See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: Reaction of terminal alkynes with I2 in the presence of a base (morpholine here acts as a base). Step 1: Identify the reagents. The substrate is phenylacetylene (H-C≡C-Ph), a terminal alkyne. The reagents are I2 and morpholine (H-N with an oxygen-containing ring, i.e., a secondary amine that acts as a base). Step 2: Understand the role of morpholine. Morpholine is a weak base/nucleophile. In this reaction, it deprotonates the terminal alkyne C-H bond. Terminal alkynes are weakly acidic (pKa ~25) and can be deprotonated by a suitable base to form an acetylide anion. Step 3: Mechanism. Morpholine abstracts the terminal proton of phenylacetylene to generate the phenylacetylide anion (Ph-C≡C⁻). This acetylide anion then attacks I2 (electrophilic iodine), displacing iodide to give Ph-C≡C-I (iodophenylacetylene). Step 4: Product identification. The product is Ph-C≡C-I, which is option (c). The triple bond is preserved; only the terminal H is replaced by I. Why other options fail: - Option (a): Would require addition across the triple bond, not substitution; also requires specific conditions for vinyl iodide formation. - Option (b): Would require full addition of I2 across the triple bond AND further reaction; not consistent with these mild base conditions. - Option (d): I-C≡C-H would mean iodination occurred at the internal (phenyl-bearing) carbon, which is not the acidic site; the terminal C-H is the one deprotonated. Therefore, the correct answer is C.