ionic reactions—nucleophilic substitution ... · 232 chapter 6 / ionic reactions—nucleophilic...

Chapter6

Ionic Reactions—Nucleophilic Substitution andIonic Reactions—Nucleophilic Substitution andElimination Reactions of Alkyl HalidesElimination Reactions of Alkyl Halides

Breaking Bacterial Cell Walls with Organic ChemistryEnzymes, nature’s quintessential chemists, catalyze most reactions of life. Enzymes cat-alyze metabolic reactions, the flow of genetic information, and the synthesis of moleculesthat provide biological structure. They also help defend us against infections and disease.Although mechanisms have been elucidated for the action of many enzymes, those forwhich mechanistic secrets have been unlocked constitute only a fraction of all of theenzymes involved in life’s processes. It is widely accepted, however, that all reactionscatalyzed by enzymes occur on the basis of rational chemical reactivity. The mechanismsutilized by enzymes are essentially those that we learn about in organic chemistry. Onesuch example involves lysozyme.

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Lysozyme is an enzyme in nasal mucus that fights infection by degrading bacterial cellwalls. Lysozyme employs a mechanism that generates an unstable, positively chargedcarbon intermediate (called a carbocation) within the molecular architecture of the bacterialcell wall. Lysozyme elegantly stabilizes this carbocation by providing a nearby negativelycharged site from its own structure. This facilitates cleavage of the cell wall, yet does notinvolve bonding of lysozyme itself with the carbocation intermediate in the cell wall.Carbocation intermediates are central to a number of organic reaction types. One of theseis called unimolecular nucleophilic substitution and it is one of the important re-(S 1),N

actions whose mechanism you will study in this chapter. (We shall revisit the mechanismof lysozyme in detail in Section 24.10)

6.1 Introduction6.2 Physical Propertiesof Organic Halides6.3 NucleophilicSubstitution Reactions6.4 Nucleophiles6.5 Leaving Groups6.6 Kinetics of aNucleophilic SubstitutionReaction: An S 2N

Reaction6.7 A Mechanism forthe ReactionS 2N

6.8 Transition StateTheory: Free EnergyDiagrams6.9 The Stereochemistryof ReactionsS 2N

6.10 The Reaction oftert-Butyl Chloride withHydroxide Ion: An S 1N

Reaction6.11 A Mechanism forthe ReactionS 1N

6.12 Carbocations6.13 TheStereochemistry of S 1N

Reactions6.14 Factors Affectingthe Rates of andS 1 S 2N N

Reactions6.15 Organic Synthesis:Functional GroupTransformations Using

ReactionsS 2N

6.16 EliminationReactions of Alkyl Halides6.17 The E2 Reaction6.18 The E1 Reaction6.19 Substitution versusElimination

6.1 INTRODUCTION

The halogen atom of an alkyl halide is attached to an carbon. The arrange-3sp -hybridizedment of groups around the carbon atom, therefore, is generally tetrahedral. Because hal-ogen atoms are more electronegative than carbon, the carbon–halogen bond of alkyl hal-ides is polarized; the carbon atom bears a partial positive charge, the halogen atom apartial negative charge.

C Xd1 d2

The size of the halogen atom increases as we go down the periodic table: fluorine atomsare the smallest and iodine atoms the largest. Consequently, the carbon–halogen bondlength (Table 6.1) also increases as we go down the periodic table.

In the laboratory and in industry, alkyl halides are used as solvents for relatively non-polar compounds, and they are used as the starting materials for the synthesis of manycompounds. As we shall learn in this chapter, the halogen atom of an alkyl halide can beeasily replaced by other groups, and the presence of a halogen atom on a carbon chainalso affords us the possibility of introducing a multiple bond.

Compounds in which a halogen atom is bonded to an carbon are called2sp -hybridizedvinylic halides or phenyl halides. The compound has the common nameCH "CHCl2

vinyl chloride, and the group is commonly called the vinyl group. VinylicCH "CH92

halide, therefore, is a general term that refers to a compound in which a halogen is attachedto a carbon atom that is also forming a double bond to another carbon atom. Phenyl halidesare compounds in which a halogen is attached to a benzene ring (Section 2.5B). Phenylhalides belong to a larger group of compounds that we shall study later, called aryl halides.

A vinylic halide A phenyl halide or aryl halide

X

X

C C

Together with alkyl halides, these compounds comprise a larger group of compoundsknown simply as organic halides or organohalogen compounds. The chemistry of vinylic

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Table 6.1 Carbon–HalogenBond Lengths

Bond Bond Length (A)

CH39F 1.39CH39Cl 1.78CH39Br 1.93CH39 I 2.14

230

and aryl halides is, as we shall also learn later, quite different from alkyl halides, and it ison alkyl halides that we shall focus most of our attention in this chapter.

6.2 PHYSICAL PROPERTIES OF ORGANICHALIDES

Most alkyl and aryl halides have very low solubilities in water, but as we might expect,they are miscible with each other and with other relatively nonpolar solvents. Dichloro-methane also called methylene chloride), trichloromethane also called(CH Cl , (CHCl ,2 2 3

chloroform), and tetrachloromethane also called carbon tetrachloride) are often(CCl ,4

used as solvents for nonpolar and moderately polar compounds. Many chloroalkanes,including and have a cumulative toxicity and are carcinogenic, however, andCHCl CCl ,3 4

should therefore be used only in fume hoods and with great care.Methyl iodide (bp 427C) is the only monohalomethane that is a liquid at room tem-

perature and pressure. Ethyl bromide (bp 387C) and ethyl iodide (bp 727C) are both1 atmliquids, but ethyl chloride (bp 137C) is a gas. The propyl chlorides, bromides, and iodidesare all liquids. In general, higher alkyl chlorides, bromides, and iodides are all liquids andtend to have boiling points near those of alkanes of similar molecular weights.

Polyfluoroalkanes, however, tend to have unusually low boiling points (Section 2.14D).Hexafluoroethane boils at 2797C, even though its molecular weight is near(MW 5 138)that of decane bp 1747C).(MW 5 144;

Table 6.2 lists the physical properties of some common organic halides.

6.3 Nucleophilic Substitution Reactions 231


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Table 6.2 Organic Halides

Group

Fluoride

bp (7C)Density

(g mL21)

Chloride

bp (7C)Density

(g mL21)

Bromide

bp (7C)Density

(g mL21)

Iodide

bp (7C)Density

(g mL21)

Methyl 278.4 0.84260 223.8 0.9220 3.6 1.730 42.5 2.2820

Ethyl 237.7 0.7220 13.1 0.9115 38.4 1.4620 72 1.9520

Propyl 22.5 0.7823 46.6 0.8920 70.8 1.3520 102 1.7420

Isopropyl 29.4 0.7220 34 0.8620 59.4 1.3120 89.4 1.7020

Butyl 32 0.7820 78.4 0.8920 101 1.2720 130 1.6120

sec-Butyl 68 0.8720 91.2 1.2620 120 1.6020

Isobutyl 69 0.8720 91 1.2620 119 1.6020

tert-Butyl 12 0.7512 51 0.8420 73.3 1.2220 100 deca 1.570

Pentyl 62 0.7920 108.2 0.8820 129.6 1.2220 155740 1.5220

Neopentyl 84.4 0.8720 105 1.2020 127 deca 1.5313

CH2"CH9 272 0.6826 213.9 0.9120 16 1.5214 56 2.0420

CH2"CHCH29 23 45 0.9420 70 1.4020 102–103 1.8422

C6H59 85 1.0220 132 1.1020 155 1.5220 189 1.8220

C6H5CH29 140 1.0225 179 1.1025 201 1.4422 9310 1.7325

a Decomposes is abbreviated dec.

6.3 NUCLEOPHILIC SUBSTITUTION REACTIONS

There are many reactions of the general type shown here.

2NuC 1 R9 aXC !!© R9Nu 1 2CaXCNucleophile Alkyl

halide(substrate)

Product Halide ion

Following are some examples:

2 aHaOC 1 CH 9 CCl3 !!© 2aCH 9 aOH 1 C CCl3

2 aCH aOC 1 CH CH 9 CBr3 3 2 !!© 2aCH CH 9 aOCH 1 C CBr3 2 3

2 aCaIC 1 CH CH CH 9 CCl3 2 2 !!© 2aCH CH CH 9aIC 1 C CCl3 2 2

In this type of reaction a nucleophile, a species with an unshared electron pair, reactswith an alkyl halide (called the substrate) by replacing the halogen substituent. A substi-tution reaction takes place and the halogen substituent, called the leaving group, departsas a halide ion. Because the substitution reaction is initiated by a nucleophile, it is calleda nucleophilic substitution reaction.

In nucleophilic substitution reactions the carbon–halogen bond of the substrate under-goes heterolysis, and the unshared pair of the nucleophile is used to form a new bond tothe carbon atom:

Leaving group

NucleophileR 1 X

Heterolysisoccurshere.

1Nu Nu2 X 2R

One of the questions we shall want to address later in this chapter is this: When doesthe carbon–halogen bond break? Does it break at the same time that the new bond betweenthe nucleophile and the carbon forms?

d2 d22 2© ©NuC 1 RCaXC !! NuªRª aXC !! NuCR 1 CaXC

Or does the carbon–halogen bond break first?

+ 2©RCaXC !! R 1 CaXC

And then2 + ©NuC 1 R !! NuCR

We shall find that the answer depends primarily on the structure of the alkyl halide.

6.4 NUCLEOPHILES

A nucleophile is a reagent that seeks a positive center. (The word nucleophile comes fromnucleus, the positive part of an atom, plus -phile from the Green word philos meaning tolove.) When a nucleophile reacts with an alkyl halide, the positive center that the nucle-ophile seeks is the carbon atom that bears the halogen atom. This carbon atom carries apartial positive charge because the electronegative halogen pulls the electrons of the car-bon–halogen bond in its direction (see Section 2.4).

X

The electronegative halogenpolarizes the C9X bond.

This is the positivecenter that the

nucleophile seeks.

Cd2d1

232 Chapter 6 / Ionic Reactions—Nucleophilic Substitution and Elimination Reactions of Alkyl Halides


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In Section 6.15 we shall seeexamples of biological nu-cleophilic substitution.

In color-coded reactions ofthis chapter, we will usered to indicate a nucleo-phile and blue to indicate aleaving group.

You may wish to reviewSection 3.2C, “OppositeCharges Attract.”

A nucleophile is any negative ion or any neutral molecule that has at least one unsharedelectron pair. For example, both hydroxide ions and water molecules can act as nucleo-philes by reacting with alkyl halides to produce alcohols.

H2O

H3O1

2 21 1H9O H9O9RR9X X

211 X

1 21 X

General Reaction for Nucleophilic Substitution of an Alkyl Halide by Hydroxide Ion

General Reaction for Nucleophilic Substitution of an Alkyl Halide by Water

H

Nucleophile

Nucleophile Alkylhalide

Alkyloxoniumion

1H9O

H

H9O9RR9X

Alkyl halide Alcohol Leaving group

H9O9R

In this last reaction the first product is an alkyloxonium ion, 1

H

R9O9H, which thenloses a proton to a water molecule to form an alcohol.

Problem 6.1Write the following as net ionic reactions and designate the nucleophile, substrate,and leaving group in each reaction.(a) CH3I 1 CH3CH2ONa !!© CH3OCH2CH3 1 NaI(b) NaI 1 CH3CH2Br !!© CH3CH2I 1 NaBr(c) + 2©2 CH OH 1 (CH ) CCl !! (CH ) COCH 1 CH OH 1 Cl3 3 3 3 3 3 3 2

(d) CH3CH2CH2Br 1 NaCN !!© CH3CH2CH2CN 1 NaBr(e) C6H5CH2Br 1 2 NH3 !!© C6H5CH2NH2 1 NH4Br

6.5 LEAVING GROUPS

Alkyl halides are not the only substances that can act as substrates in nucleophilic substi-tution reactions. We shall see later that other compounds can also react in the same way.To be reactive—that is, to be able to act as the substrate in a nucleophilic substitutionreaction—a molecule must have a good leaving group. In alkyl halides the leaving groupis the halogen substituent—it leaves as a halide ion. To be a good leaving group thesubstituent must be able to leave as a relatively stable, weakly basic molecule or ion.(We shall see why in Section 6.14E.) Because halide ions are relatively stable and are veryweak bases, they are good leaving groups. Other groups can function as good leavinggroups as well. We can write more general equations for nucleophilic substitution reactionsusing L to represent a leaving group:

2 2©NuC 1 R9L !! R9Nu 1 CL

or

1 2©NuC 1 R9L !! R9Nu 1 CL

6.5 Leaving Groups 233


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Specific Examples

2 2©a aHaOC 1 CH 9 C !! CH 9 aOH 1 C CCl Cl3 3

1 2©a aH NC 1 CH 9 C !! CH 9NH 1 C CBr Br3 3 3 3

Later we shall also see reactions where the substrate bears a formal positive charge anda reaction like the following takes place:

1 1©NuC 1 R9L !! R9Nu 1 CL

In this case, when the leaving group departs with an electron pair, its formal charge goesto zero.

Specific Example

H

1CH39O 1

H

CH39O9H 1

H

O9H1

H

CH39O9CH3

Nucleophilic substitution reactions are more understandable and useful if we know some-thing about their mechanisms. How does the nucleophile replace the leaving group? Doesthe reaction take place in one step, or is more than one step involved? If more than onestep is involved, what kinds of intermediates are formed? Which steps are fast and whichare slow? In order to answer these questions we need to know something about the ratesof chemical reactions.

6.6 KINETICS OF A NUCLEOPHILICSUBSTITUTION REACTION: AN SN2REACTION

To understand how the rate of a reaction might be measured let us consider an actualexample: the reaction that takes place between methyl chloride and hydroxide ion in aque-ous solution.

607C2 2©CH 9Cl 1 OH CH 9OH 1 Cl993 3H O2

Although methyl chloride is not highly soluble in water, it is soluble enough to allowus to carry out our kinetic study. The presence of hydroxide ion in the aqueous solutioncan be assured by simply adding sodium hydroxide. We carry out the reaction at a specifictemperature because reaction rates are known to be temperature dependent (Section 6.8).



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The rate of the reaction can be determined experimentally by measuring the rate atwhich methyl chloride or hydroxide ion disappears from the solution, or the rate at whichmethanol or chloride ion appears in the solution. We can make any of these measurementsby withdrawing a small sample from the reaction mixture soon after the reaction beginsand analyzing it for the concentrations of CH3Cl or OH2 and CH3OH or Cl2. We areinterested in what are called initial rates, because as time passes the concentrations of thereactants change. Since we also know the initial concentrations of reactants (because wemeasured them when we made up the solution), it will be easy to calculate the rate atwhich the reactants are disappearing from the solution or the products are appearing in thesolution.

We perform several such experiments keeping the temperature the same, but varyingthe initial concentrations of the reactants. The results that we might get are shown inTable 6.3.

Notice that the experiments show that the rate depends on the concentration of methylchloride and on the concentration of hydroxide ion. When we doubled the concentration

of methyl chloride in experiment 2, the rate doubled. When we doubled the concentrationof hydroxide ion in experiment 3, the rate doubled. When we doubled both concentrationsin experiment 4 the rate increased by a factor of four.

We can express these results as a proportionality,

2Rate ~ [CH Cl][OH ]3

and this proportionality can be expressed as an equation through the introduction of aproportionality constant (k) called the rate constant:

2Rate 5 k[CH Cl][OH ]3

For this reaction at this temperature we find that (Verify this24 21 21k 5 4.9 3 10 L mol s .for yourself by doing the calculation.)

This reaction is said to be second order overall.* It is reasonable to conclude, therefore,that for the reaction to take place a hydroxide ion and a methyl chloride molecule mustcollide. We also say that the reaction is bimolecular. (By bimolecular we mean that twospecies are involved in the step whose rate is being measured.) We call this kind of reactionan SN2 reaction, meaning Substitution, Nucleophilic, bimolecular.

6.7 A MA MECHANISM FOR THE SN2 REACTION

A modern mechanism for the SN2 reaction—one based on ideas proposed by Edward D.Hughes and Sir Christopher Ingold in 1937—is outlined below.

Nu C L +– –

LNu C

Bonding orbital

Antibonding orbital

According to this mechanism, the nucleophile approaches the carbon bearing the leavinggroup from the back side, that is, from the side directly opposite the leaving group. The

6.7 A Mechanism for the SN2 Reaction 235


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Table 6.3 Rate Study of Reaction of CH3Cl withOH2 at 607C

ExperimentNumber

Initial[CH3Cl]

Initial[OH2]

Initial Rate(mol L21 s21)

1 0.0010 1.0 4.9 3 1027

2 0.0020 1.0 9.8 3 1027

3 0.0010 2.0 9.8 3 1027

4 0.0020 2.0 19.6 3 1027

* In general the overall order of a reaction is equal to the sum of the exponents a and b in the rate equation

a bRate 5 k[A] [B]

If in some other reaction, for example, we found that

2Rate 5 k[A] [B]

then we would say that the reaction is second order with respect to [A], first order with respect to [B], and thirdorder overall.

Ingold (top) and Hughes ofthe University College,London, were pioneers inthis field. Their work pro-vided the foundation onwhich our modern under-standing of nucleophilicsubstitution and elimina-tion is built.

orbital that contains the electron pair of the nucleophile begins to overlap with an empty(antibonding) orbital of the carbon atom bearing the leaving group. As the reaction pro-gresses the bond between the nucleophile and the carbon atom strengthens, and the bondbetween the carbon atom and the leaving group weakens. As this happens, the carbonatom has its configuration turned inside out, it becomes inverted,* and the leaving groupis pushed away. The formation of the bond between the nucleophile and the carbon atomprovides most of the energy necessary to break the bond between the carbon atom and theleaving group. We can represent this mechanism with methyl chloride and hydroxide ionas shown in the “A Mechanism for the SN2 Reaction” box below.

The Hughes– Ingold mechanism for the SN2 reaction involves only one step. There areno intermediates. The reaction proceeds through the formation of an unstable arrangementof atoms called the transition state.

The transition state is a fleeting arrangement of the atoms in which the nucleophile andthe leaving group are both partially bonded to the carbon atom undergoing attack. Becausethe transition state involves both the nucleophile (e.g., a hydroxide ion) and the substrate(e.g., a molecule of methyl chloride), this mechanism accounts for the second-order re-action kinetics that we observe. (Because bond formation and bond breaking occur simul-taneously in a single transition state, the SN2 reaction is an example of what is called aconcerted reaction.)

The transition state has an extremely brief existance. It lasts only as long as the timerequired for one molecular vibration, about The structure and energy of the tran-21210 s.sition state are highly important aspects of any chemical reaction. We shall, therefore,examine this subject further in Section 6.8.

A Mechanism for the SN2 Reaction............Reaction:2 2©HO 1 CH Cl !! CH OH 1 Cl3 3

Mechanism:

O2

Transition state

H CCl

H

H

Cd2 d2

1d1H

H

OH Cl

HH ‡

OH

H

CH

H

Cl

The negative hydroxideion pushes a pair of

electrons into the partiallypositive carbon from

the back side. Thechlorine begins to move

away with the pair ofelectrons that have

bonded it to the carbon.

In the transition state, abond between oxygen

and carbon is partiallyformed and the bondbetween carbon andchlorine is partially

broken. Theconfiguration of the

carbon atom begins toinvert.

Now the bond betweenthe oxygen and carbon

has formed and thechloride ion hasdeparted. The

configuration of thecarbon has inverted.

d2 2



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* Considerable evidence had appeared in the years prior to Hughes and Ingold’s 1937 publication indicating thatin reactions like this an inversion of configuration of the carbon bearing the leaving group takes place. The firstobservation of such an inversion was made by the Latvian chemist Paul Walden in 1896, and such inversionsare called Walden inversions in his honor. We shall study this aspect of the SN2 reaction further in Section 6.9.

Orbitals in SN2 Reactions

SN2 Mechanism

6.8 TRANSITION STATE THEORY: FREE-ENERGYDIAGRAMS

A reaction that proceeds with a negative free-energy change is said to be exergonic; onethat proceeds with a positive free-energy change is said to be endergonic. The reactionbetween methyl chloride and hydroxide ion in aqueous solution is highly exergonic;at 607C (The reaction is also exothermic,21(333 K), DG7 5 2100 kJ mol . DH7 5

21275 kJ mol .)

2 2 21©CH 9Cl 1 OH !! CH 9OH 1 Cl DG7 5 2100 kJ mol3 3

The equilibrium constant for the reaction is extremely large:

DG7 5 22.303 RT log Keq

2DG7log K 5eq 2.303 RT

212 (2100 kJ mol )log K 5eq 21 212.303 3 0.00831 kJ K mol 3 333 K

log K 5 15.7eq

15K 5 5.0 3 10eq

An equilibrium constant as large as this means that the reaction goes to completion.Because the free-energy change is negative, we can say that in energy terms the reaction

goes downhill. The products of the reaction are at a lower level of free energy than thereactants.

However, considerable experimental evidence exists showing that if covalent bondsare broken in a reaction, the reactants must go up an energy hill first, before they cango downhill. This will be true even if the reaction is exergonic.

We can represent this graphically by plotting the free energy of the reacting particlesagainst the reaction coordinate. Such a graph is given in Fig. 6.1. We have chosen as ourexample a generalized SN2 reaction.

The reaction coordinate is a quantity that measures the progress of the reaction. Itrepresents the changes in bond orders and bond distances that must take place as thereactants are converted to products. In this instance the Y—Z distance could be used asthe reaction coordinate because as the reaction progresses the Y—Z distance becomeslonger.

In our illustration (Fig. 6.1), we can see that an energy barrier exists between thereactants and products. The height of this barrier (in kilojoules per mole) above the levelof reactants is called the the free energy of activation, ‡DG .

6.8 Transition State Theory: Free-Energy Diagrams 237


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Free

ene

rgy

Reaction coordinate

Free-energychange

Reactants

Products

∆G°

X:– + Y Z

X Y + :Z–

∆G‡Free energy of

activation

Transition stateX Y Zδ – δ –

Figure 6.1 A free-energy diagram for ahypothetical SN2 reaction that takes placewith a negative DG7.

The top of the energy hill corresponds to the transition state. The difference in freeenergy between the reactants and the transition state is the free energy of activation,

The difference in free energy between the reactants and products is the free-energy‡DG .change for the reaction, For our example, the free-energy level of the products is7DG .lower than that of the reactants. In terms of our analogy, we can say that the reactants inone energy valley must traverse an energy hill (the transition state) in order to reach thelower energy valley of the products.

If a reaction in which covalent bonds are broken proceeds with a positive free-energychange (Fig. 6.2), there will still be a free energy of activation. That is, if the productshave greater free energy than reactants, the transition state will have a free energy evenhigher. will be larger than DG7.) In other words, in the uphill (endergonic) reaction‡(DGan even larger energy hill lies between the reactants in one valley and the products in ahigher one.

Just as the overall free-energy change for a reaction contains enthalpy and entropy com-ponents (Section 3.9):

DG7 5 DH7 2 TDS7

the free energy of activation has similar components:

‡ ‡ ‡DG 5 DH 2 TDS

The enthalpy of activation is the difference in bond energies between the reactants‡(DH )and the transition state. It is, in effect, the energy necessary to bring the reactants closetogether and to bring about the partial breaking of bonds that must happen in the transitionstate. Some of this energy may be furnished by the bonds that are partially formed. Theentropy of activation is the difference in entropy between the reactants and the‡(DS )transition state. Most reactions require the reactants to come together with a particularorientation. (Consider, e.g., the specific orientation required in the SN2 reaction.) Thisrequirement for a particular orientation means that the transition state must be more orderedthan the reactants and that will be negative. The more highly ordered the transition‡DSstate, the more negative will be. When a three-dimensional plot of free energy versus‡DSthe reaction coordinate is made, the transition state is found to resemble a mountain passor col (Fig. 6.3) rather than the top of an energy hill as we have shown in Figs. 6.1 and6.2. (A plot such as that seen in Figs. 6.1 and 6.2 is simply a two-dimensional slice throughthe three-dimensional energy surface for the reaction.) That is, the reactants and productsappear to be separated by an energy barrier resembling a mountain range. While an infinitenumber of possible routes lead from reactants to products, the transition state lies at the



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top of txtbase of txtFigure 6.2 A free-energy diagram for a

hypothetical reaction with a positive free-en-ergy change.

Free

ene

rgy

Reaction coordinate

Transition state

∆G°

∆G‡

X Y Zδ – δ –

X Y + :Z–

X: – + Y Z

Reactants

Products

top of the route that requires the lowest energy climb. Whether the pass is a wide or narrowone depends on A wide pass means that there is a relatively large number of orien-‡DS .tations of reactants that allow a reaction to take place. A narrow pass means just theopposite.

The existence of an activation energy explains why most chemical reactions‡(DG )occur much more rapidly at higher temperatures. For many reactions taking place nearroom temperature, a 107C increase in temperature will cause the reaction rate to double.

This dramatic increase in reaction rate results from a large increase in the number ofcollisions between reactants that together have sufficient energy to surmount the barrier atthe higher temperature. The kinetic energies of molecules at a given temperature are notall the same. Figure 6.4 shows the distribution of energies brought to collisions at twotemperatures (that do not differ greatly), labeled T1 and T2 . Because of the way energiesare distributed at different temperatures (as indicated by the shapes of the curves), increas-ing the temperature by only a small amount causes a large increase in the number ofcollisions with larger energies. In Fig. 6.4 we have designated a particular minimal free

6.8 Transition State Theory: Free-Energy Diagrams 239


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(High-energy region)(High-energy region)

Transitionstate

Reactants (low-energy region)

Figure 6.3 Mountain pass orcol analogy for the transitionstate. (Adapted with permissionfrom Leffler, J. E.; Grunwald, E.Rates and Equilibria of OrganicReactions; Wiley: New York,1963; p 65.)

∆G‡

Frac

tion

of

colli

sion

sw

ith

a gi

ven

ener

gy

Energy

T1

T2

Figure 6.4 The distribution of energies attwo different temperatures, T1 and T2

The number of collisions with ener-(T . T ).2 1

gies greater than the free energy of activationis indicated by the appropriately shaded areaunder each curve.

energy as being required to bring about a reaction between colliding molecules. The num-ber of collisions having sufficient energy to allow reaction to take place at a given tem-perature is proportional to the area under the portion of the curve that represents freeenergies greater than or equal to At the lower temperature (T1) this number is rela-‡DG .tively small. At the higher temperature (T2), however, the number of collisions that takeplace with enough energy to react is very much larger. Consequently, a modest temperatureincrease produces a large increase in the number of collisions with energy sufficient tolead to a reaction.

There is also an important relationship between the rate of a reaction and the magnitudeof the free energy of activation. The relationship between the rate constant (k) and is‡DGan exponential one.

‡2DG /RTk 5 k e0

In this equation, e is 2.718, the base of natural logarithms, and k0 is the absolute rateconstant, which equals the rate at which all transition states proceed to products. At 257C,

Because of this exponential relationship, a reaction with a lower12 21k 5 6.2 3 10 s .0

free energy of activation will occur very much faster than a reaction with a higherone.

Generally speaking, if a reaction has a less than mol21, it will take place‡DG 84 kJreadily at room temperature or below. If is greater than mol21, heating will be‡DG 84 kJrequired to cause the reaction to occur at a reasonable rate.

A free-energy diagram for the reaction of methyl chloride with hydroxide ion is shownin Fig. 6.5. At 607C, which means that at this temperature, the‡ 21DG 5 103 kJ mol ,reaction reaches completion in a matter of a few hours.

6.9 THE STEREOCHEMISTRY OF SN2 REACTIONS

As we learned earlier (Section 6.7), in an SN2 reaction the nucleophile attacks from theback side, that is, from the side directly opposite the leaving group. This mode ofattack (see below) causes a change in the configuration of the carbon atom that is thetarget of nucleophilic attack. (The configuration of an atom is the particular arrangementof groups around that atom in space, Section 5.6.) As the displacement takes place, the



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top of txtbase of txtFigure 6.5 A free-energy diagram for

the reaction of methyl chloride with hy-droxide ion at 607C.

Free

ene

rgy

Reaction coordinate

Free energy ofactivation

Free-energychange

∆G° =– 100 kJ mol–1

∆G ‡ = 103kJ mol–1

Transition state

Reactants

Products

HO– + CH3Cl

CH3OH + Cl–

HO CH3 Clδ – δ –

SN2 Collision

SN2 Stereochemistry.

configuration of the carbon atom under attack inverts—it is turned inside out in muchthe same way that an umbrella is turned inside out, or inverts, when caught in a strongwind.

HO 2 C C

H

HH

H

C 1 Cl2

HH

H

Cl HO HO

An inversion of configuration

Cld2 d2

‡

HH

With a molecule such as methyl chloride, however, there is no way to prove that attackby the nucleophile inverts the configuration of the carbon atom because one form of methylchloride is identical to its inverted form. With a cyclic molecule such as cis-1-chloro-3-methylcyclopentane, however, we can observe the results of a configuration inversion.When cis-1-chloro-3-methylcyclopentane reacts with hydroxide ion in an SN2 reaction theproduct is trans-3-methylcyclopentanol. The hydroxide ion ends up being bonded on theopposite side of the ring from the chlorine it replaces:

cis-1-Chloro-3-methylcyclopentane

SN2OH2 1 Cl 21


HH

H3C Cl

trans-3-Methylcyclopentanol

H

H

H3C

OH

Presumably, the transition state for this reaction is like that shown here.

H3CH

Leaving group departsfrom the top side.

Nucleophile attacksfrom the bottom side.

H

d2OH

d2Cl

Problem 6.2Use chair conformational structures (Sect. 4.12) and show the nucleophilic substitu-tion reaction that would take place when trans-1-bromo-4-tert-butylcyclohexanereacts with iodide ion. (Show the most stable conformation of the reactant and theproduct.)

We can also observe an inversion of configuration with an acyclic molecule when theSN2 reaction takes place at a stereocenter. Here, too, we find that SN2 reactions alwayslead to inversion of configuration.

6.9 The Stereochemistry of SN2 Reactions 241


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SN2 Inversion

Transition state for an SN2reaction

A compound that contains one stereocenter, and therefore exists as a pair of enantio-mers, is 2-bromooctane. These enantiomers have been obtained separately and are knownto have the configurations and rotations shown here.

HC

Br

C6H13

CH3

C

C6H13

CH3

(R)-(2)-2-Bromooctane]D

25 5 234.25°(S)-(1)-2-Bromooctane

]D25 5 134.25°

Br H

a[ a[

The alcohol 2-octanol is also chiral. The configurations and rotations of the 2-octanolenantiomers have also been determined:

HC

OH

C6H13

CH3

C

C6H13

CH3

(R)-(2)-2-Octanol

D25 5 29.90°

(S)-(1)-2-Octanol

D25 5 19.90°

HO H

[ []a ]a

When (R)-(2)-2-bromooctane reacts with sodium hydroxide, the only substitution prod-uct that is obtained from the reaction is (S)-(+)-2-octanol.

The Stereochemistry of an SN2 Reaction............ This reaction is SN2 and takes place with complete inversion of configuration.

d2 d2

(R)-(2a

)-2-Bromooctane25 5 234.25°

Enantiomeric purity 5 100%[ D]D] a[

(S)-(1)-2-Octanol25 5 19.90°

Enantiomeric purity 5 100%

HO 2 C C

HH

C 1 Br2

HBr HO HO


Br

‡CH3 CH3

C6H13 C6H13C6H13

CH3

Problem 6.3 ➤ SN2 reactions that involve breaking a bond to a stereocenter can be used to re-late configurations of molecules because the stereochemistry of the reaction isknown. (a) Illustrate how this is true by assigning configurations to the 2-chloro-butane enantiomers based on the following data. [The configuration of (2)-2-butanolis given in Section 5.7C.]

(1)-2-Chlorobutane2OH

©99S 2N(2)-2-Butanol

25[a] 5 136.007D

Enantiomerically pure

25[a] 5 213.527D

Enantiomerically pure



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(b) When optically pure (1)-2-chlorobutane is allowed to react with potassium iodidein acetone in an SN2 reaction, the 2-iodobutane that is produced has a minus rotation.What is the configuration of (2)-2-iodobutane? Of (1)-2-iodobutane?

6.10 THE REACTION OF TERT-BUTYLCHLORIDE WITH HYDROXIDE ION:AN SN1 RREACTION

When tert-butyl chloride reacts with sodium hydroxide in a mixture of water and ace-tone, the kinetic results are quite different. The rate of formation of tert-butyl alcohol isdependent on the concentration of tert-butyl chloride, but it is independent of the con-centration of hydroxide ion. Doubling the tert-butyl chloride concentration doubles therate of the reaction, but changing the hydroxide ion concentration (within limits) hasno appreciable effect. tert-Butyl chloride reacts by substitution at virtually the samerate in pure water (where the hydroxide ion is as it does in 0.05M aqueous so-2710 M)dium hydroxide (where the hydroxide ion concentration is 500,000 times larger). (Weshall see in Section 6.11 that the important nucleophile in this reaction is a moleculeof water.)

Thus the rate equation for this substitution reaction is first order with respect to tert-butyl chloride and first order overall.

acetone2 2©(CH ) C9Cl 1 OH (CH ) C9OH 1 Cl993 3 3 3H O2

Rate ~ [(CH ) CCl]3 3

Rate 5 k[(CH ) CCl]3 3

We can conclude, therefore, that hydroxide ions do not participate in the transition stateof the step that controls the rate of the reaction, and that only molecules of tert-butylchloride are involved. This reaction is said to be unimolecular. We call this type ofreaction an SN1 reaction (Substitution, Nucleophilic, unimolecular).

How can we explain an SN1 reaction in terms of a mechanism? To do so we shall needto consider the possibility that the mechanism involves more than one step. But what kindof kinetic results should we expect from a multistep reaction? Let us consider this pointfurther.

6.10A Multistep Reactions and the Rate-Determining Step

If a reaction takes place in a series of steps, and if one step is intrinsically slower than allthe others, then the rate of the overall reaction will be essentially the same as the rate ofthis slow step. This slow step, consequently, is called the rate-limiting step or the rate-determining step.

Consider a multistep reaction such as the following:

slow©Step 1 Reactant intermediate 199

fast©Step 2 Intermediate 1 intermediate 299

fast©Step 3 Intermediate 2 product99

6.10 The Reaction of tert-Butyl Chloride with Hydroxide Ion: An SN1 Reaction 243


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When we say that the first step in this example is intrinsically slow, we mean that therate constant for step 1 is very much smaller than the rate constant for step 2 or forstep 3:

Step 1 Rate 5 k [reactant]1

Step 2 Rate 5 k [intermediate 1]2

Step 3 Rate 5 k [intermediate 2]3

k1 ,, k2 or k3

When we say that steps 2 and 3 are fast, we mean that because their rate constants arelarger, they could (in theory) take place rapidly if the concentrations of the two interme-diates ever became high. In actuality, the concentrations of the intermediates are alwaysvery small because of the slowness of step 1, and steps 2 and 3 actually occur at the samerate as step 1.

An analogy may help clarify this. Imagine an hourglass modified in the way shown inFig. 6.6. The opening between the top chamber and the one just below is considerablysmaller than the other two. The overall rate at which sand falls from the top to the bottomof the hourglass is limited by the rate at which sand passes through this small orifice. Thisstep, in the passage of sand, is analogous to the rate-determining step of the multistepreaction.



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Figure 6.6 A modified hourglass that servesas an analogy for a multistep reaction. Theoverall rate is limited by the rate of the slowstep. Narrow

opening(rate

determining)

Largeopenings

Sand

Sand

Reactant

Intermediate 1

Intermediate 2

Product

Slow(ratedetermining)

Fast

Fast

6.11 A MECHANISM FOR THE SN1 REACTION

The mechanism for the reaction of tert-butyl chloride with water (Section 6.10) apparentlyinvolves three steps. Two distinct intermediates are formed. The first step is the slowstep—it is the rate-determining step. In it a molecule of tert-butyl chloride ionizes andbecomes a tert-butyl cation and a chloride ion. Carbocation formation in general takesplace slowly because it is usually a highly endothermic process and is uphill in terms offree energy.

A Mechanism for the SN1 Reaction............ Reaction:1 2©(CH ) CCl 1 2 H O !! (CH ) COH 1 H O 1 Cl3 3 2 3 3 3

Mechanism:

Step 1

H3C C

CH3

Clslow

H2O

CH3

H3C C1

CH3

CH3

1 Cl2

Aided by the polarsolvent a chlorinedeparts with theelectron pair thatbonded it to the

carbon.

This slow stepproduces the

relatively stable 3°carbocation and a

chloride ion. Althoughnot shown here, the

ions are solvated (andstabilized) by water

molecules.

Step 2

H3C 1 1

A water molecule actingas a Lewis base donatesan electron pair to thecarbocation (a Lewisacid). This gives thecationic carbon eight

electrons.

O H

H

fastH3C

CH3

CH3

C

The product is atert-butyloxonium ion

(or protonatedtert-butyl alcohol).

O H

H

1

Step 3

H3C

CH3

CH3

C

A water molecule acting asa Brønsted base accepts a

proton from thetert-butyloxonium ion.

O H

H

1 1 O H

H

fastH3C

CH3

CH3

C

The products are tert-butylalcohol and a hydronium ion.

O

H

O H

H

1H1

C

CH3

CH3

In the second step the intermediate tert-butyl cation reacts rapidly with water to producea tert-butyloxonium ion (another intermediate); this, in the third step, rapidly transfers aproton to a molecule of water producing tert-butyl alcohol.

The first step requires heterolytic cleavage of the carbon–chlorine bond. Because noother bonds are formed in this step, it should be highly endothermic and it should have ahigh free energy of activation. That it takes place at all is largely because of the ionizing

6.11 A Mechanism for the SN1 Reaction 245


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ability of the solvent, water. Experiments indicate that in the gas phase (i.e., in the absenceof a solvent), the free energy of activation is about 630 kJ In aqueous solution,21mol !however, the free energy of activation is much lower—about Water mole-2184 kJ mol .cules surround and stabilize the cation and anion that are produced (cf. Section 2.14E).

Even though the tert-butyl cation produced in step 1 is stabilized by solvation, it is stilla highly reactive species. Almost immediately after it is formed, it reacts with one of thesurrounding water molecules to form the tert-butyloxonium ion, (CH3)3COH2

1. (It mayalso occasionally react with a hydroxide ion, but water molecules are far more plentiful.)

A free-energy diagram for the SN1 reaction of tert-butyl chloride and water is given inFig. 6.7.

The important transition state for the SN1 reaction is the transition state of the rate-determining step [TS(1)]. In it the carbon–chlorine bond of tert-butyl chloride is largelybroken and ions are beginning to develop:

CH39C Cld2d1

CH3

CH3

The solvent (water) stabilizes these developing ions by solvation.

6.12 CARBOCATIONS

Beginning in the 1920s much evidence began to accumulate implicating simple alkylcations as intermediates in a variety of ionic reactions. However, because alkyl cations arehighly unstable and highly reactive they were, in all instances studied before 1962, veryshort-lived, transient species that could not be observed directly.* However, in 1962George A. Olah (now at the University of Southern California) and co-workers publishedthe first of a series of papers describing experiments in which alkyl cations were preparedin an environment in which they were reasonably stable and in which they could beobserved by a number of spectroscopic techniques.



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top of txtbase of txtFigure 6.7 A free-

energy diagram for the SN1reaction of tert-butyl chlo-ride with water. The freeenergy of activation for thefirst step, is much‡DG (1),larger than or‡DG (2)

TS(1) represents‡DG (3).transition state (1), andso on.

Step 1 Step 2 Step 3

Free

ene

rgy

Reaction coordinate

TS(1)

TS(2)

TS(3)

∆G‡ (2)

∆G‡ (3)

(CH3)3CCl+ 2 H2O

(CH3)3C+ + Cl–

+ 2 H2O

(CH3)3COH+ Cl– + H3O+

(CH3)3COH2+ Cl– + H2O

+

∆G‡ (1)

SN1 Mechanism

* As we shall learn later, carbocations bearing aromatic groups can be much more stable; one of these had beenstudied as early as 1901.

6.12A The Structure of Carbocations

Considerable experimental evidence indicates that the structure of carbocations is trigonalplanar like that of BF3 (Section 1.16D). Just as the trigonal planar structure of BF3 canbe accounted for on the basis of sp2 hybridization so, too (Fig. 6.8), can the trigonal planarstructure of carbocations.

The central carbon atom in a carbocation is electron deficient; it has only six electronsin its outside energy level. In our model (Fig. 6.8) these six electrons are used to formsigma covalent bonds to hydrogen atoms (or to alkyl groups). The p orbital contains noelectrons.

6.12B The Relative Stabilities of Carbocations

A large body of experimental evidence indicates that the relative stabilities of carbocationsare related to the number of alkyl groups attached to the positively charged trivalent carbonatom. Tertiary carbocations are the most stable, and the methyl cation is the least stable.The overall order of stability is as follows:

.

.

R9C1

R

R

3°(moststable)

.

.

R9C1

2°

.

.

R9C1

1°

H9C1

Methyl(least

stable)

H

H

H

HH

R

This order of stability of carbocations can be explained on the basis of a law of physicsthat states that a charged system is stabilized when the charge is dispersed or delocalized.Alkyl groups, when compared to hydrogen atoms, are electron releasing. This meansthat alkyl groups will shift electron density toward a positive charge. Through electron re-lease, alkyl groups attached to the positive carbon atom of a carbocation delocalize thepositive charge. In doing so, the attached alkyl groups assume part of the positive chargethemselves and thus stabilize the carbocation. We can see how this occurs by inspectingFig. 6.9.

6.12 Carbocations 247


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Knowledge of carbocationstructure is an importanttool for understanding avariety of reaction pro-cesses.

CH3

H3CH

H

H

H3C

C+

sp2–sp3

bonds

sp2–s bonds

(a) (b)

Vacant p orbitals

C+

Figure 6.8 (a) A stylized orbital structure of the methyl cation. The bonds are sigma (s)bonds formed by overlap of the carbon atom’s three orbitals with the 1s orbitals of the2sphydrogen atoms. The p orbital is vacant. (b) A dashed line–wedge representation of the tert-butyl cation. The bonds between carbon atoms are formed by overlap of sp3 orbitals of themethyl groups with sp2 orbitals of the central carbon atom.

Relative CarbocationStability.

In the tert-butyl cation (see below) three electron-releasing methyl groups surround thecentral carbon atom and assist in delocalizing the positive charge. In the isopropyl cationthere are only two attached methyl groups that can serve to delocalize the charge. In theethyl cation there is only one attached methyl group, and in the methyl cation there is noneat all. As a result, the delocalization of charge and the order of stability of the carbocationsparallel the number of attached methyl groups.

tert-Butyl cation(3°) (most stable)

ismorestablethan

H9C1CH3

CH3

1d

1d

Isopropyl cation(2°)

ismorestablethan

CH3

CH3

Cd11d

1d

CH3

Cd1

1d

Ethyl cation(1°)

Methyl cation(least stable)

ismorestablethan

CH3 Cd11d

H H

H

H

H

The relative stability of carbocations is 37 . 27 . 17 . methyl. This trend is alsoreadily seen in electrostatic potential maps for these carbocations (Fig. 6.10).



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Vacant p orbital

Orbitals overlap hereH

H

H

C C+

2e–

Figure 6.9 How a methyl group helps stabilize the positive charge of a carbocation. Elec-tron density from one of the carbon–hydrogen sigma bonds of the methyl group flows intothe vacant p orbital of the carbocation because the orbitals can partly overlap. Shifting elec-tron density in this way makes the sp2-hybridized carbon of the carbocation somewhat lesspositive, and the hydrogens of the methyl group assume some of the positive charge. Delocal-ization (dispersal) of the charge in this way leads to greater stability. This interaction of abond orbital with a p orbital is called hyperconjugation.

Figure 6.10 Electrostaticpotential maps for (a) tert-butyl (37), (b) isopropyl (27),(c) ethyl (17), and (d) methylcarbocations show the trendfrom greater to lesser delocali-zation (stabilization) of thepositive charge in these struc-tures. Less blue color indicatesgreater delocalization of thepositive charge. (The struc-tures are mapped on the samescale of electrostatic potentialto allow direct comparison.)

(a) tert-Butyl (3°) (b) Isopropyl (2°) (c) Ethyl (1°) (d) Methyl

6.13 THE STEREOCHEMISTRY OF SN1REACTIONS

Because the carbocation formed in the first step of an SN1 reaction has a trigonal planarstructure (Section 6.12A), when it reacts with a nucleophile, it may do so from either thefront side or the back side (see below). With the tert-butyl cation this makes no differencebecause the same product is formed by either mode of attack.

front side

attack

back side

attack

Same product

C

CH3

CH3CH3

H2OH2O1 C

CH3

H3CCH3

OH2

1C

CH3

CH3

OH2

H3C

1

With some cations, however, different products arise from the two reaction possibilities.We shall study this point next.

6.13A Reactions That Involve Racemization

A reaction that transforms an optically active compound into a racemic form is said toproceed with racemization. If the original compound loses all of its optical activity in thecourse of the reaction, chemists describe the reaction as having taken place with completeracemization. If the original compound loses only part of its optical activity, as would bethe case if an enantiomer were only partially converted to a racemic form, then chemistsdescribe this as proceeding with partial racemization.

Racemization takes place whenever the reaction causes chiral molecules to be convertedto an achiral intermediate.

Examples of this type of reaction are SN1 reactions in which the leaving group departsfrom a stereocenter. These reactions almost always result in extensive and sometimescomplete racemization. For example, heating optically active (S)-3-bromo-3-methylhexanewith aqueous acetone results in the formation of 3-methyl-3-hexanol as a racemicform.

H2O

acetone

(S)-3-Bromo-3-methylhexane

(optically active)

CH3CH2

CH3CH2CH2

H3CC Br

(S)-3-Methyl-3-hexanol(optically inactive, a racemic form)

(R)-3-Methyl-3-hexanol

CH3CH2 CH2CH3

CH3CH2CH2 CH2CH2CH3

H3C CH3

C C1OH HO 1 HBr

The reason: The SN1 reaction proceeds through the formation of an intermediate car-bocation and the carbocation, because of its trigonal planar configuration, is achiral. Itreacts with water at equal rates from either side to form the enantiomers of 3-methyl-3-hexanol in equal amounts.

6.13 The Stereochemistry of SN1 Reactions 249


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SN1 Stereochemistry

The Stereochemistry of an SN1 Reaction............The carbocation hasa trigonal planar structureand is achiral.

2Br2

slowC9Br

CH3CH2CH2CH2CH2CH3

CH3CH2

H3CC1

CH2CH3H3C

Front side and back sideattack take place at equalrates, and the product isformed as a racemicmixture.

Enantiomers

Back sideattack

Front sideattack

fast fast

C9O1

CH3CH2CH2

CH3CH2

H3CCH2CH2CH3

C1

HOH HOH

H3O1

H3O1

HOH

H

H

O9C1

CH2CH2CH3

CH2CH3

CH3

H

H

A racemic mixture

C9O

CH3CH2CH2 1

1

CH3CH2

H3C H

O9C

CH2CH2CH3

CH2CH3

CH3H

CH2CH3H3C

The SN1 reaction of (S)-3-bromo-3-methylhexane proceeds with racemization becausethe intermediate carbocation is achiral and attack by the nucleophile can occur fromeither side.

Problem 6.4 ➤ Keeping in mind that carbocations have a trigonal planar structure, (a) write a struc-ture for the carbocation intermediate and (b) write structures for the alcohol (oralcohols) that you would expect from the following reaction:

H2O

SN1(CH3)3C

I

CH3

6.13B Solvolysis

The SN1 reaction of an alkyl halide with water is an example of solvolysis. A solvolysisis a nucleophilic substitution in which the nucleophile is a molecule of the solvent (solvent

cleavage by the solvent). Since the solvent in this instance is water, we could also+ lysis:call the reaction a hydrolysis. If the reaction had taken place in methanol we would callit a methanolysis.

Examples of Solvolysis

(CH3)3C9Br 1 H2O (CH3)3C9OH 1 HBr

O O

(CH3)3C9Cl 1 CH3OH

(CH3)3C9Cl 1 HCOH

(CH3)3C9OCH3 1 HCl

(CH3)3C9OCH 1 HCl



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These reactions all involve the initial formation of a carbocation and the subsequentreaction of that cation with a molecule of the solvent. In the last example the solvent isformic acid (HCO2H) and the following steps take place:

Step 1

Step 2

Step 3

(CH3)3C Cl

HH O

(CH3)3C1 1 Cl 2slow

(CH3)3

(CH3)3

C1 1 OH C

O

Hfast

2

Cl

1

C

CO

1

O

Hfast

HH O

(CH3)3

C

CO1

HO C HC9O9

1 H9Cl

(CH3)3C;HO C

O9C(CH3)3 O9C(CH3)3

Problem 6.5What product(s) would you expect from the methanolysis of the cyclohexane deriv-ative given as the reactant in Problem 6.4?

6.14 FACTORS AFFECTING THE RATES OF SN1AND SN2 REACTIONS

Now that we have an understanding of the mechanisms of SN2 and SN1 reactions, our nexttask is to explain why methyl chloride reacts by an SN2 mechanism and tert-butyl chlorideby an SN1 mechanism. We would also like to be able to predict which pathway—SN1 orSN2—would be followed by the reaction of any alkyl halide with any nucleophile undervarying conditions.

The answer to this kind of question is to be found in the relative rates of the reactionsthat occur. If a given alkyl halide and nucleophile react rapidly by an SN2 mechanism butslowly by an SN1 mechanism under a given set of conditions, then an SN2 pathway willbe followed by most of the molecules. On the other hand, another alkyl halide and anothernucleophile may react very slowly (or not at all) by an SN2 pathway. If they react rapidlyby an SN1 mechanism, then the reactants will follow an SN1 pathway.

Experiments have shown that a number of factors affect the relative rates of SN1 andSN2 reactions. The most important factors are

1. The structure of the substrate.2. The concentration and reactivity of the nucleophile (for bimolecular reactions only).3. The effect of the solvent.4. The nature of the leaving group.

6.14A The Effect of the Structure of the Substrate

SN2 Reactions Simple alkyl halides show the following general order of reactivity inSN2 reactions:

methyl . primary . secondary .. (tertiary—unreactive)

6.14 Factors Affecting the Rates of SN1 and SN2 Reactions 251


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SN2 Order of Reactivity.

Methyl halides react most rapidly and tertiary halides react so slowly as to be unreactiveby the SN2 mechanism. Table 6.4 gives the relative rates of typical SN2 reactions.

Neopentyl halides, even though they are primary halides, are very unreactive.

CH39C9CH29X

CH3

CH3

A neopentyl halide

The important factor behind this order of reactivity is a steric effect. A steric effect isan effect on relative rates caused by the space-filling properties of those parts of a moleculeattached at or near the reacting site. One kind of steric effect—the kind that is importanthere—is called steric hindrance. By this we mean that the spatial arrangement of theatoms or groups at or near the reacting site of a molecule hinders or retards a reaction.

For particles (molecules and ions) to react, their reactive centers must be able to comewithin bonding distance of each other. Although most molecules are reasonably flexible,very large and bulky groups can often hinder the formation of the required transition state.



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top of txtbase of txtTable 6.4 Relative Rates of Reactions of

Alkyl Halides in SN2 ReactionsSubstituent Compound Relative Rate

Methyl CH3X 3017 CH3CH2X 127 (CH3)2CHX 0.02

Neopentyl (CH3)3CCH2X 0.0000137 (CH3)3CX ,0

In some cases they can prevent its formation altogether.An SN2 reaction requires an approach by the nucleophile to a distance within bonding

range of the carbon atom bearing the leaving group. Because of this, bulky substituentson or near that carbon atom have a dramatic inhibiting effect (Fig. 6.11). They cause thefree energy of the required transition state to be increased and, consequently, they increasethe free energy of activation for the reaction. Of the simple alkyl halides, methyl halidesreact most rapidly in SN2 reactions because only three small hydrogen atoms interfere with

Nu C XH

H

H

Nu C XH

H

CHH

H

Nu

C

C

CC

CX

HH

HH

H

H

H HH

H

H

Nu C X

CH

H H

CHH

H

Methyl

(30)

1°

(1)

2°

(0.02)

Relative rate

Neopentyl

(0.00001)

HC X

NuCH HH

H

C

HH

H

3°

(~ 0)

CH

H

Figure 6.11 Steric effects in the SN2 reaction.

SN2 Steric Hindrance

The steric effects inthese structures canbest be appreciated bybuilding models.

the approaching nucleophile. Neopentyl and tertiary halides are the least reactive becausebulky groups present a strong hindrance to the approaching nucleophile. (Tertiary sub-strates, for all practical purposes, do not react by an SN2 mechanism.)

SN1 Reactions The primary factor that determines the reactivity of organic substratesin an SN1 reaction is the relative stability of the carbocation that is formed.

Except for those reactions that take place in strong acids, which we shall study later,the only organic compounds that undergo reaction by an SN1 path at a reasonable rate arethose that are capable of forming relatively stable carbocations. Of the simple alkyl halidesthat we have studied so far, this means (for all practical purposes) that only tertiary halidesreact by an SN1 mechanism. (Later we shall see that certain organic halides, called allylichalides and benzylic halides, can also react by an SN1 mechanism because they can formrelatively stable carbocations; see Sections 13.4 and 15.15.)

Tertiary carbocations are stabilized because three alkyl groups release electrons to thepositive carbon atom and thereby disperse its charge (see Section 6.12B).

Formation of a relatively stable carbocation is important in an SN1 reaction becauseit means that the free energy of activation for the slow step for the reaction (i.e.,

will be low enough for the overall reaction to take place at a1 2©R9X !! R 1 X )reasonable rate. If you examine Fig. 6.7 again, you will see this step (step 1) is uphill interms of free energy (DG7 for this step is positive). It is also uphill in terms of enthalpy(DH7 is also positive), and, therefore, this step is endothermic. According to a postulatemade by G. S. Hammond (then at the California Institute of Technology) and J. E. Leffler(Florida State University) the transition state for a step that is uphill in energy shouldshow a strong resemblance to the product of that step. Since the product of this step(actually an intermediate in the overall reaction) is a carbocation, any factor that stabilizesit—such as dispersal of the positive charge by electron-releasing groups—should alsostabilize the transition state in which the positive charge is developing.

Reactant Transition stateresembles product of stepbecause DG° is positive

H2O H2Od2d1CH3

CH3

CH3

C Cl CH3

CH3

CH3

C Cl

Product of stepstabilized by threeelectron-releasing

groups

CH3

CH3

CH3

C1 1 Cl2Step 1

‡

For a methyl, primary, or secondary halide to react by an SN1 mechanism it would haveto ionize to form a methyl, primary, or secondary carbocation. These carbocations, how-ever, are much higher in energy than a tertiary carbocation, and the transition states leadingto these carbocations are even higher in energy. The activation energy for an SN1 reactionof a simple methyl, primary, or secondary halide, consequently, is so large (the reactionis so slow) that, for all practical purposes, an SN1 reaction does not compete with thecorresponding SN2 reaction.

The Hammond–Leffler postulate is quite general and can be better understoodthrough consideration of Fig. 6.12. One way that the postulate can be stated is to say thatthe structure of a transition state resembles the stable species that is nearest it in freeenergy. For example, in a highly endergonic step (blue curve) the transition state lies closeto the products in free energy and we assume, therefore, that it resembles the productsof that step in structure. Conversely, in a highly exergonic step (red curve) the transitionstate lies close to the reactants in free energy, and we assume it resembles the reactants



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SN1 Reactivity and Carbo-cation Stability.

in structure as well. The great value of the Hammond–Leffler postulate is that it givesus an intuitive way of visualizing those important, but fleeting, species that we call tran-sition states. We shall make use of it in many future discussions.

Problem 6.6 ➤ The relative rates of ethanolysis of four primary alkyl halides are as follows:CH3CH2Br, 1.0; CH3CH2CH2Br, 0.28; (CH3)2CHCH2Br, 0.030; (CH3)3CCH2Br,0.00000042.(a) Are each of these reactions likely to be SN1 or SN2?(b) Provide an explanation for the relative reactivities that are observed.

6.14B The Effect of the Concentration and Strength of the Nucleophile

Since the nucleophile does not participate in the rate-determining step of an SN1 reaction,the rates of SN1 reactions are unaffected by either the concentration or the identity of thenucleophile. The rates of SN2 reactions, however, depend on both the concentration andthe identity of the attacking nucleophile. We saw in Section 6.6 how increasing the con-centration of the nucleophile increases the rate of an SN2 reaction. We can now examinehow the rate of an SN2 reaction depends on the identity of the nucleophile.

We describe nucleophiles as being good or poor. When we do this we are really de-scribing their relative reactivities in SN2 reactions. A good nucleophile is one that reactsrapidly with a given substrate. A poor nucleophile is one that reacts slowly with the samesubstrate under the same reaction conditions.

The methoxide ion, for example, is a good nucleophile. It reacts relatively rapidly withmethyl iodide to produce dimethyl ether.

rapid2 2©CH O 1 CH I CH OCH 1 I993 3 3 3

Methanol, on the other hand, is a poor nucleophile. Under the same conditions it reactsvery slowly with methyl iodide.

CH3OH 1 CH3I CH3OCH3 1 I2

H

1very slow



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top of txtbase of txtFigure 6.12 Energy diagrams for highly

exergonic and highly endergonic steps of re-actions. (Adapted from Pryor, W. A. FreeRadicals; McGraw-Hill: New York, 1966; p156. Reprinted by permission.)

Free

ene

rgy

Reaction coordinate

Transitionstate

Transitionstate

Reactants Products

Reactants Products

Highlyexergonic

step

Highlyendergonic

step

The relative strengths of nucleophiles can be correlated with two structural features:

1. A negatively charged nucleophile is always a more reactive nucleophile than itsconjugate acid. Thus HO2 is a better nucleophile than H2O and RO2 is better thanROH.

2. In a group of nucleophiles in which the nucleophilic atom is the same, nucleo-philicities parallel basicities. Oxygen compounds, for example, show the followingorder of reactivity:

RO2 . HO2 .. RCO22 . ROH . H2O

This is also their order of basicity. An alkoxide ion (RO2) is a slightly stronger base thana hydroxide ion (HO2), a hydroxide ion is a much stronger base than a carboxylate ion(RCO2

2), and so on.

6.14C Solvent Effects on SN2 Reactions: Polar Protic andAprotic Solvents

The relative strengths of nucleophiles do not always parallel their basicities when thenucleophilic atoms are not the same. When we examine the relative nucleophilicity ofcompounds or ions within the same group of the periodic table, we find that in hydroxylicsolvents such as alcohols and water the nucleophile with the larger nucleophilic atom isbetter. Thiols (R9SH) are stronger nucleophiles than alcohols (ROH); RS2 ions are betterthan RO2 ions; and the halide ions show the following order:

I2 . Br2 . Cl2 . F2

This effect is related to the strength of the interactions between the nucleophile and itssurrounding layer of solvent molecules. A molecule of a solvent such as water or analcohol—called a protic solvent (Section 3.11)—has a hydrogen atom attached to anatom of a strongly electronegative element (oxygen). Molecules of protic solvents can,therefore, form hydrogen bonds to nucleophiles in the following way:

Molecules of the proticsolvent, water, solvate

a halide ion by forminghydrogen bonds to it.

O

H

H

HH

HH

O

H

O

X2

O

H

A small nucleophile, such as a fluoride ion, because its charge is more concentrated, ismore strongly solvated than a larger one. Hydrogen bonds to a small atom are strongerthan those to a large atom. For a nucleophile to react, it must shed some of its solventmolecules because it must closely approach the carbon bearing the leaving group. A largeion, because the hydrogen bonds between it and the solvent are weaker, can shed some ofits solvent molecules more easily and so it will be more nucleophilic.

The greater reactivity of nucleophiles with large nucleophilic atoms is not entirelyrelated to solvation. Larger atoms are more polarizable (their electron clouds are moreeasily distorted); therefore, a larger nucleophilic atom can donate a greater degree ofelectron density to the substrate than a smaller nucleophile whose electrons are more tightlyheld.

While nucleophilicity and basicity are related, they are not measured in the same way.Basicity, as expressed by pKa, is measured by the position of an equilibrium involving an



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Relative NucleophileStrength.

electron-pair donor (base), a proton, the conjugate acid, and the conjugate base. Nucleo-philicity is measured by relative rates of reaction, by how rapidly an electron-pair donorreacts at an atom (usually carbon) bearing a leaving group. For example, the hydroxideion (OH2) is a stronger base than a cyanide ion (CN2); at equilibrium it has the greateraffinity for a proton (the pKa of H2O is ,16, while the pKa of HCN is ,10). Nevertheless,cyanide ion is a stronger nucleophile; it reacts more rapidly with a carbon bearing a leavinggroup than does hydroxide ion.

The relative nucleophilicities of some common nucleophiles in protic solvents are asfollows:

Relative Nucleophilicity in Protic Solvents

SH2 . CN2 . I2 . OH2 . N32 . Br2 . CH3CO2

2 . Cl2 . F2 . H2O

Polar Aprotic Solvents Aprotic solvents are those solvents whose molecules do nothave a hydrogen atom that is attached to an atom of a strongly electronegative element.Most aprotic solvents (benzene, the alkanes, etc.) are relatively nonpolar, and they do notdissolve most ionic compounds. (In Section 11.21 we shall see how they can be inducedto do so, however.) In recent years a number of polar aprotic solvents have come intowide use by chemists; they are especially useful in SN2 reactions. Several examples arethe following:

(CH3)2N9P9N(CH3)2

N(CH3)2

NSC N

N,N-Dimethylformamide(DMF)

Dimethyl sulfoxide(DMSO)

H CH3C

CH3

CH3

Dimethylacetamide(DMA)

Hexamethylphosphoramide(HMPA)

CH3

CH3

CH3 CH3

O O O O

All of these solvents (DMF, DMSO, DMA, and HMPA) dissolve ionic compounds,and they solvate cations very well. They do so in the same way that protic solvents solvatecations: by orienting their negative ends around the cation and by donating unsharedelectron pairs to vacant orbitals of the cation:

O

O

OO

OH2O

H2O

OH2

OH2

OH2

OH2

Na1 Na1

S

S

S(CH3)2

A sodium ion solvated bymolecules of the aprotic

solvent dimethyl sulfoxide

A sodium ion solvatedby molecules of theprotic solvent water

CH3H3C

CH3H3C

O S(CH3)2

(CH3)2S

(CH3)2S

However, because they cannot form hydrogen bonds and because their positive centers arewell shielded from any interaction with anions, aprotic solvents do not solvate anions toany appreciable extent. In these solvents anions are unencumbered by a layer of solventmolecules and they are therefore poorly stabilized by solvation. These “naked” anions are



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highly reactive both as bases and nucleophiles. In DMSO, for example, the relative orderof reactivity of halide ions is the same as their relative basicity:

2 2 2 2F . Cl . Br . I

This is the opposite of their strength as nucleophiles in alcohol or water solutions:

I2 . Br2 . Cl2 . F2

The rates of SN2 reactions generally are vastly increased when they are carried outin polar aprotic solvents. The increase in rate can be as large as a millionfold.

Problem 6.7

Classify the following solvents as being protic or aprotic: formic acid, HCOH;

O

acetone,

CH3CCH3;

O

acetonitrile, CH3C#N; formamide, HCNH2;

O

sulfur dioxide, SO2; ammonia,NH3; trimethylamine, N(CH3)3; ethylene glycol, HOCH2CH2OH.

Problem 6.8Would you expect the reaction of propyl bromide with sodium cyanide (NaCN),that is,

©CH CH CH Br 1 NaCN !! CH CH CH CN 1 NaBr3 2 2 3 2 2

to occur faster in DMF or in ethanol? Explain your answer.

Problem 6.9Which would you expect to be the stronger nucleophile in a protic solvent:or CH3O2? (b) H2O or H2S? (c) (CH3)3P or (CH3)3N?2(a) CH CO3 2

6.14D Solvent Effects on SN1 Reactions: The Ionizing Abilityof the Solvent

Because of its ability to solvate cations and anions so effectively, the use of a polar proticsolvent will greatly increase the rate of ionization of an alkyl halide in any SN1 reaction.It does this because solvation stabilizes the transition state leading to the intermediatecarbocation and halide ion more than it does the reactants; thus the free energy of activationis lower. The transition state for this endothermic step is one in which separated chargesare developing, and thus it resembles the ions that are ultimately produced.

(CH3)3C9Cld1 d2 ‡

© ©!! [(CH ) CªCl] !!3 3 (CH3)3C1 1 Cl2

Reactant Transition stateSeparated charges are

developing.

Products

A rough indication of a solvent’s polarity is a quantity called the dielectric constant.The dielectric constant is a measure of the solvent’s ability to insulate opposite chargesfrom each other. Electrostatic attractions and repulsions between ions are smaller in sol-vents with higher dielectric constants. Table 6.5 gives the dielectric constants of somecommon solvents.

Water is the most effective solvent for promoting ionization, but most organic com-pounds do not dissolve appreciably in water. They usually dissolve, however, in alcohols,



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Polar Protic Solvents andSN1 Rates.

Polar Aprotic Solvents andSN2 Rates.

and quite often mixed solvents are used. Methanol–water and ethanol–water are commonmixed solvents for nucleophilic substitution reactions.

Problem 6.10 ➤ When tert-butyl bromide undergoes solvolysis in a mixture of methanol and water,the rate of solvolysis (measured by the rate at which bromide ions form in the mixture)increases when the percentage of water in the mixture is increased. this(a) Explainoccurrence. an explanation for the observation that the rate of the SN2(b) Providereaction of ethyl chloride with potassium iodide in methanol and water decreaseswhen the percentage of water in the mixture is increased.

6.14E The Nature of the Leaving Group

The best leaving groups are those that become the most stable ions after they depart. Sincemost leaving groups leave as a negative ion, the best leaving groups are those ions thatstabilize a negative charge most effectively. Because weak bases do this best, the bestleaving groups are weak bases. The reason that stabilization of the negative charge isimportant can be understood by considering the structure of the transition states. In eitheran SN1 or SN2 reaction the leaving group begins to acquire a negative charge as thetransition state is reached.



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top of txtbase of txtTable 6.5 Dielectric Constants of Common Solvents

Solvent FormulaDielectricConstant

Increasingsolventpolarity

Water

Formic acid

Dimethyl sulfoxide (DMSO)

N,N-Dimethylformamide (DMF)

Acetone

Acetic acid

Acetonitrile

Methanol

Hexamethylphosphoramide(HMPA)

Ethanol

CH3SCH3

HCOH

O

H2O

O

HCN(CH3)2

O

CH3CCH3

O

CH3COH

O

CH3C#N

CH3OH

[(CH3)2N]3P"O

CH3CH2OH

80

59

49

37

21

6

36

33

30

24

Good leaving groups areweak bases.

C X

SN1 Reaction (rate-limiting step)

Transition state

d2d1

C1 1 X2C X‡

SN2 Reaction

C X

Transition state

d2d2

C 1 X2CNu NuXNu 2

‡

Stabilization of this developing negative charge at the leaving group stabilizes the transitionstate (lowers its free energy); this lowers the free energy of activation, and thereby increasesthe rate of the reaction. Of the halogens, an iodide ion is the best leaving group and afluoride ion is the poorest:

I2 . Br2 . Cl2 .. F2

The order is the opposite of the basicity:

F2 .. Cl2 . Br2 . I2

Other weak bases that are good leaving groups that we shall study later are alkanesul-fonate ions, alkyl sulfate ions, and the p-toluenesulfonate ion.

An alkanesulfonate ionO

2O R

O

S

An alkyl sulfate ionO

2O O R

O

S

p-Toluenesulfonate ionO

2O CH3

O

S

These anions are all the conjugate bases of very strong acids.The trifluoromethanesulfonate ion (CF3SO3

2, commonly called the triflate ion) is oneof the best leaving groups known to chemists. It is the anion of CF3SO3H, an exceedinglystrong acid—one that is much stronger than sulfuric acid.

CF3SO32

Triflate ion(a “super” leaving group)

Strongly basic ions rarely act as leaving groups. The hydroxide ion, for example, isa strong base and thus reactions like the following do not take place:

X2 X 1 OH2R ROHThis reaction does nottake place because the

leaving group is a stronglybasic hydroxide ion.



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However, when an alcohol is dissolved in a strong acid it can react with a halide ion.Because the acid protonates the 9OH group of the alcohol, the leaving group no longer

needs to be a hydroxide ion; it is now a molecule of water—a much weaker base than ahydroxide ion.

X2 X 1 H2O1

R ROH

HThis reaction takes place

because the leavinggroup is a weak base.

Problem 6.11 ➤ List the following compounds in order of decreasing reactivity toward CH3O2 in anSN2 reaction carried out in CH3OH: CH3F, CH3Cl, CH3Br, CH3I, CH3OSO2CF3,14CH3OH.

Very powerful bases such as hydride ions (H:2) and alkanide ions (R:2) virtually neveract as leaving groups. Therefore, reactions such as the following are not feasible:

Nu CH3CH2

CH3

12 H CH3CH2 Nu 1H 2

or

Nu 12 Nu 1 2CH3 CH3 CH3

These arenot leaving

groups.

6.14F Summary: SN1 versus SN2

Reactions of alkyl halides by an SN1 mechanism are favored by the use of substrates thatcan form relatively stable carbocations, by the use of weak nucleophiles, and by the useof highly ionizing solvents. SN1 mechanisms, therefore, are important in solvolysis reac-tions of tertiary halides, especially when the solvent is highly polar. In a solvolysis thenucleophile is weak because it is a neutral molecule (of the solvent) rather than an anion.

If we want to favor the reaction of an alkyl halide by an SN2 mechanism, we shoulduse a relatively unhindered alkyl halide, a strong nucleophile, a polar aprotic solvent, anda high concentration of the nucleophile. For substrates, the order of reactivity in SN2reactions is

methyl 1° 2°. .

CH39X . R9CH29X . R9CH9X

R

Tertiary halides do not react by an SN2 mechanism.



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Table 6.6 Factors Favoring SN1 versus SN2 ReactionsFactor SN1 SN2

Substrate 37 (requires formation of a rela-tively stable carbocation)

Methyl . 17 . 27 (requires unhinderedsubstrate)

Nucleophile Weak Lewis base, neutral mole-cule, nucleophile may be the sol-vent (solvolysis)

Strong Lewis base, rate favored byhigh concentration of nucleophile

Solvent Polar protic (e.g., alcohols, water) Polar aprotic (e.g., DMF, DMSO)Leaving group I . Br . Cl . F for both SN1 and SN2

(the weaker the base after the group departs, the better the leaving group)

The effect of the leaving group is the same in both SN1 and SN2 reactions: alkyl iodidesreact fastest; fluorides react slowest. (Because alkyl fluorides react so slowly, they areseldom used in nucleophilic substitution reactions.)

R9I . R9Br . R9Cl SN1 or SN2

These factors are summarized in Table 6.6.

6.15 ORGANIC SYNTHESIS: FUNCTIONALGROUP TRANSFORMATIONS USING SN2REACTIONS

In Chapter 4 we had an introduction to the synthesis of organic molecules and retrosyn-thetic analysis. Now that we have studied nucleophilic substitution reactions, these reac-tions give us new tools to add to our toolbox.

SN2 reactions are highly useful in organic synthesis because they enable us to convertone functional group into another—a process that is called a functional group transfor-mation or a functional group interconversion. With the SN2 reactions shown in Fig.6.13, the functional group of a methyl, primary, or secondary alkyl halide can be trans-formed into that of an alcohol, ether, thiol, thioether, nitrile, ester, and so on. (Note: Theuse of the prefix thio- in a name means that a sulfur atom has replaced an oxygen atomin the compound.) In Section 4.18C we saw how carbon–carbon bonds can be formed byalkylation of alkynide anions. This was an SN2 reaction, too.

Alkyl chlorides and bromides are also easily converted to alkyl iodides by nucleophilicsubstitution reactions.

I2

I (1 Cl2 or Br2)RBrR

ClR

6.15 Organic Synthesis: Functional Group Transformations Using SN2 Reactions 261


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OH2

R9O2

SH2

R9S2

CN2

R99C#C2

R9CO2

R3N9

N32

R9X

(R 5 Me, 1°, or 2°)(X 5 Cl, Br, or I)

(2X2)

R9NR3 X2

19

R9OCR9

R9C#C9R9

R9C#N

R9SR9

R9SH

R9OR9

R9OH

R9N3

Quaternary ammonium halide

Ester

Alkyne

Nitrile

Thioether

Thiol

Ether

Alcohol

Alkyl azide

OO

Figure 6.13 Functionalgroup interconversions ofmethyl, primary, and secondaryalkyl halides using SN2 reac-tions.

Methods for FunctionalGroup Preparation.

262


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top of txtbase of txtThe Chemistry of...

Biological Methylation: A Biological NucleophilicSubstitution Reaction

The cells of living organisms synthesize many of the compounds they need fromsmaller molecules. Often these biosyntheses resemble the syntheses organic chemistscarry out in their laboratories. Let us examine one example now.

Many reactions taking place in the cells of plants and animals involve the transfer ofa methyl group from an amino acid called methionine to some other compound. Thatthis transfer takes place can be demonstrated experimentally by feeding a plant or animalmethionine containing a radioactive carbon atom (14C) in its methyl group. Later, othercompounds containing the “labeled” methyl group can be isolated from the organism.Some of the compounds that get their methyl groups from methionine are the following.The radioactively labeled carbon atom is shown in green.

2O2CCHCH2CH2SCH3

CHCH2NHCH3

NH31

HO

HO

OH

Methionine

CH3

CH3 CH3

CH2CH2OH

CH3

N1

CholineAdrenalineNicotine

N

N

Choline is important in the transmission of nerve impulses, adrenaline causes bloodpressure to increase, and nicotine is the compound contained in tobacco that makessmoking tobacco addictive. (In large doses nicotine is poisonous.)

The transfer of the methyl group from methionine to these other compounds does nottake place directly. The actual methylating agent is not methionine; it is S-adenosyl-methionine,* a compound that results when methionine reacts with adenosine triphos-phate (ATP):

2O2CCHCH2CH2SCH3 1

NH31

Methionine

Triphosphate group

O O

O

O2 O2 O2

P

O

OHPO

O

P

CH2

OH

O

H HH

OH

H

ATP

Nucleophile

Leavinggroup

Adenine

* The prefix S is a locant meaning “on the sulfur atom” and should not be confused with the (S) used to defineabsolute configuration. Another example of this kind of locant is N, meaning “on the nitrogen atom.”

S-Adenosylmethionine Triphosphate ion

O2 O2 O2

2O O

O

P

O

OHPO

O

P

CH3

2O2CCHCH2CH2 S

NH31

1 CH2

OH

O

H HH

OH

H

Adenine 1

NH2

Adenine 5N

N

N

N

This reaction is a nucleophilic substitution reaction. The nucleophilic atom is the sulfuratom of methionine. The leaving group is the weakly basic triphosphate group of aden-osine triphosphate. The product, S-adenosylmethionine, contains a methylsulfonium

group,&

1CH 9 ES 9 .3

S-Adenosylmethionine then acts as the substrate for other nucleophilic substitutionreactions. In the biosynthesis of choline, for example, it transfers its methyl group to anucleophilic nitrogen atom of 2-(N,N-dimethylamino)ethanol:

N

2-(N,N-Dimethylamino)ethanol

CH3

CH3

CH3

CH2CH2OH 2O2CCHCH2CH2 S1

NH31

CH2

OH

O

H HH

OH

H

Adenine1

N1

Choline

CH3

CH3

CH3

CH2CH2OH 2O2CCHCH2CH2 S

NH31

CH2

OH

O

H HH

OH

H

Adenine1

These reactions appear complicated only because the structures of the nucleophiles andsubstrates are complex. Yet conceptually they are simple and they illustrate many of theprinciples we have encountered thus far in Chapter 6. In them we see how nature makesuse of the high nucleophilicity of sulfur atoms. We also see how a weakly basic group(e.g., the triphosphate group of ATP) functions as a leaving group. In the reaction of 2-(N,N-dimethylamino)ethanol we see that the more basic (CH3)2N9 group acts as thenucleophile rather than the less basic 9OH group. And when a nucleophile attacks S-adenosylmethionine, we see that the attack takes place at the less hindered CH3 9 grouprather than at one of the more hindered 9CH2 9 groups.

263


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One other aspect of the SN2 reaction that is of great importance in synthesis is itsstereochemistry (Section 6.9). SN2 reactions always occur with inversion of configura-tion at the atom that bears the leaving group. This means that when we use SN2 reactionsin syntheses we can be sure of the configuration of our product if we know the configurationof our reactant. For example, suppose we need a sample of the following nitrile with the(S) configuration.

(S)-2-Methylbutanenitrile

C

CH2CH3

H

CH3

CN

If we have available (R)-2-bromobutane, we can carry out the following synthesis:

N C 2 1 BrSN2

C(inversion)

CH3

HCH2CH3

(R)-2-Bromobutane

N C

(S)-2-Methylbutanenitrile

C

CH3CH2

CH3

H1 Br2

Problem 6.12 ➤ Starting with (S)-2-bromobutane, outline syntheses of each of the following com-pounds:

(R)-CH3CHCH2CH3

OCH2CH3

O

(a)

(R)-CH3CHCH2CH3

OCCH3

(b)

(R)-CH3CHCH2CH3

SH

(c)

(R)-CH3CHCH2CH3

SCH3

(d)

6.15A The Unreactivity of Vinylic and Phenyl Halides

As we learned in Section 6.1, compounds that have a halogen atom attached to one carbonatom of a double bond are called vinylic halides; those that have a halogen atom attachedto a benzene ring are called phenyl halides.

A vinylic halide Phenyl halide

XC

X

C

Study Problem ➤ (a) What is the leaving group when 2-(N,N-dimethylamino)ethanol reacts with S-adenosylmethionine? (b) What would the leaving group have to be if methionineitself were to react with 2-(N,N-dimethylamino)ethanol? (c) Of what special sig-nificance is this difference?

Vinylic halides and phenyl halides are generally unreactive in SN1 or SN2 reactions.Vinylic and phenyl cations are highly unstable and do not form readily. This explains theunreactivity of vinylic and phenyl halides in SN1 reactions. The carbon–halogen bond ofa vinylic or phenyl halide is stronger than that of an alkyl halide (we shall see why later),and the electrons of the double bond or benzene ring repel the approach of a nucleophilefrom the back side. These factors explain the unreactivity of a vinylic or phenyl halide inan SN2 reaction.

6.16 ELIMINATION REACTIONS OF ALKYLHALIDES

Another characteristic of alkyl halides is that they undergo elimination reactions. In anelimination reaction the fragments of some molecule (YZ) are removed (eliminated) fromadjacent atoms of the reactant. This elimination leads to the introduction of a multiplebond:

elimination

(2YZ)CC

Z

C

Y

C

6.16A Dehydrohalogenation

A widely used method for synthesizing alkenes is the elimination of HX from adjacentatoms of an alkyl halide. Heating the alkyl halide with a strong base causes the reactionto take place. The following are two examples:

C2H5ONa

C2H5OH, 55°C

C2H5ONa

C2H5OH, 55°CCH39C9Br

CH3

CH3CHCH3 CH2"CH9CH3 1 NaBr 1 C2H5OH

Br

CH3 CH3

(79%)

CH39C"CH2 1 NaBr 1 C2H5OH(91%)

Reactions like these are not limited to the elimination of hydrogen bromide. Chloroal-kanes also undergo the elimination of hydrogen chloride, iodoalkanes undergo the elimi-nation of hydrogen iodide, and, in all cases, alkenes are produced. When the elements ofa hydrogen halide are eliminated from a haloalkane in this way, the reaction is often calleddehydrohalogenation.

A base

B2 X2ab CC 1 B 11 H

X

C

H

C

Dehydrohalogenation

In these eliminations, as in SN1 and SN2 reactions, there is a leaving group and anattacking particle (the base) that possesses an electron pair.

Chemists often call the carbon atom that bears the substituent (e.g., the halogen atomin the previous reaction) the alpha (a) carbon atom and any carbon atom adjacent to it

6.16 Elimination Reactions of Alkyl Halides 265


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a beta (b) carbon atom. A hydrogen atom attached to the b carbon atom is called a bhydrogen atom. Since the hydrogen atom that is eliminated in dehydrohalogenation isfrom the b carbon atom, these reactions are often called b eliminations. They are alsooften referred to as 1,2 eliminations.

We shall have more to say about dehydrohalogenation in Chapter 7, but we can examineseveral important aspects here.

6.16B Bases Used in Dehydrohalogenation

Various strong bases have been used for dehydrohalogenations. Potassium hydroxide dis-solved in ethanol is a reagent sometimes used, but the sodium salts of alcohols, such assodium ethoxide, often offer distinct advantages.

The sodium salt of an alcohol (a sodium alkoxide) can be prepared by treating an alcoholwith sodium metal:

2 R9 aOH 12 Na !!© 2 2 1R9OaC Na 1H2

Alcohol Sodiumalkoxide

This reaction involves the displacement of hydrogen from the alcohol and is, therefore, anoxidation–reduction reaction. Sodium, an alkali metal, is a very powerful reducing agentand always displaces hydrogen atoms that are bonded to oxygen atoms. The vigorous (attimes explosive) reaction of sodium with water is of the same type.

©2 HaOH 1 2 Na !! 2 12 HaOC Na 1 H2

Sodiumhydroxide

Sodium alkoxides can also be prepared by allowing an alcohol to react with sodiumhydride (NaH). The hydride ion (HC2) is a very strong base.

H2O9H 1 Na1R O 2Na1 1 H9HR

Sodium (and potassium) alkoxides are usually prepared by using an excess of the al-cohol, and the excess alcohol becomes the solvent for the reaction. Sodium ethoxide isfrequently employed in this way.

2 CH CH aOH3 2 1 2 Na !!© 2 12 CH CH aOC Na3 2 1 H2

Ethanol(excess)

Sodium ethoxide

Potassium tert-butoxide is another highly effective dehydrohalogenating reagent.

2 CH3C9OH 1 2 K

CH3

CH3

Potassium tert-butoxidetert-Butyl alcohol(excess)

2K1 1 H22 CH3C9O

CH3

CH3

6.16C Mechanisms of Dehydrohalogenations

Elimination reactions occur by a variety of mechanisms. With alkyl halides, two mecha-nisms are especially important because they are closely related to the SN2 and SN1 reactions



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that we have just studied. One mechanism is a bimolecular mechanism called the E2reaction; the other is a unimolecular mechanism called the E1 reaction.

6.17 THE E2 REACTION

When isopropyl bromide is heated with sodium ethoxide in ethanol to form propene, thereaction rate depends on the concentration of isopropyl bromide and on the concentrationof ethoxide ion. The rate equation is first order in each reactant and second order overall.

2Rate ~ [CH CHBrCH ] [C H O ]3 3 2 5

2Rate 5 k[CH CHBrCH ] [C H O ]3 3 2 5

From this we infer that the transition state for the rate-determining step must involveboth the alkyl halide and the alkoxide ion. The reaction must be bimolecular. Considerableexperimental evidence indicates that the reaction takes place in the following way:

A Mechanism for the E2 Reaction............ Reaction:2 2©C H O 1 CH CHBrCH !! CH "CHCH 1 C H OH 1 Br2 5 3 3 2 3 2 5

Mechanism:

C2H5 O2

BrHab ab

b

b

b

HH

CH3

C C

H

C2H5 O2d

2dBrH

H

C C

H

H

H

H

CH3

C C 1 C2H5 OH 1 Br2

‡

The basic ethoxide ion begins toremove a proton from the

bond to it. At the same time, theusing its electron pair to form a

electron pair of thebegins to move in to become the

p bond of a double bond, and thebromine begins to depart with the

electrons that bonded it to the carbon.aa

aacarbon

C—H bond

Transition statePartial bonds now exist between the

oxygen and thebetween the

bromine. The carbon–carbon bondis developing double bond character.

hydrogen andcarbon and the

Now the double bond of the alkene isfully formed and the alkene has atrigonal planar geometry at each

carbon atom. The other products are amolecule of ethanol and a bromide ion.

HCH3

When we study the E2 reaction further in Section 7.6C, we shall find that the orientationsof the hydrogen atom being removed and the leaving group are not arbitrary and that theorientation where they are all in the same plane, as shown above, is required.

6.17 The E2 Reaction 267


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E2 Elimination Mechanism

6.18 THE E1 REACTION

Eliminations may take a different pathway from that given in Section 6.17. Treating tert-butyl chloride with 80% aqueous ethanol at 257C, for example, gives substitution productsin 83% yield and an elimination product (2-methylpropene) in 17% yield.

CH3

CH3C Cl

CH3

80% C2H5OH

20% H2O25°C

CH3

CH3C OH

CH3

1

CH3

CH3C

CH3

OCH2CH3

tert-Butyl chloride

tert-Butyl alcohol tert-Butyl ethyl ether

SN1

(83%)

CH2

CH3

C

CH3

E1

2-Methylpropene(17%)

The initial step for both reactions is the formation of a tert-butyl cation. This is alsothe rate-determining step for both reactions; thus both reactions are unimolecular.

slowCl 22CH3C9Cl

CH3

CH3

CH3C1 1

CH3

CH3

(solvated) (solvated)

Whether substitution or elimination takes place depends on the next step (the fast step).If a solvent molecule reacts as a nucleophile at the positive carbon atom of the tert-butyl cation, the product is tert-butyl alcohol or tert-butyl ethyl ether and the reactionis SN1.

CH3

CH3

SN1reaction

fastCH3C

1

CH3

CH3

CH3C1OH 1

H

O

CH3

CH3 H

CH3C OSol

Sol

Sol 1 H9O9Sol

H9O9Sol(Sol 5 H CH3CH2 )or

If, however, a solvent molecule acts as a base and abstracts one of the b hydrogen atomsas a proton, the product is 2-methylpropene and the reaction is E1.

E1 reactions almost always accompany SN1 reactions.

Sol O

H

H CH2 C1

CH3

Cfast

CH3

Sol O

H

1 H 1 CH2

CH3

CH3

2-Methylpropene

E1reaction



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A Mechanism for the E1 Reaction............ Reaction:1 2©(CH ) CCl 1 H O !! CH "C(CH ) 1 H O 1 Cl3 3 2 2 3 2 3

Mechanism:

Step 1

CH3

H3C C Cl Cl

CH3

slow

H2OH3C C1 1

2

Aided by thepolar solvent a

chlorinedeparts withthe electron

pair thatbonded it tothe carbon.

This slow stepproduces the

relatively stable 3°carbocation and achloride ion. Theions are solvated

(and stabilized) bysurrounding water

molecules.

CH3

CH3

Step 2

OH

H

1 C1CH

H

H

OH

H

H

H

H

C C

CH3

CH3

11

A molecule of waterremoves one of

the hydrogens fromthe carbon of the

carbocation. Anelectron pair moves into form a double bond

between the andcarbon atoms.

This step produces thealkene and a

hydronium ion.

ab

b

b

a

CH3

CH3

6.19 SUBSTITUTION VERSUS ELIMINATION

Because the reactive part of a nucleophile or a base is an unshared electron pair, allnucleophiles are potential bases and all bases are potential nucleophiles. It should not besurprising, then, that nucleophilic substitution reactions and elimination reactions oftencompete with each other.

6.19A SN2 versus E2

Since eliminations occur best by an E2 path when carried out with a high concentrationof a strong base (and thus a high concentration of a strong nucleophile), substitutionreactions by an SN2 path often compete with the elimination reaction. When the nucleophile

6.19 Substitution versus Elimination 269


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This section draws togetherthe various factors that in-fluence the competition be-tween substitution andelimination.

(base) attacks a b hydrogen atom, elimination occurs. When the nucleophile attacks thecarbon atom bearing the leaving group, substitution results.

(a)

(b)

eliminationE2

substitutionSN2

X2

CH

C

C1 Nu

C

X

H 1

Nu 2

1 X 2

(a)

(b)

CH

CNu

When the substrate is a primary halide and the base is ethoxide ion, substitution ishighly favored because the base can easily approach the carbon bearing the leaving group.

2 1CH CH O Na 1 CH CH Br3 2 3 2

C H OH2 5©99557C

(2NaBr)CH CH OCH CH3 2 2 3 1 CH2"CH2

SN2(90%)

E2(10%)

With secondary halides, however, the elimination reaction is favored because sterichindrance makes substitution more difficult.

SN2(21%)

E2(79%)

C2H5OH

55°C(2NaBr)

C2H5O2Na1 1 CH3CHCH3

Br

CH3CHCH3 1 CH2"CHCH3

O

C2H5

With tertiary halides an SN2 reaction cannot take place, so the elimination reaction ishighly favored, especially when the reaction is carried out at higher temperatures. Anysubstitution that occurs must take place through an SN1 mechanism.

SN1(9%)

Mainly E2(91%)

C2H5OH

25°C(2NaBr)

C2H5O2Na1 1 CH3CCH3

Br

CH3CCH3 1 CH2"CCH3

O

C2H5

CH3 CH3 CH3

E2 1 E1(100%)

C2H5OH

55°C(2NaBr)

C2H5O2Na1 1 CH3CCH3

Br

CH2"CCH3 1 C2H5OH

CH3 CH3

Increasing the temperature favors eliminations (E1 and E2) over substitutions. The reason:Eliminations have higher free energies of activation than substitutions because eliminationshave a greater change in bonding (more bonds are broken and formed). By giving more



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molecules enough energy to surmount the energy barriers, increasing the temperature in-creases the rates of both substitutions and eliminations; however, because the energy bar-riers for eliminations are higher, the proportion of molecules able to cross them is signif-icantly higher.

Increasing the reaction temperature is one way of favorably influencing an eliminationreaction of an alkyl halide. Another way is to use a strong sterically hindered base suchas the tert-butoxide ion. The bulky methyl groups of the tert-butoxide ion appear to inhibitits reacting by substitution, so elimination reactions take precedence. We can see an ex-ample of this effect in the following two reactions. The relatively unhindered methoxideion reacts with octadecyl bromide primarily by substitution; the bulky tert-butoxide iongives mainly elimination.

SN2(99%)

SN2(15%)

CH3OH

65°CCH3O

2 1 CH3(CH2)15CH2CH29Br

CH3(CH2)15CH"CH2 1 CH3(CH2)15CH2CH2OCH3

E2(1%)

E2(85%)

(CH3)3COH

40°CCH39C9O2 1 CH3(CH2)15CH2CH29Br

CH3

CH3

CH3(CH2)15CH"CH2 1 CH3(CH2)15CH2CH29O9C9CH3

CH3

CH3

Another factor that affects the relative rates of E2 and SN2 reactions is the relativebasicity and polarizability of the base/nucleophile. Use of a strong, slightly polarizablebase such as amide ion (NH2

2) or alkoxide ion (especially a hindered one) tends to increasethe likelihood of elimination (E2). Use of a weakly basic ion such as a chloride ion (Cl2)or an acetate ion (CH3CO2

2) or a weakly basic and highly polarizable one such as Br2,I2, or RS2 increases the likelihood of substitution (SN2). Acetate ion, for example, reactswith isopropyl bromide almost exclusively by the SN2 path:

CH3

SN2(,100%)

CH3C9O2 1 CH3CH9Br

O CH3

CH3C9O9CHCH3 1 Br2

O

The more strongly basic ethoxide ion (Section 6.16B) reacts with the same compoundmainly by an E2 mechanism.

6.19B Tertiary Halides: SN1 versus E1

Because the E1 reaction and the SN1 reaction proceed through the formation of a commonintermediate, the two types respond in similar ways to factors affecting reactivities. E1reactions are favored with substrates that can form stable carbocations (i.e., tertiary hal-ides); they are also favored by the use of poor nucleophiles (weak bases) and they aregenerally favored by the use of polar solvents.

6.19 Substitution versus Elimination 271


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It is usually difficult to influence the relative partition between SN1 and E1 productsbecause the free energy of activation for either reaction proceeding from the carbocation(loss of a proton or combination with a molecule of the solvent) is very small.

In most unimolecular reactions the SN1 reaction is favored over the E1 reaction, espe-cially at lower temperatures. In general, however, substitution reactions of tertiary halidesdo not find wide use as synthetic methods. Such halides undergo eliminations much tooeasily.

Increasing the temperature of the reaction favors reaction by the E1 mechanism at theexpense of the SN1 mechanism. If the elimination product is desired, however, it is moreconvenient to add a strong base and force an E2 reaction to take place instead.

6.20 OVERALL SUMMARY

The most important reaction pathways for the substitution and elimination reactionsof simple alkyl halides can be summarized in the way shown in Table 6.7. Let us exam-ine several sample exercises that will illustrate how the information in Table 6.7 canbe used.



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Table 6.7 Overall Summary of SN1, SN2, E1, and E2 Reactions

CH3X RCH2X

R

RCHX

R

R

R9C9X

Methyl 17 27 37

Bimolecular reactions only SN1/E1 or E2

GivesSN2reactions

Gives mainly SN2 exceptwith a hindered strongbase [e.g., (CH3)3CO2]and then gives mainlyE2

Gives mainly SN2 withweak bases (e.g., I2,CN2, RCO2

2) andmainly E2 with strongbases (e.g., RO2)

No SN2 reaction. In solvolysisgives SN1/E1, and at lowertemperatures SN1 is favored.When a strong base (e.g.,RO2) is used E2predominates.

➤ Sample Problem

Give the product (or products) that you would expect to be formed in each of thefollowing reactions. In each case give the mechanism (SN1, SN2, E1, or E2) by whichthe product is formed and predict the relative amount of each (i.e., would the productbe the only product, the major product, or a minor product?).

507C2 ©(a) CH CH CH Br 1 CH O 993 2 2 3 CH OH3

507C2 ©(b) CH CH CH Br 1 (CH ) CO 993 2 2 3 3 (CH ) COH3 3

CH3CH2

CH3

H

150°C

CH3OHBr HS2(c) C

Overall Summary.

507C2 ©(d) (CH CH ) CBr 1 OH 993 2 3 CH OH3

257C©(e) (CH CH ) CBr 993 2 3 CH OH3

Answer:(a) The substrate is a 17 halide. The base/nucleophile is CH3O2, a strong base (butnot a hindered one) and a good nucleophile. According to Table 6.7 we should expectan SN2 reaction mainly, and the major product should be CH3CH2CH2OCH3. A minorproduct might be CH3CH"CH2 by an E2 pathway.

(b) Again the substrate is a 17 halide, but the base/nucleophile, (CH3)3CO2, is a stronghindered base. We should expect, therefore, the major product to be CH3CH"CH2 byan E2 pathway, and a minor product to be CH3CH2CH2OC(CH3)3 by an SN2 pathway.

(c) The reactant is (S)-2-bromobutane, a 27 halide, and one in which the leaving groupis attached to a stereocenter. The base/nucleophile is HS2, a strong nucleophile, but aweak base. We should expect mainly an SN2 reaction, causing an inversion of config-uration at the stereocenter and producing the (R) stereoisomer:

CH2CH3

H

CH3

CHS

(d) The base/nucleophile is OH2, a strong base and a strong nucleophile. However,the substrate is a 37 halide; therefore, we should not expect an SN2 reaction. The majorproduct should be CH3CH"C(CH2CH3)2 via an E2 reaction. At this higher tempera-ture, and in the presence of a strong base, we should not expect an appreciable amountof the SN1 product, CH3OC(CH2CH3)3.

(e) This is solvolysis; the only base/nucleophile is the solvent, CH3OH, which is aweak base (therefore, no E2 reaction) and a poor nucleophile. The substrate is tertiary(therefore, no SN2 reaction). At this lower temperature we should expect mainly an SN1pathway leading to CH3OC(CH2CH3)3. A minor product, by an E1 pathway, would beCH3CH"C(CH2CH3)2.

...............

Key Terms and ConceptsKey Terms and Concepts

Nucleophilic substitution reaction Section 6.3

Nucleophile Sections 6.3, 6.4, 6.14B

Substrate Section 6.3

Leaving group Sections 6.5, 6.14E

SN2 reaction Sections 6.6–6.9, 6.14, 6.19A

Transition state Section 6.7

Inversion of configuration Section 6.9

SN1 reaction Sections 6.10, 6.11, 6.13, 6.14, 6.19B

Carbocation Sections 6.11, 6.12

Solvolysis Section 6.13B

Steric effect Section 6.14A

Key Terms and Concepts 273


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Steric hindrance Section 6.14A

Aprotic solvent Section 6.14C

Polar protic solvent Section 6.14D

Elimination reaction Section 6.16

E1 reaction Sections 6.18, 6.19B

E2 reaction Sections 6.17, 6.19A

ADDITIONAL

PROBLEMS

6.13 Show how you might use a nucleophilic substitution reaction of propyl bromide tosynthesize each of the following compounds. (You may use any other compoundsthat are necessary.)(a) CH CH CH OH3 2 2

(b) CH CH CH I3 2 2

(c) CH CH OCH CH CH3 2 2 2 3

(d) CH CH CH 9S9CH3 2 2 3

CH3COCH2CH2CH3

O

(e)

(f) CH3CH2CH2N3

CH39N9CH2CH2CH3 Br21

CH3

CH3

(g)

(h) CH CH CH CN3 2 2

(i) CH CH CH SH3 2 2

6.14 Which alkyl halide would you expect to react more rapidly by an SN2 mechanism?Explain your answer.(a) CH3CH2CH2Br or (CH3)2CHBr(b) CH3CH2CH2CH2Cl or CH3CH2CH2CH2I(c) (CH3)2CHCH2Cl or CH3CH2CH2CH2Cl(d) (CH3)2CHCH2CH2Cl or CH3CH2CH(CH3)CH2Cl(e) C6H5Br or CH3CH2CH2CH2CH2CH2Cl

6.15 Which SN2 reaction of each pair would you expect to take place more rapidly in aprotic solvent?(a) (1) CH3CH2CH2Cl 1 CH3CH2O2 !!© CH3CH2CH2OCH2CH3 1 Cl2

or(2) CH3CH2CH2Cl 1 CH3CH2OH !!© CH3CH2CH2OCH2CH3 1 HCl

(b) (1) CH3CH2CH2Cl 1 CH3CH2O2 !!© CH3CH2CH2OCH2CH3 1 Cl2

or(2) CH3CH2CH2Cl 1 CH3CH2S2 !!© CH3CH2CH2SCH2CH3 1 Cl2

(c) (1) CH3CH2CH2Br 1 (C6H5)3N !!© CH3CH2CH2N(C6H5)31 1 Br2

or(2) CH3CH2CH2Br 1 (C6H5)3P !!© CH3CH2CH2P(C6H5)3

1 1 Br2

(d) (1) CH3CH2CH2Br (1.0M) 1 CH3O2 (1.0M) !!© CH3CH2CH2OCH3 1 Br2

or(2) CH3CH2CH2Br (1.0M) 1 CH3O2 (2.0M) !!© CH3CH2CH2OCH3 1 Br2

6.16 Which SN1 reaction of each pair would you expect to take place more rapidly?Explain your answer.

(a) (1) (CH3)3CCl 1 H2O !!© (CH3)3COH 1 HClor(2) (CH3)3CBr 1 H2O !!© (CH3)3COH 1 HBr



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* Problems marked with an asterisk are “challenge problems.”

(b) (1) (CH3)3CCl 1 H2O !!© (CH3)3COH 1 HClor(2) (CH3)3CCl 1 CH3OH !!© (CH3)3COCH3 1 HCl

(c)EtOH

2 2©(1) (CH ) CCl (1.0M) 1 CH CH O (1.0M) (CH ) COCH CH 1 Cl99 3 3 2 33 3 3 2

orEtOH


(d)EtOH


orEtOH


(e) (1) (CH3)3CCl 1 H2O !!© (CH3)3COH 1 HClor(2) C6H5Cl 1 H2O !!© C6H5OH 1 HCl

6.17 With methyl, ethyl, or cyclopentyl halides as your organic starting materials andusing any needed solvents or inorganic reagents, outline syntheses of each of thefollowing. More than one step may be necessary and you need not repeat stepscarried out in earlier parts of this problem.(a) CH3I (d) CH3CH2OH (g) CH3CN (j) CH3OCH2CH3

(b) CH3CH2I (e) CH3SH (h) CH3CH2CN (k) Cyclopentene(c) CH3OH (f) CH3CH2SH (i) CH3OCH3

6.18 Listed below are several hypothetical nucleophilic substitution reactions. None issynthetically useful because the product indicated is not formed at an appreciablerate. In each case provide an explanation for the failure of the reaction to take placeas indicated.(a) 2 2©CH CH CH 1 OH 3!! CH CH OH 1 CH3 2 3 3 2 3

(b) 2 2©CH CH CH 1 OH 3!! CH CH CH OH 1 H3 2 3 3 2 2

(c) □ 2 2©1 OH 3!! CH CH CH CH OH2 2 2 2

(d) CH3CH29C9Br 1 CN2

CH3

CH3

CH3CH29C9CN 1 Br2

CH3

CH3

(e) ©NH 1 CH OCH 3!! CH NH 1 CH OH3 3 3 3 2 3

(f) 1 1©NH 1 CH OH 3!! CH NH 1 H O3 3 2 3 3 2

6.19 You have the task of preparing styrene (C6H5CH"CH2) by dehydrohalogenationof either 1-bromo-2-phenylethane or 1-bromo-1-phenylethane using KOH in etha-nol. Which halide would you choose as your starting material to give the betteryield of the alkene? Explain your answer.

6.20 Your task is to prepare isopropyl methyl ether, CH3OCH(CH3)2 , by one of thefollowing reactions. Which reaction would give the better yield? Explain yourchoice.(1) CH3ONa 1 (CH3)2CHI !!© CH3OCH(CH3)2

(2) (CH3)2CHONa 1 CH3I !!© CH3OCH(CH3)2

6.21 Which product (or products) would you expect to obtain from each of the followingreactions? In each part give the mechanism (SN1, SN2, E1, or E2) by which eachproduct is formed and predict the relative amount of each product (i.e., would theproduct be the only product, the major product, a minor product, etc.?).

(a)507C

2 ©CH CH CH CH CH Br 1 CH CH O 993 2 2 2 2 3 2 CH CH OH3 2

Additional Problems 275


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(b)507C

2 ©CH CH CH CH CH Br 1 (CH ) CO 993 2 2 2 2 3 3 (CH ) COH3 3

(c)507C

2 ©(CH ) CBr 1 CH O 993 3 3 CH OH3

(d)507C

2(CH ) CBr 1 (CH ) CO 99:3 3 3 3 (CH ) COH3 3

50°C1 I2

CH3

acetone

25°C(f ) (CH3)3C CH3OH

(e) (CH3)3C

Cl

Cl

(g)507C

2 ©3-Chloropentane 1 CH O 993 CH OH3

(h)507C

2 ©3-Chloropentane 1 CH CO 993 2 CH CO H3 2

(i)257C

2 ©HO 1 (R)-2-bromobutane 99

( j)257C

©(S)-3-Bromo-3-methylhexane 99CH OH3

(k)507C

2 ©(S)-2-Bromooctane 1 I 99CH OH3

6.22 Write conformational structures for the substitution products of the following deu-terium-labeled compounds:

(b)D

D

?H2O

CH3OH?

H

H

CH3

Cl

(d)

I2

CH3OH

I2

CH3OH(a)

H

I2

CH3OH

? ?

Cl

HH

Cl

(c)

D

H

Cl

H

D

6.23 Although ethyl bromide and isobutyl bromide are both primary halides, ethyl bro-mide undergoes SN2 reactions more than 10 times faster than isobutyl bromide does.When each compound is treated with a strong base/nucleophile (CH3CH2O2), iso-butyl bromide gives a greater yield of elimination products than substitution prod-ucts, whereas with ethyl bromide this behavior is reversed. What factor accountsfor these results?

6.24 Consider the reaction of I2 with CH3CH2Cl. you expect the reaction(a) Wouldto be SN1 or SN2? The rate constant for the reaction at 607C is

is the reaction rate if and25 21 21 2 215 3 10 L mol s . (b) What [I ] 5 0.1 mol Land21 2 21[CH CH Cl] 5 0.1 mol L ? (c) If [I ] 5 0.1 mol L [CH CH Cl] 53 2 3 2



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and21 2 21 210.2 mol L ? (d) If [I ] 5 0.2 mol L [CH CH Cl] 5 0.1 mol L ? (e) If3 2

and2 21 21[I ] 5 0.2 mol L [CH CH Cl] 5 0.2 mol L ?3 2

6.25 Which reagent in each pair listed here would be the more reactive nucleophile in aprotic solvent?

2 1(a) CH NH or CH NH (e) H O or H O3 3 2 2 31(f) NH or NH3 4

2(g) H S or HS2CH3O2 or CH3CO2

O

(b)(c) CH SH or CH OH3 3

(d) (C H ) N or (C H ) P6 5 3 6 5 3 CH3CO2 or OH2

O

(h)

6.26 Write mechanisms that account for the products of the following reactions:OH2

H2O(a) HOCH2CH2Br H2C

O

N

H

CH2

OH2

H2O(b) H2NCH2CH2CH2CH2Br

6.27 Many SN2 reactions of alkyl chlorides and alkyl bromides are catalyzed by theaddition of sodium or potassium iodide. For example, the hydrolysis of methylbromide takes place much faster in the presence of sodium iodide. Explain.

6.28 Explain the following observations: When tert-butyl bromide is treated with sodiummethoxide in a mixture of methanol and water, the rate of formation of tert-butylalcohol and tert-butyl methyl ether does not change appreciably as the concentrationof sodium methoxide is increased. However, increasing the concentration of sodiummethoxide causes a marked increase in the rate at which tert-butyl bromide disap-pears from the mixture.

6.29 the general problem of converting a tertiary alkyl halide to an alkene,(a) Considerfor example, the conversion of tert-butyl chloride to 2-methylpropene. What ex-perimental conditions would you choose to ensure that elimination is favored oversubstitution? the opposite problem, that of carrying out a substitution(b) Considerreaction on a tertiary alkyl halide. Use as your example the conversion of tert-butylchloride to tert-butyl ethyl ether. What experimental conditions would you employto ensure the highest possible yield of the ether?

6.30 1-Bromobicyclo[2.2.1]heptane is extremely unreactive in either SN2 or SN1 reac-tions. Provide explanations for this behavior.

6.31 When ethyl bromide reacts with potassium cyanide in methanol, the major productis CH3CH2CN. Some CH3CH2NC is formed as well, however. Write Lewis struc-tures for the cyanide ion and for both products, and provide a mechanistic expla-nation of the course of the reaction.

6.32 Starting with an appropriate alkyl halide, and using any other needed reagents,outline syntheses of each of the following. When alternative possibilities exist fora synthesis you should be careful to choose the one that gives the better yield.(a) Butyl sec-butyl ether (g) (S)-2-Pentanol(b) CH3CH2SC(CH3)3 (h) (R)-2-Iodo-4-methylpentane(c) Methyl neopentyl ether (i) (CH3)3CCH"CH2

(d) Methyl phenyl ether ( j) cis-4-Isopropylcyclohexanol(e) C6H5CH2CN (k) (R)-CH3CH(CN)CH2CH3

(f) CH3CO2CH2C6H5 (l) trans-1-Iodo-4-methylcyclohexane



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6.33 Give structures for the products of each of the following reactions:

(a) 1 NaI (1 mol) C5H8FI 1 NaBrH

Br

F

Hacetone

(b) ©1,4-Dichlorohexane (1 mol) 1 NaI (1 mol) C H ClI 1 NaCl99 6 12acetone

(c) ©1,2-Dibromoethane (1 mol) 1 NaSCH CH SNa !! C H S 1 2 NaBr2 2 4 8 2

(d)(2H ) heat2

© ©4-Chloro-1-butanol 1 NaH C H ClONa C H O 1 NaCl99 994 8 4 8Et O Et O2 2

(e)(2NH ) CH I3 3

© ©Propyne 1 NaNH C H Na C H 1 NaI99 993 3 4 62 liq. NH3

6.34 When tert-butyl bromide undergoes SN1 hydrolysis, adding a “common ion” (e.g.,NaBr) to the aqueous solution has no effect on the rate. On the other hand, when(C6H5)2CHBr undergoes SN1 hydrolysis, adding NaBr retards the reaction. Giventhat the (C6H5)2CH1 cation is known to be much more stable than the (CH3)3C1

cation (and we shall see why in Section 15.12A), provide an explanation for thedifferent behavior of the two compounds.

6.35 When the alkyl bromides (listed here) were subjected to hydrolysis in a mixture ofethanol and water (80% C2H5OH/20% H2O) at 557C, the rates of the reactionshowed the following order:

(CH3)3CBr . CH3Br . CH3CH2Br . (CH3)2CHBr

Provide an explanation for this order of reactivity.

6.36 The reaction of 17 alkyl halides with nitrite salts produces both RNO2 and RONO.Account for this behavior.

6.37 What would be the effect of increasing solvent polarity on the rate of each of thefollowing nucleophilic substitution reactions?(a) NuC 1 R9L !!© R9Nu1 1 CL2

(b) R9L1 !!© R1 1 CL

6.38 Competition experiments are those in which two reactants at the same concentration(or one reactant with two reactive sites) compete for a reagent. Predict the majorproduct resulting from each of the following competition experiments.

DMFCl9CH29C9CH29CH29Cl 1 I2

CH3

CH3

(a)

acetoneCl9C9CH29CH29Cl 1 H2O

CH3

CH3

(b)

6.39 In contrast to SN2 reactions, SN1 reactions show relatively little nucleophile selec-tivity. That is, when more than one nucleophile is present in the reaction medium,SN1 reactions show only a slight tendency to discriminate between weak nucleo-philes and strong nucleophiles, whereas SN2 reactions show a marked tendency todiscriminate. an explanation for this behavior. how your(a) Provide (b) Showanswer accounts for the fact that CH3CH2CH2CH2Cl reacts with NaCN in0.01 Methanol to yield primarily CH3CH2CH2CH2CN, whereas under the same conditions(CH3)3CCl reacts to give primarily (CH3)3COCH2CH3.



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*6.40 In the gas phase, the homolytic bond dissociation energy (Section 10.2A) for thecarbon–chlorine bond of tert-butyl chloride is the ionization po-211328 kJ mol ;tential for a tert-butyl radical is and the electron affinity of chlorine211715 kJ mol ;is Using these data, calculate the enthalpy change for the gas phase212330 kJ mol .ionization of tert-butyl chloride to a tert-butyl cation and a chloride ion (this is theheterolytic bond dissociation energy of the carbon–chlorine bond).

*6.41 The reaction of chloroethane with water in the gas phase to produce ethanoland hydrogen chloride has a and a21DH7 5 126.6 kJ mol DS7 5 14.81 J

at 257C. of these terms, if either, favors the reaction going21 21K mol (a) Whichto completion? DG7 for the reaction. What can you now say about(b) Calculatewhether the reaction will proceed to completion? the equilibrium(c) Calculateconstant for the reaction. aqueous solution the equilibrium constant is very(d) Inmuch larger than the one you just calculated. How can you account for this fact?

*6.42 When (S)-2-bromopropanoic acid [(S)-CH3CHBrCO2H] reacts with concentratedsodium hydroxide the product formed (after acidification) is (R)-2-hydroxypropa-noic acid [(R)-CH3CHOHCO2H, commonly known as (R)-lactic acid]. This is, ofcourse, the normal stereochemical result for an SN2 reaction. However, when thesame reaction is carried out with a low concentration of hydroxide ion in the pres-ence of Ag2O (where Ag1 acts as a Lewis acid), it takes place with overall retentionof configuration to produce (S)-2-hydroxypropanoic acid. The mechanism of thisreaction involves a phenomenon called neighboring group participation. Write adetailed mechanism for this reaction that accounts for the net retention of config-uration when Ag1 and a low concentration of hydroxide are used.

*6.43 The phenomenon of configuration inversion in a chemical reaction was discoveredin 1896 by Paul von Walden (Section 6.7). Walden’s proof of configuration inver-sion was based on the following cycle:

Ag2O

Ag2O

H2O

H2OPCl5

KOH

KOHPCl5

HO2CCH2CH(OH)CO2H(2)-Malic acid

HO2CCH2CH(OH)CO2H(1)-Malic acid

HO2CCH2CHClCO2H(2)-Chlorosuccinic acid

HO2CCH2CHClCO2H(1)-Chlorosuccinic acid

The Walden cycle

your answer on the preceding problem, which reactions of the Walden(a) Basingcycle are likely to take place with overall inversion of configuration and which arelikely to occur with overall retention of configuration? acid with a neg-(b) Malicative optical rotation is now known to have the (S) configuration. What are theconfigurations of the other compounds in the Walden cycle? also(c) Waldenfound that when (1)-malic acid is treated with thionyl chloride (rather than PCl5),the product of the reaction is (1)-chlorosuccinic acid. How can you explain thisresult? that the reaction of (2)-malic acid and thionyl chloride has(d) Assumingthe same stereochemistry, outline a Walden cycle based on the use of thionyl chlo-ride instead of PCl5 .



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*6.44 (R)-(3-Chloro-2-methylpropyl) methyl ether (A) on reaction with azide ion 2(N )3

in aqueous ethanol gives (S)-(3-azido-2-methylpropyl) methyl ether (B). CompoundA has the structure ClCH2CH(CH3)CH2OCH3.(a) Draw wedge-dashed wedge-line formulas of both A and B.(b) Is there a change of configuration during this reaction?

*6.45 Predict the structure of the product of this reaction:

NaOH in

aqueous EtOHC6H10S

H Cl

HS H

The product has no infrared absorption in the 1620 to 1680 region.21cm

*6.462t-BuO in

©cis-4-Bromocyclohexanol racemic C H O (compound C)99 6 10t-BuOH

Compound C has infrared absorption in the 1620 to and in the 3590 to211680 cmregions. Draw and label the (R) and (S) enantiomers of product C.213650 cm

LEARNING GROUP

PROBLEMS

1. Consider the solvolysis reaction of (1S,2R)-1-bromo-1,2-dimethylcyclohexane in 80%H2O/20% CH3CH2OH at room temperature.(a) Write the structure of all chemically reasonable products from this reaction andpredict which would be the major one.(b) Write a detailed mechanism for formation of the major product.(c) Write the structure of all transition states involved in formation of the majorproduct.

2. Consider the following sequence of reactions, taken from the early steps in a synthesisof v-fluorooleic acid, a toxic natural compound from an African shrub. (v-fluorooleicacid, also called “ratsbane,” has been used to kill rats and also as an arrow poison intribal warfare. Two more steps beyond those below are required to complete its syn-thesis.)

(i) l-Bromo-8-fluorooctane 1 sodium acetylide (the sodium salt of ethyne)©!! compound A (C H F)10 17

©(ii) Compound A 1 NaNH !! compound B (C H FNa)2 10 16

©(iii) Compound B 1 I9(CH ) 9Cl !! compound C (C H ClF)2 7 17 30

©(iv) Compound C 1 NaCN !! compound D (C H NF)18 30

(a) Elucidate the structure of compounds A, B, C, and D above.



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(b) Write the mechanism for each of the reactions above.(c) Write the structure of the transition state for each reaction.

ionic reactions—nucleophilic substitution ... · 232 chapter 6 / ionic reactions—nucleophilic...

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