Jones M., Fleming S.A. Organic Chemistry

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9.11 Hydroboration in Synthesis: Alcohol Formation 399

We are not yet up to a full discussion of the mechanism of this reaction, which

is outlined in Figure 9.67, but a number of important points are within our powers.

–

repeat the

steps above

two times

Boric acid

A boronic ester

–

B(OH)

–

The initial product is another trigonal

boron compound; notice that we

have traded a B

R bond for a

OR bond

3 R

FIGURE 9.67 The mechanism

of alcohol formation from an

alkylborane.

The first step, addition of peroxidate ion ( ) to the borane, is just

the reaction of a filled orbital on oxygen with the empty 2p orbital on boron. It’s

another nucleophile–electrophile reaction. Notice that the boron becomes oxidized

in this process. A discussion of the later steps is deferred for now, but at least you

can see that the boron–oxygen bonds in the B(OR)

(boronic ester) should be espe-

cially strong ones and the reaction will be favored thermodynamically. Finally, in

excess hydroxide the boronic ester equilibrates with boric acid and the alcohol.

PROBLEM 9.20 Why are boron–oxygen bonds especially strong?

PROBLEM 9.21 Write a mechanism for the formation of boric acid, B(OH)

,from

hydroxide ion, HO



, and a boronic ester, B(OR)

In this case,questions of mechanism are secondary to the utility of this reaction.Note

one very important thing: In the overall reaction we are able to produce the less substi-

tuted alcohol from the alkene. The hydroxyl group enters at the position to which the

boron was attached:the less substituted end of the double bond.This reaction is an over-

all anti-Markovnikov addition (Fig. 9.68) because the regioselectivity of the product is

the opposite of the Markovnikov product. But Markovnikov’s rule was not broken.

C 3CH

CH CH

/ HO

B(OH)

–

Three molecules of alcohol

This alcohol is the product of

overall anti-Markovnikov

addition of water to the alkene

A trialkylborane

FIGURE 9.68 Hydroboration/oxidation accomplishes

the anti-Markovnikov addition of water to an alkene.

400 CHAPTER 9 Additions to Alkenes 1

Our previous method for alcohol synthesis, hydration of alkenes, necessarily

produced the more substituted alcohol. So now we have complementary synthetic

methods for producing both possible alcohols from a given alkene (Fig. 9.69).

Direct hydration gives the more substituted product (Markovnikov addition) and

the indirect hydroboration/oxidation method gives the less substituted alcohol (anti-

Markovnikov addition). Be sure to add these reactions to your file-card collection

of synthetically useful reactions.

ETHYL ALCOHOL

Ethyl alcohol, or ethanol, can be produced by several reac-

tions described in this chapter. Although it has other uses,

most ethyl alcohol is made for the production of alcoholic

beverages, a use deeply rooted in history. Indeed, the rise

of civilization has been attributed to the discovery of beer,

and the necessity to give up a nomadic life to attend to the

cultivation of the ingredients. In 2005, the average adult

American consumed slightly more than 31 gal of beer and

more than 4.5 gal of wine and spirits.

Ethyl alcohol

Ethyl alcohol acts as a depressant in humans, interfering

with neurotransmission. It works by binding to one side of

the synapse, changing its shape slightly and making it able

to bind γ-aminobutyric acid (GABA) more efficiently.

Bound GABA widens the synaptic channel, allowing chlo-

ride ions to migrate in, thus changing the voltage across the

synapse. The nerve cell is less able to fire, and neurotrans-

mission is inhibited. Your body cleanses itself of alcohol by

converting it into acetaldehyde with the enzyme alcohol

dehydrogenase. A second enzyme, aldehyde dehydrogenase,

completes the conversion into acetic acid. There are indi-

viduals, indeed whole races, who are short of aldehyde

dehydrogenase and thus overly susceptible to the deleterious

effects of alcohol. By contrast, alcoholics have been found to

contain elevated levels of this enzyme.

1. BH

2. HOOH / HO

–

Anti-Markovnikov addition

CCH

CH CH

Markovnikov addition

HOSO

FIGURE 9.69 Two addition reactions that can produce alcohols with different regiochemistry.

Summary

Hydroboration/oxidation is a process that yields an alcohol that is the prod-

uct of overall anti-Markovnikov addition. The mechanism of hydroboration

is complex, but several lines of evidence have led to the picture we have

of a concerted reaction with an unsymmetrical transition state in which

one of the alkene’s carbon atoms becomes partially positively charged

(Figs. 9.55–9.60). The synthetic utility of this reaction is not complex at all.

For unsymmetrical alkenes, hydroboration/oxidation leads to the less substi-

tuted alcohol.

alcohol

dehydrogenase

aldehyde

dehydrogenase

OH H

Acetic acidAcetaldehyde

9.12 Special Topic: Rearrangements in Biological Processes 401

PROBLEM 9.22 Devise syntheses for the following molecules from 2-methyl-

2-butene. You may also use any inorganic reagent as well as organic compounds

containing no more than one carbon.

(a)

(b)

(c)

(d)

(e)

(f)

(g)

OCH

9.12 Special Topic: Rearrangements

in Biological Processes

The hydride shifts and alkyl migrations discussed in Section 9.9 are not arcane exam-

ples fit only for the nether regions of an organic text (and for especially vexing prob-

lems).They are going on all the time as part of many of the enzyme-driven reactions

in your body. So, as you read this, hydrides are migrating somewhere deep inside you.

Here is an example of a biological reaction in which an intramolecular hydride

shift must be occurring. Remember the section in Chapter 3 (p. 142) on methyl

transfer to an alkene? We are a bit further along now and can flesh out the

rather sketchy mechanistic description given earlier. In particular, Figure 9.70

shows the ultimate product of the reaction introduced in Figure 3.82. It looks as

COO

S -Adenosylmethionine

COO

10-Methylstearate

COO

Oleate

WEB 3D

FIGURE 9.70 The final product

of the methylation of oleate is

10-methylstearate.

402 CHAPTER 9 Additions to Alkenes 1

9.13 Summary

New Concepts

deuteride shift

elimination

enzymatic reduction

COO

S -Adenosylmethionine

COO

10-Methylstearate

COO

FIGURE 9.71 This deuterium labeling

experiment shows that there is a

“hydride” shift at the center of this

reaction.

The central topic of this chapter is the addition of mol-

ecules to alkenes. These reactions begin by the addition of the

positive end of the dipole (the electrophile) to the alkene (the

nucleophile) in such a way as to form the more stable carbocation.

The anion (X



) then adds to the cation to form the final

addition product. The direction of addition—the regiochemistry

HX of the reaction—is determined by the relative stabilities of the

possible intermediate carbocations. The electrophile will always

add to give the more stable carbocation. Resonance and induc-

tive effects are important factors that influence stability.

Alkyl groups stabilize carbocations by a resonance effect in

which a filled σ orbital of the alkyl group overlaps with the

if intermediate A has somehow been reduced to give the final product.But a clever

labeling experiment (Fig. 9.71) shows that the process is far more complex and

that a hydride shift is involved. It should be no surprise to you that the secondary

carbocationic intermediate A′ undergoes an intramolecular hydride shift to give

the more stable tertiary carbocation B.A proton is then lost from the methyl group

(E1 reaction, p. 298) to give an alkene. It is this alkene that is reduced to give the

final product.

9.13 Summary 403

Key Terms

alkene hydrohalogenation (p. 365)

anti-Markovnikov addition (p. 399)

cationic polymerization (p. 386)

concerted reaction (p. 389)

hydration (p. 381)

hydride shift (p. 387)

hydroboration (p. 390)

hyperconjugation (p. 377)

inductive effects (p. 379)

Markovnikov’s rule (p. 375)

rearrangement (p. 387)

syn addition (p. 394)

Wagner–Meerwein rearrangement

(p. 387)

Reactions, Mechanisms, and Tools

Many acids ( , generally, ) will

add directly to a π bond. The first step is addition of a proton to

give the more stable carbocation. In the second step of the reac-

tion, the negative end of the original dipole adds to com-

plete the addition process. The regiochemistry of the addition is

determined by the formation of the more stable carbocation in

the original addition (Fig. 9.20).

Other molecules are not strong enough acids to proto-

nate alkenes. Water ( ) is an excellent example. However,

such molecules will add across the double bond if the reaction is

acid catalyzed. Enough acid catalyst is added to give the proto-

nated alkene, which is then attacked by water. The catalyst is

regenerated in the last step and recycles to carry the reaction

further (Fig. 9.34).

If the concentration of alkene is high enough, the nu-

cleophilic alkene can compete with other nucleophiles in the

HOH

HXHBr,

HCl, HI, HOSO

system (for example, X



), and dimerization or even polymeriza-

tion of the alkene can take place. This reaction is not mysteri-

ous; the alkene is merely acting as a nucleophile toward the

electrophilic carbocation (Fig. 9.43).

In hydroboration, the boron atom of “BH

” adds to an alkene

to give an alkylborane. The mechanism involves a single step in

which the new carbon–hydrogen and carbon–boron bonds are

both made. Subsequent further hydroborations can give dialkyl-

and trialkylboranes. Treatment of these boranes with basic

hydrogen peroxide leads to alcohols (Figs. 9.66 and 9.67).

Hydroboration is especially important because it gives access to

the less substituted alcohol (anti-Markovnikov addition).

Additions that go through carbocationic intermediates can

be complicated by rearrangements of hydride (H



) or alkyl

groups (R



) to produce more stable carbocations from less

stable ones (Figs. 9.46 and 9.47).

empty 2p orbital on carbon. This effect is known by the curious

name of “hyperconjugation” (Figs. 9.24–9.26).

In this chapter, we continue to use stereochemical experi-

ments to determine reaction mechanisms. Generally, one-step

(concerted) reactions preserve the stereochemical relationships

of groups in the reacting molecules. Multistep processes intro-

duce the possibility of rotation around carbon–carbon bonds

and loss of the initial stereochemistry.

Remember: No experiment can prove a mechanism. We are

always at the mercy of the next experiment, the results of which

might contradict our current mechanistic ideas. Experiments can

certainly disprove a mechanism, but they can never prove one.

Syntheses

In this chapter, we see the synthesis of alkyl halides and alco-

hols from alkenes through addition reactions. Hydroboration

allows us to do addition reactions in the anti-Markovnikov

sense. The S

2 and S

1 reactions from earlier chapters allow us

to do further transformations of the alcohols and halides.The

new synthetic methods are summarized below.

1. Alcohols

–

OHBH

More highly substituted alcohol is formed

(Markovnikov addition)

Less highly substituted alcohol is formed

(anti-Markovnikov addition)

2. Alkylboranes

repeat

The number of hydroborations depends on the bulk of the

alkene; boron becomes attached to the less substituted

end of the alkene

404 CHAPTER 9 Additions to Alkenes 1

9.14 Additional Problems

Common Errors

Keeping clear the difference between resonance and equilibrium

is a constant difficulty for many students. Resonance forms are

simply different electronic descriptions for a single molecule.

The key word is “electronic.” In resonance, only electrons are

allowed to move to produce the various representations of the

molecule. Atoms may not change their positions. If they do, we

are not talking about resonance, but about equilibrium. Be care-

ful! This point is trickier than it sounds.

Remember also that the two-dimensional paper surface can

fool us. Molecules are three-dimensional.To have delocalization

of electrons, orbitals must overlap. Sometimes it looks in two

dimensions as though resonance forms exist when, in fact, they do

not in the real world of three dimensions.Two excellent examples

appear in Figure 9.18 (p. 373) and in Problem 9.4 (p. 373).

The grammar of chemistry—our use of arbitrary

conventions—is important because we need to be precise in

communicating with each other. The double-headed resonance

arrow is reserved for resonance and never used for anything else.

There are several schemes for representing resonance forms,

ranging from drawing them all out in full, to the summary

structures of Figure 9.14 (p. 370).

Both mechanistic analysis and synthesis are gaining in

complexity. There is much stereochemical detail to keep track of

in many mechanistic analyses, for example. Perhaps the most

complex mechanism considered so far is that for hydroboration.

If you understand why it was necessary to modify the standard

mechanism for the addition of Lewis acids to alkenes to

accommodate the experimental observations, you are in fine

shape so far.

There are no really complex synthetic procedures yet.

However, even very simple steps taken in sequence can lead to

difficulty. This area will rapidly proliferate and become more

difficult, so do not be lulled by the deceptive simplicity of the

reactions so far.

This chapter and Chapter 10 continue our cataloging of the

standard reactions of organic chemistry. To the S

1, S

2, E1,

and E2 reactions we now add a variety of alkene addition

reactions. Although there are several different mechanisms

for additions, many take place through a three-step sequence

of protonation, addition, and deprotonation. The following

new problems allow you to practice the basics of addition

reactions and to extend yourself to some more complex mat-

ters. Even simple additions become complicated when they

occur in intramolecular fashion, for example. These problems

also allow you to explore the influence of resonance and

inductive effects, and to use the regiochemistry and stereo-

chemistry of addition to help work out the probable mecha-

nisms of reactions.

Your sophistication in synthesis is also growing, and the

variety of addition reactions encountered in this chapter adds

to the ways you have available to make differently substituted

molecules. You are still not quite ready to undertake multistep

syntheses, but you are now very close. In anticipation of these

tougher problems, be sure to start working synthetic questions

backward. Always ask the question, What molecule is

the immediate precursor of the target? Don’t start, even with

simple problems, thinking of how an ultimate starting

material might be transformed into product. That approach

will work in one- or two-step synthesis, but becomes almost

impossible to do efficiently when we come to longer multistep

syntheses.

PROBLEM 9.23 See if you can write the π molecular

orbitals of allyl from memory. Show the electronic configura-

tion (orbital occupancy) for the allyl cation, radical, and

anion.

PROBLEM 9.24 The molecular orbitals for pentadienyl are

shown on the next page.

Pentadienyl

3. Alkyl Halides

More highly substituted halide is formed

(Markovnikov addition)

(X = Br, Cl, or I)

4. Allyl Halides

1,2- and 1,4-Addition compete

(1,2) (1,4)

9.14 Additional Problems 405

Compare these molecular orbitals with those of allyl. What

rules can you develop to predict the molecular orbitals of any

odd-carbon, fully conjugated (2p orbital on each carbon) chain?

It is probably easiest to work from the schematic top views at

the right of the figure.

PROBLEM 9.25 Apply your rules to the molecular orbitals of

heptatrienyl.

PROBLEM 9.26 Draw a third resonance form for the allyl

cation that involves a three-membered ring. Be careful not

to move any atoms! What is the relative importance of this

resonance form? Explain.

Heptatrienyl

––

+++

–

+ – 0 + –

= + – + – +

+ 0 – 0 +

+ + 0 – –

+ + + + +

Views from the top of pentadienyl;

beneath every (

+) is a (–) and

beneath every (

–) is a (+); nodes

are shown as (0)

PROBLEM 9.27 Write resonance forms for the following

species. You will first have to write good Lewis structures, as the

lone pairs have been deliberately left out. You will also have to

write a full Lewis structure for the nitro (NO

) group.

PROBLEM 9.28 How many signals would appear in the

NMR spectra of the compounds in Problem 9.27 (a) and (b)?

PROBLEM 9.29 How many signals would appear in the

NMR spectra of the following compounds?

PROBLEM 9.30 Which of the following pairs represent

resonance forms and which do not? Explain your choices.

PROBLEM 9.31 Analyze the relative importance of the follow-

ing two resonance forms. What would you say to someone who

claimed that form B must be unimportant because the positive

charge is on the more electronegative atom? You will have to

add electron dots to make good Lewis structures first.

(a)

(b)

(c)

(d)

–

––

(a) (b)

–

(d)(c)

N(CH

)

N(CH

)

406 CHAPTER 9 Additions to Alkenes 1

PROBLEM 9.32 Predict the regiochemistry of the following addi-

tion reactions of a generic acid, . Explain your predictions.

PROBLEM 9.33 What would be the structure of the polymer

produced from each of the following two monomers through

cationic polymerization?

PROBLEM 9.34 At 78 °C, hydrogen bromide adds to

1,3-butadiene to give the two products shown. Carefully write

a mechanism for this reaction, and rationalize the formation

of two products.

PROBLEM 9.35 Reaction of cis-2-butene with hydrogen

bromide gives racemic 2-bromobutane.

(a) Show carefully the origin of the two enantiomers of

2-bromobutane.

(b) In addition, small amounts of two alkenes isomeric with

cis-2-butene can be isolated. What are these two alkenes

and how are they formed?

PROBLEM 9.36 Indene can be hydrated to give a different

product (A or B) under the following two conditions.

Reagents for the first route: H



Reagents for the second route: (1) BH

, (2) HOOH/HO



One of these two routes leads to a compound, A (C

O), which

has a

C NMR spectrum consisting of nine signals. The other

route leads to a different compound, B, which also has the formula

O, and has a

C NMR spectrum consisting of only five

signals. What are the structures of the two compounds A and B,

and which route leads to which compound? Explain your reason-

ing in mechanistic terms for each compound and make very clear

how you used the

C data in the structure determination. (You do

not have to sketch a mechanism for the oxidation step of route 2.)

Indene

–78 ⬚C

HBr

(81%)

(18%)

(b)

(a) (c)

(e)(d)

In the following four problems, reactions appear that are quite

similar to those discussed in this chapter, but are not exactly the

same. Each raises one or more slightly different mechanistic points.

In this book, we aim to develop an ability to understand new obser-

vations in the light of old knowledge. In a sense, you are being put

in the place of the research worker presented with new experimen-

tal results. In each case, explain the products mechanistically.

PROBLEM 9.37

PROBLEM 9.38

PROBLEM 9.39

PROBLEM 9.40

(CH

)

(a)

CH(CH

)

CCO

not

OCH

(CH

)

(a)

(b)

OCH

(b)

9.14 Additional Problems 407

PROBLEM 9.41

Given the following experimental observation

and the knowledge that the initial hydroboration step involves a

syn addition, what can you say about the overall stereochemistry

of the migration and oxidation steps of the organoborane (see

Fig. 9.67, p. 399). Is there overall retention or inversion of con-

figuration?

PROBLEM 9.42 Show the reactions you would use to synthe-

size 1-butanethiol from 1-butene.

PROBLEM 9.43 Show how you would make 1-azidobutane if

your only source of carbon is 1-butene.

PROBLEM 9.44 Show how you would make 1-isopropoxyhexane

starting with 1-hexene and using any other reagents you need.

PROBLEM 9.45 See the syntheses of 2-chloro-3-methylbutane

and 2-bromo-3-methylbutane suggested in the answer to

Problem 9.22 (b) and (e), given in the Study Guide. Carefully

consider the last step in the proposed syntheses, the reaction of

3-methyl-2-butanol with hydrogen bromide or hydrogen

chloride. Do you see any potential problems with this step?

What else might happen?

PROBLEM 9.46 Propose syntheses of the following molecules

starting from methylenecyclohexane, alcohols, iodides contain-

ing no more than four carbon atoms, and any inorganic materi-

als (no carbons) you want.

Methylene-

cyclohexane

(a)

(b)

(c)

(d)

(e)

(f)

OCH

1. BH

2. H

/HO

–

PROBLEM 9.47 Predict the major product(s) for the hydro-

boration/oxidation reaction (1. BH

/THF; 2. H

/NaOH)

with each of the following alkenes:

(a) 1-pentene (b) 2-methyl-2-pentene

PROBLEM 9.48 Which of the products in the previous

problem are chiral and which are achiral?

PROBLEM 9.49 Predict the possible products in the reaction

between HBr with the following alkenes:

(a) cyclopentene (b) trans-2-butene

(e) 2-hexene

PROBLEM 9.50 Predict the major product(s) for the reaction

between (E)-2-pentene and the following reagents:

(a) H



O (b) 1. BH

/THF; 2. H

/NaOH

(e) H



/CH

PROBLEM 9.51 Predict the major product(s) for the

reaction between (E)-3-methyl-2-pentene and the following

reagents:

(a) H



O (b) 1. BH

/THF; 2. H

/NaOH

(e) H



/CH

In each of the following reactions, more than a simple addition

takes place. Keep your wits about you, think in simple stages,

and work out mechanisms.

PROBLEM 9.52 Provide a mechanism for the following

change:

PROBLEM 9.53 Provide a mechanism for the following

change:

408 CHAPTER 9 Additions to Alkenes 1

PROBLEM 9.54 Provide a mechanism for the following

change: Hint for the hard part and for the next few problems: Think

Wagner–Meerwein, and use the CH

groups as markers (p. 387).

PROBLEM 9.55 Provide mechanisms for the following trans-

formations. Once again, use the CH

groups to track where the

atoms go.

PROBLEM 9.56 What two products do you anticipate for the

reaction shown below? Apply what you already know about the

chemistry of alcohols in acid.There are two reasonable products,

each with the formula C

O, but with very different structures.

In fact, these compounds are formed in only minor amounts. The

major product is pinacolone, shown in Problem 9.57.

PROBLEM 9.57 What is the mechanism of pinacolone

formation in the reaction in Problem 9.56?

Pinacolone

(CH

)

CCC

Pinacolone

Easy

HCl

Hard!!

PROBLEM 9.58 Provide a mechanism for the following

transformation of morphine into apomorphine. Caution: This

problem is very hard.

Use Organic Reaction Animations (ORA) to answer the fol-

lowing questions:

PROBLEM 9.59 Select the animation titled “Alkene hydro-

halogenation” and observe the intermediate in the reaction.

What is the hybridization of the central carbon? Select the

LUMO representation of the intermediate. What do we learn

from this calculated information? What options are available

for the nucleophile in the second step of this reaction?

PROBLEM 9.60 Observe the “Alkene polymerization”

animation. What experimental conditions are necessary for

polymerization to be the major pathway? The animation shows

formation of the dimer and the trimer. How many alkene

molecules need to add before we can call the product a polymer?

PROBLEM 9.61 The “Alkene polymerization” animation does

not show a final product. Do you suppose an alkene polymer

would have a charge? Milk jugs are made from alkene poly-

mers. What do you know about the properties of a milk jug

that help you answer the question? What are possible final steps

in the formation of the polymer?

PROBLEM 9.62 The “Alkene hydroboration” reaction was a

challenge to calculate. Observe the first part of the hydrobora-

tion animation, the reaction of the alkene with the borane. Why

is it difficult to calculate this part of the reaction? Observe the

second part of the animation, the oxidation of the boron. Notice

that there is a 1,2-alkyl shift. What molecular orbital relation-

ship do you think is the key for that shift?

Morphine

Apomorphine

O/H

150 C