See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Step 1: Analyze reaction (a). The starting material is tert-butoxide anion, (CH3)3C-O^-, and the product is tert-butyl ethyl ether, (CH3)3C-OEt. This is an alkylation of the alkoxide. The tert-butoxide is a nucleophile and needs to be alkylated with an ethyl group. The reagent needed is EtBr (ethyl bromide), because the alkoxide attacks the less hindered ethyl bromide in an SN2 reaction. Therefore (a) matches (q) EtBr. Step 2: Analyze reaction (b). The starting material is a tertiary alkyl bromide (2-bromo-2-methylbutane) and the product is a trisubstituted alkene (2-methylbut-2-ene). This is an elimination (E2 or E1) reaction. A strong, bulky base promotes elimination. EtO^- (ethoxide) is a strong base that promotes E2 elimination of the tertiary bromide to give the more substituted alkene (Zaitsev product). Therefore (b) matches (p) EtO^-. Step 3: Analyze reaction (c). The starting material is 2-methylbut-2-ene (an alkene) and the product is 2-ethoxy-2-methylbutane (an ether where OEt is on the more substituted carbon, Markovnikov regiochemistry). This is acid-catalyzed addition of ethanol to an alkene. Under acidic conditions, EtOH/H^+ protonates the alkene to form the more stable tertiary carbocation, which is then captured by ethanol as nucleophile. Therefore (c) matches (r) EtOH/H^+. Step 4: Analyze reaction (d). The starting material is ethyl chloride (Et-Cl, CH3CH2Cl) and the product is butane (CH3CH2CH2CH3). This is a carbon-carbon bond forming reaction coupling two ethyl groups. The Wurtz reaction uses alkyl halides with sodium metal in ether solvent: 2 RX + 2Na --> R-R + 2NaX. The reagent Et-Cl/Na in ether performs the Wurtz coupling of two ethyl chloride molecules to give butane. Therefore (d) matches (s) Et-Cl/Na ether. Why other options fail: - (a) cannot be EtO^- because the substrate is already an alkoxide; it needs an electrophilic ethyl source (EtBr), not another nucleophile. - (b) cannot be EtBr (electrophile) because the substrate is an alkyl bromide needing a base for elimination. - (c) cannot use EtBr or EtO^- because it requires electrophilic activation of the alkene followed by nucleophilic trapping by ethanol. - (d) cannot use EtO^- or EtOH/H^+ as these do not form C-C bonds from alkyl chlorides. Therefore, the correct answer is {"A": ["Q"], "B": ["P"], "C": ["R"], "D": ["Q"]}.