Sorrell 2nd Chapt 1

THE STRUCTURES OFORGANIC MOLECULES

1.1 STRUCTURAL COMPONENTS OF ORGANIC MOLECULES

1.2 STRUCTURAL FORMULAS AND CONDENSED STRUCTURES

1.3 SYSTEMATIC NOMENCLATURE: IUPAC NAMES

1.4 CONSTITUTIONAL ISOMERS AND HYDROCARBON SUBSTITUENTS

1.5 NAMING ORGANIC COMPOUNDS

CHAPTER SUMMARY

The organic art of Hans Hoffman. Organic gardening. Frank Lloyd Wright’s organicstyle of architecture. Organic wines. The adjective organic became increasingly popularduring the latter half of the twentieth century because of its association with the notionof life or naturalness regarding its subject. Similarly, the term organic chemistry was usedduring the early nineteenth century to describe the study of substances related to liv-ing, or once-living, organisms, a concept that grew out of the prevailing belief of thatday in vitalism, the life-giving force of the universe. Those substances such as rocks andminerals that had no obvious relationship to living things were referred to as inorganic.

With passage of time, the differentiation between living and nonliving substancesbecame less clear as newly discovered chemical reactions could be used to interconvertmaterials. Consequently, scientists came to care more about whether chemical compoundshad high or low melting points, were soluble or not in water, or were liquids, solids, orgasses. Using these criteria, chemists realized that molecules containing carbon sharemany similar properties. Gradually, organic chemistry came to be defined as the chemistryof substances containing carbon, the definition we currently recognize and use.

The field of organic chemistry is extensive, encompassing structures and reactionsof literally millions of molecules. It is an important area for study and research formany reasons, two of which warrant mention.

1. The structures of organic molecules are well understood, so new types of com-pounds are readily prepared by rational approaches. This ability to make new sub-stances with predictable structures and properties is one aspect that sets organicchemistry apart from other branches of chemistry, physics, and biology.

2. The essential processes of biology are chemical reactions of organic molecules.Studying biochemistry requires one to understand organic chemistry, a primaryreason why many students—perhaps you—have chosen to study the subject. Learn-ing organic chemistry is a natural prerequisite for entry into many medical, dental,pharmacy, and graduate schools.

This chapter presents some central concepts about the structures of organic mol-ecules and ways to depict them. In addition, basic rules of nomenclature are presentedso that you are able to communicate with others about these substances. When you

C H A P T E R

1

1

have finished this chapter, you should be able to draw structural formulas, and youshould be familiar with the general strategies required to interpret the names of manyof the molecules that you will encounter in this course.

1.1 STRUCTURAL COMPONENTS OF ORGANIC MOLECULES

1.1a A CARBON ATOM FORMS FOUR BONDS TO NEIGHBORING ATOMS

IN STABLE MOLECULES

If you are unfamiliar with the electronic structures of atoms and how bonds areformed, you may be puzzled about the diversity of structures you see in this chapter.Chemists in the nineteenth century knew little about bonding, yet they performed andpredicted the products of reactions readily. You can therefore assume that much can belearned without a detailed knowledge of bonding models, which are presented in thenext chapter (Chapter 2).

For now, it is sufficient to know that a carbon atom forms four bonds in a stable, isolable com-pound. This pattern exists because the shell of electrons around the nucleus of the carbonatom (and the other second-period elements N, O, and F) will be filled, hence stable,when eight electrons are present. Each bond has two electrons, so four bonds provideeight electrons. The bonds may be single, double, or triple, as illustrated in Figure 1.1.

2 CHAPTER 1 The Structures of Organic Molecules

H C C HHH

H

H

C H

H

H

C H

H

H

C

H

H H

H

C C

Methane Ethane

Contains a carbon–carbon single bond

Ethylene

Contains a carbon–carbon double bond

Acetylene

Contains a carbon–carbon triple bond

Figure 1.1Representative structures ofcarbon-containing substances.

H

H

H

C N CH

H

H

C N

Ammonia

Nitrogen:normally forms three bonds

Oxygen:normally forms two bonds

Aminomethane Acetonitrile

OH H HOH

H

H

C OC

Water Methanol Formaldehyde

N

NH H

H

H

H

H

C

H

H

C

H

Br

Halogen atom (F, Cl, Br, I):normally forms one bond H

H

H

C

Chloromethane Bromoethene

Cl

Figure 1.2Representative structures ofsubstances with elements inaddition to carbon andhydrogen. Notice that eachcarbon atom forms fourbonds and each hydrogenatom forms one.

In a stable organic molecule, nitrogen usually forms three bonds to neighboringatoms, and oxygen normally forms two bonds. Except in rare instances, hydrogen andthe halogens [fluorine (F), chlorine (Cl), bromine (Br), and iodine (I)] form onebond to another atom. Figure 1.2 illustrates these trends.

1.1b A FUNCTIONAL GROUP IS A REACTIVE CENTER THAT CONTAINS

HETEROATOMS OR A MULTIPLE CARBON–CARBON BOND

Given the general bonding properties of atoms that constitute a variety of molecules, asurvey of organic compounds would reveal features that are common among them.Four principal characteristics would emerge:

1. All contain carbon, and bonds usually exist between two or more carbon atoms.

2. Most contain hydrogen, often, but not always bonded to carbon.

3. Many contain elements besides C and H. These are called heteroatoms, and themost common are oxygen, nitrogen, sulfur, phosphorus, silicon, and the halogens.Heteroatoms can form bonds with each other or with carbon. The bonds may be sin-gle, double, or triple.

4. Most contain a functional group, which is defined as any group of atoms other thanthose with only carbon–carbon and carbon–hydrogen single bonds. Functional groupsare normally the reactive portions of a molecule, and they range in complexity fromthe carbon–carbon double bond to groups with multiply bonded heteroatoms.

For the sake of understanding the structures and reactions of organic compounds,you will need to learn the identities of common functional groups. Many, but not all,of the functional groups you will encounter in this text are listed and illustrated inTable 1.1. The condensed structures and the use of suffixes will be discussed shortly.

EXERCISE 1.1

For each of the following compounds, name each functional group that is present.

1.2 STRUCTURAL FORMULAS AND CONDENSED STRUCTURES

1.2a CONDENSED STRUCTURES ELIMINATE THE NEED TO

SHOW EVERY ATOM AND BOND IN A MOLECULE

Drawing every bond between pairs of atoms, as in the structural formulas alreadyshown, becomes tedious and cumbersome, especially in complex molecules. Fortu-nately, easier ways suffice to represent structures of organic compounds. A commonway to simplify a structure is to express groups of atoms that are bonded together as asingle entity, generating a condensed structure.

HO

H

H

C

H

H

C

H

H

C O HC

O H

H

C

H

H

C C N

H

H

N

C

C

C

C

CH

H HH

C

OC

H

H

H

H H

H H

H

H H

C

O

CC

C CC C

H

H

H H

H

HHH

H SH

1.2 Structural Formulas and Condensed Structures 3


C C C C

C C C C

C N

COCl

OH OH

ClC

O

HC

O

C

O

N

H

H

OHC

O

OC

O

SH SH

OHS

*

*

*

*

*

*

O

O

C

O

O O

CHO

NH2

COOH

CONH2

CO2H

COO CO2

CO

CN

SO3H

Structure a Condensed structure Name Suffix b

Acid chloride

or

or

-oyl chloride

Alcohol orPhenol

-ol

Aldehyde -al (-carbaldehyde)

Alkene -ene

Alkyne -yne

Amide -amide(-carboxamide)

Amine -amine

Carboxylic acid -oic acid(-carboxylic acid)

Ester -oate (-carboxylate)

Ether ether

Benzene benzenec

Ketone -one

Nitrile -nitrile(-carbonitrile)

Sulfonic acid -sulfonic acid

Thiol -thiol

NH2

Incr

easi

ng

pri

ori

ty

aFunctional groups with an * must appear at the end of a carbon chain or be attached to a ring. b The suffixes in parentheses are used in names of cyclic compounds. c The benzene ring is a functional group, but its ring is also the carbon skeleton in many aromatic compounds as explained in Section 1.3a.

H

H

H

H H

C

C

C

C

C

C

Table 1.1 Functional groups and their IUPAC nomenclature suffixes.


As an illustration, consider the following structure:

All of the atoms have the normal number of bonds (C, four; O, two; and H, one).Starting with the fragments that have only carbon and hydrogen atoms, which arecalled hydrocarbon fragments, we condense them to methyl (CH3), methylene (CH2), andmethine (CH) groups, which are shown in Table 1.2.

This first level of simplification gives the following condensed structures:

Notice that we can show the bonds between the hydrocarbon fragments or we can leavethem out. Because we know each hydrogen atom forms only one bond, it is acceptedthat the hydrogen atoms are attached only to the carbon atom on their immediate left.

The second level of condensed structures puts functional groups into their con-densed formats. These structures are listed in Table 1.1. For this molecule, therefore,we would draw one of the following condensed structures:

CH3CH2CH2COOH or CH3CH2CH2CO2H

O HC

O

CH2CH2CH3 O HC

O

CH3CH2CH2or

H

H

H

C

H

H

C

H

H

C O HC

O

Name Structure Condensed form

Methyl

Methylene

Methine

H

H

C H

H

H

C

C H CH

CH3

CH2

Table 1.2 Hydrocarbon fragments.

EXAMPLE 1.1

Represent the following structural formula in a condensed format:

H

H

C

Cl

H

C

H

H

C H

H

H

N

First, convert the hydrocarbon units to their condensed representations.

Next, express the functional group with its condensed notation. Any of the followingwould be acceptable:

Notice that we write H2N rather than NH2 when the amino group appears on the left.The same holds true for alcohols (HO, not OH), carboxylic acids, (HOOC, notCOOH), aldehydes (OHC, not HOC or CHO), and thiols (HS). We can also write H3Cfor a methyl group, but CH3 is preferred.

EXERCISE 1.2

Represent each of the compounds in Exercise 1.1 in a condensed format. For cycliccompounds, connect the condensed hydrocarbon units with bonds to form the ap-propriate size ring.

EXERCISE 1.3

Draw structural formulas for each of the following compounds, showing every bond:

1.2b HYDROCARBON FRAGMENTS MAY BE REPRESENTED

BY ANGLED LINES AND POLYGONS

The propensity of carbon to form four bonds, coupled with the ubiquity of hydrogenin organic compounds, has led to an even more abbreviated notation than the oneshown in Section 1.2a. Using lines for the carbon–carbon bonds and ignoring all car-bon–hydrogen bonds, we use an intersection of lines to define the position of each car-bon atom. Heteroatoms and hydrogen atoms attached to heteroatoms are written.

As an example, consider the following molecule:

We can draw a variety of condensed structures according to the procedure outlinedin Section 1.2a, and several are shown in the following scheme. Make note that the ori-

H

H

H

C

H

H

C

H

H

C

H

H

C HC

O

2-Pentanone

CH3CH2CHO

HSCH2CH2OCH3 CH3CHBrCH2COOCH3

CH2

CH2

CHOHH2C

H2Ca.

c.

b.

d.

Cl

CHCH2H2N CH3CHClCH2H2N CH3 H2NCH2CHClCH3

ClH

H

N CHCH2 CH3


entation of the structure on the page is not crucial, a point that will become clearer insubsequent chapters when you learn more about three-dimensional representations.

To condense the structures further, we omit the atom labels (C and H) and drawlines for each carbon–carbon bond. The meeting point of two lines, which we draw tooccur at an angle, represents the position of a carbon atom. Any heteroatoms (in this ex-ample, oxygen) are always shown. We know that each carbon atom forms four bonds, sowe can mentally include the requisite number of hydrogen atoms at each position tocomplete the structure. Again, the orientation of the structures shown below is arbitrary.

At the beginning, you may want to place a dot at each carbon atom position. Thismeasure will help you keep track of the carbon atoms while you learn to convert be-tween line and condensed structures.

There are instances in which it is helpful to include some of the carbon or hydro-gen atom labels. A specific case occurs when a functional group appears at the end ofthe chain.

To represent compounds with a ring, we draw a regular polygon in which each ver-tex of the ring represents a carbon atom position. Functional groups attached to a ringare usually written out.

O

H

Br

CHO CN

Hydrogen atom shown. Condensed functional group notation used;other carbon atom positions indicated by angles.

CCH CH2

CH3CH3

H3C

OOO

� �

O

1

1

O2

1

O

2O

1

O

1 4

2 3

At the end of the chain, onlyone bond to the carbon atomis drawn; the three hydrogenatoms are not drawn.

Only two bonds to the indicatedcarbon atoms are shown. The twohydrogen atoms on each carbonatom are not shown.

All four bonds of this carbonatom are drawn. A double bondcounts as two bonds.

C

O

CH2 CH3CH2CH3 C

O

CH2 CH3CH2CH3

CH3C

O

CH3CH2CH2 CH3C

O

CH3(CH2)2 CH3CH2CH2COCH3

The orientation of a structure on the page is not important;all of the structures and formulas shown here are equivalent.


EXAMPLE 1.2

Draw the line structure for the condensed representation given here:

First, indicate each carbon atom position, including the bonds between carbon atoms,leaving the heteroatoms and functional groups in place (below, a.).

Finish by expanding the carboxylic acid group to show all atoms, and then add zero,one, two or three hydrogen atoms to each carbon atom to give each a total of four bonds(below, b.).

EXERCISE 1.4

Draw the structural formulas for each of the following condensed representations:

EXERCISE 1.5

Draw a condensed structure for each of the following compounds:

1.3 SYSTEMATIC NOMENCLATURE: IUPAC NAMES

Nomenclature is not the most interesting topic in any new subject you study, but to un-derstand what others are talking about, it is crucial to gain knowledge of the basic con-cepts. When you were learning to talk, your parents made sounds that you tried toimitate, even though it took months to construct words and sentences yourself. Like-wise, if you have studied a foreign language, you probably spent the first part of thecourse learning to imitate your instructor. In either case, it was more important at thebeginning for you to understand what was said than to make statements yourself. If

C

C

C

CH

H CC

H

H

C

H

OC

O

CH3

H

H

C

CNCHCH3CH2CH2CH2

Cl

C

a. b.

O

CH3

NHO

CHO

a. b.

C

C

C

C

C

CCCl CO2H

CCCl

C

C

H HH

H HC

CH

C

CH

O

O

H

H

a. b.

Cl CO2H


your parents said, Don’t touch the stove, that command was crucial to your safety. It isunlikely that you had to warn your parents about getting burned.

So it is in your study of organic chemistry: It is initially more important for you tounderstand the names that appear in this book or are used by your instructor. For thatreason, the emphasis in this chapter will be on interpretation of names. As you master this“new language”, you will learn much about the structures of organic molecules. Thisknowledge will later be applied to assimilate information about chemical reactionswhen you encounter them. The process of naming molecules will follow naturally fromlearning how to interpret names.

When learning nomenclature, you can easily become overwhelmed by hundreds ofdetailed rules. Focus instead on the basic pattern that underlies most of the compoundnames that you will encounter in this book. Because systematic names share the sameformal construction, which is presented in Section 1.3a, you can easily learn how to in-terpret many names within a short period of time. Some of the more complex types ofmolecules as well as more detailed rules will be presented in later chapters.

1.3a THE SYSTEMATIC NAME OF AN ORGANIC MOLECULE

IS BASED ON THE NAME OF THE LONGEST CARBON

CHAIN OR ON THE SIZE OF A RING

In 1949, the International Union of Pure and Applied Chemistry (IUPAC) formulatedrules for naming organic compounds. These IUPAC rules are most commonly usedtoday. Insofar as possible, this book will use IUPAC rules and names.

Before the IUPAC rules were accepted, trivial or common names were widely em-ployed, and many of these were incorporated into the IUPAC system. Because trivialnames are still used (and are common in the older literature), you will need to learnmany of them. This situation is analogous to that for languages, which can be eitherformal or colloquial. In written documents, the formal system is followed more strictly,but in speaking, colloquial words are used because they are often easier to say.

The IUPAC name of an organic compound can be divided and allocated amongfour fields, a process illustrated in the chart below and discussed in subsequent sections.

• The first field includes the names and positions of substituents.

• The second field contains the compound root word.

• The third field is what we call the multiple-bond index.

• The fourth field defines the principal functional group.

Substituents

4-Chloro- pent an 2-one

2,3-Dimethyl- hex 2-ene

3,3-Diphenyl cyclobut an ol

2-Chloro-5-nitro phenol

3-Hydroxy but an al

Compound root Mulitple-bondindex

Principal Functionalgroup

First field Second field Third field Fourth field

4-Chloro-2-pentanone:

2,3-Dimethyl-2-hexene:

3,3-Diphenylcyclobutanol:

2-Chloro-5-nitrophenol

3-Hydroxybutanal

Examples:

1.3 Systematic Nomenclature: IUPAC Names 9

Start by identifying the root word, which derives from the longest continuous carbonchain in the case of aliphatic compounds. The term aliphatic originates from a Greekword for “fat”, the source of many hydrocarbons isolated in the 1800s. Today, we use theword aliphatic to refer to those organic compounds that have chains of carbon atoms.These chains can be straight or branched, but they lack rings. The root word in the nameof an aliphatic compound is based on the identity of the corresponding alkanes, hydro-carbon molecules with only single bonds between atoms. The names and structures ofthe 20 simplest alkanes—those with unbranched chains—are shown in Table 1.3.

The root words for compounds with rings are only slightly more complicated. Ali-cyclic (aliphatic cyclic) compounds, which have rings and a preponderance of singlebonds, use the root word of the alkane with the same number of carbon atoms, but theprefix “cyclo“ is inserted just in front of this root, as illustrated in Figure 1.3.

Benzene is a cyclic compound that is the simplest member of the family of sub-stances known as aromatic compounds or arenes. Their distinguishing feature is thepresence of apparently alternating single and double bonds within a six-memberedring. Figure 1.4a shows several ways to draw benzene.

Many aromatic compounds are known by the common names given in Figure 1.4.The IUPAC system incorporated many of these common names, so “toluene”, “phenol”,


StructureAlkane name Root word

Decane Dec-

Nonane Non-

Octane Oct-

Heptane Hept-

Hexane Hex-

Pentane Pent-

Butane But-

Propane Prop-

Ethane Eth-

Methane Meth-

StructureAlkane name Root word

Eicosane Eicos-

Nonadecane Nonadec-

Octadecane Octadec-

Heptadecane Heptadec-

Hexadecane Hexadec-

Pentadecane Pentadec-

Tetradecane Tetradec-

Tridecane Tridec-

Dodecane Dodec-

Undecane Undec-

CH3

CH3

CH3

CH3

CH3

CH3

CH3 CH2

(CH2)6 CH3

(CH2)7 CH3

(CH2)8 CH3

(CH2)5 CH3

(CH2)3 CH3

(CH2)4 CH3

CH3 CH2 CH3

CH2 CH3

CH3 CH3

CH4

CH3

CH3

CH3

CH3

CH3

CH3

CH3 (CH2)12 CH3

(CH2)10 CH3

(CH2)11 CH3

(CH2)16 CH3

(CH2)17 CH3

(CH2)18 CH3

(CH2)15 CH3

(CH2)13 CH3

(CH2)9 CH3

(CH2)14 CH3

CH3

CH3

CH4

Table 1.3 Structures and names of alkanes with 1–20 carbon atoms.

Cyclopropane Cyclobutane Cyclopentane Cyclohexane Cycloheptane

There is a CH2 group at each vertex of the polygon.

Figure 1.3Structures and names of cycloalkanes with three-to-seven carbon atoms.

and so on are considered systematic names. The names of other arenes will be coveredlater, but for now concentrate on knowing the names and structures shown in Figure 1.4band on recognizing the root word “benz”, used for many of benzene’s derivatives such asbenzaldehyde (C6H5CHO), benzoic acid (C6H5COOH), and benzonitrile (C6H5CN).

1.3b THE MULTIPLE-BOND INDEX SPECIFIES THE NUMBER AND TYPE

OF MULTIPLE BONDS BETWEEN CARBON ATOMS

In an IUPAC name, the multiple-bond index follows directly after the compound root andindicates whether double or triple bonds are present. If no carbon–carbon multiplebond is present, then the suffix “ane” (or “an”) follows the root.

The suffix “ene” refers to the presence of a double bond between two carbonatoms (C=C), and “yne” refers to a triple bond between carbon atoms (C ∫C). A com-pound that has a carbon–carbon double bond is called an alkene, as a general class,and one with a triple bond is an alkyne. If the molecule has two double bonds, the word “diene” follows the compound root; for three double bonds, “triene”; and so on.Table 1.4 matches these suffixes with specific examples.


H

Ha.

H H

H H

C

C

C

C

C

C

H

H

H H

H H

HN

HH HO

CH3HO

Hb.

C

Benzene, C6H6, is the simplest of the “aromatic compounds.” All of the structuresshown above are equivalent. For the two structures at the right, a hydrogen atomis attached at each vertex of the hexagon. The significance of the circle within thering will be discussed in more detail in Chapters 2 and 17.

Aniline AnisolePhenolToluene

Just as for benzene, there is a hydrogen atom at each vertex of the hexagon thatdoes not have another atom already attached.

Figure 1.4The structures of benzene (a.) and of aniline, anisole, phenol, and toluene (b.).

Suffix Meaning Example

Two double bonds-diene

One C–C triple bond-yne

One C–C double bond-ene

No C–C double or triple bonds

1,3-Butadiene

1-Butyne

1-Butene

Butane-ane CH2 CH3CH2CH3

CH CH2CHCH2

CH2 CH3CHCH2

CH2 CH3H C C

2 3 41

2 3 41

2 3 41

Table 1.4 Suffixes for the multiple-bond index illustrated for compounds with four carbonatoms (root = but—).

A multiple bond can appear anywhere within a chain of carbon atoms, so a nu-meral is often included in the name to specify at which carbon atom the double ortriple bond begins. Figure 1.5 shows some examples. A molecule with no other func-tional group is numbered from the end that gives the lower possible number to the po-sition of the multiple bond. For a diene or triene, position numbers are given for eachdouble bond. In a cycloalkene or cycloalkyne, the double or triple bond is assumed tostart at C1, unless otherwise specified. (Incidentally, carbon atom positions are de-scribed using one of the following equivalent phrases: “at the 2-position”, “at C2”, or “atthe number 2 carbon atom”. When the numerals are subscripts [e.g., C2H5], they referto the total numbers of atoms.)


EXERCISE 1.6

Based on the structures in Tables 1.3 and 1.4 and Figure 1.5, draw structural formulasand condensed structures for each of the following alkenes and alkynes:

a. 3-Hexene b. 4-Octyne c. 1-Butene-3-yne d. Cyclobutene

1.3c THE IDENTITY OF THE PRINCIPAL FUNCTIONAL GROUP APPEARS

AT THE END OF THE COMPOUND’S NAME

In the IUPAC system, the identity of the principal functional group appears at the endof the name. Common suffixes for the functional groups were given in Table 1.1. Otherfunctional groups may be present in the molecule as well, and these will be specified assubstituents. What constitutes the “principal” group is based on a priority ranking, andTable 1.1 is organized so that the highest priority group is at the top, and the lowest isat the bottom. (Even though benzene is listed at the bottom of Table 1.1, its presencein a molecule is reflected by the compound root word.)

1

2

CH2 CHCH2CH3 CH2

5 4 3 2 1CH2CH2CH3 CH3C C

1 2 3 4 5 6

1

21524

3

6H

H

C

C

CH2

CH2

H2C

H2C

1-Pentene 2-Hexyne

Cyclohexene Cyclooctyne

Shown are two representations of this six-memberedring compound. By convention, the double bond startsat carbon atom 1 (C1), so its position does not need tobe specified in the name.

Cycloalkynes with fewer than eight carbon atomsin the ring are not stable at room temperature.

Figure 1.5Examples of numbering in compounds with carbon–carbon double and triple bonds.

Some functional groups, by their very nature, must be at the end of the carbonchain because the carbon atom within the group already has three bonds to hydrogenor heteroatoms. Examples include aldehydes, carboxylic acids, and nitriles (an asterisknext to each name in Table 1.1 identifies these groups). Incidentally, the carbon atomat the end of a chain, whether part of a functional group or not, is called the terminalcarbon atom.

Some functional groups can be located at practically any position of a chain, so anumeral is added in those cases to specify its position, as was the case for multiplebonds. Ketones and alcohols are the most common functional groups in this category.This numeral immediately precedes the functional group suffix if a multiple bond ispresent; otherwise it appears in front of the root word. The following examples illus-trate names that have been separated into their parts (root, multiple-bond index, andprincipal functional group).

When a functional group that is required to be at the end of chain is attached to aring, then a different suffix is required (Table 1.1) because the ring’s root word does notinclude the carbon atom of the functional group. Common groups in this category in-clude the aldehyde (suffix = carbaldehyde) and carboxylic acid (suffix = carboxylic acid)groups. The following structures illustrate how these suffixes are treated. Notice that thecarbon atom in the ring at which this functional group is attached is designated C1.

UNN 1.12 GOES HERE

C

O

CH2 OHCH2CH2CH35 4 3 2 1

CH2CH2CH31 2

CH25

CH364

C

O

3

H C

O

1CH3CHCH

2 3 4

CH2 CH2 OHC C4 3 2 1

CH35

Pentanoic acid

pent 5 carbon atoms an no multiple bonds (C– C single bonds only)oic acid carboxylic acid functional group, its C atom defines C1

2-Butenal

but 4 carbon atoms en double bond, at C2 al aldehyde functional group, its C atom defines C1

3-Hexanone

hex 6 carbon atoms an no multiple bonds (C– C single bonds only) one carbonyl group (ketone) at C3

3-Pentyne-1-ol

pent 5 carbon atoms yne triple bond, at C3 ol OH group (alcohol), at C1


1

23

4

5 6

OHC

O

1

23

45

HC

O

Cyclohexanecarboxylic acid 2-Cyclopentenecarbaldehyde

cyclohex a ring of 6 carbon atoms ane no multiple bonds (C– C single bonds only)

carboxylic acid carboxylic acid functional group attached to the ring at C1

cyclopent a ring of 5 carbon atoms ene double bond, at C2

carbaldehyde aldehyde functional group attached to the ring at C1

EXERCISE 1.7

What is the principal functional group in each of the following compounds? Draw thefull and condensed structures of this generalized functional group, ignoring the com-pound root and substituents.

Example: 3-Chloropentanoic acid.

a. 2,2-Dimethyl-3-hexanone b 3-Methoxybenzaldehdye c. 2-Methyl-2-butanol

A practical and common way to designate the part of the molecule other than its func-tional group is by use of R, which may be thought of as “the Remainder of the molecule”.Similarly, Ar is used to designate an aromatic ring, usually a derivative of benzene. Thesesymbols are used when you want to focus attention on a particular functional group, andthe identity of the remaining structure is not crucial. Most of the time, R and Ar are usedto represent portions of the molecule that contain only carbon and hydrogen, but othersubstituents (Section 1.3d) are sometimes also included, depending on the context.

This introduction to functional groups would be incomplete without specific ref-erence to the carbonyl group, which is the structural unit with a carbon–oxygen dou-ble bond, C=O. (“Carbonyl” is pronounced carbon-EEL.) Looking at Table 1.1, you willsee several functional groups that contain the carbonyl group. The ketone functionalgroup consists of a carbonyl group attached to two carbon-containing fragments. Thealdehyde functional group has a carbonyl group attached to a carbon-containing frag-ment and a hydrogen atom. Several important functional groups are shown below inabbreviated format with the carbonyl group highlighted in color. Notice the use of Rand R¢, which means that the two “remainder” groups may not be the same.

EXERCISE 1.8

Draw the following structures in the more general form using the label R or Ar in com-bination with the principal functional group.

O

NH2 HC C

CH3

CHOa. d.b. c.

OH

CH3

CR R�

O

ketone

CR H

O

aldehyde

CR

HO

R�O

O

carboxylic acid

CR

O

ester

CR NH2

O

amide

H

CH3

O

CHO

� R

� R

� Ar

R OHRAr COOH

Cl

COOH

OH

Suffix = -oic acid = carboxylic acid = �C

O

OH

COOH


1.3d THE IDENTITIES AND POSITIONS OF SUBSTITUENTS ARE SPECIFIED

IN THE FIRST FIELD OF A COMPOUND’S NAME

Substituents are atoms or groups that appear in place of hydrogen atoms attached to thecarbon skeleton. The identities of the substituents constitute the first field of an IUPACname, and their positions of attachment are specified by numerals. The carbon chainis numbered so that the functional group with the highest priority (Table 1.1) has thelowest number. If a substituent is attached to a heteroatom instead of the carbon skele-ton, its placement is denoted by the italicized N, O, or S, for nitrogen, oxygen, or sul-fur, respectively.

Common examples of substituents are given in Table 1.5. When more than one ofthe same type of substituent is present, the name of the atom or group follows a prefixfor that number of items: di = two, tri = three, tetra = four, and so on. (When the sub-stituents are more complex than those listed in Table 1.5, you will instead see and usethe prefixes bis, tris, tetrakis, etc. Meanings are the same, e.g., di = bis = two.) Further-more, there must be a corresponding numeral for each atom or group. Thus,

2, 2-Dichloro . . . (correct)

but not 2-Dichloro . . . (incorrect: not enough specifying numerals)

or 2,2-Chloro . . . (incorrect: prefix specifying two chlorine atoms is missing)

Exceptions to the numeral rule occur when there is no chance for ambiguity, forexample, when there is only one carbon atom that can be substituted by the specifiedgroup. Then, numerals are not needed, but a prefix is still required. For example,

EXERCISE 1.9

Correct any of the following names that are inconsistent.

a. 2,3,4-Hydroxyhexanal b. 2,2,4,4-Tetrachloropentane c. Triiodomethane

FF

H

H

CDifluoromethane


Table 1.5 A summary of prefixes for common substituents.

O

OR

Cl

Br

R

NH2

COOH

C N

F

OH

I

SH

O

NO2

CH3

C

O

H

C

O

Substituent Prefix Substituent Prefix

Alkyl- (see text)

Alkoxy- (see text)

Acetyl-

Amino-

Bromo-

Carboxy-

Chloro-

Cyano-

Fluoro-

Hydroxy-

Formyl-

Iodo-

Nitro-

Mercapto-

Oxo-

Phenoxy-

The following examples illustrate how the IUPAC format is applied to interpretingcompound names.

EXAMPLE 1.3

Draw the structural formula and condensed structure of 5,5-dichloro-3-hexanone.

First, identify the compound root and draw the carbon skeleton. Root = hex (six carbonatoms).

Next, translate the multiple-bond index. The suffix that directly follows the root wordis “an” meaning that there are no carbon–carbon double or triple bonds. The suffix is“one”, which means that the compound is a ketone, and the numeral “3” indicates thatthe carbon–oxygen double bond of the ketone is at C3:

Next, we place the substituents, in this case chlorine atoms, at the indicated positions:

Finish by adding hydrogen atoms to give each carbon atom a total of four bonds.

EXAMPLE 1.4

Draw the structural formula and condensed structure of 3-hydroxybutanoic acid.

The compound root is “but” (four carbon atoms), and the multiple-bond index is “an”,which means that there are no carbon–carbon double or triple bonds. The suffix is “oicacid”, so the compound is a carboxylic acid. The carboxylic acid functional group hasto be at the end of the chain, and any principal functional group at the end defines the1-position with its carbon atom.

Next, we place the substituent, in this case the hydroxy group (OH), at the 3-position:

C OHC C

OOH

4 3 2 1C

C C OHC CC C C C

O

4 3 2 1

H

H

C

H

H

C

H

H

C

H

H

C HH

Cl

Cl

CC

O

CH2CH2CH3 CH3C

O Cl

Cl

C�

C C C CC

O

1 2 3 4 5 6

Cl

Cl

C

C C C C CC

O

1 2 3 4 5 6

C C C C C C


The last step is to add hydrogen atoms to give each carbon atom a total of four bonds.

EXERCISE 1.10

Draw the structural formula for each of the following compounds:

a. 4-Fluorobutanal b. 3-Mercapto-2-pentanol c. Trichloroethanenitrile

EXAMPLE 1.5

Draw the structural formula and condensed structure of N-chloro-2-propenamide.

The compound root is “prop” (three carbon atoms), and the multiple-bond index is “en”,which means that the compound has a carbon–carbon double bond, which starts at C2.

The suffix is “amide”, so the compound is an amide (also called a carboxamide), whichhas a carbonyl group attached to a carbon-containing fragment and a nitrogen atom.The amide functional group must be at the end of the chain, and its carbon atom de-fines C1. Next, place the substituents. A chlorine atom is attached to the nitrogenatom, as indicated by the letter N at the beginning of the name. A numeral would ap-pear if the chlorine atom were attached to one of the carbon atoms in the chain.

The last step is to add hydrogen atoms to give each carbon atom a total of four bondsand to the nitrogen atom to give it three bonds.

EXERCISE 1.11

Draw the structural formula for each of the following compounds:

a. 3-Oxopentanoic acid b. 2-Nitropentanal c. 4-Hydroxy-2-hexyne

1.3e THE PLACEMENT OF SUBSTITUENTS IN CYCLIC COMPOUNDS

DEPENDS ON DEFINING A STARTING POINT FOR NUMBERING

In an aliphatic compound, numbering starts at the end of the chain that gives the prin-cipal functional group the lower possible number. Numbering is done in such a waythat the functional group, multiple bonds, and substituents have the lowest numericalvalues possible. For example,

H

H

C

H

C

H

C ClN

O

ClCH NHH2C C

O

==

C N ClC C

O

3 2 1C NC C

O

3 2 1

C CC C C C3 2 1

H

H

H

C

OH

C

OHH

H

C

H

OHC

O O

CCHCH3 CH2 OH==


A ring, whether it is alicyclic or aromatic, has no “end”. Therefore, numberingstarts with the position at which the principal functional group or a substituent is at-tached and proceeds around the ring. The direction of numbering is the one with thelower number at the first point of difference. Placement of the other groups is definedrelative to the starting point, as illustrated by the examples in Figure 1.6.

UNN 1.24 GOES HERE


Numbering from the right endgives the principal functionalgroup a lower number

CH2CH CH3C51

42

33

24

15

CH H

3-Pentenal

O

Numbering from the right endgives the principal functionalgroup a lower number

CH2 CH2C CH3 H2C51

42

33

24

15

CH

4-Penten-2-one

O

The principal functional groupis at the same position whetherthe chain is numbered left toright or right to left; numberingfrom the left end gives thedouble bond a lower number,however.

CH2 CH3CH51

42

33

24

15

C

1-Penten-3-one

O

EXERCISE 1.12

Draw the structure for each of the following compounds:

a. 2-Bromobenzoic acid b. 3-Fluoro-2-cyclohexenone c. 3-Chloro-4-nitrophenol

1.4 CONSTITUTIONAL ISOMERS ANDHYDROCARBON SUBST ITUENTS

1.4a STRUCTURES OF CARBON-CONTAINING SUBSTITUENTS ARE DIVERSE

Until now, substituents in the examples have mainly been groups containing het-eroatoms. To deal with substituents that contain carbon, you also have to begin tolearn about isomerism. Isomers exist whenever a molecular formula can be repre-

Cl OH1

23

4

5

NH2

Cl OH

6

5

26

4

3 1

NH2

Br

1

54

23 Br

1

23

54

Br

4

56

7

32

1 Cl

Cl1

6

5

2

34

CH3

NO2

Cl1

2

3

6

54

CH3

NO2

2-Amino-5-chlorocyclohexanol

The principal functional group (alcohol)defines C1. Numbering is clockwisebecause 1,2,5 is lower than 1,3,6.

The presence of the double bond definesC1. Numbering is counterclockwise becausecarbon atoms in multiple bonds are numberedconsecutively.

3-Bromocylopentene

1-Bromo-4-chlorocycloheptane

When only two substituents are presentand there is no principal functionalgroup, the substituent that appearsfirst alphabetically defines C1.

The methyl group of toluene, which is thecompound root, defines C1. Numbering isclockwise because 1,2,4 is lower than 1,4,6.

2-Chloro-4-nitrotoluene

Figure 1.6Examples of numberingpatterns in cyclic compounds.

sented by different arrangements of the constituent atoms, and Figure 1.7 summarizesthree types of constitutional isomers, which are isomers having different connectivitiesbetween neighboring atoms. The three categories of constitutional isomers are skele-tal, positional, and functional isomers.

Related to the concept of isomerism is the classification of carbon atoms with foursingle bonds as primary, secondary, tertiary, and quaternary. A carbon atom attached toonly one other carbon atom is a primary carbon atom, which we designate 1°. A carbonatom attached to two other carbon atoms is secondary (2°); to three other carbonatoms is tertiary (3°); and to four carbon atoms, quaternary (4°).

A carbon atom with four single bonds that is attached to no other carbon atom is des-ignated methyl, methylene, or methine according to the number of attached hydrogenatoms (three, two, and one, respectively).

Notice that within a carbon chain, a CH3 group can be designated as either methyl or 1°,depending on the context. Likewise, a CH2 group can be labeled either methylene or2°; and a CH group can be identified as either methine or 3°.

The terms primary, secondary, and tertiary can be applied to other substances, too.For amines and amides, these words are used to indicate how many carbon-containinggroups are attached to the nitrogen atom.

CH3 CH2

CH3

H

C CH3

Methine or 3°Methyl or 1°

Methyl or 1° Methylene or 2°

SH

Methyl

H

H

H

C F

Methylene

Methine

F

H

H

C

Methyl

OCl

Cl

H

C H

H

H

C

HH

H

H

C

Methane

RH

H

H

C

1°

RH

H

R

C

2°

RH

R

R

C

3°

RR

R

R

C

4°

1.4 Constitutional Isomers 19

CH2 CH3CH2CH3 CH3CHCH3

CH2 OHCH2CH3 CH3CHCH3

OH

CH3

CH2 OHCH3CH3OCH3

Butane: C4H10 2-Methylpropane: C4H10

1-Propanol: C3H6O 2-Propanol: C3H6O

Dimethyl ether: C2H6O Ethanol: C2H6O

Skeletal isomers

Positional isomers

Functional isomersFigure 1.7Types of constitutionalisomers

EXERCISE 1.13

For the carbon atoms with four single bonds in the following structures, classify eachas 1°, 2°, 3°, or 4°. First, expand each structure to show all of the hydrogen atoms.

1.4b THE NAMES OF CARBON-CONTAINING SUBSTITUENTS ARE DERIVED

FROM THE NAMES OF THE PARENT HYDROCARBONS

The simplest carbon-containing substituents are alkyl groups, formed by removing a hy-drogen atom from an alkane. For example, removing a hydrogen atom from methanecreates the methyl (often abbreviated as Me) group. Removing a hydrogen atom fromethane generates the ethyl (abbreviated as Et) group. Taking a hydrogen atom fromthe end of any alkane chain yields a “straight-chain” alkyl group.

If an alkane has three carbon atoms or more, then isomeric alkyl groups can be made.Taking a hydrogen atom from the C1 of propane gives the propyl (Pr) group. Removinga hydrogen atom from C2 yields the 2-propyl or isopropyl (iPr) group. The IUPAC systemincludes both propyl and isopropyl. Notice that the prefix “iso” is not italicized.

Among substances with four carbon atoms are examples of alkane isomers: butaneand isobutane (IUPAC name: 2-methylpropane). Isobutane is an example of an alkanewith a branched-carbon chain. While alkanes are normally named according to IUPACrules, terms like “isobutane” are still commonly used.

CH3

Butane Isobutane(2-Methylpropane)

CH2CH2CH3 CH3

CH3CCH3

H

H

Propyl 2-Propyl (or Isopropyl)

CH2CH2CH3

CH3CCH3

CH4

CH3CH3

CH3CH2CH3CH3

CH2CH3

CH2CH2CH3

CH3CH2CH2CH2CH2CH3 CH3CH2CH2CH2CH2CH2

HexaneEthane

Methane Propane

HexylEthyl

Methyl Propyl

O

O

Br

a. b.

UNN 1.28 GOES HERE


O

R

R

C N

R

3° Amide

O

R

H

C N

R

2° Amide

O

R

H

C N

H

1° Amide3° Amine2° Amine1° AmineAmmonia

HHN

H

HRN

H

HRN

R

RRN

R

The result of this carryover is that alkyl groups generated from alkanes with fourcarbon atoms are frequently called butyl, isobutyl (the isomer derived from removinga hydrogen atom from the primary carbon atom of isobutane), sec-butyl (sec stands for“secondary” to indicate removal of a hydrogen atom from the 2° carbon atom of bu-tane), and tert-butyl (t-Bu; tert stands for “tertiary”). Figure 1.8 shows the structures ofthese isomeric alkyl groups. Notice that he prefixes ”sec-“ and “tert-“ are italicized.

EXAMPLE 1.6

Draw the structural formula of 3-ethyl-2-hexanone.

The compound root is “hex” (six carbon atoms), the multiple-bond index is “an”,which means that there are no carbon–carbon double or triple bonds, and the suffix is“one.” The compound is a ketone, and the carbonyl group is at C2.

We place the substituent, an ethyl group, at C3. The structure is completed by addinghydrogen atoms until each carbon atom has a total of four bonds.

EXAMPLE 1.7

Draw the structure of 3-tert-butyl-2-cyclohexenecarboxylic acid.

The compound root is “cyclohex” (six carbon atoms in a ring), and the multiple-bondindex is “en”, which means a carbon–carbon double bond is present, starting at C2.The suffix is “-carboxylic acid,” so the compound is a carboxylic acid, with the –COOHgroup attached to the ring. Its point of attachment defines C1 of the ring.

At C3, we place the substituent, a tert-butyl group. The structure is completed by addinghydrogen atoms until each carbon atom has a total of four bonds.

C H

O

15

24

3

6

O

C C C C CC

O O6 5 4 3 2 1

H3C CH2 CH3CH2 C C

H

CH3H2CCH3H2C

C C C C CC

O

6 5 4 3 2 1


H

Butyl 2-Butyl or sec-Butyl 2-Methyl-1-propyl(or Isobutyl)

CH2 CH2CH2CH3 CH3CCH2CH3

CH3

2-Methyl-2-Propyl(or tert-Butyl)

CH3CCH3

CH3

CH2CCH3

H

Figure 1.8Structures of the alkyl groups derived from the butane isomers.

EXERCISE 1.14

Draw the structural formula for each of the following compounds containing alkylgroup substituents:

a. Isobutylbenzene b. 3-tert-Butylhexanol c. 3-Ethylcyclopentanone

If an alkyl group has substituents itself, its name is set apart from the root name ofthe parent compound by enclosing it in parentheses. We deduce its identity by findingits root name and its substituents, just as we do for the parent compound. The follow-ing example illustrates this procedure.

EXAMPLE 1.8

Draw the structure of 3-(2- chloroethyl)-2-heptanol.

The compound root is “hept” (seven carbon atoms), the multiple-bond index is “an”,which means no carbon–carbon double or triple bonds, and the suffix is “ol.” The com-pound is an alcohol, a functional group that can be attached to any carbon atom; the“2” in front of the root word tells us the OH group is at C2. At position 3, we attach thealkyl substituent, “X” ( = 2- chloroethyl).

Now, interpret the name of “X”. Its root is “eth” (2 carbon atoms), and it is attached to themain chain at its C1 by convention. A chlorine atom is attached at C2 of this side chain.

Finally, we add hydrogen atoms to every carbon atom to give each a total of four bonds.

EXERCISE 1.15

Draw the structure of 2-(1,1-dimethylpropyl)hexanoic acid.

CH3 CH2 CH3CH2 CH2 CH CH

CH2 CH2Cl

OH

C C C C CCC

C

C2

1

OH

C C C C CCC

C

C2

1

OH

Cl

OH

C C C C

X

CC6

C7 5 4 3 2 1

OH

C C C C CC6

C7 5 4 3 2 1

C

OC H

H H

H

H

H

HH

H

CH3H3CCH3

CCH3H3C

CH3

O

OHC

O

15

243

6


Related to the alkyl groups are alkoxy groups, which comprise an alkyl group at-tached to the main compound framework through an oxygen atom. The names ofthese substituents are generated by inserting “oxy” after the alkyl root name.

EXAMPLE 1.9

Draw the structure of 3-methoxy-4-pentenal

The compound root is “pent” (5 carbon atoms), the multiple-bond index is “en”, whichmeans a carbon–carbon double is present starting at C4, and the suffix is “al.” The com-pound is an aldehyde, a functional group that must be at the end of the chain.

A methoxy group (methyl group + oxygen atom) is attached at C3.

1.4c NAMES OF UNSATURATED SUBSTITUENTS ARE DERIVED

FROM THE NAMES OF THE CORRESPONDING ALKENES, ALKYNES, AND ARENES

Carbon-containing substituents are not limited to those with only four single bonds tothe carbon atoms: multiple bonds can also be present. First, realize that a carbon atomthat forms a multiple bond cannot be designated as 1°, 2°, 3°, or 4°. Instead, as shownbelow, these carbon atoms are identified according to the functional group of whichthey are part. A carbon atom in a carbon–carbon double bond is termed an alkenyl (orvinyl) carbon atom; in a triple bond, an alkynyl carbon atom; and in a benzene ring, anaryl carbon atom. The following examples illustrate the use of these terms, along withthe designations for the carbon atoms that have only single bonds.

The simplest substituents with multiple bonds have names incorporated into theIUPAC system from common names: vinyl, allyl, phenyl, and benzyl, which are sum-marized in Figure 1.9. These are treated in a name in the same fashion as alkyl groups.

UNN 1.40 GOES HERE

OCH3

C C C C HC

OOCH3 O

5 4 3 2 1CH2 CH2CH HC C

H

C C C C HC

O

5 4 3 2 1

(CH3)2CH

CH3

CH3CH2

O(CH3)2CH

O

OCH3

CH3CH2

Methyl

Ethyl

Isopropyl

Methoxy

Ethoxy

Isopropoxy


O

Aldehydecarbon atom

H

H

C

H

C

H

C C HH

1°

2°

1°

C

H H

H H

CH CH2C

C

C C

C N CH

Nitrilecarbon atom

1°3°

Alkynylcarbon atoms

Alkenylcarbon atoms

CCH3 C

CH3

H

Aryl carbon atoms (6)

EXAMPLE 1.10

Draw the structure of 3-vinylcyclohexanone.

The compound root is “cyclohex” (six carbon atoms), and the multiple-bond index is“an”, which means no carbon–carbon multiple bonds. The principal functional groupis a ketone (one), which in a ring defines C1.

At position 3, we attach the substituent, which is the vinyl group. We finish by addinghydrogen atoms to give every carbon atom a total of four bonds.

EXAMPLE 1.11

Draw the structure of 3-phenylcycloheptene.

The compound root is “cyclohept” (seven carbon atoms in a ring), the multiple-bondindex is “ene”, which means a carbon–carbon double is present, starting at C1.

A phenyl (Ph) group is attached at C3. Hydrogen atoms are added to give each carbonatom four bonds.

21

5

7 3

46

O

6

1

4

32

5

HCH2

H2C

C

H2C

HH

CC

H

HC

CH2

O

H

CHC

O

6

1

4

3

2

5


H

H H

H

C C

Ethene

H

H

H

C C

Propene

H

H

H

C C

Vinyl or Ethenyl

H

H

H

C C

CH2CH3

Allyl or 2-Propenyl

Benzene ToluenePhenyl Benzyl

H

HH

HH

H

HH

HH

H CH3 CH2

HH

HH

H

HH

HH

H

23

1

Figure 1.9Common names ofunsaturated hydrocarbonsubstituents and thehydrocarbons from which theyderive. Notice that benzyl isnot derived from benzene butrather from toluene.

EXERCISE 1.16

Draw the structural formula of 3-allyl-5-chlorobenzoic acid.

1.5 NAMING ORGANIC COMPOUNDS

Having seen how IUPAC names can be interpreted by translating each portion of aname, you are now in a position to write the name of a compound given its structure.The steps in the naming process follow:

1. Identify the principal functional group, which is defined by the priority listing givenin Table 1.1, and choose the suffix that will appear at the end of the name.

2. The longest carbon chain that also contains the principal functional group is then iden-tified, and its name becomes the compound’s root. If the compound is cyclic, thenthe ring connected to the principal functional group is chosen as the root.

3. The chain or ring is numbered in such a way that the principal functional group isattached at (or is part of) the carbon atom with the lowest possible number. In acyclic compound, the next priority comprises the positions of multiple bonds.

Numbering can be a problem when no high priority group is present. In such a case,the numbering scheme chosen is the one with the lower number at the first point ofdifference. In the following example, 1, 2, 4 < 1, 3, 4 (2 lower than 3) and 1, 2, 4 < 1,2, 5 (4 lower than 5) so the name is 2-bromo-1-chloro-4-methylcyclohexane.

COOHExample:

OCH3

Example:CHOBr

223

45

146

135

OCH3

COOH = 6 carbon atoms = hex (compound root)Example:

= 5 carbon atoms = cyclopent (compound root)Example:CHOBr

COOH = oic acid (suffix)Example:

OCH3

= carbaldehyde (suffix)Example:CHOBr

2

3

1 CCH2C C

CH2

CH2H2C H

H

H

H

H HH H

CCH2C C

CH2

CH2H2C H

Ph

H H

1.5 Naming Organic Compounds 25

4. Once the order of numbering is defined, then the designation of the appropriatemultiple-bond index is inserted between the root word and the functional groupsuffix. If necessary, a numeral is included to indicate at which carbon atom thedouble or triple bond begins.

5. Finally, the substituents are specified by appending the appropriate prefixes andnumerals to the name.

When more than one substituent is present, their names are arranged in alpha-betical order, ignoring any italicized words such as sec- or tert- as well as prefixes hav-ing to do with how many of a given substituent are present.

Example: tert-Butyl comes before chloro because butyl comes before chloroalphabetically.

Example: Ethyl comes before dimethyl because ethyl comes before methylalphabetically (i.e., the “di“ prefix is ignored).

The following examples illustrate the procedure used to name organic molecules.

EXAMPLE 1.12

Give an acceptable systematic name for the following compound:

This compound has the ketone functional group, so the suffix is “one.” The longest car-bon chain has six carbon atoms, so the root is “hex”. There are no carbon–carbon dou-ble or triple bonds, so the multiple-bond index is “an.” The name so far is hex/an/one = hexanone.

The position of the ketone functional group must be specified by a numeral. Withrespect to numbering, rule 3 states that the chain is numbered to give the functional

OCH3

= 5-Methoxyhexanoic acidExample:

= 4-Bromo-2-cyclopentenecarbaldehydeExample:

COOH

OCH3

CHOBr

2

23

45

1

46135

= No multiple bonds = an= hex/an/oic acid

Example:

= One double bond = en (starts at C2)= 2-Cyclopent/ene/carbaldehyde

Example:

COOH

OCH3

CHOBr

2

23

45

1

46135

CH3

Cl

Br2

34

1

1,2,4

CH3

Cl

Br1

4

3

5

2

1,2,5

CH3

Cl

Br3

21

4

1,3,4

Example:


group the lowest possible number. Therefore, numbering begins at the right end ofthis structure instead of at the left end.

Once the numbering is done, the position of the methyl group is established, which isC4. The name is therefore 4-methyl-2-hexanone.

EXAMPLE 1.13

Give an acceptable systematic name for the following compound:

This compound has a nitrile functional group, so the name ends in “nitrile.” Thelongest carbon chain, including the nitrile carbon atom, has seven carbon atoms, so theroot is “hept”. There are no carbon–carbon double or triple bonds, so the multiple-bond index is “ane”. The name so far is hept/ane/nitrile = heptanenitrile.

The principal functional group is one that must be at the end of the chain (Table1.1), so numbering begins at the right end of the structure.

The positions of attachment for the bromine atom and the methyl group are subse-quently established, and bromo precedes methyl in the name because of alphabeticalorder, even though the methyl group is attached to a carbon atom with a lower num-ber. The name is 5-bromo-3-methylheptanenitrile.

EXAMPLE 1.14

Give an acceptable systematic name for the compound shown here.

This compound has the carboxylic acid as its principal functional group, which is at-tached to a six-membered ring. The root of “cyclohex” must be compounded with thesuffix “carboxylic acid”. There is a carbon–carbon double bond, so the multiple-bondindex is “ene”. The name so far is cyclohexenecarboxylic acid.

The point of attachment of the principal functional group defines where the num-bering begins, and we number in the clockwise direction so as to give the position ofthe double bond the lower possible number (2 instead of 5, which would be its positionif we numbered counterclockwise).

The methyl group is attached at C2, so the name is 2-methyl-2-cyclohexenecarboxylic acid.

CH3

COOH

4

1 1

2 6

6

3

5

CH3

COOH

4

2

5

3

CH3

CO2H

CH CH2 CH CH2 CNCH2CH3

CH3Br

567 4 3 2 1

CH CH2 CH CH2 CNCH2CH3

CH3Br

OO

5 3 1246 2 4 6531

CH3 CH3

1.5 Naming Organic Compounds 27

EXAMPLE 1.15


This compound has both amine and nitro functional groups. The nitro group is alwayslisted as a substituent, so the amino group is the principal functional group. Aromaticcompounds with an amino group are named as derivatives of aniline (Fig. 1.4), and thepoint of attachment of the amino group defines the C1 position of the ring.

Numbering is counterclockwise because 3,4– is lower than 4,5– at the first point ofdifference. Isopropyl comes before nitro because of alphabetical order, so the name ofthis compound is 3-isopropyl-4-nitroanilline.

EXAMPLE 1.16


This compound has no principal functional group or multiple bond. The longest car-bon chain is 6 carbon atoms, which is hexane. There are two substituents and chlorowill precede methyl in the name because of alphabetical order. The chain can be num-bered from either end, however.

Because 2,4 is lower than 3,5 at the first point of difference, the correct name is 4-chloro-2-methylhexane.

EXERCISE 1.17

Give an acceptable systematic name for each of the following compounds:

CC

H CH3

COOHH3C

CH2

O

C(CH3)3 CH3CH2 CH2CH CH CH2CH3

OH

CH3

a. b. c.

ClCH3ClCH3

5 3 1246 2 4 6531

3-Chloro-5-methylhexane 4-Chloro-2-methylhexane

Cl

NO2

H2N1

4

5

3

6

2 NO2

H2N1

4

3

5

2

6

NO2

H2N


Chapter Summary 29

Section 1.1 Structural components of organic molecules

• Carbon always forms four bonds to other atoms in stable molecules. Nitrogennormally forms three bonds, oxygen forms two bonds, and hydrogen and thehalogens (F, Cl, Br, and I) form one bond. These trends are summarized in Fig-ures 1.1 and 1.2.

• Multiple bonds—double and triple—count as two and three bonds, respectively.

• Organic compounds comprise carbon atoms (always), hydrogen atoms (often),and heteroatoms (sometimes). Heteroatoms are elements besides C and H, andtheir presence often creates a functional group, the reactive portion of a mole-cule. Common functional groups are listed in Table 1.1.

Section 1.2 Structural formulas and condensed structures

• Condensed formulas eliminate the need to show every atom and bond. Methyl(CH3), methylene (CH2), and methine (CH) groups are condensed hydrocar-bon units.

• Carbon atom positions can be represented by angled meetings of lines or as thevertices of a regular polygon.

• The carbon–oxygen double bond, called the carbonyl group, is a constituent ofseveral functional groups including ketone, aldehyde, and carboxylic acid.

Section 1.3 Systematic nomenclature: IUPAC names

• IUPAC names are systematic and read as follows from left to right:

a. The nature and positions of substituents (Section 1.3d).

b. The root word (Section 1.3a).

c. The nature and number of double or triple carbon–carbon bonds (Section1.3b).

d. The identity of the principal functional group (Section 1.3c).

• For aliphatic compounds, the root word is the longest carbon chain related instructure to the corresponding alkane, a molecule with the general formulaCH3(CH2)nCH3. The chain is numbered to give the principal functional groupthe lowest possible number.

• For alicyclic compounds, the root word is based on the size of the ring.

• For aromatic compounds, the root word contains “benz”, short for benzene, ora common name such as toluene, anisole, aniline, or phenol, among others.

• The abbreviation “R” is used to generalize the identity of a substituent, espe-cially one that contains carbon and hydrogen atoms. The abbreviation “Ar” islikewise used as a shorthand notation for an aromatic ring.

• The positions of substituents are specified by numerals that precede the nameof the substituent group. Multiple substituents of the same type are identified bya prefix (di-, tri-, tetra-, etc.) that tells how many such groups are present.

Section 1.4 Constitutional isomers and hydrocarbon substituents

• Isomers are compounds that have the same chemical formula but a differentspatial arrangement of atoms. Figure 1.7 summarizes the types of constitutionalisomers.

• A carbon atom that forms four single bonds is classified as primary (1°), sec-ondary (2°), tertiary (3°), or quaternary (4°) according to the number of othercarbon atoms attached. Carbon atoms with four single bonds can also be

CHAPTER SUMMARY


Introductionorganic chemistry

Section 1.1bheteroatomfunctional group

Section 1.2acondensed structurehydrocarbon

Section 1.3aIUPAC rulesaliphatic compoundsalkanealicyclicaromatic compoundarene

Section 1.3balkenealkyne

Section 1.3cRArcarbonyl group

Section 1.4aisomersconstitutional isomersskeletal isomerspositional isomersfunctional isomersprimary carbon atomsecondary carbon atomtertiary carbon atomquaternary carbon atom

Section 1.4balkyl groupalkoxy group

Section 1.4calkenyl carbon atomvinyl carbon atomalkynyl carbon atomaryl carbon atom

KEY TERMS

designated methyl, methylene, and methine (CH) groups according to thenumber of hydrogen atoms attached (3, 2, and 1, respectively).

• A nitrogen atom that forms three single bonds in amines and amides is classifiedas primary (1°), secondary (2°), or tertiary (3°) according to the number of car-bon atoms attached.

• Removing the hydrogen atom from an alkane generates an alkyl group.

• An alkoxy group is a substituent in which an alkyl group is connected to themain chain or ring via an oxygen atom.

• Substituents with carbon–carbon double and triple bonds are formed from un-saturated hydrocarbons and are called alkenyl, alkynyl, and aryl groups.

Section 1.5 Naming organic compounds

• IUPAC names are generated by identifying the longest carbon chain or ring, theprincipal functional group that is present, the kind and number of carbon–car-bon multiple bonds present, and the types and positions of substituents at-tached to the chain or ring.

1.18. Draw structural formulas for the following alkanes, showing all of the atoms inthe longest carbon chain:

a. The five isomers of C6H14.

b. The nine isomers of C7H16.

1.19. Draw a condensed structure for each answer in Exercise 1.18.

ADDITIONAL EXERCISES

1.20. Show all of the hydrogen atoms in each of the following condensed represen-tations of some naturally occurring organic compounds. You may use CH3 to rep-resent the methyl groups instead of showing all three hydrogen atoms with theirbonds.

1.21. Each of the following molecules has a single functional group. Give its identityaccording to those listed in Table 1.1. Represent each molecule in the form thatmakes use of “R” along with the functional group (see Exercise 1.8).

1.22. Draw two structures that exemplify each of the following types of compounds:

a. A ketone with the formula C5H10O.

b. A chloro ketone with four carbon atoms.

c. An aromatic amine.

d. An aldehyde with six carbon atoms.

e. A hydroxy aldehyde.

f. An alicyclic carboxylic acid.

OH

C C OCH3

CH3

CH3

CH3

CH3

OCH3

CH3CH3

CH3

C CCH2 OH

CH3CHCH3

SH

C NH2

H

CH3

CH3 CH C CH

CH3

CH3

CH2

CH2H2C

H2C

C

O

C

O

H

a. b. c. d.

e. f. g. h.

O

O

O

O

O

OH

COOH

Pinene

Oil of turpentine

Carvone

Spearmint oil

Thromboxane A2

A potent aggregator of blood platelets

Progesterone

The precursor for allsteroidal hormones

Additonal Exercises 31

1.23. For the compounds shown in Exercise 1.22, identify the carbon atoms as 1°, 2°, 3°,4° (Section 1.4a) or alkenyl, alkynyl, aryl, or specific functional group (Section 1.4c).

1.24. Shown below are condensed representations of some organic compounds thathave been used as drugs. Show all of the hydrogen atoms and identify the follow-ing functional groups: alcohol, aldehyde, amide, amine, carboxylic acid, ether, carboxylic acid ester, ketone, and thiol (some of these groups appear more thanonce; some do not appear). You may use CH3 to represent the methyl groups in-stead of showing all three hydrogen atoms with their bonds.

1.25. For the compounds shown in Exercise 1.24, identify the carbon atoms as 1°, 2°, 3°,4° (Section 1.4a) or alkenyl, alkynyl, aryl, or specific functional group (Section 1.4c).

1.26. The root word for an aliphatic compound is the longest carbon chain that alsocontains the principal functional group. Together with the other rules you havelearned, explain why each of the following names is incorrect:

a. Methylheptane b. 3-Propylhexane

c. 2,2-Dimethyl-3-ethylbutane d. 2-Dimethylpentane

e. 2-Isopropyl-1-propanol f. Dichloroheptane

1.27. Draw a structure for each of the seven isomers of C6H10 that have a triple bond.You may use condensed structures. Name each compound.

1.28. Draw every alcohol with the molecular formula C5H12O. Label each carbonatom as 1°, 2°, 3°, or 4°. You may use condensed structures. Name each com-pound.

1.29. Draw a structural formula for each of the following alcohols:

a. 2-Phenylethanol b. 1,3-Dibromo-2-pentanol

c. 3-Chloropropanol d. 2-Methyl-3-buten-2-ol

e. 2,2,2-Trifluoroethanol f. 2-Amino-2-methylbutanol

g. 2,3-Butadienol h. 4-Hexynol

1.30. Draw a structural formula for each of the following carboxylic acids:

a. 2-Aminobenzoic acid b. 2,2-Difluorobutanoic acid

c. 2,3-Dibromopropanoic acid d. 4-Isobutylbenzoic acid

e. 3-Methoxycycloheptanecarboxylic acid f. 3-Mercapto-4-hexenoic acid

g. 5-Hydroxy-3-heptenoic acid h. 2,5-Dimethylbenzoic acid

NHS

O

O

ON

COOH

Captopril

antihypertensive drug

ProzacNovocain

local anesthetic antidepressant drug

O

CF3NH2

CH3N

H


1.31. Within each group of compounds, which of the following structures are identi-cal and which are constitutional isomers?

1.32. Give an acceptable systematic name for each of the following aliphatic or ali-cyclic compounds:

1.33. Draw a structural formula for each of the following amines and amides:

a. N,N-Dimethylaniline

b. 1,2-Diaminocyclohexane

c. 1-Amino-3-phenylbutane

d. 3-Hydroxycyclopentanecarboxamide

e. N-Methylbutanamide

f. 2,4-Dimethylaniline

Br

OOH

CH2 CHOC CCH2

COOHOHCH CHCH2CH3 CH2

a. b. c.

d. f.e.

C6H5O

H

H

C

H

H

C

H

H

C

H

H

C HH

H

H

C

H

H

C

Cl

H

C

H

H

C HH

H

H

C

H

H

C O

H

C

H

H

C

H

HH

H

H

C

H

H

C O

H

C

H H

H

C HH

H

H

C O

H

C

H

H

C

H

H

C

H

HH

H

H

C

H

H

C

H

C

Cl H

H

C HH

H

H

C

H

H

C

CCH

H

C

H H

HH

H

H

C

CCH

H

C

HH

H

C HH

H

H

C

CC

H

H

C

HH

H

C HH

H

Cl

CH3 CHCH2

OH

CH3 CH3CHHO CH2

CH3

CHHO CH3

CH2 CH3

CHHO CH3CH2

CH3

a.

b.

c.

d.

Additonal Exercises 33

1.34. Draw a structural formula for each of the following aldehydes and ketones:

a. 3-Methyl-2-butanone

b. 1-Chloro-3-hexene-2-one

c. 3-Methoxypentanal

d. 3-Isopropylcyclohexanecarbaldehyde

e. 4-Bromobenzaldehyde

f. 3-Ethoxy-2-hexanone

g. 2-Cyclohexenone

h. 3-tert-Butylcyclobutanone

1.35. Give an acceptable systematic name for each of the following aromatic com-pounds:

1.36. Draw a structural formula for each of the following compounds:

a. 1-Chloro-2-hexyne b. 1,3-Dicyanobenzene

c. 6-Bromohexanoic acid d. 2-Nitrobenzaldehyde

e. 4-Nitrotoluene f. 2-Allyl-6-chlorophenol

g. 4-Pentynol h. 3,5-Heptadienal

i. 1-Decanethiol j. 1-Nitropropane

k. 2, 3, 4-Hexanetriol l. 2-Butyl-3-chlorobenzonitrile

1.37. Give an acceptable systematic name for each of the following compounds:

1.38. Draw a structural formula for each of the following compounds that have a com-plex substituent. Follow the procedure shown in Example 1.8: First interpret thename of the parent compound, then interpret the name of the substituent givenin parentheses. By convention, the carbon atom of the substituent attached to theparent chain or ring is designated C1.

a. 4-(1,1-Dimethylethyl)-4-octanol

b. 3-Chloro-2-(1-hydroxyethyl)-6-nitrophenol

c. 3-(2-Fluoro-2-propenyl)-5-hepten-2-one

d. 4-(1-Methylethyl)-5-methyl-3-hexenal

OH

COOH

a.

OCH3 CHOBr

Br

b. c.d.

F

NH2

CH3

a.

OH

Cl

b. c. d.

O

COOH CHO

Br NO2

F F


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