Jones M., Fleming S.A. Organic Chemistry

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2.14 Nuclear Magnetic Resonance Spectroscopy 89

about ethane? There are two carbons here, but they are in identical environments and

thus equivalent: one signal again. How about propane? Three carbons are here, two

methyl groups and a single methylene.Surely,we must observe a different signal for that

than for the CH

groups. And we do—there are two signals in the

C NMR

spectrum for propane,one for the CH

and another for the two equivalent CH

groups.

Counting the number of signals is trivial if (and only if) we can discern whether

the molecular environments of the carbons are the same or not. Here is a slightly

harder example.The molecule tetrahydrofuran produces two

C signals (Fig. 2.56).

It is tempting to say that because there are only methylene (CH

) groups in this

molecule,there should only be one signal.But those methylenes are not all the same.

Two of them are adjacent to the oxygen, and two are not. So there are two signals,

one for each set of two different methylenes.

We will pick these points up again as we discuss the various structural types in

the next few chapters, but for the moment, work through the following Problem

Solving box and then try a few problems on your own.

One signal

Another signal

FIGURE 2.56 Tetrahydrofuran has

two different methylene groups.

PROBLEM SOLVING

Nearly everyone has initial problems with seeing differences in the atoms making

up any molecule. Why, for example, are there three different methylene (CH

)

groups in heptane? Maybe it is intuitively obvious (maybe!) that the central

methylene group is different from the others, but almost everyone worries about

the other methylene groups.

The best technique to use when you are in doubt is to list all the groups to

which the carbons in question are attached. And be very detailed in making that

list—do not just look at nearest neighbors. Here is an example using heptane:

1234567

Carbon 1 methyl: Attached to C CCCCC,same as carbon 7

methyl

Carbon 2 methylene: Attached to C on one side and to C CCCC

on the other, same as carbon 6 methylene

Carbon 3 methylene: Attached to C C and to C C C C, same as

carbon 5 methylene

Carbon 4 methylene: Attached to C C C and to C C C

Carbon 5 methylene: Attached to C C C C and to C C, same as

carbon 3 methylene

Carbon 6 methylene: Attached to C CCCC and to C, same as

carbon 2 methylene

Carbon 7 methyl: Attached to C CCCCC,same as carbon 1

methyl

Therefore, heptane will have four signals in the

C NMR. One signal for the

equivalent methyl groups (carbons 1 and 7), one signal for the methylenes of

carbons 2 and 6, one signal for the methylenes of carbons 3 and 5, and one signal

for the carbon 4 methylene.

OOOOO

OOOO

OOOOO

PROBLEM 2.34 How many

C signals will we see in an NMR spectrum of the

molecules in Figures 2.45–2.47?

90 CHAPTER 2 Alkanes

PROBLEM 2.35 How many

C signals will an NMR spectrometer detect in the

molecules of Figure 2.49?

PROBLEM 2.36 How many

C signals will an NMR spectrometer detect in the

molecules of Figure 2.52?

Hydrogen NMR (

H NMR) spectroscopy is similar to

C spectroscopy.The areas

of the signals we observe in

H NMR spectroscopy are in the ratio of the numbers of

different hydrogens giving rise to those signals. For example, the signal recorded for

six hydrogens of one kind will give a signal three times as big as a signal for two hydro-

gens and so on.These ratios from the signals are more reliably determined in

H NMR

than they are in

C NMR. There are other very useful complications introduced by

the abundance of the NMR active isotope (

H) in organic molecules,but we can leave

them for Chapter 15. We will use some of the molecules introduced in this chapter

to work through a bit of hydrogen NMR. We’ll find a few more subtleties, but the

overall picture will not be very different from what we have seen for

PROBLEM 2.37 What will be the ratios of the signals in the

H NMR spectra for

propane (Fig. 2.55), and THF (Fig. 2.56)?

PROBLEM 2.38 How many signals would be seen in the

H NMR spectrum for

heptane (Fig. 2.44)? What will the relative sizes of these signals be?

PROBLEM 2.39 How many signals would be seen in the

H NMR spectrum for

cis-1,2-dimethylcyclopropane (Fig. 2.51)? Caution: This question is a bit tricky—

look carefully at a model, or use the website to see cis-1,2-dimethylcyclopropane

in three dimensions.

2.15 Acids and Bases Revisited: More Chemical

Reactions

In Section 1.7 (p.41), we introduced acids and bases. Now we know quite a bit more

about structure and can return to the important subject of acids and bases in greater

depth. In particular, we know about carbocations and carbanions, which play an

important role in acid–base chemistry in organic chemistry. The Lewis definition

of acids and bases is far more inclusive than the Brønsted

definition, which focus-

es solely on proton donation (Brønsted acid) and acceptance (Brønsted base).The

archetypal Brønsted acid–base reaction is the reaction between KOH and HCl to

transfer a proton from HCl to HO



. This reaction is a competition between the

hydroxide and the chloride for a proton. In this case, the stronger base hydroxide

wins easily (Fig. 2.57).

These acids and bases were named after Johannes Nicolaus Brønsted (1879–1947).

HCl

HOH

–

FIGURE 2.57 The two Brønsted

bases, HO



and Cl



, compete for the

proton, H



WEB 3D

2.15 Acids and Bases Revisited: More Chemical Reactions 91

In the Lewis description of acids and bases, any reaction between an empty

orbital and a reactive pair of electrons in a filled orbital is an acid–base reaction.

Another way of putting this is to point out that the interaction between an empty

orbital (Lewis acid) and a filled orbital (Lewis base) is stabilizing (Fig. 2.58).

Consider the methyl cation (Fig. 2.59). That empty 2p orbital makes



powerful Lewis acid,very reactive toward all manner of Lewis bases with their elec-

tron pairs ready to react. Figure 2.59 gives three examples of such Lewis acid–Lewis

base reactions, each of which fits the orbital interaction diagram of Figure 2.58.

In Figure 2.60,the curved arrow formalism shows the electron flow as the new bond

is formed in this illustrated nucleophile–electrophile (Lewis base–Lewis acid) reaction.

WORKED PROBLEM 2.40 Write arrow formalisms and products for the following

reactions. Identify the participants as acids and bases. Sketch in the empty orbital

for each Lewis acid, and fill in the complete Lewis dot structures for the Lewis

bases. Remember that charge must be conserved in each of these reactions.

Energy

Empty orbital

(Lewis acid)

Filled orbital

(Lewis base)

Stabilization

FIGURE 2.58 The interaction of a

filled orbital (Lewis base) with an

empty orbital (Lewis acid) is

stabilizing.

–

ClH

OHH

FIGURE 2.59 Three Lewis bases

reacting with the methyl cation, a

Lewis acid.

ClH

–

FIGURE 2.60 The curved arrow

formalism for the formation of

methyl chloride from the methyl

cation (Lewis acid) and a chloride

ion (Lewis base).

(a) (CH

)





(b) (CH

)





*(d) H

BHO



(continued)

ANSWER (d) Hydroxide must be the Lewis base, but where is the empty orbital?

Where is the Lewis acid? Recall the structure of the sp

hybridized borane

(p. 56). Boron is neutral, but there is an empty 2p orbital on boron, and borane is

very definitely a Lewis acid. Reaction with the Lewis base hydroxide looks just

like the reaction of the methyl cation with hydroxide (Fig. 2.59) except that the

product is negatively charged.

92 CHAPTER 2 Alkanes

Lewis base–Lewis acid reactions, even the very simple ones, form the basis for

understanding much of organic chemistry. Essentially all polar reactions can be

looked at in exactly these terms—the ultimately stabilizing reaction of a filled orbital

with an empty orbital.Figure 2.58 really does describe much of the next 1000 pages

or so! But,of course, the details will change and there are surely devils in those details.

Those all-too-prolific details will be easier to manage if you keep Figure 2.58 in mind

at all times—search for the Lewis base and the Lewis acid in every reaction, and

you will be able to generalize.

2.16 Special Topic: Alkanes as Biomolecules

In truth, there really isn’t very much biochemistry of alkanes.The carbon–carbon

and carbon–hydrogen bonds of alkanes are just too strong to enter easily into

chemical reactions. These molecules are not totally inert, as we shall see shortly,

but saturated alkane chains serve more as frameworks in our bodies, holding less

strongly bonded atoms in the proper place for reactions to occur, than as reactive

species themselves. However, as we have seen already (p. 68), in the presence of

oxygen and activating energy (a spark or a lighted match, for example), combus-

tion occurs and heat is given off. We use that energy, that exothermicity of

reaction, to warm our homes and propel our automobiles. Gasoline is largely a

mixture of saturated hydrocarbons. Combustion is a very biological process. Life

depends on it.

In the movie Mad Max: Beyond Thunderdome, the trading village Bartertown

is powered by the methane produced by herds of pigs kept in the depths of

Undertown. How did those piggies produce that methane? Bacteria in the stom-

achs of both pigs and cattle are capable of producing methane from plant material,

and the belching and flatulence of those animals is a major source of planetary

methane. Contrary to conventional wisdom, it is mostly belching that produces

the methane.

Methane may also have been essential to the formation of life.The early atmos-

phere on Earth was relatively rich in methane and ammonia (as are the current

atmospheres of most of the outer planets). Given an energy source such as ultraviolet

–

radiation or lightning, methane and ammonia react to form hydrogen cyanide

(HCN), a molecule that polymerizes to give adenine, a common component of

ribonucleic acid (RNA) and other molecules of biological importance. In the pres-

ence of water and an energy source, methane and ammonia react to give amino acids,

the building blocks of proteins.

2.17 Summary 93

In this chapter, a bonding scheme for the alkanes is developed.

We continue, as in Chapter 1, to form bonds through the over-

lap of atomic orbitals. We describe a model in which the atomic

orbitals of carbon are combined to form new, hybrid atomic

orbitals. The four orbitals resulting from a combination of three

2p orbitals and one 2s orbital are called sp

hybrid atomic

orbitals, reflecting the 75% p character and 25% s character in

the hybrid. These hybrid orbitals solve the problems encoun-

tered in forming bonds between pure atomic orbitals.The sp

hybrids are asymmetric, and have a fat and a thin lobe. Overlap

between a hydrogen 1s orbital and the fat lobe provides a

stronger bond than that between hydrogen 1s and carbon 2p

orbitals. In addition, these hybrid orbitals are directed toward

the corners of a tetrahedron, which keeps the electrons in the

bonds as far apart as possible, thus minimizing destabilizing

interactions. Other hybridization schemes are sp

, in which the

central atom is bonded to three other atoms, and sp, in which

the central atom is bonded to two other atoms. Simple mol-

ecules in which a central carbon is hybridized sp

are planar,

with the three attached groups at the corners of an equilateral

triangle. Molecules with sp hybridized carbons are linear.

Even simple molecules have complicated structures.

Methane is a perfect tetrahedron, but one need only substitute a

methyl group for one hydrogen of methane for complexity to

arise. In ethane, for example, we must consider the consequences

of rotation about the carbon–carbon bond.The minimum ener-

gy conformation for ethane is the arrangement in which all car-

bon–hydrogen bonds are staggered. About 3 kcal/mol above this

energy minimum form is the eclipsed form, the transition state,

the high-energy point (not an energy minimum, but an energy

maximum) separating two staggered forms of ethane. The

3 kcal/mol constitutes the rotational barrier between the two stag-

gered forms.This barrier is small compared to the available ther-

mal energy at room temperature, and rotation in ethane is fast.

This chapter covers again the concept of Lewis acids and

bases. The familiar Brønsted bases compete for a proton, but

Lewis bases are far more versatile. A Lewis base is defined as

anything with a reactive pair of electrons, and a Lewis acid is

anything that reacts with a Lewis base. We are paid back for

these very general definitions with an ability to see as similar all

manner of seemingly different reactions. These concepts will

run through the entire book.

2.17 Summary

New Concepts

alkanes (p. 51)

alkyl compounds (p. 69)

Brønsted acid (p. 90)

Brønsted base (p. 90)

butyl group (p. 75)

sec-butyl group (p. 75)

tert-butyl group (p. 75)

carbanion (p. 62)

carbocation (p. 62)

cis (p. 84)

conformation (p. 65)

conformational analysis (p. 73)

conformational isomers (p. 70)

cycloalkanes (p. 51)

dihedral angle (p. 65)

eclipsed ethane (p. 65)

ethyl compounds (p. 69)

hybridization (p. 52)

hybrid orbitals (p. 53)

hydride (p. 62)

hydrocarbon (p. 51)

isobutyl group (p. 75)

isomers (p. 70)

isopropyl group (p. 73)

methane (p. 51)

methine group (p. 75)

methyl anion (p. 62)

methyl cation (p. 62)

methyl compounds (p. 60)

methylene group (p. 72)

methyl radical (p. 63)

natural product (p. 86)

Newman projection (p. 65)

NMR spectrum (p. 88)

nuclear magnetic resonance (NMR)

spectroscopy (p. 88)

primary carbon (p. 76)

propyl group (p. 73)

quaternary carbon (p. 76)

reactive intermediates (p. 62)

saturated hydrocarbons (p. 83)

secondary carbon (p. 76)

sigma bond (p. 54)

spectroscopy (p. 88)

sp hybrid (p. 53)

hybrid (p. 56)

staggered ethane (p. 65)

steric requirements (p. 78)

structural isomers (p. 73)

substituent (p. 60)

tertiary carbon (p. 76)

trans (p. 84)

transition state (TS) (p. 67)

unsaturated hydrocarbons (p. 83)

van der Waals forces (p. 87)

Key Terms

94 CHAPTER 2 Alkanes

In this chapter, new molecules are built up by first constructing

a generic substituted molecule such as ( ).

The new hydrocarbon is then generated by letting X  CH

.In

principle, each different carbon–hydrogen bond in a molecule

could yield a new hydrocarbon when X  CH

. In practice, this

procedure is not so simple. The problem lies in seeing which

hydrogens are really different. The two-dimensional drawings

are deceptive. One really must see the molecule in three dimen-

sions before the different carbon–hydrogen bonds can be identi-

fied with certainty.

The coding or drawing procedures can get quite abstract. It

is vital to be able to keep in mind the real, three-dimensional

structures of molecules even as you write them in two-dimen-

Xmethyl

sional code. The Newman projection, an enormously useful

device for representing molecules three-dimensionally, is

described in this chapter.

The naming convention for alkanes is introduced. There

are several common or trivial names that are often used and

which, therefore, must be learned.

Both

H and

C NMR spectroscopy are introduced

in this chapter. At this point, the NMR spectrometer

functions basically as a machine to determine the numbers

of different hydrogens or carbons in a molecule. That function

is important, however, both in these early stages of your

study of chemistry and in the “real world” of structure

determination.

Reactions, Mechanisms, and Tools

It is easy to become confused by the abstractions used to repre-

sent molecules on paper or blackboards. Do not trust the flat

surface! It is easy to be fooled by a “carbon” that does not really

exist, or to be bamboozled by the complexity introduced by ring

structures. One good way to minimize such problems (none of

us ever really becomes free of the necessity to think about struc-

tures presented on the flat surface) is to work lots of isomer

problems, and play with models!

Common Errors

PROBLEM 2.41 Provide the IUPAC name for the following

compounds:

2.18 Additional Problems

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

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

PROBLEM 2.42 Write Lewis structures for ethane, ethylene,

acetylene, ethanol, ethanethiol, tetraethylammonium ion,

diethylphosphine, the imine of diethyl ketone, diethylborane,

tetraethylborate ion, diethyl ether, diethyl sulfide, acetalde-

hyde, acetone, acetic acid, ethyl acetate, acetamide, acetyl

chloride, propanenitrile, ethyl fluoride, and ethyl chloride.

Show nonbonding electrons as dots and electrons in bonds

as lines.

PROBLEM 2.43 Draw all the isomers of “bromopentane,”

Br. Give systematic names to all of them.

PROBLEM 2.44 How many signals would a

C NMR spectrome-

ter “see” for each of the molecules in your answer to Problem 2.43?

PROBLEM 2.45 Draw all the isomers of “chlorohexane,”

Cl. Give proper systematic names to all of them. Hint:

There are 17 isomers.

2.18 Additional Problems 95

PROBLEM 2.46

Use your model set to look down the

bond of 2-chlorobutane. Draw the Newman

projection for all possible staggered conformations. Circle the

one you think is the most stable and explain why you’ve chosen

that one.

PROBLEM 2.47 Use your model set to look down the

bond of 2-bromo-3-methylbutane. Draw the

Newman projection for all possible staggered conformations.

Circle the one you think is the most stable and explain why

you’ve chosen that one. Determine the number of gauche

interactions in each projection.

PROBLEM 2.48 Draw the Newman projections for the differ-

ent eclipsed and staggered conformations of 2,3-dichlorobutane.

Label each projection as either eclipsed or staggered. In each

staggered projection, determine the number and type of gauche

interactions.

PROBLEM 2.49 Draw Newman projections of the eclipsed

and staggered conformations of 1,2-dichloropropane by looking

down the bond.C(1)

C(2)

C(3)

C(2)

C(3)

ClCH

(e)

(f)

(a)

(d)

HC C

(b)

C(CH

)

(c)

H H

PROBLEM 2.50 Draw Newman projections constructed by

looking down the bond of 2-methylpentane.

Repeat this process looking down the bond. In

each case, indicate which conformations will be the most stable.

PROBLEM 2.51 Although a chlorine is about the same size as

a methyl group, the conformation of 1,2-dichloroethane in

which the two chlorines are eclipsed is higher in energy than

the conformation of butane in which the two methyl groups are

eclipsed. Explain. Hint: Size isn’t everything!

PROBLEM 2.52 What is the approximate hybridization of the

indicated carbon in the following compounds?

C(2)

C(3)

C(1)

C(2)

PROBLEM 2.54 Use the boiling point data in Table 2.4 to

estimate the boiling point of pentadecane, C

PROBLEM 2.55 Rationalize the differences in the melting

points for the isomeric pentanes shown below.

(a) Xanturil is a diuretic

(b) Viquidil is a vasodilator and

antiarrhythmic found in cinchona bark

Pentane

mp –129.7 ⬚C

Isopentane

mp –159.9 ⬚C

Neopentan

mp –16.5 ⬚C

PROBLEM 2.53 Indicate the hybridization for each carbon,

nitrogen, and oxygen in the molecules shown below.

Put a circle around the sp

-hybridized atoms, a triangle

around sp

-hybridized atoms and a box around the

sp-hybridized atoms.

PROBLEM 2.56 Imagine replacing two hydrogens of a mol-

ecule with some group X. For example, methane (CH

) would

become CH

. (a) What compounds would be produced by

replacing two hydrogens of ethane? (b) What compounds

would be produced by replacing two hydrogens of propane?

hydrogens of butane?

PROBLEM 2.57 Write systematic names for all the pentanes

and hexanes.

PROBLEM 2.58 How many signals would a

C NMR spec-

trometer “see” for each of the molecules in your answer to

Problem 2.57?

96 CHAPTER 2 Alkanes

PROBLEM 2.59 Draw and write systematic names for all the

isomers of nonane (C

). (Hint: There is a rumor that

there are 35 isomers!) This problem is hard to get completely

right.

You may want to continue isomer drawing and naming on

your own. But be careful, this is no trivial task. For example,

there are 75 structural isomers of decane and over 6500 for

tetradecane.

PROBLEM 2.60 Write systematic names for the following

compounds:

PROBLEM 2.63 One of the octane isomers of Problem 2.28,

3-ethyl-2-methylpentane, could have been reasonably named

otherwise. Draw this compound and find the other reasonable

name. The answer will tell you why the name given is preferred.

PROBLEM 2.64 Name the following compound:

(a)

(b)

(d)

(c)

PROBLEM 2.61 Draw structures for the following compounds:

(a) 4-ethyl-4-fluoro-2-methylheptane

(b) 2,7-dibromo-4-isopropyloctane

(d) 2-chloro-7-iodo-5-isobutyldecane

PROBLEM 2.62 Although you should be able to draw struc-

tures for the following compounds, each is named incorrectly

below. Give the correct systematic name for each of the follow-

ing compounds:

(a) 2-methyl-2-ethylpropane

(b) 2,6-diethylheptane

(d) 5-fluoro-8-methyl-3-tert-butylnonane

PROBLEM 2.65 Show the curved arrow formalism (electron

pushing or arrow pushing) for the protonation of water in sul-

furic acid (HOSO

OH  H

PROBLEM 2.66 Show the curved arrow formalism (electron

pushing or arrow pushing) for the protonation of ammonia

(NH

) in hydrochloric acid.

PROBLEM 2.67 Show the curved arrow formalism for the pro-

tonation of an alcohol (ROH) in sulfuric acid.

Use Organic Reaction Animations (ORA) to answer the

following questions:

PROBLEM 2.68 Choose the reaction titled “S

2 with cyanide”

and click on the Play button. As you watch the reaction occur,

notice that the methyl group on the left rotates during the

process. Explain why there is rotation.

PROBLEM 2.69 Choose the reaction “Alkene hydrohalogena-

tion” and click on the Play button. Notice that there are two

steps to this reaction. Identify the electrophile and the nucleo-

phile in the first step. Identify the electrophile and the nucleo-

phile in the second step.

PROBLEM 2.70 Choose the reaction “Bimolecular nucleophilic

substitution: S

2” and click on the Play button. Notice that the

2 is a one-step reaction. What is the nucleophile in this

reaction? You can click on the highest occupied molecular

orbital (HOMO) button to observe the electron density of the

nucleophile. What is the electrophile? The electrophile is shown

in the LUMO movie.

Alkenes and Alkynes

3.1 Preview

3.2 Alkenes: Structure and

Bonding

3.3 Derivatives and Isomers of

Alkenes

3.4 Nomenclature

3.5 The Cahn–Ingold–Prelog

Priority System

3.6 Relative Stability of Alkenes:

Heats of Formation

3.7 Double Bonds in Rings

3.8 Physical Properties of Alkenes

3.9 Alkynes: Structure and

Bonding

3.10 Relative Stability of Alkynes:

Heats of Formation

3.11 Derivatives and Isomers of

Alkynes

3.12 Triple Bonds in Rings

3.13 Physical Properties of Alkynes

3.14 Acidity of Alkynes

3.15 Molecular Formulas and

Degrees of Unsaturation

3.16 An Introduction to Addition

Reactions of Alkenes and

Alkynes

3.17 Mechanism of the Addition of

Hydrogen Halides to Alkenes

3.18 The Regiochemistry of the

Addition Reaction

3.19 A Catalyzed Addition to

Alkenes: Hydration

3.20 Synthesis: A Beginning

3.21 Special Topic: Alkenes and

Biology

3.22 Summary

3.23 Additional Problems

HOT CHEMISTRY The smallest alkyne is acetylene It is a gas and its

most common use is for welding. An acetylene/oxygen mixture burns at the very high

temperature of 3200 °C (5800 °F).

(HC

CH).

98 CHAPTER 3 Alkenes and Alkynes

Butane

)

a saturated alkane

Cyclobutane

)

a cycloalkane

FIGURE 3.1 The chemical properties

of saturated alkanes, C

2n2

,are

very similar to those of the

cycloalkanes, C

On the atomic and subatomic levels, weird electrical forces are crackling

and flaring, and amorphous particles ...are spinning simultaneously

forward, backward, sideways, and forever at speeds so uncalculable that

expressions such as “arrival,” “departure,” “duration,” and “have a nice

day” become meaningless. It is on such levels that magic occurs.

—TOM ROBBINS,

SKINNY LEGS AND ALL

3.1 Preview

Not all hydrocarbons have the formula C

2n2

. Indeed, we have already seen some

that do not: the ring compounds, C

(Fig. 3.1). We noted in Chapter 2 (p. 83)

that the chemical properties of these ring compounds closely resemble those of the

Thomas Eugene Robbins is an American author born in 1936 in Blowing Rock, NC.

acyclic saturated hydrocarbons. There is another family of hydrocarbons that also

has the formula C

,many of whose chemical properties are sharply different from

those of the ring compounds and saturated chains we have seen so far. These com-

pounds are different from other hydrocarbons in that they contain carbon–carbon

double bonds. One bond is a sigma bond similar to the sigma bond of alkanes, the

other is a weaker bond formed by overlapping pi orbitals (see Section 3.2).They are

called alke

nes to distinguish them from the saturated alkanes.

There are also hydrocarbons of the formula C

2n2

, whose chemical proper-

ties resemble those of the alkenes but not those of the alkanes or cycloalkanes.These

compounds are the acetylenes, or alky

nes.

Alkenes and alkynes contain fewer hydrogen atoms than alkanes with the same

number of carbons. Because they are not “saturated” with hydrogen, they are called

unsaturated hydrocarbons. The structures and some of the properties of the fami-

lies of alkenes and alkynes are discussed in this chapter.

Our study of the structures of alkenes and alkynes allows us to begin to exam-

ine chemical reactivity in a serious way.

ESSENTIAL SKILLS AND DETAILS

1. The hybridization model for sp

and sp bonding. It is critical that you be able to use the

hybridization model to derive structures for the alkenes (double bonds) and the alkynes

(triple bonds).

2. Reactivity—the addition reaction. In this chapter, we encounter reactivity, as addition

reactions to alkenes appear. It is most important to see these reactions not as the first of

a near-endless series of different processes, but as the initial exemplars of classes of

reactions. Be sure you can relate what you learn about these addition reactions to the

discussion of acids and bases in Chapters 1 and 2.

3. Priorities! The Cahn–Ingold–Prelog priority system is essential for determining E and

Z, the arrangement of groups around a double bond.

4. Character counts. It is important to see the relationship between the amount of s

character of an orbital and the stability of an electron in that orbital.