Hydrocarbons · Lesson 6 of 14
Naming, Isomerism, and Preparation of Alkenes
“Name and compare alkenes, then choose preparations that create their double bonds and geometry.”
• Select and number a parent chain containing the double bond. • Distinguish chain, position, and geometrical isomerism. • Test the substituents on both double-bonded carbons for geometrical isomerism. • Compare selective reduction, elimination, and dehalogenation preparations.
Include the double bond in the parent
Choose the longest suitable chain containing the double bond and number it to give that bond the lower position. Replace -ane by -ene. The locant identifies the first of the double-bonded carbons: but-1-ene and but-2-ene are different compounds. Multiple double bonds require endings such as -diene or -tetraene and a locant for each bond.
Problem
Name .
- 1.A path through both propyl arms has seven carbons but omits the double bond. It is unsuitable as the alkene parent in this exercise.
- 2.Choose a five-carbon path beginning at CH2, crossing the double bond, then continuing along a propyl arm. The other propyl arm attaches at carbon 2.
- 3.The name is 2-propylpent-1-ene. Its double bond is not sacrificed merely to maximise an unrelated path.
More extended source examples are 2,8-dimethyldeca-3,6-diene, octa-1,3,5,7-tetraene, and 4-ethyl-2,6-dimethyldec-4-ene. These names encode two, four, and one double bonds respectively. Counting all carbons and hydrogens after drawing each structure provides a useful independent check.
Problem
Find sigma and pi counts for the four naming examples.
- 1.For the acyclic dimethyldecadiene, twelve carbons and two double bonds give twenty-two hydrogens. Eleven carbon connections plus twenty-two C–H bonds give 33 sigma bonds and 2 pi bonds.
- 2.Octatetraene has eight carbons and ten hydrogens: 7 + 10 = 17 sigma bonds and 4 pi bonds. The eight-carbon monoalkene 2-propylpent-1-ene has 7 + 16 = 23 sigma bonds and 1 pi bond.
- 3.The fourteen-carbon monoalkene 4-ethyl-2,6-dimethyldec-4-ene has 13 + 28 = 41 sigma bonds and 1 pi bond. Every connected carbon pair contributes one sigma bond, even if the connection is double.
Separate connectivity from configuration
But-1-ene and but-2-ene are position isomers; either compared with 2-methylpropene changes the carbon skeleton and gives chain isomerism. The five acyclic monoalkene structural isomers with five carbons are pent-1-ene, pent-2-ene, 2-methylbut-1-ene, 3-methylbut-1-ene, and 2-methylbut-2-ene. This count concerns connectivity; geometrical forms must be considered separately where possible.
Geometrical isomerism requires two different groups at each double-bonded carbon. In familiar cases having a matching pair across the bond, cis places that pair on the same side and trans on opposite sides. If either carbon carries two identical groups, interchanging them makes no new configuration. More generally, alkenes with four different groups can also have geometrical isomers even when simple cis/trans names are not suitable.
The source cardboard activity uses two connections to resist relative rotation of stiff pieces. It illustrates a constraint, not literal mechanical nails inside a bond. Cis/trans isomers can differ in polarity, solubility, and phase-change temperatures. Cis-but-2-ene has a small dipole moment, about 0.33 D, and boils at about 277 K. In trans-but-2-ene, the dipole contributions cancel; it boils at about 274 K. Symmetry and favourable crystal packing often give trans isomers higher melting points, but that is not an absolute rule for every pair.
Problem
Test , , , and for geometrical isomerism.
- 1.The first fails because one double-bond carbon carries two methyl groups. The second fails because one carbon has two hydrogens and the other has two bromines.
- 2.In the third, the pairs are phenyl/hydrogen and methyl/hydrogen. In the fourth, they are methyl/hydrogen and methyl/chlorine. Both pass the two-different-groups test.
- 3.The same test allows cis/trans forms of 1,2-dichloroethene and 3,4-dimethylhex-3-ene. Draw the matching chlorines or methyl groups on the same side for cis and opposite sides for trans.
Make an alkene by reduction or elimination
Partial alkyne reduction can control alkene geometry. Hydrogen with a poisoned palladium catalyst, called Lindlar catalyst, gives a cis alkene from a suitable internal alkyne. Sodium in liquid ammonia gives the trans alkene. Ordinary active platinum, palladium, or nickel may continue reduction to the alkane; therefore the catalyst and amount of hydrogen matter.
Heating an alkyl halide with alcoholic potassium hydroxide removes halogen from the alpha carbon and hydrogen from a neighbouring beta carbon. This is beta elimination, or dehydrohalogenation. In the source comparison, iodides eliminate more readily than bromides, then chlorides; more substituted alkyl groups generally eliminate more readily under comparable conditions.
A vicinal dihalide has halogens on adjacent carbons. Zinc can remove both as zinc halide, giving an alkene by dehalogenation. An alcohol can instead lose water when heated with concentrated sulphuric acid. This acidic dehydration removes hydroxyl from one carbon and hydrogen from an adjacent carbon. All three routes create a pi bond, but they remove different groups.
Finding a double bond is not sufficient to establish cis–trans isomerism. Propene, including propene obtained by partial reduction of propyne, has a terminal carbon with two identical hydrogens and cannot form this pair of configurations.
Quiz
Which pair illustrates position isomerism?
Which alkene can have cis and trans forms?
Which system selectively gives cis-but-2-ene from but-2-yne?
What is eliminated in dehydrohalogenation?
Why is trans-but-2-ene essentially non-polar?
Practice Problems
- Draw the five acyclic structural monoalkene isomers of C5H10 and identify any geometrical pairs.
- Draw both configurations of 1,2-dichloroethene and 3,4-dimethylhex-3-ene.
- Choose an alkyne and conditions to produce either configuration of but-2-ene.
- Distinguish dehalogenation, dehydrohalogenation, and dehydration using equations.
- Count sigma and pi bonds in octa-1,3,5,7-tetraene.
- Explain why ethane conformers interconvert readily while cis/trans configurations do not interconvert by simple free rotation.
Key Takeaways
• The alkene parent must include the double bond and locate it correctly. • Structural isomers differ in connectivity; geometrical isomers differ in configuration. • Each double-bond carbon needs two different groups for geometrical isomerism. • Selective reduction controls geometry; elimination creates a double bond by removing groups.