See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: NaNH2 is a very strong base (pKa of NH3 ~ 38) capable of deprotonating terminal alkynes, and also acts as a base for elimination reactions. Step 1: Identify the starting material. The compound is Ph-CCl(CH3), i.e., 1-chloro-1-phenylethan-1-yl (a secondary benzylic chloride with structure Ph-C(Cl)-CH3, where the carbon bearing Cl also bears a methyl group and a phenyl group — it is actually Ph-CHCl-CH3 based on structural drawing, a secondary benzylic chloride: 1-chloro-1-phenylethane). Wait, re-examining the structure: Ph-C(-CH3)(-Cl) with a vertical bond to Cl, meaning the carbon has Ph, CH3, Cl, and one implicit H — so it is Ph-CHCl-CH3, i.e., (1-chloroethyl)benzene or 1-chloro-1-phenylethane. Step 2: First equivalent of NaNH2 acts as a base and causes dehydrohalogenation (E2 elimination) of Ph-CHCl-CH3. Removal of the beta H (from CH3) gives Ph-CH=CH2 (styrene). This accounts for 1 equivalent of NaNH2 (producing NaCl and NH3). Step 3: Second equivalent of NaNH2 performs another elimination — but styrene (Ph-CH=CH2) is a vinyl system. NaNH2 can abstract an allylic/vinylic proton and cause isomerization or further elimination. NaNH2 is strong enough to deprotonate a vinylic position. The base abstracts a vinylic proton from styrene, generating a vinyl carbanion, which leads to the formation of Ph-C≡CH (phenylacetylene) via elimination of H(-). So: Ph-CH=CH2 + NaNH2 → Ph-C≡CH + NaH + NH3 (or via base-induced 1,2-elimination of H from vinylic position). This accounts for the 2nd equivalent. Step 4: Third equivalent of NaNH2 deprotonates the terminal alkyne Ph-C≡CH (pKa ~25), since NaNH2 is strong enough (pKa of NH3 ~38). This gives Ph-C≡C(-)Na(+) (sodium phenylacetylide) and NH3. This accounts for the 3rd equivalent. Overall: Ph-CHCl-CH3 + 3NaNH2 → Ph-C≡C(-)Na(+) + NaCl + 2NH3 + NaH (or equivalent byproducts). Why other options fail: (a) Ph-CH=CH2 is only the intermediate after first elimination, not the final product. (b) Ph-C≡CH is the product after two equivalents of NaNH2, but the third equivalent deprotonates it. (c) Ph-CH2-CH3 would require reduction, not elimination; NaNH2 is a base, not a reducing agent. (d) Ph-C≡C(-)Na(+) is the correct final product after all three equivalents of NaNH2 are consumed. Therefore, the correct answer is D.