Industrial methods for producing acetylene
In the chemical industry, acetylene is produced by two main routes that require completely different reaction conditions.
The first method is methane pyrolysis. This industrial method is based on the extremely rapid heating of the starting gas. The process requires maintaining very high temperatures in the range of 1500–1600 °C. Under such harsh thermal conditions, methane molecules undergo decomposition, which is accompanied by the simultaneous formation of target acetylene. It is worth noting that the yield of the finished product in this reaction is up to 15%.
The second classic method is the hydrolysis of calcium carbide. This is a heterogeneous reaction proceeding according to the equation: $CaC_2 + 2H_2O = C_2H_2 \uparrow + Ca(OH)_2$ During this process, solid calcium carbide reacts with two moles of water. The reaction is accompanied by the active evolution of gaseous acetylene and the formation of a calcium hydroxide precipitate.
Cooling, isolation, and purification
After high-temperature synthesis (e.g., pyrolysis), the resulting gas mixture requires separation and isolation of the target product. The technological process begins by subjecting the red-hot gas mixture to rapid cooling.
Further isolation of pure acetylene is based on its specific physicochemical properties. The gas is separated from accompanying impurities by dissolving in water, which is carried out under elevated pressure. The efficiency of this method is due to the fact that the volume solubility coefficient ($K_V$) of acetylene is significantly higher than that of most other hydrocarbons. At a temperature of $15^\circ C$, this indicator reaches a value of $1.15$, which makes it possible to reliably extract the alkyne from the multi-component gas phase.
Laboratory synthesis: double elimination
To obtain more complex acetylene homologues in laboratory practice, the reaction of double elimination (or splitting off) is used. Dihalogen derivatives of alkanes serve as the starting substrate in this case.
The key reagent ensuring the elimination process is an alcoholic alkali solution (typically sodium hydroxide, $NaOH$, is used). Under the action of this alcoholic reagent, halogen and hydrogen atoms are successively detached from the dihaloalkane molecule, leading to the formation of a triple carbon-carbon bond.
A typical reaction equation using the preparation of propyne as an example is as follows: $CH_3-CH(Br)-CH_2(Br) + 2NaOH_{(alc)} \rightarrow CH_3-C \equiv CH + 2NaBr + 2H_2O$ As a result of the interaction of one mole of dibromopropane with two moles of alcoholic alkali, propyne, two moles of sodium bromide, and two molecules of water are formed.
Multi-step synthesis: from alkene to alkyne
In practice, to obtain an alkyne with a specific triple bond location (e.g., internal), a multi-step transformation chain is often required. A characteristic example is the synthesis of but-2-yne from a terminal alkene—but-1-ene.
This process includes four sequential stages:
- Hydrohalogenation: hydrogen bromide ($HBr$) is added to the starting but-1-ene. A monohalogen derivative is formed—2-bromobutane ($C_2H_5-CHBr-CH_3$).
- Dehydrohalogenation (obtaining an alkene): 2-bromobutane is treated with an alcoholic alkali solution ($NaOH$). Hydrogen bromide is eliminated, yielding an alkene with a new double bond position—but-2-ene ($CH_3-CH=CH-CH_3$), as well as by-products ($NaBr$ and water).
- Halogenation: molecular bromine ($Br_2$) is added to the resulting but-2-ene. The double bond is broken, and a dihaloalkane is formed—2,3-dibromobutane ($CH_3-CH(Br)-CH(Br)-CH_3$).
- Double dehydrohalogenation (obtaining an alkyne): the resulting dibromoalkane is exposed to two moles of an alcoholic alkali solution. Double elimination of hydrogen bromide occurs, resulting in the synthesis of the target alkyne—but-2-yne ($CH_3-C \equiv C-CH_3$), along with salt and water.
Briefly: answers to topical questions
What types of dihaloalkanes can be used as substrates for obtaining alkynes?
To obtain alkynes, halogen derivatives of hydrocarbons are used under the action of an alcoholic alkali solution. The given examples involve dihaloalkanes with halogen atoms on adjacent carbon atoms: $CH_3-CH(Br)-CH_2(Br)$, from which propyne is obtained, and $CH_3-CH(Br)-CH(Br)-CH_3$, used to obtain but-2-yne.
What rule determines the regioselectivity of hydrogen halide elimination during the formation of but-2-ene?
Zaitsev's rule determines the regioselectivity of hydrogen halide elimination during the formation of but-2-ene. It is used in elimination reactions for asymmetric haloalkanes when different numbers of hydrogen atoms are attached to two adjacent carbon atoms. According to this rule, hydrogen is predominantly eliminated from the less hydrogenated carbon atom (from the C atom that has fewer H atoms). For example, during the reaction of 2-chlorobutane with an alcoholic alkali solution, hydrogen leaves the third carbon atom, resulting in the formation of but-2-ene rather than but-1-ene.
By means of which reactions can acetylene homologues be obtained from terminal alkynes?
Acetylene homologues can be obtained using acetylide formation reactions and their subsequent alkylation. Acetylene and its homologues with a terminal triple bond possess weak acidic properties and are capable of replacing a hydrogen atom with a metal. The following stages are used for synthesis:
- Formation of acetylides — interaction of an alkyne with an ammoniacal solution of silver oxide or copper(I) chloride with precipitation.
- Alkylation — the reaction of the resulting acetylides with halogen derivatives of hydrocarbons.
This process allows lengthening the carbon chain and synthesizing new homologues.
What is the mechanism of the double dehydrohalogenation reaction of dihaloalkanes?
The reaction mechanism consists of double elimination (splitting off). The dehydrohalogenation process is the elimination of a halogen atom and a hydrogen atom from two adjacent carbon atoms. An important condition for the double dehydrohalogenation reaction of dihaloalkanes to proceed with the formation of alkynes is the use of an alcoholic alkali solution (e.g., $NaOH$ or $KOH$). As a result of the sequential elimination of two molecules of hydrogen halide, a triple bond is formed between carbon atoms, and salt and water are also produced.
For elimination reactions of dihaloalkanes, remember the solvent rule: "Water gives alcohol, and alcohol gives a bond." To obtain a triple bond in alkynes, always use an alcoholic alkali solution.
Class and exam questions
Why is rapid cooling required during methane pyrolysis?
The gas mixture is rapidly cooled to fix the formed acetylene and prevent its further thermal destruction at ultra-high temperatures.
Which reagent is necessary for the synthesis of alkynes from dihaloalkanes?
For the double elimination of hydrogen halides and the formation of a triple bond, exclusively an alcoholic alkali solution (e.g., alcoholic NaOH) is used.
What is the method for isolating acetylene from mixtures based on?
Acetylene is separated by dissolving in water under pressure, since its volume solubility coefficient (1.15 at 15 °C) significantly exceeds the solubility of other hydrocarbons.
What else to study on this topic
- The mechanism of the double elimination reaction (E2)
- Physicochemical properties of calcium carbide and its hydroxide
- Thermodynamics of methane pyrolysis at 1500–1600 °C
- The effect of pressure on the volume solubility coefficient of gases
- Regioselectivity at the stages of hydrohalogenation and dehydrohalogenation
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