Consider the compound $\ce{Alkyl-(4 oxocyclopent-2-ene)carboxylate (A)}$. Now consider the operations:

$\ce{1. (A) + (i)BH3/THF (ii) H+}$

$\ce{2. (A) + H2/Pt}$

$\ce{3. (A) + H2/Pd-C}$


In $1.$ ester group reduced to an alcohol.

In $2.$ double bond reduced to a single bond, and

In $3.$ double bond reduced to a single bond, ester group reduced to an alcohol.

However, on the next page addition of $\ce{(i)BH3/THF (ii) H+}$ selectively reduced carboxylic acid when both ester and carboxylic acid were present.

My questions:

$(i)$ I know that $\ce{(i)BH3/THF (ii) H+}$ can reduce carboxylic acid, but I am unaware if it can reduce ester.

$(ii)$ I only knew that, the reduction of ester in presence of carbonyl group could be done by first protecting the carbonyl group (converting it to cyclic ketal by reacting with ethylene glycol) and then using $\ce{LiAlH4}$. Is reaction $3.$ correct?

$(iii)$ Why aren't product $2.$ and $3.$ same? I've always seen the two work same.


My chemistry teacher gave me few notes for Chemistry Olympiad preparation, so I can not cite source. I don't know how to make compounds, so will completely reside on IUPAC nomenclature.

  • 1
    $\begingroup$ Man reduction is highly specific and its trends are difficult. However, I am sure there should be a reason to this. All I can add is that Pd-C is stronger than Pt as an adsorbent. Nice question! $\endgroup$ – Haha Hahaha Jan 11 at 16:34
  • 1
    $\begingroup$ I'm not sure that differences in Pt vs Pd hydrogenation can be easily explained. I'm also not entirely convinced that H2 with Pd/C reduces esters. Maybe in a few isolated cases, or maybe with higher pressures, but I doubt it's anywhere near as facile as hydrogenation of C=C bonds. As far as I know, this doesn't occur sufficiently regularly to use it in a predictive sense. Also, your IUPAC name is slightly ambiguous: is this a CO2R group sticking off carbon-1 of your cyclopentene? $\endgroup$ – orthocresol Jan 11 at 16:53
  • 1
    $\begingroup$ Yes, it's CO2R group. What's the correct IUPAC name? $\endgroup$ – Zenix Jan 11 at 17:01
  • 1
    $\begingroup$ @Zenix: from which book is this question? I feel that these questions are interesting! $\endgroup$ – Rahul Verma Jan 17 at 12:54

Your Answer

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

Browse other questions tagged or ask your own question.