See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1: Identify the starting material. 1,2-Dibromopropane is CH3-CHBr-CH2Br. Step 2: Determine the product. Treatment with NaNH2 (a strong base) followed by ethyl bromide (CH3CH2Br) gives a pentyne. A pentyne has 5 carbons. Starting with 3 carbons (propane backbone) and adding an ethyl group (2 carbons) gives 5 carbons total, consistent with a pentyne. Step 3: Reaction with NaNH2 (dehydrohalogenation). NaNH2 is a strong base that promotes elimination of HBr to form alkynes. - First equivalent of NaNH2: eliminates HBr from 1,2-dibromopropane to give an alkenyl bromide (vinyl bromide), specifically CH3-CBr=CH2 or CH3-CH=CHBr (a 1-bromo or 2-bromopropene). - Second equivalent of NaNH2: eliminates a second HBr to give propyne (CH3-C≡CH), a terminal alkyne. - Third equivalent of NaNH2: deprotonates the terminal alkyne C-H (pKa ~25) to form the acetylide anion CH3-C≡C^- Na^+. NaNH2 (pKa of NH3 ~38) is strong enough to deprotonate terminal alkynes. Step 4: Alkylation. The propyne acetylide anion (CH3-C≡C^-) reacts with ethyl bromide (CH3CH2Br) via SN2 to give CH3-C≡C-CH2CH3, which is pent-2-yne (a pentyne). This confirms the product. Step 5: Count total moles of NaNH2 used. Two moles are used for double dehydrohalogenation (removing two HBr molecules from 1,2-dibromopropane to form propyne), and one mole is used to form the acetylide anion. Total = 2 + 1 = 3 moles. Why other options fail: - (a) One: Only one elimination would occur, not enough to reach alkyne and form acetylide. - (b) Two: Would form propyne but not the acetylide needed for alkylation. - (d) Four: Excess beyond what is needed; only 3 equivalents are required for this sequence. Therefore, the correct answer is C.