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CONSTITUTIONAL AND GEOMETRIC ISOMERS

Goals

Identify and define conformer, constitutional isomer, geometric isomer, and isomer.

Differentiate conformer, constitutional isomers, and geometric isomers.

Build organic molecules using molecular models.

Compare and contrast the rotation around carbon-carbon single bond and

carbon-carbon double bonds.

Materials

molecular model kit

Discussion

Organic chemistry is the study of compounds that contain the element carbon.

Organic compounds always contain the element carbon, usually hydrogen, and sometimes

the heteroatoms oxygen, sulfur, nitrogen, phosphorus, or a halogen. Organic compounds

number in the millions, and more are synthesized every day. Within this vast number of

compounds, there are specific groups of atoms called functional groups that give compounds

similar properties. The identification of functional groups allows us to classify organic

compounds according to their structure, to name compounds within each family, and to

predict their physical properties and chemical reactivity.

IUPAC Guidelines for Hydrocarbon Nomenclature

Alkane (suffix –ANE)

1. Find the longest continuous chain of carbon atoms (the parent chain or backbone) and

assign a parent name using the root word describing the number of carbons in that

chain and a suffix describing the functional group found on the chain. For alkanes, the

ending is –ANE.

1 carbon = methane 6 carbons = hexane

2 carbons = ethane 7 carbons = heptane

3 carbons = propane 8 carbons = octane

4 carbons = butane 9 carbons = nonane

5 carbons = pentane 10 carbons = decane

2. Locate any substituents, atoms or groups that are not part of the parent chain but are

attached to the parent chain. Name the substituents. If it is an alkyl substituent (one

having only carbon and hydrogen), use the name of the root that describes the number

of carbons in the substituent, followed by the suffix “-yl” (e.g. a three-carbon

hydrocarbon substituent is named “propyl”).

3. Number the parent chain by starting at one end of the chain and counting to the other

end. Note that starting at one end may give a numbering pattern for the substituents

that is different than if you started counting from the other end of the chain. If there is

a choice of numbering pattern, the correct pattern will be the one that gives the

placement of the substituents the lowest possible set of numbers on the chain.

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4. The completed name for the alkane will first include the names of the substituents (in

alphabetical order of substituent name). Following the names and assignment of the

substituents, end with the parent name.

The location of a substituent on the parent chain is indicated by using the

carbon number of the parent chain to which it is attached, followed by the

name of the substituent, which is separated by a hyphen (e.g. 2-methyl).

If there is more than one of the same substituent on the molecule, use Greek

prefixes in front of the name of the substituent. Note the location of each of

the substituents. (di- = 2, tri- =3, tetra- = 4, penta- = 5, hexa- = 6, hepta- = 7,

octa- = 8, nona- = 9, deca- = 10)

In the case of 2,2-dimethylbutane or 2,3-dimethylbutane, the prefix “di-” and

two carbon numbers are included to describe where they are attached to the

parent chain. Do not use the Greek prefixes for alphabetizing purposes.

The numbers are separated from other numbers by commas and the numbers

are separated from letters in the name by a hyphen.

Alkene (suffix –ENE) and Alkyne (suffix –YNE)

1. Find the longest continuous chain of carbon atoms containing the multiple-bond and

assign a parent name using the root word describing the number of carbons in that

chain and a suffix describing the functional group found on the chain. For alkenes, the

ending is –ENE. For alkynes, the ending is –YNE.

2. Number the parent chain by starting at one end of the carbon chain to give the

multiple-bond the lowest number. If there is more than one possible numbering

pattern, the correct pattern will be the one that gives the parent chain carbons that

have substituents attached the lowest possible numbering pattern for the substituents.

3. Locate any substituents, atoms or groups that are not part of the parent chain but

attached to the parent chain. Name the substituents as for alkanes, above.

Structural Formulas

We first used Lewis structures to represent simple molecular structures. In Lewis

structures, we used lines to depict bonding electrons between atoms and we used dots on

individual atoms to represent the nonbonding electrons. Lewis structures can be simplified to

complete structures (or expanded structural formulas), where all bonds are represented with

lines and the lone pairs are omitted. Drawing a complete structural formula can be quite time

consuming with larger molecules but it is the clearest representation of bonding in a

molecule.

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To further simply the representation, a condensed structure may be used. In the

condensed structure, most single bonds are omitted. The main carbon chain (or backbone) is

written horizontally and the hydrogen atoms that are attached to these carbons are usually

written to the right of the associated carbon atom. Other groups that are branching from the

main chain are added through connecting vertical lines.

The most efficient notation for drawing organic molecules is the skeletal structure (or

line formula). This notation represents the carbon backbone with zigzagging straight lines,

omitting all hydrogen atoms that are connected to the carbon atoms. In the zigzag frame, the

termini (or ends) represent a CH3 group and each vertex corresponds to a carbon atom. When

using the line formula, it is important to remember that carbon is tetravalent (or has four

bonds). This means that you will need to assume there are enough hydrogen atoms around

each carbon to give it four bonds. With the skeletal structure, all non-carbon atoms and the

hydrogen atoms that are directly bonded to them must be drawn in.

Additionally, using the line-dash-wedge notation, we can accurately and efficiently

represent three-dimensional structures of organic molecules. This notation uses lines to

depict bonds that lie in the plane of the paper, dashes for bonds behind the plane of the paper

(or pointing away from you), and wedges to represent bonds in front of the plane of the paper

(or bonds pointing toward you). Every tetrahedral center should have two straight lines

(bonds in the plane of the paper), one wedge (bond in front of the paper) and one set of

dashes (bond in back of the paper). For butane, which has carbon atoms with tetrahedral

molecular geometry, the carbon backbone should be drawn in the plane of the paper using a

zigzag line. The terminal carbon atoms have bonds to three hydrogen atoms and the atoms at

the vertices have bonds to two hydrogen atoms.

or

butane

CH3CH2CH2CH3

butane

butane

2H

3H

3H

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Expanded Structure Condensed Structure Skeletal Structure

2-methylpropane

butenone

propynol

(2-propyn-1-ol)

The molecular formula of a compound gives an account of the atoms that are in a

compound. It’s very much like an inventory list. However, this list really doesn’t tell you

much about how the compound is put together. As we have seen with Lewis structures and

VSEPR theory, the arrangement of the atoms within a molecule will affect the identity of the

compound and its physical and chemical properties. In our previous lab discussions, we

examined numerous ways in which we can represent the structures of a compound. These

representations allowed us to quickly determine the assembly of the atoms within a molecule.

CH3CHCH3

CH3

CH3CHCH2CH2Br

Br

CH3CCH

O

CH2

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In this experiment, we will be examining sets of compounds that have the exact same

chemical formula, but have different chemical and physical properties because of the

differences in the placement of their atoms. Use the molecular model kits to your full

advantage; they will help you to note the changes in shape as you rotate bonds and the

structural changes that occur when you break and rearrange covalent bonds.

Conformations

The first thing that we will examine is conformations. Conformations can be thought

of as snapshots of molecules as they are moving about. A molecule is not stationary,

stagnant, and inflexible. Instead a molecule is constantly in motion – tumbling, spinning, and

frolicking about. Especially when a molecule has carbon-carbon single bonds, there is free

rotation around every one of these bonds. Hence, a single molecule can exist in many

different conformations (or rotational forms). Although different conformations may appear

as though they are different molecules, they are in fact the same molecule. Anytime a

structure can be converted into another structure by merely rotating about one or more of the

carbon-carbon single bonds, the two structures are the same molecule, just in different

conformations (i.e. they are conformers of one another). It is easy to verify this when you

build a model.

Example of two conformations

Constitutional Isomers

Now that we’ve examined a way to verify whether two molecules are the same

through simple rotations around a single bond, let’s examine those situations where this is

not the case. Some molecules have exactly the same chemical formula – the same number

and type of atoms; however, these atoms are joined together differently. The order in which

the atoms are connected is known as the connectivity. Constitutional (or structural) isomers

are molecules with the same chemical formula but a different connectivity. These structural

isomers are distinctly different molecules with different physical properties and a different

chemical name (please note the many uses of the word “different”). When determining

whether molecules are constitutional isomers, first check that the molecular formula for these

molecules is the same. Then verify whether the framework of covalent bonds (the skeletal

structure) is different. Unlike conformers, there is no way to interconvert between two

constitutional isomers by merely rotating bonds; one must break chemical bonds in order to

change the connectivity.

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Examples of two constitutional isomers for molecular formula C4H10

Stereoisomers

Stereoisomers are another category of isomers. Again, we are examining molecules

with the same molecular formulas. In the case of stereoisomers, the atoms are bonded in the

same order – this means that they have the same connectivity – but they differ in the

arrangement of the atoms in space.

Geometric isomers

Cis-trans isomers (or geometric isomers) are a type of stereoisomer. Geometric

isomers arise when there is a structural feature, such as a double bond or a ring, which

restricts the rotation and movement of the molecule. Geometric isomers are compounds with

the same chemical formula and same connectivity but a different three-dimensional spatial

arrangement because of the restricted carbon-carbon bond. Geometric isomers have different

chemical and physical properties.

In this lab, we will be focusing only on cis-trans isomers that occur due to

carbon-carbon double bonds. In an alkene, there is no free rotation around the carbon-carbon

double bond. As a result, the groups of atoms attached to each double-bonded carbon atom

are locked into a particular, fixed arrangement. The fixed three-dimensional orientation of

these groups or atoms creates the possibility for geometric isomers. Cis-trans isomers are

possible with the alkene functional group because the double bond restricts rotation between

two carbon atoms. The carbon atoms involved in the double bond lie in one plane (they are

trigonal planar). The atoms that are connected to them can both project in the same direction

on the plane of the double bond (cis-) or in opposite directions on the plane of the double

bond (trans-). Not all alkenes can form cis-trans isomers. If either one of the carbon atoms

that is involved in the double bond is attached to two groups that are identical, cis-trans

isomerism cannot occur.

Examples of geometric isomers

The difference in the spatial arrangement will be apparent when you compare the models.

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CONSTITUTIONAL AND GEOMETRIC ISOMERS

WORKSHEET

Name_________________________

Date__________________________

Section________________________

Instructor______________________

Conformations of Molecules

1. Construct the molecule 1,2-dichloroethane. Include all of the hydrogen atoms (white

spheres) and chlorine atoms (green polyhedrons). Rotate the bonds between carbon #1

and carbon #1 to find the different conformation of 1,2-dichloroethane.

Draw two different conformations for 1,2-dichloroethane.

In the actual molecule, the relative size of the atoms is larger and the distance between atoms

is smaller than suggested in the ball and stick models. In addition, the atoms repel each other

when they come close. The most stable conformation experiences the least repulsive forces.

Using this information, predict which conformation of 1,2-dichloroethane (your left diagram

or your right diagram) is the most stable and explain why.

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2. Do not include the hydrogen atoms when building the following molecule. Using the

molecular models, construct each of the following carbon (black polyhedrons) chains

diagrammed below:

(a) Describe at least two specific things that these three structures have in common.

(b) Would these structures be identified as conformers or isomers?

Constitutional Isomers

3. The structure given in (2) has the molecular formula C5H12. Using the molecular model

kit, construct all the different constitutional isomers for with the molecular formula

C5H12.

(a) Draw each different compound using the condensed structure and skeletal structure.

Then name each isomer.

(b) How many constitutional isomers are possible for this molecule?

(c) In your own words, describe why these structures are called constitutional isomers.

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4. There are two different molecules with the formula C2H6O. Build a model of each.

(a) Draw the Lewis Structure for each isomer.

(b) Identify the molecular geometry around each central atom.

(c) Draw the molecule using the skeletal structure.

(d) Draw a 3D-sketch for each molecule using the line-dash-wedge system.

(e) Indicate whether each molecule is polar or non-polar.

(f) Identify the dominant intermolecular force.

Lewis Structure of Molecule Line-dash-wedge Drawing/Sketch

Isomer #1

C2H6O VE _________

Molecular geometry at each central atom(s):

Line formula:

Polar or Non-polar?

Dominant intermolecular force:

Isomer #2

C2H6O VE _________

Molecular geometry at each central atom(s):

Line formula:

Polar or Non-polar?

Dominant intermolecular force:

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Cis-Trans Isomers

5. Construct the following molecules and have your instructor approve your structures.

1-butene

2-methyl-2-pentene

2-butene

2,3-dichloro-2-butene

For each molecule:

a. Determine whether the molecule can form cis-trans isomers.

b. Draw each structure using the condensed structure and skeletal structure.

For the condensed structural formula, make each C=C trigonal planar in

appearance.

c. For each molecule that has cis-trans isomers, draw both the cis- and trans-isomers

using the condensed structure and skeletal structure. Name each geometric isomer

as the cis- or trans-isomer.

1-butene

Can 1-butene form cis-trans isomers?

Condensed structure Skeletal structure

2-methyl-2-pentene

Can 2-methyl-2-pentene form cis-trans isomers?

Condensed structure Skeletal structure

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2-butene

Can 2-butene form cis-trans isomers?

Condensed structure Skeletal structure

2,3-dichloro-2-butene

Can 2,3-dichloro-2-butene form cis-trans isomers?

Condensed structure Skeletal structure

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Questions and Problems

1. Use structures A through H to give the best answer(s) for the following questions. If

there is no letter(s) corresponding to the description, write “NONE”. A blank answer will

be marked incorrect. There may be more than one answer for each.

___________ cis-isomer

___________ constitutional isomer of A

___________ trans-isomers

___________ alkene (neither cis nor trans)

___________ cycloalkane

2. Draw two constitutional isomers with the chemical formula C4H10F2.

___________ acyclic alkane

___________ constitutional isomer of B

___________ constitutional isomer of F