Answer: A, C, D
Amide forms the acetylide, which is methylated at carbon to give CHX3CHX2−C≡C−CHX3, pent-2-yne. Lindlar reduction stops at the alkene, giving but-1-ene. Markovnikov hydration gives butan-2-one. Hydroboration with a hindered borane followed by oxidation is anti-Markovnikov and gives the enol that tautomerises to butanal, not butan-2-one.