not retain its position in the WK nor in the Clemmensen. The product will not survive the acidic conditions of the Clemmensen. I have reduced polycyclic, aliphatic ketones by the Clemmensen method. The answer is e) none of the above. If the OP is interested in what can go wrong by these two methods, I will provide suggestions. $\endgroup$
    – user55119
    Commented May 9, 2019 at 23:47

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