A Summary of $\mathrm{S}_{\mathrm{N}} 2$ Reaction Characteristics - 11.4 The $\mathrm{S}_{\mathrm{N}} 1$ Reaction
Most nucleophilic substitutions take place by the $S_{N} 2$ pathway just discussed. The reaction is favored when carried out with an unhindered substrate and a negatively charged nucleophile in a polar aprotic solvent, but is disfavored when carried out with a hindered substrate and a neutral nucleophile in a protic solvent. You might therefore expect the reaction of a tertiary substrate (hindered) with water (neutral, protic) to be among the slowest of substitution reactions. Remarkably, however, the opposite is true. The reaction of the tertiary halide 2-bromo-2-methylpropane $\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CBr}$ with $\mathrm{H}_{2} \mathrm{O}$ to give the alcohol 2-methyl-2-propanol is more than 1 million times faster than the corresponding reaction of $\mathrm{CH}_{3} \mathrm{Br}$ to give methanol.

What's going on here? A nucleophilic substitution reaction is occurring-a hydroxyl group is replacing a halogen-yet the reactivity order seems backward. These reactions can't be taking place by the $S_{N} 2$ mechanism we've been discussing, so we must therefore conclude that they are occurring by an alternative substitution
mechanism. This alternative mechanism is called the $\mathbf{S}_{\mathbf{N}} \mathbf{1}$ reaction, for substitution, nucleophilic, unimolecular.

In contrast to the $\mathrm{S}_{\mathrm{N}} 2$ reaction of $\mathrm{CH}_{3} \mathrm{Br}$ with $\mathrm{OH}^{-}$, the $\mathrm{S}_{\mathrm{N}} 1$ reaction of $\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CBr}$ with $\mathrm{H}_{2} \mathrm{O}$ has a rate that depends only on the alkyl halide concentration and is independent of the $\mathrm{H}_{2} \mathrm{O}$ concentration. In other words, the process is a first-order reaction; the concentration of the nucleophile does not appear in the rate equation.


A Summary of $\mathrm{S}_{\mathrm{N}} 2$ Reaction Characteristics - 11.4 The $\mathrm{S}_{\mathrm{N}} 1$ Reaction
$$
\begin{aligned}
\text { Reaction rate } & =\text { Rate of disappearance of alkyl halide } \\
& =k \times[\mathrm{RX}]
\end{aligned}
$$

To explain this result, we need to know more about kinetics measurements. Many organic reactions occur in several steps, one of which usually has a higher-energy transition state than the others and is therefore slower. We call this step with the highest transition-state energy the rate-limiting step, or rate-determining step. No reaction can proceed faster than its rate-limiting step, which acts as a kind of traffic jam, or bottleneck. In the $\mathrm{S}_{\mathrm{N}} 1$ reaction of $\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CBr}$ with $\mathrm{H}_{2} \mathrm{O}$, the fact that the nucleophile concentration does not appear in the firstorder rate equation means that it is not involved in the rate-limiting step and must therefore be involved in some other, non-rate-limiting step. The mechanism shown in FIGURE 11.9 accounts for these observations.


FIGURE 11.9 MECHANISM -
The mechanism of the $\mathrm{S}_{\mathrm{N}} 1$ reaction of 2-bromo-2-methylpropane with $\mathrm{H}_{2} \mathrm{O}$ involves three steps. Step 1 -the spontaneous, unimolecular dissociation of the alkyl bromide to yield a carbocation-is rate-limiting.
(1) Spontaneous dissociation of the alkyl bromide occurs in a slow, rate-limiting step to generate a carbocation intermediate plus bromide ion.
2) The carbocation intermediate reacts with water as a nucleophile in a fast step to yield protonated alcohol as product.
(3) Loss of a proton from the protonated alcohol intermediate then gives the neutral alcohol product.

Unlike what occurs in an $\mathrm{S}_{\mathrm{N}} 2$ reaction, where the leaving group is displaced while the incoming nucleophile approaches, an $\mathrm{S}_{\mathrm{N}} 1$ reaction takes place by loss of the leaving group before the nucleophile approaches. 2-Bromo-2-methylpropane spontaneously dissociates to the tert-butyl carbocation $\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}^{+}$, plus $\mathrm{Br}^{-}$in a slow, rate-limiting step, and the intermediate carbocation is then immediately trapped by the nucleophile water in
a faster second step. Thus, water is not a reactant in the step whose rate is measured. The energy diagram is shown in FIGURE 11.10.

FIGURE 11.10 An energy diagram for an $\mathbf{S}_{\mathbf{N}} \mathbf{1}$ reaction. The rate-limiting step is the spontaneous dissociation of the alkyl halide to give a carbocation intermediate. Reaction of the carbocation with a nucleophile then occurs in a second, faster step.


FIGURE 11.9 MECHANISM -
Because an $\mathrm{S}_{\mathrm{N}} 1$ reaction occurs through a carbocation intermediate, its stereochemical outcome is different from that of an $S_{N} 2$ reaction. Carbocations, as we've seen, are planar, $s p^{2}$-hybridized, and achiral. Thus, if we carry out an $\mathrm{S}_{\mathrm{N}} 1$ reaction on one enantiomer of a chiral reactant and go through an achiral carbocation intermediate, the product loses its optical activity (Section 8.12). That is, the symmetrical intermediate carbocation can react with a nucleophile equally well from either side, leading to a racemic, 50 : 50 mixture of enantiomers (FIGURE 11.11).

FIGURE 11.11 Stereochemistry of the $\mathbf{S}_{\mathbf{N}} \mathbf{1}$ reaction. Because the reaction goes through an achiral intermediate, an enantiomerically pure reactant gives an optically inactive racemic product.

The conclusion that $\mathrm{S}_{\mathrm{N}} 1$ reactions on enantiomerically pure substrates should give racemic products is nearly, but not exactly, what is found. In fact, few $\mathrm{S}_{\mathrm{N}} 1$ displacements occur with complete racemization. Most give a minor ( $0-20 \%$ ) excess of inversion. The reaction of ( $R$ )-6-chloro-2,6-dimethyloctane with $\mathrm{H}_{2} \mathrm{O}$, for example, leads to an alcohol product that is approximately $80 \%$ racemized and $20 \%$ inverted $(80 \% R, S+20 \% S$ is equivalent to $40 \% R+60 \% S$ ).

(R)-6-Chloro-2,6-dimethyloctane

60\% S (inversion)

40\% R (retention)


FIGURE 11.9 MECHANISM -
(R)-6-Chloro-2,6-dimethyloctane

60\% S (inversion)

40\% R (retention)

This lack of complete racemization in $\mathrm{S}_{\mathrm{N}} 1$ reactions is due to the fact that ion pairs are involved. According to this explanation, first proposed by Saul Winstein at UCLA, dissociation of the substrate occurs to give a structure in which the two ions are still loosely associated and in which the carbocation is effectively shielded from reaction on one side by the departing anion. If a certain amount of substitution occurs before the two ions fully diffuse apart, then a net inversion of configuration will be observed (FIGURE 11.12).

Inversion
Racemization
FIGURE 11.12 Ion pairs in an $\mathbf{S}_{\mathbf{N}} \mathbf{1}$ reaction. The leaving group shields one side of the carbocation intermediate from reaction with the nucleophile, thereby leading to some inversion of configuration rather than complete racemization.

PROBLEM What product(s) would you expect from reaction of (S)-3-chloro-3-methyloctane with acetic acid?
11-8 Show the stereochemistry of both reactant and product.
PROBLEM Among the many examples of $\mathrm{S}_{\mathrm{N}} 1$ reactions that occur with incomplete racemization, the optically 11-9 pure tosylate of 2,2-dimethyl-1-phenyl-1-propanol $\left([\alpha]_{D}=-30.3\right)$ gives the corresponding acetate $\left([\alpha]_{D}=+5.3\right)$ when heated in acetic acid. If complete inversion had occurred, the optically pure acetate would have had $[\alpha]_{D}=+53.6$. What percentage racemization and what percentage inversion occurred in this reaction?

PROBLEM Assign configuration to the following substrate, and show the stereochemistry and identity of the
11-10 product you would obtain by $S_{N} 1$ reaction with water (reddish brown $=B r$ ):


FIGURE 11.9 MECHANISM - 11.5 Characteristics of the $\mathrm{S}_{\mathrm{N}} 1$ Reaction
Just as the $S_{N} 2$ reaction is strongly influenced by the structure of the substrate, the leaving group, the nucleophile, and the solvent, the $S_{N} 1$ reaction is similarly influenced. Factors that lower $\Delta G^{\ddagger}$, either by lowering the energy level of the transition state or by raising the energy level of the ground state, favor faster $\mathrm{S}_{\mathrm{N}} 1$ reactions. Conversely, factors that raise $\Delta G^{\ddagger}$, either by raising the energy level of the transition state or by lowering the energy level of the reactant, slow down the $\mathrm{S}_{\mathrm{N}} 1$ reaction.


The Substrate -
According to the Hammond postulate (Section 7.10), any factor that stabilizes a high-energy intermediate also stabilizes the transition state leading to that intermediate. Because the rate-limiting step in an $\mathrm{S}_{\mathrm{N}} 1$ reaction is the spontaneous, unimolecular dissociation of the substrate to yield a carbocation, the reaction is favored whenever a stabilized carbocation intermediate is formed. The more stable the carbocation intermediate, the faster the $\mathrm{S}_{\mathrm{N}} 1$ reaction.

We saw in Section 7.9 that the stability order of alkyl carbocations is $3^{\circ}>2^{\circ}>1^{\circ}>$ methyl. To this list we should also add the resonance-stabilized allyl and benzyl cations. Just as allylic radicals are unusually stable because the unpaired electron can be delocalized over an extended $\pi$ orbital system (Section 10.4), so allylic and benzylic carbocations are unusually stable. (The word benzylic means "next to an aromatic ring.") As FIGURE 11.13 indicates, an allylic cation has two resonance forms. In one form, the double bond is on the "left"; in the other form it's on the "right." A benzylic cation has five resonance forms, all of which contribute to the overall resonance hybrid.

FIGURE 11.13 Resonance forms of allylic and benzylic carbocations. The positive charge is delocalized over the $\pi$ system in both. Electron-poor atoms are indicated by blue arrows.

Because of resonance stabilization, a primary allylic or benzylic carbocation is about as stable as a secondary
alkyl carbocation, and a secondary allylic or benzylic carbocation is about as stable as a tertiary alkyl carbocation. This stability order of carbocations is the same as the order of $\mathrm{S}_{\mathrm{N}} 1$ reactivity for alkyl halides and tosylates.


Carbocation stability -
We should also note parenthetically that primary allylic and benzylic substrates are particularly reactive in $\mathrm{S}_{\mathrm{N}} 2$ reactions as well as in $\mathrm{S}_{\mathrm{N}} 1$ reactions. Allylic and benzylic $\mathrm{C}-\mathrm{X}$ bonds are about $50 \mathrm{~kJ} / \mathrm{mol}$ ( $12 \mathrm{kcal} / \mathrm{mol}$ ) weaker than the corresponding saturated bonds and are therefore more easily broken.

PROBLEM Rank the following substances in order of their expected $\mathrm{S}_{\mathrm{N}} 1$ reactivity: 11-11

PROBLEM 3-Bromo-1-butene and 1-bromo-2-butene undergo $\mathrm{S}_{\mathrm{N}} 1$ reaction at nearly the same rate, even 11-12 though one is a secondary halide and the other is primary. Explain.


The Leaving Group -
We said during the discussion of $S_{N} 2$ reactivity that the best leaving groups are those that are most stable; that is, those that are the conjugate bases of strong acids. An identical reactivity order is found for the $\mathrm{S}_{\mathrm{N}} 1$ reaction because the leaving group is directly involved in the rate-limiting step. Thus, the $\mathrm{S}_{\mathrm{N}} 1$ reactivity order is

$$
\mathrm{HO}^{-}<\mathrm{Cl}^{-}<\mathrm{Br}^{-}<\mathrm{I}^{-} \approx \mathrm{TosO}^{-}<\mathrm{H}_{2} \mathrm{O}
$$


Leaving group reactivity -
Note that in the $S_{N} 1$ reaction, which is often carried out under acidic conditions, neutral water is sometimes the leaving group. This occurs, for example, when an alkyl halide is prepared from a tertiary alcohol by reaction with HBr or HCl (Section 10.5). As shown in FIGURE 11.14, the alcohol is first protonated and then spontaneously loses $\mathrm{H}_{2} \mathrm{O}$ to generate a carbocation, which reacts with halide ion to give the alkyl halide. Knowing that an $\mathrm{S}_{\mathrm{N}} 1$ reaction is involved in the conversion of alcohols to alkyl halides explains why the reaction works well only for tertiary alcohols. Tertiary alcohols react fastest because they give the most stable carbocation intermediates.


FIGURE 11.14 MECHANISM -
The mechanism of the $\mathbf{S}_{\mathbf{N}} 1$ reaction of a tertiary alcohol with HBr to yield an alkyl halide. Neutral water is the leaving group (step 2).
(1) The -OH group is first protonated by HBr .
(2) Spontaneous dissociation of the protonated alcohol occurs in a slow, rate-limiting step to yield a carbocation intermediate plus water.

Carbocation
(3) The carbocation intermediate reacts with bromide ion in a fast step to yield the neutral substitution product.

(1)

© ||

(3)


The Nucleophile -
The nature of the nucleophile plays a major role in the $S_{N} 2$ reaction but does not affect an $S_{N} 1$ reaction. Because the $\mathrm{S}_{\mathrm{N}} 1$ reaction occurs through a rate-limiting step in which the added nucleophile has no part, the nucleophile can't affect the reaction rate. The reaction of 2-methyl-2-propanol with HX, for instance, occurs at the same rate regardless of whether X is $\mathrm{Cl}, \mathrm{Br}$, or I. Furthermore, neutral nucleophiles are just as effective as negatively charged ones, so $\mathrm{S}_{\mathrm{N}} 1$ reactions frequently occur under neutral or acidic conditions.

2-Methyl-2-propanol
(Same rate for $\mathrm{X}=\mathrm{Cl}, \mathrm{Br}, \mathrm{I}$ )


The Solvent -
What about the solvent? Do solvents have the same effect in $\mathrm{S}_{\mathrm{N}} 1$ reactions that they have in $\mathrm{S}_{\mathrm{N}} 2$ reactions? The answer is both yes and no. Yes, solvents have a large effect on $\mathrm{S}_{\mathrm{N}} 1$ reactions, but no, the reasons for the effects on $S_{N} 1$ and $S_{N} 2$ reactions are not the same. Solvent effects in the $S_{N} 2$ reaction are due largely to stabilization or destabilization of the nucleophile reactant, while solvent effects in the $\mathrm{S}_{\mathrm{N}} 1$ reaction are due largely to stabilization or destabilization of the transition state.

The Hammond postulate says that any factor stabilizing the intermediate carbocation should increase the rate of an $S_{N} 1$ reaction. Solvation of the carbocation-the interaction of the ion with solvent molecules-has such an
effect. Solvent molecules orient around the carbocation so that the electron-rich ends of the solvent dipoles face the positive charge (FIGURE 11.15), thereby lowering the energy of the ion and favoring its formation.

FIGURE 11.15 Solvation of a carbocation by water. The electron-rich oxygen atoms of solvent molecules orient around the positively charged carbocation and thereby stabilize it.

The properties of a solvent that contribute to its ability to stabilize ions by solvation are related to the solvent's polarity. $\mathrm{S}_{\mathrm{N}} 1$ reactions take place much more rapidly in strongly polar solvents, such as water and methanol, than in less polar solvents, such as ether and chloroform. In the reaction of 2-chloro-2-methylpropane, for example, a rate increase of 100,000 is observed upon going from ethanol (less polar) to water (more polar). The rate increases when going from a hydrocarbon solvent to water are so large they can't be measured accurately.


The Solvent -
It should be emphasized again that both the $\mathrm{S}_{\mathrm{N}} 1$ and the $\mathrm{S}_{\mathrm{N}} 2$ reaction show solvent effects, but that they do so for different reasons. $\mathrm{S}_{\mathrm{N}} 2$ reactions are disfavored in protic solvents because the ground-state energy of the nucleophile is lowered by solvation. $\mathrm{S}_{\mathrm{N}} 1$ reactions are favored in protic solvents because the transition-state energy leading to carbocation intermediate is lowered by solvation.


A Summary of $\mathrm{S}_{\mathrm{N}} 1$ Reaction Characteristics -
The effects on $\mathrm{S}_{\mathrm{N}} 1$ reactions of the four variables-substrate, leaving group, nucleophile, and solvent-are summarized in the following statements:

| Substrate | The best substrates yield the most stable carbocations. As a result, $\mathrm{S}_{\mathrm{N}} 1$ reactions are best
 for tertiary, allylic, and benzylic halides. |
| :--- | :--- |
| Leaving | Good leaving groups increase the reaction rate by lowering the energy level of the
 group
 transition state for carbocation formation. |
| Nucleophile | The nucleophile must be nonbasic to prevent a competitive elimination of HX (Section
 11.7), but otherwise does not affect the reaction rate. Neutral nucleophiles work well. |
| Solvent | Polar solvents stabilize the carbocation intermediate by solvation, thereby increasing the
 reaction rate. |


Predicting the Mechanism of a Nucleophilic Substitution Reaction -
Predict whether each of the following substitution reactions is likely to be $\mathrm{S}_{\mathrm{N}} 1$ or $\mathrm{S}_{\mathrm{N}} 2$ :


Strategy -
Look at the substrate, leaving group, nucleophile, and solvent. Then decide from the summaries at the ends of Section 11.3 and Section 11.5 whether an $S_{N} 1$ or an $S_{N} 2$ reaction is favored. $S_{N} 1$ reactions are favored by tertiary, allylic, or benzylic substrates, by good leaving groups, by nonbasic nucleophiles, and by protic solvents. $\mathrm{S}_{\mathrm{N}} 2$ reactions are favored by primary substrates, by good leaving groups, by good nucleophiles, and by polar aprotic solvents.


Solution -
(a) This is likely to be an $\mathrm{S}_{\mathrm{N}} 1$ reaction because the substrate is secondary and benzylic, the nucleophile is weakly basic, and the solvent is protic.
(b) This is likely to be an $\mathrm{S}_{\mathrm{N}} 2$ reaction because the substrate is primary, the nucleophile is a good one, and the solvent is polar aprotic.

PROBLEM Predict whether each of the following substitution reactions is likely to be $S_{N} 1$ or $S_{N} 2$ :


11-13 (a) -
(b)


11-13 (a) - 11.6 Biological Substitution Reactions
Both $\mathrm{S}_{\mathrm{N}} 1$ and $\mathrm{S}_{\mathrm{N}} 2$ reactions are common in biological chemistry, particularly in the pathways for biosynthesis of the many thousands of plant-derived substances called terpenoids, which we saw briefly in the Chapter 8 Chemistry Matters and will discuss in Section 27.5. Unlike what typically happens in the laboratory, however, the substrate in a biological substitution reaction is usually an organodiphosphate rather than an alkyl halide. Thus, the leaving group is the diphosphate ion, abbreviated $\mathrm{PP}_{\mathrm{i}}$, rather than a halide ion. In fact, it's useful to think of the diphosphate group as the "biological equivalent" of a halide. The dissociation of an organodiphosphate in a biological reaction is typically assisted by complexation to a divalent metal cation such as $\mathrm{Mg}^{2+}$ to help neutralize charge and make the diphosphate a better leaving group.


An organodiphosphate -
Diphosphate ion
As an example, two $S_{N} 1$ reactions occur during the biosynthesis of geraniol, a fragrant alcohol found in roses and used in perfumery. Geraniol biosynthesis begins with dissociation of dimethylallyl diphosphate to give an allylic carbocation, which reacts with isopentenyl diphosphate (FIGURE 11.16). From the viewpoint of isopentenyl diphosphate, the reaction is an electrophilic alkene addition, but from the viewpoint of dimethylallyl diphosphate, the process is an $S_{N} 1$ reaction in which the carbocation intermediate reacts with a double bond as the nucleophile.

FIGURE 11.16 Biosynthesis of geraniol from dimethylallyl diphosphate. Two $\mathrm{S}_{\mathrm{N}} 1$ reactions occur, both with diphosphate ion as the leaving group.

Following this initial $\mathrm{S}_{\mathrm{N}} 1$ reaction, loss of the pro- $R$ hydrogen gives geranyl diphosphate, itself an allylic diphosphate that dissociates a second time. Reaction of the geranyl carbocation with water in a second $\mathrm{S}_{\mathrm{N}} 1$ reaction, followed by loss of a proton, then yields geraniol.


An organodiphosphate -
As another example, $\mathrm{S}_{\mathrm{N}} 2$ reactions are involved in almost all biological methylations, which transfer a $-\mathrm{CH}_{3}$ group from an electrophilic donor to a nucleophile. The donor is $S$-adenosylmethionine (abbreviated SAM), which contains a positively charged sulfur (a sulfonium ion, Section 5.12), and the leaving group is the neutral $S$-adenosylhomocysteine molecule. In the biosynthesis of epinephrine (adrenaline) from norepinephrine, for
instance, the nucleophilic nitrogen atom of norepinephrine attacks the electrophilic methyl carbon atom of $S$-adenosylmethionine in an $\mathrm{S}_{\mathrm{N}} 2$ reaction, displacing $S$-adenosylhomocysteine (FIGURE 11.17). In effect, $S$-adenosylmethionine is simply a biological equivalent of $\mathrm{CH}_{3} \mathrm{Cl}$.

FIGURE 11.17 The biosynthesis of epinephrine from norepinephrine occurs by an $\mathbf{S}_{\mathbf{N}} \mathbf{2}$ reaction with $\boldsymbol{S}$-adenosylmethionine.
PROBLEM Review the mechanism of geraniol biosynthesis shown in Figure 11.16, and propose a mechanism
11-14 for the biosynthesis of limonene from linalyl diphosphate.

Linalyl diphosphate
Limonene


An organodiphosphate - 11.7 Elimination Reactions: Zaitsev's Rule
We said at the beginning of this chapter that two kinds of reactions can take place when a nucleophile/Lewis base reacts with an alkyl halide. The nucleophile can either substitute for the halide by reaction at carbon or can cause elimination of HX by reaction at a neighboring hydrogen:

Elimination reactions are more complex than substitution reactions for several reasons. One is the problem of regiochemistry. What product results by loss of HX from an unsymmetrical halide? In fact, elimination reactions almost always give mixtures of alkene products, and the best we can usually do is to predict which will be the major product.

According to Zaitsev's rule, formulated in 1875 by the Russian chemist Alexander Zaitsev, base-induced elimination reactions generally (although not always) give the more stable alkene product-that is, the alkene with more alkyl substituents on the double-bond carbons. In the following two cases, for example, the more highly substituted alkene product predominates.


ZAITSEV'S RULE -
In the elimination of HX from an alkyl halide, the more highly substituted alkene product predominates.


2-Bromo-2-methylbutane -
2-Methyl-2-butene
(70\%)

2-Methyl-1-butene
(30\%)

Another factor that complicates a study of elimination reactions is that they can take place by different mechanisms, just as substitutions can. We'll consider three of the most common mechanisms-the E1, E2, and E1cB reactions-which differ in the timing of $\mathrm{C}-\mathrm{H}$ and $\mathrm{C}-\mathrm{X}$ bond-breaking.

In the E1 reaction, the $\mathrm{C}-\mathrm{X}$ bond breaks first to give a carbocation intermediate, which undergoes subsequent base abstraction of $\mathrm{H}^{+}$to yield the alkene. In the E 2 reaction, base-induced $\mathrm{C}-\mathrm{H}$ bond cleavage is simultaneous with $\mathrm{C}-\mathrm{X}$ bond cleavage, giving the alkene in a single step. In the E1cB reaction (cB for "conjugate base"), base abstraction of the proton occurs first, giving a carbanion ( $\mathrm{R}:^{-}$) intermediate. This anion, the conjugate base of the reactant "acid," then undergoes loss of $\mathrm{X}^{-}$in a subsequent step to give the alkene. All three mechanisms occur frequently in the laboratory, but the E1cB mechanism predominates in biological pathways.

E1 Reaction: $C-X$ bond breaks first to give a carbocation intermediate, followed by base removal of a proton to yield the alkene.

E2 Reaction: $\mathrm{C}-\mathrm{H}$ and $\mathrm{C}-\mathrm{X}$ bonds break simultaneously, giving the alkene in a single step without intermediates.

E1cB Reaction: $\mathrm{C}-\mathrm{H}$ bond breaks first, giving a carbanion intermediate that loses $\mathrm{X}^{-}$to form the alkene.


Predicting the Product of an Elimination Reaction -
What product would you expect from reaction of 1-chloro-1-methylcyclohexane with KOH in ethanol?


Strategy -
Treatment of an alkyl halide with a strong base such as KOH yields an alkene. To find the products in a specific case, locate the hydrogen atoms on each carbon next to the leaving group, and then generate the potential alkene products by removing HX in as many ways as possible. The major product will be the one that has the most highly substituted double bond-in this case, 1-methylcyclohexene.


Solution -
PROBLEM Ignoring double-bond stereochemistry, what products would you expect from elimination reactions
11-15 of the following alkyl halides? Which product will be the major product in each case?
(a)

(b)

(c)

PROBLEM What alkyl halides might the following alkenes have been made from?
11-16 (a)

(b)


Solution - 11.8 The E2 Reaction and the Deuterium Isotope Effect
The E2 reaction (for elimination, bimolecular) occurs when an alkyl halide is treated with a strong base, such as hydroxide ion or alkoxide ion ( $\mathrm{RO}^{-}$). It is the most commonly occurring pathway for elimination and can be formulated as shown in FIGURE 11.18.


FIGURE 11.18 MECHANISM -
Mechanism of the E2 reaction of an alkyl halide. The reaction takes place in a single step through a transition state in which the double bond begins to form at the same time the H and X groups are leaving.
(1) Base (B:) attacks a neighboring hydrogen and begins to remove the H at the same time as the alkene double bond starts to form and the X group starts to leave.
(2) Neutral alkene is produced when the $\mathrm{C}-\mathrm{H}$ bond is fully broken and the X group has departed with the $\mathrm{C}-\mathrm{X}$ bond electron pair.

(1)

Transition state

Like the $\mathrm{S}_{\mathrm{N}} 2$ reaction, the E2 reaction takes place in one step without intermediates. As the base begins to abstract $\mathrm{H}^{+}$from a carbon next to the leaving group, the $\mathrm{C}-\mathrm{H}$ bond begins to break, a $\mathrm{C}=\mathrm{C}$ bond begins to form, and the leaving group begins to depart, taking with it the electron pair from the $\mathrm{C}-\mathrm{X}$ bond. Among the pieces of evidence supporting this mechanism is the fact that E2 reactions show second-order kinetics and follow the rate law: rate $=k \times[\mathrm{RX}] \times[\mathrm{Base}]$. That is, both the base and alkyl halide take part in the rate-limiting step.


FIGURE 11.18 MECHANISM -
A second piece of evidence in support of the E2 mechanism is provided by a phenomenon known as the deuterium isotope effect. For reasons that we won't go into, a carbon-hydrogen bond is weaker by about $5 \mathrm{~kJ} / \mathrm{mol}(1.2 \mathrm{kcal} / \mathrm{mol})$ than the corresponding carbon-deuterium bond. Thus, a $\mathrm{C}-\mathrm{H}$ bond is more easily broken than an equivalent $\mathrm{C}-\mathrm{D}$ bond, and the rate of $\mathrm{C}-\mathrm{H}$ bond cleavage is faster. For instance, the baseinduced elimination of HBr from 1-bromo-2-phenylethane proceeds 7.11 times faster than the corresponding elimination of DBr from 1-bromo-2, 2-dideuterio-2-phenylethane. This result tells us that the $\mathrm{C}-\mathrm{H}$ (or $\mathrm{C}-\mathrm{D}$ ) bond is broken in the rate-limiting step, consistent with our picture of the E 2 reaction as a one-step process. If it were otherwise, we wouldn't observe a rate difference.


(H)-Faster reaction <br> (D)-Slower reaction -
Yet a third piece of mechanistic evidence involves the stereochemistry of E2 eliminations. As shown by a large number of experiments, E2 reactions occur with periplanar geometry, meaning that all four reacting atoms-the hydrogen, the two carbons, and the leaving group-lie in the same plane. Two such geometries are possible: syn periplanar geometry, in which the H and the X are on the same side of the molecule, and anti periplanar geometry, in which the H and the X are on opposite sides of the molecule. Of the two, anti periplanar geometry is energetically preferred because it allows the substituents on the two carbons to adopt a staggered relationship, whereas syn geometry requires that the substituents be eclipsed.

Anti periplanar geometry (staggered, lower energy)

Syn periplanar geometry
(eclipsed, higher energy)
What's so special about periplanar geometry? Because the $s p^{3} \sigma$ orbitals in the reactant $\mathrm{C}-\mathrm{H}$ and $\mathrm{C}-\mathrm{X}$ bonds must overlap and become $p \pi$ orbitals in the alkene product, there must also be some overlap in the transition state. This can occur most easily if all the orbitals are in the same plane to begin with-that is, if they're periplanar (FIGURE 11.19).

FIGURE 11.19 The transition state for the E2 reaction of an alkyl halide with base. Overlap of the developing $p$ orbitals in the transition state requires periplanar geometry of the reactant.

You can think of E2 elimination reactions with periplanar geometry as being similar to $\mathrm{S}_{\mathrm{N}} 2$ reactions with $180^{\circ}$ geometry. In an $\mathrm{S}_{\mathrm{N}} 2$ reaction, an electron pair from the incoming nucleophile pushes out the leaving group on the opposite side of the molecule. In an E2 reaction, an electron pair from a neighboring $\mathrm{C}-\mathrm{H}$ bond also pushes out the leaving group on the opposite side of the molecule.

$\mathrm{S}_{\mathrm{N}} 2$ reaction (backside attack)


(H)-Faster reaction <br> (D)-Slower reaction -
$\mathrm{S}_{\mathrm{N}} 2$ reaction (backside attack)

E2 reaction (anti periplanar)

Anti periplanar geometry for E2 eliminations has specific stereochemical consequences that provide strong evidence for the proposed mechanism. To take just one example, meso-1,2-dibromo-1,2-diphenylethane undergoes E2 elimination on treatment with base to give only the $E$ alkene. None of the isomeric $Z$ alkene is formed because the transition state leading to the $Z$ alkene would have to have syn periplanar geometry and
would thus be higher in energy.


WORKED EXAMPLE 11.4 -
Predicting the Double-Bond Stereochemistry of the Product in an E2 Reaction
What stereochemistry do you expect for the alkene obtained by E2 elimination of (1S,2S)-1,2-dibromo-1,2-diphenylethane?


Strategy -
Draw ( $1 S, 2 S$ )-1,2-dibromo-1,2-diphenylethane so that you can see its stereochemistry and so that the -H and -Br groups to be eliminated are anti periplanar. Then carry out the elimination while keeping all substituents in approximately the same positions, and see what alkene results.


Solution -
Anti periplanar elimination of HBr gives (Z)-1-bromo-1,2-diphenylethylene.

PROBLEM What stereochemistry do you expect for the alkene obtained by E2 elimination of 11-17 ( $1 R, 2 R$ )-1,2-dibromo-1,2-diphenylethane? Draw a Newman projection of the reacting conformation.

PROBLEM What stereochemistry do you expect for the trisubstituted alkene obtained by E2 elimination of the
11-18 following alkyl halide on treatment with KOH ? (reddish brown = Br.)


Solution - 11.9 The E2 Reaction and Cyclohexane Conformation
Anti periplanar geometry for E2 reactions is particularly important in cyclohexane rings, where chair geometry forces a rigid relationship between the substituents on neighboring carbon atoms (Section 4.8). The anti periplanar requirement for E2 reactions overrides Zaitsev's rule and can be met in cyclohexanes only if the hydrogen and the leaving group are trans diaxial (FIGURE 11.20). If either the leaving group or the hydrogen is
equatorial, E2 elimination can't occur.


Equatorial chlorine: H and Cl are not anti periplanar -
FIGURE 11.20 The geometric requirement for an E2 reaction in a substituted cyclohexane. The leaving group and the hydrogen must both be axial for anti periplanar elimination to occur.

The elimination of HCl from the isomeric menthyl and neomenthyl chlorides shown in FIGURE 11.21 gives a good illustration of this trans-diaxial requirement. Neomenthyl chloride undergoes elimination of HCl on reaction with ethoxide ion 200 times faster than menthyl chloride. Furthermore, neomenthyl chloride yields 3 -menthene as the major alkene product, whereas menthyl chloride yields 2-menthene.

Neomenthyl chloride
(b)

Menthyl chloride

$\downarrow \mid$ Ring-flip

FIGURE 11.21 Dehydrochlorination of menthyl and neomenthyl chlorides. (a) Neomenthyl chloride loses HCl directly from its more stable conformation, but (b) menthyl chloride must first ring-flip to a higher energy conformation before HCl loss can occur. The abbreviation "Et" represents an ethyl group.

The difference in reactivity between the isomeric menthyl chlorides is due to the difference in their conformations. Neomenthyl chloride has the conformation shown in FIGURE 11.21a, with the methyl and isopropyl groups equatorial and the chlorine axial-a perfect geometry for E2 elimination. Loss of the hydrogen atom at C 4 occurs easily to yield the more substituted alkene product, 3-menthene, as predicted by Zaitsev's rule.

Menthyl chloride, by contrast, has a conformation in which all three substituents are equatorial (FIGURE 11.21b). To achieve the necessary geometry for elimination, menthyl chloride must first ring-flip to a higherenergy chair conformation, in which all three substituents are axial. E2 elimination then occurs with loss of
the only trans-diaxial hydrogen available, leading to the non-Zaitsev product 2 -menthene. The net effect of the simple change in chlorine stereochemistry is a 200 -fold change in reaction rate and a complete change of product. The chemistry of the molecule is controlled by its conformation.


Equatorial chlorine: H and Cl are not anti periplanar -
PROBLEM Which isomer would you expect to undergo E2 elimination faster, 11-19 trans-1-bromo-4-tert-butylcyclohexane or cis-1-bromo-4-tert-butylcyclohexane? Draw each molecule in its more stable chair conformation, and explain your answer.


Equatorial chlorine: H and Cl are not anti periplanar - 11.10 The E1 and E1cB Reactions
The E1 Reaction
Just as the E2 reaction is analogous to the $\mathrm{S}_{\mathrm{N}} 2$ reaction, the $\mathrm{S}_{\mathrm{N}} 1$ reaction has a close analog called the E1 reaction (for elimination, unimolecular). The E1 reaction can be formulated as shown in FIGURE 11.22, with the elimination of HCl from 2-chloro-2-methylpropane.

FIGURE 11.22 MECHANISM
Mechanism of the E1 reaction. Two steps are involved, the first of which is ratelimiting, and a carbocation intermediate is present.

1 Spontaneous dissociation of the tertiary alkyl chloride yields an intermediate carbocation in a slow, rate-limiting step.
(1) $\| \begin{aligned} & \text { Rate- } \\ & \text { limiting }\end{aligned}$

Carbocation

2 Loss of a neighboring $\mathrm{H}^{+}$in a fast step yields the neutral alkene product. The electron pair from the $\mathrm{C}-\mathrm{H}$ bond goes to form the alkene $\pi$ bond.

(2) $\downarrow$ Fast

E1 eliminations begin with the same unimolecular dissociation to give a carbocation that we saw in the $\mathrm{S}_{\mathrm{N}} 1$ reaction, but the dissociation is followed by loss of $\mathrm{H}^{+}$from the adjacent carbon rather than by substitution. In fact, the E1 and $\mathrm{S}_{\mathrm{N}} 1$ reactions normally occur together whenever an alkyl halide is treated in a protic solvent with a nonbasic nucleophile. Thus, the best E1 substrates are also the best $\mathrm{S}_{\mathrm{N}} 1$ substrates, and mixtures of substitution and elimination products are usually obtained. For example, when 2-chloro-2-methylpropane is warmed to $65^{\circ} \mathrm{C}$ in $80 \%$ aqueous ethanol, a 64 : 36 mixture of 2-methyl-2-propanol ( $\mathrm{S}_{\mathrm{N}} 1$ ) and 2-methylpropene (E1) results.


Equatorial chlorine: H and Cl are not anti periplanar - 11.10 The E1 and E1cB Reactions
2-Chloro-2-methylpropane

2-Methyl-2-propanol (64\%)
$+$

2-Methylpropene (36\%)

Much evidence has been obtained in support of the E1 mechanism. For example, E1 reactions show first-order kinetics, consistent with a rate-limiting, unimolecular dissociation process. Furthermore, E1 reactions show no
deuterium isotope effect because rupture of the $\mathrm{C}-\mathrm{H}($ or $\mathrm{C}-\mathrm{D}$ ) bond occurs after the rate-limiting step rather than during it. Thus, we can't measure a rate difference between a deuterated and nondeuterated substrate.

A final piece of evidence involves the stereochemistry of elimination. Unlike the E2 reaction, where anti periplanar geometry is required, there is no geometric requirement on the E1 reaction because the halide and the hydrogen are lost in separate steps. We might therefore expect to obtain the more stable (Zaitsev's rule) product from E1 reaction, which is just what we find. To return to a familiar example, menthyl chloride loses HCl under E1 conditions in a polar solvent to give a mixture of alkenes in which the Zaitsev product, 3-menthene, predominates (FIGURE 11.23).

FIGURE 11.23 Elimination reactions of menthyl chloride. E2 conditions (1, strong base in 100\% ethanol) lead to 2-menthene through an anti periplanar elimination, whereas E1 conditions ( 2 , dilute base in $80 \%$ aqueous ethanol) lead to a mixture of 2 -menthene and 3-menthene.


The E1cB Reaction -
In contrast to the E1 reaction, which involves a carbocation intermediate, the E1cB reaction takes place through a carbanion intermediate. Base-induced abstraction of a proton in a slow, rate-limiting step gives an anion, which expels a leaving group on the adjacent carbon. The reaction is particularly common in substrates that have a poor leaving group, such as -OH , two carbons removed from a carbonyl group, as in $\mathrm{HOC}-\mathrm{CH}-\mathrm{C}=\mathrm{O}$. The poor leaving group disfavors the alternative E1 and E2 possibilities, and the carbonyl group makes the adjacent hydrogen unusually acidic by resonance stabilization of the anion intermediate. We'll look at this acidifying effect of a carbonyl group in Section 22.5.


The E1cB Reaction - 11.11 Biological Elimination Reactions
All three elimination reactions-E2, E1, and E1cB-occur in biological pathways, but the E1cB mechanism is particularly common. The substrate is usually an alcohol rather than an alkyl halide, and the H atom removed is usually adjacent to a carbonyl group, just as in laboratory reactions. Thus, 3-hydroxy carbonyl compounds are frequently converted to unsaturated carbonyl compounds by elimination reactions. A typical example occurs during the biosynthesis of fats and oils when a 3-hydroxybutyryl thioester is dehydrated to the corresponding unsaturated (crotonyl) thioester. The base in this reaction is a histidine amino acid in the enzyme, and the loss of the -OH group is assisted by simultaneous protonation.


3-Hydroxybutyryl -
thioester


3-Hydroxybutyryl - 11.12 A Summary of Reactivity: $\mathrm{S}_{\mathrm{N}} 1, \mathrm{~S}_{\mathrm{N}} 2$, E1, E1cB, and E2
$\mathrm{S}_{\mathrm{N}} 1, \mathrm{~S}_{\mathrm{N}} 2$, E1, E1cB, E2-how can you keep it all straight and predict what will happen in any given case? Will substitution or elimination occur? Will the reaction be bimolecular or unimolecular? There are no rigid answers to these questions, but it's possible to recognize some trends and make some generalizations.

- Primary alkyl halides $\mathrm{S}_{\mathrm{N}} 2$ substitution occurs if a good nucleophile is used, E2 elimination occurs if a strong, sterically hindered base is used, and E1cB elimination occurs if the leaving group is two carbons away from a carbonyl group.
- Secondary alkyl halides $\mathrm{S}_{\mathrm{N}} 2$ substitution occurs if a weakly basic nucleophile is used in a polar aprotic solvent, E2 elimination predominates if a strong base is used, and E1cB elimination takes place if the leaving group is two carbons away from a carbonyl group. Secondary allylic and benzylic alkyl halides can also undergo $\mathrm{S}_{\mathrm{N}} 1$ and E1 reactions if a weakly basic nucleophile is used in a protic solvent.
- Tertiary alkyl halides E2 elimination occurs when a base is used, but $\mathrm{S}_{\mathrm{N}} 1$ substitution and E1 elimination occur together under neutral conditions, such as in pure ethanol or water. E1cB elimination takes place if the leaving group is two carbons away from a carbonyl group.


Predicting the Product and Mechanism of Reactions -
Tell whether each of the following reactions is likely to be $\mathrm{S}_{\mathrm{N}} 1, \mathrm{~S}_{\mathrm{N}} 2$, E1, E1cB, or E2, and predict the product of each:

(b)


Strategy -
Look carefully in each reaction at the structure of the substrate, the leaving group, the nucleophile, and the solvent. Then decide from the preceding summary which kind of reaction is likely to be favored.


Solution -
(a) A secondary, nonallylic substrate can undergo an $\mathrm{S}_{\mathrm{N}} 2$ reaction with a good nucleophile in a polar aprotic solvent but will undergo an E2 reaction on treatment with a strong base in a protic solvent. In this case, E2 reaction is likely to predominate.

(b) A secondary benzylic substrate can undergo an $\mathrm{S}_{\mathrm{N}} 2$ reaction on treatment with a nonbasic nucleophile in a polar aprotic solvent and will undergo an E2 reaction on treatment with a base. Under protic conditions, such as aqueous formic acid $\left(\mathrm{HCO}_{2} \mathrm{H}\right)$, an $\mathrm{S}_{\mathrm{N}} 1$ reaction is likely, along with some E1 reaction.

PROBLEM Tell whether each of the following reactions is likely to be $\mathrm{S}_{\mathrm{N}} 1, \mathrm{~S}_{\mathrm{N}} 2$, E1, E1cB, or E2:
11-20 (a)

(b)

(c)

(d)


Green Chemistry -
Organic chemistry in the 20th century changed the world, giving us new medicines, food preservatives, insecticides, adhesives, textiles, dyes, building materials, composites, and all manner of polymers. But these advances did not come without a cost: Almost every chemical process produces waste that must be dealt with, including reaction solvents and toxic by-products that might evaporate into the air or be leached into groundwater if not disposed of properly. Even apparently harmless by-products must be safely buried or otherwise sequestered. As always, there's no such thing as a free lunch. With the good also comes the bad.

It may never be possible to make organic chemistry completely benign, but awareness of the environmental problems caused by many chemical processes has grown dramatically in recent years, giving rise to a movement called green chemistry. Green chemistry is the design and implementation of chemical products and processes that reduce waste and attempt to eliminate the generation of hazardous substances. There are 12 principles of green chemistry:

Prevent waste - Waste should be prevented rather than treated or cleaned up after it has been created.
Maximize atom economy - Synthetic methods should maximize the incorporation of all materials used in a process into the final product so that waste is minimized.
Use less hazardous processes - Synthetic methods should use reactants and generate wastes with minimal toxicity to health and the environment.
Design safer chemicals - Chemical products should be designed to have minimal toxicity.
Use safer solvents - Minimal use should be made of solvents, separation agents, and other auxiliary substances in a reaction.
Design for energy efficiency - Energy requirements for chemical processes should be minimized, with reactions carried out at room temperature if possible.
Use renewable feedstocks - Raw materials should come from renewable sources when feasible.
Minimize derivatives - Syntheses should be designed with minimal use of protecting groups to avoid extra steps and reduce waste.
Use catalysis - Reactions should be catalytic rather than stoichiometric.
Design for degradation - Products should be designed to be biodegradable at the end of their useful lifetimes.
Monitor pollution in real time - Processes should be monitored in real time for the formation of hazardous substances.
Prevent accidents - Chemical substances and processes should minimize the potential for fires, explosions, or other accidents.


Green Chemistry -
The foregoing 12 principles may not all be met in most real-world applications, but they provide a worthy goal and they can make chemists think more carefully about the environmental implications of their work. Real success stories have occurred, and more are in progress. Approximately 7 million pounds per year of ibuprofen ( 6 billion tablets!) are now made by a "green" process that produces approximately $99 \%$ less waste than the process it replaces. Only three steps are needed, the anhydrous HF solvent used in the first step is recovered and reused, and the second and third steps are catalytic.


Key Terms -
- anti periplanar
- benzylic
- bimolecular
- deuterium isotope effect
- E1 reaction
- E1cB reaction
- E2 reaction
- first-order reaction
- ion pair
- kinetics
- leaving group
- periplanar
- rate-determining step
- rate-limiting step
- second-order reaction
- $\mathrm{S}_{\mathrm{N}} 1$ reaction
- $\mathrm{S}_{\mathrm{N}} 2$ reaction
- solvation
- syn periplanar
- tosylate
- unimolecular
- Zaitsev’s rule
- nucleophilic substitution reaction


Summary -
The reaction of an alkyl halide or tosylate with a nucleophile/base results either in substitution or in elimination. The resultant nucleophilic substitution and base-induced elimination reactions are two of the most widely occurring and versatile reaction types in organic chemistry, both in the laboratory and in biological pathways.

Nucleophilic substitutions are of two types: $\mathbf{S}_{\mathbf{N}} \mathbf{2}$ reactions and $\mathbf{S}_{\mathbf{N}} \mathbf{1}$ reactions. In the $\mathrm{S}_{\mathrm{N}} 2$ reaction, the entering nucleophile approaches the halide from a direction $180^{\circ}$ away from the leaving group, resulting in an umbrellalike inversion of configuration at the carbon atom. The reaction is kinetically second-order and is strongly inhibited by increasing steric bulk of the reactants. Thus, $\mathrm{S}_{\mathrm{N}} 2$ reactions are favored for primary and secondary substrates.

In the $\mathrm{S}_{\mathrm{N}} 1$ reaction, the substrate spontaneously dissociates to a carbocation in a slow rate-limiting step, followed by a rapid reaction with the nucleophile. As a result, $\mathrm{S}_{\mathrm{N}} 1$ reactions are kinetically first-order and take place with substantial racemization of configuration at the carbon atom. They are most favored for tertiary substrates. Both $\mathrm{S}_{\mathrm{N}} 1$ and $\mathrm{S}_{\mathrm{N}} 2$ reactions occur in biological pathways, although the leaving group is typically a diphosphate ion rather than a halide.


Summary -
Eliminations of alkyl halides to yield alkenes occur by three mechanisms: E2 reactions, E1 reactions, and E1cB reactions, which differ in the timing of $\mathrm{C}-\mathrm{H}$ and $\mathrm{C}-\mathrm{X}$ bond-breaking. In the E 2 reaction, $\mathrm{C}-\mathrm{H}$ and $\mathrm{C}-\mathrm{X}$ bondbreaking occur simultaneously when a base abstracts $\mathrm{H}^{+}$from one carbon while the leaving group departs from the neighboring carbon. The reaction takes place preferentially through an anti periplanar transition state in which the four reacting atoms-hydrogen, two carbons, and leaving group-are in the same plane. The reaction shows second-order kinetics and a deuterium isotope effect, and occurs when a secondary or tertiary substrate is treated with a strong base. These elimination reactions usually give a mixture of alkene products in which the more highly substituted alkene predominates (Zaitsev's rule).

In the E1 reaction, $\mathrm{C}-\mathrm{X}$ bond-breaking occurs first. The substrate dissociates to yield a carbocation in the
slow rate-limiting step before losing $\mathrm{H}^{+}$from an adjacent carbon in a second step. The reaction shows firstorder kinetics and no deuterium isotope effect and occurs when a tertiary substrate reacts in polar, nonbasic solution.

In the E1cB reaction, $\mathrm{C}-\mathrm{H}$ bond-breaking occurs first. A base abstracts a proton to give a carbanion, followed by loss of the leaving group from the adjacent carbon in a second step. The reaction is favored when the leaving group is two carbons removed from a carbonyl, which stabilizes the intermediate anion by resonance. Biological elimination reactions typically occur by this E1cB mechanism.

In general, substrates react in the following way:


Summary -
In general, substrates react in the following way:

$$
\begin{array}{lll}
\begin{array}{l}
\mathrm{RCH}_{2} \mathrm{X} \\
\text { (primary) }
\end{array} & \longrightarrow & {\text { Mostly } \mathrm{S}_{\mathrm{N}} 2 \text { substitution }} \\
\begin{array}{l}
\mathrm{R}_{2} \mathrm{CHX} \\
\text { (secondary) }
\end{array} & \longrightarrow & \begin{array}{l}
\mathrm{S}_{\mathrm{N}} 2 \text { substitution with nonbasic nucleophiles } \\
\mathrm{E} 2 \text { elimination with strong bases }
\end{array} \\
\begin{array}{ll}
\mathrm{R}_{3} \mathrm{CX} \\
\text { (tertiary) }
\end{array} & \longrightarrow & \begin{array}{l}
\text { Mostly E2 elimination } \\
\text { (S } \mathrm{S}_{\mathrm{N}} 1 \text { substitution and E1 elimination in nonbasic solvents) }
\end{array}
\end{array}
$$


Summary of Reactions -
1. Nucleophilic substitutions
a. $\mathrm{S}_{\mathrm{N}} 1$ reaction of $3^{\circ}$, allylic, and benzylic halides (Section 11.4 and Section 11.5)

b. $\mathrm{S}_{\mathrm{N}} 2$ reaction of $1^{\circ}$ and simple $2^{\circ}$ halides (Section 11.2 and Section 11.3)

2. Eliminations
a. E1 reaction (Section 11.10)

b. E1cB reaction (Section 11.10)

c. E2 reaction (Section 11.8)


Visualizing Chemistry -
PROBLEM Write the product you would expect from reaction of each of the following alkyl halides with (1) $\mathrm{Na}^{+}$
11-21 ${ }^{-} \mathrm{SCH}_{3}$ and (2) $\mathrm{Na}^{+-} \mathrm{OH}($ green $=\mathrm{Cl})$ :
(a)

(b)

(c)

PROBLEM From what alkyl bromide was the following alkyl acetate made by $\mathrm{S}_{\mathrm{N}} 2$ reaction? Write the reaction,
11-22 showing all stereochemistry.

PROBLEM Assign $R$ or $S$ configuration to the following molecule, write the product you would expect from $\mathrm{S}_{\mathrm{N}} 2$
11-23 reaction with NaCN , and assign $R$ or $S$ configuration to the product (green $=\mathrm{Cl}$ ):

PROBLEM Draw the structure and assign $Z$ or $E$ stereochemistry to the product you expect from E2 reaction of
11-24 the following molecule with $\mathrm{NaOH}($ green $=\mathrm{Cl})$ :


Mechanism Problems -
PROBLEM Predict the product(s) and show the mechanism for each of the following reactions. What do the
11-25 mechanisms have in common? Why?
(a) $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OTos}+\mathrm{NaBr}$ DMF
(b)

(c)

(d)

PROBLEM Show the mechanism for each of the following reactions. What do the mechanisms have in 11-26 common? Why?
(a)

(b)

(c)

PROBLEM Predict the product(s) for each of the following elimination reactions. In each case show the
11-27 mechanism. What do the mechanisms have in common? Why?
(a)

(b)

(c)

PROBLEM Predict the product(s) for each of the following elimination reactions. In each case show the
11-28 mechanism. What do the mechanisms have in common? Why?
(a)

(b)

(c)

PROBLEM Predict the product(s) for each of the following elimination reactions. In each case show the
11-29 mechanism. What do the mechanisms have in common? Why?
(a)

(b)

(c)

PROBLEM Predict the product of each of the following reactions, and indicate if the mechanism is likely to be 11-30 $S_{N} 1, S_{N} 2$, E1, E2, or E1cB.
(a)

(c)

$+$

(b)

(d)

PROBLEM We saw in Section 8.7 that bromohydrins are converted into epoxides when treated with base.
11-31 Propose a mechanism, using curved arrows to show the electron flow.

$\xrightarrow[\text { Ethanol }]{\mathrm{NaOH}}$

PROBLEM The following tertiary alkyl bromide does not undergo a nucleophilic substitution reaction by either
11-32 $\mathrm{S}_{\mathrm{N}} 1$ or $\mathrm{S}_{\mathrm{N}} 2$ mechanisms. Explain.

PROBLEM Metabolism of $S$-adenosylhomocysteine (Section 11.6) involves the following sequence. Propose a
11-33 mechanism for the second step.


Mechanism Problems -
PROBLEM Metabolism of $S$-adenosylhomocysteine (Section 11.6) involves the following sequence. Propose a
11-33 mechanism for the second step.

PROBLEM Reaction of iodoethane with $\mathrm{CN}^{-}$yields a small amount of isonitrile, $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{~N} \equiv \mathrm{C}$, along with the 11-34 nitrile $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{C} \equiv \mathrm{N}$ as the major product. Write electron-dot structures for both products, assign formal charges as necessary, and propose mechanisms to account for their formation.

PROBLEM One step in the urea cycle for ridding the body of ammonia is the conversion of argininosuccinate
11-35 to the amino acid arginine plus fumarate. Propose a mechanism for the reaction, and show the structure of arginine.

PROBLEM Methyl esters $\left(\mathrm{RCO}_{2} \mathrm{CH}_{3}\right)$ undergo a cleavage reaction to yield carboxylate ions plus iodomethane
11-36 on heating with LiI in dimethylformamide:

The following evidence has been obtained: (1) The reaction occurs much faster in DMF than in ethanol. (2) The corresponding ethyl ester $\left(\mathrm{RCO}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\right)$ cleaves approximately 10 times more slowly than the methyl ester. Propose a mechanism for the reaction. What other kinds of experimental evidence could you gather to support your hypothesis?

PROBLEM $\mathrm{S}_{\mathrm{N}} 2$ reactions take place with inversion of configuration, and $\mathrm{S}_{\mathrm{N}} 1$ reactions take place with
11-37 racemization. The following substitution reaction, however, occurs with complete retention of configuration. Propose a mechanism. (Hint: two inversions = retention.)

PROBLEM Propose a mechanism for the following reaction, an important step in the laboratory synthesis of 11-38 proteins:


Nucleophilic Substitution Reactions -
PROBLEM Draw all isomers of $\mathrm{C}_{4} \mathrm{H}_{9} \mathrm{Br}$, name them, and arrange them in order of decreasing reactivity in the 11-39 $\mathrm{S}_{\mathrm{N}} 2$ reaction.

PROBLEM The following Walden cycle has been carried out: Explain the results, and indicate where inversion 11-40 occurs.

PROBLEM Which compound in each of the following pairs will react faster in an $\mathrm{S}_{\mathrm{N}} 2$ reaction with $\mathrm{OH}^{-}$?
11-41 (a) $\mathrm{CH}_{3} \mathrm{Br}$ or $\mathrm{CH}_{3} \mathrm{I}$ (b) $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{I}$ in ethanol or in dimethyl sulfoxide (c) $\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCl}$ or $\mathrm{CH}_{3} \mathrm{Cl}$
(d) $\mathrm{H}_{2} \mathrm{C}=\mathrm{CHBr}$ or $\mathrm{H}_{2} \mathrm{C}=\mathrm{CHCH}_{2} \mathrm{Br}$

PROBLEM Which reactant in each of the following pairs is more nucleophilic? Explain.
11-42
(a) ${ }^{-} \mathrm{NH}_{2}$ or $\mathrm{NH}_{3}$
(b) $\mathrm{H}_{2} \mathrm{O}$ or $\mathrm{CH}_{3} \mathrm{CO}_{2}^{-}$
(c) $\mathrm{BF}_{3}$ or $\mathrm{F}^{-}$
(d) $\left(\mathrm{CH}_{3}\right)_{3} \mathrm{P}$ or $\left(\mathrm{CH}_{3}\right)_{3} \mathrm{~N}$
(e) $\mathrm{I}^{-}$or $\mathrm{Cl}^{-}$
(f) ${ }^{-} \mathrm{C} \equiv \mathrm{N}$ or ${ }^{-} \mathrm{OCH}_{3}$


Nucleophilic Substitution Reactions -
PROBLEM What effect would you expect the following changes to have on the rate of the $\mathrm{S}_{\mathrm{N}} 2$ reaction of
11-43 1-iodo-2-methylbutane with cyanide ion?
(a) The $\mathrm{CN}^{-}$concentration is halved, and the 1-iodo-2-methylbutane concentration is doubled.
(b) Both the $\mathrm{CN}^{-}$and the 1-iodo-2-methylbutane concentrations are tripled.

PROBLEM What effect would you expect the following changes to have on the rate of the reaction of ethanol
11-44 with 2 -iodo-2-methylbutane?
(a) The concentration of the halide is tripled.
(b) The concentration of the ethanol is halved by adding diethyl ether as an inert solvent.

PROBLEM How might you prepare each of the following using a nucleophilic substitution reaction at some
11-45 step?
(a)

(b)

(c) $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CN}$
(d) $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}$


Nucleophilic Substitution Reactions -
PROBLEM Which reaction in each of the following pairs would you expect to be faster?
11-46 (a) The $\mathrm{S}_{\mathrm{N}} 2$ displacement by $\mathrm{I}^{-}$on $\mathrm{CH}_{3} \mathrm{Cl}$ or on $\mathrm{CH}_{3} \mathrm{OTos}$
(b) The $\mathrm{S}_{\mathrm{N}} 2$ displacement by $\mathrm{CH}_{3} \mathrm{CO}_{2}^{-}$on bromoethane or on bromocyclohexane
(c) The $\mathrm{S}_{\mathrm{N}} 2$ displacement on 2-bromopropane by $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{O}^{-}$or by $\mathrm{CN}^{-}$
(d) The $\mathrm{S}_{\mathrm{N}} 2$ displacement by $\mathrm{HC} \equiv \mathrm{C}^{-}$on bromomethane in benzene or in acetonitrile

PROBLEM Predict the product and give the stereochemistry resulting from reaction of each of the following
11-47 nucleophiles with $(R)$-2-bromooctane:
(a) ${ }^{-} \mathrm{CN}$
(b) $\mathrm{CH}_{3} \mathrm{CO}_{2}^{-}$
(c) $\mathrm{CH}_{3} \mathrm{~S}^{-}$

PROBLEM ( $R$ )-2-Bromooctane undergoes racemization to give ( $\pm$ )-2-bromooctane when treated with NaBr in
11-48 dimethyl sulfoxide. Explain.


Elimination Reactions -
PROBLEM Propose structures for compounds that fit the following descriptions:
11-49 (a) An alkyl halide that gives a mixture of three alkenes on E2 reaction
(b) An organohalide that will not undergo nucleophilic substitution
(c) An alkyl halide that gives the non-Zaitsev product on E2 reaction
(d) An alcohol that reacts rapidly with HCl at $0^{\circ} \mathrm{C}$

PROBLEM What products would you expect from the reaction of 1-bromopropane with each of the following?
11-50 (a) $\mathrm{NaNH}_{2}$
(b) $\mathrm{KOC}\left(\mathrm{CH}_{3}\right)_{3}$
(c) NaI
(d) NaCN
(e) $\mathrm{NaC} \equiv \mathrm{CH}$
(f) Mg , then $\mathrm{H}_{2} \mathrm{O}$

PROBLEM 1-Chloro-1,2-diphenylethane can undergo E2 elimination to give either cis- or
11-51 trans-1,2-diphenylethylene (stilbene). Draw Newman projections of the reactive conformations leading to both possible products, and suggest a reason why the trans alkene is the major product.

1-Chloro-1,2-diphenylethane

trans-1,2-Diphenylethylene

PROBLEM Predict the major alkene product of the following E1 reaction:
11-52

PROBLEM There are eight diastereomers of 1,2,3,4,5,6-hexachlorocyclohexane. Draw each in its more stable
11-53 chair conformation. One isomer loses HCl in an E2 reaction nearly 1000 times more slowly than the others. Which isomer reacts so slowly, and why?


General Problems -
PROBLEM The following reactions are unlikely to occur as written. Tell what is wrong with each, and predict
11-54 the actual product.
(a)

(b)

(c)

PROBLEM Arrange the following carbocations in order of increasing stability.
11-55 (a)

$+$

(b)

(c)

PROBLEM Order each of the following sets of compounds with respect to $\mathrm{S}_{\mathrm{N}} 1$ reactivity:
11-56 (a)

(b) $\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCl}$
$\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CBr}$
$\left(\mathrm{CH}_{3}\right)_{3} \mathrm{COH}$
(c)

PROBLEM Order each of the following sets of compounds with respect to $\mathrm{S}_{\mathrm{N}} 2$ reactivity:
11-57 (a)

$\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Cl}$

(b)

(c) $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OCH}_{3}$
$\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OTos}$
$\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Br}$
PROBLEM Predict the major product(s) of each of the following reactions. Identify those reactions where you 11-58 would expect the product mixture to rotate plane-polarized light.
(a)

(b)

(c)

PROBLEM Reaction of the following $S$ tosylate with cyanide ion yields a nitrile product that also has $S$
11-59 stereochemistry. Explain.

( $S$ stereochemistry)
PROBLEM Ethers can often be prepared by $\mathrm{S}_{\mathrm{N}} 2$ reaction of alkoxide ions, $\mathrm{RO}^{-}$, with alkyl halides. Suppose you
11-60 wanted to prepare cyclohexyl methyl ether. Which of the following two possible routes would you choose? Explain.


General Problems -
PROBLEM How can you explain the fact that trans-1-bromo-2-methylcyclohexane yields the non-Zaitsev
11-61 elimination product 3-methylcyclohexene on treatment with base?

trans-1-Bromo-2-methylcyclohexane
3-Methylcyclohexene
PROBLEM Predict the product(s) of the following reaction, indicating stereochemistry where necessary:
11-62

PROBLEM Alkynes can be made by dehydrohalogenation of vinylic halides in a reaction that is essentially
11-63 an E2 process. In studying the stereochemistry of this elimination, it was found that (Z)-2-chloro-2-butenedioic acid reacts 50 times as fast as the corresponding $E$ isomer. What conclusion can you draw about the stereochemistry of eliminations in vinylic halides? How does this result compare with eliminations of alkyl halides?

PROBLEM Based on your answer to Problem 11-63, predict the product(s) and show the mechanism for each
11-64 of the following reactions.
(a)

(b)

(c)

PROBLEM (S)-2-Butanol slowly racemizes on standing in dilute sulfuric acid. Explain.
11-65

PROBLEM Reaction of HBr with ( $R$ )-3-methyl-3-hexanol leads to racemic 3-bromo-3-methylhexane. Explain.
11-66

3-Methyl-3-hexanol

PROBLEM Treatment of 1-bromo-2-deuterio-2-phenylethane with strong base leads to a mixture of
11-67 deuterated and nondeuterated phenylethylenes in an approximately 7:1 ratio. Explain.

PROBLEM Propose a structure for an alkyl halide that gives only ( $E$ )-3-methyl-2-phenyl-2-pentene on E2 11-68 elimination. Make sure you indicate the stereochemistry.


General Problems -
PROBLEM Although anti periplanar geometry is preferred for E2 reactions, it isn't absolutely necessary. The
11-69 following deuterated bromo compound reacts with strong base to yield an undeuterated alkene. A syn elimination has occurred. Make a molecular model of the reactant, and explain the result.

PROBLEM The reaction of 1-chlorooctane with $\mathrm{CH}_{3} \mathrm{CO}_{2}^{-}$to give octyl acetate is greatly accelerated by adding 11-70 a small quantity of iodide ion. Explain.

PROBLEM Compound $\mathbf{X}$ is optically inactive and has the formula $\mathrm{C}_{16} \mathrm{H}_{16} \mathrm{Br}_{2}$. On treatment with strong base,
11-71 $\mathbf{X}$ gives hydrocarbon $\mathbf{Y}, \mathrm{C}_{16} \mathrm{H}_{14}$. Compound $\mathbf{Y}$ absorbs 2 equivalents of hydrogen when reduced over a palladium catalyst and reacts with ozone to give two fragments. One fragment, $\mathbf{Z}$, is an aldehyde with formula $\mathrm{C}_{7} \mathrm{H}_{6} \mathrm{O}$. The other fragment is glyoxal, $(\mathrm{CHO})_{2}$. Write the reactions involved, and suggest structures for $\mathbf{X}, \mathbf{Y}$, and $\mathbf{Z}$. What is the stereochemistry of $\mathbf{X}$ ?

PROBLEM When a primary alcohol is treated with $p$-toluenesulfonyl chloride at room temperature in the
11-72 presence of an organic base such as pyridine, a tosylate is formed. When the same reaction is carried out at higher temperature, an alkyl chloride is often formed. Explain.


General Problems -
PROBLEM The amino acid methionine is formed by a methylation reaction of homocysteine with
11-73 $N$-methyltetrahydrofolate. The stereochemistry of the reaction has been probed by carrying out the transformation using a donor with a "chiral methyl group," which contains protium (H), deuterium (D), and tritium (T) isotopes of hydrogen. Does the methylation reaction occur with inversion or retention of configuration?

$N$-Methyltetrahydrofolate
Tetrahydrofolate
PROBLEM Amines are converted into alkenes by a two-step process called the Hofmann elimination. $\mathrm{S}_{\mathrm{N}} 2$
11-74 reaction of the amine with an excess of $\mathrm{CH}_{3} \mathrm{I}$ in the first step yields an intermediate that undergoes E2 reaction when treated with silver oxide as base. Pentylamine, for example, yields 1-pentene. Propose a structure for the intermediate, and explain why it readily undergoes elimination.
$\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2} \xrightarrow[\text { 2. } \mathrm{Ag}_{2} \mathrm{O}, \mathrm{H}_{2} \mathrm{O}]{\text { 1. Excess } \mathrm{CH}_{3} \mathrm{I}} \quad \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}=\mathrm{CH}_{2}$
PROBLEM The antipsychotic drug flupentixol is prepared by the following scheme:
11-75


General Problems -
(a) What alkyl chloride $\mathbf{B}$ reacts with amine $\mathbf{A}$ to form $\mathbf{C}$ ?
(b) Compound $\mathbf{C}$ is treated with $\mathrm{SOCl}_{2}$, and the product is allowed to react with magnesium metal to give a Grignard reagent $\mathbf{D}$. What is the structure of $\mathbf{D}$ ?
(c) We'll see in Section 19.7 that Grignard reagents add to ketones, such as E, to give tertiary alcohols, such as $\mathbf{F}$. Because of the newly formed chirality center, compound $\mathbf{F}$ exists as a pair of enantiomers. Draw both, and assign $R, S$ configurations.
(d) Two stereoisomers of flupentixol are subsequently formed from $\mathbf{F}$, but only one is shown. Draw the other isomer, and identify the type of stereoisomerism.


CHAPTER 12 -
Structure Determination: Mass Spectrometry and Infrared Spectroscopy

FIGURE 12.1 More than a thousand different chemical compounds have been isolated from coffee. Their structures were determined using various spectroscopic techniques. (credit: modification of work "Coffee" by Rafael Saldaña/Flickr, CC BY 2.0)


CHAPTER CONTENTS -
12.1 Mass Spectrometry of Small Molecules: Magnetic-Sector Instruments
12.2 Interpreting Mass Spectra
12.3 Mass Spectrometry of Some Common Functional Groups
12.4 Mass Spectrometry in Biological Chemistry: Time-of-Flight (TOF) Instruments
12.5 Spectroscopy and the Electromagnetic Spectrum
12.6 Infrared Spectroscopy
12.7 Interpreting Infrared Spectra
12.8 Infrared Spectra of Some Common Functional Groups

WHY THIS CHAPTER? Finding the structures of new molecules, whether small ones synthesized in the laboratory or large proteins and nucleic acids found in living organisms, is central to progress in chemistry and biochemistry. We can only scratch the surface of structure determination in this book, but after reading this and the following two chapters, you should have a good idea of the range of structural techniques available and of how and when each is used.

Every time a reaction is run, the products must be identified, and every time a new compound is found in nature, its structure must be determined. Determining the structure of an organic compound was a difficult and time-consuming process until the mid-20th century, but powerful techniques and specialized instruments are now routinely used to simplify the problem. In this and the next two chapters, we'll look at four such techniques-mass spectrometry (MS), infrared (IR) spectroscopy, ultraviolet spectroscopy (UV), and nuclear magnetic resonance spectroscopy (NMR)-and we'll see the kind of information that can be obtained from each.

| Mass spectrometry | What is the size and formula? |
| :--- | :--- |
| Infrared spectroscopy | What functional groups are present? |
| Ultraviolet spectroscopy | Is a conjugated $\pi$ electron system present? |
| Nuclear magnetic resonance spectroscopy | What is the carbon-hydrogen framework? |


CHAPTER CONTENTS - 12.1 Mass Spectrometry of Small Molecules: Magnetic-Sector Instruments
At its simplest, mass spectrometry (MS) is a technique for measuring the mass, and therefore the molecular weight (MW), of a molecule. In addition, it's often possible to gain structural information about a molecule by measuring the masses of the fragments produced when molecules are broken apart.

More than 20 different kinds of commercial mass spectrometers are available depending on the intended application, but all have three basic parts: an ionization source in which sample molecules are given an electrical charge, a mass analyzer in which ions are separated by their mass-to-charge ratio, and a detector in which the separated ions are observed and counted.

Among the most common mass spectrometers used for routine purposes in the laboratory is the electronimpact, magnetic-sector instrument shown schematically in FIGURE 12.2. A small amount of sample is vaporized into the ionization source, where it is bombarded by a stream of high-energy electrons. The energy of the electron beam can be varied but is commonly around 70 electron volts (eV), or $6700 \mathrm{~kJ} / \mathrm{mol}$. When a highenergy electron strikes an organic molecule, it dislodges a valence electron from the molecule, producing a cation radical-cation because the molecule has lost an electron and now has a positive charge; radical because the molecule now has an odd number of electrons.

Electron bombardment transfers so much energy that most of the cation radicals fragment after formation. They break apart into smaller pieces, some of which retain the positive charge and some of which are neutral. The fragments then flow through a curved pipe in a strong magnetic field, which deflects them into different paths according to their mass-to-charge ratio $(\mathrm{m} / \mathrm{z})$. Neutral fragments are not deflected by the magnetic field and are lost on the walls of the pipe, but positively charged fragments are sorted by the mass spectrometer onto a detector, which records them as peaks at the various $m / z$ ratios. Since the number of charges $z$ on each ion is usually 1 , the value of $m / z$ for each ion is simply its mass $m$. Masses up to approximately 2500 atomic mass units (amu) can be analyzed by this type of instrument.


CHAPTER CONTENTS - 12.1 Mass Spectrometry of Small Molecules: Magnetic-Sector Instruments
FIGURE 12.2 Representation of an electron-ionization, magnetic-sector mass spectrometer. Molecules are ionized by collision with highenergy electrons, causing some of the molecules to fragment. Passage of the charged fragments through a magnetic field then sorts them according to their mass.

Another common type of mass spectrometer uses what is called a quadrupole mass analyzer, which has a set of four solid rods is arranged parallel to the direction of the ion beam, with an oscillating electrostatic field is generated in the space between the rods. For a given field, only one $m / z$ value will make it through the quadrupole region. The others will crash into the rods or the walls of the instrument and never reach the detector FIGURE 12.3.

FIGURE 12.3 Representation of a quadrupole mass analyzer. Only ions of a certain $m / z$ will reach the detector; other ions will collide with the rods.

The mass spectrum of a compound is typically presented as a bar graph, with masses ( $m / z$ values) on the $x$ axis and intensity, or relative abundance of ions of a given $\mathrm{m} / \mathrm{z}$ striking the detector, on the $y$ axis. The tallest peak, assigned an intensity of $100 \%$, is called the base peak, and the peak that corresponds to the unfragmented cation radical is called the parent peak, or the molecular ion ( $M^{+}$, or simply $M$ ). FIGURE 12.4 shows the mass spectrum of propane.

FIGURE 12.4 Mass spectrum of propane ( $\mathrm{C}_{3} \mathrm{H}_{8}$; $\mathbf{M W}=44$ ).
Mass spectral fragmentation patterns are usually complex, and the molecular ion is often not the base peak. The mass spectrum of propane in FIGURE 12.4, for instance, shows a molecular ion at $m / z=44$ that is only about $30 \%$ as high as the base peak at $m / z=29$. In addition, many other fragment ions are present.


CHAPTER CONTENTS - 12.2 Interpreting Mass Spectra
What kinds of information can we get from a mass spectrum? The most obvious information is the molecular weight of the sample, which in itself can be invaluable. If we were given samples of hexane (MW = 86), 1-hexene ( $\mathrm{MW}=84$ ), and 1-hexyne ( $\mathrm{MW}=82$ ), for example, mass spectrometry would easily distinguish them.

Some instruments, called double-focusing mass spectrometers, have two magnetic sectors in their mass analyzers, giving these spectrometers have such high resolution that they provide mass measurements accurate to 5 ppm , or about 0.0005 amu , making it possible to distinguish between two formulas with the same nominal mass. For example, both $\mathrm{C}_{5} \mathrm{H}_{12}$ and $\mathrm{C}_{4} \mathrm{H}_{8} \mathrm{O}$ have MW $=72$, but they differ slightly beyond the decimal point: $\mathrm{C}_{5} \mathrm{H}_{12}$ has an exact mass of 72.0939 amu , whereas $\mathrm{C}_{4} \mathrm{H}_{8} \mathrm{O}$ has an exact mass of 72.0575 amu . A highresolution instrument can easily distinguish between them. Note, however, that exact mass measurements refer to molecules with specific isotopic compositions. Thus, the sum of the exact atomic masses of the specific isotopes in a molecule is measured -1.00783 amu for ${ }^{1} \mathrm{H}, 12.00000 \mathrm{amu}$ for ${ }^{12} \mathrm{C}, 14.00307 \mathrm{amu}$ for ${ }^{14} \mathrm{~N}, 15.994$ 91 amu for ${ }^{16} \mathrm{O}$, and so on-rather than the sum of the average atomic masses of elements, as found on a periodic table.


CHAPTER CONTENTS - 12.2 Interpreting Mass Spectra
Unfortunately, not every compound shows a molecular ion in its electron-impact mass spectrum. Although $\mathrm{M}^{+}$ is usually easy to identify if it's abundant, some compounds, such as 2,2-dimethylpropane, fragment so easily that no molecular ion is observed (FIGURE 12.5). In such cases, alternative "soft" ionization methods that don't use electron bombardment can prevent or minimize fragmentation (see Section 12.4).

FIGURE 12.5 Mass spectrum of 2,2-dimethylpropane ( $\mathbf{C}_{\mathbf{5}} \mathbf{H}_{\mathbf{1 2}}$; MW = 72). No molecular ion is observed when electron-impact ionization is used. What do you think is the formula and structure of the $M^{+}$peak at $m / z=57$ ?

Knowing the molecular weight makes it possible to narrow considerably the choices of molecular formula. For example, if the mass spectrum of an unknown compound shows a molecular ion at $m / z=110$, the molecular formula is likely to be $\mathrm{C}_{8} \mathrm{H}_{14}, \mathrm{C}_{7} \mathrm{H}_{10} \mathrm{O}, \mathrm{C}_{6} \mathrm{H}_{6} \mathrm{O}_{2}$, or $\mathrm{C}_{6} \mathrm{H}_{10} \mathrm{~N}_{2}$. There are always a number of molecular formulas possible for all but the lowest molecular weights, and a computer can easily generate a list of the choices.


CHAPTER CONTENTS - 12.2 Interpreting Mass Spectra
A further point about mass spectrometry, noticeable in the spectra of both propane (FIGURE 12.4) and 2,2-dimethylpropane (FIGURE 12.5), is that the peak for the molecular ion is not at the highest $\mathrm{m} / \mathrm{z}$ value. There is also a small peak at $\mathrm{M}+1$ due to the presence of different isotopes in the molecules. Although ${ }^{12} \mathrm{C}$ is the most abundant carbon isotope, a small amount ( $1.10 \%$ natural abundance) of ${ }^{13} \mathrm{C}$ is also present. Thus, a certain percentage of the molecules analyzed in the mass spectrometer are likely to contain a ${ }^{13} \mathrm{C}$ atom, giving rise to the observed $\mathrm{M}+1$ peak. In addition, a small amount of ${ }^{2} \mathrm{H}$ (deuterium; $0.015 \%$ natural abundance) is present, making a further contribution to the $\mathrm{M}+1$ peak.

Mass spectrometry would be useful even if molecular weight and formula were the only information that could be obtained, but in fact it provides much more. For one thing, the mass spectrum of a compound serves as a kind of "molecular fingerprint." Every organic compound fragments in a unique way depending on its structure, and the likelihood of two compounds having identical mass spectra is small. Thus, it's sometimes possible to identify an unknown by computer-based matching of its mass spectrum to one of the more than 785,061 searchable spectra recorded in a database called the Registry of Mass Spectral Data.

It's also possible to derive structural information about a molecule by interpreting its fragmentation pattern. Fragmentation occurs when the high-energy cation radical flies apart by spontaneous cleavage of a chemical bond. One of the two fragments retains the positive charge and is a carbocation, while the other fragment is a neutral radical.


CHAPTER CONTENTS - 12.2 Interpreting Mass Spectra
Not surprisingly, the positive charge often remains with the fragment that is best able to stabilize it. In other words, a relatively stable carbocation is often formed during fragmentation. For example, 2,2-dimethylpropane tends to fragment in such a way that the positive charge remains with the tert-butyl group. 2,2-Dimethylpropane therefore has a base peak at $m / z=57$, corresponding to $\mathrm{C}_{4} \mathrm{H}_{9}{ }^{+}$(FIGURE 12.5).

$m / z=57$
Because mass-spectral fragmentation patterns are usually complex, it's often difficult to assign structures to fragment ions. Most hydrocarbons fragment in many ways, as demonstrated by the mass spectrum of hexane in FIGURE 12.6. The hexane spectrum shows a moderately abundant molecular ion at $m / z=86$ and fragment ions at $m / z=71,57,43$, and 29 . Since all the carbon-carbon bonds of hexane are electronically similar, all break to a similar extent, giving rise to the observed mixture of ions.


CHAPTER CONTENTS - 12.2 Interpreting Mass Spectra
FIGURE 12.6 Mass spectrum of hexane $\left(\mathbf{C}_{\mathbf{6}} \mathbf{H}_{\mathbf{1 4}} \mathbf{;} \mathbf{M W}=\mathbf{8 6}\right)$. The base peak is at $m / z=57$, and numerous other ions are present.
FIGURE 12.7 shows how the hexane fragments might arise. The loss of a methyl radical $\left(\mathrm{CH}_{3}, \mathrm{M}=15\right)$ from the hexane cation radical $\left(\mathrm{M}^{+}=86\right)$ gives rise to a fragment of mass $86-15=71$; the loss of an ethyl radical $\left(\mathrm{C}_{2} \mathrm{H}_{5}\right.$, $M=29$ ) accounts for a fragment of mass $86-29=57$; the loss of a propyl radical ( $\mathrm{C}_{3} \mathrm{H}_{7}, \mathrm{M}=43$ ) accounts for a fragment of mass $86-43=43$; and the loss of a butyl radical accounts for a fragment of mass 29 . With practice, it's sometimes possible to analyze the fragmentation pattern of an unknown compound and work backward to a structure that is compatible with the data.

FIGURE 12.7 Fragmentation of hexane in a mass spectrometer.
We'll see in the next section and in later chapters that specific functional groups, such as alcohols, ketones, aldehydes, and amines, show specific kinds of mass spectral fragmentations that can be interpreted to provide structural information.


Using Mass Spectra to Identify Compounds -
Assume that you have two unlabeled samples, one of methylcyclohexane and the other of ethylcyclopentane. How could you use mass spectrometry to tell them apart? The mass spectra of both are shown in FIGURE 12.8.

FIGURE 12.8 Mass spectra of unlabeled samples A and B for Worked Example 12.1.


Strategy -
Look at the possible structures and decide on how they differ. Then think about how any of these differences in structure might give rise to differences in mass spectra. Methyl cyclohexane, for instance, has a $-\mathrm{CH}_{3}$ group,
and ethylcyclopentane has a $-\mathrm{CH}_{2} \mathrm{CH}_{3}$ group, which should affect the fragmentation patterns.


Solution -
Both mass spectra show molecular ions at $\mathrm{M}^{+}=98$, corresponding to $\mathrm{C}_{7} \mathrm{H}_{14}$, but they differ in their fragmentation patterns. Sample $A$ has its base peak at $m / z=69$, corresponding to the loss of a $\mathrm{CH}_{2} \mathrm{CH}_{3}$ group (29 mass units), but $\mathbf{B}$ has a rather small peak at $m / z=69$. Sample $\mathbf{B}$ shows a base peak at $m / z=83$, corresponding to the loss of a $\mathrm{CH}_{3}$ group ( 15 mass units), but sample $\mathbf{A}$ has only a small peak at $\mathrm{m} / \mathrm{z}=83$. We can therefore be reasonably certain that $\mathbf{A}$ is ethylcyclopentane and $\mathbf{B}$ is methylcyclohexane.

PROBLEM The sex hormone testosterone contains only C, H, and O and has a mass of 288.2089 amu , as 12-1 determined by high-resolution mass spectrometry. What is the likely molecular formula of testosterone?

PROBLEM Two mass spectra are shown in Figure 12.9. One spectrum is that of 2-methyl-2-pentene; the other 12-2 is of 2-hexene. Which is which? Explain.
(a)

(b)

FIGURE 12.9 Mass spectra for Problem 12-2.


Solution - 12.3 Mass Spectrometry of Some Common Functional Groups
As each functional group is discussed in future chapters, mass-spectral fragmentations characteristic of that group will be described. As a preview, though, we'll point out some distinguishing features of several common functional groups.


Alcohols -
Alcohols undergo fragmentation in a mass spectrometer by two pathways: alpha cleavage and dehydration. In the $\alpha$-cleavage pathway, a $\mathrm{C}-\mathrm{C}$ bond nearest the hydroxyl group is broken, yielding a neutral radical plus a resonance-stabilized, oxygen-containing cation. This type of fragmentation is seen in the spectrum of 2-pentanol in FIGURE 12.10.

In the dehydration pathway, water is eliminated, yielding an alkene radical cation with a mass 18 amu less than $\mathrm{M}^{+}$. For simplicity, we have drawn the dehydration below as an E2-type process. Often the hydrogen that is lost is not beta to the hydroxyl. Only a small peak from dehydration is observed in the spectrum of 2-pentanol (FIGURE 12.10).


Amines -
The nitrogen rule of mass spectrometry says that a compound with an odd number of nitrogen atoms has an odd-numbered molecular weight. The logic behind the rule comes from the fact that nitrogen is trivalent, thus requiring an odd number of hydrogen atoms. The presence of nitrogen in a molecule is often detected simply by observing its mass spectrum. An odd-numbered molecular ion usually means that the unknown compound has one or three nitrogen atoms, and an even-numbered molecular ion usually means that a compound has either zero or two nitrogen atoms.

Aliphatic amines undergo a characteristic $\alpha$ cleavage in a mass spectrometer, similar to that observed for alcohols. A C-C bond nearest the nitrogen atom is broken, yielding an alkyl radical and a resonance-stabilized, nitrogen-containing cation.

The mass spectrum of triethylamine has a base peak at $m / z=86$, which arises from an alpha cleavage resulting in the loss of a methyl group (FIGURE 12.11).

FIGURE 12.11 Mass spectrum of triethylamine.


Halides -
The fact that some elements have two common isotopes gives their mass spectra a distinctive appearance. Chlorine, for example, exists as two isotopes, ${ }^{35} \mathrm{Cl}$ and ${ }^{37} \mathrm{Cl}$, in roughly a $3: 1$ ratio. In a sample of chloroethane, three out of four molecules contain a ${ }^{35} \mathrm{Cl}$ atom and one out of four has a ${ }^{37} \mathrm{Cl}$ atom. In the mass spectrum of chloroethane (FIGURE 12.12 we see the molecular ion (M) at $m / z=64$ for ions that contain a ${ }^{35} \mathrm{Cl}$ and another peak at $m / z=66$, called the $M+2$ peak, for ions containing a ${ }^{37} C l$. The ratio of the relative abundance of $M: M+$ 2 is about $3: 1$, a reflection of the isotopic abundances of chlorine.

FIGURE 12.12 Mass spectrum of chloroethane.
In the case of bromine, the isotopic distribution is $50.7 \%{ }^{79} \mathrm{Br}$ and $49.3 \%{ }^{81} \mathrm{Br}$. In the mass spectrum of 1-bromohexane (FIGURE 12.13) the molecular ion appears at $\mathrm{m} / \mathrm{z}=164$ for ${ }^{79} \mathrm{Br}$-containing ions and the $\mathrm{M}+2$ peak is at $m / z=166$ for ${ }^{81} \mathrm{Br}$-containing ions. The ions at $\mathrm{m} / \mathrm{z}=135$ and 137 are informative as well. The two nearly equally large peaks tell us that the ions at those $m / z$ values still contain the bromine atom. The peak at $m / z=85$, on the other hand, does not contain bromine because there is not a large peak at $m / z=87$.

FIGURE 12.13 Mass spectrum of 1-bromohexane.


Carbonyl Compounds -
Ketones and aldehydes that have a hydrogen on a carbon three atoms away from the carbonyl group undergo a characteristic mass-spectral cleavage called the McLafferty rearrangement. The hydrogen atom is transferred to the carbonyl oxygen, a $\mathrm{C}-\mathrm{C}$ bond between the alpha and beta carbons is broken, and a neutral alkene fragment is produced. The charge remains with the oxygen-containing fragment.

In addition, ketones and aldehydes frequently undergo $\alpha$ cleavage of the bond between the carbonyl carbon and the neighboring carbon to yield a neutral radical and a resonance-stabilized acyl cation. Because the carbon neighboring the carbonyl carbon is called the alpha carbon, the reaction is called an alpha cleavage.

(To be more general about neighboring positions in carbonyl compounds, Greek letters are used in alphabetical order: alpha, beta, gamma, delta, and so on.)

The mass spectrum of butyrophenone illustrates both alpha cleavage and the McLafferty rearrangement (FIGURE 12.14). Alpha cleavage of the propyl substituent results in the loss of $\mathrm{C}_{3} \mathrm{H}_{7}=43$ mass units from the parent ion at $m / z=148$ to give the fragment ion at $m / z=105$. A McLafferty rearrangement of butyrophenone results in the loss of ethylene, $\mathrm{C}_{2} \mathrm{H}_{4}=28$ mass units, from the parent leaving the ion at $\mathrm{m} / \mathrm{z}=120$.

FIGURE 12.14 Mass spectrum of butyrophenone.


Identifying Fragmentation Patterns in a Mass Spectrum -
The mass spectrum of 2-methyl-3-pentanol is shown in FIGURE 12.15. What fragments can you identify?

FIGURE 12.15 Mass spectrum of 2-methyl-3-pentanol, for Worked Example 12.2.


Strategy -
Calculate the mass of the molecular ion, and identify the functional groups in the molecule. Then write the fragmentation processes you might expect, and compare the masses of the resultant fragments with the peaks present in the spectrum.


Solution -
2-Methyl-3-pentanol, an open-chain alcohol, has $\mathrm{M}^{+}=102$ and might be expected to fragment by $\alpha$ cleavage and by dehydration. These processes would lead to fragment ions of $m / z=84,73$, and 59 . Of the three expected fragments, dehydration is not observed (no $m / z=84$ peak), but both $\alpha$ cleavages take place ( $m / z=73,59$ ).

PROBLEM What are the masses of the charged fragments produced in the following cleavage pathways?
12-3 (a) Alpha cleavage of 2-pentanone $\left(\mathrm{CH}_{3} \mathrm{COCH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\right)$
(b) Dehydration of cyclohexanol (hydroxycyclohexane)
(c) McLafferty rearrangement of 4-methyl-2-pentanone $\left[\mathrm{CH}_{3} \mathrm{COCH}_{2} \mathrm{CH}\left(\mathrm{CH}_{3}\right)_{2}\right]$
(d) Alpha cleavage of triethylamine $\left[\left(\mathrm{CH}_{3} \mathrm{CH}_{2}\right)_{3} \mathrm{~N}\right]$

PROBLEM List the masses of the parent ion and of several fragments you might expect to find in the mass
12-4 spectrum of the following molecule:


Solution - 12.4 Mass Spectrometry in Biological Chemistry: Time-of-Flight (TOF) Instruments
MS analyses of sensitive biological samples rarely use magnetic sector ionization. Instead, they typically use either electrospray ionization ( $E S I$ ) or matrix-assisted laser desorption ionization (MALDI), typically linked to a time-of-flight (TOF) mass analyzer. Both ESI and MALDI are soft ionization methods that produce charged molecules with little fragmentation, even with sensitive biological samples of very high molecular weight.

In an ESI source, as a sample solution exits the tube, it is subjected to a high voltage that causes the droplets to become charged. The sample molecules gain one or more protons from charged solvent molecules in the droplet. The volatile solvent quickly evaporates, giving variably protonated sample molecules $\left(M+H_{n}{ }^{n+}\right)$. In a MALDI source, the sample is adsorbed onto a suitable matrix compound, such as 2,5-dihydroxybenzoic acid, which is ionized by a short burst of laser light. The matrix compound then transfers the energy to the sample and protonates it, forming $\mathrm{M}+\mathrm{H}_{n}{ }^{n+}$ ions.

Following ion formation, the variably protonated sample molecules are electrically focused into a small packet with a narrow spatial distribution, and the packet is given a sudden kick of energy by an accelerator electrode. As each molecule in the packet is given the same energy, $E=m v^{2} / 2$, it begins moving with a velocity that depends on the square root of its mass, $v=\sqrt{2 E / m}$. Lighter molecules move faster, and heavier molecules move slower. The analyzer itself-the drift tube-is simply an electrically grounded metal tube inside which the different charged molecules become separated as they move at different velocities and take different amounts of time to complete their flight.


Solution - 12.4 Mass Spectrometry in Biological Chemistry: Time-of-Flight (TOF) Instruments
The Time of Flight technique is considerably more sensitive than the magnetic sector alternative, and protein samples of up to 100 kilodaltons ( $100,000 \mathrm{amu}$ ) can be separated with a mass accuracy of 3 ppm. FIGURE 12.16 shows a MALDI-TOF spectrum of chicken egg-white lysozyme, MW = 14,306.7578 daltons. Biochemists generally use the unit dalton, abbreviated Da, instead of amu, although the two are equivalent ( 1 dalton = 1 amu ).

FIGURE 12.16 MALDI-TOF mass spectrum of chicken egg-white lysozyme. The peak at $14,306.7578$ daltons (amu) is due to the monoprotonated protein, $\mathrm{M}+\mathrm{H}^{+}$, and the peak at $28,614.2188$ daltons is due to an impurity formed by dimerization of the protein. Other peaks at lower $m / z$ values are various protonated species, $M+H_{n}{ }^{n+}$.


Solution - 12.5 Spectroscopy and the Electromagnetic Spectrum
Infrared, ultraviolet, and nuclear magnetic resonance spectroscopies differ from mass spectrometry in that they are nondestructive and involve the interaction of molecules with electromagnetic energy rather than with an ionizing source. Before beginning a study of these techniques, however, let's briefly review the nature of radiant energy and the electromagnetic spectrum.

Visible light, X rays, microwaves, radio waves, and so forth are all different kinds of electromagnetic radiation. Collectively, they make up the electromagnetic spectrum, shown in FIGURE 12.17. The electromagnetic spectrum is arbitrarily divided into regions, with the familiar visible region accounting for only a small portion, from $3.8 \times 10^{-7} \mathrm{~m}$ to $7.8 \times 10^{-7} \mathrm{~m}$ in wavelength. The visible region is flanked by the infrared and ultraviolet regions.

FIGURE 12.17 The electromagnetic spectrum covers a continuous range of wavelengths and frequencies, from radio waves at the lowfrequency end to gamma ( $\gamma$ ) rays at the high-frequency end. The familiar visible region accounts for only a small portion near the middle of the spectrum.

Electromagnetic radiation is often said to have dual behavior. In some respects, it has the properties of a particle, called a photon, yet in other respects it behaves as an energy wave. Like all waves, electromagnetic radiation is characterized by a wavelength, a frequency, and an amplitude (FIGURE 12.18). The wavelength, $\boldsymbol{\lambda}$ (Greek lambda), is the distance from one wave maximum to the next. The frequency, $\boldsymbol{\nu}$ (Greek nu), is the
number of waves that pass by a fixed point per unit time, usually given in reciprocal seconds $\left(\mathrm{s}^{-1}\right)$, or hertz, $\mathbf{H z}$ ( $1 \mathrm{~Hz}=1 \mathrm{~s}^{-1}$ ). The amplitude is the height of a wave, measured from midpoint to peak. The intensity of radiant energy, whether a feeble glow or a blinding glare, is proportional to the square of the wave's amplitude.

(c)

Infrared radiation

$$
\left(\nu=3.75 \times 10^{14} \mathrm{~s}^{-1}\right)
$$


Solution - 12.5 Spectroscopy and the Electromagnetic Spectrum
(c)

Infrared radiation

$$
\left(\nu=3.75 \times 10^{14} \mathrm{~s}^{-1}\right)
$$

FIGURE 12.18 Electromagnetic waves are characterized by a wavelength, a frequency, and an amplitude. (a) Wavelength ( $\lambda$ ) is the distance between two successive wave maxima. Amplitude is the height of the wave measured from the center. (b)-(c) What we perceive as different kinds of electromagnetic radiation are simply waves with different wavelengths and frequencies.
Multiplying the wavelength of a wave in meters ( m ) by its frequency in reciprocal seconds ( $\mathrm{s}^{-1}$ ) gives the speed of the wave in meters per second ( $\mathrm{m} / \mathrm{s}$ ). The rate of travel of all electromagnetic radiation in a vacuum is a constant value, commonly called the "speed of light" and abbreviated $c$. Its numerical value is defined as exactly $2.99792458 \times 10^{8} \mathrm{~m} / \mathrm{s}$, usually rounded off to $3.00 \times 10^{8} \mathrm{~m} / \mathrm{s}$.

$$
\begin{aligned}
& \text { Wavelength } \times \text { Frequency }=\text { Speed } \\
& \qquad \begin{array}{c}
\lambda(\mathrm{m}) \times v\left(\mathrm{~s}^{-1}\right)=c(\mathrm{~m} / \mathrm{s}) \\
\lambda=\frac{c}{v} \text { or } v=\frac{c}{\lambda}
\end{array}
\end{aligned}
$$

Just as matter comes only in discrete units called atoms, electromagnetic energy is transmitted only in discrete amounts called quanta. The amount of energy $\varepsilon$ corresponding to 1 quantum of energy ( 1 photon) of a given frequency $\nu$ is expressed by the Planck equation

$$
\varepsilon=h v=\frac{h c}{\lambda}
$$


Solution - 12.5 Spectroscopy and the Electromagnetic Spectrum
$$
\varepsilon=h v=\frac{h c}{\lambda}
$$

where $h=$ Planck's constant $\left(6.62 \times 10^{-34} \mathrm{~J} \cdot \mathrm{~s}=1.58 \times 10^{-34} \mathrm{cal} \cdot \mathrm{s}\right)$.
The Planck equation says that the energy of a given photon varies directly with its frequency $\nu$ but inversely with its wavelength $\lambda$. High frequencies and short wavelengths correspond to high-energy radiation such as gamma rays; low frequencies and long wavelengths correspond to low-energy radiation such as radio waves. Multiplying $\varepsilon$ by Avogadro's number $N_{\mathrm{A}}$ gives the same equation in more familiar units, where $E$ represents the energy of Avogadro's number (one "mole") of photons of wavelength $\lambda$ :

$$
E=\frac{N_{A} h c}{\lambda}=\frac{1.20 \times 10^{-4} \mathrm{~kJ} / \mathrm{mol}}{\lambda(\mathrm{~m})} \text { or } \frac{2.86 \times 10^{-5} \mathrm{kcal} / \mathrm{mol}}{\lambda(\mathrm{~m})}
$$

When an organic compound is exposed to a beam of electromagnetic radiation, it absorbs energy of some wavelengths but passes, or transmits, energy of other wavelengths. If we irradiate the sample with energy of many different wavelengths and determine which are absorbed and which are transmitted, we can measure the absorption spectrum of the compound.

An example of an absorption spectrum-that of ethanol exposed to infrared radiation-is shown in FIGURE
12.19. The horizontal axis records the wavelength, and the vertical axis records the intensity of the various energy absorptions in percent transmittance. The baseline corresponding to $0 \%$ absorption (or $100 \%$ transmittance) runs along the top of the chart, so a downward spike means that energy absorption has occurred at that wavelength.


Solution - 12.5 Spectroscopy and the Electromagnetic Spectrum
FIGURE 12.19 An infrared absorption spectrum for ethanol, $\mathbf{C H}_{\mathbf{3}} \mathbf{C H}_{\mathbf{2}} \mathbf{O H}$. A transmittance of $100 \%$ means that all the energy is passing through the sample, whereas a lower transmittance means that some energy is being absorbed. Thus, each downward spike corresponds to an energy absorption.

The energy a molecule gains when it absorbs radiation must be distributed over the molecule in some way. With infrared radiation, the absorbed energy causes bonds to stretch and bend more vigorously. With ultraviolet radiation, the energy causes an electron to jump from a lower-energy orbital to a higher-energy one. Different radiation frequencies affect molecules in different ways, but each provides structural information when the results are interpreted.

There are many kinds of spectroscopies, which differ according to the region of the electromagnetic spectrum used. We'll look at three: infrared spectroscopy, ultraviolet spectroscopy, and nuclear magnetic resonance spectroscopy. Let's begin by seeing what happens when an organic sample absorbs infrared energy.


Correlating Energy and Frequency of Radiation -
Which is higher in energy, FM radio waves with a frequency of $1.015 \times 10^{8} \mathrm{~Hz}(101.5 \mathrm{MHz})$ or visible green light with a frequency of $5 \times 10^{14} \mathrm{~Hz}$ ?


Strategy -
Remember the equations $\varepsilon=h \nu$ and $\varepsilon=h c / \lambda$, which say that energy increases as frequency increases and as wavelength decreases.


Solution -
Since visible light has a higher frequency than radio waves, it is higher in energy.

PROBLEM Which has higher energy, infrared radiation with $\lambda=1.0 \times 10^{-6} \mathrm{~m}$ or an X ray with $\lambda=3.0 \times 10^{-9} \mathrm{~m}$ ? 12-5 Radiation with $\nu=4.0 \times 10^{9} \mathrm{~Hz}$ or with $\lambda=9.0 \times 10^{-6} \mathrm{~m}$ ?

PROBLEM It's useful to develop a feeling for the amounts of energy that correspond to different parts of 12-6 the electromagnetic spectrum. Calculate the energies in $\mathrm{kJ} / \mathrm{mol}$ of each of the following kinds of radiation:
(a) A gamma ray with $\lambda=5.0 \times 10^{-11} \mathrm{~m}$ (b) An X ray with $\lambda=3.0 \times 10^{-9} \mathrm{~m}$
(c) Ultraviolet light with $\nu=6.0 \times 10^{15} \mathrm{~Hz}$ (d) Visible light with $\nu=7.0 \times 10^{14} \mathrm{~Hz}$
(e) Infrared radiation with $\lambda=2.0 \times 10^{-5} \mathrm{~m}$ (f) Microwave radiation with $\nu=1.0 \times 10^{11} \mathrm{~Hz}$


Solution - 12.6 Infrared Spectroscopy
In infrared (IR) spectroscopy, the IR region of the electromagnetic spectrum covers the range from just above the visible $\left(7.8 \times 10^{-7} \mathrm{~m}\right)$ to approximately $10^{-4} \mathrm{~m}$, but only the midportion from $2.5 \times 10^{-6} \mathrm{~m}$ to $2.5 \times 10^{-5} \mathrm{~m}$
is used by organic chemists (FIGURE 12.20. Wavelengths within the IR region are usually given in micrometers $\left(1 \mu \mathrm{~m}=10^{-6} \mathrm{~m}\right)$, and frequencies are given in wavenumbers rather than in hertz. The wavenumber $\widetilde{v}$ is the reciprocal of wavelength in centimeters and is therefore expressed in units of $\mathrm{cm}^{-1}$.

$$
\text { Wavenumber: } \widetilde{v}\left(\mathrm{~cm}^{-1}\right)=\frac{1}{\lambda(\mathrm{~cm})}
$$

Thus, the useful IR region is from 4000 to $400 \mathrm{~cm}^{-1}$, corresponding to energies of $48.0 \mathrm{~kJ} / \mathrm{mol}$ to $4.80 \mathrm{~kJ} / \mathrm{mol}$ (11.5-1.15 kcal/mol).

FIGURE 12.20 The infrared and adjacent regions of the electromagnetic spectrum.
Why does an organic molecule absorb some wavelengths of IR radiation but not others? All molecules have a certain amount of energy and are in constant motion. Their bonds stretch and contract, atoms wag back and forth, and other molecular vibrations occur. Some of the kinds of allowed vibrations are shown below: