Jones M., Fleming S.A. Organic Chemistry

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3.19 A Catalyzed Addition to Alkenes: Hydration 139

is “no,” and in fact, we can never know for certain that we are absolutely right.

A mechanistic hypothesis survives only pending the next clever test.Someone—you

perhaps—may develop a new test of the ideas outlined in this section that reveals

an error, or at least an area that needs further thinking and working out of details.

It’s sad but true: It is easy to disprove a scientific hypothesis—just find some con-

tradictory data. It is impossible ever to prove a mechanistic hypothesis in chemistry

beyond a shadow of a doubt.

We all have to get used to this ambiguity—it is part of the human condition.

We can be quite certain we are right in our mechanistic ideas, and the more time

that goes by, and the more tests our ideas pass, the more confident we can be that

we are basically right. However, there have been some big surprises in science, as

conventional wisdom is reversed every once in a while, and there will be more. To

be honest, we sort of like it when that happens.

PROBLEM 3.30 Predict the products of the following reactions:

HBr

(HO SO

OH)

PROBLEM SOLVING

All “which direction does HX add” problems can be approached—and solved—

in the same way. Always draw out the two possible carbocations formed by initial

protonation of the double bond. For an unsymmetrical (ends different) double

bond, there are always two and only two possibilities. Next evaluate those two

carbocations—ask yourself which of them is more stable? It is that one that will

lead to the major product. The answer to Problem 3.30 is typical. Use it as a

prototype, a template for answering all similar questions.

3.19 A Catalyzed Addition to Alkenes: Hydration

Not every HX is a strong enough acid to protonate an alkene. Water, HOH

(here X  OH), is an example of such a weak acid. Yet it would be very useful to be

able to add the elements of water to an alkene,to “hydrate”an alkene as in Figure 3.80,

AlcoholAlkene

????

H O H

FIGURE 3.80 The highly desirable

addition of water to an alkene.

140 CHAPTER 3 Alkenes and Alkynes

because the products ( , called alcohols) are useful in themselves and

are also versatile starting materials for the synthesis of other molecules. The

solution to this problem is to use a small amount of an acid that is strong

enough to protonate the alkene. The added acid acts as a catalyst for the addi-

tion. A catalyst is a material that increases the rate of a reaction without being

consumed.

Let’s work through the mechanism for the hydration of alkenes.The first step

is protonation of the alkene by the catalyst, H



. As in the reaction of HCl, this

addition leads to a carbocation intermediate (Fig. 3.81). Now there is no halide

Catalyst is

used up here

1. Protonation of

the alkene

Carbocation intermediate

H OH

3. Deprotonation

by water

2. Addition

of water

Oxonium ion

intermediate

CCH

Alcohol

Catalyst is

regenerated here

FIGURE 3.81 The acid-catalyzed

addition of water to 2,3-dimethyl-

2-butene—a hydration reaction.

ion (I



,Br



,Cl



) to capture the carbocation, but there are lots of water mol-

ecules available. Capture by water leads not to the final product, the alcohol, but

to an intermediate oxonium ion, in which a trivalent oxygen bears a positive

charge. Water then deprotonates the oxonium ion to give the alcohol, and regen-

erate the catalyst, H



You can see from this last step why only a catalytic amount of acid is

necessary. Although it is consumed in the first protonation step, it is regener-

ated in the last step and can recycle to start the hydration reaction again

(Fig. 3.81).

WORKED PROBLEM 3.31 Only one alcohol is produced in the following reaction.

What is it, and why is only one alcohol, not two, produced?

C C A single alcohol product

(continued)

3.20 Synthesis: A Beginning 141

ANSWER Formation of the more stable tertiary carbocation rather than the much

less stable primary carbocation inevitably leads to the single product observed,

tert-butyl alcohol.

tert-Butyl alcohol

CCH

Much less stable Not formed

More stable

PROBLEM SOLVING

The order of carbocation stability is: tertiary (3°, R



) more stable than secondary

(2°, R



), more stable than primary (1°, RH



), more stable than methyl



). We’ll explain this stability order later, but you should remember that

simple primary and methyl carbocations are too high in energy to be formed in

common organic reactions. A primary or methyl carbocation is a mechanistic stop

sign. Nothing you write after it can be correct. Don’t go through the stop sign!

What does “simple” mean? It means that no special factors such a resonance

stabilization can be present—a “simple” carbocation is one that is not

significantly delocalized by resonance.

STOP

3.20 Synthesis: A Beginning

Organic chemistry is not only understanding how reactions occur. A very impor-

tant part of chemistry is the use of those reactions to make molecules.The construc-

tion of target molecules from smaller pieces is called synthesis, and we are now able

to do a surprising amount of this kind of thing. Let’s just take stock of what kinds

of transformations we can do—of what kinds of molecules we can make out of other

kinds of molecules.

The addition of hydrogen halides to alkenes gives us a way to make alkyl bro-

mides, chlorides, and iodides (fluorides don’t work very well) as long as we have the

appropriate starting alkene. We will have to be careful when using unsymmetrical

alkenes, because there is a choice to be made here. As you saw in Section 3.18, the

addition of HX will only lead to the more substituted alkyl halide, as the regiochem-

istry of the addition is controlled by the stability of the intermediate carbocation.

142 CHAPTER 3 Alkenes and Alkynes

We can also make alcohols through the acid-catalyzed addition of water to

alkenes (Section 3.19).

In each chapter, the synthetic methods developed and discussed are collected on

the Summary page. In this chapter, these reactions appear on pages 143 and 144.

PROBLEM 3.32 Devise syntheses of the following three compounds, starting in

each case with any alkene that contains four carbons or fewer. You do not have to

write mechanisms, although at this point it may be very helpful for you to do so.

(CH

)

COH (CH

)

CHOH CH

CHOHCH

3.21 Special Topic: Alkenes and Biology

In sharp contrast to the alkanes, the unsaturated alkenes are highly active biologi-

cally. As we saw in Section 3.2,alkenes contain π bonds that are substantially weak-

er than almost all σ bonds of alkanes. Accordingly, their reactivity is higher, and

Nature “uses” this reactivity to accomplish things. One thing that can be accom-

plished is the formation of carbon–carbon single bonds. It is extraordinarily fortu-

nate for us that Nature is able to do that reaction easily, because our tissues are

composed mainly of carbon–carbon single bonds. We’ll just sketch one example of

alkene-mediated carbon–carbon single bond formation here.We will return to more

details of this and related transformations in later chapters.

For its own purposes, the tuberculosis-causing organism Mycobacterium tubercu-

losis needs a molecule called 10-methylstearate.It makes this compound from a mod-

ified amino acid, S-adenosylmethionine, and oleate, a salt of oleic acid (Fig. 3.82).

Oleate

S-Adenosylmethionine

carbon–carbon bond making

and

carbon–sulfur bond breaking

Further reactions lead to 10-methylstearate

C COO

–

COO

–

COO

–

COO

–

FIGURE 3.82 The reaction of a

nucleophilic alkene (Lewis base) to

produce a carbon–carbon single

bond, shown in red.

3.22 Summary 143

The first interesting step in what turns out to be, ultimately, a reaction chock full of

interesting steps is the transfer of the methyl group to the alkene in oleate.The dou-

ble bond acts as the trigger for the whole sequence. Note also that here the π bond

acts as a Lewis base, much as it does in its reaction with HCl (p. 132), where it acts

as a Brønsted base.

This chapter deals mainly with the structural consequences of

and sp hybridizations. The doubly bonded carbons in alkenes

are hybridized sp

and the triply bonded carbons in alkynes are

hybridized sp. In these molecules, atoms are bound not only by

the σ bonds we saw in Chapter 2 but by π bonds as well.

These π bonds, composed of overlapping 2p orbitals, have

substantial impact upon the shape (stereochemistry) of the mol-

ecule. Alkenes can exist in cis (Z) or trans (E) forms—they have

sides. Unlike the alkanes, which contain only σ bonds which

have very low barriers to rotation, there is a substantial barrier

(⬃66 kcal/mol) to rotation about a π bond. The linear alkynes

do not have cis/trans isomers and the question of rotation does

not arise.

Double and triple bonds can be contained in rings if the

ring size is large enough to accommodate the strain incurred by

the required bond angles.The smallest trans cycloalkene stable

at room temperature is trans-cyclooctene. The smallest stable

cycloalkyne is cyclooctyne.

More substituted alkenes and alkynes are more stable than

their less substituted relatives.

3.22 Summary

New Concepts

acetylenes (p. 98)

acetylide (p. 129)

activation energy (ΔG

‡

) (p. 136)

alcohol (p. 140)

alkenes (p. 98)

alkynes (p. 98)

allyl (p. 108)

Bredt’s rule (p. 122)

bridgehead position (p. 121)

Cahn–Ingold–Prelog priority system (p.111)

catalyst (p. 140)

cycloalkenes (p. 110)

degree of unsaturation (Ω) (p. 131)

double bond (p. 98)

ethene (p. 99)

ethylene (p. 99)

heat of formation (p. 115)

HOMO (p. 133)

LUMO (p. 133)

olefins (p. 123)

oxonium ion (p. 140)

pi (π) orbitals (p. 104)

polyenes (p. 110)

propargyl (p. 127)

regiochemistry (p. 137)

triple bond (p. 114)

vinyl (p. 107)

1¢H °

Key Terms

The Cahn–Ingold–Prelog priority system is introduced. It is

used in the (Z/E) naming system. We will encounter it again

very soon in Chapter 4.

In this chapter, we continue the idea of constructing larger

hydrocarbons out of smaller ones by replacing the different

available hydrogens with an X group. When X is methyl (CH

a larger hydrocarbon is produced from a smaller one.

Addition reactions between alkenes and acids, , are

introduced. Many acids ( , , , HOSO

OH,

generally, ) add directly to alkenes. The first step is addi-

tion of a proton to give the more stable carbocation. In the sec-

ond step of the reaction, the negative end of the original H

ClH

dipole usually adds to complete the addition process. The regio-

chemistry of the addition is determined by the formation of the

more stable carbocation in the original addition.

Other molecules are not strong enough acids to

protonate alkenes. Water (HOH) is an excellent example.

However, such molecules will add if the reaction is acid-cat-

alyzed. Enough acid catalyst is added to give the protonated

alkene, which is attacked by water. The catalyst is regenerated in

the last step and recycles to carry the reaction further.

The removal by a base of a hydrogen from the terminal

position of acetylenes to give acetylides is mentioned. Terminal

alkynes are moderately acidic molecules.

Reactions, Mechanisms, and Tools

1. Acetylides

–

Base

–

C CH+ +BHCH

Syntheses

144 CHAPTER 3 Alkenes and Alkynes

2. Alkyl Halides

X = Cl, Br, I, sometimes F

Addition gives the compound

with the halide, X, on the

more substituted carbon

CCH

The concept of cis/trans (Z/E) isomerism is a continuing

problem for students. Errors of omission (the failure to find a

possible isomer, for example) and of commission (failure to

recognize that a certain double bond, so easily drawn on

paper, really cannot exist) are all too common. Unlike σ

bonds, π bonds have substantial barriers to rotation

(⬃66 kcal/mol). The barrier arises because of the shape of the

p orbitals making up the π bond. Rotation decreases overlap

and raises energy. Accordingly, double bonds have sides,

which are by no means easily interchanged. Substituents

cannot switch sides, and are locked in position by the high

barrier to rotation. It is vital to see why there are two isomers

of 2-butene, cis- and trans-2-butene, but only one isomer of

2-methyl-2-butene (Fig. 3.83).

This relatively easy idea has more complicated implications.

For example, the need to maintain planarity (p/p overlap) in a π

bond leads to the difficulty of accommodating a trans double

bond in a small ring, or at the bridgehead position of bridged

bicyclic molecules. In each case, the geometry of the molecule

results in a twisted (poorly overlapping) pair of p orbitals.

Common Errors

In this chapter, we encounter the determination of molecular

formulas. The following three review problems (problems

3.33–3.35) deal with this subject.

PROBLEM 3.33 Alkanes combine with oxygen to produce car-

bon dioxide and water according to the following scheme:

This process is generally referred to as combustion. An impor-

tant use of this reaction is the quantitative determination of ele-

mental composition (elemental analysis). Typically, a small sam-

ple of the compound is completely burned and the water and

carbon dioxide produced are collected and weighed. From the

weight of water the amount of hydrogen in the compound can

be determined. Similarly, the amount of carbon dioxide formed

allows us to determine the amount of carbon in the original

compound. Oxygen, if present, is usually determined by differ-

ence. The determination of the relative molar proportions of

2n + 2

+ (3n + 1)/2 O

(n + 1)H

O + nCO

carbon and hydrogen in a compound is the first step in deriving

its molecular formula.

If combustion of 5.00 mg of a hydrocarbon gives 16.90 mg

of carbon dioxide and 3.46 mg of water, what are the weight

percents of carbon and hydrogen in the sample?

PROBLEM 3.34 Calculate the weight percents for each

element in the following compounds: (a) C

and

(b) C

ClNO.

PROBLEM 3.35 A compound containing only carbon,

hydrogen, and oxygen was found to contain 70.58% carbon

and 5.92% hydrogen by weight. Calculate the empirical

formula for this compound. If the compound has a molecular

weight of approximately 135 g/mol, what is the molecular

formula?

PROBLEM 3.36 How many degrees of unsaturation are

there in compounds of the formula C

? Write at least eight

isomers including ones with no π bonds and ones with

no rings.

3.23 Additional Problems

These are different molecules

These are the same molecule

trans-2-Butene

cis-2-Butene

2-Methyl-2-butene

FIGURE 3.83

3. Alcohols

The addition reaction of water

is catalyzed by acid

CCH

3.23 Additional Problems 145

PROBLEM 3.37

Determine the degrees of unsaturation for the

following compounds.

PROBLEM 3.43 Write systematic names for the following

compounds:

PROBLEM 3.44 Draw structures for the following compounds:

(a) (Z)-3-fluoro-2-methyl-3-hexene

(b) 2,6-diethyl-1,7-octadien-4-yne

(d) 5-chloro-4-iodo-6-methyl-1-heptyne

PROBLEM 3.45 Although you should be able to draw structures

for the following compounds, they are named incorrectly. Give

the correct systematic name for each compound:

(a) 5-chloro-2-methyl-6-vinyldecane

(b) 3-isopropyl-1-pentyne

(d) (E)-3-propyl-2,5-hexadiene

PROBLEM 3.46 Find all the compounds of the formula

in which the dipole moment is zero. It is easiest to

divide this long problem into sections. First, work out the num-

ber of degrees of unsaturation so that you can see what kinds of

molecules to look for. Then find the acyclic molecules. Next

work out the compounds containing a ring.

PROBLEM 3.47 Find all the compounds of the formula C

in which there are four different carbon atoms. There are only

nine compounds possible, but two of them may seem quite

strange at this point. Hint: One of these molecules contains a

carbon that is part of two double bonds, and the other is a com-

pound containing only three-membered rings.

PROBLEM 3.48 Find all the isomers of the formula C

Cl.

As in Problem 3.46, it is easiest to divide this problem into sec-

tions. First, find the degrees of unsaturation so that you know

what types of molecule to examine. There are 23 isomers to

find. Hint: See the hint for Problem 3.47.

(a)

(b)

(c)

(d)

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

PROBLEM 3.38 Provide the IUPAC name for the following

compounds:

PROBLEM 3.39 Are the two structures below the same

molecule? Draw the Newman projection for each looking down

the bond. Which form do you think is more

stable? Why?

PROBLEM 3.40 Draw and write the systematic names for all

of the acyclic isomers of 1-heptene (C

). There are 36 iso-

mers, including 1-heptene.

PROBLEM 3.41 Draw and write the systematic names for all of

the acyclic isomers of 1-octyne (C

) containing a triple

bond. There are 32 isomers, including 1-octyne.

PROBLEM 3.42 Draw and write the systematic names for all

of the isomers of dichlorobutene (C

). There are 27 iso-

mers, according to usually reliable sources.

(a) (b)

C(2)

C(3)

(a) (b) (c)

(d) (e)

146 CHAPTER 3 Alkenes and Alkynes

PROBLEM 3.49 Find the compound of the formula C

composed of two six-membered rings, that has only three dif-

ferent carbons.

PROBLEM 3.50 Develop a bonding scheme for the compound

CO in which both carbon and oxygen are hybridized sp

PROBLEM 3.51 (a) Develop a bonding scheme for a ring made

up of six CH groups, (CH)

. Each carbon is hybridized sp

. (b) This

part is harder.The molecule you built in (a) exists, but all the car-

bon–carbon bond lengths are equal. Is this consistent with the struc-

ture you wrote in (a), which almost certainly requires two different

carbon–carbon bond lengths? How would you resolve the problem?

PROBLEM 3.52 In this chapter, we developed a picture of the

methyl cation (



), a planar species in which carbon is

hybridized sp

. In Chapter 2, we briefly saw compounds formed

from several rings, bicyclic compounds. Explain why it is possi-

ble to form carbocation (a) in the bicyclic compound shown

below, but very difficult to form carbocation (b). It may be

helpful to use models to see the geometries of these molecules.

PROBLEM 3.53 Show the curved arrow formalism (electron

pushing or arrow pushing) for the following reactions. Which side

of the equilibrium is favored in each reaction? Use the pK

table in

the endpaper of the book to answer this part of the problem.

(

)(

)

PROBLEM 3.55 Explain the formation of two products in the

following reaction:

–

RCC

HOH

–

RCC

–

RCCH

–

RCC

PROBLEM 3.54 Predict the products of the following reactions:

HCl

C +

PROBLEM 3.56 Explain the formation of the two products in

the reaction shown.

PROBLEM 3.57 In principle, there is a second “double addi-

tion” possible in Problem 3.56. Explain why only one compound

is formed. Hint: Think “resonance.”

Use Organic Reaction Animations (ORA) to answer the

following questions:

PROBLEM 3.58 Choose the reaction titled “Alkene hydration”

and click on the HOMO button. Observe the calculated high-

est occupied molecular orbital for 2-methylpropene. One hydro-

gen on each methyl group is not involved in the HOMO. Why

is that the case?

PROBLEM 3.59 In the “Alkene hydration” reaction, click on

Play and observe the last step of the reaction, which is removal

of an acidic hydrogen. It is shown in the following figure:

Show the electron pushing for the reaction going from left to

right. Show the electron pushing for the reaction going from

right to left. Which direction is favored? How can the neutral

alcohol be obtained from this reaction?

PROBLEM 3.60 Choose the reaction “Acetylide addition” and

click on the HOMO button and then play the reaction. Stop

the reaction before the acetylide reacts with the bromoethane.

Observe the calculated highest occupied molecular orbital of

the propyne anion. Are the π orbitals involved in the anion?

Why or why not?

RO ROH

HCl

This compound is formed

C CH

This compound

is not formed

C CCH

CCCCH

C +C

C CH

HCl

CCCCH

Stereochemistry

147

4.1 Preview

4.2 Chirality

4.3 The (R/S) Convention

4.4 Properties of Enantiomers:

Physical Differences

4.5 The Physical Basis of Optical

Activity

4.6 Properties of Enantiomers:

Chemical Differences

4.7 Interconversion of

Enantiomers by Rotation

about a Single Bond:

gauche-Butane

4.8 Diastereomers and Molecules

Containing More than One

Stereogenic Atom

4.9 Physical Properties of

Diastereomers: Resolution, a

Method of Separating

Enantiomers from Each Other

4.10 Determination of Absolute

Configuration (R or S)

4.11 Stereochemical Analysis

of Ring Compounds

(a Beginning)

4.12 Summary of Isomerism

4.13 Special Topic: Chirality without

“Four Different Groups

Attached to One Carbon”

4.14 Special Topic: Stereochemistry

in the Real World:

Thalidomide, the

Consequences of Being

Wrong-Handed

4.15 Summary

4.16 Additional Problems

MIRROR IMAGES This photo shows a woman and her mirror image. In this chapter

we will see that some molecules exist in mirror image forms.

148 CHAPTER 4 Stereochemistry

cis-1,2-Dimethylcyclopropane

Two representations of

trans-1,2-dimethylcyclopropane

cis-2-Butene

or (Z )-2-butene

trans-2-Butene

or (E)-2-butene

C C

WEB 3D

FIGURE 4.1 cis- and trans-1,2-

Dimethylcyclopropane and cis-

and trans-2-butene are pairs of

stereoisomeric molecules. In the two

cyclopropanes, the thicker bond

indicates the side of the ring closest

to you.

I held them in every light. I turned them in every attitude. I surveyed their

characteristics. I dwelt upon their peculiarities. I pondered upon their

conformation.

—EDGAR ALLAN POE,

BERENICE

Edgar Allan Poe (1809–1849) was an American poet and writer probably most famous for his macabre

stories and poems. As the quotation shows, he was also an early devotee of stereochemistry.

4.1 Preview

If you worked Problem 2.29 (p. 85), you saw that there are two isomers of

trans-1,2-dimethylcyclopropane. We will see in this chapter why two trans-1,2-

dimethylcyclopropanes exist and continue with a detailed discussion of stereochemistry,

the structural and chemical consequences of the arrangement of atoms in space.We have

already seen several examples of stereoisomeric molecules, compounds that differ

only in the spatial arrangement of their constituent parts. The compounds cis-

and trans-1,2-dimethylcyclopropane are stereoisomers (as are all cis/trans pairs).

Substituents on both ring compounds and alkenes can be attached in two ways: cis

and trans (Fig. 4.1). We will soon see that the two trans-1,2-dimethylcyclopropanes

are stereoisomers of a different kind.

In this chapter, we will examine subtle questions of stereoisomerism. Most mol-

ecules of Nature are chiral. The word “chiral”is derived from the Greek and means

“handed.” Chiral molecules are related to their mirror images in the same way that

your left hand is related to your right. Your hands are not identical—they cannot be

superimposed. They are mirror images of one another. Many molecules of Nature

(for example, the amino acids) are handed in the same way: One might say that our

amino acids are all left-handed. The source of the ubiquity of left-handedness of

the amino acids is one of the great remaining questions in chemistry.Was it chance?

Did an accidental left-handed start determine all that followed? When we encounter

extraterrestrial civilizations will their amino acids (assuming they have any) also be

left-handed or will we find right-handed amino acids? As it now appears we will

find no cohabitants within our own solar system, and as physical contact with our

galactic neighbors is unlikely, you might consider how you would convey to an invis-

ible resident of the planet Altair 4 our concepts of left and right.

Chirality, or handedness, is a complicated but important subject that is

absolutely essential to organic and biological chemistry. When we return to our

examination of reaction mechanisms—of the way molecules come together to