Question

The final product (B) formed in the following reaction is:

Propyne ⟶(I) NaNH₂ (1 eq.) ⟶(II) C₂H₅I A ⟶(I) Na/liquid NH₃, (II) Hg(OAc)₂/H₂O, (III) NaBH₄ B

Answer 1

pentan-3-ol

Answer 2

Step 1: Propyne (CH₃–C≡CH) is deprotonated by NaNH₂ to give the acetylide anion, which on alkylation with C₂H₅I yields CH₃–C≡C–CH₂–CH₃ (pent-2-yne).

Step 2: Pent-2-yne undergoes oxymercuration (Hg(OAc)₂/H₂O) to give the enol, which tautomerizes to pentan-3-one (3-pentanone).

Step 3: Reduction of the ketone with NaBH₄ affords pentan-3-ol.