Introduction to Organic and Biochemistry 8th Ed. (Bettelheim)

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Introduction to Organic and Biochemistry, Eighth EditionFrederick A. Bettelheim, William H. Brown, Mary K. Campbell, Shawn O. Farrell, Omar J. Torres

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The bark of the Pacific yew contains paclitaxel, a substance that has proven effective in treating certain types of ovarian and breast cancer (see Chemical Connections 1A).

Tom

and

Pat

Lee

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1.1 What Is Organic Chemistry?

Organic chemistry is the chemistry of the compounds of carbon. As you study Chapters 1–11 (organic chemistry) and 12–23 (biochemis-try), you will see that organic compounds are everywhere around us.

They are in our foods, flavors, and fragrances; in our medicines, toiletries, and cosmetics; in our plastics, films, fibers, and resins; in our paints, var-nishes, and glues; and, of course, in our bodies and the bodies of all other living organisms.

Perhaps the most remarkable feature of organic compounds is that they involve the chemistry of carbon and only a few other elements—chiefly, hydrogen, oxygen, and nitrogen. While the majority of organic compounds contain carbon and just these three elements, many also contain sulfur, a halogen (fluorine, chlorine, bromine, or iodine), and phosphorus.

As of the writing of this text, there are 118 known elements. Organic chemistry concentrates on carbon, just one of the 118. The chemistry of the other 117 elements comes under the field of inorganic chemistry. As we see in Figure 1.1, carbon is far from being among the most abundant elements

Key Questions

1.1 What Is Organic Chemistry?

1.2 Where Do We Obtain Organic Compounds?

1.3 How Do We Write Structural Formulas of Organic Compounds?

1.4 What Is a Functional Group?

Organic Chemistry 1

Sign in to OWL at www.cengage.com/owl to view tutorials and simulations, develop problem-solving skills, and complete online homework assigned by your professor.

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2 Chapter 1 Organic Chemistry

in the Earth’s crust. In terms of elemental abundance, approximately 75% of the Earth’s crust is composed of just two elements: oxygen and silicon. These two elements are the components of silicate minerals, clays, and sand. In fact, carbon is not even among the ten most abundant elements. Instead, it is merely one of the elements making up the remaining 0.9% of the Earth’s crust. Why, then, do we pay such special attention to just one element from among 117?

The first reason is largely historical. In the early days of chemistry, sci-entists thought organic compounds were those produced by living organ-isms and that inorganic compounds were those found in rocks and other nonliving matter. At that time, they believed that a “vital force,” possessed only by living organisms, was necessary to produce organic compounds. In other words, chemists believed that they could not synthesize any organic compound by starting with only inorganic compounds. This theory was very easy to disprove if, indeed, it was wrong. It required only one experiment in which an organic compound was made from inorganic compounds. In 1828, Friedrich Wöhler (1800–1882) carried out just such an experiment. He heated an aqueous solution of ammonium chloride and silver cyanate, both inorganic compounds and—to his surprise— obtained urea, an “organic” compound found in urine.

Silvercyanate

Silverchloride

Urea

Urea

AgNCO AgCl1 1CH2N NH2

O

NH4Cl heat

Ammoniumchloride

Although this single experiment of Wöhler’s was sufficient to disprove the “doctrine of vital force,” it took several years and a number of additional ex-periments for the entire scientific community to accept the fact that organic compounds could be synthesized in the laboratory. This discovery meant that the terms “organic” and “inorganic” no longer had their original mean-ings because, as Wöhler demonstrated, organic compounds could be obtained from inorganic materials. A few years later, August Kekulé (1829–1896) put forth a new definition—organic compounds are those containing carbon—and his definition has been accepted ever since.

A second reason for the study of carbon compounds as a separate disci-pline is the sheer number of organic compounds. Chemists have discovered or synthesized more than 10 million of them, and an estimated 10,000 new ones are reported each year. By comparison, chemists have discovered or synthesized an estimated 1.7 million inorganic compounds. Thus, approxi-mately 85% of all known compounds are organic compounds.

A third reason—and one particularly important for those of you going on to study biochemistry—is that biochemicals, including carbohydrates, lipids, proteins, enzymes, nucleic acids (DNA and RNA), hormones, vita-mins, and almost all other important chemicals in living systems are or-ganic compounds. Furthermore, their reactions are often strikingly similar to those occurring in test tubes. For this reason, knowledge of organic chem-istry is essential for an understanding of biochemistry.

One final point about organic compounds. They generally differ from inorganic compounds in many of their properties, some of which are shown in Table 1.1. Most of these differences stem from the fact that the bonding in organic compounds is almost entirely covalent, while most inorganic com-pounds have ionic bonds.

FIGURE 1.1 Abundance of the elements in the Earth’s crust.

Oxygen49.5%

Silicon25.7%

Aluminum7.4%

Iron 4.7%Calcium 3.4%

Sodium 2.6%Potassium 2.4% Hydrogen 0.9%

Magnesium 1.9%

Titanium 0.6%Others 0.9%




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1.2 Where Do We Obtain Organic Compounds? 3

Of course, the entries in Table 1.1 are generalizations, but they are largely true for the vast majority of compounds of both types.

1.2 Where Do We Obtain Organic Compounds?

Chemists obtain organic compounds in two principal ways: isolation from nature and synthesis in the laboratory.

A. Isolation from Nature

Living organisms are “chemical factories.” Each terrestrial, marine, and freshwater plant (flora) and animal (fauna)—even microorganisms such as bacteria—make thousands of organic compounds by a process called biosyn-thesis. One way, then, to get organic compounds is to extract, isolate, and purify them from biological sources. In this book, we will encounter many compounds that are or have been isolated in this way. Some important examples include vitamin E, the penicillins, table sugar, insulin, quinine, and the anticancer drug paclitaxel (Taxol, see Chemical Connections 1A). Nature also supplies us with three other important sources of organic compounds: natural gas, petroleum, and coal. We will discuss them in Section 2.4.

B. Synthesis in the Laboratory

Ever since Wöhler synthesized urea, organic chemists have sought to de-velop more ways to synthesize the same compounds or design derivatives of those found in nature. In recent years, the methods for doing so have become so sophisticated that there are few natural organic compounds, no matter how complicated, that chemists cannot synthesize in the laboratory.

Compounds made in the laboratory are identical in both chemical and physical properties to those found in nature—assuming, of course, that each is 100% pure. There is no way that anyone can tell whether a sample of any particular compound was made by chemists or obtained from nature. As a consequence, pure ethanol made by chemists has exactly the same physical and chemical properties as pure ethanol prepared by distilling wine. The same is true for ascorbic acid (vitamin C). There is no advantage, therefore,

Table 1.1 A Comparison of Properties of Organic and Inorganic Compounds

Organic Compounds Inorganic Compounds

Bonding is almost entirely covalent.

Most have ionic bonds.

Many are gases, liquids, or solids with low melting points (less than 360°C).

Most are solids with high melting points.

Most are insoluble in water. Many are soluble in water.Most are soluble in organic solvents such as diethyl ether, toluene, and dichloromethane.

Almost all are insoluble in organic solvents.

Aqueous solutions do not conduct electricity.

Aqueous solutions form ions that conduct electricity.

Almost all burn and decompose. Very few burn.Reactions are usually slow. Reactions are often very fast.

Organic and inorganic compounds differ in their properties because they differ in their structure and composition—not because they obey different natural laws. One set of natural laws applies to all compounds.




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in paying more money for vitamin C obtained from a natural source than for synthetic vitamin C, because the two are identical in every way.

Organic chemists, however, have not been satisfied with merely dupli-cating nature’s compounds. They have also designed and synthesized com-pounds not found in nature. In fact, the majority of the more than 10 million known organic compounds are purely synthetic and do not exist in living organisms. For example, many modern drugs—Valium, albuterol, Prozac,

Taxol: A Story of Search and Discovery

Chemical Connections 1A

Pacific yew bark being stripped for Taxol extraction

In the early 1960s, the National Cancer Institute under-took a program to analyze samples of native plant materi-als in the hope of discovering substances that would prove effective in the fight against cancer. Among the materials tested was an extract of the bark of the Pacific yew, Taxusbrevifolia, a slow-growing tree found in the old-growth for-ests of the Pacific Northwest. This biologically active ex-tract proved to be remarkably effective in treating certain types of ovarian and breast cancer, even in cases where other forms of chemotherapy failed. The structure of the cancer-fighting component of yew bark was determined in 1962, and the compound was named paclitaxel (Taxol).

NH

OH

O

O

O

OH

OO

O

H

OO

HO

O

O

O

Paclitaxel(Taxol)

Unfortunately, the bark of a single 100-year-old yew tree yields only about 1 g of Taxol, not enough for effective treatment of even one cancer patient. Furthermore, iso-lating Taxol means stripping the bark from trees, which kills them. In 1994, chemists succeeded in synthesizing Taxol in the laboratory, but the cost of the synthetic drug was far too high to be economical. Fortunately, an alter-native natural source of the drug was found. Researchers in France discovered that the needles of a related plant, Taxus baccata, contain a compound that can be converted in the laboratory to Taxol. Because the needles can be gathered without harming the plant, it is not necessary to kill trees to obtain the drug.

Taxol inhibits cell division by acting on microtubules, a key component of the scaffolding of cells. Before cell division can take place, the cell must disassemble these microtubule units, and Taxol prevents this disassembly. Because cancer cells divide faster than normal cells, the drug effectively controls their spread.

The remarkable success of Taxol in the treatment of breast and ovarian cancer has stimulated research efforts to isolate and/or synthesize other substances that will act upon the human body in the same way and that may be even more effective anticancer agents than Taxol.

Pete

K. Z

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Test your knowledge with Problems 1.39 and 1.40.




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1.3 How Do We Write Structural Formulas of Organic Compounds? 5

Zantac, Zoloft, Lasix, Viagra, and Enovid—are synthetic organic compounds not found in nature. Even the over-the-counter drugs aspirin and ibuprofen are synthetic organic compounds not found in nature.

1.3 How Do We Write Structural Formulas of Organic Compounds?

A structural formula shows all the atoms present in a molecule as well as the bonds that connect the atoms to each other. The structural formula for ethanol, whose molecular formula is C2H6O, for example, shows all nine at-oms and the eight bonds that connect them:

H9C9C9O9H

H

H

H

H

Ethanol

The Lewis model of bonding enables us to see how carbon forms four cova-lent bonds that may be various combinations of single, double, and triple bonds. Furthermore, the valence-shell electron-pair repulsion (VSEPR) model tells us that the most common bond angles about carbon atoms in covalent compounds are approximately 109.5°, 120°, and 180°, for tetrahedral, trigonal planar, and linear geometries, respectively.

Table 1.2 shows several covalent compounds containing carbon bonded to hydrogen, oxygen, nitrogen, and chlorine. From these examples, we see the following:

• Carbon normally forms four covalent bonds and has no unshared pairs of electrons.

• Nitrogen normally forms three covalent bonds and has one unshared pair of electrons.

• Oxygen normally forms two covalent bonds and has two unshared pairs of electrons.

• Hydrogen forms one covalent bond and has no unshared pairs of electrons.

• A halogen (fl uorine, chlorine, bromine, and iodine) normally forms one covalent bond and has three unshared pairs of electrons.

The vitamin C in an orange is identical to its synthetic tablet form.

Geo

rge

Sem

ple

Table 1.2 Single, Double, and Triple Bonds in Compounds of Carbon. Bond angles and geometries for carbon are predicted using the VSEPR model.

Hydrogen cyanide(bond angle 180°)

H9C#N

Formaldehyde(bond angles

120°)

C"O

H

H

H

H

H

H

H9C9C9H

Ethane(bond angles

109.5°)

H

H

H

H

H9C9C9Cl

Chloroethane(bond angles

109.5°)

H

H

H9C9O9H

Methanol(bond angles

109.5°)

H

H

H

H9C9N9H

Methylamine(bond angles

109.5°)

Acetylene(bond angles

180°)

H9C#C9H

Ethylene(bond angles

120°)

C"C

H

H

H

H

Methyleneimine(bond angles 120°)

C"N

H

H

H




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Following are structural formulas for acetic acid, CH3COOH, and ethyl-amine, CH3CH2NH2.

H

H O

H9C9C9O9H

Acetic acidH

H

H

H

H

H9C9C9N9H

Ethylamine

(a) Complete the Lewis structure for each molecule by adding unshared pairs of electrons so that each atom of carbon, oxygen, and nitrogen has a complete octet.

(b) Using the VSEPR model, predict all bond angles in each molecule.

Strategy and Solution(a) Each carbon atom must be surrounded by eight valence electrons to

have a complete octet. Each oxygen must have two bonds and two unshared pairs of electrons to have a complete octet. Each nitrogen must have three bonds and one unshared pair of electrons to have a complete octet.

(b) To predict bond angles about a carbon, nitrogen, or oxygen atom, count the number of regions of electron density (lone pairs and bond-ing pairs of electrons about it). If four regions of electron density sur-round the atom, the predicted bond angles are 109.5°. If three regions surround it, the predicted bond angles are 120°. If two regions surround it, the predicted bond angle is 180°.

CCH

H

H

O H

4 (109.5˚)

Acetic acid

3 (120˚)

O

a

CCH

H H

H H H

N H

4 (109.5˚)

Ethylamine

A

Problem 1.1The structural formulas for ethanol, CH3CH2OH, and propene, CH3CHwCH2, are:

H

H

H

H

H9C9C9O9H

EthanolH

H

H H

H9C9C"C9H

Propene

(a) Complete the Lewis structure for each molecule showing all valence electrons.

(b) Using the VSEPR model, predict all bond angles in each molecule.

Example 1.1 Writing Structural Formulas




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1.4 What Is a Functional Group? 7

1.4 What Is a Functional Group?As noted earlier in this chapter, more than 10 million organic compounds have been discovered and synthesized by organic chemists. It might seem an almost impossible task to learn the physical and chemical properties of so many com-pounds. Fortunately, the study of organic compounds is not as formidable a task as you might think. While organic compounds can undergo a wide variety of chemical reactions, only certain portions of their structures undergo chemi-cal transformations. We call the atoms or groups of atoms of an organic mol-ecule that undergo predictable chemical reactions a functional group. As we will see, the same functional group, in whatever organic molecule it occurs, undergoes the same types of chemical reactions. Therefore, we do not have to study the chemical reactions of even a fraction of the 10 million known organic compounds. Instead, we need to identify only a few characteristic functional groups and then study the chemical reactions that each undergoes.

Functional groups are also important because they are the units by which we divide organic compounds into families of compounds. For example, we group those compounds that contain an iOH (hydroxyl) group bonded to a tetrahedral carbon into a family called alcohols; compounds containing a iCOOH (carboxyl group) belong to a family called carboxylic acids. Table 1.3 introduces six of the most common functional groups. A complete list of all functional groups that we will study appears on the inside back cover of the text.

Functional group An atom or group of atoms within a molecule that shows a characteristic set of predictable physical and chemical behaviors

CH3CH2OHAlcohol

CH3CH2NH2Amine

CH3CHAldehyde

Family Example NameFunctional

Group

O

CH3CCH3Ketone

O

CH3COHCarboxylic acid

O

CH3COCH2CH3

Ethanol

Ethylamine

Acetaldehyde

Acetone

Acetic acid

Ethyl acetateEster

9OH9NH2

9C9H

O

9C9

O

9C9OH

O

9C9OR

O O

Table 1.3 Six Common Functional Groups

At this point, our concern is simply pattern recognition—that is, how to recognize and identify one of these six common functional groups when you see it and how to draw structural formulas of molecules containing it. We will have more to say about the physical and chemical properties of these and several other functional groups in Chapters 2–11.

Functional groups also serve as the basis for naming organic compounds. Ideally, each of the 10 million or more organic compounds must have a unique name different from the name of every other organic compound. We will show how these names are derived in Chapters 2–11 as we study indi-vidual functional groups in detail.

To summarize, functional groups:

• Are sites of predictable chemical behavior—a particular functional group, in whatever compound it is found, undergoes the same types of chemical reactions.




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• Determine in large measure the physical properties of a compound.• Serve as the units by which we classify organic compounds into

families.• Serve as a basis for naming organic compounds.

A. Alcohols

As previously mentioned, the functional group of an alcohol is an iOH (hydroxyl) group bonded to a tetrahedral carbon atom (a carbon having bonds to four atoms). In the general formula of an alcohol (shown below on the left), the symbol R— indicates either a hydrogen or another carbon group. The important point of the general structure is the iOH group bonded to a tetrahedral carbon atom.

OC HR

Functional group Structural formulaCondensed

structural formula

OC

H

HR

CH

HR

H

H CH3CH2OH

An alcohol(Ethanol)

R 5 H orcarbon group

a

Here we represent the alcohol as a condensed structural formula, CH3CH2OH. In a condensed structural formula, CH3 indicates a carbon bonded to three hydrogens, CH2 indicates a carbon bonded to two hydro-gens, and CH indicates a carbon bonded to one hydrogen. Unshared pairs of electrons are generally not shown in condensed structural formulas.

Alcohols are classified as primary (1°), secondary (2°), or tertiary (3°), depending on the number of carbon atoms bonded to the carbon bearing the iOH group.

H

H

CH39C9OH

A 1° alcoholCH3

H

CH39C9OH

A 2° alcoholCH3

CH3

CH39C9OH

A 3° alcohol

Alcohol A compound containing an i OH (hydroxyl) group bonded to a tetrahedral carbon atom

Hydroxyl group An iOH group bonded to a tetrahedral carbon atom

Draw Lewis structures and condensed structural formulas for the two alcohols with molecular formula C3H8O. Classify each as primary, secondary, or tertiary.

Strategy and SolutionBegin by drawing the three carbon atoms in a chain. The oxygen atom of the hydroxyl group may be bonded to the carbon chain at two different positions on the chain: either to an end carbon or to the middle carbon.

C9C9CCarbon chain

C9C9C9OH

OH

C9C9CThe two locations for the 9OH group

Finally, add seven more hydrogens, giving a total of eight as shown in the molecular formula. Show unshared electron pairs on the Lewis structures but not on the condensed structural formulas.

Example 1.2 Drawing Structural Formulas of Alcohols




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B. Amines

The functional group of an amine is an amino group—a nitrogen atom bonded to one, two, or three carbon atoms. In a primary (1°) amine, nitro-gen is bonded to two hydrogens and one carbon group. In a secondary (2°) amine, it is bonded to one hydrogen and two carbon groups. In a tertiary (3°) amine, it is bonded to three carbon groups. The second and third structural formulas can be written in a more abbreviated form by collecting the CH3 groups and writing them as 1CH3 22NH and 1CH3 23N, respectively. The latter are known as condensed structural formulas.

Methylamine(a 1° amine)

Dimethylamine(a 2° amine)

Trimethylamine(a 3° amine)

CH3NH2 CH3NH

CH3

(CH3)2NHor CH3NCH3

CH3

(CH3)3Nor

Amine An organic compound in which one, two, or three hydrogens of ammonia are replaced by carbon groups; RNH2, R2NH, or R3N

Amino group A iNH2, RNH2, R2NH, or R3N group

OC

H

H

C

H

H

CH

H

H

H CH3CH2CH2OHA primary alcohol

(1-Propanol)

Lewis structures Condensed structural formulas

Ball-and-stick models

a

HC

H

H

C

H

H

CH

H OH

H

CH3CHCH3

A secondary alcohol(2-Propanol)

COC

The secondary alcohol 2-propanol, whose common name is isopropyl alcohol, is the cooling, soothing component in rubbing alcohol.

Problem 1.2Draw Lewis structures and condensed structural formulas for the four alcohols with the molecular formula C4H10O. Classify each alcohol as pri-mary, secondary, or tertiary. (Hint: First consider the connectivity of the four carbon atoms; they can be bonded either four in a chain or three in a chain with the fourth carbon as a branch on the middle carbon. Then consider the points at which the iOH group can be bonded to each carbon chain.)

2-Propanol (isopropyl alcohol) is used to disinfect cuts and scrapes.

Cha

rles

D. W

inte

rs

Draw condensed structural formulas for the two primary amines with the molecular formula C3H9N.

Strategy and SolutionFor a primary amine, draw a nitrogen atom bonded to two hydrogens and one carbon.

Example 1.3 Drawing Structural Formulas of Amines




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C. Aldehydes and Ketones

Both aldehydes and ketones contain a CwO (carbonyl) group. The aldehyde functional group contains a carbonyl group bonded to a hydrogen. Formaldehyde, CH2O, the simplest aldehyde, has two hydrogens bonded to its carbonyl carbon. In a condensed structural formula, the aldehyde group may be written showing the carbon-oxygen double bond as CHwO or, alternatively, it may be written iCHO. The functional group of a ketone is a carbonyl group bonded to two carbon atoms. In the general structural formula of each functional group, we use the symbol R to represent other groups bonded to carbon to complete the tetravalence of carbon.

Carbonyl group A Cw O group

Aldehyde A compound containing a carbonyl group bonded to a hydrogen; a iCHO group

Ketone A compound containing a carbonyl group bonded to two carbon groups

C NH2C C

NH2

CC

NH2

C CH3CH2CH2NH2 CH3CHCH3

The three carbons may bebonded to nitrogen in two ways

Add seven hydrogens to give each carbon four bonds and give the correct molecular formula

Problem 1.3Draw structural formulas for the three secondary amines with the molecular for-mula C4H11N.

Functionalgroup

Acetaldehyde(an aldehyde)

Functionalgroup

Acetone(a ketone)

H

O

C H

O

C

O

C

O

CC CC CH3 CH3CH3

R

R

R

R

R

R

Draw condensed structural formulas for the two aldehydes with the molecular formula C4H8O.

Strategy and SolutionFirst draw the functional group of an aldehyde and then add the remain-ing carbons. These may be bonded in two ways. Then add seven hydro-gens to complete the tetravalence of each carbon.

orCH3CHCH

O

CH3

CH3CHCHO

CH3

Example 1.4 Drawing Structural Formulas of Aldehydes




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D. Carboxylic Acids

The functional group of a carboxylic acid is a iCOOH (carboxyl: carbonyl 1 hydroxyl) group. In a condensed structural formula, a carboxyl group may also be written iCO2H.

O

RCOH

O

CH3COHFunctional

groupAcetic acid

(a carboxylic acid)

Carboxyl group A iCOOH group

Carboxylic acid A compound containing a iCOOH group

CH3CH2CH2CHOCH3CH2CH2CH

O

or

Problem 1.4Draw condensed structural formulas for the three ketones with the molecular formula C5H10O.

Draw a condensed structural formula for the single carboxylic acid with the molecular formula C3H6O2.

Srategy and SolutionThe only way the carbon atoms can be written is three in a chain, and the iCOOH group must be on an end carbon of the chain.

CH3CH2CO2HCH3CH2COH or

O

Problem 1.5Draw condensed structural formulas for the two carboxylic acids with the molecular formula C4H8O2.

Example 1.5 Drawing Structural Formulas of Carboxylic Acids

E. Carboxylic Esters

A carboxylic ester, commonly referred to as simply an ester, is a deriv-ative of a carboxylic acid in which a carbon group replaces the hydrogen of the carboxyl group. The ester group is written iCOOR or iCO2R in this text.

Carboxylic ester A derivative of a carboxylic acid in which the H of the carboxyl group is replaced by a carbon group




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9C9O9C9 CH39C9O9CH3 CH3COOCH3or

O

Methyl acetate(an ester)

COC

Functional group

a

The molecular formula of methyl acetate is C3H6O2. Draw the structural formula of another ester with the same molecular formula.

Strategy and SolutionThere is only one other ester with this molecular formula. Its structural formula is:

H9C9O9CH29CH3

O

Ethyl formate

Problem 1.6Draw structural formulas for the four esters with the molecular formula C4H8O2.

Example 1.6 Drawing Structural Formulas of Esters

• Nitrogen normally forms three bonds and has one un-shared pair of electrons. Its bonds may be three single bonds, one single bond and one double bond, or one triple bond.

• Oxygen normally forms two bonds and has two un-shared pairs of electrons. Its bonds may be two single bonds or one double bond.

Section 1.4 What Is a Functional Group? • A functional group is a site of chemical reactivity; a

particular functional group, in whatever compound it is found, always undergoes the same types of chemical reactions.

• In addition, functional groups are the characteristic structural units by which we both classify and name organic compounds. Important functional groups in-clude the hydroxyl group of 1°, 2°, and 3° alcohols; the amino group of 1°, 2°, and 3° amines; the carbonyl group of aldehydes and ketones; the carboxyl group of carboxylic acids; and the ester group.

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Section 1.1 What Is Organic Chemistry? • Organic chemistry is the study of compounds containing

carbon.

Section 1.2 Where Do We Obtain Organic Compounds?• Chemists obtain organic compounds either by isolation

from plant and animal sources or by synthesis in the laboratory.

Section 1.3 How Do We Write Structural Formulas of Organic Compounds?• Carbon normally forms four bonds and has no unshared

pairs of electrons. Its four bonds may be four single bonds, two single bonds and one double bond, or one single bond and one triple bond.

Summary




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an atom of each element. (Hint: Use the Periodic Table.)

(a) Carbon (b) Oxygen (c) Nitrogen (d) Fluorine

1.16 What is the relationship between the number of electrons in the valence shell of each of the following atoms and the number of covalent bonds it forms?

(a) Carbon (b) Oxygen (c) Nitrogen (d) Hydrogen

1.17 Write Lewis structures for these compounds. Show all valence electrons. None of them contains a ring of atoms. (Hint: Remember that carbon has four bonds, nitrogen has three bonds and one unshared pair of electrons, oxygen has two bonds and two unshared pairs of electrons, and each halogen has one bond and three unshared pairs of electrons.)

(a) H2O2 (b) N2H4

Hydrogen peroxide Hydrazine (c) CH3OH (d) CH3SH Methanol Methanethiol (e) CH3NH2 (f) CH3Cl Methylamine Chloromethane

1.18 Write Lewis structures for these compounds. Show all valence electrons. None of them contains a ring of atoms.

(a) CH3OCH3 (b) C2H6

Dimethyl ether Ethane (c) C2H4 (d) C2H2

Ethylene Acetylene (e) CO2 (f) CH2O Carbon dioxide Formaldehyde (g) H2CO3 (h) CH3COOH Carbonic acid Acetic acid

1.19 Write Lewis structures for these ions. (a) HCO3

2 (b) CO3

22

Bicarbonate ion Carbonate ion (c) CH3COO2 (d) Cl2

Acetate ion Chloride ion

1.20 Why are the following molecular formulas impossible?

(a) CH5 (b) C2H7

Section 1.3 Review of the VSEPR Model1.21 Explain how to use the valence-shell electron-pair

repulsion (VSEPR) model to predict bond angles and geometry about atoms of carbon, oxygen, and nitrogen.

1.22 Suppose you forget to take into account the pres-ence of the unshared pair of electrons on nitrogen in the molecule NH3. What would you then predict for the HiNiH bond angles and the geometry (bond angles and shape) of ammonia?

Interactive versions of these problems may be assigned in OWL.

Orange-numbered problems are applied.

References to previous chapters are given in parentheses.

Problems

Section 1.1 What Is Organic Chemistry?1.7 Answer true or false. (a) All organic compounds contain one or more

atoms of carbon. (b) The majority of organic compounds are built

from carbon, hydrogen, oxygen, and nitrogen. (c) By number of atoms, carbon is the most abun-

dant element in the Earth’s crust. (d) Most organic compounds are soluble in water.

Section 1.2 Where Do We Obtain Organic Compounds?1.8 Answer true or false. (a) Organic compounds can only be synthesized in

living organisms. (b) Organic compounds synthesized in the labora-

tory have the same chemical and physical prop-erties as those synthesized in living organisms.

(c) Chemists have synthesized many organic com-pounds that are not found in nature.

1.9 Is there any difference between vanillin made syn-thetically and vanillin extracted from vanilla beans?

1.10 Suppose that you are told that only organic sub-stances are produced by living organisms. How would you rebut this assertion?

1.11 What important experiment did Wöhler carry out in 1828?

Section 1.3 How Do We Write Structural Formulas of Organic Compounds?1.12 Answer true or false. (a) In organic compounds, carbon normally has four

bonds and no unshared pairs of electrons. (b) When found in organic compounds, nitrogen nor-

mally has three bonds and one unshared pair of electrons.

(c) The most common bond angles about carbon in organic compounds are approximately 109.5° and 180°.

1.13 List the four principal elements that make up organic compounds and give the number of bonds each typically forms.

1.14 Think about the types of substances in your imme-diate environment and make a list of those that are organic—for example, textile fibers. We will ask you to return to this list later in the course and to refine, correct, and possibly expand it.

1.15 How many electrons are in the valence shell of each of the following atoms? Write a Lewis structure for

Problems 13




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( j) The molecular formula of the smallest carboxylic acid is C2H4O2.

1.27 What is meant by the term f unctional group?1.28 List three reasons why functional groups are

important in organic chemistry.1.29 Draw Lewis structures for each of the following

functional groups. Show all valence electrons in each functional group.

(a) A carbonyl group (b) A carboxyl group (c) A hydroxyl group (d) A primary amino group (e) An ester group1.30 Complete the following structural formulas by add-

ing enough hydrogens to complete the tetravalence of each carbon. Then write the molecular formula of each compound.

C

C

O

C9C"C9C9C(a)

C9C9C9C9OH(b)

O

C9C9C9C(c)

O

C9C9C9H(d)

C9C9C9C9NH2(e)

C

C

O

C9C9C9OH(f )

NH2

OH

OH

C9C9C9C9C(g)

C9C9C9C9OH(h)

O

C"C9C9OH(i)

1.31 What is the meaning of the term tertiary (3°) when it is used to classify alcohols?

1.32 Draw a structural formula for the one tertiary (3°) alcohol with the molecular formula C4H10O.

1.33 What is the meaning of the term tertiary (3°) when it is used to classify amines?

1.23 Suppose you forget to take into account the presence of the two unshared pairs of electrons on the oxygen atom of ethanol, CH3CH2OH. What would you then predict for the CiOiH bond angle and the geom-etry of ethanol?

1.24 Use the VSEPR model to predict the bond angles and geometry about each highlighted atom. (Hint: Remember to take into account the presence of unshared pairs of electrons.)

(c) CCH

H

H

C H

(a) CCH

H

H

H

H

O H

(b) CCH

H H

Cl

1.25 Use the VSEPR model to predict the bond angles about each highlighted atom.

H9C9O9H(a)

O

H

H9C9N9H(b)

H

H

H9O9N"O(c)

Section 1.4 What Is a Functional Group?1.26 Answer true or false. (a) A functional group is a group of atoms in an or-

ganic molecule that undergoes a predictable set of chemical reactions.

(b) The functional group of an alcohol, an aldehyde, and a ketone have in common the fact that each contains a single oxygen atom.

(c) A primary alcohol has one iOH group, a second-ary alcohol has two iOH groups, and a tertiary alcohol has three iOH groups.

(d) There are two alcohols with the molecular formula C3H8O.

(e) There are three amines with the molecular formula C3H9N.

(f) Aldehydes, ketones, carboxylic acids, and esters all contain a carbonyl group.

(g) A compound with the molecular formula of C3H6O may be either an aldehyde, a ketone, or a carboxylic acid.

(h) Bond angles about the carbonyl carbon of an aldehyde, a ketone, a carboxylic acid, and an ester are all approximately 109.5°.

(i) The molecular formula of the smallest aldehyde is C3H6O, and that of the smallest ketone is also C3H6O.




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1.46 Draw structural formulas for the three tertiary (3°) amines with the molecular formula C5H13N.

1.47 Which of these covalent bonds are polar, and which are nonpolar? (Hint: Review Section 3.7B.)

(a) CiC (b) CwC (c) CiH (d) CiO (e) OiH (f) CiN (g) NiH (h) NiO1.48 Of the bonds in Problem 1.47, which is the most

polar? Which is the least polar?1.49 Using the symbol d1 to indicate a partial positive

charge and d2 to indicate a partial negative charge, indicate the polarity of the most polar bond (or bonds if two or more have the same polarity) in each of the following molecules.

CH3OH(a) CH3NH2(b)

HSCH2CH2NH2(c)

O

CH3CCH3(d)

O

HCH(e)

O

CH3COH(f )

Looking Ahead1.50 Following is a structural formula and a ball-and-

stick model of benzene, C6H6.

C C

CH H

H H

C

C

C

H

H

(a) Predict each HiCiC and CiCiC bond angle in benzene.

(b) Predict the shape of a benzene molecule.1.51 Following is a structural formula for naphthalene. It

was first obtained by heating coal to a high tempera-ture in the absence of air (oxygen). At one time, it was used in “mothballs.”

Naphthalene

H

H

H

H

CC C

CCC

CC

CC

H&

H

H H

&

& &

(a) Predict the shape of naphthalene. (b) Is a naphthalene molecule polar or nonpolar? 1.52 Identify the functional group(s) in each compound.

O

CH3CH2CCH3(a)2-Butanone

(a solvent forpaints and lacquers)

1.34 Draw condensed structural formulas for all com-pounds with the molecular formula C4H8O that con-tain a carbonyl group (there are two aldehydes and one ketone).

1.35 Draw structural formulas for each of the following: (a) The four primary (1°) alcohols with the molecu-

lar formula C5H12O. (b) The three secondary (2°) alcohols with the

molecular formula C5H12O. (c) The one tertiary (3°) alcohol with the molecular

formula C5H12O.1.36 Draw structural formulas for the six ketones with

the molecular formula C6H12O.1.37 Draw structural formulas for the eight carboxylic

acids with the molecular formula C6H12O2.1.38 Draw structural formulas for each of the following: (a) The four primary (1°) amines with the molecular

formula C4H11N. (b) The three secondary (2°) amines with the mo-

lecular formula C4H11N. (c) The one tertiary (3°) amine with the molecular

formula C4H11N.

Chemical Connections1.39 (Chemical Connections 1A) How was Taxol

discovered?1.40 (Chemical Connections 1A) In what way does Taxol

interfere with cell division?

Additional Problems1.41 Use the VSEPR model to predict the bond angles

about each atom of carbon, nitrogen, and oxygen in these molecules. (Hint: First, add unshared pairs of electrons as necessary to complete the valence shell of each atom and then predict the bond angles.)

CH3CH2CH2OH(a)

O

CH3CH2CH(b)

CH3CH"CH2(c) CH3C#CCH3(d)

O

CH3COCH3(e)

CH3

CH3NCH3(f )

1.42 Silicon is immediately below carbon in Group 4A of the Periodic Table. Predict the CiSiiC bond angles in tetramethylsilane, 1CH3 24Si.

1.43 Phosphorus is immediately below nitrogen in Group 5A of the Periodic Table. Predict the CiPiC bond angles in trimethylphosphine, 1CH3 23P.

1.44 Draw the structure for a compound with the molecu-lar formula:

(a) C2H6O that is an alcohol (b) C3H6O that is an aldehyde (c) C3H6O that is a ketone (d) C3H6O2 that is a carboxylic acid1.45 Draw structural formulas for the eight aldehydes

with the molecular formula C6H12O.

Problems 15




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Tying It Together1.56 Following is the structural formula of acetylsalicylic

acid, better known by its common name aspirin.

Acetylsalicylic acid(Aspirin)

OH H

H

H

C C H

O

H

H

CC C OH

CCC

C

O

(a) Name the two oxygen-containing functional groups in aspirin.

(b) What is the molecular formula of aspirin? 1.57 Aspirin is prepared by the reaction of salicylic

acid with acetic anhydride as shown in the follow-ing equation. The stoichiometry of the reaction is given in the equation. Acetic acid is a by-product of the reaction and must be separated and removed so that aspirin can then be sold as a pure product. How many grams of aspirin can be prepared from 120 grams of salicylic acid? Assume that there is an excess of acetic anhydride.

Salicylic acidH

H O

H

H

CC C OH

CCC

C

OH Acetic anhydride

CH3 CH3C O C

OO

1

Acetylsalicylic acid(Aspirin)

OH H

H

H

C C H

O

H

H

CC C OH

CCC

C

O Acetic acid

CH3 C OH

O

1

1.58 Following is the structural formula of acetamide, a molecule that contains a functional group called an amide. In an amide, the iOH of a carboxyl group is replaced by an amino group.

Acetamide(an amide)

H H

H

H C C N H

O

O O

HOCCH2CH2CH2CH2COH(b)Hexanedioic acid

(the second componentof nylon-66)

O

H2NCH2CH2CH2CH2CHCOH(c)

Lysine(one of the 20 amino acid

building blocks of proteins)

NH2

O

HOCH2CCH2OH(d)Dihydroxyacetone

(a component of severalartificial tanning lotions)

1.53 Consider molecules with the molecular formula C4H8O2. Write the structural formula for a molecule with this molecular formula that contains:

(a) A carboxyl group (b) An ester group (c) A ketone group and a 2° alcohol (d) An aldehyde and a 3° alcohol1.54 Urea, (NH2)2CO, is used in plastics and in fertilizers.

It is also the primary nitrogen-containing substance excreted by humans.

UreaH H

O

NNH HC

(a) Complete the Lewis structure of urea, showing all valence electrons.

(b) Predict the bond angle about each C and N. (c) Which is the most polar bond in the molecule? (d) Is urea polar or nonpolar?1.55 The compound drawn here is lactic acid, a natural

compound found in sour milk.

Lactic acid

H O

H H

H C

H

C C O H

O

(a) What is the molecular formula of lactic acid? (b) Name the two functional groups in lactic acid. (c) Predict the bond angles about each carbon atom. (d) Which bonds are polar, and which are nonpolar? (e) Would you predict that lactic acid is polar or

nonpolar?




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electrons and that the oxygen bears a negative charge. Compare the Lewis structure of this oxy-gen with the oxygen atom in the hydroxide ion.

(c) Notice that the nitrogen atom of structure (b) has four bonds and bears a positive charge. Compare the Lewis structure and bonding of this nitrogen with the nitrogen in the ammonium ion, NH4

1. (d) If the acetamide hybrid is best represented by

contributing structure (a), predict the HiNiH bond angle.

(e) If, on the other hand, the acetamide hybrid is best represented by contributing structure (b), predict the HiNiH bond angle.

(f) Proteins are molecules that can be described as polyamides (Chapter 14). Linus Pauling, in his pioneering studies on the structure of proteins, discovered that the actual HiNiH bond angle in each amide bond of a protein is 120°. What does this fact tell you about the relative impor-tance of contributing structures (a) and (b) in the resonance hybrid?

(a) Complete the Lewis structure for acetamide, showing all valence electrons.

(b) Use the valence-shell electron-pair repul-sion (VSEPR) model to predict all bond angles in acetamide.

(c) Which is the most polar bond in acetamide?1.59 The amide group, in acetamide as well as in all

other amides, is best represented as a resonance hybrid. Following are two contributing structures for the hybrid.

H H

H

H C C N H

O

H H

H

H C C N H

O1

2

(a) (b)

(a) Show by the use of curved arrows how contribut-ing structure (a) is converted into contributing structure (b).

(b) Notice that structure (b) contains an oxygen atom with one bond and three unshared pairs of

Problems 17




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A8

Chapter 1 Organic Chemistry

1.1 Following are Lewis structures showing all valence electrons and all bond angles.

CH C O H

H

H

H

H109.5˚ 109.5˚

(a)

a

CH C C H

H

H

H H

109.5˚ 120˚(b)

1.2 Of the four alcohols with the molecular formula C4H10O, two are 1°, one is 2°, and one is 3°. For the Lewis structures of the 3° alcohol and one of the 1° alcohols, some CiCH3 bonds are drawn longer to avoid crowding in the formulas.

H H

H9C9C9C9C9O9H

HH

H H

HH

CH3CH2CH2CH2OHPrimary (1°)

H H

H9C9C9C9C9H

HH

O

H

H OH

HH

CH3CH2CHCH3

Secondary (2°)

H

H9C9C9C9O9H

H

HH

HH HC

H

CH3CHCH2OHPrimary (1°)

CH3

H

H9C9C9OH

H

H

H

H HC

H HC

CH3COH

CH3

Tertiary (3°)

CH3

1.3 The three secondary (2°) amines with the molecular formula C4H11N are:

CH3CHNHCH3CH3CH2CH2NHCH3 CH3CH2NHCH2CH3

CH3

1.4 The three ketones with the molecular formula C5H10O are:

CH3CH2CCH2CH3CH3CH2CH2CCH3 CH3CCHCH3

O O O

CH3

1.5 The two carboxylic acids with the molecular formula C4H8O2 are:

CH3CH2CH2COH

O

CH3CHCOH

O

CH3

and'

&

'

1.6 The four esters with the molecular formula C4H8O2 are:

HCOCH2CH2CH3

(1)

O

HCOCHCH3

(2)

O

CH3

CH3COCH2CH3

(3)

O

CH3CH2COCH3

(4)

O

1.7 (a) T (b) T (c) F (d) F(c) False: Carbon isn’t even close. Silicon and oxygen are the two most abundant elements in the Earth’s crust.(d) False: Most organic compounds are insoluble in water.1.9 Assuming that each is pure, there are no differences in their chemical or physical properties.1.11 Wöhler heated ammonium chloride and silver cyanate, both inorganic compounds, and obtained urea, an organic compound1.13 The four most principal elements that make up organic compounds and the number of bonds each typically forms are: H forms one bond.C forms four bonds.O forms two bonds.N forms three bonds.1.15 Following are Lewis dot structures for each element:

DCD(a) B DOD(b) DND F(c) (d)

(4) (5)(6) (7)

Answers

09761_ans_ptg01_A08-A57.indd 809761_ans_ptg01_A08-A57.indd 8 06/10/11 6:14 PM06/10/11 6:14 PM



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A9 Answers

(b) The three secondary (2°) alcohols with the molecular formula C5H12O are:

CH3CHCHCH3

OHOH OH

CH3

CH3CH2CHCH2CH3CH3CHCH2CH2CH3

(c) The one tertiary (3°) alcohol with the molecular formula C5H12O is:

CH3CH2C OH

CH3

CH3

1.37 The eight carboxylic acids with the molecular formula C6H12O2 are:

a five-carbon a four-carbon chain with a chain with a six- one-carbon two carbons as carbon chain branch branches

CH3

CH3CHCH2CH2CO2H

CH3

CH3CH2CHCH2CO2H

CH3

CH3CH2CH2CHCO2H

CH3

CH3CHCHCO2H

CH3

CH3CH2CHCO2H

CH3

CH2CH3

CH3CH2CCO2H

CH3

CH3

CH3CCH2CO2H

CH3

CH3CH2CH2CH2CH2CO2H

1.39 Taxol was discovered during a survey of indigenous plants for those containing phytochemicals that exhibited anti-tumor activity. It was sponsored by the National Cancer Institute with the goal of discovering new chemicals for fighting cancer.1.41 The arrows point to atoms and show bond angles about each atom.

CH3!CH2!CH2!OH(a)

109.5°

CH3!CH2!C!H(b)

O109.5° 120°

CH3!CH"CH2(c)

109.5° 120°

CH3!C#C!CH3(d)

109.5° 180°

CH3!C!O!CH3(e)

O109.5°109.5° 120°

CH3!N!CH3(f)

CH3

109.5°

1.43 Predict 109.5° for CiPiC bond angles.

CH3!P!CH3

CH3

109.5°

1.17 H O O(a)

Hydrogen peroxide

H H N N(b)

Hydrazine

H

HH

H C O(c)

Methanol

H

H

H

H C S(d)

Methanethiol

H

H

H

H C N(e)

Methylamine

H

H

H

H

H C Cl(f)

Chloromethane

H

H

1.19 Following is a Lewis structure for each ion.

H O(a) C O

O

O(b) C O

O

(c) CCH3 O

O

(d) Cl1.21 To use the VSEPR model to predict bond angles and the geometry about atoms of carbon, nitrogen, and oxygen: (1) Write the Lewis structure for the target molecule showing all valence electrons. (2) Determine the number of regions of electron density around an atom of C, O, or N. (3) If you find four regions of electron density, predict bond angles of 109.5°. If you find three regions, predict bond angles of 120°. If you find two regions, predict bond angles of 180°.1.23 You would find two regions of electron density around oxygen and, therefore, predict 180° for the CiOiH bond angle. The actual bond angle is approximately 109.5°.1.25 (a) 120° about C and 109.5° about O(b) 109.5° about N (c) 120° about N1.27 A functional group is a group of atoms that undergoes a predictable set of chemical reactions.

1.29

O

9C9O9H(a) (b) 9O9H(c)

O

9C9

9N9H(d)

H

O

9C9O9(e)

1.31 When applied to alcohols, tertiary (3°) means that the carbon bearing the –OH group is bonded to three other carbon atoms.1.33 When applied to amines, tertiary (3°) means that the amine nitrogen is bonded to three other carbon groups.1.35 (a) The four primary (1°) alcohols with the molecular formula C5H12O are:

CH3CCH2OH

CH3

CH3

CH3CH2CHCH2OH

CH3

CH3CHCH2CH2OH

CH3CH2CH2CH2CH2OH

CH3




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Answers A10

1.45 The eight aldehydes with the molecular formula C6H12O are below. The aldehyde functional group is written CHO.

a five-carbon a four-carbon chain with a chain with a six- one-carbon two carbons as carbon chain branch branches

CH3

CH3CHCH2CH2CHO

CH3

CH3CH2CHCH2CHO

CH3

CH3CH2CH2CHCHO

CH3

CH3CHCHCHO

CH3

CH3CH2CHCHO

CH3

CH2CH3

CH3CH2CCHO

CH3

CH3

CH3CCH2CHO

CH3

CH3CH2CH2CH2CH2CHO

1.47 (a) nonpolar covalent (b) nonpolar covalent (c) nonpolar covalent (d) polar covalent (e) polar covalent(f) polar covalent (g) polar covalent (h) polar covalent1.49 Under each formula is given the difference in electronegativity between the atoms of the most polar bond.

H

H

d2 d1

H!C!O!H

O!H (3.5–2.1 5 1.4)

(a)

H

H H

d2 d1

d1

H!C!N!H(b)

N!H (3.0–2.1 5 0.9)

H H

H H H

d2 d1

d1

H!S!C!C!N!H(c)

N!H (3.0–2.1 5 0.9)

OH H

H H

d2

d1

H!C!C!C!H(d)

C"O (3.5–2.5 5 1.0)

H

Hd2d1C"O(e)

C"O (3.5–2.5 5 1.0)

Od2 d1

(f)

O!H (3.5–2.1 5 1.4)

H

H

H!C!C!O!H

1.51 (a) All bond angles are approximately 120°, and the molecule is planar.(b) Naphthalene is nonpolar.1.53 The following all have the molecular formula C4H8O2.

CH3CH2CH2C9 OH(a) Two carboxylic acids:

O

CH3CHC9 OH

CH3

O

CH3CH2C9 OCH3(b) Two esters:

O

CH3C9 OCH2CH3

O

CH3CHCCH3

OH

O

(c) One ketone 1 2° alcohol:

CH3C9 C9 H

CH3

OHO

(d) One aldehyde 1 3° alcohol:

1.55 (a) The molecular formula of lactic acid is C3H6O3.(b) The two functional groups are a 2° hydroxyl group and a carboxyl group.(c) Predict bond angles of 120° about the carbonyl carbon and bond angles of 109.5° about the two other carbons and about the oxygen of each hydroxyl group.(d) The CwO, OiH, and OiH bonds are polar covalent.(e) Predict that lactic acid is a polar molecule.1.57 Convert grams of salicylic acid (138 g/mol) to moles of salicylic acid. From the balanced equation, see that one mole of salicylic acid gives one mole of aspirin (180 g/mol). Finally, convert moles of aspirin to grams of aspirin. Doing this math gives 157 grams of aspirin.

120 3180138

5 157 grams of aspirin

1.59 (a) Curved arrows show the repositioning of two pairs of electrons.

(a)

OH

H H

H!C!C!N!H

(b)

OHH

H H

H!C!C" N! H1

2

(b) Hydroxide ion shows 6 1 1 1 1 5 8 valence electrons. In both the hydroxide ion and the oxygen atom of contributing structure B, the oxygen in question has a complete octet. That is, each has a single bond and three nonbonding electron pairs and bears a negative charge.

H!O2

(c) The ammonium ion contains 5 1 4 2 1 5 8 valence electrons. In both the ammonium ion and one nitrogen atom of contributing structure B, the nitrogen atom has a complete octet; it has four bonds and bears a positive charge.

HH

H

H!N!H1

(d) If the hybrid is best represented by contributing structure A, then predict the HiNiH bond angles to be approximately 109.5°. (e) If the hybrid is best represented by contributing structure B, then predict the HiNiH bond angles to be 120°.(f) The discovery that the actual HiNiH bond angles about each amide nitrogen atom in a protein are 120° suggests that contributing structure B makes a greater contribution to the hybrid than contributing structure A.




Introduction to Organic and Biochemistry 8th Ed. (Bettelheim)

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