As part of a synthesis I was considering the following reaction:


How selective is the Clemmensen reduction? I know that this reduction favours ketones adjacent to benzene rings but will it also reduce the other ketone, and/or affect any other parts of the molecule. Are there any other methods that would achieve this transformation more effectively.

  • $\begingroup$ Use a weak base/TMSCl to form a 6-cyclic (O-silyl)ketal with the outer carbonyl then use Wolff-Kishner ? Probably not good yield. $\endgroup$
    – J. LS
    Apr 15, 2015 at 19:24

1 Answer 1


A Clemmensen reduction of a 1,3-diketone probably isn't the best idea. The reduction is likely to proceed via a ketyl radical.

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In the case of a 1,3-diketone, formation of a dihydroxy cyclopropane, which undergoes opening/rearrangement cannot be excluded.

This reaction has for instance been observed for dimedone, where 2,4,4-trimethyl cyclopentanone was formed.


Which other viable procedures exist?

If you insist on using the given starting material, the following might (!) work:

  1. regioselective (asymmetric) catalytic hydrogenation of the acetophenone-type carbonyl using a 1,2-diamine-substituted Noyori catalyst

  2. Barton-McCombie reaction of the resulting alcohol

If you are rather interested in the product, you might want to rethink the strategy:

The intended product is an arylbutanone, related to the raspberry ketone. A possible retrosynthesis would be:

Retrosynthesis of an arylbutan-2-one

If you get your hands on the necessary benzaldehyde, an aldol reaction with acetone, followed by catalytic hydrogenation might give the right product.


5-tert.-butyl-2-hydroxybenzaldehyde is commerically available, but not really cheap.

However, 4-tert.-butylphenol is cheap as dirt. In case you need significant quantities of your final product, you might want to consider the synthesis of the benzaldehyde. The substitution pattern cries for a Reimer-Tiemann reaction.

  • $\begingroup$ Are there any suitable alternative methods for this transformation then? $\endgroup$
    – bon
    Apr 16, 2015 at 21:02
  • $\begingroup$ @bon I've added a possible strategy to my answer. $\endgroup$ Apr 17, 2015 at 8:06

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