Alkanes

We begin our study of organic chemistry with the hydrocarbons, the simplest organic compounds, which are composed of carbon and hydrogen atoms only. As we noted, there are several different kinds of hydrocarbons. They are distinguished by the types of bonding between carbon atoms and the properties that result from that bonding.

The bonding in the hydrogen molecule is fairly straightforward, but the situation is more complicated in organic molecules with tetravalent carbon atoms. Take methane, CH₄, for instance. As we’ve seen, carbon has four valence electrons (2s² 2p²) and forms four bonds. Because carbon uses two kinds of orbitals for bonding, 2s and 2p, we might expect methane to have two kinds of C–H bonds. In fact, though, all four C–H bonds in methane are identical and are spatially oriented toward the corners of a regular tetrahedron as shown in Figure 20.1a. How can we explain this?

An answer was provided in 1931 by Linus Pauling, who showed mathematically how an s orbital and three p orbitals on an atom can combine, or hybridize, to form four equivalent atomic orbitals with tetrahedral orientation. Shown in Figure 20.1b., these tetrahedrally oriented orbitals are called sp³ hybrid orbitals. Note that the superscript 3 in the name sp³ tells how many of each type of atomic orbital combine to form the hybrid, not how many electrons occupy it.

The concept of hybridization explains how carbon forms four equivalent tetrahedral bonds but not why it does so. The shape of the hybrid orbital suggests the answer to why. When an s orbital hybridizes with three p orbitals, the resultant sp³ hybrid orbitals are unsymmetrical about the nucleus. One of the two lobes is larger than the other and can therefore overlap more effectively with an orbital from another atom to form a bond. As a result, sp³ hybrid orbitals form stronger bonds than do unhybridized s or p orbitals.

The asymmetry of sp³ orbitals arises because, as noted previously, the two lobes of a p orbital have different algebraic signs, + and –, in the wave function. Thus, when a p orbital hybridizes with an s orbital, the positive p lobe adds to the s orbital but the negative p lobe subtracts from the s orbital. The resultant hybrid orbital is therefore unsymmetrical about the nucleus and is strongly oriented in one direction.

When each of the four identical sp³ hybrid orbitals of a carbon atom overlaps with the 1s orbital of a hydrogen atom, four identical C–H bonds are formed and methane results. Each C–H bond in methane has a strength of 439 kJ/mol (105 kcal/mol) and a length of 109 pm. Because the four bonds have a specific geometry, we also can define a property called the bond angle. The angle formed by each H–C–H is 109.5°, the so-called tetrahedral angle. Methane thus has the structure shown in Figure 20.1c.

The flat representations shown do not accurately portray bond angles or molecular geometry. Methane has a tetrahedral shape that chemists often portray with wedges indicating bonds coming out toward you and dashed lines indicating bonds that go back away from you. An ordinary solid line indicates a bond in the plane of the page. Recall that the VSEPR theory correctly predicts a tetrahedral shape for the methane molecule (Figure 20.1d.).

Hydrocarbons with only carbon-to-carbon single bonds (C–C) and existing as a continuous chain of carbon atoms also bonded to hydrogen atoms are called alkanes (or saturated hydrocarbons). Saturated, in this case, means that each carbon atom is bonded to four other atoms (hydrogen or carbon)—the most possible; there are no double or triple bonds in the molecules. The word saturated has the same meaning for hydrocarbons as it does for the dietary fats and oils: the molecule has no carbon-to-carbon double bonds (C=C).

The three simplest alkanes—methane (CH₄), ethane (C₂H₆), and propane (C₃H₈) and they are shown in Figure 20.1e.

Methane (CH₄), ethane (C₂H₆), and propane (C₃H₈) are the beginning of a series of compounds in which any two members in a sequence differ by one carbon atom and two hydrogen atoms—namely, a CH₂ unit. The first 10 members of this series are given in Table 20.1a.

Table source: “12.2: Structures and Names of Alkanes” In Basics of GOB Chemistry (Ball et al.), CC BY-NC-SA 4.0.

Consider the series in Figure 20.1f. The sequence starts with C₃H₈, and a CH₂ unit is added in each step moving up the series. Any family of compounds in which adjacent members differ from each other by a definite factor (here a CH₂ group) is called a homologous series. The members of such a series, called homologs, have properties that vary in a regular and predictable manner. The principle of homology gives organization to organic chemistry in much the same way that the periodic table gives organization to inorganic chemistry. Instead of a bewildering array of individual carbon compounds, we can study a few members of a homologous series and from them deduce some of the properties of other compounds in the series.

The principle of homology allows us to write a general formula for alkanes: C_nH_{2n + 2}. Using this formula, we can write a molecular formula for any alkane with a given number of carbon atoms. For example, an alkane with eight carbon atoms has the molecular formula C₈H_{(2 × 8) + 2} = C₈H₁₈.

The video What is Organic Chemistry below gives an introduction to alkanes.

Watch What Is Organic Chemistry?: Crash Course Organic Chemistry #1 – YouTube (10 min)

Physical Properties of Alkanes

Because alkanes have relatively predictable physical properties and undergo relatively few chemical reactions other than combustion, they serve as a basis of comparison for the properties of many other organic compound families. Let’s consider their physical properties first.

Table 20.1b. describes some of the properties of some of the first 10 straight-chain alkanes. Because alkane molecules are nonpolar, they are insoluble in water, which is a polar solvent, but are soluble in nonpolar and slightly polar solvents. Consequently, alkanes themselves are commonly used as solvents for organic substances of low polarity, such as fats, oils, and waxes. Nearly all alkanes have densities less than 1.0 g/mL and are therefore less dense than water (the density of H₂O is 1.00 g/mL at 20°C). These properties explain why oil and grease do not mix with water but rather float on its surface as demonstrated in Figure 20.1g.

Table source: “12.6: Physical Properties of Alkanes” In Basics of GOB Chemistry (Ball et al.), CC BY-NC-SA 4.0.

Table 20.1b. indicates that the first four members of the alkane series are gases at ordinary temperatures. Natural gas is composed chiefly of methane, which has a density of about 0.67 g/L. The density of air is about 1.29 g/L. Because natural gas is less dense than air, it rises. When a natural-gas leak is detected and shut off in a room, the gas can be removed by opening an upper window. On the other hand, bottled gas can be either propane (density 1.88 g/L) or butanes (a mixture of butane and isobutane; density about 2.5 g/L). Both are much heavier than air (density 1.2 g/L). If bottled gas escapes into a building, it collects near the floor. This presents a much more serious fire hazard than a natural-gas leak because it is more difficult to rid the room of the heavier gas.

Also shown in Table 20.1b are the boiling points of the straight-chain alkanes increase with increasing molar mass. This general rule holds true for the straight-chain homologs of all organic compound families. Larger molecules have greater surface areas and consequently interact more strongly; more energy is therefore required to separate them. For a given molar mass, the boiling points of alkanes are relatively low because these nonpolar molecules have only weak dispersion forces to hold them together in the liquid state.

Spotlight on Everyday Chemistry: Characteristics of Alkanes

Hydrocarbons are the simplest organic compounds, but they have interesting physiological effects. These effects depend on the size of the hydrocarbon molecules and where on or in the body they are applied. Alkanes of low molar mass—those with from 1 to approximately 10 or so carbon atoms—are gases or light liquids that act as anesthetics. Inhaling (“sniffing”) these hydrocarbons in gasoline or aerosol propellants for their intoxicating effect is a major health problem that can lead to liver, kidney, or brain damage or to immediate death by asphyxiation by excluding oxygen.

Swallowed, liquid alkanes do little harm while in the stomach. In the lungs, however, they cause “chemical” pneumonia by dissolving fatlike molecules from cell membranes in the tiny air sacs (alveoli). The lungs become unable to expel fluids, just as in pneumonia caused by bacteria or viruses. People who swallow gasoline or other liquid alkane mixtures should not be made to vomit, as this would increase the chance of getting alkanes into the lungs. (There is no home-treatment antidote for gasoline poisoning; call a poison control center.)

Liquid alkanes with approximately 5–16 carbon atoms per molecule wash away natural skin oils and cause drying and chapping of the skin, while heavier liquid alkanes (those with approximately 17 or more carbon atoms per molecule) act as emollients (skin softeners). Such alkane mixtures as mineral oil and petroleum jelly can be applied as a protective film. Water and aqueous solutions such as urine will not dissolve such a film, which explains why petroleum jelly protects a baby’s tender skin from diaper rash.

Attribution & References

Except where otherwise noted, this page is adapted by Adrienne Richards from:

• "12.0: Prelude to Organic Chemistry - Alkanes and Halogenated Hydrocarbons," "12.2: Structures and Names of Alkanes," "12.6: Physical Properties of Alkanes" and "12.7: Chemical Properties of Alkanes" In Basics of General, Organic, and Biological Chemistry (Ball et al.) by David W. Ball, John W. Hill, and Rhonda J. Scott (LibreTexts), licensed under CC BY-NC-SA 4.0. / A derivative of Introduction to Chemistry: General, Organic, and Biological (V. 1.0), CC BY-NC-SA 3.0. / Content from individual pages has been remixed into this one page. Tables and images are attributed individually.
• Except where otherwise noted, content from the 2nd paragraph under Alkanes heading through Figure 20.1 c  is adapted from "1.6 sp3 Hybrid Orbitals and the Structure of Methane" In Organic Chemistry (OpenStax) by John McMurray, licensed under CC BY-NC-SA 4.0. Access for free at Organic Chemistry (OpenStax)
• Figure 20.1a is adapted from "1.4 Development of Chemical Bonding Theory" In Organic Chemistry (OpenStax) by John McMurray, licensed under CC BY-NC-SA 4.0. Access for free at Organic Chemistry (OpenStax)

Types of Formulas to Represent Hydrocarbons

We use several kinds of formulas to describe organic compounds. A molecular formula shows only the kinds and numbers of atoms in a molecule. The general formula for alkanes: C_nH_{2n+ 2}. Using this formula, we can write a molecular formula for any alkane with a given number of carbon atoms. Table 20.2a. shows the molecular formulas of the first 10 straight-chain alkanes.

Table source: “12.2: Structures and Names of Alkanes” In Basics of GOB Chemistry (Ball et al.), CC BY-NC-SA 4.0.

For example in Table 20.2a. above, the molecular formula C₄H₁₀ tells us there are 4 carbon atoms and 10 hydrogen atoms in a molecule, but it doesn’t distinguish between butane and isobutane. A structural formula shows all the carbon and hydrogen atoms and the bonds attaching them. Thus, structural formulas identify the specific isomers (learned in the next section) by showing the order of attachment of the various atoms.

Unfortunately, structural formulas are difficult to type/write and take up a lot of space. Chemists often use condensed structural formulas to alleviate these problems. The condensed structural formulas for the first 10 alkanes are demonstrated in Table 20.2a. Parentheses in condensed structural formulas indicate that the enclosed grouping of atoms is attached to the adjacent carbon atom. The condensed formulas show hydrogen atoms right next to the carbon atoms to which they are attached, as illustrated in Figure 20.2a. for butane:

The ultimate condensed formula is a line structure formula. In this type of structure, carbon atoms are not symbolized with a C, but represented by each end of a line or bend in a line. Hydrogen atoms are not drawn if they are attached to a carbon. Other atoms besides carbon and hydrogen are represented by their elemental symbols. For example in Figure 20.2b., we can represent pentane (CH₃CH₂CH₂CH₂CH₃) and isopentane [(CH₃)₂CHCH₂CH₃] as follows:

In summary, the following Figure 20.2c. demonstrates the structural (expanded), condensed and line structure formulas.

For a summary chart of the formulae used in organic chemistry, infographic 20.2a. demonstrates the different ways to represent organic compounds. Molecular and empirical formulas are explained as the simplest followed by condensed, structural and skeletal (or line) formulae.

 

Example 20.2a

Draw the line structures for these two molecules:

Solution

Each carbon atom is converted into the end of a line or the place where lines intersect. All hydrogen atoms attached to the carbon atoms are left out of the structure (although we still need to recognize they are there):

Example & image source: General Chemistry 1 & 2 , CC BY 4.0.

Exercise 20.2a

Draw the line structures for these two molecules:

Check your answer

Example & image source: General Chemistry 1 & 2 , CC BY 4.0.

 

Example 20.2b

Identify the molecular formula of the molecule represented here:

Solution

There are eight places where lines intersect or end, meaning that there are eight carbon atoms in the molecule. Since we know that carbon atoms tend to make four bonds, each carbon atom will have the number of hydrogen atoms that are required for four bonds. This compound contains 16 hydrogen atoms for a molecular formula of C₈H₁₆.

Location of the hydrogen atoms:

Example & image source: General Chemistry 1 & 2 , CC BY 4.0.

Exercise 20.2b

Identify the molecular formula of the molecule represented here:

Check your AnswerC₉H₂₀

Example & image source: General Chemistry 1 & 2 , CC BY 4.0.

Isomers

Hydrocarbons with the same formula, including alkanes, can have different structures. For example, two alkanes have the formula C₄H₁₀: They are called n-butane and 2-methylpropane (or isobutane), and have the following structural formulas as shown in Figure 20.3a:

The compounds n-butane and 2-methylpropane are structural isomers (the term constitutional isomers is also commonly used). Constitutional isomers have the same molecular formula but different spatial arrangements of the atoms in their molecules. The n-butane molecule contains an unbranched chain, meaning that no carbon atom is bonded to more than two other carbon atoms. We use the term normal, or the prefix n, to refer to a chain of carbon atoms without branching. The compound 2–methylpropane has a branched chain (the carbon atom in the center of the structural formula is bonded to three other carbon atoms).

Recall from section 20.2, Table 20.3a. which shows the molecular formula and condensed structural formulas for the first 10 straight-chain alkanes. Table 20.3a also shows the number of isomers for each alkane. The number of isomers increases rapidly as the number of carbon atoms increases.

Table source: “12.2: Structures and Names of Alkanes” In Basics of GOB Chemistry (Ball et al.), CC BY-NC-SA 4.0.

Identifying isomers from structural formulae is not as easy as it looks. Structural formulae that look different may actually represent the same isomers. For example, the three structures in Figure 20.3b all represent the same molecule, n-butane, and hence are not different isomers. They are identical because each contains an unbranched chain of four carbon atoms.

When identifying isomers, it is useful to trace the carbon backbone with your finger or a pencil and count carbons until you need to lift your hand or pencil to get the another carbon. Try this with each of the above arrangements of four carbons above in Figure 20.3b. Butane has a continuous chain of four carbons no matter how the bonds are rotated – you can connect the carbons in a line without lifting your finger from the page. In a later portion of this chapter, you will learn how to systematically name compounds by counting the number of carbons in the longest continuous chain and identifying any functional groups present.

Adding one more carbon to the butane chain gives pentane, which has the formula, C₅H₁₂. Pentane and its two branched-chain isomers are shown below in Figure 20.3c. The compound at the far left is pentane because it has all five carbon atoms in a continuous chain. The compound in the middle is isopentane; like isobutane, it has a one CH₃ branch off the second carbon atom of the continuous chain. The compound at the far right, discovered after the other two, was named neopentane (from the Greek neos, meaning “new”). Although all three have the same molecular formula, they have different properties, including boiling points: pentane, 36.1°C; isopentane, 27.7°C; and neopentane, 9.5°C. The names isopentane and neopentane are common names for these molecules. As mentioned above, we will learn the systematic rules for naming compounds next.

A summary of isomers is illustrated in infographic 20.3a, which includes structural isomerism and stereoisomerism.

Table 20.3b. Stems That Indicate the Number of Carbon Atoms in Organic Molecules

Stem	Number
meth-	1
eth-	2
prop-	3
but-	4
pent-	5
hex-	6
hept-	7
oct-	8
non-	9
dec-	10

Table source: “12.5: IUPAC Nomenclature” In Basics of GOB Chemistry (Ball et al.), CC BY-NC-SA 4.0.

An alkyl group is a group of atoms that results when one hydrogen atom is removed from an alkane. The group is named by replacing the -ane suffix of the parent hydrocarbon with -yl. For example, the -CH₃ group derived from methane (CH₄) results from subtracting one hydrogen atom and is called a methyl group. The alkyl groups we will use most frequently are listed in Table 20.3c. Alkyl groups are not independent molecules; they are parts of molecules that we consider as a unit to name compounds systematically.

Table 20.3c. Common Alkyl Groups

Parent Alkane		Alkyl Group		Condensed Structural Formula
methane		methyl		CH3–
ethane		ethyl		CH3CH2–
propane		propyl		CH3CH2CH2–
		isopropyl		(CH3)2CH–
butane		butyl*		CH3CH2CH2CH2–

Table 20.3c. note: *There are four butyl groups, two derived from butane and two from isobutane. We will introduce the other three where appropriate. (Image credits:  Introduction to Chemistry: GOB (V. 1.0). , CC BY-NC-SA 3.0).

Simplified IUPAC rules for naming alkanes are as follows (demonstrated in Example 20.3a.).

1. Name alkanes according to the LCC (longest continuous chain) of carbon atoms in the molecule (rather than the total number of carbon atoms). This LCC, considered the parent chain, determines the base name, to which we add the suffix –ane to indicate that the molecule is an alkane.
2. If the hydrocarbon is branched, number the carbon atoms of the LCC. Numbers are assigned in the direction that gives the lowest numbers to the carbon atoms with attached substituents. Hyphens are used to separate numbers from the names of substituents; commas separate numbers from each other. (The LCC need not be written in a straight line; for example, the LCC in the following has five carbon atoms.)
   

   

3. Place the names of the substituent groups in alphabetical order before the name of the parent compound. If the same alkyl group appears more than once, the numbers of all the carbon atoms to which it is attached are expressed. If the same group appears more than once on the same carbon atom, the number of that carbon atom is repeated as many times as the group appears. Moreover, the number of identical groups is indicated by the Greek prefixes di-, tri-, tetra-, and so on. These prefixes are not considered in determining the alphabetical order of the substituents. For example, ethyl is listed before dimethyl; the di- is simply ignored. The last alkyl group named is prefixed to the name of the parent alkane to form one word.

When these rules are followed, every unique compound receives its own exclusive name. The rules enable us to not only name a compound from a given structure but also draw a structure from a given name. The best way to learn how to use the IUPAC system is to put it to work, not just memorize the rules. It’s easier than it looks.

Example 20.3a

Name the molecule whose structure is shown here:

Solution

The longest carbon chain runs horizontally across the page and contains six carbon atoms (this makes the base of the name hexane, but we will also need to incorporate the name of the branch). In this case, we want to number from right to left (as shown by the blue numbers) so the branch is connected to carbon 3 (imagine the numbers from left to right—this would put the branch on carbon 4, violating our rules). The branch attached to position 3 of our chain contains two carbon atoms (numbered in red)—so we take our name for two carbons eth- and attach -yl at the end to signify we are describing a branch. Putting all the pieces together, this molecule is 3-ethylhexane.

Example & image source: General Chemistry 1 & 2 , CC BY 4.0.

Exercise 20.3a

Name the following molecule:

Check Your Answer4-propyloctane

Exercise & image source: General Chemistry 1 & 2 , CC BY 4.0.

For a summary on naming organic compounds, infographic 20.3b looks at the rules for decoding the types of organic compounds and how to name them.

Attribution & References

Except where otherwise noted, this page is adapted by Adrienne Richards from the following sources

• “12.3: The Structure of Organic Molecules – Alkanes and Their Isomers” & “12.6: Naming Alkanes” In Map: Fundamentals of General Organic and Biological Chemistry (McMurry et al.) by LibreTexts, licensed under CC BY-NC-SA 3.0 . / A derivative of 12.5: IUPAC Nomenclature, 12.2: Structures and Names of Alkanes, & 12.3: Branched-Chain Alkanes In Basics of GOB (Ball et al.), CC BY-NC-SA 4.0 a LibreTexts version of Introduction to Chemistry: GOB (v. 1.0), CC BY-NC 3.0
• “18.1 Hydrocarbons” In General Chemistry 1 & 2 by Rice University, a derivative of Chemistry (Open Stax) by Paul Flowers, Klaus Theopold, Richard Langley & William R. Robinson and is licensed under CC BY 4.0. ​Access for free at Chemistry (OpenStax)
• “12.2: Structures and Names of Alkanes” In Basics of General, Organic, and Biological Chemistry (Ball et al.) by David W. Ball, John W. Hill, and Rhonda J. Scott via LibreTexts, licensed under CC BY-NC-SA 4.0. /  A derivative of Introduction to Chemistry: General, Organic, and Biological (V. 1.0), CC BY-NC-SA 3.0.

Cylcoalkanes

The hydrocarbons we have encountered so far have been composed of molecules with open-ended chains of carbon atoms. When a chain contains three or more carbon atoms, the atoms can join to form ring or cyclic structures. The simplest of these cyclic hydrocarbons has the formula C₃H₆. Each carbon atom has two hydrogen atoms attached (Figure 20.4a.) and is called cyclopropane.

Spotlight on Everyday Chemistry: Cyclopropane (C₃H₆) as an Anesthetic

With its boiling point of −33°C, cyclopropane is a gas at room temperature. It is also a potent, quick-acting anesthetic with few undesirable side effects in the body. It is no longer used in surgery, however, because it forms explosive mixtures with air at nearly all concentrations. A line structure of cyclopropane is shown in Figure 20.4b.

The cycloalkanes—cyclic hydrocarbons with only single bonds—are named by adding the prefix cyclo- to the name of the open-chain compound having the same number of carbon atoms as there are in the ring. Thus the name for the cyclic compound C₄H₈ is cyclobutane. The carbon atoms in cyclic compounds can be represented by line structure formulas that result in regular geometric figures. Keep in mind, however, that each corner of the geometric figure represents a carbon atom plus as many hydrogen atoms as needed to give each carbon atom four bonds. Figure 20.4 demonstrates the line structure formulas of both cyclopropane and cyclohexane.

Some cyclic compounds have substituent groups attached. Example 20.4a interprets the name of a cycloalkane with a single substituent group.

Example 20.4a

Draw the structure for each compound.

1. cyclopentane
2. methylcyclobutane

Solution

1. The name cyclopentane indicates a cyclic (cyclo) alkane with five (pent-) carbon atoms. It can be represented as a pentagon.
   

   

• The name methylcyclobutane indicates a cyclic alkane with four (but-) carbon atoms in the cyclic part. It can be represented as a square with a CH₃ group attached.
  

  

Exercise 20.4a

Draw the structure for each compound.

1. cycloheptane
2. ethylcyclohexane

Check Your Answer

b. See the image: Ethylcyclohexane | C8H16 | CID 15504 - PubChem (nih.gov).
Image Source: a. Image by Rhododendronbusch, PDM

The properties of cyclic hydrocarbons are generally quite similar to those of the corresponding open-chain compounds. So cycloalkanes (with the exception of cyclopropane, which has a highly strained ring) act very much like noncyclic alkanes. Cyclic structures containing five or six carbon atoms, such as cyclopentane and cyclohexane, are particularly stable. We will see later that some carbohydrates (sugars) form five- or six-membered rings in solution.

The cyclopropane ring is strained because the C–C–C angles are 60°, and the preferred (tetrahedral) bond angle is 109.5°. (This strain is readily evident when you try to build a ball-and-stick model of cyclopropane; see Figure 20.4a.) Cyclopentane and cyclohexane rings have little strain because the C–C–C angles are near the preferred angles. Cyclohexane is shown in Figure 20.4b.

Substituted Cycloalkanes

We’ll see numerous instances in future chapters where the chemistry of a given functional group is affected by being in a ring rather than an open chain. Because cyclic molecules are encountered in most pharmaceuticals and in all classes of biomolecules, including proteins, lipids, carbohydrates, and nucleic acids, it’s important to understand the behaviour of cyclic structures.

Although we’ve only discussed open-chain compounds up to now, most organic compounds contain rings of carbon atoms. Chrysanthemic acid in Figure 20.4c, for instance, whose esters occur naturally as the active insecticidal constituents of chrysanthemum flowers, contains a three-membered (cyclopropane) ring.

Prostaglandins, potent hormones that control an extraordinary variety of physiological functions in humans, contain a five-membered (cyclopentane) ring. An example of a prostaglandin is in Figure 20.4d.

Steroids, such as cortisone, contain four rings joined together—three six-membered (cyclohexane) and one five-membered.

Substituted cycloalkanes are named by rules similar to those we saw for open-chain alkanes. For most compounds, there are only two steps.

1. Find the parent. Count the number of carbon atoms in the ring and the number in the largest substituent. If the number of carbon atoms in the ring is equal to or greater than the number in the substituent, the compound is named as an alkyl-substituted cycloalkane. If the number of carbon atoms in the largest substituent is greater than the number in the ring, the compound is named as a cycloalkyl-substituted alkane. Refer to Figure 20.4f. for examples of substituted cycloalkanes.
   

   

2. Number the substituents and write the name. For an alkyl- or halo-substituted cycloalkane, choose a point of attachment as carbon 1 and number the substituents on the ring so that the second substituent has as low a number as possible. If ambiguity still exists, number so that the third or fourth substituent has as low a number as possible, until a point of difference is found as shown in the example within Figure 20.4g.

(a) When two or more different alkyl groups are present that could potentially take the same numbers, number them by alphabetical priority, ignoring numerical prefixes such as di- and tri-. An example is demonstrated in Figure 20.4h.

(b) If halogens are present, treat them just like alkyl groups as shown in Figure 20.4i.

Some additional examples that follow the IUPAC nomenclature system for cycloalkanes are demonstrated in Figure 20.4j.

Cis-Trans Isomerism of Cycloalkanes

In many respects, the chemistry of cycloalkanes is like that of open-chain alkanes: both are nonpolar and fairly inert. There are, however, some important differences. One difference is that cycloalkanes are less flexible than open-chain alkanes. In contrast with the relatively free rotation around single bonds in open-chain alkanes, there is much less freedom in cycloalkanes. Cyclopropane, for example, must be a rigid, planar molecule because three points (the carbon atoms) define a plane. No bond rotation can take place around a cyclopropane carbon–carbon bond without breaking open the ring as shown in Figure 20.4k.

 

Larger cycloalkanes have increasing rotational freedom, and very large rings (C₂₅ and up) are so floppy that they are nearly indistinguishable from open-chain alkanes. The common ring sizes (C₃–C₇), however, are severely restricted in their molecular motions.

Because of their cyclic structures, cycloalkanes have two faces when viewed edge-on, a “top” face and a “bottom” face. As a result, isomerism is possible in substituted cycloalkanes. For example, there are two different 1,2-dimethylcyclopropane isomers, one with the two methyl groups on the same face of the ring and one with the methyl groups on opposite faces as shown in Figure 20.4l. Both isomers are stable compounds, and neither can be converted into the other without breaking and reforming chemical bonds.

Unlike the constitutional isomers butane and isobutane, which have their atoms connected in a different order, the two 1,2-dimethylcyclopropanes have the same order of connections but differ in the spatial orientation of the atoms. Such compounds, with atoms connected in the same order but differing in three-dimensional orientation, are called stereochemical isomers, or stereoisomers. As we saw previously, the term stereochemistry is used generally to refer to the three-dimensional aspects of structure and reactivity. Figure 20.4m. demonstrates the difference between the types of isomers.

The 1,2-dimethylcyclopropanes are members of a subclass of stereoisomers called cis–trans isomers as shown in Figure 20.4n. The prefixes cis– (Latin “on the same side”) and trans– (Latin “across”) are used to distinguish between them. Cis–trans isomerism is a common occurrence in substituted cycloalkanes and in many cyclic biological molecules.

Example 20.4b

Naming Cycloalkanes

Name the following substances, including the cis– or trans– prefix:

Strategy

In these views, the ring is roughly in the plane of the page, a wedged bond protrudes out of the page, and a dashed bond recedes into the page. Two substituents are cis if they are both out of or both into the page, and they are trans if one is out of and one is into the page.

Solution

(a) trans-1,3-Dimethylcyclopentane

(b) cis-1,2-Dichlorocyclohexane

Spotlight on Everyday Chemistry: Carp and The Earthy Flavour Geosmin

Serving carp is common at Christmas in Europe. Thanks to the cycloalkane compound geosmin (Figure 20.4o), it gives the carp an earthly flavour. For more information see the infographic Compound Interest: The Chemistry Advent Calendar 2023 (compoundchem.com).

Attribution & References

Except where otherwise noted, this page is adapted by Adrienne Richards from

• “12.9: Cycloalkanes” In Basics of General, Organic, and Biological Chemistry (Ball et al.) by David W. Ball, John W. Hill, and Rhonda J. Scott via LibreTexts, licensed under CC BY-NC-SA 4.0. / A derivative of Introduction to Chemistry: General, Organic, and Biological (V. 1.0), CC BY-NC-SA 3.0.
• “Ch. 4 Why This Chapter?“, “4.1 Naming Cycloalkanes” and “4.2 Cis–Trans Isomerism in Cycloalkanes” In Organic Chemistry (OpenStax) licensed under CC BY-NC-SA 4.0. Access for free at Organic Chemistry (OpenStax)

Halogenated Alkanes

Many organic compounds are closely related to the alkanes. As we noted previously, alkanes react with halogens to produce halogenated hydrocarbons, the simplest of which have a single halogen atom substituted for a hydrogen atom of the alkane. Even more closely related are the cycloalkanes, compounds in which the carbon atoms are joined in a ring, or cyclic fashion.

Halogens are found in column 7A of the periodic table and include Fluorine, Chlorine, Bromine and Iodine.  Refer to Appendix A: Key Element Information for more details about halogens.

Alkyl halides are encountered less frequently than their oxygen-containing relatives and are not often involved in the biochemical pathways of terrestrial organisms, but some of the kinds of reactions they undergo—nucleophilic substitutions and eliminations—are encountered frequently. Thus, alkyl halide chemistry is a relatively simple model for many mechanistically similar but structurally more complex reactions found in biomolecules.

Now that we’ve covered the chemistry of hydrocarbons, it’s time to start looking at more complex substances that contain elements in addition to C and H. We’ll begin by discussing the chemistry of organohalides, compounds that contain one or more halogen atoms.

Halogen-substituted organic compounds are widespread in nature, and more than 5000 organohalides have been found in algae and various other marine organisms. Chloromethane, for instance, is released in large amounts by ocean kelp, as well as by forest fires and volcanoes. Halogen-containing compounds also have an array of industrial applications, including their use as solvents, inhaled anesthetics in medicine, refrigerants, and pesticides as shown in Figure 20.5b.

Still other halo-substituted compounds are used as medicines and food additives. The nonnutritive sweetener sucralose, marketed as Splenda, contains three chlorine atoms, for instance. Sucralose is about 600 times as sweet as sucrose, so only 1 mg is equivalent to an entire teaspoon of table sugar. Refer to Figure 20.5c. for the structural formula of sucralose.

A large variety of organohalides are known. The halogen might be bonded to an alkynyl group (C≡C−X), a vinylic group (C═C–XC═C–X">C═C–X), an aromatic ring (Ar−X), or an alkyl group.

The reactions of alkanes with halogens produce halogenated hydrocarbons, compounds in which one or more hydrogen atoms of a hydrocarbon have been replaced by halogen atoms (F, Cl, Br and I):

The replacement of only one hydrogen atom gives an alkyl halide (or haloalkane) as shown in Figure 20.5d. above.

Indigenous Perspectives: The Impact of Fluorocarbons on Inuit People of Canada

The long-chained fluorocarbon ending with a -SO₃H group is known as perfluorooctanesulfonic acid also known as PFOS (Figure 20.5e.). PFOS and their relatives contain a long fluorocarbon backbone that is extremely resistant to decomposition. In addition, the PFOS family of compounds are volatile and can spread throughout the Earth’s surface. Research has shown that through global distillation, these PFOS have accumulated in the Arctic. This is a problematic health concern for the Inuit in the north. Several food sources that are part of the Inuit diet have been tested and contain PFOS. Exposure to PFOS have numerous health implications such as cancer, endocrine delays and others as described in the article below. There is no end in sight from exposure to these harmful fluorocarbons as they are expected to remain in the arctic for hundreds and possibly thousands of years (Anderson & Rayner-Canham, 2022).

For more detailed information regarding the fluorocarbons and their impact on the Inuit see the following link: PFOS | Chem 13 News Magazine | University of Waterloo (uwaterloo.ca).

Naming Haloalkanes

Although commonly called alkyl halides, halogen-substituted alkanes are named systematically as haloalkanes, treating the halogen as a substituent on a parent alkane chain. There are three steps:

1. Find the longest chain, and name it as the parent. If a double or triple bond is present, the parent chain must contain it.
2. Number the carbons of the parent chain beginning at the end nearer the first substituent, whether alkyl or halo. Assign each substituent a number according to its position on the chain. Figure 20.5f. provides two examples on how to number the carbon chain with either an alkyl or halo first appearing in the chain.
   

   

3. If different halogens are present, number each one and list them in alphabetical order when writing the name as shown in Figure 20.5g.
   

   

4. If the parent chain can be properly numbered from either end by step 2, begin at the end nearer the substituent that has alphabetical precedence. Figure 20.5h. shows an example of alphabetical precedence.
   

   

5. In addition to their systematic names, many simple alkyl halides are also named by identifying first the alkyl group and then the halogen. For example in Figure 20.5i., CH₃I can be called either iodomethane or methyl iodide. Such names are well entrenched in the chemical literature and in daily usage, but they won’t be used in this book.

In summary, the common names of alkyl halides consist of two parts: the name of the alkyl group plus the stem of the name of the halogen, with the ending -ide. The IUPAC system uses the name of the parent alkane with a prefix indicating the halogen substituents, preceded by number indicating the substituent’s location. The prefixes are fluoro-, chloro-, bromo-, and iodo-. Thus CH₃CH₂Cl has the common name ethyl chloride and the IUPAC name chloroethane. Alkyl halides with simple alkyl groups (one to four carbon atoms) are often called by common names. Those with a larger number of carbon atoms are usually given IUPAC names.

Example 20.5d

Give the IUPAC name for each compound.

 

Solution

1. The parent alkane has five carbon atoms in the longest continuous chain; it is pentane. A bromo (Br) group is attached to the second carbon atom of the chain. The IUPAC name is 2-bromopentane.
2. The parent alkane is hexane. Methyl (CH₃₎ and bromo (Br) groups are attached to the second and fourth carbon atoms, respectively. Listing the substituents in alphabetical order gives the name 4-bromo-2-methylhexane.

Exercise 20.5b

Give the IUPAC name for each compound.

 

 

Check Your Answer

1. 2-chloro, 3-methylbutane
2. 1-bromo, 2-chloro, 4-methylpentane

For an overview on how to name alkyl halides, watch the video Naming Alkyl Halides – IUPAC Nomenclature below.

Watch Naming Alkyl Halides – IUPAC Nomenclature – YouTube (12 min)

One or more interactive elements has been excluded from this version of the text. You can view them online here: https://ecampusontario.pressbooks.pub/orgbiochemsupplement/?p=628#oembed-3

Video source: The Organic Chemistry Tutor. (2018, April 21). Naming Alkyl Halides – IUPAC Nomenclature – YouTube [Video]. YouTube.

A wide variety of interesting and often useful compounds have one or more halogen atoms per molecule. For example, methane (CH₄) can react with chlorine (Cl₂), replacing one, two, three, or all four hydrogen atoms with Cl atoms. Several halogenated products derived from methane and ethane (CH₃CH₃) are listed in Table 20.5a., along with some of their uses.

Table 20.5a. Some Halogenated Hydrocarbons

Formula	Derived from	Common Name	IUPAC Name	Some Important Uses
CH3Cl	CH4	methyl chloride	chloromethane	refrigerant; the manufacture of silicones, methyl cellulose, and synthetic rubber
CH2Cl2	CH4	methylene chloride	dichloromethane	laboratory and industrial solvent
CHCl3	CH4	chloroform	trichloromethane	industrial solvent
CCl4	CH4	carbon tetrachloride	tetrachloromethane	dry-cleaning solvent and fire extinguishers (but no longer recommended for use)
CBrF3	CH4	halon-1301	bromotrifluoromethane	fire extinguisher systems
CCl3F	CH4	chlorofluorocarbon-11 (CFC-11)	trichlorofluoromethane	foaming plastics
CCl2F2	CH4	chlorofluorocarbon-12 (CFC-12)	dichlorodifluoromethane	refrigerant
CH3CH2Cl	CH3CH3	ethyl chloride	chloroethane	local anesthetic
ClCH2CH2Cl	CH3CH3	ethylene dichloride	1,2-dichloroethane	solvent for rubber
CCl3CH3	CH3CH3	methylchloroform	1,1,1-trichloroethane	solvent for cleaning computer chips and molds for shaping plastics

Table source: “12.8: Halogenated Hydrocarbons” In Basics of GOB Chemistry (Ball et al.), CC BY-NC-SA 4.0.

Alkyl halides can be both harmful and beneficial to your health. In the next examples, To Your Health and Spotlight in Chemistry you can learn more about these interesting compounds.

Spotlight on Everyday Chemistry: Halogenated Hydrocarbons Risks and Benefits

Once widely used in consumer products, many chlorinated hydrocarbons are suspected carcinogens (cancer-causing substances) and also are known to cause severe liver damage. An example is carbon tetrachloride (CCl₄), once used as a dry-cleaning solvent and in fire extinguishers but no longer recommended for either use. Even in small amounts, its vapor can cause serious illness if exposure is prolonged. Moreover, it reacts with water at high temperatures to form deadly phosgene (COCl₂) gas, which makes the use of CCl₄ in fire extinguishers particularly dangerous.

Ethyl chloride, in contrast, is used as an external local anesthetic. When sprayed on the skin, it evaporates quickly, cooling the area enough to make it insensitive to pain. It can also be used as an emergency general anesthetic.

Bromine-containing compounds are widely used in fire extinguishers and as fire retardants on clothing and other materials. Because they too are toxic and have adverse effects on the environment, scientists are engaged in designing safer substitutes for them, as for many other halogenated compounds.

Halomon (IUPAC name (3S,6R)-6-Bromo-3-(bromomethyl)-2,3,7-trichloro-7-methyloct-1-ene) is a pentahalogenated alkene. Halomon as shown in Figure 20.5j., has been isolated from the red alga Portieria hornemannii and found to have anticancer activity against several human tumor cell lines.

Links to Enhanced Learning

For more examples, visit Haloalkanes and Names and Properties of Alkyl Halides by LibreTextsChemistry.

Attribution & References

Except where otherwise noted, this page is adapted by Adrienne Richards from:

• “18.1 Hydrocarbons” In General Chemistry 1 & 2 by Rice University, a derivative of Chemistry (Open Stax) by Paul Flowers, Klaus Theopold, Richard Langley & William R. Robinson and is licensed under CC BY 4.0. ​Access for free at Chemistry (OpenStax)
• “12.8: Halogenated Hydrocarbons” In Basics of General, Organic, and Biological Chemistry (Ball et al.) by David W. Ball, John W. Hill, and Rhonda J. Scott via LibreTexts, licensed under CC BY-NC-SA 4.0. / A derivative of Introduction to Chemistry: GOB (v. 1.0), CC BY-NC 3.0.
• “10.1 Names and Structures of Alkyl Halides” and “Chemistry Matters—Naturally Occurring Organohalides” In Organic Chemistry (OpenStax) by John McMurray, CC BY-NC-SA 4.0. Access for free at Organic Chemistry (OpenStax)

References cited in-text

Anderson, C. C., & Rayner-Canham, G. (2022, Fall). PFOS: The newest Arctic pollutant. Chem 13 News Magazine.

Reactions of Alkanes

Alkane molecules are nonpolar and therefore generally do not react with ionic compounds such as most laboratory acids, bases, oxidizing agents, or reducing agents. Consider butane as an example in Figure 20.6a.

Neither positive ions nor negative ions are attracted to a nonpolar molecule. In fact, the alkanes undergo so few reactions that they are sometimes called paraffins, from the Latin parum affinis, meaning “little affinity.”

However, heat or light can initiate the breaking of C–H or C–C single bonds in reactions called combustion and substitution.

Watch Alkanes: Crash Course Organic Chemistry #6 – YouTube (12 min)

Recall that organic functional groups can be converted into other functional groups through reactions.  A map of some of the more common reactions to convert functional groups can be found in Section 19.6 – General Reactions of Carbonin Infographic 19.6a.

Combustion

Alkanes are relatively stable molecules, but heat or light will activate reactions that involve the breaking of C–H or C–C single bonds. Combustion is one such reaction:

Alkanes burn in the presence of oxygen, a highly exothermic oxidation-reduction reaction that produces carbon dioxide and water. As a consequence, alkanes are excellent fuels. For example, methane, CH₄, is the principal component of natural gas. Butane, C₄H₁₀, used in camping stoves and lighters is an alkane. Gasoline is a liquid mixture of continuous- and branched-chain alkanes, each containing from five to nine carbon atoms, plus various additives to improve its performance as a fuel. Kerosene, diesel oil, and fuel oil are primarily mixtures of alkanes with higher molecular masses. The main source of these liquid alkane fuels is crude oil, a complex mixture that is separated by fractional distillation. Fractional distillation takes advantage of differences in the boiling points of the components of the mixture (see Figure 20.6b.). You may recall that boiling point is a function of intermolecular interactions.

If the reactants of combustion reactions are adequately mixed, and there is sufficient oxygen, the only products are carbon dioxide (CO₂), water (H₂O), and energy—heat for cooking foods, heating homes, and drying clothes. Because conditions are rarely ideal, other unwanted by-products are frequently formed. When the oxygen supply is limited, carbon monoxide (CO) is a by-product:

\[2CH_4 + 3O_2 \rightarrow​ 2CO + 4H_2O\label{2} \]

This reaction is responsible for dozens of deaths each year from unventilated or improperly adjusted gas heaters. (Similar reactions with similar results occur with kerosene heaters.)

Spotlight on Everyday Chemistry: Fuel

We use fuel (or petrol in the UK) in our vehicles everyday.  Fuel comes from fossil fuels. Read more about how fuel works in Infographic 20.6a.

Substitution

In a substitution reaction (ex. halogenation), another typical reaction of alkanes, one or more of the alkane’s hydrogen atoms is replaced with a different atom or group of atoms. No carbon-carbon bonds are broken in these reactions, and the hybridization of the carbon atoms does not change. For example in Figure 20.6c., the reaction between ethane and molecular chlorine demonstrates a substitution reaction.

The C–Cl portion of the chloroethane molecule is an example of a functional group, the part or moiety of a molecule that imparts a specific chemical reactivity. The types of functional groups present in an organic molecule are major determinants of its chemical properties and are used as a means of classifying organic compounds as detailed in the remaining sections of this chapter.

A wide variety of interesting and often useful compounds have one or more halogen atoms per molecule. For example, methane (CH₄) can react with chlorine (Cl₂), replacing one, two, three, or all four hydrogen atoms with Cl atoms. With more chlorine, a mixture of products is obtained: CH₃Cl, CH₂Cl₂, CHCl₃, and CCl₄. Fluorine (F), the lightest halogen, combines explosively with most hydrocarbons. Iodine (I) is relatively unreactive. Fluorinated and iodinated alkanes are produced by indirect methods.

Several halogenated products derived from methane and ethane (CH₃CH₃) are listed in Table 20.6a., along with some of their uses.

Table 20.6a. Some Halogenated Hydrocarbons

Formula	Common Name	IUPAC Name	Some Important Uses
Derived from CH4
CH3Cl	methyl chloride	chloromethane	refrigerant; the manufacture of silicones, methyl cellulose, and synthetic rubber
CH2Cl2	methylene chloride	dichloromethane	laboratory and industrial solvent
CHCl3	chloroform	trichloromethane	industrial solvent
CCl4	carbon tetrachloride	tetrachloromethane	dry-cleaning solvent and fire extinguishers (but no longer recommended for use)
CBrF3	halon-1301	bromotrifluoromethane	fire extinguisher systems
CCl3F	chlorofluorocarbon-11 (CFC-11)	trichlorofluoromethane	foaming plastics
CCl2F2	chlorofluorocarbon-12 (CFC-12)	dichlorodifluoromethane	refrigerant
Derived from CH3CH3
CH3CH2Cl	ethyl chloride	chloroethane	local anesthetic
ClCH2CH2Cl	ethylene dichloride	1,2-dichloroethane	solvent for rubber
CCl3CH3	methylchloroform	1,1,1-trichloroethane	solvent for cleaning computer chips and molds for shaping plastics

Table source: Map: Fundamentals of GOB Chemistry (McMurry et al.), CC BY-NC-SA 3.0.

Spotlight on Everyday Chemistry: Chlorofluorocarbons (CFC’s), The Ozone Layer and Susan Solomon

Chlorofluorocarbons and the Ozone Layer

Alkanes substituted with both fluorine (F) and chlorine (Cl) atoms have been used as the dispersing gases in aerosol cans, as foaming agents for plastics, and as refrigerants. Two of the best known of these chlorofluorocarbons (CFCs) are listed in Table 20.6a.

Chlorofluorocarbons contribute to the greenhouse effect in the lower atmosphere. They also diffuse into the stratosphere, where they are broken down by ultraviolet (UV) radiation to release Cl atoms. These in turn break down the ozone (O₃) molecules that protect Earth from harmful UV radiation as shown in Figure 20.6d. Worldwide action has reduced the use of CFCs and related compounds. The CFCs and other Cl- or bromine (Br)-containing ozone-destroying compounds are being replaced with more benign substances. Hydrofluorocarbons (HFCs), such as CH₂FCF₃, which have no Cl or Br to form radicals, are one alternative. Another is hydrochlorofluorocarbons (HCFCs), such as CHCl₂CF₃. HCFC molecules break down more readily in the troposphere, and fewer ozone-destroying molecules reach the stratosphere.

Thanks to Susan Solomon as described in infographic 20.6b, she confirmed that ozone could react with CFC’s in the stratosphere breaking it down..

For more information on reactions of alkanes, watch Radical Reactions & Hammond’s Postulate below.

Watch Radical Reactions & Hammond’s Postulate: Crash Course Organic Chemistry #19 – YouTube (12 min)

Cracking (Elimination) – Making Alkenes

Ethylene and propylene, the simplest alkenes, are the two most important organic chemicals produced industrially. Approximately 220 million tons of ethylene and 138 million tons of propylene are produced worldwide each year for use in the synthesis of polyethylene, polypropylene, ethylene glycol, acetic acid, acetaldehyde, and a host of other substances (Figure 20.6e.).

Ethylene, propylene, and butene are synthesized from (C₂–C₈) alkanes by a process called steam cracking at temperatures up to 900 °C. This process is shown in Figure 20.6f. below.

The cracking process is complex, although it undoubtedly involves radical reactions. The high-temperature reaction conditions cause spontaneous breaking of C−C and C−H bonds, with the resultant formation of smaller fragments. We might imagine, for instance, that a molecule of butane splits into two ethyl radicals, each of which then loses a hydrogen atom to generate two molecules of ethylene as demonstrated in Figure 20.6g.

Other Alkane Reactions

Two additional alkane reactions include dehydrogenation and isomerization. Dehydrogenation is an elimination reaction where a hydrogen is lost from an alkane to create an alkene under high temperatures. The results are a by-product of hydrogen gas and an alkene. This reaction is unpredictable as the location of the carbon-carbon double bond is random. The dehydrogenation processis used in the production of motor fuels and petrochemicals (Hein et al., 2013, p. 464). The dehydrogenation reaction of butane is shown in Figure 20.6h.

Isomerization occurs when there is a rearrangement of the molecular structure under heat, pressure and exposure to a catalyst. Again, this process is used in the production of motor fuels and petrochemicals (Hein et al., 2013, p. 464). For example, in Figure 20.6i., the isomerization of butane is demonstrated.

For more details on reactions involving alkanes refer to themap of some of the more common reactions to convert functional groups in Section 19.6 – General Reactions of Carbonin Infographic 19.6a.

Attribution & References

Except where otherwise noted, this page is adapted by Adrienne Richards from

• “12.8: Reactions of Alkanes” In Map: Fundamentals of General Organic and Biological Chemistry (McMurry et al.) by LibreTexts, licensed under CC BY-NC-SA 3.0 . / A derivative of
  • “12.7: Chemical Properties of Alkanes“, “12.8: Halogenated Hydrocarbons” In Basics of GOB (Ball et al.), CC BY-NC-SA 4.0 a LibreTexts version of Introduction to Chemistry: GOB (v. 1.0), CC BY-NC 3.0
  • “20.1: Hydrocarbons” In Chemistry 1e (OpenStax) a LibreText version of Chemistry / Chemistry 2e (OpenStax) by by Paul Flowers, Klaus Theopold, Richard Langley & William R. Robinson is licensed CC BY 4.0. Access for free at Chemistry 2e (Open Stax)
• Combustion section also includes content from “18.1 Hydrocarbons” In General Chemistry 1 & 2 by Rice University. / A derivative of Chemistry (Open Stax) by Paul Flowers, Klaus Theopold, Richard Langley & William R. Robinson and is licensed under CC BY 4.0. ​Access for free at Chemistry (OpenStax)
• “Cracking – Making Alkenes” is adapted from “7.1 Industrial Preparation and Use of Alkenes” In Organic Chemistry (OpenStax) by John McMurray, licensed under CC BY-NC-SA 4.0. Access for free at Organic Chemistry (OpenStax)

References cited in-text

Hein, M., Pattison, S., Arena, S., & Best, L. (2013). Introduction to general, organic, and biochemistry (11th ed.). John Wiley & Sons, Inc.

20.1 Characteristics of Alkanes

Organic chemistry is the chemistry of carbon compounds, and inorganic chemistry is the chemistry of all the other elements. Carbon atoms can form stable covalent bonds with other carbon atoms and with atoms of other elements, and this property allows the formation the tens of millions of organic compounds. Hydrocarbons contain only hydrogen and carbon atoms.

Hydrocarbons in which each carbon atom is bonded to four other atoms are called alkanes or saturated hydrocarbons. They have the general formula C_nH_{2n + 2}. Any given alkane differs from the next one in a series by a CH₂ unit. Any family of compounds in which adjacent members differ from each other by a definite factor is called a homologous series.

20.2 Isomers and Nomenclature of Alkanes

Carbon atoms in alkanes can form straight chains or branched chains. Two or more compounds having the same molecular formula but different structural formulas are isomers of each other. There are no isomeric forms for the three smallest alkanes; beginning with C₄H₁₀, all other alkanes have isomeric forms.

A structural formula shows all the carbon and hydrogen atoms and how they are attached to one another. A condensed structural formula shows the hydrogen atoms right next to the carbon atoms to which they are attached. A line formula is a formula in which carbon atoms are implied at the corners and ends of lines. Each carbon atom is understood to be attached to enough hydrogen atoms to give each carbon atom four bonds.

The IUPAC System of Nomenclature provides rules for naming organic compounds. An alkyl group is a unit formed by removing one hydrogen atom from an alkane.

The physical properties of alkanes reflect the fact that alkane molecules are nonpolar. Alkanes are insoluble in water and less dense than water.

20.3 Cycloalkanes

Cycloalkanes are hydrocarbons whose molecules are closed rings rather than straight or branched chains. A cyclic hydrocarbon is a hydrocarbon with a ring of carbon atoms.

20.4 Halogenated Alkanes

Alkanes react with halogens by substituting one or more halogen atoms for hydrogen atoms to form halogenated hydrocarbons. An alkyl halide (haloalkane) is a compound resulting from the replacement of a hydrogen atom of an alkane with a halogen atom.

20.5 Reactions with Alkanes

Alkanes are generally unreactive toward laboratory acids, bases, oxidizing agents, and reducing agents. They do burn (undergo combustion reactions) and undergo substitution reactions with halogens to create halogenated hydrocarbons.

Attribution & References

20.1 Characteristics of Alkanes

20.2 Alkane Formulas

20.4 Cycloalkanes

20.2a A Brief Guide to Types of Organic Chemistry Formulae

Infographic of a guide to formulae in organic chemistry.

• Molecular formula: The molecular formula of an organic compound simply shows the number of each type of atom present. It tells you nothing about the bonding within the compound. Example:   .
• Empirical formula: The empirical formula of an organic compound gives the simplest possible whole number ratio of the different types of atoms within the compound. Example:  .
• Condensed formula: In condensed formulae, each formulae atom is listed separately, with atoms attached to it following. In cyclic parts of molecules, like benzene, carbons are grouped.
• Displayed formula: A displaced formula shows all of the atoms and all of the bonds present in an organic compound. The bonds are represented as lines.
• Structural formula: Similar to displayed formula; not all bonds are shown, although all atoms are still indicated using script numbers. Carbon hydrogen bonds are often simplified.
• Skeletal formula: In skeletal formulae, most hydrogen atoms are omitted. Line ends or vertices represent carbon. Functional groups and atoms other than carbon or hydrogen are still shown.

Read more about “A Brief Guide to Types of Organic Chemistry Formulae” by Andy Brunning / Compound Interest, CC BY-NC-ND

20.3a A Brief Guide to Types of Isomerism in Organic Chemistry

Infographic about the types of isomerism in organic chemistry.

A guide to the five main types of isomerism that can be exhibited by organic compounds. An isomer of a molecule is a molecule with the same molecular formula but a different structural or spatial arrangement of atoms. This variation can lead to a difference in physical or chemical properties.

Structural isomerism:

• Chain: different arrangement of molecule’s carbon skeleton. The position of the carbon atoms in the molecule can be rearranged to give ‘branched’ carbon chains coming off the main chain. The name of the molecule changes to reflect this, but the molecular formula is still the same. Examples: Butane, 2-Methyl propane.
• Position: The differing position of the same functional group in the molecule. The molecular formula remains the same; the type of functional group also remains the same, but its position in the molecule changes. The name of the molecule changes to reflect the new position of the functional group. Examples: But-2-ene, But-1-ene.
• Functional: Differing positions of atoms give a different functional group. Also referred to as functional group isomerism, these isomers have the same molecular formula but the atoms are rearranged to five a different functional group. The name of the molecule changes to reflect the new functional group. Examples: But-2-ene, Cyclobutane.

Stereoisomerism:

• Geometric: Different substituents around a bond with restricted rotation. Commonly exhibited by alkenes, the presence of two different substituents on both carbon atoms at wither end of the double bond can give rise to two different, non-superimposable isomers due to the restricted rotation of the bond. Examples: (E)-1,2-Dicholorethene ,  (Z)-1,2-Dichloroethene.
• Optical: Non-superimposable mirror images of the same molecule. Optical isomers differ by the placement of different substituents, around one or more atoms in a molecule. Different arrangements of these substituents can be impossible to superimpose – these are optical isomers.

Read more about “A Brief Guide to Types of Isomerism in Organic Chemistry” by Andy Brunning / Compound Interest, CC BY-NC-ND

20.3b A Basic Guide to Decoding Organic Compound Names

Infographic of basic guide to decoding organic compound names. The names of organic molecules can be long and look lie a confusing mix of words and numbers. However, they follow a particular set of rules which allows their structure to be decoded from their name.

• Organic compound representation: Organic molecules are usually represented using skeletal formula. In these diagrams, the line ends and vertices represent carbon atoms. Hydrogen atoms are ‘implied’ – that is, they are not usually shown, but each carbon must have four bonds, and it’s assumed they have the required number of hydrogens for this to be the case. Atoms other than carbon or hydrogen are always shown, and hydrogen atoms are shown if they are bonded to one of these ‘heteroatoms’.
• Functional groups: A molecule’s functional group is the group of atoms that give it its chemical properties and reactivity. It’s usually indicated by a suffix at the end of the name, with a number indicating its position if this is required for clarity. There are many different functional groups. Different functional groups have different suffixes. Examples of funcitonal groups: Alchohols (-ol) e.g. ethanol, aldehydes (-al) e.g. ethanal, ketones (-one) e.g. propane, and amine (-amine) e.g. ethanamine.
• Bond types: Carbon atoms can be lined by single bonds, double bonds, or even triple bonds. The name of the molecules reflects the bonds present.
  

  Types of molecule bonds

  Phrase present in the name	Type of molecule bond present
  -an-	molecule contains only single bonds.
  -en-	molecule contains at least 1 double bond.
  -yn-	molecule contains at least 1 triple bond.
  Double or triple bonds	numbers indicate their position.

• Parent chain: Part of the organic molecule’s name denotes how many carbons make up its ‘parent chain’. This is defined as the longest continuously connected chain of carbon atoms including the functional group in the molecule. Carbons not included are dealt with as ‘side chains’. Examples: Butane, Hexane.

• Side chains: Molecules can have one or more carbons that aren’t part of the parent chain, referred to as a ‘side chain’. The number of carbons in the side chain is used to name it, in the same way as for the parent chain, but the ending -yl is then added. A number is added to show the location of the side chain on the parent chain. If there is more than one of the same side chain at different points, the prefixes di- (2), tri- (3), or tetra- (4) are used in the name. Examples: 2-Methylbutane, 3-Methylpentane, 2-4-Dimethylpentane, 4-Ethylnonan-1-Ol, 3,5,7-Trimethyldecane.
• Stereoisomerism: Chemical names sometimes contain a letter in brackets; for examples, (Z), (E), (R), or (S). These refer to stereoisomerism: when a molecules has the same chemical formula as another, but a different arrangement in 3D space. This can be due to a different arrangement of atoms around a double fond, or when a molecule has two different arrangements of four different groups of atoms around a central carbon which are non-superimposable mirror images.

Read more about “A Basic Guide to Decoding Organic Compound Names” by Andy Brunning / Compound Interest, CC BY-NC-ND

20.6a The Chemistry of Petrol & The Tetraethyl Lead Story

Petrol and diesel are obtained by fractional distillation of crude oil. Diesel is removed from crude oil at a higher boiling point, and contains a larger amount of energy per litre, meaning more miles can be covered with the same volume of fuel. Petrol: 35-200 degree Celsius, 5-12 carbons, and 33,7 megajoules per litre. Diesel: 250-300 degrees Celsius, 10-15 carbons, 36.9 megajoules per litre.

In the engine, a mixture of air and fuels is compressed and burned. Combustion forces the piston down, then the piston pushes back up to expel exhaust gases and the cycle begins again. In diesel engines, the fuel is injected after the air has been compressed before combustion.

Engine knocking happens when the combustion of the fuel doesn’t occur in sync with the engine cycle, causing lower engine efficiency and engine damage. Octane ratings measure how well fuel avoids this problem; higher values indicate less knocking. Isooctane and n-heptane are refences.

Compounds added to petrol to boost octane rating: Tetraethyl lead, was banned in most countries due to releasing toxic lead fumes. Anti-knocking agents used in unleaded petrol: methyl tertiary-butyl ether (MTBE), ethanol, benzene, and toluene.

Read more about “The Chemistry of Petrol & The Tetraethyl Lead Story” by Andy Brunning / Compound Interest, CC BY-NC-ND

20.6b Today in Chemistry History: Susan Solomon, ozone depletion, and CFCs

Susan Solomon was born in 1956. They provided first direct evidence of chlorine compounds breaking down ozone. Solomon’s work confirmed that ozone could react with chlorofluorocarbons on the surface of polar stratospheric clouds. Her work informed the Montreal Protocol, legislation which regulates chemicals that damage the ozone layer.

Ozone and Chlorofluorocarbons: In the stratosphere, CFCs are broken down by UV radiation, releasing highly reactive chlorine radicals. These reacts with and break down ozone molecules. The chlorine radicals are regenerated, so they can go on the to react with thousands of ozone molecules.

Read more about “Today in Chemistry History: Susan Solomon, ozone depletion, and CFCs” by Andy Brunning / Compound Interest, CC BY-NC-ND

Attribution & References

Compound Interest infographics are created by Andy Brunning and licensed under CC BY-NC-ND

Except where otherwise noted, content on this page has been created as a textual summary of the infographics used within our OER. Please refer to the original website (noted below each description) for further details about the image.

As we know, a scientific theory is a strongly supported explanation for observed natural laws or large bodies of experimental data. For a theory to be accepted, it must explain experimental data and be able to predict behavior. For example, VSEPR theory has gained widespread acceptance because it predicts three-dimensional molecular shapes that are consistent with experimental data collected for thousands of different molecules. However, VSEPR theory does not provide an explanation of chemical bonding.

There are successful theories that describe the electronic structure of atoms. We can use quantum mechanics to predict the specific regions around an atom where electrons are likely to be located: A spherical shape for an s orbital, a dumbbell shape for a p orbital, and so forth. However, these predictions only describe the orbitals around free atoms. When atoms bond to form molecules, atomic orbitals are not sufficient to describe the regions where electrons will be located in the molecule. A more complete understanding of electron distributions requires a model that can account for the electronic structure of molecules. One popular theory holds that a covalent bond forms when a pair of electrons is shared by two atoms and is simultaneously attracted by the nuclei of both atoms. In the following sections, we will discuss how such bonds are described by valence bond theory and hybridization.

Valence bond theorydescribes a covalent bond as the overlap of half-filled atomic orbitals (each containing a single electron) that yield a pair of electrons shared between the two bonded atoms. We say that orbitals on two different atoms overlap when a portion of one orbital and a portion of a second orbital occupy the same region of space. According to valence bond theory, a covalent bond results when two conditions are met: (1) an orbital on one atom overlaps an orbital on a second atom and (2) the single electrons in each orbital combine to form an electron pair. The mutual attraction between this negatively charged electron pair and the two atoms’ positively charged nuclei serves to physically link the two atoms through a force we define as a covalent bond. The strength of a covalent bond depends on the extent of overlap of the orbitals involved. Orbitals that overlap extensively form bonds that are stronger than those that have less overlap.

The energy of the system depends on how much the orbitals overlap. Figure 21.1a. illustrates how the sum of the energies of two hydrogen atoms (the coloured curve) changes as they approach each other. When the atoms are far apart there is no overlap, and by convention we set the sum of the energies at zero. As the atoms move together, their orbitals begin to overlap. Each electron begins to feel the attraction of the nucleus in the other atom. In addition, the electrons begin to repel each other, as do the nuclei. While the atoms are still widely separated, the attractions are slightly stronger than the repulsions, and the energy of the system decreases. (A bond begins to form.) As the atoms move closer together, the overlap increases, so the attraction of the nuclei for the electrons continues to increase (as do the repulsions among electrons and between the nuclei). At some specific distance between the atoms, which varies depending on the atoms involved, the energy reaches its lowest (most stable) value. This optimum distance between the two bonded nuclei is the bond distance between the two atoms. The bond is stable because at this point, the attractive and repulsive forces combine to create the lowest possible energy configuration. If the distance between the nuclei were to decrease further, the repulsions between nuclei and the repulsions as electrons are confined in closer proximity to each other would become stronger than the attractive forces. The energy of the system would then rise (making the system destabilized), as shown at the far left of Figure 21.1a.

The bond energy is the difference between the energy minimum (which occurs at the bond distance) and the energy of the two separated atoms. This is the quantity of energy released when the bond is formed. Conversely, the same amount of energy is required to break the bond. For the H₂ molecule shown in Figure 21.1a, at the bond distance of 74 pm the system is 7.24 × 10⁻¹⁹ J lower in energy than the two separated hydrogen atoms. This may seem like a small number. However, we know from our earlier description of thermochemistry that bond energies are often discussed on a per-mole basis. For example, it requires 7.24 × 10⁻¹⁹ J to break one H–H bond, but it takes 4.36 × 10⁵ J to break 1 mole of H–H bonds. A comparison of some bond lengths and energies is shown in Table 21.1a. We can find many of these bonds in a variety of molecules, and this table provides average values. For example, breaking the first C–H bond in CH₄ requires 439.3 kJ/mol, while breaking the first C–H bond in H–CH₂C₆H₅ (a common paint thinner) requires 375.5 kJ/mol.

Table Source: Table 8.1 in “8.1 Valence Bond Theory” in Chemistry (OpenStax), CC BY 4.0

In addition to the distance between two orbitals, the orientation of orbitals also affects their overlap (other than for two s orbitals, which are spherically symmetric). Greater overlap is possible when orbitals are oriented such that they overlap on a direct line between the two nuclei. Figure 21.1b. illustrates this for two p orbitals from different atoms; the overlap is greater when the orbitals overlap end to end rather than at an angle.

The overlap of two s orbitals (as in H₂), the overlap of an s orbital and a p orbital (as in HCl), and the end-to-end overlap of two p orbitals (as in Cl₂) all produce sigma bonds (σ bonds), as illustrated in Figure 21.1c. A σ bond is a covalent bond in which the electron density is concentrated in the region along the internuclear axis; that is, a line between the nuclei would pass through the center of the overlap region. Single bonds in Lewis structures are described as σ bonds in valence bond theory.

A pi bond (π bond) is a type of covalent bond that results from the side-by-side overlap of two p orbitals, as illustrated in Figure 21.1d. In a π bond, the regions of orbital overlap lie on opposite sides of the internuclear axis. Along the axis itself, there is a node, that is, a plane with no probability of finding an electron.

While all single bonds are σ bonds, multiple bonds consist of both σ and π bonds. As the Lewis structures in suggest, O₂ contains a double bond, and N₂ contains a triple bond. The double bond consists of one σ bond and one π bond, and the triple bond consists of one σ bond and two π bonds. Between any two atoms, the first bond formed will always be a σ bond, but there can only be one σ bond in any one location. In any multiple bond, there will be one σ bond, and the remaining one or two bonds will be π bonds. These bonds are described in more detail later in this chapter.

As seen in Table 21.1a., an average carbon-carbon single bond is 347 kJ/mol, while in a carbon-carbon double bond, the π bond increases the bond strength by 267 kJ/mol. Adding an additional π bond causes a further increase of 225 kJ/mol. We can see a similar pattern when we compare other σ and π bonds. Thus, each individual π bond is generally weaker than a corresponding σ bond between the same two atoms. In a σ bond, there is a greater degree of orbital overlap than in a π bond.

Example 21.1a

 

Butadiene, C₆H₆, is used to make synthetic rubber. Identify the number of σ and π bonds contained in this molecule.

Solution

There are six σ C–H bonds and one σ C–C bond, for a total of seven from the single bonds. There are two double bonds that each have a π bond in addition to the σ bond. This gives a total nine σ and two π bonds overall.

Exercise 21.1a

Identify each illustration as depicting a σ or π bond:

1. side-by-side overlap of a 4p and a 2p orbital
2. end-to-end overlap of a 4p and 4p orbital
3. end-to-end overlap of a 4p and a 2p orbital

Check Your Answer

(a) is a π bond with a node along the axis connecting the nuclei while (b) and (c) are σ bonds that overlap along the axis.

The video below shows the connection of sigma and pi bonds to organic chemistry and molecule shapes.

Watch Hybrid Orbitals explained – Valence Bond Theory | Orbital Hybridization sp3 sp2 sp (12 mins) on YouTube

Attribution & References

Except where otherwise noted, this page is adapted by Samantha Sullivan Sauer from “5.1 Valence Bond Theory” In General Chemistry 1 & 2 by Rice University, a derivative of Chemistry (Open Stax) by Paul Flowers, Klaus Theopold, Richard Langley & William R. Robinson and is licensed under CC BY 4.0. ​Access for free at Chemistry (OpenStax)

Thinking in terms of overlapping atomic orbitals is one way for us to explain how chemical bonds form in diatomic molecules. However, to understand how molecules with more than two atoms form stable bonds, we require a more detailed model. As an example, let us consider the water molecule, in which we have one oxygen atom bonding to two hydrogen atoms. Oxygen has the electron configuration 1s²2s²2p⁴, with two unpaired electrons (one in each of the two 2p orbitals). Valence bond theory would predict that the two O–H bonds form from the overlap of these two 2p orbitals with the 1s orbitals of the hydrogen atoms. If this were the case, the bond angle would be 90°, as shown in Figure 21.2a.Note that orbitals may sometimes be drawn in an elongated “balloon” shape rather than in a more realistic “plump” shape in order to make the geometry easier to visualize., because p orbitals are perpendicular to each other. Experimental evidence shows that the bond angle is 104.5°, not 90°. The prediction of the valence bond theory model does not match the real-world observations of a water molecule; a different model is needed.

Quantum-mechanical calculations suggest why the observed bond angles in H₂O differ from those predicted by the overlap of the 1s orbital of the hydrogen atoms with the 2p orbitals of the oxygen atom. The mathematical expression known as the wave function, ψ, contains information about each orbital and the wavelike properties of electrons in an isolated atom. When atoms are bound together in a molecule, the wave functions combine to produce new mathematical descriptions that have different shapes. This process of combining the wave functions for atomic orbitals is called hybridization and is mathematically accomplished by the linear combination of atomic orbitals, LCAO, (a technique that we will encounter again later). The new orbitals that result are called hybrid orbitals. The valence orbitals in an isolated oxygen atom are a 2s orbital and three 2p orbitals. The valence orbitals in an oxygen atom in a water molecule differ; they consist of four equivalent hybrid orbitals that point approximately toward the corners of a tetrahedron (Figure 21.2b.). Consequently, the overlap of the O and H orbitals should result in a tetrahedral bond angle (109.5°). The observed angle of 104.5° is experimental evidence for which quantum-mechanical calculations give a useful explanation: Valence bond theory must include a hybridization component to give accurate predictions.

The following ideas are important in understanding hybridization:

1. Hybrid orbitals do not exist in isolated atoms. They are formed only in covalently bonded atoms.
2. Hybrid orbitals have shapes and orientations that are very different from those of the atomic orbitals in isolated atoms.
3. A set of hybrid orbitals is generated by combining atomic orbitals. The number of hybrid orbitals in a set is equal to the number of atomic orbitals that were combined to produce the set.
4. All orbitals in a set of hybrid orbitals are equivalent in shape and energy.
5. The type of hybrid orbitals formed in a bonded atom depends on its electron-pair geometry as predicted by the VSEPR theory.
6. Hybrid orbitals overlap to form σ bonds. Unhybridized orbitals overlap to form π bonds.

Watch Orbitals: Crash Course Chemistry #25 – YouTube (10 min)

In the following sections, we shall discuss the common types of hybrid orbitals.

sp Hybridization

The beryllium atom in a gaseous BeCl₂ molecule is an example of a central atom with no lone pairs of electrons in a linear arrangement of three atoms. There are two regions of valence electron density in the BeCl₂ molecule that correspond to the two covalent Be–Cl bonds. To accommodate these two electron domains, two of the Be atom’s four valence orbitals will mix to yield two hybrid orbitals. This hybridization process involves mixing of the valence s orbital with one of the valence p orbitals to yield two equivalent sp hybrid orbitals that are oriented in a linear geometry (Figure 21.2c.). In this figure, the set of sp orbitals appears similar in shape to the original p orbital, but there is an important difference. The number of atomic orbitals combined always equals the number of hybrid orbitals formed. The p orbital is one orbital that can hold up to two electrons. The sp set is two equivalent orbitals that point 180° from each other. The two electrons that were originally in the s orbital are now distributed to the two sp orbitals, which are half filled. In gaseous BeCl₂, these half-filled hybrid orbitals will overlap with orbitals from the chlorine atoms to form two identical σ bonds.

We illustrate the electronic differences in an isolated Be atom and in the bonded Be atom in the orbital energy-level diagram in Figure 21.2d. These diagrams represent each orbital by a horizontal line (indicating its energy) and each electron by an arrow. Energy increases toward the top of the diagram. We use one upward arrow to indicate one electron in an orbital and two arrows (up and down) to indicate two electrons of opposite spin.

When atomic orbitals hybridize, the valence electrons occupy the newly created orbitals. The Be atom had two valence electrons, so each of the sp orbitals gets one of these electrons. Each of these electrons pairs up with the unpaired electron on a chlorine atom when a hybrid orbital and a chlorine orbital overlap during the formation of the Be–Cl bonds.

Any central atom surrounded by just two regions of valence electron density in a molecule will exhibit sp hybridization. Other examples include the mercury atom in the linear HgCl₂ molecule, the zinc atom in Zn(CH₃)₂, which contains a linear C–Zn–C arrangement, and the carbon atoms in HCCH and CO₂.

sp² Hybridization

The valence orbitals of a central atom surrounded by three regions of electron density consist of a set of three sp² hybrid orbitals and one unhybridized p orbital. This arrangement results from sp² hybridization, the mixing of one s orbital and two p orbitals to produce three identical hybrid orbitals oriented in a trigonal planar geometry (Figure 21.2e.).

Although quantum mechanics yields the “plump” orbital lobes as depicted in Figure 21.2e., sometimes for clarity these orbitals are drawn thinner and without the minor lobes, as in Figure 21.2f., to avoid obscuring other features of a given illustration. We will use these “thinner” representations whenever the true view is too crowded to easily visualize.

The observed structure of the borane molecule, BH_3, suggests sp² hybridization for boron in this compound. The molecule is trigonal planar, and the boron atom is involved in three bonds to hydrogen atoms (Figure 21.2g.). We can illustrate the comparison of orbitals and electron distribution in an isolated boron atom and in the bonded atom in BH₃ as shown in the orbital energy level diagram in Figure 21.2h. We redistribute the three valence electrons of the boron atom in the three sp² hybrid orbitals, and each boron electron pairs with a hydrogen electron when B–H bonds form.

Any central atom surrounded by three regions of electron density will exhibit sp² hybridization. This includes molecules with a lone pair on the central atom, such as ClNO (Figure 21.2i.), or molecules with two single bonds and a double bond connected to the central atom, as in formaldehyde, CH₂O, and ethene, H₂CCH₂.

sp³ Hybridization

The valence orbitals of an atom surrounded by a tetrahedral arrangement of bonding pairs and lone pairs consist of a set of four sp³ hybrid orbitals. The hybrids result from the mixing of one s orbital and all three p orbitals that produces four identical sp³ hybrid orbitals (Figure 21.2j.). Each of these hybrid orbitals points toward a different corner of a tetrahedron.

A molecule of methane, CH₄, consists of a carbon atom surrounded by four hydrogen atoms at the corners of a tetrahedron. The carbon atom in methane exhibits sp³ hybridization. We illustrate the orbitals and electron distribution in an isolated carbon atom and in the bonded atom in CH₄ in Figure 21.2k. The four valence electrons of the carbon atom are distributed equally in the hybrid orbitals, and each carbon electron pairs with a hydrogen electron when the C–H bonds form.

In a methane molecule, the 1s orbital of each of the four hydrogen atoms overlaps with one of the four sp³ orbitals of the carbon atom to form a sigma (σ) bond. This results in the formation of four strong, equivalent covalent bonds between the carbon atom and each of the hydrogen atoms to produce the methane molecule, CH₄.

The structure of ethane, C₂H_6, is similar to that of methane in that each carbon in ethane has four neighbouring atoms arranged at the corners of a tetrahedron—three hydrogen atoms and one carbon atom (Figure 21.2l.). However, in ethane an sp³ orbital of one carbon atom overlaps end to end with an sp³ orbital of a second carbon atom to form a σ bond between the two carbon atoms. Each of the remaining sp³ hybrid orbitals overlaps with an s orbital of a hydrogen atom to form carbon–hydrogen σ bonds. The structure and overall outline of the bonding orbitals of ethane are shown in Figure 21.2l. The orientation of the two CH₃ groups is not fixed relative to each other. Experimental evidence shows that rotation around σ bonds occurs easily.

An sp³ hybrid orbital can also hold a lone pair of electrons. For example, the nitrogen atom in ammonia is surrounded by three bonding pairs and a lone pair of electrons directed to the four corners of a tetrahedron. The nitrogen atom is sp³ hybridized with one hybrid orbital occupied by the lone pair.

The molecular structure of water is consistent with a tetrahedral arrangement of two lone pairs and two bonding pairs of electrons. Thus we say that the oxygen atom is sp³ hybridized, with two of the hybrid orbitals occupied by lone pairs and two by bonding pairs. Since lone pairs occupy more space than bonding pairs, structures that contain lone pairs have bond angles slightly distorted from the ideal. Perfect tetrahedra have angles of 109.5°, but the observed angles in ammonia (107.3°) and water (104.5°) are slightly smaller. Other examples of sp³ hybridization include CCl₄, PCl₃, and NCl₃.

sp³d and sp³d² Hybridization

To describe the five bonding orbitals in a trigonal bipyramidal arrangement, we must use five of the valence shell atomic orbitals (the s orbital, the three p orbitals, and one of the d orbitals), which gives five sp³d hybrid orbitals. With an octahedral arrangement of six hybrid orbitals, we must use six valence shell atomic orbitals (the s orbital, the three p orbitals, and two of the d orbitals in its valence shell), which gives six sp³d² hybrid orbitals. These hybridizations are only possible for atoms that have d orbitals in their valence subshells (that is, not those in the first or second period).

In a molecule of phosphorus pentachloride, PCl₅, there are five P–Cl bonds (thus five pairs of valence electrons around the phosphorus atom) directed toward the corners of a trigonal bipyramid. We use the 3s orbital, the three 3p orbitals, and one of the 3d orbitals to form the set of five sp³d hybrid orbitals (Figures 21.2m. and 21.2n.) that are involved in the P–Cl bonds. Other atoms that exhibit sp³d hybridization include the sulfur atom in SF₄ and the chlorine atoms in ClF₃ and in ClF₄⁺. (The electrons on fluorine atoms are omitted for clarity.)

The sulfur atom in sulfur hexafluoride, SF₆, exhibits sp³d² hybridization. A molecule of sulfur hexafluoride has six bonding pairs of electrons connecting six fluorine atoms to a single sulfur atom. There are no lone pairs of electrons on the central atom. To bond six fluorine atoms, the 3s orbital, the three 3p orbitals, and two of the 3d orbitals form six equivalent sp³d² hybrid orbitals, each directed toward a different corner of an octahedron. Other atoms that exhibit sp³d² hybridization include the phosphorus atom in PCl₆⁻, the iodine atom in the interhalogens IF₆⁺, IF₅, ICl₄⁻, IF₄⁻ and the xenon atom in XeF₄.

Assignment of Hybrid Orbitals to Central Atoms

The hybridization of an atom is determined based on the number of regions of electron density that surround it. The geometrical arrangements characteristic of the various sets of hybrid orbitals are shown in Figure 16. These arrangements are identical to those of the electron-pair geometries predicted by VSEPR theory. VSEPR theory predicts the shapes of molecules, and hybrid orbital theory provides an explanation for how those shapes are formed. To find the hybridization of a central atom, we can use the following guidelines:

1. Determine the Lewis structure of the molecule.
2. Determine the number of regions of electron density around an atom using VSEPR theory, in which single bonds, multiple bonds, radicals, and lone pairs each count as one region.
3. Assign the set of hybridized orbitals from Figure 21.2p that corresponds to this geometry.

Exercise 21.2a

Explore the electron and molecule geometry of various real molecules. Turn on the electron geometry, molecule geometry, lone pairs and bond angles for the molecule being shown. Compare the results to Figure 21.2p. Be sure to look at H₂O, CH₄, and CO₂.

Practice using the following PhET simulation: Molecule Shapes

One or more interactive elements has been excluded from this version of the text. You can view them online here: https://ecampusontario.pressbooks.pub/orgbiochemsupplement/?p=352

Activity source: Simulation by PhET Interactive Simulations, University of Colorado Boulder, licensed under CC-BY-4.0

It is important to remember that hybridization was devised to rationalize experimentally observed molecular geometries. The model works well for molecules containing small central atoms, in which the valence electron pairs are close together in space. However, for larger central atoms, the valence-shell electron pairs are farther from the nucleus, and there are fewer repulsions. Their compounds exhibit structures that are often not consistent with VSEPR theory, and hybridized orbitals are not necessary to explain the observed data. For example (Figure 21.2q), we have discussed the H–O–H bond angle in H₂O, 104.5°, which is more consistent with sp³ hybrid orbitals (109.5°) on the central atom than with 2p orbitals (90°). Sulfur is in the same group as oxygen, and H₂S has a similar Lewis structure. However, it has a much smaller bond angle (92.1°), which indicates much less hybridization on sulfur than oxygen. Continuing down the group, tellurium is even larger than sulfur, and for H₂Te, the observed bond angle (90°) is consistent with overlap of the 5p orbitals, without invoking hybridization. We invoke hybridization where it is necessary to explain the observed structures.

Example 21.2a

Assigning Hybridization
Ammonium sulfate is important as a fertilizer. What is the hybridization of the sulfur atom in the sulfate ion, SO₄²⁻?

Solution

The Lewis structure of sulfate shows there are four regions of electron density. The hybridization is sp³.

Exercise 21.2b

What is the hybridization of the selenium atom in SeF₄?

Check Your AnswerThe selenium atom is sp³d hybridized.

Example 21.2b

Assigning Hybridization
Urea, NH₂C(O)NH₂, is sometimes used as a source of nitrogen in fertilizers. What is the hybridization of each nitrogen and carbon atom in urea?

Solution

The Lewis structure of urea is

The nitrogen atoms are surrounded by four regions of electron density, which arrange themselves in a tetrahedral electron-pair geometry. The hybridization in a tetrahedral arrangement is sp³ (Figure 21.2p). This is the hybridization of the nitrogen atoms in urea.

The carbon atom is surrounded by three regions of electron density, positioned in a trigonal planar arrangement. The hybridization in a trigonal planar electron pair geometry is sp² (Figure 21.2p), which is the hybridization of the carbon atom in urea.

Exercise 21.2c

Acetic acid, H₃CC(O)OH, is the molecule that gives vinegar its odour and sour taste. What is the hybridization of the two carbon atoms in acetic acid?

 

Check Your Answer

H₃C, sp³; C(O)OH, sp²

Attribution & References

Except where otherwise noted, this page is adapted by Samantha Sullivan Sauer from “5.2 Hybrid Atomic Orbitals” In General Chemistry 1 & 2 by Rice University, a derivative of Chemistry (Open Stax) by Paul Flowers, Klaus Theopold, Richard Langley & William R. Robinson and is licensed under CC BY 4.0. ​Access for free at Chemistry (OpenStax)

The hybrid orbital model appears to account well for the geometry of molecules involving single covalent bonds. Is it also capable of describing molecules containing double and triple bonds? We have already discussed that multiple bonds consist of σ and π bonds. Next we can consider how we visualize these components and how they relate to hybrid orbitals. The Lewis structure of ethene, C₂H₄, (Figure 21.3a.) shows us that each carbon atom is surrounded by one other carbon atom and two hydrogen atoms.

The three bonding regions form a trigonal planar electron-pair geometry. Thus we expect the σ bonds from each carbon atom are formed using a set of sp² hybrid orbitals that result from hybridization of two of the 2p orbitals and the 2s orbital (Figure 21.3b.). These orbitals form the C–H single bonds and the σ bond in the C=C double bond (Figure 21.3c.). The π bond in the C=C double bond results from the overlap of the third (remaining) 2p orbital on each carbon atom that is not involved in hybridization. This unhybridized p orbital (lobes shown in red and blue in Figure 21.3c.) is perpendicular to the plane of the sp² hybrid orbitals. Thus the unhybridized 2p orbitals overlap in a side-by-side fashion, above and below the internuclear axis (Figure 21.3c.) and form a π bond.

In an ethene molecule, the four hydrogen atoms and the two carbon atoms are all in the same plane. If the two planes of sp² hybrid orbitals tilted relative to each other, the p orbitals would not be oriented to overlap efficiently to create the π bond. The planar configuration for the ethene molecule occurs because it is the most stable bonding arrangement. This is a significant difference between σ and π bonds; rotation around single (σ) bonds occurs easily because the end-to-end orbital overlap does not depend on the relative orientation of the orbitals on each atom in the bond. In other words, rotation around the internuclear axis does not change the extent to which the σ bonding orbitals overlap because the bonding electron density is symmetric about the axis. Rotation about the internuclear axis is much more difficult for multiple bonds; however, this would drastically alter the off-axis overlap of the π bonding orbitals, essentially breaking the π bond.

In molecules with sp hybrid orbitals, two unhybridized p orbitals remain on the atom (Figure 21.3d.). We find this situation in acetylene, H−C≡C−H, which is a linear molecule. The sp hybrid orbitals of the two carbon atoms overlap end to end to form a σ bond between the carbon atoms (Figure 21.3e.). The remaining sp orbitals form σ bonds with hydrogen atoms. The two unhybridized p orbitals per carbon are positioned such that they overlap side by side and, hence, form two π bonds. The two carbon atoms of acetylene are thus bound together by one σ bond and two π bonds, giving a triple bond.

Hybridization involves only σ bonds, lone pairs of electrons, and single unpaired electrons (radicals). Structures that account for these features describe the correct hybridization of the atoms. However, many structures also include resonance forms. Remember that resonance forms occur when various arrangements of π bonds are possible. Since the arrangement of π bonds involves only the unhybridized orbitals, resonance does not influence the assignment of hybridization.

For example, molecule benzene has two resonance forms (Figure 21.3f.). We can use either of these forms to determine that each of the carbon atoms is bonded to three other atoms with no lone pairs, so the correct hybridization is sp². The electrons in the unhybridized p orbitals form π bonds. Neither resonance structure completely describes the electrons in the π bonds. They are not located in one position or the other, but in reality are delocalized throughout the ring. Valence bond theory does not easily address delocalization. Bonding in molecules with resonance forms is better described by molecular orbital theory in the next section.

Example 21.3a

Assignment of Hybridization Involving Resonance
Some acid rain results from the reaction of sulfur dioxide with atmospheric water vapor, followed by the formation of sulfuric acid. Sulfur dioxide, SO₂, is a major component of volcanic gases as well as a product of the combustion of sulfur-containing coal. What is the hybridization of the S atom in SO₂?

Solution

The resonance structures of SO₂ are

The sulfur atom is surrounded by two bonds and one lone pair of electrons in either resonance structure. Therefore, the electron-pair geometry is trigonal planar, and the hybridization of the sulfur atom is sp².

Watch 3D Structure and Bonding: Crash Course Organic Chemistry #4 – YouTube (14 min)

Attribution & References

Except where otherwise noted, this page is adapted by Samantha Sullivan Sauer from “5.3 Multiple Bonds” In General Chemistry 1 & 2 by Rice University, a derivative of Chemistry (Open Stax) by Paul Flowers, Klaus Theopold, Richard Langley & William R. Robinson and is licensed under CC BY 4.0. ​Access for free at Chemistry (OpenStax)

For almost every covalent molecule that exists, we can now draw the Lewis structure, predict the electron-pair geometry, predict the molecular geometry, and come close to predicting bond angles. However, one of the most important molecules we know, the oxygen molecule O₂, presents a problem with respect to its Lewis structure as shown in Figure 21.4a. We would write the following Lewis structure for O₂:

This electronic structure adheres to all the rules governing Lewis theory. There is an O=O double bond, and each oxygen atom has eight electrons around it. However, this picture is at odds with the magnetic behavior of oxygen. By itself, O₂ is not magnetic, but it is attracted to magnetic fields. Thus, when we pour liquid oxygen past a strong magnet, it collects between the poles of the magnet and defies gravity. Such attraction to a magnetic field is called paramagnetism, and it arises in molecules that have unpaired electrons. And yet, the Lewis structure of O₂ indicates that all electrons are paired. How do we account for this discrepancy?

Magnetic susceptibility measures the force experienced by a substance in a magnetic field. When we compare the weight of a sample to the weight measured in a magnetic field (Figure 21.4b.), paramagnetic samples that are attracted to the magnet will appear heavier because of the force exerted by the magnetic field. We can calculate the number of unpaired electrons based on the increase in weight.

Experiments show that each O₂ molecule has two unpaired electrons. The Lewis-structure model does not predict the presence of these two unpaired electrons. Unlike oxygen, the apparent weight of most molecules decreases slightly in the presence of an inhomogeneous magnetic field. Materials in which all of the electrons are paired are diamagnetic and weakly repel a magnetic field. Paramagnetic and diamagnetic materials do not act as permanent magnets. Only in the presence of an applied magnetic field do they demonstrate attraction or repulsion.

Water, like most molecules, contains all paired electrons. Living things contain a large percentage of water, so they demonstrate diamagnetic behavior. If you place a frog near a sufficiently large magnet, it will levitate.

Watch Diamagnetism: How to Levitate a Frog – YouTube (5 min)

Molecular orbital theory (MO theory) provides an explanation of chemical bonding that accounts for the paramagnetism of the oxygen molecule. It also explains the bonding in a number of other molecules, such as violations of the octet rule and more molecules with more complicated bonding (beyond the scope of this text) that are difficult to describe with Lewis structures. Additionally, it provides a model for describing the energies of electrons in a molecule and the probable location of these electrons. Unlike valence bond theory, which uses hybrid orbitals that are assigned to one specific atom, MO theory uses the combination of atomic orbitals to yield molecular orbitals that are delocalized over the entire molecule rather than being localized on its constituent atoms. MO theory also helps us understand why some substances are electrical conductors, others are semiconductors, and still others are insulators. Table 21.4a. summarizes the main points of the two complementary bonding theories. Both theories provide different, useful ways of describing molecular structure.

Table Source: Table 8.2 in “8.4 Molecular Orbital Theory” in Chemistry (OpenStax),CC BY 4.0

Molecular orbital theory describes the distribution of electrons in molecules in much the same way that the distribution of electrons in atoms is described using atomic orbitals. Using quantum mechanics, the behavior of an electron in a molecule is still described by a wave function, Ψ, analogous to the behavior in an atom. Just like electrons around isolated atoms, electrons around atoms in molecules are limited to discrete (quantized) energies. The region of space in which a valence electron in a molecule is likely to be found is called a molecular orbital (Ψ²). Like an atomic orbital, a molecular orbital is full when it contains two electrons with opposite spin.

We will consider the molecular orbitals in molecules composed of two identical atoms (H₂ or Cl₂, for example). Such molecules are called homonuclear diatomic molecules. In these diatomic molecules, several types of molecular orbitals occur.

The mathematical process of combining atomic orbitals to generate molecular orbitals is called the linear combination of atomic orbitals (LCAO). The wave function describes the wavelike properties of an electron. Molecular orbitals are combinations of atomic orbital wave functions. Combining waves can lead to constructive interference, in which peaks line up with peaks, or destructive interference, in which peaks line up with troughs (Figure 21.4c.). In orbitals, the waves are three dimensional, and they combine with in-phase waves producing regions with a higher probability of electron density and out-of-phase waves producing nodes, or regions of no electron density.

There are two types of molecular orbitals that can form from the overlap of two atomic s orbitals on adjacent atoms. The two types are illustrated in Figure 21.4d. The in-phase combination produces a lower energy σ_s molecular orbital (read as “sigma-s”) in which most of the electron density is directly between the nuclei. The out-of-phase addition (which can also be thought of as subtracting the wave functions) produces a higher energy molecular orbital (read as “sigma-s-star”) molecular orbital in which there is a node between the nuclei. The asterisk signifies that the orbital is an antibonding orbital. Electrons in a σ_s orbital are attracted by both nuclei at the same time and are more stable (of lower energy) than they would be in the isolated atoms. Adding electrons to these orbitals creates a force that holds the two nuclei together, so we call these orbitals bonding orbitals. Electrons in the  orbitals are located well away from the region between the two nuclei. The attractive force between the nuclei and these electrons pulls the two nuclei apart. Hence, these orbitals are called antibonding orbitals. Electrons fill the lower-energy bonding orbital before the higher-energy antibonding orbital, just as they fill lower-energy atomic orbitals before they fill higher-energy atomic orbitals.

Watch 14.2/S2.2.16 Describe σ and π bonds IB Chemistry [HL IB Chemistry] (2 mins) on YouTube

In p orbitals, the wave function gives rise to two lobes with opposite phases, analogous to how a two-dimensional wave has both parts above and below the average. We indicate the phases by shading the orbital lobes different colours. When orbital lobes of the same phase overlap, constructive wave interference increases the electron density. When regions of opposite phase overlap, the destructive wave interference decreases electron density and creates nodes. When p orbitals overlap end to end, they create σ and σ* orbitals (Figure 21.4e.). If two atoms are located along the x-axis in a Cartesian coordinate system, the two p_x orbitals overlap end to end and form σ_px (bonding) and (antibonding) (read as “sigma-p-x” and “sigma-p-x star,” respectively). Just as with s-orbital overlap, the asterisk indicates the orbital with a node between the nuclei, which is a higher-energy, antibonding orbital.

The side-by-side overlap of two p orbitals gives rise to a pi (π) bonding molecular orbital and a π* antibonding molecular orbital, as shown in Figure 21.4f. In valence bond theory, we describe π bonds as containing a nodal plane containing the internuclear axis and perpendicular to the lobes of the p orbitals, with electron density on either side of the node. In molecular orbital theory, we describe the π orbital by this same shape, and a π bond exists when this orbital contains electrons. Electrons in this orbital interact with both nuclei and help hold the two atoms together, making it a bonding orbital. For the out-of-phase combination, there are two nodal planes created, one along the internuclear axis and a perpendicular one between the nuclei.

In the molecular orbitals of diatomic molecules, each atom also has two sets of p orbitals oriented side by side (p_y and p_z), so these four atomic orbitals combine pairwise to create two π orbitals and two π* orbitals. The π_py and orbitals are oriented at right angles to the π_pz and orbitals. Except for their orientation, the π_py and π_pz orbitals are identical and have the same energy; they are degenerate orbitals. The and antibonding orbitals are also degenerate and identical except for their orientation. A total of six molecular orbitals results from the combination of the six atomic p orbitals in two atoms: σ_px and , π_py and , π_pz and .

Example 21.4a

Molecular Orbitals
Predict what type (if any) of molecular orbital would result from adding the wave functions so each pair of orbitals shown overlap. The orbitals are all similar in energy.

Solution

1. is an in-phase combination, resulting in a σ_3p orbital
2. will not result in a new orbital because the in-phase component (bottom) and out-of-phase component (top) cancel out. Only orbitals with the correct alignment can combine.
3. is an out-of-phase combination, resulting in a orbital.

Exercise 21.4a

Label the molecular orbital shown as σ or π, bonding or antibonding and indicate where the node occurs.

Check Your Understanding

The orbital is located along the internuclear axis, so it is a σ orbital. There is a node bisecting the internuclear axis, so it is an antibonding orbital.

Images in solution: Chemistry (OpenStax), CC BY 4.0)

Spotlight on Everyday Chemistry: Scientist Walter Kohn

Walter Kohn (Figure 21.4g.) is a theoretical physicist who studies the electronic structure of solids. His work combines the principles of quantum mechanics with advanced mathematical techniques. This technique, called density functional theory, makes it possible to compute properties of molecular orbitals, including their shape and energies. Kohn and mathematician John Pople were awarded the Nobel Prize in Chemistry in 1998 for their contributions to our understanding of electronic structure. Kohn also made significant contributions to the physics of semiconductors.

Kohn’s biography has been remarkable outside the realm of physical chemistry as well. He was born in Austria, and during World War II he was part of the Kindertransport program that rescued 10,000 children from the Nazi regime. His summer jobs included discovering gold deposits in Canada and helping Polaroid explain how its instant film worked. Although he is now an emeritus professor, he is still actively working on projects involving global warming and renewable energy.

Computational Chemistry in Drug Design

While the descriptions of bonding described in this chapter involve many theoretical concepts, they also have many practical, real-world applications. For example, drug design is an important field that uses our understanding of chemical bonding to develop pharmaceuticals. This interdisciplinary area of study uses biology (understanding diseases and how they operate) to identify specific targets, such as a binding site that is involved in a disease pathway. By modeling the structures of the binding site and potential drugs, computational chemists can predict which structures can fit together and how effectively they will bind (see Figure 21.4h.). Thousands of potential candidates can be narrowed down to a few of the most promising candidates. These candidate molecules are then carefully tested to determine side effects, how effectively they can be transported through the body, and other factors. Dozens of important new pharmaceuticals have been discovered with the aid of computational chemistry, and new research projects are underway.

Bond Order

The filled molecular orbital diagram shows the number of electrons in both bonding and antibonding molecular orbitals. The net contribution of the electrons to the bond strength of a molecule is identified by determining the bond order that results from the filling of the molecular orbitals by electrons.

When using Lewis structures to describe the distribution of electrons in molecules, we define bond order as the number of bonding pairs of electrons between two atoms. Thus, a single bond has a bond order of 1, a double bond has a bond order of 2, and a triple bond has a bond order of 3. We define bond order differently when we use the molecular orbital description of the distribution of electrons, but the resulting bond order is usually the same. The MO technique is more accurate and can handle cases when the Lewis structure method fails, but both methods describe the same phenomenon.

In the molecular orbital model, an electron contributes to a bonding interaction if it occupies a bonding orbital, and it contributes to an antibonding interaction if it occupies an antibonding orbital. The bond order is calculated by subtracting the destabilizing (antibonding) electrons from the stabilizing (bonding) electrons. Since a bond consists of two electrons, we divide by two to get the bond order. We can determine bond order with the following equation:

The order of a covalent bond is a guide to its strength; a bond between two given atoms becomes stronger as the bond order increases (Table 11.5 a and b in Chapter 11.5 Strengths of Ionic and Covalent Bonds). If the distribution of electrons in the molecular orbitals between two atoms is such that the resulting bond would have a bond order of zero, a stable bond does not form. We next look at some specific examples of MO diagrams and bond orders.

Bonding in Diatomic Molecules

A dihydrogen molecule (H₂) forms from two hydrogen atoms. When the atomic orbitals of the two atoms combine, the electrons occupy the molecular orbital of lowest energy, the σ_1s bonding orbital. A dihydrogen molecule, H₂, readily forms because the energy of a H₂ molecule is lower than that of two H atoms. The σ_1s orbital that contains both electrons is lower in energy than either of the two 1s atomic orbitals.

A molecular orbital can hold two electrons, so both electrons in the H₂ molecule are in the σ_1s bonding orbital; the electron configuration is . We represent this configuration by a molecular orbital energy diagram (Figure 21.4j.) in which a single upward arrow indicates one electron in an orbital, and two (upward and downward) arrows indicate two electrons of opposite spin.

A dihydrogen molecule contains two bonding electrons and no antibonding electrons so we have

Because the bond order for the H–H bond is equal to 1, the bond is a single bond.

A helium atom has two electrons, both of which are in its 1s orbital. Two helium atoms do not combine to form a dihelium molecule, He₂, with four electrons, because the stabilizing effect of the two electrons in the lower-energy bonding orbital would be offset by the destabilizing effect of the two electrons in the higher-energy antibonding molecular orbital. We would write the hypothetical electron configuration of He₂ as as in Figure 21.4k. The net energy change would be zero, so there is no driving force for helium atoms to form the diatomic molecule. In fact, helium exists as discrete atoms rather than as diatomic molecules. The bond order in a hypothetical dihelium molecule would be zero.

A bond order of zero indicates that no bond is formed between two atoms.

The Diatomic Molecules of the Second Period

Eight possible homonuclear diatomic molecules might be formed by the atoms of the second period of the periodic table: Li₂, Be₂, B₂, C₂, N₂, O₂, F₂, and Ne₂. However, we can predict that the Be₂ molecule and the Ne₂ molecule would not be stable. We can see this by a consideration of the molecular electron configurations (Table 21.4b.).

Table 21.4b. Electron Configuration and Bond Order for Molecular Orbitals in Homonuclear Diatomic Molecules of Period Two Elements

Molecule	Electron Configuration	Bond Order
Li2		1
Be2 (unstable)		0
B2		1
C2		2
N2		3
O2		2
F2		1
Ne2 (unstable)		0

Table Source: Table 8.3 in “8.4 Molecular Orbital Theory” in Chemistry (OpenStax), CC BY 4.0

We predict valence molecular orbital electron configurations just as we predict electron configurations of atoms. Valence electrons are assigned to valence molecular orbitals with the lowest possible energies. Consistent with Hund’s rule, whenever there are two or more degenerate molecular orbitals, electrons fill each orbital of that type singly before any pairing of electrons takes place.

As we saw in valence bond theory, σ bonds are generally more stable than π bonds formed from degenerate atomic orbitals. Similarly, in molecular orbital theory, σ orbitals are usually more stable than π orbitals. However, this is not always the case. The MOs for the valence orbitals of the second period are shown in Figure 21.4l. Looking at Ne₂ molecular orbitals, we see that the order is consistent with the generic diagram shown in the previous section. However, for atoms with three or fewer electrons in the p orbitals (Li through N) we observe a different pattern, in which the σ_p orbital is higher in energy than the π_p set. Obtain the molecular orbital diagram for a homonuclear diatomic ion by adding or subtracting electrons from the diagram for the neutral molecule.

This switch in orbital ordering occurs because of a phenomenon called s-p mixing. s-p mixing does not create new orbitals; it merely influences the energies of the existing molecular orbitals. The σ_s wavefunction mathematically combines with the σ_p wavefunction, with the result that the σ_s orbital becomes more stable, and the σ_p orbital becomes less stable (Figure 21.4m.). Similarly, the antibonding orbitals also undergo s-p mixing, with the σ_s* becoming more stable and the σ_p* becoming less stable.

s-p mixing occurs when the s and p orbitals have similar energies. When a single p orbital contains a pair of electrons, the act of pairing the electrons raises the energy of the orbital. Thus the 2p orbitals for O, F, and Ne are higher in energy than the 2p orbitals for Li, Be, B, C, and N. Because of this, O₂, F₂, and N₂ only have negligible s-p mixing (not sufficient to change the energy ordering), and their MO diagrams follow the normal pattern, as shown in Figure 21.4n. All of the other period 2 diatomic molecules do have s-p mixing, which leads to the pattern where the σ_p orbital is raised above the π_p set.

Using the MO diagrams shown in Figure 21.4m., we can add in the electrons and determine the molecular electron configuration and bond order for each of the diatomic molecules. As shown in Table 21.4b., Be₂ and Ne₂ molecules would have a bond order of 0, and these molecules do not exist.

The combination of two lithium atoms to form a lithium molecule, Li₂, is analogous to the formation of H₂, but the atomic orbitals involved are the valence 2s orbitals. Each of the two lithium atoms has one valence electron. Hence, we have two valence electrons available for the σ_2s bonding molecular orbital. Because both valence electrons would be in a bonding orbital, we would predict the Li₂ molecule to be stable. The molecule is, in fact, present in appreciable concentration in lithium vapor at temperatures near the boiling point of the element. All of the other molecules in Table 21.4b. with a bond order greater than zero are also known.

The O₂ molecule has enough electrons to half fill the (, ) level. We expect the two electrons that occupy these two degenerate orbitals to be unpaired, and this molecular electronic configuration for O₂ is in accord with the fact that the oxygen molecule has two unpaired electrons (Figure 21.4n.). The presence of two unpaired electrons has proved to be difficult to explain using Lewis structures, but the molecular orbital theory explains it quite well. In fact, the unpaired electrons of the oxygen molecule provide a strong piece of support for the molecular orbital theory.

Band Theory

When two identical atomic orbitals on different atoms combine, two molecular orbitals result (see Figure 21.4d.). The bonding orbital is lower in energy than the original atomic orbitals because the atomic orbitals are in-phase in the molecular orbital. The antibonding orbital is higher in energy than the original atomic orbitals because the atomic orbitals are out-of-phase.

In a solid, similar things happen, but on a much larger scale. Remember that even in a small sample there are a huge number of atoms (typically > 10²³ atoms), and therefore a huge number of atomic orbitals that may be combined into molecular orbitals. When N valence atomic orbitals, all of the same energy and each containing one (1) electron, are combined, N/2 (filled) bonding orbitals and N/2 (empty) antibonding orbitals will result. Each bonding orbital will show an energy lowering as the atomic orbitals are mostly in-phase, but each of the bonding orbitals will be a little different and have slightly different energies. The antibonding orbitals will show an increase in energy as the atomic orbitals are mostly out-of-phase, but each of the antibonding orbitals will also be a little different and have slightly different energies. The allowed energy levels for all the bonding orbitals are so close together that they form a band, called the valence band. Likewise, all the antibonding orbitals are very close together and form a band, called the conduction band. Figure 21.4o. shows the bands for three important classes of materials: insulators, semiconductors, and conductors.

In order to conduct electricity, electrons must move from the filled valence band to the empty conduction band where they can move throughout the solid. The size of the band gap, or the energy difference between the top of the valence band and the bottom of the conduction band, determines how easy it is to move electrons between the bands. Only a small amount of energy is required in a conductor because the band gap is very small. This small energy difference is “easy” to overcome, so they are good conductors of electricity. In an insulator, the band gap is so “large” that very few electrons move into the conduction band; as a result, insulators are poor conductors of electricity. Semiconductors conduct electricity when “moderate” amounts of energy are provided to move electrons out of the valence band and into the conduction band. Semiconductors, such as silicon, are found in many electronics.

Semiconductors are used in devices such as computers, smartphones, and solar cells. Solar cells produce electricity when light provides the energy to move electrons out of the valence band. The electricity that is generated may then be used to power a light or tool, or it can be stored for later use by charging a battery. As of December 2014, up to 46% of the energy in sunlight could be converted into electricity using solar cells.

Example 21.4b

Molecular Orbital Diagrams, Bond Order, and Number of Unpaired Electrons
Draw the molecular orbital diagram for the oxygen molecule, O₂. From this diagram, calculate the bond order for O₂. How does this diagram account for the paramagnetism of O₂?

Solution

We draw a molecular orbital energy diagram similar to that shown in Figure 21.4m. Each oxygen atom contributes six electrons, so the diagram appears as shown in Figure 21.4n.

We calculate the bond order as

Oxygen’s paramagnetism is explained by the presence of two unpaired electrons in the (π_2py, π_2pz)* molecular orbitals.

Exercise 21.4b

The main component of air is N₂. From the molecular orbital diagram of N₂, predict its bond order and whether it is diamagnetic or paramagnetic.

Check Your Learning

N₂ has a bond order of 3 and is diamagnetic.

Example 21.4c

Ion Predictions with MO Diagrams
Give the molecular orbital configuration for the valence electrons in C₂²⁻. Will this ion be stable?

Solution

Looking at the appropriate MO diagram, we see that the π orbitals are lower in energy than the σ_p orbital. The valence electron configuration for C₂ is . Adding two more electrons to generate the C₂²⁻ anion will give a valence electron configuration of . Since this has six more bonding electrons than antibonding, the bond order will be 3, and the ion should be stable.

Exercise 21.4c

How many unpaired electrons would be present on a Be₂²⁻ ion? Would it be paramagnetic or diamagnetic?

Check Your Learning

two, paramagnetic

Links to Enhanced Learning

Practice labeling and filling molecular orbitals with this interactive tutorial from the University of Sydney.

Key Equations

• 

Attribution & References

Except where otherwise noted, this page is adapted by Samantha Sullivan Sauer from “5.4 Moledular Orbital Theory” In General Chemistry 1 & 2 by Rice University, a derivative of Chemistry (Open Stax) by Paul Flowers, Klaus Theopold, Richard Langley & William R. Robinson and is licensed under CC BY 4.0. ​Access for free at Chemistry (OpenStax)

21.1 Valence Bond Theory

Valence bond theory describes bonding as a consequence of the overlap of two separate atomic orbitals on different atoms that creates a region with one pair of electrons shared between the two atoms. When the orbitals overlap along an axis containing the nuclei, they form a σ bond. When they overlap in a fashion that creates a node along this axis, they form a π bond.

21.2 Hybrid Atomic Orbitals

We can use hybrid orbitals, which are mathematical combinations of some or all of the valence atomic orbitals, to describe the electron density around covalently bonded atoms. These hybrid orbitals either form sigma (σ) bonds directed toward other atoms of the molecule or contain lone pairs of electrons. We can determine the type of hybridization around a central atom from the geometry of the regions of electron density about it. Two such regions imply sp hybridization; three, sp² hybridization; four, sp³ hybridization; five, sp³d hybridization; and six, sp³d² hybridization. Pi (π) bonds are formed from unhybridized atomic orbitals (p or d orbitals).

21.3 Multiple Bonds

Multiple bonds consist of a σ bond located along the axis between two atoms and one or two π bonds. The σ bonds are usually formed by the overlap of hybridized atomic orbitals, while the π bonds are formed by the side-by-side overlap of unhybridized orbitals. Resonance occurs when there are multiple unhybridized orbitals with the appropriate alignment to overlap, so the placement of π bonds can vary.

21.4 Molecular Orbital Theory

Attribution & References

Except where otherwise noted, this page is adapted by Samantha Sullivan Sauer from “5.1 Valence Bond Theory“, “5.2 Hybrid Atomic Orbitals“, “5.3 Multiple Bonds“, and “5.4 Moledular Orbital Theory“,  In General Chemistry 1 & 2 by Rice University, a derivative of Chemistry (Open Stax) by Paul Flowers, Klaus Theopold, Richard Langley & William R. Robinson and is licensed under CC BY 4.0. ​Access for free at Chemistry (OpenStax)

21.2 Hybrid Atomic Orbitals

21.3 Multiple Bonds

21.4 Molecular Orbital Theory

Attribution & References

Except where otherwise noted, this page (including images in solutions) is adapted by Samantha Sullivan Sauer from

• “5.1 Valence Bond Theory“, “5.2 Hybrid Atomic Orbitals“, “5.3 Multiple Bonds“, &”5.4 Moledular Orbital Theory” In General Chemistry 1 & 2 by Rice University, a derivative of Chemistry (Open Stax) by Paul Flowers, Klaus Theopold, Richard Langley & William R. Robinson and is licensed under CC BY 4.0. ​Access for free at Chemistry (OpenStax)
• All other images in solutions are from the source listed above: Chemistry (OpenStax), CC BY 4.0

Learning Objectives

By the end of this section, you will be able to:

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

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 22.2a.

In 1,2-dichloroethane (part (a) of Figure 22.2b.), 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 Figure 22.2b.), 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 22.2c.).

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 according to Figure 22.2d.

Note, however, that the presence of a double bond does not necessarily lead to cis-trans isomerism (Figure 22.2e.). 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.

Advanced Note: E/Z Isomerization

If a molecule has a C=C bond with one non-hydrogen group attached to each of the carbons, cis/trans nomenclature described above is enough to describe it. However, if you have three different groups (or four), then the cis/trans approach is insufficient to describe the different isomers, since we do not know which two of the three groups are being described. For example, if you have a C=C bond, with a methyl group and a bromine on one carbon, and an ethyl group on the other, it is neither trans nor cis, since it is not clear whether the ethyl group is trans to the bromine or the methyl. 

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. Refer to Figure 22.2f. to compare cis-1,2-dimethylcyclopropane and trans-1,2-dimethylcyclopropane.

For a look at the types of isomerism in organic chemistry. Refer back to Infographic 20.3a.

Example 22.2a

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:

   

• 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.
• 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.
• This compound meets rule 2; it has two nonidentical groups on each carbon atom and exists as both cis and trans isomers:

 

Exercise 22.2a

Which compounds can exist as cis-trans isomers?

a. CH₂=CHCH₂CH₂CH₃

b. CH₃CH=CHCH₂CH₃

c. CH₃CH₂CH=CHCH₂CH₃

d.

e.

Check Your Answers:

a) no cis-trans isomers b) cis-trans isomers are possible c) cis-trans isomers are possible d) no cis-trans isomers e) no cis-trans isomers

Attribution & References

Except where otherwise noted, this page is adapted by Adrienne Richards and Samantha Sullivan Sauer from “13.3: The Structure of Alkenes- Cis-Trans Isomerism” In Map: Fundamentals of General Organic and Biological Chemistry (McMurry et al.) , CC BY-NC-SA 3.0. / A derivative of Basics of GOB (Ball et al.), CC BY-NC-SA 4.0 a LibreTexts version of Introduction to Chemistry: GOB (v. 1.0), CC BY-NC 3.0

Learning Objectives

By the end of this section, you will be able to:

• Write equations for the addition reactions of alkenes and alkynes with hydrogen, halogens, and water
• Describe Markovnikov’s Rule as it applies to addition reactions
• Describe chemical tests to test for the presence of unsaturated hydrocarbons.

Addition Reactions

Alkenes are valued mainly for addition reactions, in which one of the bonds in the double bond is broken. Each of the carbon atoms in the bond can then attach another atom or group while remaining joined to each other by a single bond. Examples of addition reactions include hydrogenation, halogenation and hydration.

Hydrogenation of Alkenes

Perhaps the simplest addition reaction is hydrogenation—a reaction with hydrogen (H₂) in the presence of a catalyst such as nickel (Ni) or platinum (Pt). An example of the addition reaction between ethylene and hydrogen is shown in Figure 22.3a.

The product is an alkane having the same carbon skeleton as the alkene.

Halogenation of Alkenes

Alkenes also readily undergo halogenation—the addition of halogens. Indeed, the reaction with bromine (Br₂) can be used to test for alkenes. Bromine solutions are brownish red. When we add a Br₂ solution to an alkene, the colour of the solution disappears because the alkene reacts with the bromine as shown in Figure 22.3b.

Hydration of Alkenes

Another important addition reaction is that between an alkene and water to form an alcohol. This reaction, called hydration as shown in Figure 22.3c., requires a catalyst—usually a strong acid, such as sulfuric acid (H₂SO₄).

The hydration reaction is discussed in a later chapter, where we deal with this reaction in the synthesis of alcohols.

Example 22.3a

Write the equation for the reaction between CH₃CH=CHCH₃ and each substance.

1. H₂ (Ni catalyst)
2. Br₂
3. H₂O (H₂SO₄ catalyst)

Solution

In each reaction, the reagent adds across the double bond.

a.

b.

c.

Image credit: Intro Chemistry: GOB (V. 1.0)., CC BY-NC-SA 3.0.

 

Exercise 22.3a

Write the equation for each reaction.

1. CH₃CH₂CH=CH₂ with H₂ (Ni catalyst)
2. CH₃CH=CH₂ with Cl₂
3. CH₃CH₂CH=CHCH₂CH₃ with H₂O (H₂SO₄ catalyst)

Check Your Answers:

Source: Exercise 22.3a is adapted from Fundamentals of GOB Chem, CC BY-NC-SA 4.0 with answer images drawn by Samantha Sullivan Sauer / Biovia Draw, CC BY-NC 4.0

Markovnikov’s Rule

In the addition any unsymmetrical hydrogen-based molecule (e.g. HX or H₂O) to an alkene, the H attaches to the carbon with fewer alkyl substituents and the X attaches to the carbon with more alkyl substituents. Examples of the rule is demonstrated in Figure 22.3d.

When both double-bonded carbon atoms have the same degree of substitution, a mixture of addition products results as demonstrated in Figure 22.3e.

Because carbocations are involved as intermediates in these electrophilic addition reactions, Markovnikov’s rule can be restated in the following way:

Markovnikov’s rule restated In the addition of HX or H₂O to an alkene, the more highly substituted carbocation is formed as the intermediate rather than the less highly substituted one.

For example, in Figure 22.3f., addition of H⁺ to 2-methylpropene yields the intermediate tertiary carbocation rather than the alternative primary carbocation, and addition to 1-methylcyclohexene yields a tertiary cation rather than a secondary one. Why should this be?

Example 22.3b

What product would you expect from reaction of HCl with 1-ethylcyclopentene?

Strategy

When solving a problem that asks you to predict a reaction product, begin by looking at the functional group(s) in the reactants and deciding what kind of reaction is likely to occur. In the present instance, the reactant is an alkene that will probably undergo an electrophilic addition reaction with HCl. Next, recall what you know about electrophilic addition reactions to predict the product. You know that electrophilic addition reactions follow Markovnikov’s rule, so H⁺ will add to the double-bond carbon that has one alkyl group (C2 on the ring) and the Cl will add to the double-bond carbon that has two alkyl groups (C1 on the ring).

Solution

The expected product is 1-chloro-1-ethylcyclopentane.

Example 22.3c

Synthesizing a Specific Compound

What alkene would you start with to prepare the following alkyl halide? There may be more than one possibility.

Strategy

When solving a problem that asks how to prepare a given product, always work backward. Look at the product, identify the functional group(s) it contains, and ask yourself, “How can I prepare that functional group?” In the present instance, the product is a tertiary alkyl chloride, which can be prepared by reaction of an alkene with HCl. The carbon atom bearing the −Cl atom in the product must be one of the double-bond carbons in the reactant. Draw and evaluate all possibilities.

Solution

There are three possibilities, all of which could give the desired product according to Markovnikov’s rule.

Exercise 22.3b

Draw the major product formed from these reactions.

Check Your Answers:

Activity source: Exercise 22.3b is created by Samantha Sullivan Sauer, using images from Biovia Draw, licensed under CC BY-NC 4.0

For more details on alkene addition reactions including the concept of Markonikov’s rule, watch Alkene Addition Reactions below.

Watch Alkene Addition Reactions: Crash Course Organic Chemistry #16 (youtube.com) (13 min).

One or more interactive elements has been excluded from this version of the text. You can view them online here: https://ecampusontario.pressbooks.pub/orgbiochemsupplement/?p=865#oembed-3

Video Source: Crash Course. (2020, Nov 11). Alkene Addition Reactions: Crash Course Organic Chemistry #16 (youtube.com) [Video]. YouTube.

Polymerization

The most important commercial reactions of alkenes are polymerizations, reactions in which small molecules, referred to in general as monomers (from the Greek monos, meaning “one,” and meros, meaning “parts”), are assembled into giant molecules referred to as polymers (from the Greek poly, meaning “many,” and meros, meaning “parts”). For more information on polymerization see Chapter 27: Polymers.

Sourcing of Alkenes

In summary, recall that organic functional groups can be converted into other functional groups through reactions.  To look at the sourcing of alkenes, refer to the map of some of the more common reactions to convert functional groups can be found in Section 19.6 – General Reactions of Carbon in Infographic 19.6a.

Reactions of Alkynes

Chemically, the alkynes are similar to the alkenes. Alkynes can undergo addition, hydration and hydrogenation (or reduction) reactions.

Halogenation of Alkynes

Since the $\text{C}\equiv \text{C}$functional group has two π bonds, alkynes typically react even more readily, and react with twice as much reagent in addition reactions. The reaction of acetylene with bromine is a typical example as shown in Figure 22.3g.

Figure 22.3g. Addition reaction involving acetylene and bromine (credit: Chemistry: Atoms First 2e (OpenStax), CC BY 4.0)

Acetylene and the other alkynes also burn readily. An acetylene torch takes advantage of the high heat of combustion for acetylene.

As a general rule, electrophiles undergo addition reactions with alkynes much as they do with alkenes. Take the reaction of alkynes with HX, for instance. The reaction often can be stopped with the addition of 1 equivalent of HX, but reaction with an excess of HX leads to a dihalide product. For example, reaction of 1-hexyne with 2 equivalents of HBr yields 2,2-dibromohexane. As the following examples indicate, the regiochemistry of addition follows Markovnikov’s rule, with halogen adding to the more highly substituted side of the alkyne bond and hydrogen adding to the less highly substituted side. Trans stereochemistry of H and X normally, although not always, occurs in the product.

Bromine and chlorine also add to alkynes to give addition products, and trans stereochemistry again results as demonstrated in Figure 22.3i. below.

The mechanism of alkyne addition is similar but not identical to that of alkene addition. When an electrophile such as HBr adds to an alkene, the reaction takes place in two steps and involves an alkyl carbocation intermediate. If HBr were to add by the same mechanism to an alkyne, an analogous vinylic carbocation would be formed as the intermediate as shown in Figure 22.3j.

A vinylic carbocation has an sp-hybridized carbon and generally forms less readily than an alkyl carbocation (Figure 22.3j.). As a rule, a secondary vinylic carbocation forms about as readily as a primary alkyl carbocation, but a primary vinylic carbocation is so difficult to form that there is no clear evidence it even exists. Thus, many alkyne additions occur through more complex mechanistic pathways.

Hydration of Alkynes

Hydration of alkynes also can take place. Alkynes don’t react directly with aqueous acid but will undergo hydration readily in the presence of mercury(II) sulfate as a Lewis acid catalyst. The reaction occurs with Markovnikov regiochemistry, so the −OH group adds to the more highly substituted carbon and the −H attaches to the less highly substituted one as demonstrated in Figure 22.3k.

Interestingly, the actual product isolated from alkyne hydration is not a vinylic alcohol, or enol (ene + ol), but is instead a ketone.

Hydrogenation of Alkynes

Lastly, hydrogenation (reduction) of alkynes is another chemical reaction that can take place. Alkynes are reduced to alkanes by addition of H₂ over a metal catalyst. The reaction in Figure 22.3l., occurs in two steps through an alkene intermediate, and measurements show that the first step in the reaction is more exothermic than the second.

Complete reduction to the alkane occurs when palladium on carbon (Pd/C) is used as catalyst, but hydrogenation can be stopped at the alkene stage if the less active Lindlar catalyst is used. The Lindlar catalyst is a finely divided palladium metal that has been precipitated onto a calcium carbonate support and then deactivated by treatment with lead acetate and quinoline, an aromatic amine. The hydrogenation occurs with syn stereochemistry, giving a cis alkene product.

The alkyne hydrogenation reaction has been explored extensively by the Hoffmann–LaRoche pharmaceutical company, where it is used in the commercial synthesis of vitamin A. The cis isomer of vitamin A produced initially on hydrogenation is converted to the trans isomer by heating as shown in Figure 22.3n.

An alternative method for the conversion of an alkyne to an alkene uses sodium or lithium metal as the reducing agent in liquid ammonia as solvent. This method is complementary to the Lindlar reduction because it produces trans rather than cis alkenes. For example, in Figure 22.3o., 5-decyne gives trans-5-decene on treatment with lithium in liquid ammonia.

Spotlight on Everyday Chemistry: 2022 Nobel Prize in Chemistry

Alkynes were involved in the concept of “click” chemistry where an azide is added to an alkyne with a copper catalyst allowing the two molecules to click together forming a cyclic molecule. The click chemistry concept was awarded the 2022 Nobel Prize in Chemistry to Carolyn R. Bertozzi, Morten Meldal and K. Barry Sharpless. For more information refer to infographic 22.3a.

Testing for Presence of Alkenes/Alkynes

There are several ways to test for the presence of carbon-carbon double bonds and triple bonds (unsaturated hydrocarbons).  One such method, as previously mentioned above, is the bromine test.  Here the organic compound containing a double or triple C-C bond is mixed with an aqueous solution of bromine (or chlorine).  With bromine, there is a visible colour change resulting when bromine is added to the double or triple bond (Figure 22.3p. and Figure 22.3q.). Before addition, the bromine is brownish-red.  After addition the solution is colourless. If the solution stays brownish-red, it is a negative result and the compound being tested is saturated.  This means there is no opportunity for addition.

A second such test to confirm the presence of a carbon-carbon double or triple bond (unsaturated hydrocarbon) is the oxidation or permanganate test (Figure 22.3s.).  This test is also known as the Baeyer test. Here, potassium permanganate, KMnO₄, is used as an oxidizing agent to convert the alkene or alkyne to a diol (two alcohol functional groups in the same molecule).  The visual colour change is from dark purple (permanganate solution) to dark green (manganate solution) then to black precipitate (manganese dioxide) (Figure 22.3r.). If the solution stays purple, it is a negative result and the compound being tested is saturated. This means there is no opportunity for oxidation. This test can give conflicting results in that any other components in the molecule or solution that are mildly reducing will also give a positive result.

Links to Enhanced Learning

• Watch E/Z Alkenes, Electrophilic Addition, & Carbocations: Crash Course Organic Chemistry Crash Course Organic Chemistry #14 by Crash Course.
• Watch Alkene Redox Reactions Crash Course Organic Chemistry #17 by Crash Course.
• Watch Alkyne Reactions & Tautomerization Crash Course Organic Chemistry #18 by Crash Course.

Attribution & References

Except where otherwise noted, this page is written and adapted by David Wegman, Adrienne Richards and Samantha Sullivan Sauer from

• “13.6: Addition Reactions of Alkenes” and “13.7: Alkene Polymers” In Map: Fundamentals of General Organic and Biological Chemistry (McMurry et al.) by LibreTexts, licensed under CC BY-NC-SA 3.0. / A derivative of “13.4: Chemical Properties of Alkenes” In Basics of GOB (Ball et al.), CC BY-NC-SA 4.0 a LibreTexts version of Introduction to Chemistry: GOB (v. 1.0), CC BY-NC 3.0.
• “Reactions of Alkynes” section is adapted from “18.1 Hydrocarbons” In General Chemistry 1 & 2 by Rice University, a derivative of Chemistry (Open Stax) by Paul Flowers, Klaus Theopold, Richard Langley & William R. Robinson and is licensed under CC BY 4.0. ​Access for free at Chemistry (OpenStax)
• “7.8 Orientation of Electrophilic Additions: Markovnikov’s Rule“, “9.3 Reactions of Alkynes: Addition of HX and X2“, “9.4 Hydration of Alkynes “, and “9.5 Reduction of Alkynes” In Organic Chemistry (OpenStax) by John McMurray, CC BY-NC-SA 4.0. Access for free at Organic Chemistry (OpenStax)
• “21.1 Hydrocarbons”  In Chemistry: Atoms First 2e (OpenStax) by Paul Flowers, Edward J. Neth, William R. Robinson, Klaus Theopold & Richard Langley, CC BY 4.0
• Testing for Presence of Alkenes/Alkynes section is adapted by Samantha Sullivan Sauer from “4.6.2.1: Chemistry of Manganese” by Jim Clark In Inorganic Chemistry II (CHEM4210) , CC BY-NC 4.0 and “Reactions of Alkenes with Bromine” by Jim Clark In Supplemental Modules (Organic Chemistry), CC BY-NC 4.0