See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1: Reaction of Styrene with Br2/CCl4 Styrene (Ph-CH=CH2) undergoes addition of Br2 across the double bond to give Ph-CHBr-CH2Br (1,2-dibromide). This is product (A). Step 2: (A) treated with (i) alc. KOH, then (ii) NaNH2 Alcoholic KOH causes dehydrohalogenation (elimination of HBr) from Ph-CHBr-CH2Br. The first elimination gives Ph-CBr=CH2 or Ph-CH=CHBr (vinyl bromide type). A second elimination with alc. KOH or NaNH2 removes another HBr to give Ph-C≡CH (phenylacetylene). NaNH2 is a strong base that facilitates the second elimination to form the terminal alkyne. So product (B) is Ph-C≡CH (phenylacetylene). Step 3: (B) treated with (i) NaNH2, then (ii) CH3Cl NaNH2 deprotonates the terminal alkyne Ph-C≡CH (the terminal alkynyl H is acidic, pKa ~25) to form Ph-C≡C⁻Na⁺ (sodium phenylacetylide). This carbanion then undergoes nucleophilic substitution with CH3Cl (methyl chloride) to give Ph-C≡C-CH3 (1-phenyl-1-propyne / phenyl methyl acetylene). This is product (C). Why other options fail: (a) Ph-C≡C-Na: This is an intermediate formed after NaNH2 treatment but before CH3Cl alkylation; the reaction proceeds further with CH3Cl. (b) Ph-CH2-C≡CH: This would require a different carbon skeleton rearrangement; the alkylation occurs at the acetylide carbon directly attached to phenyl, not at a CH2 group. (d) Ph-CH=C=CH2: This is an allene structure; it is not formed under these conditions. Therefore, the correct answer is C.