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Introduction and Review px

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1-1The Origins of Organic Chemistry

The modern definition of organic chemistry is the chemistry of carbon compounds.What is so special about carbon that a whole branch of chemistry is devoted to itscompounds? Unlike most other elements, carbon forms strong bonds to other car-bon atoms and to a wide variety of other elements. Chains and rings of carbonatoms can be built up to form an endless variety of molecules. It is this diversity ofcarbon compounds that provides the basis for life on Earth. Living creatures arecomposed largely of complex organic compounds that serve structural, chemical,or genetic functions.

The term organic literally means “derived from living organisms.” Originally,the science of organic chemistry was the study of compounds extracted from livingorganisms and their natural products. Compounds such as sugar, urea, starch, waxes,and plant oils were considered “organic,” and people accepted Vitalism, the belief thatnatural products needed a “vital force” to create them. Organic chemistry, then, wasthe study of compounds having the vital force. Inorganic chemistry was the study ofgases, rocks, and minerals, and the compounds that could be made from them.

In the nineteenth century, experiments showed that organic compounds couldbe synthesized from inorganic compounds. In 1828, the German chemist FriedrichWöhler converted ammonium cyanate, made from ammonia and cyanic acid, to ureasimply by heating it in the absence of oxygen.

Urea had always come from living organisms and was presumed to contain the vitalforce, yet ammonium cyanate is inorganic and thus lacks the vital force. Somechemists claimed that a trace of vital force from Wöhler’s hands must have contami-nated the reaction, but most recognized the possibility of synthesizing organic com-pounds from inorganics. Many other syntheses were carried out, and the vital forcetheory was eventually discarded.

Since Vitalism was disproved in the early nineteenth century, you’d think itwould be extinct by now. And you’d be wrong! Vitalism lives on today in the mindsof those who believe that “natural” (plant-derived) vitamins, flavor compounds, etc.are somehow different and more healthful than the identical “artificial” (synthesized)compounds.

NH4� �OCN

heatH2N9C9NH2

O

urea(organic)

ammonium cyanate(inorganic)

The Jarvik 7 artificial heart, composedlargely of synthetic organic materials.

WADEMC01_0131478710.QXD 11/8/04 7:56 AM Page 1

morphine

OOH OH

H

CH3N

2 Chapter 1: Introduction and Review

As chemists, we know that plant-derived compounds and the synthesized com-pounds are identical. Assuming they are pure, the only way to tell them apart is through

dating: Compounds synthesized from petrochemicals have a lower content ofradioactive and appear old because their has decayed over time. Plant-derivedcompounds are recently synthesized from in the air. They have a higher content ofradioactive Some large chemical suppliers provide isotope ratio analyses to showthat their “naturals” have high content and are plant-derived. Such a sophisticatedanalysis lends a high-tech flavor to this twenty-first-century form of Vitalism.

Even though organic compounds do not need a vital force, they are still distin-guished from inorganic compounds. The distinctive feature of organic compounds is thatthey all contain one or more carbon atoms. Still, not all carbon compounds are organic;substances such as diamond, graphite, carbon dioxide, ammonium cyanate, and sodiumcarbonate are derived from minerals and have typical inorganic properties. Most of themillions of carbon compounds are classified as organic, however.

We ourselves are composed largely of organic molecules, and we are nourished bythe organic compounds in our food. The proteins in our skin, the lipids in our cell mem-branes, the glycogen in our livers, and the DNA in the nuclei of our cells are all organiccompounds. Our bodies are also regulated and defended by complex organic compounds.

14C

14C.CO2

14C14C

14C

N

N CH3

nicotine

OH

O

CH2OH

HO

HCOH

H

O

vitamin C

glucose

OH

OH

COOH

OH O

O

HO

carmine

Four examples of organic compounds in living organisms. Tobacco contains nicotine, an addictive alkaloid. Rose hips contain vitaminC, essential for preventing scurvy. The red dye carmine comes from cochineal insects, shown on prickly pear cactus. Opium poppiescontain morphine, a pain-relieving, addictive alkaloid.

Chemists have learned to synthesize or simulate many of these complex mole-cules. The synthetic products serve as drugs, medicines, plastics, pesticides, paints,and fibers. Many of the most important advances in medicine are actually advances inorganic chemistry. New synthetic drugs are developed to combat disease, and newpolymers are molded to replace failing organs. Organic chemistry has gone full circle.It began as the study of compounds derived from “organs,” and now it gives us thedrugs and materials we need to save or replace those organs.

One of the reasons chemistssynthesize derivatives of complexorganic compounds like morphineis to discover new drugs that retainthe good properties (potent pain-relieving) but not the bad proper-ties (highly addictive).

One of nicotine’s effects is toincrease the concentration ofdopamine, a chemical in the brain’sreward system. Release of thischemical makes smokers feel goodand reinforces the need to smoke.


cloud of electrons

�

nucleus(protons and neutrons)

Á Figure 1-1Basic atomic structure. An atom hasa dense, positively charged nucleussurrounded by a cloud of electrons.

1-2 Principles of Atomic Structure 3

1-2Principles of Atomic Structure

Before we begin our study of organic chemistry, we must review some basic prin-ciples. Many of these concepts of atomic and molecular structure are crucial to yourunderstanding of the structure and bonding of organic compounds.

1-2A Structure of the AtomAtoms are made up of protons, neutrons, and electrons. Protons are positively chargedand are found together with (uncharged) neutrons in the nucleus. Electrons, which havea negative charge that is equal in magnitude to the positive charge on the proton, occu-py the space surrounding the nucleus (Figure 1-1). Protons and neutrons have similarmasses, about 1800 times the mass of an electron. Almost all the atom’s mass is in thenucleus, but it is the electrons that take part in chemical bonding and reactions.

Each element is distinguished by the number of protons in the nucleus (the atom-ic number). The number of neutrons is usually similar to the number of protons,although the number of neutrons may vary. Atoms with the same number of protonsbut different numbers of neutrons are called isotopes. For example, the most commonkind of carbon atom has six protons and six neutrons in its nucleus. Its mass number(the sum of the protons and neutrons) is 12, and we write its symbol as About 1%of carbon atoms have seven neutrons; the mass number is 13, written A very smallfraction of carbon atoms have eight neutrons and a mass number of 14. The iso-tope is radioactive, with a half-life (the time it takes for half of the nuclei to decay)of 5730 years. The predictable decay of is used to determine the age of organicmaterials up to about 50,000 years old.

1-2B Electronic Structure of the AtomAn element’s chemical properties are determined by the number of protons in thenucleus and the corresponding number of electrons around the nucleus. The electronsform bonds and determine the structure of the resulting molecules. Because they aresmall and light, electrons show properties of both particles and waves; in many ways,the electrons in atoms and molecules behave more like waves than like particles.

Electrons that are bound to nuclei are found in orbitals. The Heisenberg uncer-tainty principle states that we can never determine exactly where the electron is;nevertheless, we can determine the electron density, the probability of finding theelectron in a particular part of the orbital. An orbital, then, is an allowed energy statefor an electron, with an associated probability function that defines the distribution ofelectron density in space.

Atomic orbitals are grouped into different “shells” at different distances from thenucleus. Each shell is identified by a principal quantum number n, with for thelowest-energy shell closest to the nucleus. As n increases, the shells are farther fromthe nucleus, higher in energy, and can hold more electrons. Most of the commonelements in organic compounds are found in the first two rows of the periodic table,indicating that their electrons are found in the first two electron shells. The first shell

can hold two electrons, and the second shell can hold eight.The first electron shell contains just the 1s orbital. All s orbitals are spherically

symmetrical, meaning that they are nondirectional. The electron density is only a func-tion of the distance from the nucleus. The electron density of the 1s orbital is graphedin Figure 1-2. Notice how the electron density is highest at the nucleus and falls offexponentially with increasing distance from the nucleus. The 1s orbital might be imag-ined as a cotton boll, with the cottonseed at the middle representing the nucleus. Thedensity of the cotton is highest nearest the seed, and it becomes less dense at greaterdistances from this “nucleus.”

The second electron shell consists of the 2s and 2p orbitals. The 2s orbital is spher-ically symmetrical like the 1s orbital, but its electron density is not a simple exponentialfunction. The 2s orbital has a smaller amount of electron density close to the nucleus.

1n = 221n = 12

n = 1

14C

14C

13C.

12C.



electron density

distance from the nucleus

distance

1s

nucleus» Figure 1-2Graph and diagram of the 1s atomicorbital. The electron density ishighest at the nucleus and dropsoff exponentially with increasingdistance from the nucleusin any direction.

» Figure 1-3Graph and diagram of the 2s atomicorbital. The 2s orbital has a smallregion of high electron density closeto the nucleus, but most of theelectron density is farther from thenucleus, beyond a node, or regionof zero electron density.

electron density

node node

distancefrom thenucleus

distancefrom thenucleus

node

node

2s

nucleus

Most of the electron density is farther away, beyond a region of zero electron densitycalled a node. Because most of the 2s electron density is farther from the nucleus than thatof the 1s, the 2s orbital is higher in energy. Figure 1-3 shows a graph of the 2s orbital.

In addition to the 2s orbital, the second shell also contains three 2p atomicorbitals, one oriented in each of the three spatial directions. These orbitals are calledthe the and the according to their direction along the x, y, or z axis. The2p orbitals are slightly higher in energy than the 2s, because the average location of theelectron in a 2p orbital is farther from the nucleus. Each p orbital consists of two lobes,one on either side of the nucleus, with a nodal plane at the nucleus. The nodal plane isa flat (planar) region of space, including the nucleus, with zero electron density. Thethree 2p orbitals differ only in their spatial orientation, so they have identical energies.Orbitals with identical energies are called degenerate orbitals. Figure 1-4 shows theshapes of the three degenerate 2p atomic orbitals.

2pz,2py,2px,


1-2 Principles of Atomic Structure 5

« Figure 1-4The 2p orbitals. There are three 2porbitals, oriented at right angles toeach other. Each is labeled accordingto its orientation along the x, y,or z axis.

nucleus

distance fromthe nucleus

electrondensity

node2p

the 2 px orbital the 2px , 2py , and 2pzorbitals superimposed

at 90� angles

x

y

z

directions of axes(z comes out toward us)

nodalplane

x

y

x

z

y

z

The Pauli exclusion principle tells us that each orbital can hold a maximum of2 electrons, provided that their spins are paired. The first shell (one 1s orbital) canaccommodate 2 electrons. The second shell (one 2s orbital and three 2p orbitals) canaccommodate 8 electrons, and the third shell (one 3s orbital, three 3p orbitals, and five3d orbitals) can accommodate 18 electrons.

1-2C Electronic Configurations of AtomsAufbau means “building up” in German, and the aufbau principle tells us how to buildup the electronic configuration of an atom’s ground (most stable) state. Starting withthe lowest-energy orbital, we fill the orbitals in order until we have added the propernumber of electrons. Table 1-1 shows the ground-state electronic configurations of theelements in the first two rows of the periodic table.

TABLE 1-1 Electronic Configurations of the Elements of the First and Second Rows

Element Configuration Valence Electrons

H 1He 2Li 1Be 2B 3C 4N 5O 6F 7Ne 81s22s22px

22py22pz

21s22s22px

22py22pz

11s22s22px

22py12pz

11s22s22px

12py12pz

11s22s22px

12py1

1s22s22px1

1s22s21s22s11s21s1

Relative orbital energies

energy

2py 2pz2px

1s

2s



Lithium carbonate, a salt of lithium,is a mood-stabilizing agent used totreat the psychiatric disorder knownas mania. Mania is characterized bybehaviors such as elated mood, feel-ings of greatness, racing thoughts,and an inability to sleep. We don’tknow how lithium carbonate helpsto stabilize these patients’ moods.

» Figure 1-5First three rows of the periodic table.The organization of the periodic tableresults from the filling of atomicorbitals in order of increasing energy.For these representative elements, thenumber of the column corresponds tothe number of valence electrons.

IA

H

noblegases(VIII)

Li

Na

IIA

Be

Mg

IIIA

B

Al

IVA

C

Si

VA

N

P

VIA

O

S

VIIA

F

Cl

He

Ne

Ar

Partial periodic table

1-3Bond Formation: The Octet Rule

In 1915, G. N. Lewis proposed several new theories describing how atoms bondtogether to form molecules. One of these theories states that a filled shell of electronsis especially stable, and atoms transfer or share electrons in such a way as to attain afilled shell of electrons. A filled shell of electrons is simply the electron configurationof a noble gas, such as He, Ne, or Ar. This principle has come to be called the octetrule because a filled shell implies eight valence electrons for the elements in thesecond row of the periodic table.

1-3A Ionic BondingThere are two ways that atoms can interact to attain noble-gas configurations. Some-times atoms attain noble-gas configurations by transferring electrons from one atom toanother. For example, lithium has one electron more than the helium configuration,and fluorine has one electron less than the neon configuration. Lithium easily loses itsvalence electron, and fluorine easily gains one:

Two additional concepts are illustrated in Table 1-1. The valence electronsare those electrons that are in the outermost shell. Carbon has four valence elec-trons, nitrogen has five, and oxygen has six. Helium has two valence electrons, andneon has eight, corresponding to a filled first shell and second shell, respectively. Ingeneral (for the representative elements), the column or group number of the peri-odic table corresponds to the number of valence electrons (Figure 1-5). Hydrogenand lithium have one valence electron, and they are both in the first column (groupIA) of the periodic table. Carbon has four valence electrons, and it is in group IVAof the periodic table.

Notice in Table 1-1 that carbon’s third and fourth valence electrons are notpaired; they occupy separate orbitals. Although the Pauli exclusion principle saysthat two electrons can occupy the same orbital, the electrons repel each other, andpairing requires additional energy. Hund’s rule states that when there are two ormore orbitals of the same energy, electrons will go into different orbitals rather thanpair up in the same orbital. The first 2p electron (boron) goes into one 2p orbital,the second (carbon) goes into a different orbital, and the third (nitrogen) occupiesthe last 2p orbital. The fourth, fifth, and sixth 2p electrons must pair up with thefirst three electrons.

PROBLEM 1-1

Write the electronic configurations of the third-row elements shown in the partial periodictable in Figure 1-5.


1-4 Lewis Structures 7

Li F

electron transfer

Li�

He configuration

� �F

Ne configuration

Li� F �

ionic bond

A transfer of one electron gives each of these two elements a noble-gas configu-ration. The resulting ions have opposite charges, and they attract each other to form anionic bond. Ionic bonding usually results in the formation of a large crystal latticerather than individual molecules. Ionic bonding is common in inorganic compoundsbut relatively uncommon in organic compounds.

1-3B Covalent BondingCovalent bonding, in which electrons are shared rather than transferred, is the mostcommon type of bonding in organic compounds. Hydrogen, for example, needs a sec-ond electron to achieve the noble-gas configuration of helium. If two hydrogen atomscome together and form a bond, they “share” their two electrons, and each atom hastwo electrons in its valence shell.

We will study covalent bonding in more detail in Chapter 2.

H H each H shares two electrons(He configuration)

H H�

1-4Lewis Structures

One way to symbolize the bonding in a covalent molecule is to use Lewis structures.In a Lewis structure, each valence electron is symbolized by a dot. A bonding pair ofelectrons is symbolized by a pair of dots or by a dash We try to arrange all theatoms so they have their appropriate noble-gas configurations: two electrons forhydrogen, and octets for the second-row elements.

Consider the Lewis structure of methane

Carbon contributes four valence electrons, and each hydrogen contributes one, to givea total of eight electrons. All eight electrons surround carbon to give it an octet, andeach hydrogen atom shares two of the electrons.

The Lewis structure for ethane is more complex.

Once again, we have computed the total number of valence electrons (14) and distrib-uted them so that each carbon atom is surrounded by 8 and each hydrogen by 2. Theonly possible structure for ethane is the one shown, with the two carbon atoms sharinga pair of electrons and each hydrogen atom sharing a pair with one of the carbons. Theethane structure shows the most important characteristic of carbon—its ability to formstrong carbon–carbon bonds.

H H9C9C9HC C H or

ethane

H

H

H

H

H

H

H

H

1C2 H62

H H9C9HC H or

methane

H

H

H

H

1CH42.

1¬ 2.



Valence-shell electrons that are not shared between two atoms are callednonbonding electrons. A pair of nonbonding electrons is often called a lone pair.Oxygen atoms, nitrogen atoms, and the halogens (F, Cl, Br, I) usually havenonbonding electrons in their stable compounds. These lone pairs of nonbondingelectrons help to determine the reactivity of their parent compounds. The followingLewis structures show one lone pair of electrons on the nitrogen atom of methyl-amine and two lone pairs on the oxygen atom of ethanol. Halogen atoms usuallyhave three lone pairs, as shown in the structure of chloromethane.

A correct Lewis structure should show any lone pairs. Organic chemists oftendraw structures that omit most or all of the lone pairs. These are not true Lewis struc-tures because you must imagine the correct number of nonbonding electrons.

PROBLEM 1-2

Draw Lewis structures for the following compounds.(a) ammonia, (b) water,(c) hydronium ion, (d) propane,(e) ethylamine, (f) dimethyl ether,(g) fluoroethane, (h) 2-propanol,(i) borane, (j) boron trifluoride,Explain what is unusual about the bonding in compounds in parts (i) and (j).

BF3BH3

CH3 CH1OH2CH3CH3 CH2 FCH3 OCH3CH3 CH2 NH2

C3 H8H3 O+

H2 ONH3

H

C N

H

H

H

H

lone pair

methylamine

H

C

H

H C

H

H

O

H

lone pairs H

C

H

H Cl

ethanol chloromethane

lone pairs

1-5Multiple Bonding

In drawing Lewis structures in Section 1-4, we placed just one pair of electronsbetween any two atoms. The sharing of one pair between two atoms is called a singlebond. Many molecules have adjacent atoms sharing two or even three electron pairs.The sharing of two pairs is called a double bond, and the sharing of three pairs iscalled a triple bond. Ethylene is an organic compound with a doublebond. When we draw a Lewis structure for ethylene, the only way to show both carbonatoms with octets is to draw them sharing two pairs of electrons. The followingexamples show organic compounds with double bonds. In each case, four electrons(two pairs) are shared between two atoms to give them octets. A double dash symbolizes a double bond.

Acetylene has a triple bond. Its Lewis structure shows three pairs ofelectrons between the carbon atoms to give them octets. The following examplesshow organic compounds with triple bonds. A triple dash symbolizes atriple bond.

1‚ 2

1C2 H22

C CHH

HH

or

CCH

H

H

Hethylene

CH

H

CH

H

O

O

or

formaldehyde

CH

H

CH

H

N

N

or

formaldimine

H

H

1“ 2

1C2 H42

Acetylene, in combination with oxy-gen, burns with an intense flamethat has diverse applications. It canbe used for welding parts of abridge underwater and for repairingan oil pipeline in Siberia.

PROBLEM-SOLVING HintLewis structures are the way we writeorganic chemistry. Learning now todraw them quickly and correctly willhelp you throughout this course.


1-6 Electronegativity and Bond Polarity 9

SUMMARY Common Bonding Patterns (Uncharged)

carbon40

valence:lone pairs:

nitrogen31

oxygen22

halogens13

hydrogen10

9C9 9N9 9O9 9H 9Cl

All these Lewis structures show that carbon normally forms four bonds in neutral or-ganic compounds. Nitrogen generally forms three bonds, and oxygen usually forms two. Hy-drogen and the halogens usually form only one bond. The number of bonds an atom usuallyforms is called its valence. Carbon is tetravalent, nitrogen is trivalent, oxygen is divalent, andhydrogen and the halogens are monovalent. By remembering the usual number of bonds forthese common elements, we can write organic structures more easily. If we draw a structurewith each atom having its usual number of bonds, a correct Lewis structure usually results.

H9C9C#C9C9HH9C#C9H

H

H

H9C9C#N

H

H

H

Hacetonitriledimethylacetyleneacetylene

H C C CH H C CC HH

HCH C N

H

H

ororor

H

H

1-6Electronegativity and Bond Polarity

A bond with the electrons shared equally between the two atoms is called a nonpolarcovalent bond. The bond in and the bond in ethane are nonpolar covalentbonds. In most bonds between two different elements, the bonding electrons are at-tracted more strongly to one of the two nuclei. An unequally shared pair of bondingelectrons is called a polar covalent bond.

When carbon is bonded to chlorine, for example, the bonding electrons are attractedmore strongly to the chlorine atom. The carbon atom bears a small partial positive charge,and the chlorine atom bears an equal amount of negative charge. Figure 1-6 shows thepolar carbon–chlorine bond in chloromethane. We symbolize the bond polarity by an

ClCH

nonpolarcovalent bond

polarcovalent bond

ionic bond

H ClNa+ −

C ¬ CH2

PROBLEM 1-3

Write Lewis structures for the following molecular formulas.(a) (b) HCN (c) HONO(d) (e) (f)(g) (h) HNNH (i)(j) (two double bonds) (k) (one triple bond)

PROBLEM 1-4

Circle any lone pairs (pairs of nonbonding electrons) in the structures you drew for Problem 1-3.

C3 H4C3 H4

C3 H6C2 H3 ClHCO2 HH2 CNHCO2

N2

PROBLEM-SOLVING HintThese “usual numbers of bonds”might be single bonds, or they mightbe combined into double and triplebonds. For example, three bonds tonitrogen might be three single bonds,one single bond and one doublebond, or one triple bond In working problems, considerall possibilities.

1≠N ‚ N≠2.


» Figure 1-7The Pauling electronegativitiesof some of the elements foundin organic compounds.


chloromethanechloromethane

ClC

HH

Hm

d� d�

Á Figure 1-6Bond polarity. Chloromethane contains a polar carbon–chlorine bond with a partial negative chargeon chlorine and a partial positive charge on carbon. The electrostatic potential map shows a redregion (electron-rich) around the partial negative charge and a blue region (electron-poor) aroundthe partial positive charge. Other colors show intermediate values of electrostatic potential.

H2.2Li1.0Na0.9K

0.8

Be1.6Mg1.3

B1.8Al1.6

C2.5Si1.9

N3.0P

2.2

O3.4S

2.6

F4.0Cl3.2Br3.0I

2.7

arrow with its head at the negative end of the polar bond and a plus sign at the positiveend. The bond polarity is measured by its dipole moment defined to be the amountof charge separation ( and ) multiplied by the bond length. The symbol means “asmall amount of positive charge”; means “a small amount of negative charge.”

Figure 1-6 also shows an electrostatic potential map (EPM) for chloromethane,using color to represent the calculated charge distribution in a molecule. Red showselectron-rich regions, and blue shows electron-poor regions. Orange, yellow, and greenshow intermediate levels of electrostatic potential. In chloromethane, the red regionshows the partial negative charge on chlorine, and the blue region shows the partial pos-itive charges on carbon and the hydrogen atoms.

We often use electronegativities as a guide in predicting whether a given bond will bepolar and the direction of its dipole moment. The Pauling electronegativity scale, most com-monly used by organic chemists, is based on bonding properties, and it is useful for predict-ing the polarity of covalent bonds. Elements with higher electronegativities generally havemore attraction for the bonding electrons. Therefore, in a bond between two different atoms,the atom with the higher electronegativity is the negative end of the dipole. Figure 1-7 showsPauling electronegativities for some of the important elements in organic compounds.

Notice that the electronegativities increase from left to right across the periodictable. Nitrogen, oxygen, and the halogens are all more electronegative than carbon;sodium, lithium, and magnesium are less electronegative. Hydrogen’s electronegativi-ty is similar to that of carbon, so we usually consider bonds to be nonpolar. Wewill consider the polarity of bonds and molecules in more detail in Section 2-9.

PROBLEM 1-5

Use electronegativities to predict the direction of the dipole moments of the following bonds.(a) (b) (c) (d) (e)(f) (g) (h) (i) (j) B ¬ ClN ¬ BN ¬ SN ¬ ON ¬ Cl

C ¬ BC ¬ SC ¬ NC ¬ OC ¬ Cl

C ¬ H

d-

d+d-d+

1m2,


1-7 Formal Charges 11

1-7Formal Charges

In polar bonds, the partial charges ( and ) on the bonded atoms are real. Formalcharges provide a method for keeping track of electrons, but they may or may not cor-respond to real charges. In most cases, if the Lewis structure shows that an atom has aformal charge, it actually bears at least part of that charge. The concept of formalcharge helps us determine which atoms bear most of the charge in a charged molecule,and it also helps us to see charged atoms in molecules that are neutral overall.

To calculate formal charges, count how many electrons contribute to the chargeof each atom and compare that number with the number of valence electrons in thefree, neutral atom (given by the group number in the periodic table). The electrons thatcontribute to an atom’s charge are

1. all its unshared (nonbonding) electrons; plus2. half the (bonding) electrons it shares with other atoms, or one electron of each

bonding pair.

The formal charge of a given atom can be calculated by the formula

SOLVED PROBLEM 1-1

Compute the formal charge (FC) on each atom in the following structures.(a) Methane

SOLUTIONEach of the hydrogen atoms in methane has one bonding pair of electrons (two shared electrons).Half of two shared electrons is one electron, and one valence electron is what hydrogen needsto be neutral. Hydrogen atoms with one bond are formally neutral:

The carbon atom has four bonding pairs of electrons (eight electrons). Half of eightshared electrons is four electrons, and four electrons are what carbon (group IVA) needs to beneutral. Carbon is formally neutral whenever it has four bonds:

(b) The hydronium ion,

SOLUTIONIn drawing the Lewis structure for this ion, we use eight electrons: six from oxygen plusthree from the hydrogens, minus one because the ion has a positive charge. Each hydro-gen has one bond and is formally neutral. Oxygen is surrounded by an octet, with sixbonding electrons and two nonbonding electrons. Half the bonding electrons plus all thenonbonding electrons contribute to its charge: but oxygen (group VIA) needssix valence electrons to be neutral. Consequently, the oxygen atom has a formal chargeof

(c)

HN B H

H

H HH

�

Boron has four bonds, eight bonding electrons

Nitrogen has four bonds, eight bonding electrons

�

H3 N ¬ BH3

FC = 6 - 2 -12162 = +1.+1:

62 + 2 = 5;

two nonbonding electrons

O�

HH

H three bonds, six bonding electrons

H3 O+

FC = 4 - 0 -12182 = 0.

FC = 1 - 0 - 1 = 0.

HH CH

H

1CH42

formal charge 1FC2 = [group number] - [nonbonding electrons] -12[shared electrons]

d-d+



SOLUTIONThis is a neutral compound where the individual atoms are formally charged. The Lewis struc-ture shows that both nitrogen and boron have four shared bonding pairs of electrons. Bothboron and nitrogen have electrons contributing to their charges. Nitrogen (group V)needs five valence electrons to be neutral, so it bears a formal charge of Boron (group III)needs only three valence electrons to be neutral, so it bears a formal charge of

(d)

SOLUTIONIn this structure, both carbon and nitrogen have four shared pairs of bonding electrons. Withfour bonds, carbon is formally neutral; however, nitrogen is in group V, and it bears a formalpositive charge:

This compound might also be drawn with the following Lewis structure:

In this structure, the carbon atom has three bonds with six bonding electrons. We calculatethat electrons, so carbon is one short of the four needed to be formally neutral:

Nitrogen has six bonding electrons and two nonbonding electrons. We calculate thatso the nitrogen is uncharged in this second structure:

The significance of these two Lewis structures is discussed in Section 1-9.

Most organic compounds contain only a few common elements, usually withcomplete octets of electrons. The summary table on the facing page shows the mostcommonly occurring bonding structures, using dashes to represent bonding pairs ofelectrons. Use the rules for calculating formal charges to verify the charges shown onthese structures. A good understanding of the structures shown here will help you todraw organic compounds and their ions quickly and correctly.

FC = 5 - 2 -12162 = 0

62 + 2 = 5,

FC = 4 - 0 -12162 = +1.

62 = 3

H

H

H

H

C N�

FC = 5 - 0 - 4 = +1.

�

NCH

H

H

H

[H2 CNH2]+

Boron: FC = 3 - 0 -12182 = -1

Nitrogen: FC = 5 - 0 -12182 = +1

-1.+1.

82 = 4

1-8Ionic Structures

Some organic compounds contain ionic bonds. For example, the structure of methylam-monium chloride cannot be drawn using just covalent bonds. That wouldrequire nitrogen to have five bonds, implying ten electrons in its valence shell. The cor-rect structure shows the chloride ion ionically bonded to the rest of the structure.

1CH3 NH3 Cl2

methylammonium chloride

NCH H

HH

HH

Cl� �

cannot be drawn covalently

NCH

HH

HH

Cl

Htoo many electronsaround nitrogen


PROBLEM-SOLVING HintThis is a very important table. Workenough problems to become familiarwith these bonding patterns so youcan recognize other patterns as beingeither unusual or wrong.

1-9 Resonance 13

SUMMARY Common Bonding Patterns in Organic Compounds and Ions

B 3 B B �

C C �C�

4C

(no octet)

(no octet)

N N �N5N �

O 6 O O O ��

halogen 7 �Cl Cl Cl �

Atom ValenceElectrons

PositivelyCharged

Neutral NegativelyCharged

1-9Resonance

1-9A Resonance HybridsSome compounds’ structures are not adequately represented by a single Lewis structure.When two or more valence-bond structures are possible, differing only in the placementof electrons, the molecule will usually show characteristics of both structures. The dif-ferent structures are called resonance structures or resonance forms because they arenot different compounds, just different ways of drawing the same compound. The actualmolecule is said to be a resonance hybrid of its resonance forms. In Solved Problem1-1(d) we saw that the ion might be represented by either of the followingresonance forms:

H

H

H

H

C N�

H

H

H

H

C N�

H

H

H

H

C N�d d�

�

resonance forms of a resonance hybrid combinedrepresentation

[H2 CNH2]+

Some molecules can be drawn either covalently or ionically. For example, sodiumacetate may be drawn with either a covalent bond or an ionic bond be-tween sodium and oxygen. Because sodium generally forms ionic bonds with oxygen(as in NaOH), the ionically bonded structure is usually preferred. In general, bonds be-tween atoms with very large electronegativity differences (about 2 or more) are usuallydrawn as ionic.

PROBLEM 1-6

Draw Lewis structures for the following compounds and ions, showing appropriate formalcharges.(a) (b) (c)(d) (e) (f)(g) (h) (i)(j) (k) (l) [H2 C “ OH]+KOC1CH323[HONH3]+

1CH322 O ¬ BF3NaBH3 CNNaBH4

-CH3+CH3NaOCH3

1CH322 NH2 ClNH4 Cl[CH3 OH2]+

more common

Na9O9C9C9H

H

H

Na� � O9C9C9H

O H

Hless common

O

1NaOCOCH32



The actual structure of this ion is a resonance hybrid of the two structures. In the actualmolecule, the positive charge is delocalized (spread out) over both the carbon atom and thenitrogen atom. In the left resonance form, the positive charge is on carbon, but carbon doesnot have an octet. Nitrogen’s nonbonding electrons can move into the bond (as indicated bythe red arrow) to give the second structure with a double bond a positive charge on nitrogenand an octet on carbon. The combined representation attempts to combine the two reso-nance forms into a single picture with the charge shared by carbon and nitrogen.

Spreading the positive charge over two atoms makes the ion more stablethan it would be if the entire charge were localized only on the carbon or only on thenitrogen. We call this a resonance-stabilized cation. Resonance is most important whenit allows a charge to be delocalized over two or more atoms, as in this example.

Resonance stabilization plays a crucial role in organic chemistry, especially inthe chemistry of compounds having double bonds. We will use the concept of reso-nance frequently throughout this course. For example, the acidity of acetic acid(following) is enhanced by resonance effects. When acetic acid loses a proton, theresulting acetate ion has a negative charge delocalized over both of the oxygen atoms.Each oxygen atom bears half of the negative charge, and this delocalization stabilizesthe ion. Each of the carbon–oxygen bonds is halfway between a single bond and adouble bond, and they are said to have a bond order of 1

12.

acetic acid

CH

HH

H

C

O

� H2O

equilibrium

CH

H

H

C

O

resonance

acetate ion

� H3O�CH

H

H

C

O

O�

O

O �

We use a single double-headed arrow between resonance forms (and often en-close them in brackets) to indicate that the actual structure is a hybrid of the Lewisstructures we have drawn. By contrast, an equilibrium is represented by two arrows inopposite directions.

Some uncharged molecules actually have resonance-stabilized structures withequal positive and negative formal charges. We can draw two Lewis structures for ni-tromethane but both of them have a formal positive charge on nitrogenand a negative charge on one of the oxygens. Thus, nitromethane has a positive chargeon the nitrogen atom and a negative charge spread equally over the two oxygen atoms.The bonds are midway between single and double bonds, as indicated in thecombined representation:

Remember that individual resonance forms do not exist. The molecule does not“resonate” between these structures. It is a hybrid with some characteristics of both.An analogy is a mule, which is a hybrid of a horse and a donkey. The mule does not“resonate” between looking like a horse and looking like a donkey; it looks like a muleall the time, with the broad back of the horse and the long ears of the donkey.

1-9B Major and Minor Resonance ContributorsTwo or more correct Lewis structures for the same compound may or may not repre-sent electron distributions of equal energy. Although separate resonance forms do not

CH

H

H

N

O

O

N�

resonance forms

O

O

CH

H

H

�

�

� CH

H

H

N��

O

combined representation

212

O212

N ¬ O

1CH3 NO22,PROBLEM-SOLVING HintSecond-row elements (B, C, N, O, F)cannot have more than eightelectrons in their valence shells. Thefollowing is NOT a valid Lewisstructure:

ten electrons on N

C

H

H

N

O

O

H


1-9 Resonance 15

exist, we can estimate their relative energies as if they did exist. More stable resonanceforms are closer representations of the real molecule than less stable ones. The tworesonance forms shown earlier for the acetate ion have similar bonding, and they are ofidentical energy. The same is true for the two resonance forms of nitromethane. Thefollowing resonance forms are bonded differently, however.

These structures are not equal in estimated energy. The first structure has thepositive charge on nitrogen. The second has the positive charge on carbon, and the car-bon atom does not have an octet. The first structure is more stable because it has anadditional bond and all the atoms have octets. Many stable ions have a positive chargeon a nitrogen atom with four bonds (see Summary Table, page 13). We call the morestable resonance form the major contributor, and the less stable form is the minorcontributor. The structure of the actual compound resembles the major contributormore than it does the minor contributor.

Many organic molecules have major and minor resonance contributors.Formaldehyde can be written with a negative charge on oxygen, balancedby a positive charge on carbon. This polar resonance form is higher in estimated ener-gy than the double-bonded structure because it has charge separation, fewer bonds,and a positively charged carbon atom without an octet. The charge-separated structureis only a minor contributor, but it helps to explain why the formaldehyde bond is very polar, with a partial positive charge on carbon and a partial negativecharge on oxygen. The electrostatic potential map (EPM) also shows an electron-richregion (red) around oxygen and an electron-poor region (blue) around carbon informaldehyde.

O

C

HH

all octetsno charge separation(major contributor)

C�

HH

no octet on Ccharge separation

(minor contributor)

O �

C

HH

dipole moment

O

�

�d

d

m

C “ O

1H2 C “ O2

H

H

H

H

C N�

H

H

H

H

C N�

major contributor minor contributor

EPM of formaldehyde

In drawing resonance forms, we try to draw structures that are as low in energyas possible. The best candidates are those that have the maximum number of octetsand the maximum number of bonds. Also, we look for structures with the minimumamount of charge separation.

Only electrons can be delocalized. Unlike electrons, nuclei cannot be delocalized.They must remain in the same places, with the same bond distances and angles, in allthe resonance contributors. The following general rules will help us to draw realisticresonance structures:

1. All the resonance structures must be valid Lewis structures for the compound.2. Only the placement of the electrons may be shifted from one structure to

another. (Electrons in double bonds and lone pairs are the ones that are mostcommonly shifted.) Nuclei cannot be moved, and the bond angles mustremain the same.


PROBLEM-SOLVING HintResonance forms can be comparedusing the following criteria, beginningwith the most important:1. As many octets as possible2. As many bonds as possible3. Any negative charges on

electronegative atoms4. As little charge separation

as possible


resonance forms

C

H

H H

H H

C C� C

H

H H

H H

C C�

NOT resonance

C

H

H H

H HH HC C

C

H

H H

H H

C C

3. The number of unpaired electrons (if any) must remain the same. Most stablecompounds have no unpaired electrons, and all the electrons must remain pairedin all the resonance structures.

4. The major resonance contributor is the one with the lowest energy. Good con-tributors generally have all octets satisfied, as many bonds as possible, and as lit-tle charge separation as possible. Negative charges are more stable on moreelectronegative atoms, such as O, N, and S.

5. Resonance stabilization is most important when it serves to delocalize a chargeover two or more atoms.

SOLVED PROBLEM 1-2

For each of the following compounds, draw the important resonance forms. Indicate which structures are major and minor contributorsor whether they would have the same energy.(a)

SOLUTION

The first (minor) structure has a carbon atom with only six electrons around it. The second (major) structure has octets on all atoms and anadditional bond.

(b)

SOLUTION

Both of these structures have octets on oxygen and both carbon atoms, and they have the same number of bonds. The first structurehas the negative charge on carbon; the second has it on oxygen. Oxygen is the more electronegative element, so the second structureis the major contributor.(c)

SOLUTION

The first structure, with more bonds and less charge separation, is possible because sulfur is a third-row element with accessible d orbitals,giving it an expandable valence. For example, is a stable compound with 12 electrons around sulfur. Theoretical calculations suggestthat the last structure, with octets on all atoms, may be the major resonance contributor, however. We cannot always predict the major con-tributor of a resonance hybrid.

SF6

H9O9S9O9H

�

�

�

O

O O

H9O9S9O9H

O

O �

�2H9O9S9O9H

O�

H9O9S9O9H

O �O

H2 SO4

major contributorminor contributor

C�

C

O

C C

O �H

H H

H

H H

H

C

O�

C

H H

major contributorminor contributor

H

H

�

H9C9O9C�

H

H

H

H

H9C9O"C

H

H

[CH3 OCH2]+


1-10 Structural Formulas 17

PROBLEM 1-7

Draw the important resonance forms for the following molecules and ions.(a) (b) (c) (d)(e) (f) (g) [CH3 C1OCH322]+SO4

2-H2 C “ CH ¬ CH2-

H2 C “ CH ¬ CH2+NO2

-NO3-CO3

2-

1-10Structural Formulas

Several kinds of formulas are used by organic chemists to represent organic com-pounds. Some of these formulas involve a shorthand notation that requires someexplanation. Structural formulas actually show which atoms are bonded to which.There are two types of structural formulas, complete Lewis structures and condensedstructural formulas. In addition, there are several ways of drawing condensed structur-al formulas. As we have seen, a Lewis structure symbolizes a bonding pair of electronsas a pair of dots or as a dash Lone pairs of electrons are shown as pairs of dots.

1-10A Condensed Structural FormulasCondensed structural formulas (Table 1-2) are written without showing all the indi-vidual bonds. In a condensed structure, each central atom is shown together with theatoms that are bonded to it. The atoms bonded to a central atom are often listed afterthe central atom (as in rather than ) even if that is not their actu-al bonding order. In many cases, if there are two or more identical groups, parenthesesand a subscript may be used to represent all the identical groups. Nonbonding elec-trons are rarely shown in condensed structural formulas.

H3 C ¬ CH3CH3 CH3

1¬ 2.

PROBLEM-SOLVING HintIn drawing resonance forms for ions,see how you can delocalize the chargeover several atoms. Try to spread anegative charge over electronegativeelements like oxygen and nitrogen.Try to spread a positive charge over asmany carbons as possible, but especiallyover any atoms that can bear thepositive charge and still have an octet;for example, oxygen (with three bonds)or nitrogen (with four bonds).

PROBLEM 1-8

For each of the following compounds, draw the important resonance forms. Indicate whichstructures are major and minor contributors or whether they have the same energy.(a) (b) (c)

(d) (e) (f)

(g) (h) H9C9NH2

O

H9C9CH9C9H

O O�

H2 N ¬ C+

H ¬ CH “ CH ¬ NH2[H2 CCN]-H2 CNN

[H2 COH]+H2 C “ CH ¬ NO2[H2 CNO2]-

TABLE 1-2 Examples of Condensed Structural Formulas

Compound Lewis Structure Condensed Structural Formula

H9C9C9C9C9C9C9Hn-hexane

isobutane

ethane

CH3(CH2)4CH3

(CH3)3CH

CH3CH3

H H H H H H

H H H

H9C9H

H

H9C9C9C9H

H

H

H

H

H

H9C9C9H

H

H

H

H

H H H

(Continued)



TABLE 1-3 Condensed Structural Formulas for Double and Triple Bonds


2-butene H9C9C"C9C9H

H

H

H

H

H

H

C C HH

H

H

O

C C HH

H

H

C

H

H

O

O

CH3CHCHCH3 or CH3CH"CHCH3

CH3CHO or CH3CH

CH3COCH3

O

or CH3CCH3

acetonitrile

acetaldehyde

C C HH

H

H

O

O O

CH3COOH or CH3C9OHor CH3CO2H

acetic acid

acetone

H9C9C#N

H

H

CH3CN or CH3C#N

TABLE 1-2 Continued


diethyl ether

ethanol

isopropyl alcohol

dimethylamine

CH3CH2OCH2CH3or CH3CH29O9CH2CH3or (CH3CH2)2O

CH3CH2OH

(CH3)2CHOH

(CH3)2NH

H9C9C9O9H

H H

H H

H9C9C9O9C9C9H

H H

H H

H H

H H

H9C9N9C9H

H HH

H H

H9C9C99C9H

O9H

H H

H

H

H

When a condensed structural formula is written for a compound containingdouble or triple bonds, the multiple bonds are often drawn as they would be in aLewis structure. Table 1-3 shows examples of condensed structural formulascontaining multiple bonds. Notice that the group of an aldehyde and the

group of a carboxylic acid are actually bonded differently from what thecondensed notation suggests.¬ COOH

¬ CHO


1-10 Structural Formulas 19

As you can see from Tables 1-2 and 1-3, the distinction between a completeLewis structural formula and a condensed structural formula can be blurry. Chemistsoften draw formulas with some parts condensed and other parts completely drawn out.You should work with these different types of formulas so that you understand what allof them mean.

PROBLEM 1-9

Draw complete Lewis structures for the following condensed structural formulas.(a)(b)(c)(d)(e)(f)(g)

1-10B Line–Angle FormulasAnother kind of shorthand used for organic structures is the line–angle formula,sometimes called a skeletal structure or a stick figure. Line–angle formulas are oftenused for cyclic compounds and occasionally for noncyclic ones. In a stick figure,bonds are represented by lines, and carbon atoms are assumed to be present wherevertwo lines meet or a line begins or ends. Nitrogen, oxygen, and halogen atoms areshown, but hydrogen atoms are not usually drawn unless they are bonded to an atomthat is drawn. Each carbon atom is assumed to have enough hydrogen atoms to give ita total of four bonds. Nonbonding electrons are rarely shown. Table 1-4 shows someexamples of line–angle drawings.

1CH3 CH222 CO1CH323 CCOOHCH3 COCNCH3 CH2 CHOCH3 CH2 COCHCH2

1CH322 CHCH2 ClCH31CH223 CH1CH322

TABLE 1-4 Examples of Line–Angle Drawings

Compound Condensed Structure Line–Angle Formula

hexane CH3(CH2)4CH3

CH3CH2-hexene CHCH2CH2CH3

OH3-hexanol CH3CH2CH(OH)CH2CH2CH3

CHCH

CH2

CH2

CCH2

2-cyclohexenone

2-methylcyclohexanol

nicotinic acid(a vitamin, also called niacin)

OO

CHCH3

CH2

CH2

CH2

CHOHCH2

CH3

OH

or

OH

CHN

C

C

HC

C

H

COOH

H

COOH

N

orOH

N

O



PROBLEM 1-10

Give Lewis structures corresponding to the following line–angle structures.

(a) (b) (c) (d)

(e) (f) (g) (h)

PROBLEM 1-11

Draw condensed structural formulas corresponding to the following line–angle structures.

(a) (b) (c) (d)O

H

OH

OH

O

O

O

OCHO

OHNHO

NH

1-11Molecular Formulas and Empirical Formulas

Before we can write possible structural formulas for a compound, we need to know itsmolecular formula. The molecular formula simply gives the number of atoms of eachelement in one molecule of the compound. For example, the molecular formula for1-butanol is

1-butanol, molecular formula

Calculation of the Empirical Formula Molecular formulas can be determinedby a two-step process. The first step is the determination of an empirical formula,simply the relative ratios of the elements present. Suppose, for example, that anunknown compound was found by quantitative elemental analysis to contain 40.0%carbon and 6.67% hydrogen. The remainder of the weight (53.3%) is assumed to beoxygen. To convert these numbers to an empirical formula, we can follow a simpleprocedure.

1. Assume the sample contains 100 g, so the percent value gives the number ofgrams of each element. Divide the number of grams of each element by theatomic weight to get the number of moles of that atom in the 100-g sample.

2. Divide each of these numbers of moles by the smallest one. This step should giverecognizable ratios.

For the unknown compound, we do the following computations:

The first computation divides the number of grams of carbon by 12, thenumber of grams of hydrogen by 1, and the number of grams of oxygen by 16. We

53.3 g O

16.0 g>mol= 3.33 mol O; 3.33 mol

3.33 mol= 1

6.67 g H

1.01 g>mol= 6.60 mol H; 6.60 mol

3.33 mol= 1.98 � 2

40.0 g C

12.0 g>mol= 3.33 mol C; 3.33 mol

3.33 mol= 1

C4 H10 O

CH3 CH2 CH2 CH2 OH

C4 H10 O.


1-12 Arrhenius Acids and Bases 21

1-12Arrhenius Acids and Bases

The properties and reactions of acids and bases are central to our study of organicchemistry. We need to consider exactly what is meant by the terms acid and base.Most people would agree that is an acid and NaOH is a base. Is an acid ora base? Is ethylene an acid or a base? To answer these questions, weneed to understand the three different definitions of acids and bases: the Arrhenius def-inition, the Brønsted–Lowry definition, and the Lewis definition.

Acidic compounds were first classified on the basis of their sour taste. The Latinterms acidus (sour) and acetum (vinegar) gave rise to our modern terms acid andacetic acid. Alkaline compounds (bases) were substances that neutralize acids, such aslimestone and plant ashes (al kalai in Arabic).

The Arrhenius theory, developed at the end of the nineteenth century, definedacids as substances that dissociate in water to give ions. The stronger acids,such as sulfuric acid were assumed to dissociate to a greater degree thanweaker acids, such as acetic acid

O

CH39C9OH

O

CH39C9O�

acetic acidH2O� H3O

� �

H2SO4 HSO4�H2O� H3O

� �sulfuric acid

1CH3 COOH2.1H2 SO42,

H3 O+

1H2 C “ CH22BF3H2 SO4

compare these numbers by dividing them by the smallest number, 3.33. The finalresult is a ratio of one carbon to two hydrogens to one oxygen. This result gives theempirical formula or which simply shows the ratios of the ele-ments. The molecular formula can be any multiple of this empirical formula,because any multiple also has the same ratio of elements. Possible molecular for-mulas are etc.

Calculation of the Molecular Formula How do we know the correct molecularformula? We can choose the right multiple of the empirical formula if we know themolecular weight. Molecular weights can be determined by methods that relate thefreezing-point depression or boiling-point elevation of a solvent to the molal concen-tration of the unknown. If the compound is volatile, we can convert it to a gas and useits volume to determine the number of moles according to the gas law. Newer methodsinclude mass spectrometry, which we will cover in Chapter 11.

For our example (empirical formula ), let’s assume that the molecularweight is determined to be about 60. The weight of one unit is 30, so our un-known compound must contain twice this many atoms. The molecular formula mustbe The compound might be acetic acid.C2 H4 O2.

CH2 OCH2 O

CH2 O, C2 H4 O2, C3 H6 O3, C4 H8 O4,

CH2 O,C1 H2 O1

CH39C9OHacetic acid, C2H4O2

O

In Chapters 12, 13, and 15 we will use spectroscopic techniques to determine thecomplete structure for a compound once we know its molecular formula.

PROBLEM 1-12

Compute the empirical and molecular formulas for each of the following elemental analyses.In each case, propose at least one structure that fits the molecular formula.

C H N Cl MW(a) 40.0% 6.67% 0 0 90(b) 32.0% 6.67% 18.7% 0 75(c) 37.2% 7.75% 0 55.0% 64(d) 38.4% 4.80% 0 56.8% 125

PROBLEM-SOLVING HintIf an elemental analysis does not addup to 100%, the missing percentageis assumed to be oxygen.



1-13Brønsted–Lowry Acids and Bases

In 1923, Brønsted and Lowry defined acids and bases on the basis of the transfer ofprotons. A Brønsted–Lowry acid is any species that can donate a proton, and aBrønsted–Lowry base is any species that can accept a proton. These definitions alsoinclude all the Arrhenius acids and bases because compounds that dissociate to give

are proton donors, and compounds that dissociate to give are protonacceptors. (Hydroxide ion accepts a proton to form )

In addition to Arrhenius acids and bases, the Brønsted–Lowry definitionincludes bases that have no hydroxide ions, yet can accept protons. Consider the

H2 O.

-OHH3 O+

According to the Arrhenius definition, bases are substances that dissociate inwater to give hydroxide ions. Strong bases, such as NaOH, were assumed to dissociatemore completely than weaker, sparingly soluble bases such as

The acidity or basicity of an aqueous (water) solution is measured by the con-centration of This value also implies the concentration of because thesetwo concentrations are related by the water ion-product constant:

In a neutral solution, the concentrations of and are equal.

Acidic and basic solutions are defined by an excess of or

Because these concentrations can span a wide range of values, the acidity orbasicity of a solution is usually measured on a logarithmic scale. The pH is defined asthe negative logarithm (base 10) of the concentration.

A neutral solution has a pH of 7, an acidic solution has a pH less than 7, and a basicsolution has a pH greater than 7.

PROBLEM 1-13

Calculate the pH of the following solutions.(a) 5.00 g of HBr in 100 mL of aqueous solution(b) 1.50 g of NaOH in 50 mL of aqueous solution

The Arrhenius definition was an important contribution to understanding manyacids and bases, but it does not explain why a compound such as ammonia neutralizes acids, even though it has no hydroxide ion in its molecular formula. InSection 1-13 we discuss a more versatile theory of acids and bases that will includeammonia and a wider variety of organic acids and bases.

1NH32

pH = - log10[H3 O+]

H3 O+

[�OH] � 10�7 MM andacidic: [H3O�] � 10�7

basic: [�OH] � 10�7 MM and[H3O�] � 10�7

-OH.H3 O+

[H3O�] � [�OH] � 1.00 � 10�7 M in a neutral solution

-OHH3 O+

[H3O�][�OH]Kw � 1.00 � 10�14 (at 24°C)�

-OHH3 O+.

NaOH Na� �OH�

2 �OH�Mg(OH)2 Mg2�

Mg1OH22.


1-13 Brønsted–Lowry Acids and Bases 23

following examples of acids donating protons to bases. NaOH is a base under eitherthe Arrhenius or Brønsted–Lowry definition. The other three are Brønsted–Lowrybases but not Arrhenius bases, because they have no hydroxide ions.

When a base accepts a proton, it becomes an acid capable of returning thatproton. When an acid donates its proton, it becomes a base capable of acceptingthat proton back. One of the most important principles of the Brønsted–Lowry def-inition is this concept of conjugate acids and bases. For example, and are a conjugate acid–base pair. is the base; when it accepts a proton, it is trans-formed into its conjugate acid, Many compounds (water, for instance) canreact either as an acid or as a base. Here are some additional examples of conjugateacid–base pairs.

1-13A Acid StrengthThe strength of a Brønsted–Lowry acid is expressed as it is in the Arrhenius definition,by the extent of its ionization in water. The general reaction of an acid (HA) with wateris the following:

acid

conjugate acid–base pair

base

KaHA H2O H3O�� A� Ka [HA]

[H3O�][A�]

� �

O

H9C9OH

acid

O

H9C9O

conjugatebase

CH39O

base

� CH39O9H

conjugateacid

��

H2Oacid

�OHconjugate

base

NH3

base� NH4

�

conjugateacid

�

H2SO4

acid conjugatebase

H2Obase

� H3O�

conjugateacid

�HSO4�

NH4+.

NH3

NH3NH4+

Hprotondonor

Cl � NaOHproton

acceptor

NaCl �

NH3

�

HOH

H2SO4 � NH3

protonacceptor

protondonor

HSO4� � H

C C

H

H H

CH3

H

protondonor

Cl �

protonacceptor

C C

H

H

H H

CH3

��Cl�

HNO3 �

protondonor

C N

H

H Hproton

acceptor

C N

H

H H

�

H

��NO3



is called the acid-dissociation constant, and its value indicates the relativestrength of the acid. The stronger the acid, the more it dissociates, giving a larger valueof Acid-dissociation constants vary over a wide range. Strong acids are almostcompletely ionized in water, and their dissociation constants are greater than 1. Mostorganic acids are weak acids, with values less than Many organic compoundsare extremely weak acids; for example, methane and ethane are essentially nonacidic,with values less than

Because they span such a wide range, acid-dissociation constants are oftenexpressed on a logarithmic scale. The of an acid is defined just like the pH of asolution: as the negative logarithm (base 10) of

SOLVED PROBLEM 1-3

Calculate and for water.

SOLUTIONThe equilibrium that defines for water is

Water serves as both the acid and the solvent in this dissociation. The equilibrium expression is

We already know that the ion-product constant for water.The concentration of in water is simply the number of moles of water in 1 L

(about 1 kg).

Substitution gives

The logarithm of is so the of water is 15.7.

Strong acids generally have values around 0, and weak acids, such as mostorganic acids, have values that are greater than 4. Weaker acids have largervalues. Table 1-5 lists and values for some common inorganic and organiccompounds. Notice that the values increase as the values decrease.

PROBLEM 1-14

Ammonia appears in Table 1-5 both as an acid and as a conjugate base.(a) Explain how ammonia can act as both an acid and a base. Which of these roles does it

commonly fill in aqueous solutions?(b) Show how water can serve as both an acid and a base.(c) Show how methanol can serve as both an acid and a base. Write an equation

for the reaction of methanol with sulfuric acid.

1-13B Base StrengthThe strength of an acid is inversely related to the strength of its conjugate base. Foran acid (HA) to be strong, its conjugate base must be stable in its anionic form;1A-2

1CH3 OH2

KapKa

pKaKa

pKapKa

pKa

pKa-15.7,1.8 * 10-16

Ka =

[H3 O+][-OH]

[H2 O]=

1.00 * 10-14

55.6= 1.8 * 10-16 M

1000 g>L

18 g>mol= 55.6 mol>L

H2 O[H3 O+][-OH] = 1.00 * 10-14,

Ka =

[H3 O+][A-]

[HA]=

[H3 O+][-OH]

[H2 O]

acid (HA)

H2Osolvent

H2O H3O��

conjugate base (A�)

�OH�Ka

Ka

pKaKa

pKa = - log10 Ka

Ka.pKa

10-40.Ka

10-4.Ka

Ka.

Ka

PROBLEM-SOLVING HintIn most cases, the of an acidcorresponds to the pH where the acidis about half dissociated. At a lower(more acidic) pH, the acid is mostlyundissociated; at a higher (more basic)pH, the acid is mostly dissociated.

pKa



otherwise, HA would not easily lose its proton. Therefore, the conjugate base of astrong acid must be a weak base. On the other hand, if an acid is weak, its conjugate isa strong base.

In the reaction of an acid with a base, the equilibrium generally favors theweaker acid and base. For example, in the preceding reactions, is a weaker acidthan HCl but a stronger acid than It also follows that is a stronger basethan but a weaker base than CH3 O-.Cl-

H2 OCH3 OH.H3 O+

strong acidHCl H2O H3O

��

weak acid strong base

CH39OH H2O H3O� CH3O�

weak baseCl��

� �

TABLE 1-5 Relative Strength of Some Common Organic and Inorganic Acids and Their Conjugate Bases

Acid Conjugate Base

H �stronger Clhydrochloric

acid

H2O H3O� � Cl�

chlorideion

weakerbases

1.6 � 102 � 2.2

3.17

3.76

4.74

9.24

15.5

15.7

33

� 40

6.8 � 10�4

1.7 � 10�4

1.8 � 10�5

6.0 � 10�10

5.8 � 10�10

3.2 � 10�16

1.8 � 10�16

10�33

�10�40

HF � H2O H3O� � F�

fluorideionacid

H C

O

H

CH3

N

CH4

4

H � H2O H3O� �O H C

O

O�

hydrofluoric

formateionacid

formic

C O

O

H � H2O H3O� � CH3 C

O

O�

acetic acid acetate ion

H C � H2O H3O�

hydrocyanicacid

cyanide ion

NC��

�N � H2O H3O� � NH3

ammoniumion

ammonia

CH3 OH � H2O H3O� �

methylalcohol

CH3O�

methoxideion

H2O� H3O� � HO�

hydroxideion

water

H2O

H2O� H3O� �H3Nammonia

H2Namide

ion

H2O� H3O� � CH3�

methylanion

methane

strongerbasesweaker

strongacids

weakacids

very

not

weak

acidic

9.22

�

pKaKa



The strength of a base is measured much like the strength of an acid, by usingthe equilibrium constant of the hydrolysis reaction.

The equilibrium constant for this reaction is called the base-dissociationconstant for the base Because this constant spans a wide range of values, it isoften given in logarithmic form. The negative logarithm (base 10) of is definedas

When we multiply by we can see how the acidity of an acid is related tothe basicity of its conjugate base.

Logarithmically,

The product of and must always equal the ion-product constant of water,If the value of is large, the value of must be small; that is, the stronger an

acid, the weaker its conjugate base. Similarly, a small value of (weak acid) impliesa large value of (strong base).

The stronger an acid, the weaker its conjugate base.

The weaker an acid, the stronger its conjugate base.

Acid–base reactions favor the weaker acid and the weaker base.

PROBLEM 1-15 (PARTIALLY SOLVED)

Write equations for the following acid–base reactions. Use the information in Table 1-5 topredict whether the equilibrium will favor the reactants or the products.(a) (b)(c) (d)(e) (f)

SOLUTION TO (A)Cyanide is the conjugate base of HCN. It can accept a proton from formic acid:

Reading from Table 1-5, formic acid is a stronger acid than HCNand cyanide is a stronger base than formate. The products (weaker acid

and base) are favored.1pKa = 9.222,

1pKa = 3.762

H9C9O9H

formic acidstronger acid

cyanidestronger base

� ��C#N

weaker acid

H9C#N�

O

H9C9O

formateweaker base

O

H3 O++ CH3 O-HCl + H2 O

NaOCH3 + HCNCH3 OH + NaNH2

CH3 COO-+ CH3 OHHCOOH +

-CN

Kb

Ka

KbKa10-14.KbKa

pKa + pKb = - log 10-14= 14

(Ka)(Kb) [HA][H3O

�][A�][H3O

�][�OH]�

(Ka)(Kb) �

� 1.0 � 10�14

10�14

water ion-product constant�

[HA][�OH][A�]

Kb,Ka

Kb =

[HA][-OH]

[A-] pKb = - log10 Kb

pKb.Kb

A-.1Kb2

conjugatebase

conjugateacid

KbH2O HA�A� � �OH

PROBLEM-SOLVING HintAn acid will donate a proton to theconjugate base of any weaker acid(smaller or higher ).pKaKa

The acid–base properties of manynatural products are important fortheir isolation, distribution in thebody, and therapeutic effects. Forexample, morphine (page 2) is iso-lated from the opium poppy andcrosses into the brain as the freebase, where the nitrogen is notcharged. However, it exerts its pain-relieving effects as the chargedspecies.



SOLVED PROBLEM 1-5

Each of the compounds in Solved Problem 1-4 can also react as a base. Show the reaction of each compound with a general acid(HA), and show the Lewis structure of the conjugate acid that results.

SOLUTION(a) Ethanol can undergo protonation on its oxygen atom. Notice that one of the lone pairs of the oxygen forms the new bond.

(b) The nitrogen atom of methylamine has a pair of electrons that can bond to a proton.

(c) Acetic acid has nonbonding electrons on both of its oxygen atoms. Either of these oxygen atoms might become protonated, butprotonation of the double-bonded oxygen is favored because protonation of this oxygen gives a symmetrical, resonance-stabi-lized conjugate acid.

acetic acid(very weak base)

conjugate acid of acetic acid(very strong acid)

CH3 C

O

H �O HA CH3 C O

�O

H

H

CH3 C O H

HO

�CH3

O H

O�

C H �� A

�HA�

methylamine(moderate base)

acid

H

CH39NH2

(moderate acid)

CH39NH2 A �

�

CH3CH29O9H

(strong acid)

A� �HA�

ethanol(weak base)

acid

H

CH3CH29O9H �

O ¬ H

SOLVED PROBLEM 1-4

Each of the following compounds can act as an acid. Show the reaction of each compound with a general base and show theLewis structure of the conjugate base that results.(a) (b) (c)

SOLUTION(a) Ethanol can lose the proton to give a conjugate base that is an organic analogue of hydroxide ion.

( protons are much less acidic than protons because carbon is less electronegative than oxygen, and the negativecharge is therefore less stable on carbon.)

(b) Methylamine is a very weak acid. A very strong base can abstract a proton to give a powerful conjugate base.

(c) Acetic acid is a moderately strong acid, giving the resonance-stabilized acetate ion as its conjugate base.

acetic acid(moderate acid)

acetate ion(moderate base)

CH3 C

O

H �O CH3 C

O

O � CH3 C

O �

O � HAA �

1CH3 COOH2

CH39N9H �

methylamine(very weak acid)

very strongbase

� HA�

(powerful base)

H

CH39N9HA �

1CH3 NH22

O ¬ HC ¬ H

CH3CH29O9H A

ethanol(weak acid)

base

� HA�CH3CH29O

ethoxide(strong base)

��

O ¬ H1CH3 CH2 OH2

CH3 COOHCH3 NH2CH3 CH2 OH

1A-2,



PROBLEM 1-16

Solved Problem 1-5(c) showed protonation of the double-bonded oxygen in acetic acid. Show the product of protonation on the otheroxygen. Explain why protonation of the double-bonded oxygen is favored.

PROBLEM 1-17

(a) Rank ethanol, methylamine, and acetic acid in decreasing order of acidity.(b) Rank ethanol, methylamine and ethoxide ion in decreasing order of basicity. In each case, explain

your ranking.1CH3 CH2 O-21pKb 3.362,

1¬ OH2

1-13C Structural Effects on AcidityHow can we look at a structure and predict whether a compound will be a strong acid,a weak acid, or not an acid at all? To be a Brønsted–Lowry acid (HA), a compoundmust contain a hydrogen atom that can be lost as a proton. A strong acid must have astable conjugate base after losing the proton.

The stability of the conjugate base is a good guide to acidity. More stableanions tend to be weaker bases, and their conjugate acids tend to be stronger acids.Some of the factors that affect the stability of conjugate bases are electronegativity,size, and resonance.

Electronegativity A more electronegative element bears a negative charge moreeasily, giving a more stable conjugate base and a stronger acid. Electronegativities in-crease from left to right in the periodic table:

electronegativity increases

Electronegativity

Stability

acidity increases

Acidity

Basicity

H CH3 � H NH2 � H OH � H F

C � N � O � F

�CH3 � �NH2 � �OH � �F

�CH3 � �NH2 � �OH � �F

basicity increases

Acidity

Stability

H F � H Cl � H Br �

� � �

H I

F� Cl� Br� I�

size increases

acidity increases

Size The negative charge of an anion is more stable if it is spread over a larger region of space. Within a column of the periodic table, acidity increases down the column, as the size of the elements increases.

1A:-2


1-14 Lewis Acids and Bases 29

Resonance Stabilization The negative charge of a conjugate base may bedelocalized over two or more atoms by resonance. Depending on how electronega-tive those atoms are, and how many share the charge, resonance delocalization isoften the dominant effect helping to stabilize an anion. Consider the followingconjugate bases.

Conjugate Base Acid

acetate ion

ethoxide ion

methanesulfonate ion

CH3 C

O

O CH3 C

O

OH

O

�

CH3CH2�

ethanol (weak acid)

15.9

(moderate acid)

4.74

�1.2(strong acid)

OHCH3CH2

CH3 C

O �

O

CH3 S

O

O

O

�

O

CH3 S O O

O

O �

CH3 S

O �

methanesulfonic acid

acetic acid

CH3 S OH

O

O

pKa

Ethoxide ion is the strongest of these three bases. Ethoxide has a negative chargelocalized on one oxygen atom; acetate ion has the negative charge shared by twooxygen atoms; and the methanesulfonate ion has the negative charge spreadover three oxygen atoms. The values of the conjugate acids of these anionsshow that acids are stronger if they deprotonate to give resonance-stabilizedconjugate bases.

PROBLEM 1-18

Write equations for the following acid–base reactions. Label the conjugate acids and bases,and show any resonance stabilization. Predict whether the equilibrium favors the reactants orproducts.(a) (b)(c) (d)(e) (f)(g) CH3 SO3

-+ CH3 COOH

CH3 O-+ CH3 COOHCH3 NH3

++ CH3 O-

NaOH + H2 SCH3 OH + H2 SO4

CH3 CH2 COOH + CH3 NHCH3CH3 CH2 OH + CH3 NH-

pKa

1-14Lewis Acids and Bases

The Brønsted–Lowry definition of acids and bases depends on the transfer of aproton from the acid to the base. The base uses a pair of nonbonding electrons toform a bond to the proton. G. N. Lewis reasoned that this kind of reaction does notneed a proton. Instead, a base could use its lone pair of electrons to bond to someother electron-deficient atom. In effect, we can look at an acid–base reaction fromthe viewpoint of the bonds that are formed and broken rather than a proton that istransferred. The following reaction shows the proton transfer, with emphasis onthe bonds being broken and formed. Organic chemists routinely use curved arrowsto show the movement of the participating electrons.

Lewis bases are species with nonbonding electrons that can be donated toform new bonds. Lewis acids are species that can accept these electron pairsto form new bonds. Since a Lewis acid accepts a pair of electrons, it is called an

�B B� H A AH �


Some of the terms associated with acids and bases have evolved specific mean-ings in organic chemistry. When organic chemists use the term base, they usually meana proton acceptor (a Brønsted–Lowry base). Similarly, the term acid usually means aproton donor (a Brønsted–Lowry acid). When the acid–base reaction involves forma-tion of a bond to some other element (especially carbon), organic chemists refer to theelectron donor as a nucleophile (Lewis base) and the electron acceptor as anelectrophile (Lewis acid).

The following illustration shows electrostatic potential maps for the precedingsecond example, the reaction of with The electron-rich (red) region of attacks the electron-poor (blue) region of The product shows high electron densi-ty on the boron atom and its three fluorine atoms and low electron density on nitrogenand its three hydrogen atoms.

BF3.NH3BF3.NH3


electrophile, from the Greek words meaning “lover of electrons.” A Lewis base iscalled a nucleophile, or “lover of nuclei,” because it donates electrons to a nucleuswith an empty (or easily vacated) orbital. In this book, we sometimes use coloredtype for emphasis: blue for nucleophiles, green for electrophiles, and occasionallyred for acidic protons.

The Lewis acid–base definitions include reactions having nothing to do withprotons. Following are some examples of Lewis acid–base reactions. Notice that thecommon Brønsted–Lowry acids and bases also fall under the Lewis definition, with aproton serving as the electrophile. Curved arrows (red) are used to show the movementof electrons, generally from the nucleophile to the electrophile.

B � H�

nucleophile(Lewis base)

electrophile(Lewis acid)

B Hbond formed

H

N

H

H

nucleophile

B

F

F

F

electrophile

H

H

H

�N B

F

F

F�

bond formed

electrophile

O �CH3

nucleophile bond formed

CH3 C

H

H

H � Cl �O

H

C ClH

H

electrophileacid

NH3

nucleophilebase

bond formed(conjugated acid)

(conjugated base)

H3 H ��

O

C CH3OH N�

O

C CH3O�

PROBLEM-SOLVING HintA nucleophile donates electronsAn electrophile accepts electrons.Acidic protons may serve as electronacceptors.


PROBLEM-SOLVING HintUse one curved arrow for each pair ofelectrons participating in the reaction.

1-14 Lewis Acids and Bases 31

The curved-arrow formalism is used to show the flow of an electron pair fromthe electron donor to the electron acceptor. The movement of each pair of electrons in-volved in making or breaking bonds is indicated by its own separate arrow, as shown inthe preceding set of reactions. In this book, these curved arrows are always printed inred. In the preceding reaction of with one curved arrow shows the lonepair on oxygen forming a bond to carbon. Another curved arrow shows that the bonding pair detaches from carbon and becomes a lone pair on the product.

The curved-arrow formalism is universally used for keeping track of the flow ofelectrons in reactions. We have also used this device (in Section 1-9, for example) tokeep track of electrons in resonance structures as we imagined their “flow” in goingfrom one resonance structure to another. Remember that electrons do not “flow” inresonance structures: They are simply delocalized. Still, the curved-arrow formalismhelps our minds flow from one resonance structure to another. We will find ourselvesconstantly using these (red) curved arrows to keep track of electrons, both as reactantschange to products and as we imagine additional resonance structures of a hybrid.

PROBLEM 1-19 (PARTIALLY SOLVED)

In the following acid–base reactions,1. determine which species are acting as electrophiles (acids) and which are acting as nu-

cleophiles (bases).2. use the curved-arrow formalism to show the movement of electron pairs in these reac-

tions, as well as the imaginary movement in the resonance hybrids of the products.3. indicate which reactions are best termed Brønsted–Lowry acid–base reactions.

This reaction is a proton transfer from HCl to the group of acetaldehyde.Therefore, it is a Brønsted–Lowry acid–base reaction, with HCl acting as the acid (pro-ton donor) and acetaldehyde acting as the base (proton acceptor). Before drawing anycurved arrows, remember that arrows must show the movement of electrons from theelectron-pair donor (the base) to the electron-pair acceptor (the acid). An arrow must gofrom the electrons on acetaldehyde that form the bond to the hydrogen atom, and thebond to chlorine must break, with the chloride ion taking these electrons. Drawing thesearrows is easier once we draw good Lewis structures for all the reactants and products.

The resonance forms of the product show that a pair of electrons can be moved be-tween the oxygen atom and the pi bond. The positive charge is delocalized over thecarbon and oxygen atoms, with most of the positive charge on oxygen because all octets aresatisfied in that resonance structure.

(b)acetaldehyde

CH39C9H CH39O�

O�

�

O

CH39C9H

O9CH3

C “ O

CH3

O

C H

base

H Cl

acid

CH3

O

C H

major

�H H

CH3

O

C H�

minor

� Cl �

C “ O

acetaldehyde

CH39C9H HCl� Cl��

�(a) O

CH39C9H

O9H

CH3 O � CH

H

H

Cl CH3 O C

H

H

H � Cl �

nucleophile electrophile

Cl-

C ¬ ClCH3 Cl,CH3 O-

PROBLEM-SOLVING HintThe curved arrows we use inmechanisms show the flow ofelectrons and not the movementof atoms. We will use these curvedarrows constantly throughoutthis course.



Chapter 1 Glossary

Each chapter ends with a glossary that summarizes the most important new terms inthe chapter. These glossaries are more than just a dictionary to look up unfamiliarterms as you encounter them (the index serves that purpose). The glossary is one of thetools for reviewing the chapter. You can read carefully through the glossary to see ifyou understand and remember all the terms and associated chemistry mentioned there.Anything that seems unfamiliar should be reviewed by turning to the page numbergiven in the glossary listing.

acid-dissociation constant The equilibrium constant for the reaction of the acid withwater to generate (p. 23)

The negative logarithm of is expressed as

acids and bases (pp. 21–32)(Arrhenius definitions) acid: dissociates in water to give

base: dissociates in water to give -OHH3 O+

pKa = - log10 Ka

pKa:Ka

acid

conjugate acid–base pair

base

KaHA H2O H3O�� A� Ka [HA]

[H3O�][A�]

� �

H3 O+.(Ka)

In this case, no proton has been transferred, so this is not a Brønsted–Lowryacid–base reaction. Instead, a bond has formed between the carbon atom and theoxygen of the group. Drawing the Lewis structures helps to show that the

group (the nucleophile in this reaction) donates the electrons to form the newbond to acetaldehyde (the electrophile). This result agrees with our intuition that a negative-ly charged ion is likely to be electron-rich and therefore an electron donor.

Notice that acetaldehyde acts as the nucleophile (Lewis base) in part (a) and as theelectrophile (Lewis acid) in part (b). Like most organic compounds, acetaldehyde is bothacidic and basic. It acts as a base if we add a strong enough acid to make it donate electronsor accept a proton. It acts as an acid if the base we add is strong enough to donate an elec-tron pair or abstract a proton.

CH39NH2 � CH39Cl CH39NH29CH3 � Cl�(f )�

� H2O

O�O

CH39C9H � �OH

O

H9C9C9H(e)

H

�H9C"C9H

H

O

CH39C9H � �OH CH39C9H(d)

O�

OH

BH3 � CH39O9CH3 CH39O9CH3�

(c)

�BH3

CH3

O

C H

electrophile

O� CH3

nucleophile

CH3 C H

O �

O CH3

CH3 ¬ O-

CH3 ¬ O-

C “ O


Chapter 1 Glossary 33

(Brønsted–Lowry definitions) acid: proton donorbase: proton acceptor

(Lewis definitions) acid: electron-pair acceptor (electrophile)base: electron-pair donor (nucleophile)

conjugate acid The acid that results from protonation of a base. (p. 23)conjugate base The base that results from loss of a proton from an acid. (p. 25)covalent bonding Bonding that occurs by the sharing of electrons in the region between twonuclei. (p. 7)

single bond A covalent bond that involves the sharing of one pair of electrons. (p. 8)double bond A covalent bond that involves the sharing of two pairs of electrons. (p. 8)triple bond A covalent bond that involves the sharing of three pairs of electrons. (p. 8)

curved-arrow formalism A method of drawing curved arrows to keep track of electron move-ment from nucleophile to electrophile (or within a molecule) during the course of a reaction. (p. 31)degenerate orbitals Orbitals with identical energies. (p. 4)dipole moment A measure of the polarity of a bond (or a molecule), proportional to theproduct of the charge separation times the bond length. (p. 10)electron density The relative probability of finding an electron in a certain region of space. (p. 3)electronegativity A measure of an element’s ability to attract electrons. Elements with higherelectronegativities attract electrons more strongly. (p. 10)electrophile An electron-pair acceptor (Lewis acid). (p. 29)electrostatic potential map (EPM) A computer-calculated molecular representation that usescolors to show the charge distribution in a molecule. In most cases, the EPM uses red to showelectron-rich regions (most negative electrostatic potential) and blue to show electron-poorregions (most positive electrostatic potential). The intermediate colors orange, yellow, andgreen show regions with intermediate electrostatic potentials. (p. 10)empirical formula The ratios of atoms in a compound. (p. 20) See also molecular formula.formal charges A method for keeping track of charges, showing what charge would be on anatom in a particular Lewis structure. (p. 11)Hund’s rule When there are two or more unfilled orbitals of the same energy (degenerateorbitals), the lowest-energy configuration places the electrons in different orbitals (with parallelspins) rather than paired in the same orbital. (p. 6)ionic bonding Bonding that occurs by the attraction of oppositely charged ions. Ionic bondingusually results in the formation of a large, three-dimensional crystal lattice. (p. 7)isotopes Atoms with the same number of protons but different numbers of neutrons; atoms ofthe same element but with different atomic masses. (p. 3)Lewis acid, Lewis base See acids and bases.Lewis structure A structural formula that shows all valence electrons, with the bonds symbol-ized by dashes or by pairs of dots, and nonbonding electrons symbolized by dots. (p. 7)line–angle formula (skeletal structure, stick figure) A shorthand structural formula withbonds represented by lines. Carbon atoms are implied wherever two lines meet or a line beginsor bends. Atoms other than C and H are drawn in, but hydrogen atoms are not shown unlessthey are on an atom that is drawn. Each carbon atom is assumed to have enough hydrogens togive it four bonds. (p. 19)

lone pair A pair of nonbonding electrons. (p. 7)molecular formula The number of atoms of each element in one molecule of a compound.The empirical formula simply gives the ratios of atoms of the different elements. For example,the molecular formula of glucose is Its empirical formula is Neither themolecular formula nor the empirical formula gives structural information. (p. 20)node A region in an orbital with zero electron density. (p. 4)nodal plane A flat (planar) region of space with zero electron density. (p. 4)nonbonding electrons Valence electrons that are not used for bonding. A pair of nonbondingelectrons is often called a lone pair. (p. 7)nucleophile An electron-pair donor (Lewis base). (p. 29)

CH2 O.C6 H12 O6.

CHC

C

H

C

H

CC

H

H

HOH

O

HH

H

H

Lewis structure of 2-cyclohexenol2-cyclohexenol

equivalent line–angle formula

1¬ 2

1m2



octet rule Atoms generally form bonding arrangements that give them filled shells of electrons(noble-gas configurations). For the second-row elements, this configuration has eight valenceelectrons. (p. 6)orbital An allowed energy state for an electron bound to a nucleus; the probability functionthat defines the distribution of electron density in space. The Pauli exclusion principle statesthat up to two electrons can occupy each orbital if their spins are paired. (p. 3)organic chemistry New definition: The chemistry of carbon compounds. Old definition: Thestudy of compounds derived from living organisms and their natural products. (p. 1)pH A measure of the acidity of a solution, defined as the negative logarithm (base 10) of the

concentration: (p. 22)polar covalent bond A covalent bond in which electrons are shared unequally. A bond withequal sharing of electrons is called a nonpolar covalent bond. (p. 9)resonance hybrid A molecule or ion for which two or more valid Lewis structures can bedrawn, differing only in the placement of the valence electrons. These Lewis structures are calledresonance forms or resonance structures. Individual resonance forms do not exist, but we canestimate their relative energies. The more important (lower-energy) structures are called majorcontributors, and the less important (higher-energy) structures are called minor contributors.When a charge is spread over two or more atoms by resonance, it is said to be delocalized andthe molecule is said to be resonance stabilized. (pp. 13–16)structural formulas A complete structural formula (such as a Lewis structure) shows all theatoms and bonds in the molecule. A condensed structural formula shows each central atomalong with the atoms bonded to it. A line–angle formula (sometimes called a skeletal structureor stick figure) assumes that there is a carbon atom wherever two lines meet or a line begins orends. See Section 1-10 for examples. (p. 17)valence The number of bonds an atom usually forms. (p. 9)valence electrons Those electrons that are in the outermost shell. (p. 6)Vitalism The belief that syntheses of organic compounds require the presence of a “vitalforce.” (p. 1)

Essential Problem-Solving Skills in Chapter I

1. Draw and interpret Lewis, condensed, and line–angle structural formulas. Show whichatoms bear formal charges.

2. Draw resonance forms, and use them to predict stabilities.

3. Calculate empirical and molecular formulas from elemental compositions.

4. Predict relative acidities and basicities based on structure, bonding, and resonance ofconjugate acid–base pairs.

5. Calculate, use, and interpret values of and

6. Identify nucleophiles (Lewis bases) and electrophiles (Lewis acids), and write equationsfor Lewis acid–base reactions using curved arrows to show the flow of electrons.

pKa.Ka

pH = - log10[H3 O+]H3 O+

Study ProblemsIt’s easy to fool yourself into thinking you understand organic chemistry when you actually do not. As you read through thisbook, all the facts and ideas may make sense, yet you have not learned to combine and use those facts and ideas. An exam-ination is a painful time to learn that you do not really understand the material.

The best way to learn organic chemistry is to use it. You will certainly need to read and reread all the material in the chap-ter, but this level of understanding is just the beginning. Problems are provided so you can work with the ideas, applying them tonew compounds and new reactions that you have never seen before. By working problems, you force yourself to use the materi-al and fill in the gaps in your understanding. You also increase your level of self-confidence and your ability to do well on exams.

Several kinds of problems are included in each chapter. There are problems within the chapters, providing examples anddrill for the material as it is covered. Work these problems as you read through the chapter to ensure your understanding as you goalong. Answers to many of these in-chapter problems are found at the back of this book. Study Problems at the end of each chap-ter give you additional experience using the material, and they force you to think in depth about the ideas. Problems with red stars(*) are more difficult problems that require extra thought and perhaps some extension of the material presented in the chapter.Some of the study problems have short answers in the back of this book, and all of them have detailed answers in theaccompanying Solutions Manual.


Study Problems 35

Taking organic chemistry without working the problems is like skydiving without a parachute. Initially there is a breezysense of freedom and daring. But then, there is the inevitable jolt that comes at the end for those who went unprepared.

1-20 Define and give an example for each term.(a) isotopes (b) orbital (c) node(d) degenerate orbitals (e) valence electrons (f) ionic bonding(g) covalent bonding (h) Lewis structure (i) nonbonding electrons(j) single bond (k) double bond (l) triple bond(m) polar bond (n) formal charges (o) resonance forms(p) molecular formula (q) empirical formula (r) Arrhenius acid and base(s) Brønsted–Lowry acid and base (t) Lewis acid and base (u) electrophile(v) nucleophile

1-21 Name the element that corresponds to each electronic configuration.(a) (b) (c) (d)

1-22 There is a small portion of the periodic table that you must know to do organic chemistry. Construct this part from memory,using the following steps.(a) From memory, make a list of the elements in the first two rows of the periodic table, together with their numbers of

valence electrons.(b) Use this list to construct the first two rows of the periodic table.(c) Organic compounds often contain sulfur, phosphorus, chlorine, bromine, and iodine. Add these elements to your

periodic table.1-23 For each compound, state whether its bonding is covalent, ionic, or a mixture of covalent and ionic.

(a) NaCl (b) NaOH (c) (d)(e) (f) (g)

1-24 (a) Both and are stable compounds. Draw Lewis structures for these two compounds.(b) is a known compound, but all attempts to synthesize have failed. Draw Lewis structures for and

a hypothetical and explain why is an unlikely structure.1-25 Draw a Lewis structure for each species.

(a) (b) (c) (d)(e) (f) (g) (h)(i) (j) (k)

1-26 Draw a Lewis structure for each compound. Include all nonbonding pairs of electrons.(a) (b)(c) (d)

1-27 Draw a line–angle formula for each compound in Problem 1-26.1-28 Draw Lewis structures for

(a) two compounds of formula (b) two compounds of formula (c) three compounds of formula (d) two compounds of formula

1-29 Draw a complete structural formula and a condensed structural formula for(a) three compounds of formula (b) five compounds of formula

1-30 Some of the following molecular formulas correspond to stable compounds. When possible, draw a stable structure foreach formula.

Can you propose a general rule for the numbers of hydrogen atoms in stable hydrocarbons?1-31 Draw complete Lewis structures, including lone pairs, for the following compounds.

(h)

O O

OHSO3H(g)CHO(f)(e)

O

N

g-aminobutyric acid(a neurotransmitter)

(d) NH2 COOH(c)

N

O

Hmorpholine

(b)NH

pyrrolidine

N

(a)

pyridine

CH2 CH3 CH4 CH5

C2 H2 C2 H3 C2 H4 C2 H5 C2 H6 C2 H7

C3 H3 C3 H4 C3 H5 C3 H6 C3 H7 C3 H8 C3 H9

C3 H6 OC3 H8 O

C2 H4 OC3 H8 O2

C2 H7 NC4 H10

CH3 CH1CH32CH2 C1CH2 CH322 CHOCH2 CHCH1OH2CH2 CO2 HNCCH2 COCH2 CHOCH3 CHCHCH2 CHCHCOOH

1CH323 CNOCH3 C1NH2CH3CH3 OSO2 OCH3

CH3 NCOH2 SO4CH3 S1O2CH3CH3 CHOCH3 CN1CH324 NClN2 H2N2 H4

NCl5NCl5,NCl3NCl5NCl3

PCl5PCl3

CF4HCO2 NaNaOCH3

CH2 Cl2CH3 Li

1s22s22p63s23p51s22s22p63s23p31s22s22p41s22s22p2



1-32 Give the molecular formula of each compound shown in Problem 1-31.1-33 Compound X, isolated from lanolin (sheep’s wool fat), has the pungent aroma of dirty sweatsocks. A careful analysis

showed that compound X contains 62.0% carbon and 10.4% hydrogen. No nitrogen or halogen was found.(a) Compute an empirical formula for compound X.(b) A molecular weight determination showed that compound X has a molecular weight of approximately 117. Find the

molecular formula of compound X.(c) There are many possible structures that have this molecular formula. Draw complete structural formulas for four of them.

1-34 For each of the following structures,1. Draw a Lewis structure; fill in any nonbonding electrons.2. Calculate the formal charge on each atom other than hydrogen. All are electrically neutral except as noted.

1-35 1. From what you remember of electronegativities, show the direction of the dipole moments of the following bonds.2. In each case, predict whether the dipole moment is relatively large or small.

(a) (b) (c) (d) (e)(f) (g) (h) (i) (j)

1-36 Determine whether the following pairs of structures are actually different compounds or simply resonance forms of thesame compounds.

1-37 Draw the important resonance forms to show the delocalization of charges in the following ions.

(a) (b) (c)

(d) (e) (f)

(g) (h) (i)

(j)1-38 (a) Draw the resonance forms for (bonded ).

(b) Draw the resonance forms for ozone (bonded ).(c) Sulfur dioxide has one more resonance form than ozone. Explain why this structure is not possible for ozone.

O ¬ O ¬ OO ¬ S ¬ OSO2

CH3 ¬ CH “ CH ¬ CH “ CH ¬ CH2 ¬ C+

H2

CH3 ¬ CH “ CH ¬ CH “ CH ¬ C+

H ¬ CH3O

�

O

�

NH�O�

�

CH2

�

H C

O

CH CH CH2

�

CH3 C

O

CH2

�

O�O�

( j) CH39C9CH"CH2 CH39C"CH9CH2and��

(i) CH2"CH9CH2 CH29CH"CH2and

(h) CH2"C"O and H9C#C9OH

O�

�

(g) H9C9NH2 H9C"NH2

O

and

�

O9H

(f ) H9C9H

�O9H

H9C9Hand(e) H9C9CH3 H9C"CH2

O O9H

and

(d) CH2"C9H

OO�

�CH29C9Handand(c)

(b)

C

andO�

O

CO

O�

(a)

O

and

O

C ¬ BrO ¬ HN ¬ HC ¬ MgC ¬ BC ¬ OC ¬ NC ¬ LiC ¬ HC ¬ Cl

(c) [CH2"CH9CH2]� (e) [(CH3)3O]�(d) CH3NO2

(trimethylamine oxide)

(CH3)3NO(b)(a) CH

HN N C

H

HN N


Study Problems 37

*1-39 The following compound can become protonated on any of the three nitrogen atoms. One of these nitrogens is muchmore basic than the others, however.(a) Draw the important resonance forms of the products of protonation on each of the three nitrogen atoms.(b) Determine which nitrogen atom is the most basic.

1-40 In the following sets of resonance forms, label the major and minor contributors and state which structures would be ofequal energy. Add any missing resonance forms.

(a)

(b)

(c)

(d)

(e)

1-41 For each pair of ions, determine which ion is more stable. Use resonance forms to explain your answers.

(a)

(b)

(c)

(d)

(e)

1-42 Rank the following species in order of increasing acidity. Explain your reasons for ordering them as you do.

1-43 Rank the following species in order of increasing basicity. Explain your reasons for ordering them as you do.

1-44 The of phenylacetic acid is and the of propionic acid is 4.87.

(a) Calculate the of phenylacetic acid and the of propionic acid.(b) Which of these is the stronger acid? Calculate how much stronger an acid it is.(c) Predict whether the following equilibrium will favor the reactants or the products.

CH2COO� CH3CH2COOH� CH2COOH CH3CH2COO��

KapKa

CH2 C OH CH3

O

CH2 C OH

O

phenylacetic acid, Ka � 5.2 � 10�5 propionic acid, pKa � 4.87

pKa5.2 * 10-5,Ka

CH3 O- H2 O CH3 COO- NaOH HF NH2- HSO4

-

NH3 H2 SO4 CH3 OH CH3 COOH

�CH39C9CH3

CH39N9CH3

�CH39C9CH3

CH39CH9CH3or

CH2

�

or

CH2

�

�CH29CH3 CH29C#Nor�

CH2"CH9CH9CH3 CH2"CH9CH29CH2or� �

CH39CH9CH3 or�

CH39CH9OCH3

�

�CH39CH29C9NH2 CH39CH29C"NH2

NH2 NH2�

CH39CH9CH"CH9NO2

� �CH39CH"CH9CH9NO2

�CH39C9CH9C9CH3

O�

CH39C"CH9C9CH3

O O O

O�

�

CH39C"CH9CH9CH3

O�

�CH39C9CH"CH9CH3

CH39CH9C#N�

CH39CH"C"N �

CH3 NH C

NH

NH2



1-45 Label the reactants in these acid–base reactions as Lewis acids (electrophiles) or Lewis bases (nucleophiles). Use curvedarrows to show the movement of electron pairs in the reactions.

(a)

(b)

(c)

(d)

(e)

(f)

(g)

(h)

(i)

1-46 Predict the products of the following acid–base reactions.

(a) (b)

(c) (d)

(e) (f)

(g)

*1-47 Methyllithium is often used as a base in organic reactions.(a) Predict the products of the following acid–base reaction.

(b) What is the conjugate acid of Would you expect to be a strong base or a weak base?*1-48 In 1934, Edward A. Doisy of Washington University extracted 3000 lb of hog ovaries to isolate a few milligrams of pure

estradiol, a potent female hormone. Doisy burned 5.00 mg of this precious sample in oxygen and found that 14.54 mg ofand 3.97 mg of were generated.

(a) Determine the empirical formula of estradiol.(b) The molecular weight of estradiol was later determined to be 272. Determine the molecular formula of estradiol.

H2 OCO2

CH3 LiCH3 Li?

CH3 CH2 ¬ OH + CH3 ¬ Li :

1CH3 Li2

�HCOOH CH3O�

H2O � NH3�HO9C9OH 2 �OH

O

(CH3)3NH�

� �OHC

O

OH � �OH

CH3COOH (CH3)3N�H2SO4 CH3COO��

BF39CH29CH2 CH2"CH2� BF39CH29CH29CH29CH2

��

CH2"CH2 BF39CH29CH2BF3��

CH3 C

O

CH3 � O H� CH3 C CH2 � O H

O �

H

(CH3)3CCl � AlCl3 (CH3)3C� � �AlCl4

CH3 C

O

CH3 � H2SO4 CH3 C

O

CH3

H

� HSO4�

�

Cl �CH39NH2 CH39CH29Cl CH39NH29CH2CH3� ��

NH3

�NH3

H9C9H�

O

H9C9H

O �

�

CH3

CH39O9CH3 ��

H

CH39O9H

CH3

CH39O

H

O9H

Cl ��CH3O CH39Cl CH39O9CH3 ��


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