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If one equivalent of $\ce{H2}/\ce{Pt}$ is made to react with one equivalent of pent-4-en-2-one, what will be the product formed?

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    $\begingroup$ Generally the double bond goes, but it depends on the catalyst system used. $\endgroup$
    – Waylander
    Sep 27 '21 at 9:33
  • $\begingroup$ Is there a reason to say so? $\endgroup$
    – Ilovemath
    Sep 30 '21 at 15:17
  • $\begingroup$ For your information, actually C=C is stronger than C=O bond with bond strengths 680 & 500 KJ/mol respectively. $\endgroup$ Sep 30 '21 at 20:27
  • $\begingroup$ @Neon Prince I have edited the question. Thanks $\endgroup$
    – Ilovemath
    Oct 1 '21 at 8:10
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You have not specified the full reaction conditions - temperature and pressure can make a considerable difference - however this quote from Chemistry Libre Texts here (in turn quoting Basic Principles of Organic Chemistry by Roberts & Caserio) gives us a clear indication of the relative ease of hydrogenation of ketones vs double bonds.

Hydrogenation of aldehyde and ketone carbonyl groups is much slower than of carbon-carbon double bonds so more strenuous conditions are required. This is not surprising, because hydrogenation of carbonyl groups is calculated to be less exothermic than that of carbon-carbon double bonds....It follows that it is generally difficult to reduce a carbonyl group in the presence of a carbon-carbon double bond by hydrogenation without also saturating the double bond. Other reducing agents are more selective:

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