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I encountered the following question in an exam:

$\ce{CH3COCH3 + NaNH2}$  giving out $\ce{A}$.
$\ce{A}$ then reacts with $\ce{C2H2}$ giving out $\ce{B}$.
$\ce{B}$ then reacts with $\ce{H+}$ to give $\ce{C}$.
$\ce{C}$ then reacts with 1 equivalent of $\ce{H2}$ in presence of $\ce{Pd}$ to give $\ce{D}$.
$\ce{D}$ then is heated to $\pu{400 ^{\circ}C}$ in presence of $\ce{Al2O3}$ to give $\ce{E}$.

I reasoned it as the first reaction could be an acid-base reaction, but that didn't help much. Then I tried a nucleophilic substitution, but there aren't any good leaving groups and still we get stuck after the first step.

One of the teachers reasoned it as the first reaction being reversible and it goes back to $\ce{CH3COCH3}$ which then reacts with all other and it ends up at isoprene which is the correct answer according to the answer key, but this explanation seems very flawed to me as then what was the purpose of the first step? Can someone please check and write the steps and products in each reaction?

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Acetone and sodamide give acetone sodium enolate A, nothing further happens until acetylene is introduced. The pKas of acetone and acetylene are both around 25 so an exchange occurs reforming acetone and creating sodium acetylide. This seems an unusual way of making sodium acetylide, but I don't think this is a real world example (unless someone can find a reference).

Sodium acetylide adds to acetone giving the sodium alkoxide of tertiary alcohol B

Work up with dilute acid gives the tertiary alcohol C.

Hydrogenation over Pd catalyst reduces the triple bond to a double bond giving D.

Passing D over alumina at 400C dehydrates the tertiary alcohol to give isoprene.

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  • $\begingroup$ Nice answer to a poorly formulated reaction sequence. Given the pKa's of acetone (~20) and acetylene (~25), a practicing chemist would prepare sodium acetylide first and then add acetone. One equivalent of hydrogen in the presence of "Pd" will not be selective, also giving over reduction of the initially formed alkene. That's why we have Lindlar's catalyst. Alas! $\endgroup$
    – user55119
    May 6, 2019 at 22:45
  • $\begingroup$ Is it possible for an aldol self-condensation to happen in the first step? I went along a different line of thought. The product then undergoes DA reaction with acetylene. $\endgroup$
    – Tan Yong Boon
    May 6, 2019 at 22:58
  • $\begingroup$ An aldol, as the alkoxide, is likely reversible. Aldol, itself, is difficult to prepare in the presence of base IN the solution of acetone. A Soxhlet technique is used. You are presuming that mesityl oxide is formed. Less likely. Unsure what you had in mind for a Diels-Alder between mesityl oxide and acetylene. $\endgroup$
    – user55119
    May 7, 2019 at 2:52

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