Jones M., Fleming S.A. Organic Chemistry

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7.4 Substitution, Nucleophilic, Bimolecular: The S

2 Reaction 269

that will enable us to determine the stereochemical results. If we start with a single

enantiomer (Chapter 4, p. 151) (R) or (S), there are several possible outcomes of a

substitution reaction. We could get retention of configuration, in which the stereo-

chemistry of the starting molecule is preserved. The entering nucleophile occupies

the same stereochemical position as did the departing leaving group, and the prod-

uct has the same handedness as the starting material (Fig. 7.11).

A second possibility is inversion of configuration, in which the stereochemistry

of the starting material is reversed: (R) starting material going to (S) product and

(S) starting material going to (R) product. In this kind of reaction, the pyramid of

the starting material becomes inverted in the reaction like an umbrella in the wind

(Fig. 7.12). In this and Figure 7.11, notice how the numbers enable us to keep track

of inversions and retentions.

It is also possible that optical activity could be lost. In such a case, (R) or (S)

starting material goes to a racemic mixture—a 50:50 mixture of (R) and (S) product

(Fig. 7.13).

Finally, it is possible that some dreadful mixture of the (R) and (S) forms is pro-

duced. In such a case, it is very difficult to analyze the situation, as several reaction

mechanisms may be operating simultaneously.

retention

C C

–

FIGURE 7.11 One possible

stereochemical outcome for the S

reaction. In this scenario, the product

is formed with retention of configuration.

inversion

–

FIGURE 7.12 Another possible

stereochemical result for the S

reaction.The handedness of the

starting material could be reversed in

the product, in what is called

inversion of configuration.

(50%) (50%)

racemization

NuNu

–

FIGURE 7.13 A third possible

stereochemical result for the S

reaction.

WORKED PROBLEM 7.5 Sometimes matters appear more complicated than they

really are. It is possible, for example, that an (R) starting material could produce an

(R) product by a process involving inversion.Here is a conceivable example.Explain.

2 with

inversion

(

)

Still

(

)

–

(continued)

270 CHAPTER 7 Substitution and Elimination Reactions

In this discussion, we indicate radioactive iodine with an asterisk in the text and with the asterisk and a

yellow circle in the figures.

ANSWER The designations (R) and (S) are determined for each molecule through the

Cahn–Ingold–Prelog priority system (Chapter 3, p. 111; Chapter 4, p. 152). If the

groups attached to carbon change, as in the example shown, there is no guarantee that

inversion will change an (R) carbon to an (S) carbon. In the example shown here, the

“umbrella” is inverted, but at the same time the substituents attached to the stereo-

genic carbon have also changed from (C, Cl, H, I) to (C, Cl, H, OH). Application of

the priority system shows that both molecules are (R) despite the inversion.

For most secondary and primary substrates, substitution occurs with complete

inversion of configuration. We now are faced with three questions: (1) How do we

know the reactions go with inversion? (2) What does this result tell us about the

reaction mechanism? (3) Why is inversion preferred to retention or racemization?

The answer to the first question is easy: We measure it. Suppose we start with the

optically active iodide in Figure 7.14 and monitor the reaction with radioactive

iodide ion (



We measure two things: the incorporation of radioactive iodide and

the optical activity of . If radioactive iodide reacts with with retention

of configuration, there will be no loss of rotation at all [I

and I are not significantly

different in their contributions to the rotation of plane-polarized light (Fig. 7.14)].

I* I*

–

R I*R I

FIGURE 7.14 A mechanism involving

retention predicts that incorporation

of radioactive iodide will induce no

change in optical rotation.

What happens if the reaction goes with inversion? The observed optical rotation

of is certainly going to decrease as, for example, (R)-iodide is converted into

(S)-iodide. But it is usually not obvious how the rate of loss of optical activity com-

pares with the rate of incorporation of I

. If every replacement occurs with inversion,

the rate of loss of optical activity must be twice the rate of incorporation of I

.Every

conversion of an (R)-iodide ( ) into an (S)-iodide ( ) generates a product

molecule ( ) whose rotation exactly cancels that of a molecule of starting

In effect, the rotation of two molecules goes to zero for every inversion (Fig. 7.15).

I.I

inversion

Net rotation of this set of molecules = 4(R)

(R)

(R)(R)(R)(R)

(S)

(R)(R)

(R)(S)

…thus, the net rotation of this set of

four molecules = 2

(

)

The boxed set of two

molecules contributes

0⬚ to the rotation…

–

FIGURE 7.15 A mechanism involving inversion. For the incorporation of one atom of radioactive iodide, the

rotation of the set of molecules shown will be cut by a factor of 2.

7.4 Substitution, Nucleophilic, Bimolecular: The S

2 Reaction 271

Had the reaction occurred with retention, a frontside approach would have been

demanded. The incoming nucleophile would have had to enter the molecule from

the same side as the departing leaving group.The tetrahedral “umbrella”would have

retained its configuration. This result is not observed (Fig. 7.17).

inversion

–

FIGURE 7.16 The observation of inversion

in the S

2 reaction means that the entering

nucleophile, radioactive (or labeled) iodide, must

enter the molecule from the side opposite the

departing leaving group, here unlabeled iodide.

This change is exactly what is observed experimentally. Every incorporation of I

is accompanied by the inversion of one (R)-iodide to an (S)-iodide.

Now, what does this result tell us about the mechanism? The observation of inver-

sion tells us that the nucleophile, the incoming atom or molecule, must be approaching

the substrate from the rear of the departing atom.Only this path can lead to the observed

inversion.Figure 7.16 shows the reaction,along with an arrow formalism description of

the bonds that are forming and breaking.The arrows show the path of approach of the

incoming nucleophile and the departing leaving group. The tetrahedral “umbrella” is

inverted during the reaction.

retention

–

FIGURE 7.17 If retention had been

the experimentally observed result,

the nucleophile would have added

from the same side that the leaving

group departed.

All this analysis is useful; we now know the path of approach for the nucleophile in

the S

2 reaction, and so we know more about the mechanism of this process. But our

newfound knowledge serves only to generate the inevitable next question.Why is inver-

sion preferred? Why is frontside displacement—retention—never observed? Here comes

part of the payoff for all the work we did in earlier chapters on molecular orbitals. One

of the key lessons of those early chapters was that interactions of filled and empty orbitals

are stabilizing, because the two electrons originally in the filled orbital can be accom-

modated in the new bonding orbital, and no electrons need occupy the antibonding

molecular orbital. Now let’s look at the S

2 reaction in orbital terms.A filled nonbond-

ing orbital “n” on the nucleophile interacts with an orbital on the substrate, ,

involved in the bond from R to the leaving group, L. Because the interaction of two

filled orbitals is not stabilizing, the filled orbital n must overlap with the empty, anti-

bonding σ

orbital of .This overlapping is nothing more than the interaction of

the nucleophile’s HOMO,n,with the sigma bond’s LUMO,σ

, shown schematically in

Figure 7.18.

Filled n

(HOMO)

Empty

σ*

(LUMO)

Energy

FIGURE 7.18 An orbital interaction

scheme for the stabilizing interaction

of a filled nonbonding orbital (n) on

the nucleophile overlapping with the

empty, antibonding (σ

) orbital on

the substrate.

What does σ

look like? Figure 7.19 shows us. Note that σ

has one of its large

lobes pointing to the rear of . That is where overlap with the filled n orbital

will be most efficient.

Now look what happens if the orbitals overlap in a way that would lead to reten-

tion through a frontside substitution (Fig. 7.20).Not only is the magnitude of inter-

action reduced by the relatively poor overlap with the smaller lobe of σ

, but there

is both a bonding and an antibonding interaction,and these will tend to cancel each

other. Frontside displacement is a very unlikely process indeed.

We can now map out the progress of the bond making and breaking in the S

reaction.The reaction starts as the nucleophile , approaches the rear of the sub-

strate. Because the HOMO, the nonbonding orbital n on , overlaps with the

LUMO, the antibonding σ

orbital of , the bond from C to L is weakened as

the antibonding orbital begins to be occupied. The bond length of increases

as the new bond between Nu and C begins to form. As lengthens, the other

groups attached to C bend back, in umbrella fashion, ultimately inverting (just like

the windblown umbrella) to form the product.For a symmetrical displacement such

as the reaction of radioactive



with discussed before, the midpoint of the

reaction will be the point at which the groups attached to C are exactly halfway

inverted (Fig. 7.21).

272 CHAPTER 7 Substitution and Elimination Reactions

What is the hybridization of the central carbon at this midpoint? It must be sp

because this carbon is surrounded by three coplanar groups.There are partial bonds

to the incoming nucleophile, , and the departing leaving group, . No picture

shows the competition between the two nucleophiles ( and ) as clearly as this

one.What Lewis acid are they competing for? A carbon 2p orbital,as the figure shows.

What is this structure? At least in the symmetrical displacement of by , it

is the halfway point between the equi-energetic starting material and product.

(a) Stabilizing interaction of filled and empty orbitals (strong)

(b) Destabilizing interaction of two filled orbitals (weak

)

n The filled

nonbonding orbital on

the nucleophile, Nu

n The filled

nonbonding orbital

on the nucleophile, Nu

Note good bonding

overlap here

Note weaker bonding

overlap here

σ* The empty,

antibonding

orbital of R L

σ The filled,

bonding

orbital of R L

FIGURE 7.19 (a) The overlap of the filled nonbonding orbital (n) with the “outside” lobe of the empty σ

orbital of

takes advantage of the good overlap available with this lobe of σ

. (b) Contrast (a) with the destabilizing interaction of n and

σ, the two filled orbitals.The interaction of two filled orbitals is destabilizing, but the overlap of the lobes involved,one fat,

the other thin, is poor.Therefore, this interaction is not important, and the stabilizing interaction shown in (a) dominates.

Bonding

σ* The empty antibonding

orbital of R

n The filled

nonbonding

orbital of the

nucleophile, Nu

Antibonding

FIGURE 7.20 Frontside substitution

(retention) requires overlap with the

inside part of σ

.There is poor

overlap with the small lobes of σ

as well as offsetting bonding and

antibonding interactions.

In the transition state for this symmetrical S

2 reaction, the

umbrella is halfway inverted and the carbon is sp

hybridized

–

FIGURE 7.21 In a symmetrical

inversion process, the midpoint,

called the transition state, is the point

at which the umbrella is exactly half-

inverted.

7.4 Substitution, Nucleophilic, Bimolecular: The S

2 Reaction 273

This structure is the transition state for this reaction. Is this transition state

something one could potentially catch and isolate? No! It occupies not an energy

minimum, as stable compounds do, but it sits on an energy maximum along the

path from starting material to product.We have seen such structures before: Recall,

for example, the eclipsed form of ethane or the transition state for the interconver-

sion of two pyramidal methyl radicals (Fig. 7.22).

Actually, even our picture of the displacement of by is ever so slightly wrong. Normal iodine (I) is

not exactly the same as radioactive iodine (I*), and even this reaction is not truly symmetrical. We can ignore

the tiny differences in isotopes, however, as long as we keep the principle straight. If you want to be more cor-

rect, insert the words “except for a tiny isotope effect” wherever appropriate.

–

Transition state

Activation

energy

Energy

Reaction progress

–

Transition state

Activation

energy

Energy

Reaction progress

(a) (b)

FIGURE 7.22 Two analogous diagrams. (a) The rotation about the carbon–carbon bond in ethane; (b) the displacement



I by



I*. In each case, two equi-energetic molecules are separated by a single energy barrier, the top of which is called

the transition state, shown in brackets.

Now, let’s consider the change in energy as the S

2 reaction proceeds (Fig. 7.22b).

The starting material and products are separated by the high-energy point of

the reaction, the transition state. The energy difference between starting material

and transition state is related to the rate of the reaction. The larger this difference,

the higher is the barrier, and the slower is the reaction.So, a quantity of vital impor-

tance to any study of a chemical reaction is the energy difference between starting

material and the transition state, called the activation energy.The reaction progress

diagram is symmetrical only if the substitution reaction is symmetrical, that is, if

starting material and product are of equal energy.

This specific picture is transferable to a general second-order substitution reac-

tion, but some differences will appear. First of all, usually the starting material and

product will not be at the same energy, and the transition state cannot be exactly

halfway between starting material and product.It will lie closer to one than the other.

274 CHAPTER 7 Substitution and Elimination Reactions

The transition state cannot be exactly sp

, only nearly sp

(notice its slightly pyram-

idal shape) (Fig. 7.23).

Figure 7.23 shows both an exothermic and endothermic S

2 substitution reaction.

If we read the figure conventionally from left to right, we see the exothermic reaction,

but if we read it “backward,”from right to left,we see the endothermic reaction.Notice

that the activation energy for the forward process must be smaller than that for the

reverse reaction.Thermodynamics will determine where the equilibrium settles out,and

there may be practical consequences if one partner is substantially favored over the other,

but one diagram will suffice to examine both directions. Many of us are prisoners of

our language, which is read from left to right, and we tend to do the same for these

energy diagrams. Nature is more versatile, or perhaps less prejudiced, than we are.

Energy

Reaction progress

Slightly pyramidal transition

state—in the general case it

is not exactly at the midpoint

between starting material

and product

–

Activation

energy

forward

Activation

energy

reverse

–

FIGURE 7.23 A general S

2 reaction.

If the nucleophile (Nu) and the

leaving group (L) are different,

the transition state is not exactly

at the midpoint of the reaction, and

the hybridization of carbon is only

approximately sp

MALIC ACID

COOH

OHH

COOH

now Latvia and who died in Tübingen, Germany, in 1957.

Indeed, the inversion now known to occur universally in the

2 reaction is sometimes called Walden inversion.

MJ’s teacher of organic chemistry, William Doering, told

a wonderful story about Paul Walden. Professor Doering was

visiting Tübingen in 1956 to talk about his work and overheard

a now-famous German chemist lamenting the fact that

Tübingen had not been bombed during the war. It seems that

most of his colleagues in other “more fortunate” cities had

newly built laboratories courtesy of Allied bombing.There was

an old gentleman sitting in the corner who had been wheeled

into the department from his nursing home to hear Doering’s

seminar. Paul Walden—for indeed it was he—was 92 at the

time. He overheard the remark and replied, scathingly, “Es ist

nicht der Käfig, sondern der kleine Vogel darin!” (It is not the

cage that matters, but the little bird inside!)

The chiral malic acid figures strongly in the original discovery

of inversion of configuration in the S

2 reaction. Malic acid is

sometimes called “apple acid” because of its high concentra-

tion in apples, nectarines, and some other fruits. In fact, it was

first isolated from apple juice as early as 1785. It functions as

a molecular carrier of the carbon dioxide absorbed by plants.

The CO

appears in the CH

COOH group of malic acid.

The principal actor in the early mechanistic work on the S

reaction was Paul Walden, a chemist born in 1863 in what is

PROBLEM 7.6 Explain the following change in rate for the S

2 reaction:

Rate for R  CH

is much faster than for R  (CH

)

+ CH

+ I

7.4 Substitution, Nucleophilic, Bimolecular: The S

2 Reaction 275

The take-home lesson here is an important one: There is a strong preference for

inversion in the S

2 reaction. Indeed, there is no authenticated example of reten-

tion of configuration in this process, despite a great deal of searching by some very

clever people.

Now let’s examine the effects of structural change in the various participants in

the S

2 reaction, the R group, the nucleophile, the leaving group, and the solvent.

7.4c Effects of Substrate Structure: The R Group The structure of the

R group makes a huge difference in the rate of the S

2 reaction.We anticipated this

result when we mentioned earlier that the practically useful S

2 reaction was

restricted to methyl, primary, and secondary substrates.By implication, the rate of the

2 reaction with tertiary substrates is zero, or at least negligibly small (Table 7.1).

Why should this be? The simple answer seems to be that in a tertiary substrate the

rear of the bond is guarded by three alkyl groups, and the incoming nucleo-

phile can find no unhindered path along which to approach the fat lobe of σ

(Fig. 7.24). So the S

2 reaction is disfavored for tertiary substrates, for all of which

steric hindrance to the approaching nucleophile is prohibitively severe. Another

substitution mechanism, favorable for tertiary substrates, becomes possible. It is

called the S

1 reaction, and we will deal with its mechanism in Section 7.6.

If this steric argument is correct, secondary substrates should react more slowly

than primary substrates, and primary substrates should be slower than methyl com-

pounds. In general, this is the case (Table 7.1). In practice, the S

2 reaction is usu-

ally useful as long as there is at least one hydrogen attached to the same carbon as

the leaving group. Thus, the S

2 reaction works only for methyl, primary, and sec-

ondary substrates, all of which have at least one hydrogen attached to the carbon at

which the substitution is occurring.The small size of hydrogen opens a path at the

rear for the incoming nucleophile.

TABLE 7.1 Average Rates of

2 Substitution Reactions

for Different Groups

Average

R Relative Rate

1.3

0.033

0.013

(CH

)

CH 8.3  10

4

(CH

)

CCH

2  10

7

(CH

)

C~0

CHCH

–

No S

2 substitution

FIGURE 7.24 For tertiary substrates,

approach from the rear is hindered by

the alkyl groups, here all shown as

methyls.This steric effect makes the

2 reaction impossible.

This picture of the S

2 reaction, which emphasizes steric effects, allows us to

make a prediction. In principle, there must be some primary group so gigantic that

the S

2 reaction would be unsuccessful.

In practice, it is rather easy to find such groups. Even the neopentyl group,

(CH

)

CCH

, is large enough to slow the bimolecular displacement reaction severely,

because the tert-butyl group blocks the best pathway for rearside displacement of

the leaving group (Fig. 7.25; Table 7.1).

Very slow substitution

Neopentyl –

–

WEB 3D

FIGURE 7.25 Even neopentyl

compounds, in which a tert-butyl

group shields the rear of the

bond, are hindered enough so that

the rate of the S

2 reaction is

extremely slow.

276 CHAPTER 7 Substitution and Elimination Reactions

–

FIGURE 7.26 The S

2 reaction takes place in a

normal fashion with cyclic substrates.

–

Transition state

–

120⬚

–

Transition state

Isopropyl bromide

Cyclopropyl bromide

Wants 120⬚

but must be 60⬚!

–

FIGURE 7.27 A comparison of S

2 transition states for isopropyl bromide and cyclopropyl bromide.

slow S

displacement

–

FIGURE 7.28 The slow S

displacement of bromide by iodide in

cyclohexyl bromide. Incoming iodide

is apparently blocked somewhat by

the axial carbon–hydrogen bonds.

TABLE 7.2 Relative

Reactivities of Cycloalkyl

Bromides in the S

2 Reaction

Compound Relative Rate

Cyclopropyl bromide 10

4

Cyclobutyl bromide 8  10

3

Cyclopentyl bromide 1.6

Cyclohexyl bromide 1  10

2

Isopropyl bromide 1.0

Why should small rings be so slow in the S

2 reaction? As with any question involv-

ing rates, we need to look at the structures of the transition states for the reaction to

find an answer.In the transition state,the substrate is hybridized approximately sp

,and

that requires bond angles close to 120°. The smaller the ring, the smaller are the

angles. For a three-membered ring, the optimal 120° must be squeezed

to 60°, and for a cyclobutane, squeezed to 90°.This contraction introduces severe angle

strain in the transition state, raises its energy, and slows the rate of reaction (Fig. 7.27).

PROBLEM 7.7 Use a ring compound to design a test of the stereochemistry of the

2 reaction.

PROBLEM 7.8 Wait! The argument just presented ignores angle strain in the start-

ing material. Cyclopropane itself is strained. Won’t that strain raise the energy

of the starting material and offset the energy raising of the transition state?

Comment. Hint: Consider angle strain in both starting material and transition

state—in which will it be more important?

In cyclohexane, the reaction rate is apparently slowed by steric interactions with

the axial hydrogens. There can be no great angle strain problem here (Fig. 7.28).

The S

2 reaction also takes place when ring compounds are used as the sub-

strates,although there are some interesting effects of ring size on the rate of the reac-

tion (Table 7.2; Fig. 7.26).

7.4 Substitution, Nucleophilic, Bimolecular: The S

2 Reaction 277

WORKED PROBLEM 7.9 The axial cyclohexyl iodide shown below reacts more

quickly in S

2 displacements than the equatorial cyclohexyl iodide. Draw the

transition states for displacement of iodide from axial cyclohexyl iodide by iodide

ion and for the analogous reaction of equatorial cyclohexyl iodide. What is the

relation between these transition states? Why does the axial iodo compound react

faster than the equatorial compound? Caution! Hard, tricky question.

ANSWER A tricky problem indeed. The transition states for the two reactions are

the same! There is no energy difference between the two transition states:

–

The transition state is the same for displacement of axial and equatorial cyclohexyl iodide

from 4-iodo-tert-butylcyclohexane

(CH

)

(CH

)

C (CH

)

Energy

Reaction progress

Energy increase

in starting material

Equatorial iodo

compound

Axial iodo

compound

Transition state

Activation energy

for the axial

compound (red

arrow) is lower

than that for the

equatorial

compound

(blue arrow)

Therefore, the reason that axial cyclo-

hexyl iodide reacts more quickly than

equatorial cyclohexyl iodide cannot lie

in the transition state energies.

However, the axial iodo group raises

the energy of the starting material,

thus lowering the activation energy.

7.4d Effect of the Nucleophile Some displacing agents are more effective

than others.They are better players in the competition for the carbon 2p orbital,the

Lewis acid. In this section, we examine what makes a good nucleophile, that is,

what makes a powerful displacing agent.

Remember: Size can be important. We have already seen how the S

2 reaction

can be slowed,or even stopped altogether,by large R groups (Fig.7.25). Presumably,

large nucleophiles will also have a difficult time in getting close enough to the sub-

strate to overlap effectively with σ

(Problem 7.6, p. 275). Figure 7.29 shows two

nucleophiles, carefully chosen to minimize all differences except size.

bond rotations

Azabicyclooctane Triethylamine

WEB 3D WEB 3D

FIGURE 7.29 In both azabicyclooctane

and triethylamine, the nucleophilic

nitrogen atom is flanked by three two-

carbon chains. In the bicyclic cage

compound azabicyclooctane, they are

tied back by the CH group shown in

red and are not free to rotate. In

triethylamine, they are free to rotate

and effectively increase the bulk of the

nucleophile.

The bicyclic molecule, in which the three ethyl-like groups are tied back out of

the way, is a more effective displacing agent than is triethylamine,in which the three

278 CHAPTER 7 Substitution and Elimination Reactions

–

25 ⬚C

nitrobenzene

25 ⬚C

nitrobenzene

Relative rate = 1

Relative

rate = 252

(CH

)

–

FIGURE 7.30 The tied-back bicyclic

compound reacts faster with ethyl

iodide than does triethylamine,

which is effectively the larger

compound.The larger nucleophile

has a more difficult time in attacking

the rear of the carbon–iodine bond

in ethyl iodide.

Nucleophilicity: The competition between two nucleophiles for an

empty carbon 2p orbital

Basicity: The competition between two bases for

an empty hydrogen 1s orbital

1s Orbital Base

2p OrbitalNucleophile

Nucleophile

–

Base

–

FIGURE 7.31 Brønsted basicity and nucleophilicity are different, but related, phenomena.

ethyl groups are freely rotating.Triethylamine is effectively larger than the cage com-

pound in which the alkyl groups are tied back (Fig. 7.30).

Energy

Stabilization

Empty

Filled

Empty

Filled

Stabilization

FIGURE 7.32 Stabilization is greater when filled and empty orbitals of equal or nearly equal energy interact

than it is when orbitals that differ greatly in energy interact.The further apart two orbitals are in energy, the

smaller is the stabilization resulting from their overlap.

Actually, it is simple to refine our discussion of good nucleophiles.What do we mean

by “a good competitor for a carbon 2p orbital”? We are talking about the overlap of a

filled orbital and an empty orbital, and the resulting stabilization is a measure of how

strong the interaction is. Remember: The strongest orbital interactions, and hence the

greatest stabilizations, come from the overlap of orbitals close in energy (Fig. 7.32).

Nucleophilicity,Lewis basicity, is a measure of how well a nucleophile competes

for an empty carbon 2p orbital, and Brønsted basicity is a measure of how well a

base competes for an empty hydrogen 1s orbital. One would expect there to be a

general correlation of nucleophilicity with base strength although the two are not

exactly the same thing, because 1s and 2p orbitals are different in energy and shape.

Nevertheless, basicity and nucleophilicity are related phenomena (Fig. 7.31).