Unit One

Day 8: More About Hydrocarbons

As you work through this section, if you find that you need a bit more background material to help you understand the topics at hand, you can consult “Chemistry: The Molecular Science” (5th ed. Moore and Stanitski) Chapter 2-9 and Chapter 6-5, and/or Chapter 4.8-4.10 in the Additional Reading Materials section.

D8.1 Line Structures

Activity 1: Reflection

In your notebook, make a heading for Hydrocarbons (unless you already made one when you worked through section 7.4). Then write a few sentences describing what you know about hydrocarbons. (In section 7.4 you learned that hydrocarbons are the principal substances present in petroleum and natural gas, for example.) As you work through this section, make additional notes to review later.

Hydrocarbon molecules can have lots of carbon atoms. For example, molecules of hydrocarbons in motor oil for automobiles contain between 16 and 20 carbon atoms. Drawing  Lewis structures for such large molecules takes time, so a simpler drawing, called a line structure, is often used. In a line structure (also called a skeletal structure) for a hydrocarbon, only the C–C bonds are shown. Thus, carbon atoms are represented by end of a line or the angle between two lines; hydrogen atoms and C–H bonds are omitted. (The angles between C–C bonds are similar to the zig-zags seen in models of hydrocarbons.) In line structures, atoms other than carbon and hydrogen are represented by their elemental symbols. Figure 1 shows an example line structure; count the number of C and H atoms in the line structure to verify it represents the same molecule as the Lewis structure.

In this figure, a hydrocarbon molecule is shown in three ways. First, an expanded formula shows all individual carbon atoms, hydrogen atoms, and bonds in a branched hydrocarbon molecule. An initial C atom is bonded to three H atoms. The C atom is bonded to another C atom in the chain. This second C atom is bonded to one H atom and another C atom above the chain. The C atom bonded above the second C atom in the chain is bonded to three H atoms. The second C atom in the chain is bonded to a third C atom in the chain. This third C atom is bonded to on H atom and another C atom below the chain. This C atom is bonded to two H atoms and another C atom below the chain. This second C atom below the chain is bonded to three H atoms. The third C atom in the chain is bonded to a fourth C atom in the chain. The fourth C atom is bonded to two H atoms and a fifth C atom. The fifth C atom is bonded to two H atoms and a sixth C atom. The sixth C atom is bonded to three H atoms. Second, a condensed formula shows each carbon atom of the molecule in clusters with the hydrogen atoms bonded to it, leaving C H, C H subscript 2, and C H subscript 3 groups with bonds between them. The structure shows a C H subscript 3 group bonded to a C H group. The C H group is bonded above to a C H subscript 3 group. The C H group is also bonded to another C H group. This C H group is bonded to a C H subscript 2 group below and a C H subscript 3 group below that. This C H group is also bonded to a C H subscript 2 group which is bonded to another C H subscript 2 group. This C H subscript 2 group is bonded to a final C H subscript 2 group. The final structure in the figure is a skeletal structure which includes only line segments arranged to indicate the structure of the molecule.
Figure 1. The molecule 3-ethyl-2-methylhexane can be represented three different ways: a full Lewis structure, a condensed formula structure, and a line structure.

Exercise 1: Interpreting Line Structures

Determine the molecular formula corresponding to each line structure.

D8.2 Cycloalkanes

A cycloalkane has at least one ring of carbon atoms. Comparing a chain of C atoms with a ring shows that an additional C—C bond must be formed. This requires that two hydrogen atoms be removed to make room for the new bond. Consequently such a ring involves one more C—C bond and two less C—H bonds than the corresponding normal alkane, and the general formula for a cycloalkane containing one ring is CnH2n. We refer to this reduction in number of hydrogen atoms as degree of unsaturation, One degree of unsaturation corresponds to having two less hydrogen atoms.

The most common cycloalkane in petroleum is methylcyclohexane:

Note that in the 3D ball-and-stick model of one of the conformations of methylcyclohexane, the ring of six carbon atoms is puckered and does not lie flat in a plane.

D8.3 Alkenes

Unsaturated hydrocarbons that contain one or more double bonds are called alkenes. Carbon atoms linked by a double bond are bound together by two bonds: one σ bond and one π bond. The general formula for alkenes with one double bond is CnH2n. The formula has two less hydrogen atoms than the corresponding alkane, and hence 1 degree of unsaturation. It is possible to have a ring of carbon atoms that contains a double bond. Cyclic alkenes have one degree of unsaturation from each cyclic structure and one from each C=C double bond.

Exercise 2: Degree of Unsaturation

The presence of the double bond is signified by the suffix -ene in the name. Ethene, C2H4, commonly called ethylene, is the simplest alkene. The four carbon atoms in the chain of butene give rise to two different constitutional isomers, 1-butene and 2-butene, where the number indicates the position of the double bond.

Four structural formulas and names are shown. The first shows two red C atoms connected by a red double bond illustrated with two parallel line segments. H atoms are bonded above and below to the left of the left-most C atom. Two more H atoms are similarly bonded to the right of the C atom on the right. Beneath this structure the name ethene and alternate name ethylene are shown. The second shows three C atoms bonded together with a red double bond between the red first and second C atoms moving left to right across the three-carbon chain. H atoms are bonded above and below to the left of the C atom to the left. A single H is bonded above the middle C atom. Three more H atoms are bonded above, below, and to the right of the third C atom. Beneath this structure the name propene and alternate name propylene is shown. The third shows four C atoms bonded together, numbered one through four moving left to right with a red double bond between the red first and second carbon in the chain. H atoms are bonded above and below to the left of the C atom to the left. A single H is bonded above the second C atom. H atoms are bonded above and below the third C atom. Three more H atoms are bonded above, below, and to the right of the fourth C atom. Beneath this structure the name 1 dash butene is shown. The fourth shows four C atoms bonded together, numbered one through four moving left to right with a red double bond between the red second and third C atoms in the chain. H atoms are bonded above, below, and to the left of the left-most C atom. A single H atom is bonded above the second C atom. A single H atom is bonded above the third C atom. Three more H atoms are bonded above, below, and to the right of the fourth C atom. Beneath this structure the name 2 dash butene is shown.

The presence of the π bond make alkenes much more reactive than alkanes because a π bond is usually weaker and more easily disrupted than a σ bond. The double bond is therefore a functional group, a specific structure that has the same chemical behavior in every molecule where it occurs. For example, all alkenes can undergo a characteristic reaction in which the π bond is broken and replaced by two additional σ bonds. This reaction is called an addition reaction. Hydrogen and halogens can add to the double bond in an alkene. Here is an addition reaction involving chlorine:

Activity 2: Analyzing an Addition Reaction

  1. In the equation just above, two bonds are color-coded green in reactants and products. In your notebook, explain the significance of the color coding for chemical bonding and reactivity.
  2. Think about the reaction of chlorine with ethane, CH3CH3. Can this be an addition reaction? Explain why or why not. What bonds need to be broken and formed if chlorine reacts with ethane and how does the reaction differ from the reaction of chlorine with ethene shown above? Do you  expect the reaction of chlorine with ethane to be faster or slower than the reaction with ethene? Why?

D8.4 Geometric Isomers

Because of the additional presence of a π bond, the two carbon atoms in a C=C double bond cannot freely rotate with respect to each other. Consider the π bond in ethene, as shown in Figure 2.  If one carbon rotates by 90° around the internuclear axis, the 2p AO that forms the bonding π MO can no longer overlap with the 2p AO from the other carbon, and the π bond is broken; the σ bond is still present. Breaking the π bond requires an increase in energy (makes the molecule less stable).

Figure 2. The pi MO in ethene is shown. Move the slider to see the two p orbitals that overlap to form the pi bond and how their overlap disappears when the front of the molecule is rotated 90°. Hover over each diagram for a description.

Because rotation requires breaking the π bond, at room temperature there is not free rotation around a C=C double bond—unlike the rotation around C-C single bonds. This gives rise to geometric isomers, molecules that have the same molecular formula and the same atomic connectivity but differ in the orientation of the groups connected to a C=C bond. For example, there are three isomers of butene: 1-butene and 2-butene are constitutional (structural) isomers; for 2-butene, there are two geometric isomers, cis-2-butene and trans-2-butene. As shown in Figure 3, the isomer with both methyl groups on the same side of the double bond is called a cis isomer (both methyl groups are on the bottom side in Figure 3). The one with the methyl groups on opposite sides is called a trans isomer. Geometric isomers have different physical properties, such as boiling point, that make separating them possible.

The figure illustrates three ways to represent isomers of butene. In the first row of the figure, Lewis structural formulas show carbon and hydrogen element symbols and bonds between the atoms. The first structure in this row shows a C atom with a double bond to another C atom which is bonded down and to the right to C H subscript 2 which, in turn, is bonded to C H subscript 3. The first C atom, moving from left to right, has two H atoms bonded to it and the second C atom has one H atom bonded to it. The second structure in the row shows a C atom with a double bond to another C atom. The first C atom is bonded to an H atom up and to the left and C H subscript 3 down and to the left. The second C atom is bonded to an H atom up and to the right and C H subscript 3 down and to the right. Both C H subscript 3 structures appear in red. The third structure shows a C atom with a double bond to another C atom. The first C atom from the left is bonded up to a the left to C H subscript 3 which appears and red. It is also bonded down and to the left to an H atom. The second C atom is bonded up and to the right to an H atom and down and to the left to C H subscript 3 which appears in red. In the second row, ball-and-stick models for the structures are shown. In these representations, single bonds are represented with sticks, double bonds are represented with two parallel sticks, and elements are represented with balls. C atoms are black and H atoms are white in this image. In the third row, space-filling models are shown. In these models, atoms are enlarged and pushed together, without sticks to represent bonds. In the final row, names are provided. The molecule with the double bond between the first and second carbons is named 1 dash butene. The two molecules with the double bond between the second and third carbon atoms is called 2 dash butene. The first model, which has both C H subscript 3 groups beneath the double bond is called the cis isomer. The second which has the C H subscript 3 groups on opposite sides of the double bond is named the trans isomer.
Figure 3. These molecular models show the constitutional (structural) and geometric isomers of butene.

Exercise 3: Isomers

Exercise 4: Addition Reaction Product

Provide the IUPAC name for the alkene reactant and for the saturated product of the addition reaction shown here:

The left side of a reaction and arrow are shown with an empty product side. On the left, C H subscript 3 is bonded down and to the right to C H which has a double bond to another C H. The second C H is bonded up and to the right to C H subscript 2 which is also bonded to C H subscript 3. A plus sign is shown with a C l atom bonded to a C l atom following it. This is also followed by a reaction arrow.

D8.5 Alkynes

An alkyne is a hydrocarbon molecule with one or more triple bonds. Two carbon atoms joined by a triple bond are bound together by one σ bond and two π bonds.  The general formula any alkyne with one triple bond is CnH2n-2.  The alkyne has four less hydrogen atoms than the corresponding alkane, and hence 2 degrees of unsaturation.

The suffix -yne is used to indicate the presence of a triple bond.  The simplest alkyne is ethyne, C2H2, commonly called acetylene. The Lewis structure for ethyne is:

The structural formula and name for ethyne, also known as acetylene, are shown. In red, two C atoms are shown with a triple bond illustrated by three horizontal line segments between them. Shown in black at each end of the structure, a single H atom is bonded.

Chemically, alkynes have similar reactivity as alkenes. Since the C≡C functional group has two π bonds, alkynes can react with twice as much reagent in addition reactions. For example, acetylene can react with bromine in the following reaction:

Exercise 5: Formulas, Multiple Bonds, and Rings

Activity 3: Wrap-up

In your notebook, refer back to what you wrote about Hydrocarbons at the beginning of this day’s work. Update what you wrote and make a list of the main things you learned as you studied the unit. Your list should provide a summary you can use to review later for an exam.

Podia Question

Write a Lewis structure or a line structure for the product of each reaction shown below. (In each case assume there is an excess of bromine.) Compare the products of the second through fifth reactions with the product of the first reaction. Classify each other product as an isomer of, the same as, or neither the same nor an isomer of the first structure.

Two days before the next whole-class session, this Podia question will become live on Podia, where you can submit your answer.

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