CHAPTER 11: Unit 9. Nomenclature & Isomerism: Alkenes and Alkynes

Learning Objectives

1. Name alkenes & alkynes given formulas and write formulas for alkenes & alkynes given names.

Alkenes nomenclatures are similar to those of alkanes and cycloalkanes except the ending is -ene. Also the double bond takes precedence both in selecting and in numbering the man chain or ring. Alkynes with triple bond end with -yne.

The IUPAC rules for previously presented for naming alkanes and cycloalkanes can be used with some modification, to name alkenes and cycloalkenes.

1. replace the suffix -ane with the suffix -ene, which is used to indicate the presence of a carbon-carbon double bond.                                                                                  1-butene
2. Select the parent carbon chain the longest continuous chain of carbon atoms that contains both carbon atoms of the double bond.
3. Number the parent carbon chain beginning at the end nearest the double bond.
4. Give the position of the double bond in the chain as a single number, which is the lower-numbered carbon atom participating in the double bond. This number is placed immediately before the name of the parent carbon chain.                                                                               3-methyl-2-butene
5. Use the suffixes-diene, -triene and so on when more than one double bond is present in the molecule. A separate number must be used to locate each double bond.                                                                                           1,3-butadiene
6. A number is not needed to locate the double bond in substituted cycloalkenes with only one double bond because that bond is assumed to be between carbon 1 & 2.

Nomenclature for Alkynes

The rules for naming alkynes are identical to those used to name alkenes, except the ending -yne is used instead of -ene.

3-methyl-1-butyne

Common names of alkynes are based on acetylene, methylacetylene etc.

Acetylene

Line angle formulas are popularly used in drawing structures of alkens and alkynes( figure below)

Example 1Name each compound.SolutionThe longest chain containing the double bond has five carbon atoms, so the compound is a pentene (rule 1). To give the first carbon atom of the double bond the lowest number (rule 2), we number from the left, so the compound is a 2-pentene. There is a methyl group on the fourth carbon atom (rule 3), so the compound’s name is 4-methyl-2-pentene.The longest chain containing the double bond has four carbon atoms, so the parent compound is a butene (rule 1). (The longest chain overall has five carbon atoms, but it does not contain the double bond, so the parent name is not pentene.) To give the first carbon atom of the double bond the lowest number (rule 2), we number from the left, so the compound is a 1-butene. There is an ethyl group on the second carbon atom (rule 3), so the compound’s name is 2-ethyl-1-butene.Example 2Draw the structure for each compound.3-methyl-2-pentenecyclohexeneSolutionFirst write the parent chain of five carbon atoms: C–C–C–C–C. Then add the double bond between the second and third carbon atoms:Now place the methyl group on the third carbon atom and add enough hydrogen atoms to give each carbon atom a total of four bonds.First, consider what each of the three parts of the name means. Cyclo means a ring compound, hex means 6 carbon atoms, and –ene means a double bond.In Class Practice ProblemDraw the structure for each compound.2-ethyl-1-hexenecyclopenteneHomework ExercisesBriefly identify the important distinctions between a saturated hydrocarbon and an unsaturated hydrocarbon.Briefly identify the important distinctions between an alkene and an alkane.Classify each compound as saturated or unsaturated. Identify each as an alkane, an alkene, or an alkyne.a.b.                         CH3CH2C≡CCH3 c.AnswersUnsaturated hydrocarbons have double or triple bonds and are quite reactive; saturated hydrocarbons have only single bonds and are rather unreactive.An alkene has a double bond; an alkane has single bonds only.a. saturated; alkaneb. unsaturated; alkynec. unsaturated; alkene

PRACTICE PROBLEMS:

Additional Exercises

1. Draw the structure for each compound.a. 2-methyl-2-penteneb. 2,3-dimethyl-1-butenec. cyclohexene
2. Draw the structure for each compound.a. 5-methyl-1-hexeneb. 3-ethyl-2-pentenec. 4-methyl-2-hexene
3. Name each compound according to the IUPAC system.a.b.c. 
4. Name each compound according to the IUPAC system.a.b.c.

Answers

1.

a.

b.

c.

3.

a. 2-methyl-2-pentene

b. 3-methyl-2-heptene

c. 2,5-dimethyl-2-hexene

Constitutional Isomerism in Alkene

Constitutional isomerism is possible for alkenes just like alkanes.

Two different constitutional isomer subtypes are found among the alkene isomers: skeletal isomers and positional isomers.

Skeletal isomers are constitutional isomers that have different carbon-carbon arrangements and that contain identical functional groups if functional groups are present. For example 1-butene and 2-methyl propane.

This is the only type of constitutional isomers possible for alkanes but when functional group present, then there is more than one possible way for locating the functional groups present. Positional isomers are constitutional isomers that have the same carbon-atom arrangement and that have differing locations functional group present. For example 1-butene and 2-butene.

In Class Practice Problem

1. Briefly identify the important differences between an alkene and an alkyne. How are they similar?
2. The alkene (CH₃)₂CHCH₂CH=CH₂ is named 4-methyl-1-pentene. What is the name of (CH₃)₂CHCH₂C≡CH?
3. Do alkynes show cis-trans isomerism? Explain.

Answers

1. Alkenes have double bonds; alkynes have triple bonds. Both undergo addition reactions.
2. 4-methyl-1-pentyne
3. No; a triply bonded carbon atom can form only one other bond. It would have to have two groups attached to show cis-trans isomerism.

HOMEWORK EXERCISE

1. Draw the structure for each compound.a. acetyleneb. 3-methyl-1-hexyne
2. Draw the structure for each compound.a. 4-methyl-2-hexyneb. 3-octyne
3. Name each alkyne.a. CH₃CH₂CH₂C≡CHb. CH₃CH₂CH₂C≡CCH₃

Answers

1. a.         H–C≡C–Hb.

3.

a. 1-pentyne

b. 2-hexyne

Constitutional Isomerism in Alkene

Constitutional isomerism is possible for alkenes just like alkanes.

Two different constitutional isomer subtypes are found among the alkene isomers: skeletal isomers and positional isomers.

Skeletal isomers are constitutional isomers that have different carbon-carbon arrangements and that contain identical functional groups if functional groups are present. For example 1-butene and 2-methyl propane.

This is the only type of constitutional isomers possible for alkanes but when functional group present, then there is more than one possible way for locating the functional groups present. Positional isomers are constitutional isomers that have the same carbon-atom arrangement and that have differing locations functional group present. For example 1-butene and 2-butene.

Cis-Trans Isomers (Geometric Isomers)

Learning Objectives

1. Recognize that alkenes that can exist as cis-trans isomers.
2. Classify isomers as cis or trans.
3. Draw structures for cis-trans isomers given their names.

As noted in earlier in this chapter, there is free rotation about the carbon-to-carbon single bonds (C–C) in alkanes. In contrast, the structure of alkenes requires that the carbon atoms of a double bond and the two atoms bonded to each carbon atom all lie in a single plane, and that each doubly bonded carbon atom lies in the center of a triangle. This part of the molecule’s structure is rigid; rotation about doubly bonded carbon atoms is not possible without rupturing the bond. Look at the two chlorinated hydrocarbons in Figure 1.8 “Rotation about Bonds”.

Figure 1.8 Rotation about Bonds

In 1,2-dichloroethane (a), free rotation about the C–C bond allows the two structures to be interconverted by a twist of one end relative to the other. In 1,2-dichloroethene (b), restricted rotation about the double bond means that the relative positions of substituent groups above or below the double bond are significant.

In 1,2-dichloroethane (part (a) of Figure 1.8 “Rotation about Bonds”), there is free rotation about the C–C bond. The two models shown represent exactly the same molecule; they are not isomers. You can draw structural formulas that look different, but if you bear in mind the possibility of this free rotation about single bonds, you should recognize that these two structures represent the same molecule:

In 1,2-dichloroethene (part (b) of above figure “Rotation about Bonds”), however, restricted rotation about the double bond means that the relative positions of substituent groups above or below the double bond become significant. This leads to a special kind of isomerism. The isomer in which the two chlorine (Cl) atoms lie on the same side of the molecule is called the cis isomer (Latin cis, meaning “on this side”) and is named cis-1,2-dichloroethene. The isomer with the two Cl atoms on opposite sides of the molecule is the trans isomer (Latin trans, meaning “across”) and is named trans-1,2-dichloroethene. These two compounds are cis-trans isomers (or geometric isomers), compounds that have different configurations (groups permanently in different places in space) because of the presence of a rigid structure in their molecule.

Consider the alkene with the condensed structural formula CH₃CH=CHCH₃. We could name it 2-butene, but there are actually two such compounds; the double bond results in cis-trans isomerism (Figure 1.9 “Ball-and-Spring Models of (a) Cis-2-Butene and (b) Trans-2-Butene”).

Figure: Ball-and-Spring Models of (a) Cis-2-Butene and (b) Trans-2-Butene

Cis-trans isomers have different physical, chemical, and physiological properties.

Cis-2-butene has both methyl groups on the same side of the molecule. Trans-2-butene has the methyl groups on opposite sides of the molecule. Their structural formulas are as follows:

Note, however, that the presence of a double bond does not necessarily lead to cis-trans isomerism. We can draw two seemingly different propenes:

However, these two structures are not really different from each other. If you could pick up either molecule from the page and flip it over top to bottom, you would see that the two formulas are identical.

Thus there are two requirements for cis-trans isomerism:

1. Rotation must be restricted in the molecule.
2. There must be two nonidentical groups on each doubly bonded carbon atom.

In these propene structures, the second requirement for cis-trans isomerism is not fulfilled. One of the doubly bonded carbon atoms does have two different groups attached, but the rules require that both carbon atoms have two different groups.

In general, the following statements hold true in cis-trans isomerism:

• Alkenes with a C=CH₂ unit do not exist as cis-trans isomers.
• Alkenes with a C=CR₂ unit, where the two R groups are the same, do not exist as cis-trans isomers.
• Alkenes of the type R–CH=CH–R can exist as cis and trans isomers; cis if the two R groups are on the same side of the carbon-to-carbon double bond, and trans if the two R groups are on opposite sides of the carbon-to-carbon double bond.

Cis-trans isomerism also occurs in cyclic compounds. In ring structures, groups are unable to rotate about any of the ring carbon–carbon bonds. Therefore, groups can be either on the same side of the ring (cis) or on opposite sides of the ring (trans). For our purposes here, we represent all cycloalkanes as planar structures, and we indicate the positions of the groups, either above or below the plane of the ring.

Example

Which compounds can exist as cis-trans (geometric) isomers? Draw them.

1. CHCl=CHBr
2. CH₂=CBrCH₃
3. (CH₃)₂C=CHCH₂CH₃
4. CH₃CH=CHCH₂CH₃

Solution

All four structures have a double bond and thus meet rule 1 for cis-trans isomerism.

1. This compound meets rule 2; it has two nonidentical groups on each carbon atom (H and Cl on one and H and Br on the other). It exists as both cis and trans isomers:
2. This compound has two hydrogen atoms on one of its doubly bonded carbon atoms; it fails rule 2 and does not exist as cis and trans isomers.
3. This compound has two methyl (CH₃) groups on one of its doubly bonded carbon atoms. It fails rule 2 and does not exist as cis and trans isomers.
4. This compound meets rule 2; it has two nonidentical groups on each carbon atom and exists as both cis and trans isomers:

In Class Practice Problem

1. Which compounds can exist as cis-trans isomers? Draw them.

a.              CH₂=CHCH₂CH₂CH₃

b.              CH₃CH=CHCH₂CH₃

c.              CH₃CH₂CH=CHCH₂CH₃

d.

e.

Homework Exercises

1. What are cis-trans (geometric) isomers? What two types of compounds can exhibit cis-trans isomerism?
2. Classify each compound as a cis isomer, a trans isomer, or neither.a.b.c.d.

Answers

1. Cis-trans isomers are compounds that have different configurations (groups permanently in different places in space) because of the presence of a rigid structure in their molecule. Alkenes and cyclic compounds can exhibit cis-trans isomerism.
2. a. transb. cisc. cisd. neither