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.
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0354.jpg?height=231&width=692&top_left_y=1890&top_left_x=711)


(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.
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0355.jpg?height=221&width=736&top_left_y=193&top_left_x=689)


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).
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0355.jpg?height=560&width=1521&top_left_y=1189&top_left_x=302)


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.
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0355.jpg?height=166&width=343&top_left_y=2045&top_left_x=655)
$\mathrm{S}_{\mathrm{N}} 2$ reaction (backside attack)
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0355.jpg?height=175&width=323&top_left_y=2035&top_left_x=1134)


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.
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0356.jpg?height=576&width=1296&top_left_y=257&top_left_x=417)


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 -


Solution -

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.)
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0356.jpg?height=302&width=513&top_left_y=1920&top_left_x=421)


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.


Axial chlorine: H and Cl are anti periplanar -


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.
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0357.jpg?height=228&width=1651&top_left_y=1092&top_left_x=237)


Neomenthyl chloride
(b)
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0357.jpg?height=267&width=1567&top_left_y=1409&top_left_x=279)


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.


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.


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.
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0358.jpg?height=188&width=359&top_left_y=1343&top_left_x=1181)
(2) $\downarrow$ Fast
<img class="imgSvg" id = "m6sn3njqqzg4s80cny" src="data:image/svg+xml;base64,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"/>


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.
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0358.jpg?height=161&width=407&top_left_y=2196&top_left_x=615)


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).
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0359.jpg?height=719&width=1389&top_left_y=578&top_left_x=368)


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.
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0359.jpg?height=296&width=1291&top_left_y=1798&top_left_x=417)


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.
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0360.jpg?height=377&width=1147&top_left_y=191&top_left_x=489)


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.


WORKED EXAMPLE 11.5 -


Predicting the Product and Mechanism of Reactions -


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.
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0360.jpg?height=98&width=725&top_left_y=2206&top_left_x=700)
(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.
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0361.jpg?height=277&width=1031&top_left_y=184&top_left_x=547)


Solution -


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.


Green Chemistry -

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:


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.
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0363.jpg?height=481&width=1440&top_left_y=188&top_left_x=353)


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


Key Terms -


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.


Summary -

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).


Summary -

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.


Summary -

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.


Summary -

In general, substrates react in the following way:


Summary -

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


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)
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0364.jpg?height=174&width=860&top_left_y=1146&top_left_x=681)
b. $\mathrm{S}_{\mathrm{N}} 2$ reaction of $1^{\circ}$ and simple $2^{\circ}$ halides (Section 11.2 and Section 11.3)
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0364.jpg?height=112&width=676&top_left_y=1364&top_left_x=773)
2. Eliminations
a. E1 reaction (Section 11.10)
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0364.jpg?height=177&width=917&top_left_y=1576&top_left_x=653)
b. E1cB reaction (Section 11.10)
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0364.jpg?height=174&width=1055&top_left_y=1797&top_left_x=584)
c. E2 reaction (Section 11.8)
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0364.jpg?height=158&width=712&top_left_y=2011&top_left_x=758)


Additional Problems -


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)
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0365.jpg?height=208&width=259&top_left_y=219&top_left_x=483)
(b)
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0365.jpg?height=289&width=320&top_left_y=205&top_left_x=805)
(c)
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0365.jpg?height=276&width=383&top_left_y=212&top_left_x=1196)


Visualizing Chemistry -

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.
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0365.jpg?height=286&width=651&top_left_y=627&top_left_x=420)


Visualizing Chemistry -

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 $=\mathrm{Cl}$ ):
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0365.jpg?height=307&width=516&top_left_y=1039&top_left_x=420)


Visualizing Chemistry -

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 $=\mathrm{Cl})$ :
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0365.jpg?height=356&width=524&top_left_y=1475&top_left_x=418)


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)
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0365.jpg?height=126&width=663&top_left_y=2021&top_left_x=1124)
(c)
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0365.jpg?height=137&width=562&top_left_y=2154&top_left_x=478)
(d)
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0365.jpg?height=123&width=557&top_left_y=2156&top_left_x=1109)


Mechanism Problems -


Mechanism Problems -

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)
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0366.jpg?height=110&width=364&top_left_y=479&top_left_x=479)
(b)
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0366.jpg?height=118&width=383&top_left_y=483&top_left_x=914)
(c)
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0366.jpg?height=196&width=423&top_left_y=490&top_left_x=1366)

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)
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0366.jpg?height=155&width=394&top_left_y=806&top_left_x=481)
(b)
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0366.jpg?height=104&width=394&top_left_y=810&top_left_x=947)
(c)
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0366.jpg?height=145&width=410&top_left_y=811&top_left_x=1416)

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)
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0366.jpg?height=125&width=448&top_left_y=1092&top_left_x=481)
(b)
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0366.jpg?height=221&width=372&top_left_y=1093&top_left_x=996)
(c)
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0366.jpg?height=169&width=413&top_left_y=1103&top_left_x=1422)


Mechanism Problems -

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)
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0366.jpg?height=211&width=526&top_left_y=1420&top_left_x=474)
(c)
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(d)
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0366.jpg?height=128&width=686&top_left_y=1638&top_left_x=1126)

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.
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$\xrightarrow[\text { Ethanol }]{\mathrm{NaOH}}$
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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.
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Mechanism Problems -

PROBLEM Metabolism of $S$-adenosylhomocysteine (Section 11.6) involves the following sequence. Propose a
11-33 mechanism for the second step.
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0367.jpg?height=462&width=1332&top_left_y=192&top_left_x=424)


Mechanism Problems -

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.


Mechanism Problems -

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.
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0367.jpg?height=264&width=1267&top_left_y=990&top_left_x=418)


Mechanism Problems -

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:
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0367.jpg?height=196&width=869&top_left_y=1385&top_left_x=411)


Mechanism Problems -

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?


Mechanism Problems -

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.)
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0367.jpg?height=174&width=760&top_left_y=1946&top_left_x=414)


Mechanism Problems -

PROBLEM Propose a mechanism for the following reaction, an important step in the laboratory synthesis of 11-38 proteins:
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0367.jpg?height=196&width=996&top_left_y=2252&top_left_x=410)


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.


Nucleophilic Substitution Reactions -

PROBLEM The following Walden cycle has been carried out: Explain the results, and indicate where inversion 11-40 occurs.
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0368.jpg?height=538&width=1332&top_left_y=471&top_left_x=407)


Nucleophilic Substitution Reactions -

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 $\mathrm{H}_{2} \mathrm{C}=\mathrm{CHCH}_{2} \mathrm{Br}$


Nucleophilic Substitution Reactions -

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.


Nucleophilic Substitution Reactions -


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


Nucleophilic Substitution Reactions -

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) $\mathrm{CH}_{3} \mathrm{~S}^{-}$


Nucleophilic Substitution Reactions -

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 $0^{\circ} \mathrm{C}$


Elimination Reactions -

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 $\mathrm{H}_{2} \mathrm{O}$


Elimination Reactions -

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.
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0369.jpg?height=131&width=565&top_left_y=761&top_left_x=414)


Elimination Reactions -


Elimination Reactions -

PROBLEM Predict the major alkene product of the following E1 reaction:
11-52
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0369.jpg?height=163&width=404&top_left_y=1046&top_left_x=416)


Elimination Reactions -

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)
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0369.jpg?height=139&width=443&top_left_y=1551&top_left_x=483)
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0369.jpg?height=136&width=204&top_left_y=1550&top_left_x=955)
(b)
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0369.jpg?height=148&width=552&top_left_y=1555&top_left_x=1231)
(c)
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0369.jpg?height=175&width=705&top_left_y=1707&top_left_x=477)


General Problems -


General Problems -

PROBLEM Reaction of the following $S$ tosylate with cyanide ion yields a nitrile product that also has $S$
11-59 stereochemistry. Explain.
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0370.jpg?height=131&width=470&top_left_y=1645&top_left_x=413)
( $S$ 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.
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0370.jpg?height=373&width=744&top_left_y=1990&top_left_x=406)


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?
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0371.jpg?height=202&width=660&top_left_y=189&top_left_x=575)
trans-1-Bromo-2-methylcyclohexane
3-Methylcyclohexene
PROBLEM Predict the product(s) of the following reaction, indicating stereochemistry where necessary:
11-62
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0371.jpg?height=204&width=513&top_left_y=554&top_left_x=416)


General Problems -

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?
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0371.jpg?height=112&width=825&top_left_y=1023&top_left_x=422)


General Problems -

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)
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0371.jpg?height=147&width=466&top_left_y=1260&top_left_x=477)
(b)
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0371.jpg?height=139&width=419&top_left_y=1264&top_left_x=1078)
(c)
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0371.jpg?height=153&width=418&top_left_y=1420&top_left_x=550)


General Problems -

PROBLEM (S)-2-Butanol slowly racemizes on standing in dilute sulfuric acid. Explain.
11-65
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0371.jpg?height=104&width=410&top_left_y=1653&top_left_x=413)


General Problems -

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.
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0371.jpg?height=283&width=1129&top_left_y=2141&top_left_x=411)


General Problems -

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.
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0372.jpg?height=261&width=644&top_left_y=344&top_left_x=410)


General Problems -

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.


General Problems -

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 $\mathbf{X}$ ?


General Problems -

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.
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0372.jpg?height=144&width=744&top_left_y=1126&top_left_x=411)


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?
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0372.jpg?height=466&width=1012&top_left_y=1518&top_left_x=410)
$N$-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
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0373.jpg?height=226&width=1364&top_left_y=191&top_left_x=413)
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0373.jpg?height=389&width=1397&top_left_y=445&top_left_x=410)
(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
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0374.jpg?height=706&width=1884&top_left_y=642&top_left_x=237)


CHAPTER 12 -

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


CHAPTER CONTENTS -

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.


CHAPTER CONTENTS -

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.


CHAPTER CONTENTS -

| 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.
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0375.jpg?height=381&width=1429&top_left_y=834&top_left_x=348)


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.
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0375.jpg?height=166&width=481&top_left_y=1543&top_left_x=822)

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.
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0376.jpg?height=720&width=1220&top_left_y=193&top_left_x=447)


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.
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0376.jpg?height=551&width=933&top_left_y=1340&top_left_x=593)

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.
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0377.jpg?height=483&width=1359&top_left_y=192&top_left_x=383)

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.