Chapter 13 – Organic Chemistry

Introduction

Organic chemistry is the study of carbon based compounds. The structural and genetic materials of living organisms are organic compounds. Many of the substances that we encounter on a daily basis are organic compounds: drugs, plastics, textiles, dyes, paper, food, vitamins, etc. Because of the manner in which carbon atoms can bond to one another, the number of organic compounds is enormous. In this chapter, we explore some of the fundamental topics of the structure and reactivity of several classes of organic compounds by applying our knowledge of Lewis structures, resonance, and Lewis acid-base theory.

13.1 Hydrocarbons

Introduction

We begin our study of organic chemistry with two classes of compounds that contain only carbon and hydrogen.

Prerequisites

• •
  5.8 Formal Charge and Oxidation State (Draw Lewis structures.)

Objectives

• •
  Name and draw the two simplest alkanes.
• •
  Draw line-wedge-dash, Lewis, condensed, and skeletal structures of alkanes.
• •
  Convert skeletal structures into molecular formulas.
• •
  Distinguish between a straight or continuous chain and a branched chain.
• •
  Write the equation for the combustion of an alkane.
• •
  Define an alkene and draw and name the two simplest alkenes.
• •
  Draw the isomers of an alkene.
• •
  Define a polyene and explain why many polyenes absorb visible light.
• •
  Define alkynes and discuss their isomers.
• •
  Distinguish between saturated and unsaturated hydrocarbons.

Alkanes

13.1-1. Definition

Carbon atoms always have four bonds: four single bonds, two single and one double bond, two double bonds, or a single bond and a triple bond. Although carbon is the basic building block, organic compounds usually have hydrogen atoms as well. Oxygen and nitrogen are also common, but almost any other element can be found in organic compounds. In this section, we deal with the class of compounds known as hydrocarbons, which are compounds that contain only carbon and hydrogen. Alkanes are the simplest organic compounds. All of the carbon atoms in an alkane are sp³ hybridized and tetrahedral, and all bonds are sigma bonds, which makes alkanes relatively unreactive. They have the general formula C_nH_{2n + 2}, where n is an integer. Table 13.1 shows the three simplest alkanes.

Alkane	Structure
Methane, CH4
Ethane, C2H6
Propane, C3H8

Table 13.1: The Simplest Alkanes

13.1-2. Structures

There are times when the line-wedge-dash representation is used to show the three-dimensional structure of organic compounds, but the structures of alkanes are very dynamic because groups attached by a single bond can rotate relative to one another (much like the wheels on an axle). Consequently, other shorthand notations are frequently used. These other methods, shown in Figures 13.1a, 13.1b, 13.1c, and 13.1d, are based on the following points that are common to most organic molecules.

• •
  Carbon atoms form their backbone.
• •
  Carbon atoms always have four bonds to them.
• •
  There are usually many C–H bonds present.

Propane is C₃H₈. Figure 13.1a shows the line-wedge-dash representation of the three-dimensional structure. In a Lewis structure of an organic compound, all of the bonds and atoms are shown explicitly, but the bond angles are often shown as 90°. A condensed structure shows the atoms, but not all of the bonds. Sometimes the C–C bonds are shown, but frequently they are not. A skeletal structure does not show carbon or hydrogen atoms explicitly, nor does it show the C–H bonds. The positions of the carbon atoms in a skeletal structure are indicated by the ends and intersections of lines. Thus, the structure of propane shows three carbon atoms: one at each end and one at the intersection of the two line segments. The positions of the hydrogen atoms are inferred from the number of bonds shown to each carbon. Each carbon atom must have four bonds (a double bond counts as two bonds and a triple bond counts as three), so the number of C–H bonds to any carbon atom equals four minus the number of bonds shown to the carbon atom in the skeletal structure. For example, the skeletal structure shows only one bond to each of the terminal carbon atoms, so there must also be three C–H bonds to each of the end carbons. The structure shows two bonds to the middle carbon, so there must also be two C–H bonds.

13.1-3. Molecular Formulas for Skeletal Structures

Use the following rules to convert the skeletal structures into molecular formulas.

• •
  Carbon atoms are represented as intersections or ends of lines.
• •
  The number of hydrogen atoms attached to a carbon atom is equal to four minus the number of bonds involving the carbon atom because carbon atoms are always involved in four bonds. Note that double and triple bonds count as 2 and 3 bonds, respectively.

13.1-4. Chains

There is only one way to connect the carbon atoms in each of the alkanes discussed so far, but there are multiple ways to connect the carbon atoms in all of the alkanes with more than three carbon atoms. The way in which the carbon atoms are connected is called the connectivity. The molecules with the same formula but different connectivities are different molecules, and different molecules with the same formula are called isomers. Molecules in which no carbon is connected to more than two other carbons are called straight chains or continuous chains, while molecules that contain with at least one carbon with three or four carbon atoms attached to it are called branched chains. Butane (C₄H₁₀) exists as two isomers, one straight chain and one branched chain.

13.1-5. Combustion

Alkanes are found primarily in natural gas (mainly CH₄ and C₂H₆) and petroleum. They are relatively unreactive, but they do burn well. Burning is a rapid combustion, i.e., the reaction with oxygen. The products of the combustion of an alkane are CO₂ and H₂O. Table 13.2 shows the heats of combustion for four common alkanes.

Alkane	Use	Combustion Reaction	Heat of Combustion (kJ/mol)
methane	natural gas furnace	CH4 + 2 O2 → CO2 + 2 H2O	–891
propane	propane tank	C3H8 + 5 O2 → 3 CO2 + 4 H2O	–2220
butane	cigarette lighters	C4H10 + 6.5 O2 → 4 CO2 + 5 H2O	–2855
octane	automobile fuel	C8H18 + 12.5 O2 → 8 CO2 + 9 H2O	–5494

Table 13.2: Combustion of Some Common Alkanes

13.1-6. Octane Ratings

Each atom in a C–H or C–C bond retains one of the bonding electrons during combustion. The result is two species, called free radicals, that have unpaired electrons. Free radicals are highly reactive, but they are stabilized when the unpaired electron is on an atom that is attached to other carbon-containing groups. Thus, free radicals formed from highly branched carbon atoms are more stable and less reactive than those formed from straight chain hydrocarbons. Indeed, straight chain hydrocarbons react so fast and violently that they can cause an engine to "knock." The octane rating of a gasoline indicates the extent of knocking it causes. The reference molecules are shown in Figure 13.4. The straight chain hydrocarbon C₇H₁₆ causes substantial knocking and is assigned an octane rating of 0, while the highly branched alkane C₈H₁₈ (an isomer of octane) causes little knocking and is assigned an octane rating of 100. A gasoline with an octane rating of 87 causes the same knocking as a mixture that is 87% in the branched alkane and 13% of the straight chain alkane.

Alkenes

13.1-7. Definition

Organic compounds that contain carbon-carbon double bonds are called alkenes. The carbon atoms involved in the double bond are sp² hybridized. The two simplest alkenes are ethene (C₂H₄) and propene (C₃H₆). Alkenes in which the position of the double bond is different are different molecules. However, care must be taken to be certain that the two positions are indeed different and not just a different view of the same molecule. The two carbon atoms are each sp² hybridized, and the molecule is planar. Figure 13.5a(c) shows the pi bond (represented in red) as the overlap of the two p orbitals that are not used to make the hybrid orbitals. The electron density of the pi bond lies above and below the molecular plane. Consequently, the pi electrons can attack other substances, i.e., alkenes are weak Lewis bases. It might appear that the double bond in C₃H₆ could be placed after the first carbon or before the last carbon. However, these two positions are identical because one can be converted into the other by a rotation of the molecule. Thus, the two skeletal structures shown in Figure 13.5b(c) are identical, not isomers. Simply rotate one about the dotted line to get the other.

13.1-8. Butene

The double bond in C₄H₈ can be placed in three different positions (after the first or second carbon in a straight chain or after the first carbon of a branched chain) to produce three isomers. In addition, groups cannot rotate about double bonds like they can about single bonds because the rotation would break the pi bond. Consequently, the butene with two carbon atoms (two CH₃ groups) on the same side of the double bond is different than the one with the two carbon atoms on opposite sides. We will discuss this type of isomer in more detail later. The four isomers of C₄H₈ are derived below by starting with propene as follows.

Position of 4th C	Skeletal Structure	Comments
1		This is a straight chain alkene with double bond after the second carbon atom. In this isomer, the two CH3 groups are on the same side of the double bond.
2		This is also a straight chain alkene with double bond after the second carbon atom. In this isomer, the two CH3 groups are on opposite sides of the double bond.
3		This isomer is the only branched isomer.
4		This is the only straight chain isomer in which the double bond involves one of the end carbons.

Table 13.3: The Isomers of Butene

13.1-9. Polyenes

The names of alkenes all end in -ene, and compounds with many double bonds are called polyenes. An extended pi system results when double bonds of a polyene alternate with single bonds (as in — — — — ) and the pi electrons are delocalized over the alternating single and double bonds. Recall from our discussion of band theory (Chapter 8.6), that electronic energy levels get closer together as electrons become spread out over many bonds. Also recall from Chapter 6.5 that pi electrons can be delocalized over many carbon atoms. Thus, while most organic compounds are white or colorless because their electronic energy levels are too far apart to absorb visible light, many polyenes with alternating single and double bonds absorb visible light and are colored. Consequently, they are often used to make dyes. As we shall see later in this chapter, it is the absorption of a visible photon by a polyene that initiates the vision process. Two common polyene dyes are considered in Figures 13.7a and 13.7b.

Alkynes

13.1-10. Definition

Alkynes are hydrocarbons that contain at least one triple bond. The simplest alkyne is C₂H₂, which is ethyne, or more commonly acetylene. Carbon atoms involved in triple bonds have only two electron regions, are sp hybridized, and have 180° bond angles. The presence of π electrons makes them Lewis basic as well, and their chemistry is similar to that described above for alkenes.

13.1-11. Isomers

As with alkenes, different positions of the multiple bond can lead to different isomers. However, the positions must be different viewed from either end of the molecule. For example, there are three ways to draw butyne, but there are only two isomers. The two molecules on the left of Figure 13.9 are the same molecule as the triple bond in each is located on a terminal carbon atom. The molecule on the right is the isomer of the molecule on the left.

13.1-12. Hydrogenation and Saturation

All of the carbon atoms in alkanes are saturated because they have the maximum number of sigma bonds (four). However, carbon atoms involved in multiple (double or triple) bonds are unsaturated because their pi bonds can be replaced with sigma bonds, which increases the number of electron regions. Compounds with many multiple bonds are said to be polyunsaturated. An unsaturated compound can be saturated by hydrogenation, i.e., by reaction with hydrogen. In a hydrogenation, H₂ adds across the double bond, forming two C–H bonds, i.e., the pi bond between the carbon atom is replaced by two C–H sigma bonds.

unsaturated hydrocarbon	+	hydrogen	→	saturated hydrocarbon
	+	3 H2	→
Multiple bonds (double and triple) are unsaturated. The more multiple bonds a compound has, the higher is its degree of unsaturation. Unsaturated compounds can be saturated by hydrogenation.		One H2 is required for each double bond.		All C–C bonds are single bonds in the hydrocarbon above, so it is saturated.

13.2 Naming Simple Hydrocarbons

Introduction

Understanding the nomenclature of alkanes, alkenes, and alkynes will help you understand some of the diversity in organic molecules that will be discussed later in the chapter.

Prerequisites

• •
  5.8 Formal Charge and Oxidation State (Draw Lewis structures.)
• •
  6.4: Valence Bond Theory and Hybridization (Determine the hybridization of an atom in a Lewis structure.)

Objectives

• •
  Give the common root names and the endings for alkanes and alkenes.
• •
  Name straight chain alkanes.
• •
  Identify and name the two simplest alkyl groups.
• •
  Name branched alkanes with a single alkyl group.
• •
  Name branched alkanes with two alkyl groups.
• •
  Name straight chain alkenes and alkynes.

13.2-1. Root Names

The name of an organic compound must indicate the number of carbon atoms in the longest continuous chain, which is given by the root name of the chain as given in Table 13.4, and the type of compound, which is designated by the ending of the name as shown below.

• •
  Alkanes end in ane.
• •
  Alkenes end in ene.
• •
  Alkynes end in yne.

For example, a hexane is an alkane that contains a six-carbon chain, while a pentene is an alkene with a five-carbon chain, and butyne is an alkyne with a four-carbon chain.

# C atoms	Root Name	Name of Alkane	Name of Alkene
1	meth	methane
2	eth	ethane	ethene
3	prop	propane	propene
4	but	butane	butene
5	pent	pentane	pentene
6	hex	hexane	hexene
7	hept	heptane	heptene
8	oct	octane	octene
9	non	nonane	nonene
10	dec	decane	decene

Table 13.4: Root Names for Organic Compounds

13.2-2. Naming Organic Molecules

13.2-3. Alkyl Groups

All of the carbon atoms are not in the longest continuous chain of a branched hydrocarbon. Instead, these carbon atoms are in groups called side chains, which are attached to the longest continuous chain. Side chains can be any type of group, but we restrict our discussion to alkyl groups. Alkyl groups are formed by removing one hydrogen atom from an alkane. The resulting alkyl group binds to a longer chain at the position of the removed hydrogen. Alkyl groups are parts of compounds; they are not compounds themselves. The name of the alkyl group is obtained by adding -yl to the root name The two simplest alkyl groups are given in Table 13.5.

Formula of Alkyl Group	Name of Alkyl Group
–CH3	methyl
–C2H5	ethyl

Table 13.5: The Simplest Alkyl Groups

The dash before the formula indicates where they bond to a longer chain. Propyl, butyl, pentyl, etc., groups also exist, but there are more than one of each of these alkyl groups. For example, there are two different propyl groups that differ by the position of the removed hydrogen, which can come from a terminal carbon atom or from the middle carbon atom. We restrict our discussion of naming organic compounds to compounds that contain methyl and/or ethyl groups to avoid this complexity.

13.2-4. Naming Alkanes

Branched alkanes have side chains, so the name of a branched alkane specifies the longest continuous chain and the identity and position of the side chains (alkyl groups). Although there can be many side chains of varying lengths, we restrict our discussion of branched alkanes to those with one or two side chains, which will be either a methyl or an ethyl group. The position of the alkyl group is indicated by the number of the carbon atom to which it is attached. Thus, the carbon atoms in the longest chain must be numbered. To number the longest continuous chain, start at the end that produces the smallest number for the side chain. To name branched alkanes:

• 1
  Identify the longest chain, select the appropriate root name, and add ane to indicate an alkane.
• 2
  Number the carbon atoms of the longest continuous chain starting at the end that produces the smaller number for the carbon atom to which the alkyl group is attached.
• 3
  Identify the side chain and its position.
• 4
  Name the alkane as the number of the position of the side chain, a hyphen, and then the name of the alkyl group followed by the name obtained in step 1.

Note that the name of the alkyl group and the name of the longest chain are combined into one word. For example, 2-methylhexane has a six-carbon chain with a CH₃ group attached to the second carbon from the end (either end).

13.2-5. Naming Alkanes Exercise

Practice naming branched alkanes by naming the three organic compounds in Exercise 13.3. The root names can be found in Table 13.4.

13.2-6. Naming Alkanes with Multiple Alkyl Groups

To name alkanes with more than one side chain, simply place the names of the alkyl groups in alphabetical order separated by hyphens in front of the root name. For example, 3-ethyl-2-methylheptane is a seven carbon continuous chain with a C₂H₅ group on the third carbon and a CH₃ group attached to the second carbon. If two identical groups are attached to the same carbon, the number of the position is repeated with a comma separator, and the number of groups is given by a prefix, as in 2,2-dimethylheptane.

13.2-7. Naming Alkenes and Alkynes

To name a straight or continuous chain alkene:

• 1
  Number the chain so as to give the carbon atoms in the double or triple bond the lowest number.
• 2
  Specify the chain length with the appropriate root name and add -ene to indicate an alkene or -yne to indicate an alkyne.
• 3
  Indicate the position of the double or triple bond by the number of the carbon atom in the bond with the lower number in the chain.
• 4
  The name is the position of the bond as determined in step 3 followed by a hyphen and then the root name determined in step 2.

For example, 1-pentene is a five-carbon alkene with a double bond between carbon atoms 1 and 2, and 2-hexyne is a six-carbon alkyne with a triple bond between carbons 2 and 3.

13.3 Isomers

Introduction

We have seen that there can be several organic compounds with the same formula. Compounds with the same formula are called isomers. Isomers can have very different chemical and physical properties. In this section, we introduce two broad classifications of isomers.

Prerequisites

• •
  6.1 Molecular Shapes

Objectives

• •
  Draw the constitutional isomers with a given molecular formula.
• •
  Draw all of the constitutional isomers of an alkane.
• •
  Identify stereoisomers.
• •
  Name the two types of stereoisomers.
• •
  Distinguish between cis and trans isomers.
• •
  Describe the molecular process responsible for vision.
• •
  Identify stereocenters.
• •
  Identify enantiomers.

Constitutional

13.3-1. Constitutional Isomers

The number of constitutional isomers increases with the number of carbon atoms. C₃H₈O has three constitutional isomers, but there are six isomers of C₄H₁₀O and well over a hundred isomers of C₁₂H₂₆O. Constitutional isomers can be the same type of molecule with similar properties. For example, there are over 100 isomers of C₁₀H₂₂, but they are all alkanes. However, constitutional isomers can also be totally different types of molecules with very different properties. As shown in Figure 13.10, there are three isomers of C₃H₈O. Two are alcohols and one is an ether.

• •
  Ethers contain the C–O–C connectivity.
• •
  Alcohols contain the C–O–H connectivity.

Ethers do not hydrogen bond to one another, while alcohols do because they contain O–H bonds. Consequently, alcohols have much higher boiling points than their corresponding ether isomers. For example, the isomer of C₃H₈O that is an ether boils at –23 °C, while the two isomers that are alcohols boil at +78 °C and +83 °C. Now consider the six structures for molecules having the formula C₄H₁₀O. Figures 13.11(a) and 13.11(b) contain a
C–O–C linkage, so they are ethers, while the remaining structures all contain a hydroxyl group, which makes each of them alcohols. For C₁₂H₂₆O, there are well over one hundred isomers! The existence of so many constitutional isomers gives rise to a wealth of structural diversity in organic chemistry. While there is no simple mathematical relationship between the molecular formula and the number of constitutional isomers, you can be certain that molecules with a large number of carbon atoms will have a large number of constitutional isomers.

13.3-2. Isomers of Hexane

Follow these steps to determine the constitutional isomers of an alkane.

• 1
  Start by drawing the straight chain isomer.
• 2
  Remove one carbon atom from the straight chain produced above and place the removed carbon (methyl group) on each carbon of the resulting chain that produces a different isomer.
  • •
    The methyl group cannot be placed on a terminal carbon atom as that would lead to the same chain as in the previous step.
  • •
    It does not matter which way the carbon atoms are numbered, so placing a methyl group on each of the carbon atoms that are next to the end carbon produces the same compound.
• 3
  Remove a second carbon atom and place the two methyl groups or one ethyl group on the remaining chain. The ethyl group cannot be placed on a terminal carbon as that would produce the longest chain obtained in step 1, and it cannot be placed on the second carbon as that would produce one of the isomers produced in step 2.
• 4
  Keep removing one carbon atom until you get to a length where no new isomers can be obtained.

13.3-3. Isomers of Heptane - an Exercise

Draw the eight isomers of the alkane C₇H₁₆ by adding methyl groups to the longest chain.

Stereoisomers

13.3-4. Definitions

Stereoisomers have the same connectivities, but they differ in the spatial arrangement of their atoms. There are two types of stereoisomers.

• •
  Geometric isomers: Different spatial arrangement of atoms occurs due to the presence of a bond that cannot be twisted completely. We consider only the case of the double bond.
• •
  Enantiomers: Different spatial arrangement of atoms occurs due to the presence of a carbon atom that has four different groups attached to it.

We take up geometric isomers first, then turn to a discussion of enantiomers.

13.3-5. Bond Twisting

Sigma bonds are cylindrical, and the groups bound by them rotate around the bond to adopt different relative positions. Figure 13.12(a) shows the rotation about the sigma bond. However, groups connected by double bonds cannot rotate relative to one another without breaking the π bond because the rotation would move the two p orbitals used in the π bond away from one another and remove their overlap.

13.3-6. Cis and Trans Isomers

13.3-7. Retinal

The trans form of an alkene can be twisted into the cis form if the pi bond is broken. Indeed, it is a trans → cis conversion that initiates the vision process. Protein molecules in the eye hold retinal molecules in a structure where one double bond is in the less stable cis configuration. The retinal molecule can absorb a photon of visible light due to the presence of alternating single and double bonds. The energy of the photon breaks the pi bond at the cis double bond. Once the pi bond is broken, the molecule springs back to the more stable trans configuration. This trans → cis conversion initiates a cascade of events that results in vision.

13.3-8. Naming Cis and Trans Isomers Exercise

13.3-9. Stereocenters

The atoms attached to a stereocenter do not have to be different, but the groups do. For example, a CH₃ group is different than a C₂H₅ group, but both involve a carbon atom attached to the stereocenter. The following exercise should help.

13.3-10. Enantiomers

There is additional complexity in the structures of organic molecules that possess carbon atoms with four different groups attached to them. Such carbon atoms are said to be stereocenters. The central carbon shown in red in
molecule A in Figure 13.14 is a stereocenter because it contains four different groups: H, OH, CH₃, and Cl. Molecule B in Figure 13.14 is its mirror image. Molecules A and B are different molecules! That they are different can be seen by rotating molecule B by 180° about the C–C bond (B₁ → B₂). The rotation makes the OH and CH₃ groups of the two molecules (A and B₂) superimposable. However, the H and the Cl groups are still reversed in the two molecules. Consequently, molecules A and B are not superimposable. Note that your right and left hands are also mirror images that are not superimposable. Molecules with a stereocenter cannot be superimposed on their mirror images. Pairs of molecules that are nonsuperimposable mirror images are called enantiomers. Enantiomers are stereoisomers because their spatial arrangements are different while their connectivities are identical. Molecules A and B are enantiomers.

13.3-11. Enantiomer Exercise

13.3-12. Enantiomeric Compounds

Enantiomers have the same melting and boiling points, but they often have dramatically different biological activities because biological activities are often based on the structures of the reacting molecules. Thus, one compound may cause one effect, but its enantiomer may cause a very different effect or have no effect at all. All of the biologically active amino acids except one have stereocenters, and in each case, only one enantiomer is active. The following are a couple of examples.

13.3-13. Counting Mirror Images

A large structural diversity can occur in molecules that contain several stereocenters. In general, there are 2ⁿ stereoisomers of a molecule with n stereocenters. Thus, there are 2 stereoisomers of a molecule with only one stereocenter, 2² = 4 stereoisomers of a molecule with two stereocenters, and 2¹⁵ = 32,768 stereoisomers of a molecule with 15 stereocenters.

13.4 Functional Groups

Introduction

Organic molecules are designed with specific sizes, shapes, and functional groups to perform specific functions (e.g., a better drug or a stronger plastic). Once a molecule is designed, organic chemists develop procedures to make it. These procedures, which can involve many steps, can be developed because so many organic reactions proceed by a few well understood mechanisms. Organic mechanisms are based on the principles of the octet rule, resonance, and Lewis acid-base theory. Reactions occur at specific groups within the molecule called functional groups. Functional groups have specific reactivities that can be utilized to produce desired reactions. In this section, we introduce some of the functional groups and the fundamental types of reactions they undergo.

Objectives

• •
  Identify some common functional groups.

13.4-1. Alcohols

Alcohols can behave as Brønsted acids because they contain a hydrogen atom attached to an electronegative atom, but they are also Lewis bases because the oxygen atom contains two lone pairs. Alkyl groups are usually electron donating, so alcohols in which R is an alkyl group function more as Lewis bases than acids. Alcohols in which R is an alkyl group are named by adding -anol to the root name of the alkyl group. The position of the hydroxyl group is indicated with the number of the carbon atom to which it is attached.

13.4-2. Carbonyls

The carbon atom is sp² hybridized and is surrounded by only three electron regions. In addition, the electronegative oxygen places a partial positive charge on the carbon atom. These two properties make the carbon atom of a carbonyl a Lewis acid. In Figure 13.19, the one on the left shows the resonance form of a carbonyl that describes the bonding, and the one on the right describes the reactivity of the carbonyl—the carbon atom of the carbonyl can be attacked by the lone pair of a Lewis base. The Lewis acidity of carbonyl groups gives rise to dozens of types of chemical reactions that differ only in the species serving as the Lewis base.

13.4-3. Carboxylic Acids

The carbonyl oxygen removes electron density from the O–H bond, which makes carboxyl groups acidic. Carboxylic acids in which R is an alkyl group are named by identifying the longest continuous chain that contains the carboxyl carbon and adding "anoic" to the root name of the chain. For example, HCOOH is methanoic (or formic) acid, CH₃COOH is ethanoic (or acetic) acid, and C₂H₅COOH is propanoic acid. The conjugate base of the acid is named by changing the "ic" to "ate." Thus, C₂H₅COO^1– is the propanoate ion. Esters are named as the name of the alkyl group followed by the name of the group to which it is attached. Consider Figure 13.22, which shows the example of acetic acid (active ingredient of vinegar), which loses a proton to form the acetate ion. Adding an ethyl group to the acetate ion produces the ester ethyl acetate (active ingredient of fingernail polish remover). The carbon atom of a carbonyl is Lewis acidic, so both carboxylic acids and esters are Lewis acids. In addition, the carbonyl group is electron withdrawing (electronegative oxygen atom), so the O–H bond in a carboxyl group is weakened, which makes carboxylic acids Brønsted acids. When R is a long alkyl chain, the acid is a fatty acid. Fatty acids are found in cells. The R group in fatty acids derived from animal cells is saturated, but those obtained from vegetables are unsaturated (they contain double bonds). Stearic acid (Figure 13.23), a fatty acid derived from animals (it is saturated), is used in making soaps and candles.

13.4-4. Amines and Amides

Organic derivatives of ammonia that are not bound to a carbonyl group are called amines. Two examples of simple amines are methyl amine, CH₃NH₂, and dimethyl amine, (CH₃)₂NH, are shown in Figure 13.24. Compounds in which a nitrogen atom is bound to a carbonyl carbon are called amides. The groups around the N atom in amides are coplanar, so the N is sp² hybridized. The planarity of the amide group is accounted for with the resonance structure in Figure 13.25. Caffeine contains two amide nitrogens (red), one amine nitrogen (blue), and one nitrogen that is neither an amine nor an amide because it is involved in a double bond.

13.4-5. Amino Acids

The amine is a good Lewis base, so it can react with the carboxylic acid to produce a charge-separated species, which is the most common form of amino acids in the human body. For example, consider the structure of alanine in Figure 13.27.

13.5 Introduction to Organic Reactions

Introduction

A good amount of organic chemistry can be understood with Lewis acid-base theory. Several mechanisms for different reactions are presented in this section, but all are Lewis acid-base reactions.

Prerequisites

• •
  5.6 Determining Lewis Structures (Draw Lewis structures.)
• •
  5.8 Formal Charge and Oxidation State (Determine formal charge.)
• •
  12.1-7 Curved Arrows in a Mechanism (Represent the mechanism of a Lewis acid-base reaction with curved arrows.)
• •
  12.1 Lewis Acids and Bases (Determine Lewis acidic and basic sites.)

Objectives

• •
  Identify the Lewis acidic and basic sites involved in an organic reaction.
• •
  Describe an addition reaction.
• •
  Describe a condensation reaction and use curved arrows to show the mechanism for the reaction of a carboxylic acid and an alcohol.
• •
  Use curved arrows to show the attack of an amine on an ester and predict the product of the reaction.

13.5-1. Addition Reactions of Alkenes

Addition reactions are reactions in which two reactants combine to form a single product. We use the addition of a hydrogen halide across a C=C double bond as an example. Alkenes are weak Lewis bases due to their pi electrons, and hydrogen halides are strong Brønsted acids. Hydrogen halides can add across the double bond in two Lewis acid-base reactions.

• 1
  The pi electrons of the double bond attack the hydrogen of the hydrogen halide to produce a C–H bond, a halide ion, and a carbocation.
• 2
  A lone pair on the halide ion attacks the strong Lewis acidic carbocation to form a C–X bond.

Note that the hybridization of each carbon changes from sp² (planar) to sp³ (tetrahedral) as a result of the addition reaction.

13.5-2. Esterification

In condensation reactions, two reactants combine to form two products (one of which is often a small molecule such as water). A carboxylic acid and an alcohol can undergo a condensation reaction called esterification to produce an ester and water. The mechanism of esterification is very similar to the one we used to explain the reaction of SO₃ and H₂O in
Chapter 12. Indeed, only the identities of the Lewis acid and base have changed. The reaction can be viewed as occurring in three steps.

• 1
  In the first step, the lone pair on the oxygen of the alcohol (the Lewis base) attacks the carbonyl carbon of the acid (the Lewis acid) in a Lewis acid-base reaction. The lone pair becomes a C–O bond, and the pi electrons of the C=O double bond of the carbonyl become a lone pair on the oxygen.
• 2
  A water molecule from the solvent (shown in green) assists in a proton transfer. It first accepts a proton from the oxygen with positive formal charge and then donates a proton to the oxygen involved in the O–H bond.
• 3
  The proton transfer in the second step produces an OH₂ group with a positive formal charge on the oxygen. Electron density moves out onto the oxygen with positive formal charge to produce a water molecule. The water molecule is lost and the C–O pi bond reforms, resulting in a molecule with no formal charge. Loss of small molecules such as water is common in condensation reactions.

13.5-3. Amide Formation

Amides can be formed in a Lewis acid-base reaction of an amine (base) and an ester (acid). The other product of this condensation reaction is an alcohol. The mechanism of the reaction is identical to esterification. The reaction can be viewed as occurring in three steps.

• 1
  The lone pair on the nitrogen atom of the amine (the Lewis base) attacks the carbonyl carbon of the ester (the Lewis acid) in a Lewis acid-base reaction. The lone pair becomes a C–N bond, and the pi electrons of the C=O double bond of the carbonyl become a lone pair on the oxygen.
• 2
  A water molecule from the solvent assists in a proton transfer. It first accepts a proton from the nitrogen with positive formal charge and then donates a proton to the oxygen involved in the O–R' bond.
• 3
  The proton transfer in the second step produces a R'OH group with a positive formal charge on the oxygen. Electron density moves out onto the oxygen with positive formal charge to produce an alcohol molecule (R'OH), which is lost as the C–O pi bond reforms, resulting in a molecule with no formal charge. Loss of small molecules such as alcohols is common in condensation reactions.

13.6 Polymers

Introduction

Natural polymers, such as carbohydrates, proteins, and nucleic acids, are part of living tissue and are responsible for increasing the rates of reactions and giving structural characteristics like strength and flexibility. Others store our genetic code. When most of us think of polymers, however, we think of man-made polymers like PVC, teflon, polyethylene, or polyesters. In this section, we examine both natural and man-made polymers.

Prerequisites

• •
  7.3-10 Hydrogen Bonding (Describe the hydrogen bond and state the requirement for it.)

Objectives

• •
  Define monomer, dimer, trimer, and polymer.
• •
  Identify the monomer upon which a polymer is based.
• •
  Recognize some common polymers.
• •
  State the requirement for the reactants used to produce a condensation polymer.
• •
  Distinguish between a polyamide and a polypeptide.
• •
  Explain why proteins frequently form α-helices.
• •
  Describe a base pair, explain why they are important, and identify the hydrogen bonding sites in a base pair.
• •
  Describe the double helix structure of DNA and explain its origin.

Addition

13.6-1. Polymers

"Mer" is used to indicate a building block. How the term mer is used is demonstrated in Figure 13.33, which shows a monomer (single mer), a dimer (two mers), a trimer (three mers), and a polymer (many mers). The polymer grows as the result of successive reactions of the monomers that lengthen the polymer chain. They typically have many kinks and folds because the single bonds are flexible. Although polymers can be formed by several different mechanisms, we consider only those reactions that occur by mechanisms like those that we have already discussed. Synthetic polymers are classified as condensation or addition polymers depending on the type of reaction that is used to form them.

13.6-2. Addition Polymers

Addition polymers based on alkenes include some of the most common polymers. The polymerization process is summarized in Figure 13.34. Note that the "x" in the figure is a large number typically in the thousands.

Polymer	Structure	Uses
polyethylene		plastic bottles and bags
propylene		carpets
isobutylene		tires
poly(vinyl chloride) (PVC)		plumbing and hoses
polystyrene		insulation
tetrafluoroethylene (teflon)		cooking utensils and nonstick cooking surfaces

Table 13.6: Structures and Uses of Some Common Polymers

13.6-3. Isobutylene

The reaction of isobutylene with hydrogen halides was used as an example of an addition reaction in the previous section. We now examine the mechanism of the reaction. In the first step of that reaction, the pi electrons attacked the hydrogen of HX to produce a C–H bond and a carbocation on the carbon (the carbon with more alkyl groups). The carbocation is a strong Lewis acid, which was immediately attacked by the strongest base in the vicinity, the halide ion. However, only trace amounts of HX are used in the polymerization, so there is no halide ion close to the carbocation. Consequently, the next strongest base (another isobutylene molecule) reacts with the cation. Reaction of an isobutylene molecule with the carbocation produces a dimer that still contains a carbocation. The dimer then reacts with a third isobutylene to form a trimer, and so on to produce poly(isobutylene). A typical poly(isobutylene) chain consists of from 2,000 to 20,000 monomer units. Natural rubber is a polymer of isoprene (see Figure 13.36) that can now be made in the chemistry laboratory. Butyl rubber is a polymer made from isobutylene and isoprene. The United States produces over 500 million pounds of butyl rubber each year.

Condensation Polymers

13.6-4. Condensation Polymers

The requirement for polymerization is that the reactants must each have two reacting centers. The formation of an ester from a carboxylic acid and an alcohol was used in the previous section as an example of a condensation reaction, and the formation of a polyester from a diacid and a diol (compounds with two carboxyl and two hydroxyl groups) is demonstrated in Figures 13.37a, 13.37b, and 13.37c. The polyester poly(ethylene terephthalate) or PET is used in clothing, upholstery, tires, and other products. It is marketed as Mylar, Dacron, and Terylene. 4.5 billion pounds of PET are produced in the U.S. every year.

13.6-5. Nylon 66

Just as reacting an ester and an amine produces an amide, reacting a diester and a diamine produces a polyamide. Polyamides produced in this manner are also called nylons. Each step of the polymerization produces an amide bond, but the product always contains an amine or an ester at each end, so it can react with another diester and/or another diamide. Mechanism:

• 1
  Lone pairs on N atoms of diamines (red and blue in Figure 13.38a) attack carbonyl carbon of diester to form C–N bonds.
• 2
  Proton transfers (shown in circles in Figure 13.38a) from diamine nitrogens to oxygens of diesters produce CH₃OH groups that leave as methanol molecules.

The produced amide also contains two amines, which can each attack another diester. The product of those reactions would contain four amide linkages and two ester groups, each of which could be attacked by another diamine. Each reaction lengthens the polymer chain. One monomer building block for a polyamide is shown in Figure 13.38b. Note that it contains six carbons from the diester between amide linkages (shown in blue) and six carbons from the diamine (shown in red), so the product is named Nylon 66.

13.6-6. Polypeptides

Polyamides can also form when each reactant contains both a carboxylic acid and an amine rather than two acids or two amines, i.e., polyamides can also be produced in the reaction of amino acids. Amides produced by the reaction of amino acids are called peptides, and the polyamides made from amino acids are called polypeptides. The resulting peptide always has two reactive sites, and continued reaction with other amino acids leads to a polypeptide or eventually a protein. The structure of a polypeptide can be envisioned as a backbone of amide linkages, from which extend various side chains that characterize the various amino acids. Indeed, the amide backbone is frequently represented by thick bonds from which hang the side chains from the amino acids (methyl group from alanine and ethyl group from serine).

13.6-7. H-Bonding in Proteins

Proteins are high molecular weight polypeptides that often contain over 100 amino acids (called residues). Hydrogen bonding and dipole-dipole interactions between groups within a polypeptide are responsible for the secondary structure of a polypeptide. The strongest interaction is hydrogen bonding between an N–H group and a C=O group that is four residues away. This interaction results in the α-helix structure of proteins. The nature and position of the amino acids around the core of the α-helix are responsible for the specific function of the protein.

13.6-8. Protein Structure

An α-helix is shown in Figure 13.41. Color code: red = oxygen, blue = nitrogen, gray = carbon, white = hydrogen, and yellow = sulfur. Note only those hydrogen atoms involved in hydrogen bonding are shown.

• (a) Space-filling model
• (b) Ball-and-stick model with green ribbon to show polypeptide backbone
• (c) Ball-and-stick model in which the side chains have been removed, and hydrogen bonds are shown as dashed lines

DNA

13.6-9. Nucleotides

A nucleotide consists of a phosphate, a sugar, and an N-containing base. Nucleic acids are polymers formed from nucleotide residues. If the sugar in the nucleotide is deoxyribose, then the nucleic acid is a DeoxyriboNucleic Acid, or simply DNA.

13.6-10. Bases

Each nucleotide of DNA contains one of the following four bases. The location marked "backbone" indicates the position where each base bonds to the sugar in the nucleotide. The above bases can be combined into two base pairs due to the strong hydrogen bonding that can exist between the members of each pair.

13.6-11. Determining Base-Pairs Exercise

13.6-12. Base Pairs

There are favorable interactions in only two pairs of bases, which are called base pairs.

• 1
  adenine (A) - thymine (T)
• 2
  guanine (G) - cytosine (C)

In Figure 13.45, adenine and guanine are covalently bound to strand 1, and thymine and cytosine are covalently bound to strand 2. Hydrogen bonding of the adenine on strand 1 with the thymine on strand 2 and the guanine on
strand 1 with the cytosine on strand 2 hold the two strands of DNA in this portion in a double helix.

13.6-13. Double Helix and Function of DNA

The structural unit of DNA is called the double helix. The double helix is a consequence of hydrogen bonding between base pairs on different strands. It is the sequence of the base pairs in the double helix that forms the genetic code. Thus, your genetic code requires only four different chemical letters: A, T, C, and G. However, the human cell contains about three billion base pairs.

13.7 Exercises and Solutions

Select the links to view either the end-of-chapter exercises or the solutions to the odd exercises.

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