See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Concept: NaNH2 in liquid NH3 is a very strong base that can perform elimination reactions and also deprotonate terminal alkynes. When 3 equivalents of NaNH2 are used with a substrate containing a leaving group adjacent to an oxygen-containing ring, a ring-opening and triple-bond forming sequence occurs. Step 1 - Identify the substrate: The starting material is 2-(chloromethyl)tetrahydrofuran. It has a THF ring (cyclic ether) with a -CH2Cl group at C2. Step 2 - First equivalent of NaNH2: NaNH2 acts as a strong base and induces elimination of HCl from the -CH2Cl group in concert with ring opening. The chloride is eliminated and the C-O bond of the ring opens. This generates an intermediate vinyl/allyl system or, more precisely, the ring opens to give an alkene with a sodium alkoxide: the THF ring opens so that the oxygen becomes a terminal -ONa group and a double bond or carbanion is generated. Step 3 - Mechanism in detail: NaNH2 abstracts a proton alpha to the CH2Cl, and E2-type elimination occurs with ring opening. The C-O bond of the ring breaks heterolytically: the oxygen retains the electron pair and gets deprotonated (or directly becomes alkoxide), and the carbon chain forms a vinylic/acetylenic system. With 3 NaNH2, two successive eliminations occur on the ring-opened chain to install a triple bond, and the terminal oxygen is deprotonated to -ONa. Step 4 - Product formation: The THF ring (4 carbons + O) opens and the CH2Cl carbon becomes part of the chain. The full carbon chain is: Cl-CH2-CH(ring-O-CH2CH2CH2-). Upon ring opening and double elimination using 3 NaNH2: first equivalent opens the ring/eliminates HCl to give an alkene with -ONa; second and third equivalents perform further elimination to give a terminal alkyne with both ends functionalized. The product is NaC≡C-(CH2)3-ONa, where one end is the sodium acetylide (from two eliminations on the CH2Cl side) and the other end is sodium alkoxide from deprotonation of the terminal -OH generated by ring opening. Step 5 - Why 3 equivalents: 1st NaNH2 deprotonates and eliminates to open ring giving a vinylic chloride or allyl system with -ONa; 2nd NaNH2 eliminates HCl to give an allenic/acetylenic species; 3rd NaNH2 deprotonates the terminal alkyne C-H to give sodium acetylide NaC≡C-, and the alkoxide end is already -ONa from step 1. Thus the product NaC≡C-(CH2)3-ONa requires all three equivalents. Why other options fail: - (a) The dihydrofuran with exocyclic methylene would only require 1 equivalent of base and would not account for full ring opening. - (b) HC≡C-(CH2)3ONa: Only 2 NaNH2 equivalents would be used (no deprotonation of terminal alkyne), not 3. - (d) H-C≡C-(CH2)3OH: Neither the alkyne proton nor the alcohol is deprotonated, inconsistent with excess strong base (3 NaNH2). Therefore, the correct answer is C.