Hydrocarbons · Lesson 9 of 14
Alkyne Structure, Names, and Preparation
“Connect the linear triple bond to alkyne isomerism and methods for making ethyne.”
• Name alkynes and distinguish position and chain isomers. • Explain the sigma and two pi bonds of an alkyne. • Relate sp hybridisation to shape, bond length, and bond strength. • Write balanced preparations from calcium carbide and vicinal dihalides.
A triple bond removes four hydrogens from the alkane count
An acyclic hydrocarbon with one carbon–carbon triple bond follows for . It has four fewer hydrogens than the corresponding alkane. Ethyne, also called acetylene, is the first member. Its oxygen-fed flame reaches temperatures useful in welding and cutting. Alkynes also provide starting materials for further organic synthesis.
Select a parent chain containing the triple bond, number it to give that bond a low locant, and replace -ane by -yne. The first triply bonded carbon provides the locant. Common names sometimes describe substituted acetylene, so methylacetylene is propyne and dimethylacetylene is but-2-yne. A common name must still correspond to a definite structure.
| Systematic name | Common acetylene-based name | Triple-bond position |
|---|---|---|
| Ethyne | Acetylene | Only possible position |
| Propyne | Methylacetylene | Terminal |
| But-1-yne | Ethylacetylene | Terminal |
| But-2-yne | Dimethylacetylene | Internal |
But-1-yne and but-2-yne have the same carbon skeleton and differ in triple-bond position. For five carbons, pent-1-yne and pent-2-yne are position isomers, while 3-methylbut-1-yne has a branched skeleton. This produces chain-isomer comparisons as well. A triple-bond carbon already uses three valence connections towards its partner, so it cannot carry an arbitrary number of side groups.
Problem
List the acyclic single-triple-bond isomers of the fifth alkyne homologue.
- 1.The first alkyne is ethyne with two carbons. Counting ethyne, propyne, butyne, pentyne, then hexyne gives the fifth homologue formula C6H10, not C5H8.
- 2.The unbranched possibilities are hex-1-yne, hex-2-yne, and hex-3-yne. Five-carbon parents give 3-methylpent-1-yne, 4-methylpent-1-yne, and 4-methylpent-2-yne.
- 3.The remaining valid skeleton is 3,3-dimethylbut-1-yne. These seven structures include position-isomer pairs and chain-isomer pairs; reversing a drawing adds no new isomer.
Two perpendicular pi bonds surround a sigma framework
Each ethyne carbon has two hybrid orbitals pointing in opposite directions. Head-on overlap of one orbital from each carbon forms the C–C sigma bond. The other overlaps with a hydrogen 1s orbital. Each carbon also retains two unhybridised p orbitals perpendicular to each other and to the carbon–carbon axis. Matching parallel p orbitals overlap sideways in two perpendicular planes, forming two pi bonds.
The H–C–C arrangement is linear, with an angle of . Ethyne contains three sigma bonds in total, one C–C and two C–H, plus two pi bonds. Together the perpendicular pi systems give an overall electron distribution around the bond axis. Typical carbon–carbon bond lengths decrease from for single to about for double and for triple. Total bond enthalpies rise approximately from to to .
Problem
Count sigma and pi bonds in .
- 1.There are two carbon–carbon connections. Each contributes one sigma bond regardless of bond multiplicity.
- 2.Four carbon–hydrogen bonds contribute four more sigma bonds, giving six sigma bonds in total.
- 3.The triple bond contributes two pi bonds. The methyl carbon is sp3, while both triple-bond carbons are sp hybridised.
Make ethyne from an inorganic carbon source
On a large scale, limestone is heated to produce quicklime and carbon dioxide. Quicklime is heated with coke to make calcium carbide. Water then reacts with the carbide to release ethyne. Each equation must conserve atoms: heating calcium carbonate releases carbon dioxide, not oxygen. The sequence links a mineral starting material to an organic compound.
Problem
Why does calcium oxide require three carbon atoms to form calcium carbide in the stated equation?
- 1.Two carbon atoms are incorporated into calcium carbide, CaC2.
- 2.The oxygen atom originally present in calcium oxide leaves as carbon monoxide and needs a third carbon atom.
- 3.Thus CaO + 3C gives CaC2 + CO. Omitting the carbon monoxide product would lose oxygen and give an incorrect carbon balance.
Create the second pi bond by a second elimination
Vicinal dihalides can be converted into alkynes by successive elimination. Alcoholic potassium hydroxide first removes one hydrogen halide, producing an alkenyl halide with halogen attached to a double-bond carbon. That remaining halide is harder to eliminate than an ordinary alkyl halide. Sodamide, a stronger base, removes the second hydrogen halide equivalent to create the triple bond.
Do not place a methyl substituent on a triple-bond carbon that would then have more than four bond orders in total. Also, the fifth homologue is not automatically a five-carbon molecule: the alkyne series begins with two carbons.
Quiz
Which is the first stable alkyne?
What does a carbon–carbon triple bond contain?
Which is a pair of position isomers?
Which gas is released when calcium carbonate forms quicklime?
Why is sodamide used after an alkenyl halide forms?
Practice Problems
- Draw pent-1-yne, pent-2-yne, and 3-methylbut-1-yne and compare their isomerism.
- List the seven C6H10 isomers with one triple bond and no ring.
- Compare carbon hybridisation and geometry in ethane, ethene, and ethyne.
- Write the three balanced equations from limestone to ethyne.
- Show both elimination stages from 1,2-dibromoethane to ethyne.
- Explain why triple-bond carbons cannot carry side groups in the same way as a four-single-bond carbon.
Key Takeaways
• Alkynes with one triple bond and no ring follow the formula CnH2n−2. • A triple bond consists of one sigma and two perpendicular pi bonds. • The sp carbon centres are linear, and the carbon–carbon bond is short and strong. • Carbide hydrolysis and successive elimination provide distinct preparation routes.