Chemistry · Ch 15 — Hydrocarbons
Preparation of alkynes
Preparation of alkynes
Ethyne is manufactured industrially by two distinct routes. The first is controlled, high-temperature (1773 K) partial oxidation of methane: six molecules of methane react with two of dioxygen to give two molecules of ethyne plus carbon dioxide and hydrogen gas as byproducts. The second, and far more common, industrial (and laboratory) route starts from calcium carbide: calcium carbide reacts directly with water at room temperature, with no special catalyst needed, to give ethyne gas plus calcium hydroxide. Beyond ethyne itself, alkynes in general can also be built up in the laboratory by two further methods. The first is double dehydrohalogenation of a vicinal dihalide: the dihalide first reacts with alcoholic KOH, eliminating one H and X to form an alkenyl (vinylic) halide intermediate; this intermediate then reacts with the much stronger base sodamide (NaNH2) to eliminate the second H and X, completing the alkyne -- sodamide is needed for this second step specifically because the C-H on a vinylic (sp2) carbon is far less reactive toward elimination than an ordinary alkyl C-H, so the milder alcoholic KOH that worked for the first elimination is no longer strong enough. The second method builds a LONGER, internal alkyne from a smaller terminal one: because a terminal alkyne's own C-H is appreciably acidic (section 15.3.4), it can be deprotonated by a very strong base such as lithium amide to form a metal (lithium) acetylide, and this acetylide then reacts with a primary alkyl halide, displacing the hal …
Worked out. Route 1: controlled, high-temperature (1773 K) partial oxidation of methane converts it directly to ethyne -- 6CH4 + 2O2 gives 2H-C(triple bond)C-H + 2CO2 + 10H2. Route 2 (the common industrial and laboratory route): calcium carbide reacts with water, CaC2 + 2H2O gives C2H2 (ethyne) + Ca(OH)2, a reaction that proceeds readily at room temperature and needs no speci …
Worked out. A vicinal dihalide first reacts with alcoholic KOH to eliminate one HX and form an alkenyl (vinylic) halide; this alkenyl halide then reacts with sodamide (NaNH2) to eliminate the second HX and form the alkyne. 1,2-Dibromoethane goes through bromoethene to give ethyne. 1,2-Dichloropropane goes through 1-chloropropene to give propyne. The second dehydrohalogenation step needs the much stronger base sodamide (rather than a second dose of alcoholic KOH) because the C-H on a vinylic (sp2) carbon of the alkenyl halide intermediate is far less easily removed than an ordin …
Worked out. The hydrogen on a terminal alkyne's triply-bonded carbon (R-C(triple bond)C-H) is appreciably acidic and can be removed as a proton by a very strong base. Reacting the terminal alkyne with lithium amide (easier to handle than sodamide) forms the metal acetylide, releasing ammonia: for example ethyne plus LiNH2 gives lithium ethynide (H-C(triple bond)C-Li) plus NH3, and propyne plus LiNH2 gives lithium prop-1-yn-1-ide. This metal acetylide then reacts with a primary alkyl halide to build a longer, internal alkyne: lithium ethynide plus bromoethane gives but-1-yne plus LiBr, and lithium prop-1-yn-1-ide plus bromo …