The products of a chain reaction beginning with acetone and sodium amide 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?
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.