The product obtained in this reaction is:

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(A) The reaction involves the terminal alkyne $Ph-C \equiv C-H$ reacting with $I_2$ in the presence of a base (morpholine).$1$. The base (morpholine)

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$Ph-C \equiv C-I$

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(A) The reaction involves the terminal alkyne $Ph-C \equiv C-H$ reacting with $I_2$ in the presence of a base (morpholine).$1$. The base (morpholine) abstracts the acidic terminal proton from the alkyne to form an acetylide anion: $Ph-C \equiv C^-$.$2$. The acetylide anion then acts as a nucleophile and attacks the iodine molecule $(I-I)$,displacing an iodide ion $(I^-)$.$3$. This results in the formation of the iodoalkyne product: $Ph-C \equiv C-I$.

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