JEE Main 27-Aug-2021 Shift 1 Solved Paper

NaNH2 is a strong base that causes deprotonation of propyne and forms acetylide ion (A) which further combines with the carbon chain and forms hept - 5 - yn - 2 - ol (B) that undergoes reduction in presence of H2∕Pd−C and forms heptan-2-ol(C). Being a secondary alcohol (C), oxidises in presence of CrO3 to give corresponding ketone i.e. heptan-2-one (D).
The complete reaction take place as follows.