Jones M., Fleming S.A. Organic Chemistry

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7.6 Substitution, Nucleophilic, Unimolecular: The S

1 Reaction 289

The S

2 reaction is by no means irrelevant to you—your life literally depends on

it happening properly, as in the example studied by Floss,and not in the wrong place—

as demonstrated by the sometimes lethal mutations induced by alkylating agents.

PROBLEM 7.15 Although tert-butyl bromide does not give tert-butyl alcohol on

reaction with hydroxide ion, an alkene product (C

) is produced. Suggest a

structure for the product and an arrow formalism for this bimolecular reaction.

7.6 Substitution, Nucleophilic, Unimolecular:

The S

1 Reaction

7.6a Rate Law tert-Butyl bromide does not undergo a substitution reaction when

treated with the good nucleophiles listed in Table 7.3. The rate of the second-order

2 displacement on tert-butyl bromide by hydroxide ion—or any nucleophile—is

effectively zero. Yet, tert-butyl bromide does react with the much less powerfully

nucleophilic molecule water to form the product of substitution, tert-butyl alcohol

(Fig. 7.52).

–

OHC

HBr

–

FIGURE 7.52 Despite the far greater

nucleophilicity of hydroxide ion, the

reaction of tert-butyl bromide with

water is successful in producing tert-

butyl alcohol, whereas the reaction

with hydroxide ion is not.

The first clue to what’s happening in Figure 7.52 comes from an examination of the

rate law for the reaction.It turns out that the formation of tert-butyl alcohol in this reac-

tion is a first-order process in which the concentration of added nucleophile is irrele-

vant to the rate.The rate law leads to the name Substitution,Nucleophilic,unimolecular,

or S

1 reaction, where k is the rate constant, a fundamental constant of the reaction.

Rate  ν  k [tert-butyl bromide]

At this point, a reasonable mechanistic hypothesis might simply view the S

reaction as beginning with the unimolecular ionization shown in Figure 7.53.

ionization

(slow)

addition

(fast)

deprotonation

OHC

–

BrH

WEB 3D

FIGURE 7.53 A schematic mechanism for the S

1 reaction. A slow ionization is followed by a relatively

fast capture of the carbocation by a nucleophile, water in this case.

Unimolecular nucleophilic substitution: S

290 CHAPTER 7 Substitution and Elimination Reactions

The reaction continues as the carbocation is captured by the solvent, often water, as

in Figure 7.53, in what is called a solvolysis reaction. Solvolysis simply means that

the solvent, here water, plays the role of nucleophile in the reaction. Next, any base

present, here likely to be bromide or water, can remove a proton (deprotonate the

protonated alcohol) to give the alcohol itself (Fig. 7.53).

We can describe the S

1 reaction with an energy diagram, as we did for the S

reaction (pp. 273–274). The S

1 diagram is more complicated because of the for-

mation of the intermediate carbocation.The presence of the intermediate means that

there will be two transition states separating the starting material and the product.

The transition state for the slow step, the initial endothermic ionization, is higher

than the transition state for the second, faster step, the capture of the ion by the

nucleophile (Fig. 7.54).

This second step determines the structure of the product but not the rate of the reac-

tion. If there are several nucleophiles present, they will compete with one another in

the capture of the carbocation to produce different products, but the rate of the reac-

tion will not be affected (Fig. 7.55). The nucleophile determines the structures of the

products, but not the rate of their formation. Note that the reverse of the ionization,

recapture of the carbocation by bromide, is nothing more than another example of a

reaction of this intermediate with a nucleophile.

Energy

Reaction progress

Transition state for

ionization—higher

Transition state for ion capture

by nucleophile—lower

–

FIGURE 7.54 A generic energy

diagram for the S

1 reaction.

Notice that the transition state for

the slow step in the reaction—the

ionization—is higher in energy than

the transition state for the capture of

the ion by a nucleophile.

FIGURE 7.55 Capture of the carbocation by four

nucleophiles (



Nu,



Nu′,



Br, and H

O).The

nucleophiles have no effect on the rate of formation of

the carbocation, which is the slow step in the reaction.

slow

ionization

fast

–

Nu´

–

Nu´

–

7.6 Substitution, Nucleophilic, Unimolecular: The S

1 Reaction 291

We will now see if this simple hypothesis allows a reasonable prediction of the

stereochemical course of the reaction.

7.6b Stereochemistry of the S

1 Reaction If the S

1 reaction were as

straightforward as suggested by Figures 7.53–7.55, what would be expected if an

optically active starting material were used? The crucial intermediate in the reaction

is the planar carbocation, an achiral molecule! Once the free, planar carbocation is

formed, optical activity is inevitably lost. Addition of the nucleophile must take

place at both faces of the cation, forming equal amounts of (R) and (S) product.

This simple mechanism demands racemization, as Figure 7.56 shows.

Sometimes racemization is the actual result, and there is 50% inversion and 50%

retention in the product.Usually, however the stereochemical results are not so clean,

and an excess of inversion is found. For example, the two optically active molecules

of Figure 7.57 give 21 and 18% excess inversion in the S

1 reaction with water.

Chiral

50% Retention

50% Inversion

Achiral, planar carbocation

Racemic mixture

–

FIGURE 7.56 Loss of optical activity

occurs as soon as the free, planar

carbocation is formed. Addition of

any nucleophile will occur equally at

the two available lobes of the 2p

orbital to give a racemic mixture of

products.

(CH

)

CHCH

(39.5% retention) (60.5% inversion)

(41% retention) (59% inversion)

(CH

)

CHCH

CH(CH

)

acetone

WEB 3D

FIGURE 7.57 Incomplete

racemization is often found in S

reactions. An excess of inversion is

generally found.

Why is the product not racemic, and why is excess inversion (as opposed to reten-

tion) observed? These two questions are vitally important, as satisfactory answers

must be found if we are not to be forced to abandon the proposed mechanism for

the S

1 reaction.The answers turn out to be simple. In the mechanism outlined in

Figures 7.53–7.55, we assumed that the leaving group had diffused away from the

carbocation, leaving behind a symmetrical ion. What if reaction with solvent occurs

292 CHAPTER 7 Substitution and Elimination Reactions

before the leaving group diffuses completely away? What if reaction occurs with the

pair of ions shown in Figure 7.58? The leaving group guards the side of the carbo-

cation from which it departed, and makes addition by solvent water more difficult

at this “retention” side. The result is usually an excess of inversion.

7.6c Effects of Substrate Structure: The R Group The rate of the S

reaction is much greater for tertiary substrates than for secondary substrates, and

simple primary alkyl and methyl halides do not react at all by the S

1 process. Our

mechanistic hypothesis shows us why. If carbocation formation is the key step, only

those compounds that can form relatively stable ions can react in S

1 fashion.

Modern measurements show that the stability of carbocations increases with

substitution (p. 137). Table 7.5 shows the heats of formation of carbocations in

the gas phase. Remember: The more positive its heat of formation, the less stable

a species is (p. 115). We can’t make too much of the quantitative data in Table 7.5,

C* C

<50% Retention

>50% Inversion

Chiral,

optically

active

Inversion pathway

is unhindered

Retention pathway is

hindered by the bromide ion

–

FIGURE 7.58 Until the leaving group,

here bromide ion, completely diffuses

away, it shields the carbocation along

the retention pathway, and this results

in an excess of inversion in most S

reactions.

TABLE 7.5 Heats of Formation for Some Carbocations

Entry Cation Substitution (kcal/mol)



Methyl 261.3 Least stable



Primary 215.6



Primary 211

4 Secondary 190.9



Primary 203

6 Secondary 183

7(CH

)



Tertiary 165.8 Most stable

HCH

(CH

)

≤H °

PROBLEM 7.16 The second reaction of Figure 7.57 shows an S

1 reaction of a sec-

ondary halide. Explain why the S

1 reaction should be especially favorable for

this secondary halide.

PROBLEM 7.17 Consider the role of solvent on the stereochemistry of the S

reaction shown in Figure 7.58. A more polar solvent favors racemization. Why?

7.6 Substitution, Nucleophilic, Unimolecular: The S

1 Reaction 293

because these gas-phase measurements cannot take into account the effects of sol-

vent, which are very large indeed. Nonetheless, the data do give a good qualita-

tive picture of carbocation stability: Entries 5, 6, and 7 show that tertiary

carbocations (three alkyl groups) are more stable than secondary ones (two alkyl

groups), and secondary carbocations are more stable than primary ones (one alkyl

group). The methyl cation (



), with no alkyl groups, is the least stable of all.

Do not confuse these relative stabilities with absolute stabilities. Carbocations are

almost all very unstable species. A small atom such as carbon does not bear a pos-

itive charge with any grace, and a carbocation will react with available Lewis bases

(e.g., halide ion) very quickly. Do not say, “The tert-butyl cation is very stable,”

but rather, “For a carbocation, the tert-butyl cation is quite stable.”

PROBLEM SOLVING

WORKED PROBLEM 7.18 We have again avoided (p. 138) a detailed discussion of

why more substituted carbocations are more stable than less substituted ones.

Nevertheless you can now see the outlines of why this is so. Draw a three-

dimensional picture of the ethyl cation. (Remember: The positively charged

carbon and the three surrounding atoms are coplanar—this part of the ethyl

cation is flat.) Now consider the overlap of one of the carbon–hydrogen bonds

of the methyl group with the empty 2p

orbital. Is this interaction stabi-

lizing? Why?

ANSWER Interactions between filled and empty orbitals are stabilizing. This fun-

damental tenet of chemistry can be applied in this case. If the filled orbital of the

carbon–hydrogen bond overlaps with the empty 2p orbital of the carbocation,

the result is the stabilizing interaction called hyperconjugation.

Filled C

bond

Empty 2p

orbital

Energy

C H

σ bond

If avoiding the question drives you crazy, and we think it should at least make you a little uncomfortable, be

sure to do this problem—the answer falls right out. See also Chapter 9.

(continued)

STOP

There are many reactions in which tertiary and secondary

carbocations are intermediates but no reactions for which a simple

primary or methyl cation is required. These intermediates are

simply too unstable. In practice, this instability means that when

you find yourself writing a mechanism using a primary carbocation, think again,

it’s almost certainly wrong!

294 CHAPTER 7 Substitution and Elimination Reactions

This interaction can occur only if the appropriate carbon–hydrogen bonds exist.

The more alkyl groups, the more stabilization there is. For example, in the methyl

cation no such stabilization is possible because there are no alkyl groups to sup-

ply the needed bonds.

MethylPrimarySecondaryTertiary

faster

than

much

faster

than

orC H

L L L L

FIGURE 7.59 Relative reactivities in

the S

1 reaction.

PROBLEM 7.19 The heat of formation for cation A is 201 kcal/mol, and that of

cation B is 218.6 kcal/mol. Draw the two cations carefully and explain why A is

more stable than B.

So, tertiary compounds, which form the most stable, tertiary carbocations, com-

monly react through the S

1 mechanism, and methyl and primary compounds,

which would have to form extremely unstable carbocations, do not.Naturally enough,

secondary compounds react faster than primary ones in the S

1 reaction but more

slowly than tertiary compounds (Fig. 7.59).

7.6d Effects of the Nucleophile We have already seen that the nucleo-

phile does not affect the rate of the S

1 reaction—the reaction is first order in

substrate.Yet, the nucleophile does determine the product structure.Note the dis-

tinction between the rate-determining step (or rate-limiting step) and the

product-determining step. For example, reaction of tert-butyl bromide in water

containing a small amount of an added nucleophilic ion, such as cyanide, leads to

substantial amounts of tert-butyl cyanide. Even though water is present as solvent

and thus has an immense advantage over cyanide ion because of concentration,

the more powerfully nucleophilic cyanide still competes. Cyanide is many times

more reactive toward the carbocation than the solvent water is (Table 7.3, p. 279).

7.6 Substitution, Nucleophilic, Unimolecular: The S

1 Reaction 295

Energy

–

and

Reaction progress

Product-determining steps

–

Rate-determining step

FIGURE 7.60 The rate-determining

step and product-determining step

are sharply distinguished in the S

reaction.

In this reaction, the rate-determining step (the slow ionization) and the product-

determining step (the faster capture of the intermediate ion by nucleophiles) are

sharply separated (Fig. 7.60).

7.6e Effect of the Leaving Group The identity of the leaving group has a

large influence on the rate of the S

1 reaction because the structure of the depart-

ing group affects the ease of ionization. We have already discussed how the

structure—and hence stability—of the carbocation affects the rate of ionization, but

obviously the stability of the negatively charged counterion ( ) will be important

as well. The more stable the anion, the more easily it will be lost (Fig. 7.61). Some

good and poor leaving groups are shown in Table 7.4 (p. 281).

–

FIGURE 7.61 The rate of ionization is

affected by the stability of the leaving

group, . The more stable the

leaving group, the faster is the

ionization.

7.6f Effect of Solvent For the S

2 reaction, the effect of solvent is rather compli-

cated and depends on the charge type of the reaction. Some S

2 reactions are accelerat-

ed by a change to a more polar solvent;others are slowed down by such a change (p.286).

For most S

1 reactions the situation is simpler.Ions are usually formed in the slow step

of the reaction. In a polar solvent, represented by the dipole arrows in Figure 7.62, the

charged species formed in the S

1 reaction are stabilized. A nonpolar solvent cannot

stabilize as well. One can expect that the S

1 reaction will usually be favored by a

polar solvent.

–

FIGURE 7.62 A polar solvent

stabilizes the ions formed in

an S

1 reaction.

296 CHAPTER 7 Substitution and Elimination Reactions

7.7 Summary and Overview of the S

and S

1 Reactions

These two substitution reactions are interconnected by more than the ultimate for-

mation of a substitution product.They complement each other—when one is inef-

fective, the other is likely to work well. For example, the S

2 reaction works best for

methyl and primary substrates, which is exactly the point at which the S

1 reaction

is least effective. When the S

2 reaction fails, as with tertiary substrates, the S

reaction is most effective (Fig. 7.63).

So, at the extremes the situation is clear.The S

2 reaction dominates for methyl

and primary substrates because there is little steric hindrance to the displacement

of the leaving group and because the competing S

1 reaction is disfavored by the

necessity of forming the highly unstable methyl or primary carbocations (Fig. 7.64).

The S

1 reaction dominates for tertiary substrates because ionization can produce

the relatively stable tertiary carbocation and because the competing S

2 reaction is

strongly disfavored by the steric hindrance to the nucleophile approaching from the

rear (Fig. 7.65).

Tertiary

Secondary

Primary Methyl

Efficiency of the S

1 reaction

No S

1 No S

Efficiency of the S

2 reactionNo S

L L L L

FIGURE 7.63 The S

2 and S

reactions are complementary.

A difficult ionization to the hideously

unstable primary carbocation

+ +

A relatively easy displacement

at the unhindered primary carbon

LL L

–

FIGURE 7.64 A comparison of the S

1 and S

2 reactions for primary substrates.

An easy S

1 reaction

as the relatively stable

tertiary carbocation is

formed

The S

2 is thwarted

by the difficulty of

displacement at the

sterically hindered

tertiary carbon

–

FIGURE 7.65 For tertiary systems, the S

1 reaction dominates; the ionization produces the relatively stable

tertiary carbocation and the S

2 reaction is hindered by the three R groups that thwart approach of the

nucleophile from the rear.

7.7 Summary and Overview of the S

2 and S

1 Reactions 297

In the middle ground—secondary substrates—we might expect to be able to find

both reactions. Small R groups, use of a powerful nucleophile, and a solvent of proper

polarity for the charge type of the reaction favor the S

2 reaction. Large R groups, a

weak nucleophile, and a polar solvent favor the S

1 reaction (Fig. 7.66).

We have given a picture of these reactions in which extremes are emphasized,

in which an “either S

2 or S

1”view is developed. However, the situation is messier

than this simple explanation would suggest, especially for secondary compounds.

Imagine beginning to ionize a bond to a leaving group. As the bond stretches,

positive charge begins to develop on carbon (δ



) and rehybridization toward sp

begins as the groups attached to the carbon flatten out (Fig. 7.67).

If we simply continue the process of ionization, we have the pure S

1 reac-

tion. If we let a nucleophilic solvent molecule interact with the rear of the depart-

ing bond and ultimately form a bond to carbon, we have the S

2 reaction with

inversion (Fig. 7.68).

H H

C C

–

FIGURE 7.66 For secondary substrates, both S

1 and S

2 reactions may be possible.

–

FIGURE 7.67 The beginning of an

1 reaction.The bond to the leaving

group L



has lengthened, and the

tetrahedron has begun to flatten out.

The hybridization of the central

carbon is changing from nearly sp

the sp

of the planar carbocation.

However, solvation at the side away from the old leaving group can be converted into nucleophilic attack (S

2) to give inverted produc

–

C C

–

+ +

In the classic S

1 reaction, this ionization continues to produce the planar

carbocation. Notice, however, that the carbocation when first formed is not

symmetrical. The leaving group lurks on one side, partially shielding the cation

from solvent and other nucleophiles

–

C C

–

L L

L L L L

–

FIGURE 7.68 Especially for secondary substrates, the difference between S

2, displacement by solvent, and preferential

solvation at the rear of the departing bond is a subtle one.

How strong an interaction with the rear lobe of the orbital is needed before we call

it bonding? In polar solvents, ions are highly solvated, and in an S

1 reaction solvation

will begin to be important as soon as a δ



charge begins to build up on carbon. Solvation

will be especially easy at the rear of the departing group as the leaving group shields the

frontside. We have already seen that the leaving group can hinder reaction at one side

of the developing cation and produce some net inversion in the reaction (Fig. 7.58).

PROBLEM 7.20 We saw at least one S

1 reaction before, although it was not

labeled as such because we didn’t yet know the name. Where was it? If you get

this one, you have a great eye and memory for detail. Hint: It is in this chapter.

298 CHAPTER 7 Substitution and Elimination Reactions

Now we begin to see the difficulty of telling the two mechanisms apart in some

situations. When does “preferential solvation from the rear” become “displacement

from the rear”? It becomes painfully hard to sort out in some cases. Organic chem-

istry is frustratingly, if delightfully, messy.

This messiness raises its head in other ways. Almost all substitution reactions are

accompanied by some formation of alkene. In organic chemistry few reactions pro-

ceed in 100% yield, quantitatively producing one molecule from another. A large part

of the business of organic synthesis is the search for specificity. Synthetic chemists seek

reagents and reaction conditions that greatly favor one reaction pathway over others.

In particular, attempts to synthesize molecules through substitution reactions must

take account of competitive reactions called eliminations.A hydrogen atom and a leav-

ing group (L) are lost (eliminated) from the substrate to give an alkene (Fig. 7.69).

Sections 7.8 and 7.9 examine two general mechanisms for these elimination reactions.

7.8 The Unimolecular Elimination Reaction: E1

7.8a Rate Law If we determine the rate law for the formation of the alkene

accompanying an S

1 reaction, we find that this elimination reaction is also a first-

order process, . As in the S

1 reaction, the slow step in this reaction is

the ionization of the substrate to give a carbocation, which can do two things: It can

continue the S

1 reaction by adding a nucleophile (Nu ) or, if a proton is available,

an alkene can be formed in what is called the E1 reaction. Note that in the elimi-

nation phase of the reaction, the loss of hydrogen, the nucleophile is acting as a

Brønsted base.The carbocation is an intermediate in both processes (Fig. 7.70).

ν = k[R

++++

Substitution

reaction

Elimination

reaction

NuL

–

FIGURE 7.69 In almost all S

1 and

2 reactions, substitution is

accompanied by some elimination

of to give an alkene.H

Unimolecular elimination: E1

Here

–

Nu is acting as

Lewis base = nucleophile

Here

–

Nu is acting

as Brønsted base

THE GENERAL CASE

Favored by a highly

nucleophilic nucleophil

Favored by a highly

basic nucleophile

CCH

–

FIGURE 7.70 Competing E1 and S

1 reactions.

The ratio of E1 and S

1 products depends primarily on the nucleophile’s basicity.

Strong Brønsted bases are more effective at removing the proton than weak Brønsted

bases. That’s exactly what we mean by a strong Brønsted base—something that has a