A terminal alkyne - 9.9 An Introduction to Organic Synthesis
As mentioned in the introduction, one of the purposes of this chapter is to use alkyne chemistry as a vehicle to begin looking at some of the general strategies used in organic synthesis-the construction of complex molecules in the laboratory. There are many reasons for carrying out the laboratory synthesis of an organic compound. In the pharmaceutical industry, new molecules are designed and synthesized in the hope that some might be useful new drugs. In the chemical industry, syntheses are done to devise more economical routes to known compounds. In academic laboratories, the synthesis of extremely complex molecules is sometimes done just for the intellectual challenge involved in mastering so difficult a subject. The successful synthesis route is a highly creative work that is sometimes described by such subjective terms as elegant or beautiful.

In this book, too, we will often devise syntheses of molecules from simpler precursors, but the purpose here is to learn. The ability to plan a successful multistep synthetic sequence requires a working knowledge of the uses and limitations of many different organic reactions. Furthermore, it requires the practical ability to piece together the steps in a sequence such that each reaction does only what is desired without causing changes elsewhere in the molecule. Planning a synthesis makes you approach a chemical problem in a logical way, draw on your knowledge of chemical reactions, and organize that knowledge into a workable plan-it helps you learn organic chemistry.

There's no secret to planning an organic synthesis: all it takes is a knowledge of the different reactions and some practice. The only real trick is to work backward in what is often called a retrosynthetic direction. Don't look at a potential starting material and ask yourself what reactions it might undergo. Instead, look at the final product and ask, "What was the immediate precursor of that product?" For example, if the final product is an alkyl halide, the immediate precursor might be an alkene, to which you could add HX. If the final product is a cis alkene, the immediate precursor might be an alkyne, which you could hydrogenate using the Lindlar catalyst. Having found an immediate precursor, work backward again, one step at a time, until you get back to the starting material. You have to keep the starting material in mind, of course, so that you can work back to it, but you don't want that starting material to be your main focus.

Let's work several examples of increasing complexity.


Devising a Synthesis Route -
How would you synthesize cis-2-hexene from 1-pentyne and an alkyl halide? More than one step is needed.


Strategy -
When undertaking any synthesis problem, you should look at the product, identify the functional groups it contains, and then ask yourself how those functional groups can be prepared. Always work retrosynthetically, one step at a time.

The product in this case is a cis-disubstituted alkene, so the first question is, "What is an immediate precursor of a cis-disubstituted alkene?" We know that an alkene can be prepared from an alkyne by reduction and that the right choice of experimental conditions will allow us to prepare either a trans-disubstituted alkene (using lithium in liquid ammonia) or a cis-disubstituted alkene (using catalytic hydrogenation over the Lindlar catalyst). Thus, reduction of 2 -hexyne by catalytic hydrogenation using the Lindlar catalyst should yield cis-2-hexene.


cis-2-Hexene -
Next ask, "What is an immediate precursor of 2-hexyne?" We've seen that an internal alkyne can be prepared by alkylation of a terminal alkyne anion. In the present instance, we're told to start with 1-pentyne and an alkyl halide. Thus, alkylation of the anion of 1-pentyne with iodomethane should yield 2-hexyne.


Solution -
cis-2-Hexene can be synthesized from the given starting materials in three steps.

cis-2-Hexene


Devising a Synthesis Route -
How would you synthesize 2-bromopentane from acetylene and an alkyl halide? More than one step is needed.


Strategy -
Identify the functional group in the product (an alkyl bromide) and work the problem retrosynthetically. What is an immediate precursor of an alkyl bromide? Perhaps an alkene plus HBr. Of the two possibilities, Markovnikov addition of HBr to 1-pentene looks like a better choice than addition to 2 -pentene because the latter reaction would give a mixture of isomers.

What is an immediate precursor of an alkene? Perhaps an alkyne, which could be reduced.

$$
\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{C} \equiv \mathrm{CH} \xrightarrow[\text { Lindlar catalyst }]{\mathrm{H}_{2}} \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}=\mathrm{CH}_{2}
$$

What is an immediate precursor of a terminal alkyne? Perhaps sodium acetylide and an alkyl halide.

$$
\mathrm{Na}^{+}: \overline{\mathrm{C}} \equiv \mathrm{CH}+\mathrm{BrCH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3} \longrightarrow \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{C} \equiv \mathrm{CH}
$$


Solution -
The desired product can be synthesized in four steps from acetylene and 1-bromopropane.


Devising a Synthesis Route -
How would you synthesize 5-methyl-1-hexanol (5-methyl-1-hydroxyhexane) from acetylene and an alkyl halide?

Acetylene Alkyl 5-Methyl-1-hexanol halide


Strategy -
What is an immediate precursor of a primary alcohol? Perhaps a terminal alkene, which could be hydrated with non-Markovnikov regiochemistry by reaction with borane followed by oxidation with $\mathrm{H}_{2} \mathrm{O}_{2}$.

What is an immediate precursor of a terminal alkene? Perhaps a terminal alkyne, which could be reduced.

What is an immediate precursor of 5-methyl-1-hexyne? Perhaps acetylene and 1-bromo-3-methylbutane.


Solution -
The synthesis can be completed in four steps from acetylene and 1-bromo-3-methylbutane:

5-Methyl-1-hexanol

PROBLEM Beginning with 4-octyne as your only source of carbon, and using any inorganic reagents necessary,
9-12 how would you synthesize the following compounds?
(a) cis-4-Octene
(b) Butanal (c) 4-Bromooctane
(d) 4-Octanol
(e) 4,5-Dichlorooctane
(f) Butanoic acid

PROBLEM Beginning with acetylene and any alkyl halide needed, how would you synthesize the following 9-13 compounds?
(a) Decane
(b) 2,2-Dimethylhexane
(c) Hexanal
(d) 2-Heptanone


CRO CHEMISTRY MATTERS -
The Art of Organic Synthesis

FIGURE 9.8 Vitamin $B_{12}$ has been synthesized from scratch in the laboratory, but the bacteria growing on sludge from municipal sewage plants do a much better job. (credit: "Aeration and sludge-wasting" by U.S. Department of Agriculture/Flickr, Public Domain)

If you think some of the synthesis problems at the end of this chapter are difficult, try devising a synthesis of vitamin $\mathrm{B}_{12}$ starting only from simple substances you can buy in a chemical catalog. This extraordinary achievement was reported in 1973 as the culmination of a collaborative effort headed by Robert B. Woodward of Harvard University and Albert Eschenmoser of the Swiss Federal Institute of Technology in Zürich. More than 100 graduate students and postdoctoral associates contributed to the work, which took more than a decade to complete.

Vitamin $B_{12}$
Why put such extraordinary effort into the laboratory synthesis of a molecule so easily obtained from natural sources? There are many reasons. On a basic human level, a chemist might be motivated primarily by the challenge, much as a climber might be challenged by the ascent of a difficult peak. Beyond the pure challenge, the completion of a difficult synthesis is also valuable in that it establishes new standards and raises the field to a new level. If vitamin $B_{12}$ can be made, then why can't any molecule found in nature be made? Indeed, the decades that have passed since the work of Woodward and Eschenmoser have seen the laboratory synthesis of many enormously complex and valuable substances. Sometimes these substances-for instance, the anticancer compound paclitaxel, trade named Taxol-are not easily available in nature, so laboratory synthesis is the only
method for obtaining larger quantities.


CRO CHEMISTRY MATTERS -
But perhaps the most important reason for undertaking a complex synthesis is that, in so doing, new reactions and new chemistry are discovered. It invariably happens in a complex synthesis that a point is reached at which the planned route fails. At such a time, the only alternatives are either to quit or to devise a way around the difficulty. New reactions and new principles come from such situations, and it is in this way that the science of organic chemistry grows richer. In the synthesis of vitamin $B_{12}$, for example, unexpected findings emerged that led to the understanding of an entire new class of reactions-the pericyclic reactions that are the subject of Chapter 30 in this book. From synthesizing vitamin $\mathrm{B}_{12}$ to understanding pericyclic reactions-no one could have possibly predicted such a link at the beginning of the synthesis, but that is the way of science.


Key Terms -
- acetylide anion - Lindlar Catalyst
- alkylation - retrosynthetic
- alkyne
- tautomer
- enol


Summary -
Alkynes are less common than alkenes, both in the laboratory and in living organisms, so we haven't covered them in great detail. The real importance of this chapter is that alkyne chemistry is a useful vehicle for looking at the general strategies used in organic synthesis-the construction of complex molecules in the laboratory.

An alkyne is a hydrocarbon that contains a carbon-carbon triple bond. Alkyne carbon atoms are $s p$-hybridized, and the triple bond consists of one $s p-s p \sigma$ bond and two $p-p \pi$ bonds. There are relatively few general methods of alkyne synthesis. Two favorable ones are the alkylation of an acetylide anion with a primary alkyl halide and the twofold elimination of HX from a vicinal dihalide.

The chemistry of alkynes is dominated by electrophilic addition reactions, similar to those of alkenes. Alkynes react with HBr and HCl to yield vinylic halides and with $\mathrm{Br}_{2}$ and $\mathrm{Cl}_{2}$ to yield 1,2-dihalides (vicinal dihalides). Alkynes can be hydrated by reaction with aqueous sulfuric acid in the presence of mercury(II) catalyst. The reaction leads to an intermediate enol that immediately tautomerizes to yield a ketone. Because the addition reaction occurs with Markovnikov regiochemistry, a methyl ketone is produced from a terminal alkyne. Alternatively, hydroboration-oxidation of a terminal alkyne yields an aldehyde.

Alkynes can be reduced to yield alkenes and alkanes. Complete reduction of the triple bond over a palladium hydrogenation catalyst yields an alkane; partial reduction by catalytic hydrogenation over a Lindlar catalyst yields a cis alkene. Reduction of the alkyne with lithium in ammonia yields a trans alkene.


Summary -
Terminal alkynes are weakly acidic. The alkyne hydrogen can be removed by a strong base such as $\mathrm{Na}^{+-} \mathrm{NH}_{2}$ to yield an acetylide anion. An acetylide anion acts as a nucleophile and can displace a halide ion from a primary alkyl halide in an alkylation reaction. Acetylide anions are more stable than either alkyl anions or vinylic anions because their negative charge is in a hybrid orbital with $50 \% s$ character, allowing the charge to be closer to the nucleus.


Summary of Reactions -
No stereochemistry is implied unless specifically indicated with wedged, solid, and dashed lines.

1. Preparation of alkynes
a. Dehydrohalogenation of vicinal dihalides (Section 9.2)

b. Alkylation of acetylide anions (Section 9.8)

2. Reactions of alkynes
a. Addition of HCl and HBr (Section 9.3)

b. Addition of $\mathrm{Cl}_{2}$ and $\mathrm{Br}_{2}$ (Section 9.3)

c. Hydration (Section 9.4)
(1) Mercuric sulfate catalyzed

An enol
A methyl ketone
(2) Hydroboration-oxidation

d. Reduction (Section 9.5)
(1) Catalytic hydrogenation

(2) Lithium in liquid ammonia

A trans alkene
e. Conversion into acetylide anions (Section 9.7)

$$
\mathrm{R}-\mathrm{C} \equiv \mathrm{C}-\mathrm{H} \xrightarrow[\mathrm{NH}_{3}]{\mathrm{NaNH}_{2}} \mathrm{R}-\mathrm{C} \equiv \mathrm{C}:-\mathrm{Na}^{+}+\mathrm{NH}_{3}
$$


Visualizing Chemistry -
PROBLEM Name the following alkynes, and predict the products of their reaction with (1) $\mathrm{H}_{2}$ in the presence
9-14 of a Lindlar catalyst and (2) $\mathrm{H}_{3} \mathrm{O}^{+}$in the presence of $\mathrm{HgSO}_{4}$ :
(a)

(b)

PROBLEM From what alkyne might each of the following substances have been made? (Green = Cl.)

9-15 (a)

(b)

PROBLEM How would you prepare the following substances, starting from any compounds having four
9-16 carbons or fewer?
(a)

(b)

PROBLEM The following cycloalkyne is too unstable to exist. Explain.
9-17


Mechanism Problems -
PROBLEM Assuming that halogens add to alkynes in the same manner as they add to alkenes, propose a
9-18 mechanism for and predict the product(s) of the reaction of phenylpropyne with $\mathrm{Br}_{2}$.
PROBLEM Assuming that strong acids add to alkynes in the same manner as they add to alkenes, propose a
9-19 mechanism for each of the following reactions.
(a)

(b)

(c)

PROBLEM The mercury-catalyzed hydration of alkynes involves the formation of an organomercury enol
9-20 intermediate. Draw the electron-pushing mechanism to show how each of the following intermediates is formed.
(a)

(b)

PROBLEM The final step in the hydration of an alkyne under acidic conditions is the tautomerization of an enol
9-21 intermediate to give the corresponding ketone. The mechanism involves a protonation followed by a deprotonation. Show the mechanism for each of the following tautomerizations.
(a)

(b)

$\xrightarrow{\mathrm{H}_{3} \mathrm{O}^{+}}$

(c)

PROBLEM Predict the product(s) and show the complete electron-pushing mechanism for each of the
9-22 following dissolving metal reductions.
(a)

(b)

(c)

PROBLEM Identify the mechanisms for the following reactions as polar, radical, or both.
9-23 (a)

(b)

(c)

PROBLEM Predict the product and provide the complete electron-pushing mechanism for the following two-
9-24 step synthetic processes.
(a)

(b)

(c)

$$
\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{C} \equiv \mathrm{CH} \xrightarrow{\text { 2. } \mathrm{PhCH}_{2} \mathrm{Br}}
$$

PROBLEM Reaction of acetone with $\mathrm{D}_{3} \mathrm{O}^{+}$yields hexadeuterioacetone. That is, all the hydrogens in acetone are
9-25 exchanged for deuterium. Review the mechanism of mercuric-ion-catalyzed alkyne hydration, and then propose a mechanism for this deuterium incorporation.


Acetone -
Hexadeuterioacetone


Naming Alkynes -
PROBLEM Give IUPAC names for the following compounds:
9-26 (a)

(b) $\mathrm{CH}_{3} \mathrm{C} \equiv \mathrm{CCH}_{2} \mathrm{C} \equiv \mathrm{CCH}_{2} \mathrm{CH}_{3}$
(c)

(d)

(e)
$\mathrm{H}_{2} \mathrm{C}=\mathrm{CHCH}=\mathrm{CHC} \equiv \mathrm{CH}$
(f)

PROBLEM Draw structures corresponding to the following names:
9-27 (a) 3,3-Dimethyl-4-octyne (b) 3-Ethyl-5-methyl-1,6,8-decatriyne
(c) 2,2,5,5-Tetramethyl-3-hexyne (d) 3,4-Dimethylcyclodecyne
(e) 3,5-Heptadien-1-yne
(f) 3-Chloro-4,4-dimethyl-1-nonen-6-yne (g) 3-sec-Butyl-1-heptyne
(h) 5-tert-Butyl-2-methyl-3-octyne

PROBLEM The following two hydrocarbons have been isolated from various plants in the sunflower family.
9-28 Name them according to IUPAC rules.
(a) $\mathrm{CH}_{3} \mathrm{CH}=\mathrm{CHC} \equiv \mathrm{CC} \equiv \mathrm{CCH}=\mathrm{CHCH}=\mathrm{CHCH}=\mathrm{CH}_{2}$ (all trans)
(b) $\mathrm{CH}_{3} \mathrm{C} \equiv \mathrm{CC} \equiv \mathrm{CC} \equiv \mathrm{CC} \equiv \mathrm{CC} \equiv \mathrm{CCH}=\mathrm{CH}_{2}$


Reactions of Alkynes -
PROBLEM Terminal alkynes react with $\mathrm{Br}_{2}$ and water to yield bromo ketones. For example:
9-29

Propose a mechanism for the reaction. To what reaction of alkenes is the process analogous?
PROBLEM Predict the products of the following reactions:
9-30

PROBLEM Predict the products from reaction of 1-hexyne with the following reagents:
9-31 (a) 1 equiv HBr (b) 1 equiv $\mathrm{Cl}_{2}$ (c) $\mathrm{H}_{2}$, Lindlar catalyst (d) $\mathrm{NaNH}_{2}$ in $\mathrm{NH}_{3}$, then $\mathrm{CH}_{3} \mathrm{Br}$
(e) $\mathrm{H}_{2} \mathrm{O}, \mathrm{H}_{2} \mathrm{SO}_{4}, \mathrm{HgSO}_{4}$ (f) 2 equiv HCl

PROBLEM Predict the products from reaction of 5-decyne with the following reagents:
9-32 (a) $\mathrm{H}_{2}$, Lindlar catalyst
(b) Li in $\mathrm{NH}_{3}$
(c) 1 equiv $\mathrm{Br}_{2}$
(d) $\mathrm{BH}_{3}$ in THF, then $\mathrm{H}_{2} \mathrm{O}_{2}, \mathrm{OH}^{-}$
(e) $\mathrm{H}_{2} \mathrm{O}, \mathrm{H}_{2} \mathrm{SO}_{4}, \mathrm{HgSO}_{4}$
(f) Excess $\mathrm{H}_{2}, \mathrm{Pd} / \mathrm{C}$ catalyst

PROBLEM Predict the products from reaction of 2-hexyne with the following reagents:
9-33 (a) 2 equiv $\mathrm{Br}_{2}$
(b) 1 equiv HBr
(c) Excess HBr
(d) Li in $\mathrm{NH}_{3}$
(e) $\mathrm{H}_{2} \mathrm{O}, \mathrm{H}_{2} \mathrm{SO}_{4}, \mathrm{HgSO}_{4}$


Reactions of Alkynes -
PROBLEM Propose structures for hydrocarbons that give the following products on oxidative cleavage by
9-34 $\mathrm{KMnO}_{4}$ or $\mathrm{O}_{3}$ :
(a) $\mathrm{CO}_{2}+\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{5} \mathrm{CO}_{2} \mathrm{H}$
(b)

(c) $\mathrm{HO}_{2} \mathrm{C}\left(\mathrm{CH}_{2}\right)_{8} \mathrm{CO}_{2} \mathrm{H}$
(d)

(e)

PROBLEM Identify the reagents a-c in the following scheme: 9-35

Organic Synthesis
PROBLEM How would you carry out the following multistep conversions? More than one step may be needed 9-36 in some instances.

PROBLEM How would you carry out the following reactions?
9-37 (a)

(b)

(c)

(d)

(e)

(f)

PROBLEM Each of the following syntheses requires more than one step. How would you carry them out?
9-38 (a)

(b)

PROBLEM How would you carry out the following multistep transformation?
9-39

PROBLEM How would you carry out the following multistep conversions?


9-40 -
PROBLEM Synthesize the following compounds using 1-butyne as the only source of carbon, along with any
9-41 inorganic reagents you need. More than one step may be needed.
(a) 1,1,2,2-Tetrachlorobutane (b) 1,1-Dichloro-2-ethylcyclopropane

PROBLEM How would you synthesize the following compounds from acetylene and any alkyl halides with four 9-42 or fewer carbons? More than one step may be needed.
(a) $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{C} \equiv \mathrm{CH}$
(b) $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{C} \equiv \mathrm{CCH}_{2} \mathrm{CH}_{3}$
(c)

(d)

(e) $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CHO}$

PROBLEM How would you carry out the following reactions to introduce deuterium into organic molecules?
9-43 (a)

(c)
$\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{C} \equiv \mathrm{CH} \xrightarrow{?} \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{C} \equiv \mathrm{CD}$
(b)

(d)

PROBLEM How would you prepare cyclodecyne starting from acetylene and any required alkyl halide?

PROBLEM The sex attractant given off by the common housefly is an alkene named muscalure. Propose a
9-45 synthesis of muscalure starting from acetylene and any alkyl halides needed. What is the IUPAC name for muscalure?

Muscalure


General Problems -
PROBLEM A hydrocarbon of unknown structure has the formula $\mathrm{C}_{8} \mathrm{H}_{10}$. On catalytic hydrogenation over the
9-46 Lindlar catalyst, 1 equivalent of $\mathrm{H}_{2}$ is absorbed. On hydrogenation over a palladium catalyst, 3 equivalents of $\mathrm{H}_{2}$ are absorbed.
(a) How many degrees of unsaturation are present in the unknown structure?
(b) How many triple bonds are present? (c) How many double bonds are present?
(d) How many rings are present? (e) Draw a structure that fits the data.

PROBLEM Compound $\mathbf{A}\left(\mathrm{C}_{9} \mathrm{H}_{12}\right)$ absorbed 3 equivalents of $\mathrm{H}_{2}$ on catalytic reduction over a palladium catalyst
9-47 to give $\mathbf{B}\left(\mathrm{C}_{9} \mathrm{H}_{18}\right)$. On ozonolysis, compound $\mathbf{A}$ gave, among other things, a ketone that was identified as cyclohexanone. On treatment with $\mathrm{NaNH}_{2}$ in $\mathrm{NH}_{3}$, followed by addition of iodomethane, compound $\mathbf{A}$ gave a new hydrocarbon, $\mathbf{C}\left(\mathrm{C}_{10} \mathrm{H}_{14}\right)$. What are the structures of $\mathbf{A}, \mathbf{B}$, and $\mathbf{C}$ ?

PROBLEM Hydrocarbon $A$ has the formula $\mathrm{C}_{12} \mathrm{H}_{8}$. It absorbs 8 equivalents of $\mathrm{H}_{2}$ on catalytic reduction over
9-48 a palladium catalyst. On ozonolysis, only two products are formed: oxalic acid $\left(\mathrm{HO}_{2} \mathrm{CCO}_{2} \mathrm{H}\right)$ and succinic acid $\left(\mathrm{HO}_{2} \mathrm{CCH}_{2} \mathrm{CH}_{2} \mathrm{CO}_{2} \mathrm{H}\right)$. Write the reactions, and propose a structure for $\mathbf{A}$.


General Problems -
PROBLEM Occasionally, a chemist might need to invert the stereochemistry of an alkene-that is, to convert
9-49 a cis alkene to a trans alkene, or vice versa. There is no one-step method for doing an alkene inversion, but the transformation can be carried out by combining several reactions in the proper sequence. How would you carry out the following reactions?
(a)
(b) cis-5-Decene $\xrightarrow{?}$ trans-5-Decene
trans-5-Decene $\xrightarrow{?}$ cis-5-Decene

PROBLEM Organometallic reagents such as sodium acetylide undergo an addition reaction with ketones,
9-50 giving alcohols:

How might you use this reaction to prepare 2-methyl-1,3-butadiene, the starting material used in the manufacture of synthetic rubber?

PROBLEM The oral contraceptive agent Mestranol is synthesized using a carbonyl addition reaction like that
9-51 shown in Problem 9-50. Draw the structure of the ketone needed.


Mestranol -
PROBLEM 1-Octen-3-ol, a potent mosquito attractant commonly used in mosquito traps, can be prepared in 9-52 two steps from hexanal, $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CHO}$. The first step is an acetylide-addition reaction like that described in Problem 9-50. What is the structure of the product from the first step, and how can it be converted into 1-octen-3-ol?

1-Octen-3-ol

PROBLEM Erythrogenic acid, $\mathrm{C}_{18} \mathrm{H}_{26} \mathrm{O}_{2}$, is an acetylenic fatty acid that turns a vivid red on exposure to
9-53 light. On catalytic hydrogenation over a palladium catalyst, 5 equivalents of $\mathrm{H}_{2}$ are absorbed, and stearic acid, $\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{16} \mathrm{CO}_{2} \mathrm{H}$, is produced. Ozonolysis of erythrogenic acid gives four products: formaldehyde, $\mathrm{CH}_{2} \mathrm{O}$; oxalic acid, $\mathrm{HO}_{2} \mathrm{CCO}_{2} \mathrm{H}$; azelaic acid, $\mathrm{HO}_{2} \mathrm{C}\left(\mathrm{CH}_{2}\right)_{7} \mathrm{CO}_{2} \mathrm{H}$; and the aldehyde acid $\mathrm{OHC}\left(\mathrm{CH}_{2}\right)_{4} \mathrm{CO}_{2} \mathrm{H}$. Draw two possible structures for erythrogenic acid, and suggest a way to tell them apart by carrying out some simple reactions.


Mestranol -
PROBLEM Hydrocarbon $\mathbf{A}$ has the formula $\mathrm{C}_{9} \mathrm{H}_{12}$ and absorbs 3 equivalents of $\mathrm{H}_{2}$ to yield $\mathbf{B}, \mathrm{C}_{9} \mathrm{H}_{18}$, when
9-54 hydrogenated over a $\mathrm{Pd} / \mathrm{C}$ catalyst. On treatment of $\mathbf{A}$ with aqueous $\mathrm{H}_{2} \mathrm{SO}_{4}$ in the presence of mercury(II), two isomeric ketones, $\mathbf{C}$ and $\mathbf{D}$, are produced. Oxidation of $\mathbf{A}$ with $\mathrm{KMnO}_{4}$ gives a mixture of acetic acid $\left(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\right)$ and the tricarboxylic acid $\mathbf{E}$. Propose structures for compounds $\mathbf{A}-\mathbf{D}$, and write the reactions.

E
PROBLEM A cumulene is a compound with three adjacent double bonds. Draw an orbital picture of a
9-55 cumulene. What kind of hybridization do the two central carbon atoms have? What is the geometric relationship of the substituents on one end to the substituents on the other end? What kind of isomerism is possible? Make a model to help see the answer.
$\mathrm{R}_{2} \mathrm{C}=\mathrm{C}=\mathrm{C}=\mathrm{CR}_{2}$


A cumulene -
PROBLEM Which of the following bases could be used to deprotonate 1-butyne?


9-56 -
(a) ${ }_{\mathrm{KOH}}$
(b)

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

PROBLEM Arrange the following carbocations in order of increasing stability.
9-57
(a)

(b)


CHAPTER 10 <br> Organohalides -
FIGURE 10.1 The gases released during volcanic eruptions contain large amounts of organohalides, including chloromethane, chloroform, dichlorodifluoromethane, and many others. (credit: modification of work "Tavurvur volcano" by Taro Taylor, Richard Bartz/ Wikimedia Commons, CC BY 2.0)


CHAPTER CONTENTS -
10.1 Names and Structures of Alkyl Halides
10.2 Preparing Alkyl Halides from Alkanes: Radical Halogenation
10.3 Preparing Alkyl Halides from Alkenes: Allylic Bromination
10.4 Stability of the Allyl Radical: Resonance Revisited
10.5 Preparing Alkyl Halides from Alcohols
10.6 Reactions of Alkyl Halides: Grignard Reagents
10.7 Organometallic Coupling Reactions
10.8 Oxidation and Reduction in Organic Chemistry

WHY THIS CHAPTER? Alkyl halides are encountered less frequently than their oxygen-containing relatives and are not often involved in the biochemical pathways of terrestrial organisms, but some of the kinds of reactions they undergo-nucleophilic substitutions and eliminations-are encountered frequently. Thus, alkyl halide chemistry is a relatively simple model for many mechanistically similar but structurally more complex reactions found in biomolecules. We'll begin this chapter with a look at how to name and prepare alkyl halides, and we'll see several of their reactions. Then, in the next chapter, we'll make a detailed study of the substitution and elimination reactions of alkyl halides-two of the most important and well-studied reaction types in organic chemistry.

Now that we've covered the chemistry of hydrocarbons, it's time to start looking at more complex substances that contain elements in addition to C and H . We'll begin by discussing the chemistry of organohalides,
compounds that contain one or more halogen atoms.
Halogen-substituted organic compounds are widespread in nature, and more than 5000 organohalides have been found in algae and various other marine organisms. Chloromethane, for instance, is released in large amounts by ocean kelp, as well as by forest fires and volcanoes. Halogen-containing compounds also have an array of industrial applications, including their use as solvents, inhaled anesthetics in medicine, refrigerants, and pesticides.

Trichloroethylene (a solvent)

Halothane (an inhaled anesthetic)

Tetrafluoropropene (a refrigerant)

Bromomethane (a fumigant)


CHAPTER CONTENTS -
Trichloroethylene (a solvent)

Halothane (an inhaled anesthetic)

Tetrafluoropropene (a refrigerant)

Bromomethane (a fumigant)

Still other halo-substituted compounds are used as medicines and food additives. The nonnutritive sweetener sucralose, marketed as Splenda, contains three chlorine atoms, for instance. Sucralose is about 600 times as sweet as sucrose, so only 1 mg is equivalent to an entire teaspoon of table sugar.

Sucralose
A large variety of organohalides are known. The halogen might be bonded to an alkynyl group ( $\mathrm{C} \equiv \mathrm{C}-\mathrm{X}$ ), a vinylic group $(C=C-X)$, an aromatic ring ( $\mathrm{Ar}-\mathrm{X}$ ), or an alkyl group. In this chapter, however, we'll be primarily concerned with alkyl halides, compounds with a halogen atom bonded to a saturated, $s p^{3}$-hybridized carbon atom.


CHAPTER CONTENTS - 10.1 Names and Structures of Alkyl Halides
Although commonly called alkyl halides, halogen-substituted alkanes are named systematically as haloalkanes (Section 3.4), treating the halogen as a substituent on a parent alkane chain. There are three steps:

STEP 1
Find the longest chain, and name it as the parent. If a double or triple bond is present, the parent chain must contain it.

STEP 2
Number the carbons of the parent chain beginning at the end nearer the first substituent, whether alkyl or halo. Assign each substituent a number according to its position on the chain.

5-Bromo-2,4-dimethylheptane

2-Bromo-4,5-dimethylheptane

If different halogens are present, number each one and list them in alphabetical order when writing the name.

1-Bromo-3-chloro-4-methylpentane
STEP 3
If the parent chain can be properly numbered from either end by step 2 , begin at the end nearer the


substituent that has alphabetical precedence. -
2-Bromo-5-methylhexane
(Not 5-bromo-2-methylhexane)

In addition to their systematic names, many simple alkyl halides are also named by identifying first the alkyl group and then the halogen. For example, $\mathrm{CH}_{3} \mathrm{I}$ can be called either iodomethane or methyl iodide. Such names are well entrenched in the chemical literature and in daily usage, but they won't be used in this book.

Halogens increase in size going down the periodic table, so the lengths of the corresponding carbon-halogen bonds increase accordingly (TABLE 10.1). In addition, $C-X$ bond strengths decrease going down the periodic table. As we've been doing thus far, we'll continue using an X to represent any of the halogens $\mathrm{F}, \mathrm{Cl}, \mathrm{Br}$, or I .

| Halomethane | Bond length (pm) | Bond strength |  | Dipole moment (D) |
| :---: | :---: | :---: | :---: | :---: |
|  |  | ( $\mathrm{kJ} / \mathrm{mol}$ ) | ( $\mathrm{kcal} / \mathrm{mol}$ ) |  |
| $\mathrm{CH}_{3} \mathrm{~F}$ | 139 | 460 | 110 | 1.85 |
| $\mathrm{CH}_{3} \mathrm{Cl}$ | 178 | 350 | 84 | 1.87 |
| $\mathrm{CH}_{3} \mathrm{Br}$ | 193 | 294 | 70 | 1.81 |
| $\mathrm{CH}_{3} \mathrm{I}$ | 214 | 239 | 57 | 1.62 |


substituent that has alphabetical precedence. -
In our discussion of bond polarity in functional groups in Section 6.3, we noted that halogens are more electronegative than carbon. The $\mathrm{C}-\mathrm{X}$ bond is therefore polar, with the carbon atom bearing a slight positive charge ( $\delta+$ ) and the halogen a slight negative charge ( $\delta-$ ). This polarity results in a dipole moment for all the halomethanes (TABLE 10.1) and implies that the alkyl halide $C-X$ carbon atom should behave as an electrophile in polar reactions. We'll soon see that this is indeed the case.

PROBLEM Give IUPAC names for the following alkyl halides:
10-1 (a)
$\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{I}$
(b)

(c)

(d)

PROBLEM Draw structures corresponding to the following IUPAC names:
10-2 (a) 2-Chloro-3,3-dimethylhexane (b) 3,3-Dichloro-2-methylhexane
(c) 3-Bromo-3-ethylpentane (d) 1,1-Dibromo-4-isopropylcyclohexane
(e) 4-sec-Butyl-2-chlorononane (f) 1,1-Dibromo-4-tert-butylcyclohexane


substituent that has alphabetical precedence. - 10.2 Preparing Alkyl Halides from Alkanes: Radical Halogenation
As we saw briefly in Section 6.6, simple alkyl halides can sometimes be prepared by the radical reaction of an alkane with $\mathrm{Cl}_{2}$ or $\mathrm{Br}_{2}$ in the presence of ultraviolet light. The detailed mechanism is shown in FIGURE 10.2 for chlorination.
Initiation step
$\begin{aligned} & \text { Propagation steps } \\ & \text { (a repeating cycle) }\end{aligned}$

FIGURE 10.2 Mechanism of the radical chlorination of methane. Three kinds of steps are required in radical substitution reactions: initiation, propagation, and termination. The propagation steps are a repeating cycle, with $\mathrm{Cl} \cdot$ a reactant in step 1 and a product in step 2 , and with $\cdot \mathrm{CH}_{3}$ a product in step 1 and a reactant in step 2 . The symbol hv shown in the initiation step is the standard way of indicating irradiation with light ( $v$ is the lowercase Greek letter nu).

Radical substitution reactions require three kinds of steps: initiation, propagation, and termination. Once an initiation step has started the process by producing radicals, the reaction continues in a self-sustaining cycle. The cycle requires two repeating propagation steps in which a radical, the halogen, and the alkane yield alkyl halide product plus more radical to carry on the chain. The chain is occasionally terminated by the combination of two radicals.

Unfortunately, alkane halogenation is a poor synthetic method for preparing alkyl halides because mixtures of products invariably result. For example, chlorination of methane does not stop cleanly at the monochlorinated stage but continues to give a mixture of dichloro, trichloro, and even tetrachloro products.

The situation is even worse for chlorination of alkanes that have more than one kind of hydrogen. Chlorination of butane, for instance, gives two monochlorinated products in a $30: 70$ ratio in addition to multiply chlorinated products such as dichlorobutane, trichlorobutane, and so on.


substituent that has alphabetical precedence. - 10.2 Preparing Alkyl Halides from Alkanes: Radical Halogenation
As another example, 2-methylpropane yields 2-chloro-2-methylpropane and 1-chloro-2-methylpropane in a $35: 65$ ratio, along with more highly chlorinated products.

From these and similar reactions, it's possible to calculate a reactivity order toward chlorination for different kinds of hydrogen atoms in a molecule. Take the butane chlorination, for instance. Butane has six equivalent primary hydrogens $\left(-\mathrm{CH}_{3}\right)$ and four equivalent secondary hydrogens $\left(-\mathrm{CH}_{2}-\right)$. The fact that butane yields $30 \%$ of 1-chlorobutane product means that each one of the six primary hydrogens is responsible for $30 \% \div 6=5 \%$ of the product. Similarly, the fact that $70 \%$ of 2 -chlorobutane is formed means that each of the four secondary hydrogens is responsible for $70 \% \div 4=17.5 \%$ of the product. Thus, a secondary hydrogen reacts $17.5 \% \div 5 \%=$ 3.5 times as often as a primary hydrogen.

A similar calculation for the chlorination of 2-methylpropane indicates that each of the nine primary hydrogens accounts for $65 \% \div 9=7.2 \%$ of the product, while the single tertiary hydrogen ( $\mathrm{R}_{3} \mathrm{CH}$ ) accounts for $35 \%$ of the product. Thus, a tertiary hydrogen is $35 \% \div 7.2 \%=5$ times as reactive as a primary hydrogen toward chlorination.


substituent that has alphabetical precedence. - 10.2 Preparing Alkyl Halides from Alkanes: Radical Halogenation
The observed reactivity order of alkane hydrogens toward radical chlorination can be explained by looking at the bond dissociation energies given previously in TABLE 6.3. The data show that a tertiary $\mathrm{C}-\mathrm{H}$ bond ( $400 \mathrm{~kJ} /$ mol ; $96 \mathrm{kcal} / \mathrm{mol}$ ) is weaker than a secondary $\mathrm{C}-\mathrm{H}$ bond ( $410 \mathrm{~kJ} / \mathrm{mol} ; 98 \mathrm{kcal} / \mathrm{mol}$ ), which is in turn weaker than a primary C-H bond ( $421 \mathrm{~kJ} / \mathrm{mol} ; 101 \mathrm{kcal} / \mathrm{mol}$ ). Since less energy is needed to break a tertiary C-H bond than to break a primary or secondary $\mathrm{C}-\mathrm{H}$ bond, the resultant tertiary radical is more stable than a primary or secondary radical.

PROBLEM Draw and name all monochloro products you would expect to obtain from radical chlorination of 10-3 2-methylpentane. Which, if any, are chiral?

PROBLEM Taking the relative reactivities of $1^{\circ}, 2^{\circ}$, and $3^{\circ}$ hydrogen atoms into account, what product(s) would 10-4 you expect to obtain from monochlorination of 2 -methylbutane? What would the approximate percentage of each product be? (Don't forget to take into account the number of each kind of


hydrogen.) - 10.3 Preparing Alkyl Halides from Alkenes: Allylic Bromination
We've already seen several methods for preparing alkyl halides from alkenes, including the reactions of HX and $\mathrm{X}_{2}$ with alkenes in electrophilic addition reactions (Section 7.7 and Section 8.2). The hydrogen halides HCl , HBr , and HI react with alkenes by a polar mechanism to give the product of Markovnikov addition. Bromine and chlorine undergo anti addition through halonium ion intermediates to give 1,2-dihalogenated products.

Another laboratory method for preparing alkyl halides from alkenes is by reaction with $N$-bromosuccinimide (abbreviated NBS), in the presence of ultraviolet light, to give products resulting from substitution of hydrogen by bromine at the position next to the double bond-the allylic position. Cyclohexene, for example, gives 3-bromocyclohexene.

This allylic bromination with NBS is analogous to the alkane chlorination reaction discussed in the previous section and occurs by a radical chain-reaction pathway (FIGURE 10.3). As in alkane halogenation, a Br• radical abstracts an allylic hydrogen atom, forming an allylic radical plus HBr . The HBr then reacts with NBS to form $\mathrm{Br}_{2}$, which in turn reacts with the allylic radical to yield the brominated product and a $\mathrm{Br} \cdot$ radical that cycles back into the first step and carries on the chain.

FIGURE 10.3 Mechanism of allylic bromination of an alkene with NBS. The process is a radical chain reaction in which (1) a $\mathrm{Br} \cdot \mathrm{radical}$ abstracts an allylic hydrogen atom of the alkene and gives an allylic radical plus HBr . (2) The HBr then reacts with NBS to form $\mathrm{Br}_{2}$, which (3) reacts with the allylic radical to yield the bromoalkene product and a $\mathrm{Br} \cdot$ radical that continues the chain.


hydrogen.) - 10.3 Preparing Alkyl Halides from Alkenes: Allylic Bromination
Why does bromination with NBS occur exclusively at an allylic position rather than elsewhere in the molecule? The answer, once again, is found by looking at bond dissociation energies to see the relative stabilities of various
kinds of radicals. Although a typical secondary alkyl C-H bond has a strength of about $410 \mathrm{~kJ} / \mathrm{mol}(98 \mathrm{kcal} / \mathrm{mol})$ and a typical vinylic $\mathrm{C}-\mathrm{H}$ bond has a strength of $465 \mathrm{~kJ} / \mathrm{mol}$ ( $111 \mathrm{kcal} / \mathrm{mol}$ ), an allylic $\mathrm{C}-\mathrm{H}$ bond has a strength of only about $370 \mathrm{~kJ} / \mathrm{mol}(88 \mathrm{kcal} / \mathrm{mol})$. An allylic radical is therefore more stable than a typical alkyl radical with the same substitution by about $40 \mathrm{~kJ} / \mathrm{mol}(9 \mathrm{kcal} / \mathrm{mol})$.

We can thus expand the stability ordering to include vinylic and allylic radicals.


hydrogen.) - 10.4 Stability of the Allyl Radical: Resonance Revisited
To see why an allylic radical is so stable, look at the orbital picture in FIGURE 10.4. The radical carbon atom with an unpaired electron can adopt $s p^{2}$ hybridization, placing the unpaired electron in a $p$ orbital and giving a structure that is electronically symmetrical. The $p$ orbital on the central carbon can therefore overlap equally well with a $p$ orbital on either of the two neighboring carbons.

FIGURE 10.4 An orbital view of the allyl radical. The $p$ orbital on the central carbon can overlap equally well with a $p$ orbital on either neighboring carbon, giving rise to two equivalent resonance structures.

Because the allyl radical is electronically symmetrical, it has two resonance forms-one with the unpaired electron on the left and the double bond on the right and another with the unpaired electron on the right and the double bond on the left. Neither structure is correct by itself; the true structure of the allyl radical is a resonance hybrid of the two. (You might want to review Section 2.4 to Section 2.6 to brush up on resonance.) As noted in Section 2.5, the greater the number of resonance forms, the greater the stability of a compound, because bonding electrons are attracted to more nuclei. An allyl radical, with two resonance forms, is therefore more stable than a typical alkyl radical, which has only a single structure.

In molecular orbital terms, the stability of the allyl radical is due to the fact that the unpaired electron is delocalized, or spread out, over an extended $\pi$-orbital network rather than localized at only one site, as shown
by the computer-generated MO in FIGURE 10.4. This delocalization is particularly apparent in the so-called spin-density surface in FIGURE 10.5, which shows the calculated location of the unpaired electron. The two terminal carbons share the unpaired electron equally.

FIGURE 10.5 The spin density surface of the allyl radical locates the position of the unpaired electron and shows that it is equally shared between the two terminal carbons.


hydrogen.) - 10.4 Stability of the Allyl Radical: Resonance Revisited
FIGURE 10.5 The spin density surface of the allyl radical locates the position of the unpaired electron and shows that it is equally shared between the two terminal carbons.

In addition to its effect on stability, delocalization of the unpaired electron in the allyl radical has other chemical consequences. Because the unpaired electron is delocalized over both ends of the $\pi$ orbital system, reaction with $\mathrm{Br}_{2}$ can occur at either end. As a result, allylic bromination of an unsymmetrical alkene often leads to a mixture of products. For example, bromination of 1-octene gives a mixture of 3-bromo-1-octene and 1-bromo-2-octene. The two products are not formed in equal amounts, however, because the intermediate allylic radical is not symmetrical and reaction at the two ends is not equally likely. Reaction at the less hindered, primary end is favored.

The products of allylic bromination reactions are useful for conversion into dienes by dehydrohalogenation with base. Cyclohexene can be converted into 1,3-cyclohexadiene, for example.


Predicting the Product of an Allylic Bromination Reaction -
What products would you expect from the reaction of 4,4-dimethylcyclohexene with NBS?


Strategy -
Draw the alkene reactant, and identify the allylic positions. In this case, there are two different allylic positions; we'll label them $\mathbf{A}$ and $\mathbf{B}$. Now abstract an allylic hydrogen from each position to generate the two corresponding allylic radicals. Each of the two allylic radicals can add a Br atom at either end ( $\mathbf{A}$ or $\mathbf{A}^{\prime} ; \mathbf{B}$ or $\mathbf{B}^{\prime}$ ), to give a mixture of up to four products. Draw and name the products. In the present instance, the "two" products from reaction at position $\mathbf{B}$ are identical, so only three products are formed in this reaction.

Solution

PROBLEM Draw three resonance forms for the cyclohexadienyl radical.
10-5
Cyclohexadienyl radical

PROBLEM The major product of the reaction of methylenecyclohexane with $N$-bromosuccinimide is 10-6 1-(bromomethyl)cyclohexene. Explain.

PROBLEM What products would you expect from reaction of the following alkenes with NBS? If more than one 10-7 product is formed, show the structures of all.
(a)

(b)


Strategy - 10.5 Preparing Alkyl Halides from Alcohols
The most generally useful method for preparing alkyl halides is to make them from alcohols, which themselves can be obtained from carbonyl compounds as we'll see in Sections $\mathbf{1 7 . 4}$ and 17.5. Because of the importance of this process, many different methods have been developed to transform alcohols into alkyl halides. The simplest method is to treat the alcohol with $\mathrm{HCl}, \mathrm{HBr}$, or HI . For reasons that will be discussed in Section 11.5, this reaction works best with tertiary alcohols, $\mathrm{R}_{3} \mathrm{COH}$. Primary and secondary alcohols react much more slowly and at higher temperatures.

The reaction of HX with a tertiary alcohol is so rapid that it's often carried out simply by bubbling pure HCl or HBr gas into a cold ether solution of the alcohol. 1-Methylcyclohexanol, for example, is converted into 1-chloro-1-methylcyclohexane by treatment with HCl .


1-Methylcyclohexanol 1-Chloro-1-methylcyclohexane -
(90\%)
Primary and secondary alcohols are best converted into alkyl halides by treatment with either thionyl chloride $\left(\mathrm{SOCl}_{2}\right)$ or phosphorus tribromide $\left(\mathrm{PBr}_{3}\right)$. These reactions, which normally take place readily under mild conditions, are less acidic and less likely to cause acid-catalyzed rearrangements than the HX method.

As the preceding examples indicate, the yields of these $\mathrm{SOCl}_{2}$ and $\mathrm{PBr}_{3}$ reactions are generally high and other functional groups such as ethers, carbonyls, and aromatic rings don't usually interfere. We'll look at the mechanisms of these and other related substitution reactions in Section 11.3.

Alkyl fluorides can also be prepared from alcohols. Numerous alternative reagents are used for such reactions, including diethylaminosulfur trifluoride $\left[\left(\mathrm{CH}_{3} \mathrm{CH}_{2}\right)_{2} \mathrm{NSF}_{3}\right]$ and HF in pyridine solvent.

Cyclohexanol


Fluorocyclohexane -
(99\%)
PROBLEM How would you prepare the following alkyl halides from the corresponding alcohols?
10-8 (a)

(b)

(c)

(d)


Fluorocyclohexane - 10.6 Reactions of Alkyl Halides: Grignard Reagents
Alkyl halides, RX, react with magnesium metal in ether or tetrahydrofuran (THF) solvent to yield alkylmagnesium halides, RMgX . The products, called Grignard reagents (RMgX) after their discoverer, Francois Auguste Victor Grignard, who received the 1912 Nobel Prize in Chemistry, are examples of organometallic compounds because they contain a carbon-metal bond. In addition to alkyl halides, Grignard reagents can also be made from alkenyl (vinylic) and aryl (aromatic) halides. The halogen can be Cl, Br, or I, although chlorides are less reactive than bromides and iodides. Organofluorides rarely react with magnesium.

As you might expect from the discussion of electronegativity and bond polarity in Section 6.3, the carbon-magnesium bond is polarized, making the carbon atom of Grignard reagents both nucleophilic and basic. An electrostatic potential map of methylmagnesium iodide, for instance, indicates the electron-rich (red) character of the carbon bonded to magnesium.

A Grignard reagent is formally the magnesium salt, $\mathrm{R}_{3} \mathrm{C}^{-+} \mathrm{MgX}$, of a carbon acid, $\mathrm{R}_{3} \mathrm{C}-\mathrm{H}$, and is thus a carbon anion, or carbanion. But because hydrocarbons are such weak acids, with $\mathbf{p} \mathbf{K}_{\mathbf{a}}$ 's in the range 44 to 60 (Section 9.7), carbon anions are very strong bases. Grignard reagents must therefore be protected from atmospheric moisture to prevent their being protonated and destroyed in acid-base reactions: $\mathrm{R}-\mathrm{Mg}-\mathrm{X}+\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{R}-\mathrm{H}+$ HO-Mg-X.


Fluorocyclohexane - 10.6 Reactions of Alkyl Halides: Grignard Reagents
Grignard reagents themselves don't occur in living organisms, but they serve as useful carbon-based nucleophiles in several important laboratory reactions, which we'll look at in detail in Section 17.5. In addition, they act as a simple model for other, more complex carbon-based nucleophiles that are important in biological chemistry. We'll see many examples of these in Chapter 29.

PROBLEM How strong a base would you expect a Grignard reagent to be? Look at Table 9.1 and predict $\mathbf{1 0 - 9}$ whether the following reactions will occur as written. (The $\mathrm{p} K_{\mathrm{a}}$ of $\mathrm{NH}_{3}$ is 35.)
(a) $\mathrm{CH}_{3} \mathrm{MgBr}+\mathrm{H}-\mathrm{C} \equiv \mathrm{C}-\mathrm{H} \longrightarrow \mathrm{CH}_{4}+\mathrm{H}-\mathrm{C} \equiv \mathrm{C}-\mathrm{MgBr}$
(b) $\mathrm{CH}_{3} \mathrm{MgBr}+\mathrm{NH}_{3} \longrightarrow \mathrm{CH}_{4}+\mathrm{H}_{2} \mathrm{~N}-\mathrm{MgBr}$

PROBLEM How might you replace a halogen substituent by a deuterium atom if you wanted to prepare a 10-10 deuterated compound?


Fluorocyclohexane - 10.7 Organometallic Coupling Reactions
Many other kinds of organometallic compounds can be prepared in a manner similar to that of Grignard reagents. For instance, alkyllithium reagents, RLi, can be prepared by the reaction of an alkyl halide with lithium metal. Alkyllithiums are both nucleophiles and strong bases, and their chemistry is similar in many respects to that of alkylmagnesium halides.

One particularly valuable reaction of alkyllithiums occurs when making lithium diorganocopper compounds, $\mathrm{R}_{2} \mathrm{CuLi}$, by reaction with copper(I) iodide in diethyl ether as solvent. Often called Gilman reagents ( $\mathbf{L i R}_{\mathbf{2}} \mathbf{C u}$ ), lithium diorganocopper compounds are useful because they undergo a coupling reaction with organochlorides, bromides, and iodides (but not fluorides). One of the alkyl groups from the lithium diorganocopper reagent replaces the halogen of the organohalide, forming a new carbon-carbon bond and yielding a hydrocarbon product. Lithium dimethylcopper, for instance, reacts with 1-iododecane to give undecane in a $90 \%$ yield.

This organometallic coupling reaction is useful in organic synthesis because it forms carbon-carbon bonds, thereby allowing the preparation of larger molecules from smaller ones. As the following examples indicate, the coupling reaction can be carried out on aryl and vinylic halides as well as on alkyl halides.

trans-1-Iodo-1-nonene
trans-5-Tridecene (71\%)

Iodobenzene
Toluene (91\%)
An organocopper coupling reaction is carried out commercially to synthesize muscalure, (9Z)-tricosene, the sex attractant secreted by the common housefly. Minute amounts of muscalure greatly increase the lure of insecticide-treated fly bait and provide an effective and species-specific means of insect control.

The mechanism of the coupling reaction involves initial formation of a triorganocopper intermediate, followed
by coupling and loss of a mono-organocopper, RCu. The coupling is not a typical polar nucleophilic substitution reaction of the sort considered in the next chapter.


Fluorocyclohexane - 10.7 Organometallic Coupling Reactions
In addition to the coupling reaction of diorganocopper reagents with organohalides, related processes also occur with other organometallic reagents, particularly organopalladium compounds. One of the most commonly used procedures is the coupling reaction of an aromatic or vinyl substituted boronic acid $\left[\mathrm{R}-\mathrm{B}(\mathrm{OH})_{2}\right]$ with an aromatic or vinyl substituted organohalide in the presence of a base and a palladium catalyst. This reaction is less general than the diorganocopper reaction because it doesn't work with alkyl substrates, but it is preferred when possible because it uses only a catalytic amount of metal rather than a full equivalent and because palladium compounds are less toxic than copper compounds. For example:

Called the Suzuki-Miyaura reaction, this process is particularly useful for preparing so-called biaryl compounds, which have two linked aromatic rings. A large number of commonly used drugs fit this description, so the Suzuki-Miyaura reaction is much-used in the pharmaceutical industry. As an example, valsartan, marketed as Diovan, is widely prescribed to treat high blood pressure, heart failure, and diabetic kidney disease. Its synthesis begins with a Suzuki-Miyaura coupling of ortho-chlorobenzonitrile with para-methylbenzeneboronic acid.

Shown in a simplified form in FIGURE 10.6, the mechanism of the Suzuki-Miyaura reaction involves initial reaction of the aromatic halide with the palladium catalyst to form an organopalladium intermediate, followed by reaction of that intermediate with the aromatic boronic acid. The resultant diorganopalladium complex then decomposes to the coupled biaryl product plus regenerated catalyst.

FIGURE 10.6 Mechanism of the Suzuki-Miyaura coupling reaction of an aromatic boronic acid with an aromatic halide to give a biaryl. The reaction takes place by (1) reaction of the aromatic halide, ArX, with the catalyst to form an organopalladium intermediate, followed by (2) reaction with the aromatic boronic acid. (3) Subsequent decomposition of the diarylpalladium intermediate gives the biaryl product.

PROBLEM How would you carry out the following transformations using an organocopper coupling reaction? 10-11 More than one step is required in each case.
(a)

(b)


Fluorocyclohexane - 10.8 Oxidation and Reduction in Organic Chemistry
We've pointed out on several occasions that some of the reactions discussed in this and earlier chapters are either oxidations or reductions. As noted in Section 8.7, an organic oxidation results in a loss of electron density by carbon, caused either by bond formation between carbon and a more electronegative atom (usually $\mathrm{O}, \mathrm{N}$, or a halogen) or by bond-breaking between carbon and a less electronegative atom (usually H). Conversely, an organic reduction results in a gain of electron density by carbon, caused either by bond formation between carbon and a less electronegative atom or by bond-breaking between carbon and a more electronegative atom (Section 8.6).

Oxidation $\quad$| Decreases electron density on carbon by: |  |  |
| :--- | :--- | :--- |
|  | - forming one of these: $\mathrm{C}-\mathrm{O} \quad \mathrm{C}-\mathrm{N}$ | $\mathrm{C}-\mathrm{X}$ |
|  | $\quad$ or breaking this: $\mathrm{C}-\mathrm{H}$ |  |

Based on these definitions, the chlorination reaction of methane to yield chloromethane is an oxidation because a $\mathrm{C}-\mathrm{H}$ bond is broken and a $\mathrm{C}-\mathrm{Cl}$ bond is formed. The conversion of an alkyl chloride to an alkane via a Grignard reagent followed by protonation is a reduction, however, because a $\mathrm{C}-\mathrm{Cl}$ bond is broken and a $\mathrm{C}-\mathrm{H}$ bond is formed.

Reduction: $\mathrm{C}-\mathrm{Cl}$ bond broken and $\mathrm{C}-\mathrm{H}$ bond formed

Methane

As other examples, the reaction of an alkene with $\mathrm{Br}_{2}$ to yield a 1,2-dibromide is an oxidation because two $\mathrm{C}-\mathrm{Br}$ bonds are formed, but the reaction of an alkene with HBr to yield an alkyl bromide is neither an oxidation nor a reduction because both a $\mathrm{C}-\mathrm{H}$ and a $\mathrm{C}-\mathrm{Br}$ bond are formed.


Fluorocyclohexane - 10.8 Oxidation and Reduction in Organic Chemistry
Oxidation: Two new bonds formed between carbon and a more electronegative element

> Neither oxidation nor reduction:
> One new $\mathrm{C}-\mathrm{H}$ bond and one new $\mathrm{C}-\mathrm{Br}$ bond formed

Ethylene


Bromoethane -
A list of compounds of increasing oxidation level is shown in FIGURE 10.7. Alkanes are at the lowest oxidation level because they have the maximum possible number of $\mathrm{C}-\mathrm{H}$ bonds per carbon, and $\mathrm{CO}_{2}$ is at the highest level because it has the maximum possible number of $\mathrm{C}-\mathrm{O}$ bonds per carbon. Any reaction that converts a compound from a lower level to a higher level is an oxidation, any reaction that converts a compound from a higher level to a lower level is a reduction, and any reaction that doesn't change the level is neither an oxidation nor a reduction.

| $\mathrm{CH}_{3} \mathrm{CH}_{3}$ | $\mathrm{H}_{2} \mathrm{C}=\mathrm{CH}_{2}$ | $\mathrm{HC} \equiv \mathrm{CH}$ |  |  |
| :---: | :---: | :---: | :---: | :---: |
|  | $\mathrm{CH}_{3} \mathrm{OH}$ | $\mathrm{H}_{2} \mathrm{C}=\mathrm{O}$ | $\mathrm{HCO}_{2} \mathrm{H}$ | $\mathrm{CO}_{2}$ |
|  | $\mathrm{CH}_{3} \mathrm{Cl}$ | $\mathrm{CH}_{2} \mathrm{Cl}_{2}$ | $\mathrm{CHCl}_{3}$ | $\mathrm{CCl}_{4}$ |
|  | $\mathrm{CH}_{3} \mathrm{NH}_{2}$ | $\mathrm{H}_{2} \mathrm{C}=\mathrm{NH}$ | $\mathrm{HC} \equiv \mathrm{N}$ |  |
| Low oxidation level |  |  |  | High oxidation level |

Worked Example 10.2 shows how to compare the oxidation levels of different compounds with the same number of carbon atoms.


Comparing Oxidation Levels -
Rank the following compounds in order of increasing oxidation level:

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

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


Strategy -
Compounds that have the same number of carbon atoms can be compared by adding the number of $\mathrm{C}-\mathrm{O}, \mathrm{C}-\mathrm{N}$, and $\mathrm{C}-\mathrm{X}$ bonds in each and then subtracting the number of $\mathrm{C}-\mathrm{H}$ bonds. The larger the resultant value, the higher the oxidation level.


Solution -
The first compound (propene) has six $\mathrm{C}-\mathrm{H}$ bonds, giving an oxidation level of -6 ; the second (2-propanol) has one $\mathrm{C}-\mathrm{O}$ bond and seven $\mathrm{C}-\mathrm{H}$ bonds, giving an oxidation level of -6 ; the third (acetone) has two $\mathrm{C}-\mathrm{O}$ bonds and six $\mathrm{C}-\mathrm{H}$ bonds, giving an oxidation level of -4 ; and the fourth (propane) has eight $\mathrm{C}-\mathrm{H}$ bonds, giving an oxidation level of -8 . Thus, the order of increasing oxidation level is

PROBLEM Rank both sets of compounds in order of increasing oxidation level:
10-12 (a)

(b) $\mathrm{CH}_{3} \mathrm{CN}$
$\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{NH}_{2}$
$\mathrm{H}_{2} \mathrm{NCH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}$
PROBLEM Tell whether each of the following reactions is an oxidation, a reduction, or neither.

10-13 (a)

(b)


Naturally Occurring Organohalides -
Just forty years ago in 1980, only about 30 naturally occurring organohalides were known. It was simply assumed that chloroform, halogenated phenols, chlorinated aromatic compounds called PCBs, and other such substances found in the environment were industrial pollutants. Now, less than half a century later, the situation is quite different. More than 5000 organohalides have been found to occur naturally, and tens of thousands more surely exist. From a simple compound like chloromethane to an extremely complex one like the antibiotic vancomycin, a remarkably diverse range of organohalides exists in plants, bacteria, and animals. Many even have valuable physiological activity. The pentahalogenated alkene halomon, for instance, has been isolated from the red alga Portieria hornemannii and found to have anticancer activity against several human tumor cell lines.

Some naturally occurring organohalides are produced in massive quantities. Forest fires, volcanic eruptions, and marine kelp release up to 5 million tons of $\mathrm{CH}_{3} \mathrm{Cl}$ per year, for example, while annual industrial emissions total about 26,000 tons. Termites are thought to release as much as $10^{8} \mathrm{~kg}$ of chloroform per year. A detailed examination of the Okinawan acorn worm Ptychodera flava found that the 64 million worms living in a $1 \mathrm{~km}^{2}$ study area excreted nearly 8000 pounds per year of bromophenols and bromoindoles, compounds previously thought to be non-natural pollutants.


Naturally Occurring Organohalides -
Why do organisms produce organohalides, many of which are undoubtedly toxic? The answer seems to be that many organisms use organohalogen compounds for self-defense, either as feeding deterrents, irritants to predators, or natural pesticides. Marine sponges, coral, and sea hares, for example, release foul-tasting organohalides that deter fish, starfish, and other predators. Even humans appear to produce halogenated compounds as part of their defense against infection. The human immune system contains a peroxidase enzyme capable of carrying out halogenation reactions on fungi and bacteria, thereby killing the pathogen. And most remarkable of all, even free chlorine $-\mathrm{Cl}_{2}$-has been found to be present in humans.

FIGURE 10.8 Marine corals secrete organohalogen compounds that act as a feeding deterrent to fish. (credit: "Coral reef" by Qui Nguyen, United Nations Environment Programme/Flickr, Public Domain)

Much remains to be learned-only a few hundred of the more than 500,000 known species of marine organisms have been examined-but it's clear that organohalides are an integral part of the world around us.


Key Terms -
```
- alkyl halide
- Gilman reagent (LiR2Cu)
- allylic
- Grignard reagent (RMgX)
- carbanion
- organohalide
- delocalized
- organometallic
```


Summary -
Alkyl halides are not often found in terrestrial organisms, but the kinds of reactions they undergo are among the most important and well-studied reaction types in organic chemistry. In this chapter, we saw how to name and prepare alkyl halides, and we'll soon make a detailed study of their substitution and elimination reactions.

Simple alkyl halides can be prepared by radical halogenation of alkanes, but mixtures of products usually result. The reactivity order of alkanes toward halogenation is identical to the stability order of radicals: $\mathrm{R}_{3} \mathrm{C} \cdot>\mathrm{R}_{2} \mathrm{CH} \cdot$ $>\mathrm{RCH}_{2}$. Alkyl halides can also be prepared from alkenes by reaction with $N$-bromosuccinimide (NBS) to give the product of allylic bromination. The NBS bromination of alkenes takes place through an intermediate allylic radical, which is stabilized by resonance.

Alcohols react with HX to form alkyl halides, but the reaction works well only for tertiary alcohols, $\mathrm{R}_{3} \mathrm{COH}$. Primary and secondary alkyl halides are normally prepared from alcohols using either $\mathrm{SOCl}_{2}$, $\mathrm{PBr}_{3}$, or HF in pyridine. Alkyl halides react with magnesium in ether solution to form organomagnesium halides, called Grignard reagents (RMgX), which are both nucleophilic and strongly basic.

Alkyl halides also react with lithium metal to form organolithium reagents, RLi. In the presence of CuI, these form diorganocoppers, or Gilman reagents ( $\mathbf{L i R}_{\mathbf{2}} \mathbf{C u}$ ). Gilman reagents react with organohalides to yield coupled hydrocarbon products.


Summary of Reactions -
No stereochemistry is implied unless specifically indicated with wedged, solid, and dashed lines.

1. Preparation of alkyl halides
a. From alkenes by allylic bromination (Section 10.3)

b. From alcohols (Section 10.5)
(1) Reaction with HX

Reactivity order: $3^{\circ}>2^{\circ}>1^{\circ}$
(2) Reaction of $1^{\circ}$ and $2^{\circ}$ alcohols with $\mathrm{SOCl}_{2}$

(3) Reaction of $1^{\circ}$ and $2^{\circ}$ alcohols with $\mathrm{PBr}_{3}$

(4) Reaction of $1^{\circ}$ and $2^{\circ}$ alcohols with HF-pyridine

2. Reactions of alkyl halides
a. Formation of Grignard (organomagnesium) reagents (Section 10.6)

$$
\mathrm{R}-\mathrm{X} \underset{\text { Ether }}{\mathrm{Mg}} \mathrm{R}-\mathrm{Mg}-\mathrm{X}
$$

b. Formation of Gilman (diorganocopper) reagents (Section 10.7)

$$
\begin{gathered}
\mathrm{R}-\mathrm{X} \\
2 \mathrm{R}-\mathrm{Li}+\mathrm{CuI} \xrightarrow{\substack{\text { Pentane } \\
\text { In ether }}} \mathrm{R}-\mathrm{Li}+\mathrm{LiX} \\
{[\mathrm{R}-\mathrm{Cu}-\mathrm{R}]^{-} \mathrm{Li}^{+}+\mathrm{LiI}}
\end{gathered}
$$

c. Organometallic coupling (Section 10.7)
(1) Diorganocopper reaction

$$
\mathrm{R}_{2} \mathrm{CuLi}+\mathrm{R}^{\prime}-\mathrm{X} \xrightarrow{\text { In ether }} \mathrm{R}-\mathrm{R}^{\prime}+\mathrm{RCu}+\mathrm{LiX}
$$

(2) Palladium-catalyzed Suzuki-Miyaura reaction


Visualizing Chemistry -
PROBLEM Give IUPAC names for the following alkyl halides (green $=\mathrm{Cl}$ ):


10-14 (a) -
(b)

PROBLEM Show the product(s) of reaction of the following alkenes with NBS:


10-15 (a) -
(b)

PROBLEM The following alkyl bromide can be prepared by reaction of the alcohol ( $S$ )-2-pentanol with $\mathrm{PBr}_{3}$.
10-16 Name the compound, assign $(R)$ or $(S)$ stereochemistry, and tell whether the reaction of the alcohol results in the same stereochemistry or a change in stereochemistry (reddish brown $=\mathrm{Br}$ ).


Mechanism Problems -
PROBLEM In light of the fact that tertiary alkyl halides undergo spontaneous dissociation to yield a
10-17 carbocation plus halide ion (see Problem 10-41), propose a mechanism for the following reaction.


Naming Alkyl Halides -
PROBLEM Name the following alkyl halides:

10-18 (a)

(b)

(c)

(d)

(e) $\mathrm{ClCH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{C} \equiv \mathrm{CCH}_{2} \mathrm{Br}$

PROBLEM Draw structures corresponding to the following IUPAC names:
10-19 (a) 2,3-Dichloro-4-methylhexane (b) 4-Bromo-4-ethyl-2-methylhexane
(c) 3-Iodo-2,2,4,4-tetramethylpentane (d) cis-1-Bromo-2-ethylcyclopentane

PROBLEM Draw and name all the monochlorination products you might obtain from radical chlorination of
10-20 the following compounds. Which of the products are chiral? Are any of the products optically active?
(a) 2-methylbutane
(b) methylcyclopropane
(c) 2,2-dimethylpentane


Synthesizing Alkyl Halides -
PROBLEM How would you prepare the following compounds, starting with cyclopentene and any other
10-21 reagents needed?
(a) Chlorocyclopentane
(b) Methylcyclopentane
(c) 3-Bromocyclopentene
(d) Cyclopentanol
(e) Cyclopentylcyclopentane
(f) 1,3-Cyclopentadiene

PROBLEM Predict the product(s) of the following reactions:

10-22 (a)

(b)

(c)

(d)

(f)

(g)

PROBLEM A chemist requires a large amount of 1-bromo-2-pentene as starting material for a synthesis and
10-23 decides to carry out an NBS allylic bromination reaction. What is wrong with the following synthesis plan? What side products would form in addition to the desired product?

PROBLEM What product(s) would you expect from the reaction of 1-methylcyclohexene with NBS? Would you
10-24 use this reaction as part of a synthesis?

PROBLEM What product(s) would you expect from the reaction of 1,4-hexadiene with NBS? What is the
10-25 structure of the most stable radical intermediate?
PROBLEM What product would you expect from the reaction of 1-phenyl-2-butene with NBS? Explain.
10-26

1-Phenyl-2-butene

Oxidation and Reduction
PROBLEM Rank the compounds in each of the following series in order of increasing oxidation level:
10-27 (a)

(b)

PROBLEM Which of the following compounds have the same oxidation level, and which have different levels?
10-28

1

2

3

4

5

PROBLEM Tell whether each of the following reactions is an oxidation, a reduction, or neither:
10-29 (a)

(b)

(c)


General Problems -
PROBLEM Arrange the following radicals from most stable to least stable.

10-30 (a)

(b)

(c)

PROBLEM Alkylbenzenes such as toluene (methylbenzene) react with NBS to give products in which bromine 10-31
substitution has occurred at the position next to the aromatic ring (the benzylic position). Explain, based on the bond dissociation energies in Table 6.3.

PROBLEM Draw resonance structures for the benzyl radical, $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \cdot$, the intermediate produced in the NBS
10-32 bromination reaction of toluene (Problem 10-31).
PROBLEM Draw resonance structures for the following species:
10-33 (a)

(b)

(c)

PROBLEM (S)-3-Methylhexane undergoes radical bromination to yield optically inactive 10-34 3-bromo-3-methylhexane as the major product. Is the product chiral? What conclusions can you draw about the radical intermediate?

PROBLEM Assume that you have carried out a radical chlorination reaction on ( $R$ )-2-chloropentane and have
10-35 isolated (in low yield) 2,4-dichloropentane. How many stereoisomers of the product are formed, and in what ratio? Are any of the isomers optically active? (See Problem 10-34.)

PROBLEM How would you carry out the following syntheses?
10-36

PROBLEM The syntheses shown here are unlikely to occur as written. What is wrong with each?
10-37 (a)

(b)

PROBLEM Why do you suppose it's not possible to prepare a Grignard reagent from a bromo alcohol such
$\mathbf{1 0 - 3 8}$ as 4-bromo-1-pentanol? Give another example of a molecule that is unlikely to form a Grignard reagent.


General Problems -
PROBLEM Addition of HBr to a double bond with an ether (-OR) substituent occurs regiospecifically to give
10-39 a product in which the -Br and -OR are bonded to the same carbon. Draw the two possible carbocation intermediates in this electrophilic addition reaction, and explain using resonance why the observed product is formed.

HBr

PROBLEM Identify the reagents $\mathbf{a}-\mathbf{c}$ in the following scheme:
10-40

PROBLEM Tertiary alkyl halides, $\mathrm{R}_{3} \mathrm{CX}$, undergo spontaneous dissociation to yield a carbocation, $\mathrm{R}_{3} \mathrm{C}^{+}$, plus
10-41 halide ion. Which do you think reacts faster, $\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CBr}$ or $\mathrm{H}_{2} \mathrm{C}=\mathrm{CHC}\left(\mathrm{CH}_{3}\right)_{2} \mathrm{Br}$ ? Explain.
PROBLEM Carboxylic acids $\left(\mathrm{RCO}_{2} \mathrm{H} ; \mathrm{p} K_{\mathrm{a}} \approx 5\right)$ are approximately $10^{11}$ times more acidic than alcohols (ROH;
10-42 $\mathrm{p} K_{\mathrm{a}} \approx 16$ ). In other words, a carboxylate ion $\left(\mathrm{RCO}_{2}^{-}\right)$is more stable than an alkoxide ion ( $\mathrm{RO}^{-}$). Explain, using resonance.

PROBLEM How might you use a Suzuki-Miyaura reaction to prepare the biaryl compounds below? In each 10-43 case, show the two potential reaction partners.
(a)

(b)

(c)


General Problems -
PROBLEM How might you use a Suzuki-Miyaura reaction to prepare the biaryl compounds below? In each 10-43 case, show the two potential reaction partners.
(a)

(b)

(c)

PROBLEM The relative rate of radical bromination is $1: 82: 1640$ for $1^{\circ}: 2^{\circ}: 3^{\circ}$ hydrogens, respectively. Draw
10-44 all of the monobrominated products that you might obtain from the radical bromination of the compounds below. Calculate the relative percentage of each.
(a) methylcyclobutane
(b) 3,3-dimethylpentane
(c) 3-methylpentane

PROBLEM Choose the alcohol from each pair below that would react faster with HX to form the corresponding
10-45 alkyl halide.
(a)
or

(b)
or

(c)

PROBLEM Predict the product and provide the entire catalytic cycle for the following Suzuki-Miyaura 10-46 reactions.


CHAPTER 11 -
Reactions of Alkyl Halides: Nucleophilic Substitutions and Eliminations

FIGURE 11.1 Competition occurs throughout nature. In chemistry, competition often occurs between alternative reaction pathways, such as in the substitution and elimination reactions of alkyl halides. (credit: modification of work "Bull moose fight" by Grand Teton, National Parks Service/Flickr, Public Domain)


CHAPTER CONTENTS -
11.1 The Discovery of Nucleophilic Substitution Reactions
11.2 The $\mathbf{S}_{\mathrm{N}} 2$ Reaction
11.3 Characteristics of the $\mathbf{S}_{\mathrm{N}} \mathbf{2}$ Reaction
11.4 The $\mathrm{S}_{\mathrm{N}} 1$ Reaction
11.5 Characteristics of the $\mathrm{S}_{\mathrm{N}} 1$ Reaction
11.6 Biological Substitution Reactions
11.7 Elimination Reactions: Zaitsev's Rule
11.8 The E2 Reaction and the Deuterium Isotope Effect
11.9 The E2 Reaction and Cyclohexane Conformation
11.10 The E1 and E1cB Reactions
11.11 Biological Elimination Reactions
11.12 A Summary of Reactivity: $\mathrm{S}_{\mathrm{N}} 1, \mathrm{~S}_{\mathrm{N}} 2, \mathrm{E} 1, \mathrm{E} 1 \mathrm{CB}$, and E 2

WHY THIS CHAPTER? Nucleophilic substitution and base-induced elimination are two of the most widely occurring and versatile reactions in organic chemistry, both in the laboratory and in biological pathways. We'll look at them closely in this chapter to see how they occur, what their characteristics are, and how they can be used. We'll begin with substitution reactions.

We saw in the preceding chapter that the carbon-halogen bond in an alkyl halide is polar and that the carbon atom is electron-poor. Thus, alkyl halides are electrophiles, and much of their chemistry involves polar reactions with nucleophiles and bases. Alkyl halides do one of two things when they react with a nucleophile/ base such as hydroxide ion: they either undergo substitution of the X group by the nucleophile, or they undergo
elimination of HX to yield an alkene.

Elimination


CHAPTER CONTENTS - 11.1 The Discovery of Nucleophilic Substitution Reactions
Discovery of the nucleophilic substitution reaction of alkyl halides dates back to work carried out by the German chemist Paul Walden in 1896. Walden found that the pure enantiomeric (+)- and ( - )-malic acids could be interconverted through a series of simple substitution reactions. When Walden treated (-)-malic acid with $\mathrm{PCl}_{5}$, he isolated (+)-chlorosuccinic acid. This, on treatment with wet $\mathrm{Ag}_{2} \mathrm{O}$, gave (+)-malic acid. Similarly, reaction of $(+)$-malic acid with $\mathrm{PCl}_{5}$ gave (-)-chlorosuccinic acid, which was converted into (-)-malic acid when treated with wet $\mathrm{Ag}_{2} \mathrm{O}$. The full cycle of reactions is shown in FIGURE 11.2.

(-)-Malic acid
$[\alpha]_{D}=-2.3$
$\uparrow \mathrm{Ag}_{2} \mathrm{O}, \mathrm{H}_{2} \mathrm{O}$

(-)-Chlorosuccinic acid
(+)-Chlorosuccinic acid

(+)-Malic acid
$[\alpha]_{D}=+2.3$

FIGURE 11.2 Walden's cycle of reactions interconverting (+)- and (-)-malic acids.
At the time, the results were astonishing. The eminent chemist Emil Fischer called Walden's discovery "the most remarkable observation made in the field of optical activity since the fundamental observations of Pasteur." Because (-)-malic acid was converted into (+)-malic acid, some reactions in the cycle must have occurred with a change, or inversion, of configuration at the chirality center. But which ones, and how? (Remember from Section 5.5 that the direction of light rotation and the configuration of a chirality center aren't directly related. You can't tell by looking at the sign of rotation whether a change in configuration has occurred during a reaction.)


CHAPTER CONTENTS - 11.1 The Discovery of Nucleophilic Substitution Reactions
Today, we refer to the transformations taking place in Walden's cycle as nucleophilic substitution reactions because each step involves the substitution of one nucleophile (chloride ion, $\mathrm{Cl}^{-}$, or hydroxide ion, $\mathrm{HO}^{-}$) by another. Nucleophilic substitution reactions are one of the most common and versatile reaction types in organic chemistry.

$$
\mathrm{R}-\mathrm{X}+\mathrm{Nu}:^{-} \longrightarrow \mathrm{R}-\mathrm{Nu}+\mathrm{X}:^{-}
$$

Following the work of Walden, further investigations were undertaken during the 1920s and 1930s to clarify the mechanism of nucleophilic substitution reactions and to find out how inversions of configuration occur. Among the first series studied was one that interconverted the two enantiomers of 1-phenyl-2-propanol (FIGURE 11.3). Although this particular series of reactions involves nucleophilic substitution of an alkyl para-toluenesulfonate (called a tosylate) rather than an alkyl halide, exactly the same type of reaction is involved as that studied by Walden. For all practical purposes, the entire tosylate group acts as if it were simply a halogen substituent. (In fact, when you see a tosylate substituent in a molecule, do a mental substitution and tell yourself that you're
dealing with an alkyl halide.)

FIGURE 11.3 A Walden cycle interconverting (+) and (-) enantiomers of 1-phenyl-2-propanol. Chirality centers are marked by asterisks, and the bonds broken in each reaction are indicated by red wavy lines. The inversion of chirality occurs in step 2, where acetate ion substitutes for tosylate ion.


CHAPTER CONTENTS - 11.1 The Discovery of Nucleophilic Substitution Reactions
In the three-step reaction sequence shown in FIGURE 11.3, (+)-1-phenyl-2-propanol is interconverted with its (-) enantiomer, so at least one of the three steps must involve an inversion of configuration at the chirality center. Step 1, formation of a tosylate, occurs by breaking the $\mathrm{O}-\mathrm{H}$ bond of the alcohol rather than the $\mathrm{C}-\mathrm{O}$ bond to the chiral carbon, so the configuration around the carbon is unchanged. Similarly, step 3, hydroxide-ion cleavage of the acetate, takes place without breaking the $\mathrm{C}-\mathrm{O}$ bond at the chirality center. Thus, the inversion of stereochemical configuration must take place in step 2, the nucleophilic substitution of tosylate ion by acetate ion.

From this and nearly a dozen other series of similar reactions, researchers concluded that the nucleophilic substitution reaction of a primary or secondary alkyl halide or tosylate always proceeds with inversion of configuration. (Tertiary alkyl halides and tosylates, as we'll see shortly, give different stereochemical results and react by a different mechanism than the primary and secondary ones.)


Predicting the Stereochemistry of a Nucleophilic Substitution Reaction -
What product would you expect from a nucleophilic substitution reaction of ( $R$ )-1-bromo-1-phenylethane with cyanide ion, ${ }^{-} \mathrm{C} \equiv \mathrm{N}$, as nucleophile? Show the stereochemistry of both reactant and product, assuming that inversion of configuration occurs.


Strategy -
Draw the $R$ enantiomer of the reactant, and then change the configuration of the chirality center while replacing the ${ }^{-} \mathrm{Br}$ with $\mathrm{a}^{-} \mathrm{CN}$.

Solution

(R)-1-Bromo-1-phenylethane
(S)-2-PhenyIpropanenitrile

PROBLEM What product would you expect from a nucleophilic substitution reaction of ( $S$ )-2-bromohexane
11-1 with acetate ion, $\mathrm{CH}_{3} \mathrm{CO}_{2}^{-}$? Assume that inversion of configuration occurs, and show the stereochemistry of both the reactant and product.


Strategy - 11.2 The $\mathrm{S}_{\mathrm{N}} 2$ Reaction
In almost all chemical reactions, there is a direct relationship between the rate at which the reaction occurs and the concentrations of the reactants. When we measure this relationship, we measure the kinetics of the reaction. For example, let's look at the kinetics of a simple nucleophilic substitution-the reaction of $\mathrm{CH}_{3} \mathrm{Br}$ with $\mathrm{OH}^{-}$to yield $\mathrm{CH}_{3} \mathrm{OH}$ plus $\mathrm{Br}^{-}$.

With a given temperature, solvent, and concentration of reactants, the substitution occurs at a certain rate. If we double the concentration of $\mathrm{OH}^{-}$, the frequency of encounters between reaction partners doubles and we find that the reaction rate also doubles. Similarly, if we double the concentration of $\mathrm{CH}_{3} \mathrm{Br}$, the reaction rate again doubles. We call such a reaction, in which the rate is linearly dependent on the concentrations of two species, a second-order reaction. Mathematically, we can express this second-order dependence of the nucleophilic substitution reaction by setting up a rate equation. As either [ RX ] or [ ${ }^{-} \mathrm{OH}$ ] changes, the rate of the reaction changes proportionately.

$$
\begin{aligned}
\text { Reaction rate } & =\text { Rate of disappearance of reactant } \\
& =k \times[\mathrm{RX}] \times\left[{ }^{-} \mathrm{OH}\right]
\end{aligned}
$$

where
$[\mathrm{RX}]=\mathrm{CH}_{3} \mathrm{Br}$ concentration in molarity
$[-\mathrm{OH}]=-\mathrm{OH}$ concentration in molarity
$k=$ a constant value (the rate constant)
A mechanism that accounts for both the inversion of configuration and the second-order kinetics that are observed with nucleophilic substitution reactions was suggested in 1937 by the British chemists E. D. Hughes
and Christopher Ingold, who formulated what they called the $\mathbf{S}_{\mathbf{N}} \mathbf{2}$ reaction-short for substitution, nucleophilic, bimolecular. (Bimolecular means that two molecules, nucleophile and alkyl halide, take part in the step whose kinetics are measured.)


Strategy - 11.2 The $\mathrm{S}_{\mathrm{N}} 2$ Reaction
The essential feature of the $\mathrm{S}_{\mathrm{N}} 2$ mechanism is that it takes place in a single step, without intermediates, when the incoming nucleophile reacts with the alkyl halide or tosylate (the substrate) from a direction opposite the group that is displaced (the leaving group). As the nucleophile comes in on one side of the substrate and bonds to the carbon, the halide or tosylate departs from the other side, thereby inverting the stereochemical configuration. The process is shown in FIGURE 11.4 for the reaction of ( $S$ ) -2-bromobutane with $\mathrm{HO}^{-}$to give ( $R$ )-2-butanol.

FIGURE 11.4 MECHANISM
The mechanism of the $\mathbf{S}_{\mathbf{N}} \mathbf{2}$ reaction. The reaction takes place in a single step when the incoming nucleophile approaches from a direction $180^{\circ}$ away from the leaving halide ion, thereby inverting the stereochemistry at carbon.
(1) The nucleophile ${ }^{-} \mathrm{OH}$ uses its lone-pair electrons to attack the alkyl halide carbon $180^{\circ}$ away from the departing halogen. This leads to a transition state with a partially formed $\mathrm{C}-\mathrm{OH}$ bond and a partially broken $\mathrm{C}-\mathrm{Br}$ bond.
(2) The stereochemistry at carbon is inverted as the $\mathrm{C}-\mathrm{OH}$ bond forms fully and the bromide ion departs with the electron pair from the former $\mathrm{C}-\mathrm{Br}$ bond.

(S)-2-Bromobutane
-

Transition state

(R)-2-Butanol


Strategy - 11.2 The $\mathrm{S}_{\mathrm{N}} 2$ Reaction
(S)-2-Bromobutane
-

Transition state

(R)-2-Butanol

As shown in FIGURE 11.4, the $\mathrm{S}_{\mathrm{N}} 2$ reaction occurs when an electron pair on the nucleophile Nu : ${ }^{-}$forces out the group $\mathrm{X}:^{-}$, which takes with it the electron pair from the former $\mathrm{C}-\mathrm{X}$ bond. This occurs through a transition state in which the new $\mathrm{Nu}-\mathrm{C}$ bond is partially formed at the same time that the old $\mathrm{C}-\mathrm{X}$ bond is partially broken and in which the negative charge is shared by both the incoming nucleophile and the outgoing halide ion. The transition state for this inversion has the remaining three bonds to carbon in a planar arrangement (FIGURE 11.5).

FIGURE 11.5 The transition state of an $\mathbf{S}_{\mathbf{N}} 2$ reaction has a planar arrangement of the carbon atom and the remaining three groups. Electrostatic potential maps show that negative charge is delocalized in the transition state.

The mechanism proposed by Hughes and Ingold is fully consistent with experimental results, explaining both stereochemical and kinetic data. Thus, the requirement for a backside approach of the entering nucleophile ( $180^{\circ}$ away from the departing X group) causes the stereochemistry of the substrate to invert, much like an umbrella turning inside-out in the wind. The Hughes-Ingold mechanism also explains why second-order kinetics are observed: the $\mathrm{S}_{\mathrm{N}} 2$ reaction occurs in a single step that involves both alkyl halide and nucleophile. Two molecules are involved in the step whose rate is measured.

PROBLEM What product would you expect to obtain from $\mathrm{S}_{\mathrm{N}} 2$ reaction of $\mathrm{OH}^{-}$with $(R)$-2-bromobutane? Show
11-2 the stereochemistry of both the reactant and product.
PROBLEM Assign configuration to the following substance, and draw the structure of the product that would 11-3 result from nucleophilic substitution reaction with $\mathrm{HS}^{-}$(reddish brown $=\mathrm{Br}$ ):


Strategy - 11.3 Characteristics of the $\mathrm{S}_{\mathrm{N}} 2$ Reaction
Now that we know how $\mathrm{S}_{\mathrm{N}} 2$ reactions occur, we need to see how they can be used and what variables affect them. Some $\mathrm{S}_{\mathrm{N}} 2$ reactions are fast, and some are slow; some take place in high yield and others in low yield. Understanding the factors involved can be of tremendous value. Let's begin by recalling a few things about reaction rates in general.

The rate of a chemical reaction is determined by the activation energy $\Delta G^{\ddagger}$, the energy difference between reactant ground state and transition state. A change in reaction conditions can affect $\Delta G^{\ddagger}$ either by changing the reactant energy level or by changing the transition-state energy level. Lowering the reactant energy or raising the transition-state energy increases $\Delta G^{\ddagger}$ and decreases the reaction rate; raising the reactant energy or
decreasing the transition-state energy decreases $\Delta G^{\ddagger}$ and increases the reaction rate (FIGURE 11.6). We'll see examples of all these effects as we look at $\mathrm{S}_{\mathrm{N}} 2$ reaction variables.

FIGURE 11.6 The effects of changes in reactant and transition-state energy levels on reaction rate. (a) A higher reactant energy level (red curve) corresponds to a faster reaction (smaller $\Delta G^{\ddagger}$ ). (b) A higher transition-state energy level (red curve) corresponds to a slower reaction (larger $\Delta G^{\ddagger}$ ).


Steric Effects in the $\mathrm{S}_{\mathrm{N}} 2$ Reaction -
The first $\mathrm{S}_{\mathrm{N}} 2$ reaction variable to look at is the structure of the substrate. Because the $\mathrm{S}_{\mathrm{N}} 2$ transition state involves partial bond formation between the incoming nucleophile and the alkyl halide carbon atom, it seems reasonable that a hindered, bulky substrate should prevent easy approach of the nucleophile, making bond formation difficult. In other words, the transition state for reaction of a sterically hindered substrate, whose carbon atom is "shielded" from the approach of the incoming nucleophile, is higher in energy and forms more slowly than the corresponding transition state for a less hindered substrate (FIGURE 11.7).
(a)

(c)

(b)

(d)

FIGURE 11.7 Steric hindrance to the $\mathbf{S}_{\mathbf{N}} \mathbf{2}$ reaction. As the models indicate, the carbon atom in (a) bromomethane is readily accessible, resulting in a fast $\mathrm{S}_{\mathrm{N}} 2$ reaction. The carbon atoms in (b) bromoethane (primary), (c) 2-bromopropane (secondary), and (d) 2-bromo-2-methylpropane (tertiary) are successively more hindered, resulting in successively slower $\mathrm{S}_{\mathrm{N}} 2$ reactions.


Steric Effects in the $\mathrm{S}_{\mathrm{N}} 2$ Reaction -
As FIGURE 11.7 shows, the difficulty of nucleophile approach increases as the three substituents bonded to the halo-substituted carbon atom increase in size. Methyl halides are by far the most reactive substrates in $\mathrm{S}_{\mathrm{N}} 2$ reactions, followed by primary alkyl halides such as ethyl and propyl. Alkyl branching at the reacting center, as in isopropyl halides $\left(2^{\circ}\right)$, slows the reaction greatly, and further branching, as in tert-butyl halides $\left(3^{\circ}\right)$, effectively halts the reaction. Even branching one carbon away from the reacting center, as in 2,2-dimethylpropyl (neopentyl) halides, greatly hinders nucleophilic displacement. As a result, $\mathrm{S}_{\mathrm{N}} 2$ reactions occur only at relatively unhindered sites and are normally useful only with methyl halides, primary halides, and a few simple secondary halides. Relative reactivities for some different substrates are as follows:

Vinylic halides $\left(\mathrm{R}_{2} \mathrm{C}=\mathrm{CRX}\right)$ and aryl halides are not shown on this reactivity list because they are unreactive toward $\mathrm{S}_{\mathrm{N}} 2$ displacement. This lack of reactivity is due to steric factors: the incoming nucleophile would have to approach in the plane of the carbon-carbon double bond and burrow through part of the molecule to carry out a backside displacement.

Vinylic halide


The Nucleophile -
Another variable that has a major effect on the $S_{N} 2$ reaction is the nature of the nucleophile. Any species, either neutral or negatively charged, can act as a nucleophile as long as it has an unshared pair of electrons; that is, as long as it is a Lewis base. If the nucleophile is negatively charged, the product is neutral; if the nucleophile is neutral, the product is positively charged.

A wide array of substances can be prepared using nucleophilic substitution reactions. In fact, we've already seen examples in previous chapters. For instance, the reaction of an acetylide anion with an alkyl halide, discussed in Section 9.8, is an $\mathrm{S}_{\mathrm{N}} 2$ reaction in which the acetylide nucleophile displaces a halide leaving group.

$$
\mathrm{R}-\mathrm{C} \equiv \mathrm{C}:^{-}+\mathrm{CH}_{3} \mathrm{Br} \xrightarrow[\text { reaction }]{\mathrm{S}_{\mathrm{N}^{2}}} \mathrm{R}-\mathrm{C} \equiv \mathrm{C}-\mathrm{CH}_{3}+\mathrm{Br}^{-}
$$


An acetylide anion -
TABLE 11.1 lists some nucleophiles in the order of their reactivity, shows the products of their reactions with bromomethane, and gives the relative rates of their reactions. There are large differences in the rates at which various nucleophiles react.

TABLE 11.1 Some $S_{N} 2$ Reactions with Bromomethane


An acetylide anion -
TABLE 11.1 Some $S_{N} 2$ Reactions with Bromomethane

| $\mathrm{Nu}:^{-}+\mathrm{CH}_{3} \mathrm{Br} \rightarrow \mathrm{CH}_{3} \mathrm{Nu}+\mathrm{Br}^{-}$ |  |  |  |  |
| :---: | :---: | :---: | :---: | :---: |
| Nucleophile |  | Product |  | Relative rate of reaction |
| Formula | Name | Formula | Name |  |
| $\mathrm{H}_{2} \mathrm{O}$ | Water | $\mathrm{CH}_{3} \mathrm{OH}_{2}{ }^{+}$ | Methylhydronium ion | 1 |
| $\mathrm{CH}_{3} \mathrm{CO}_{2}{ }^{-}$ | Acetate | $\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{CH}_{3}$ | Methyl acetate | 500 |
| $\mathrm{NH}_{3}$ | Ammonia | $\mathrm{CH}_{3} \mathrm{NH}_{3}{ }^{+}$ | Methylammonium ion | 700 |
| $\mathrm{Cl}^{-}$ | Chloride | $\mathrm{CH}_{3} \mathrm{Cl}$ | Chloromethane | 1,000 |
| $\mathrm{HO}^{-}$ | Hydroxide | $\mathrm{CH}_{3} \mathrm{OH}$ | Methanol | 10,000 |
| $\mathrm{CH}_{3} \mathrm{O}^{-}$ | Methoxide | $\mathrm{CH}_{3} \mathrm{OCH}_{3}$ | Dimethyl ether | 25,000 |
| $\mathrm{I}^{-}$ | Iodide | $\mathrm{CH}_{3} \mathrm{I}$ | Iodomethane | 100,000 |
| ${ }^{-} \mathrm{CN}$ | Cyanide | $\mathrm{CH}_{3} \mathrm{CN}$ | Acetonitrile | 125,000 |
| HS ${ }^{-}$ | Hydrosulfide | $\mathrm{CH}_{3} \mathrm{SH}$ | Methanethiol | 125,000 |


An acetylide anion -
What are the reasons for the reactivity differences observed in TABLE 11.1? Why do some reactants appear to be much more "nucleophilic" than others? The answers to these questions aren't straightforward. Part of the problem is that the term nucleophilicity is imprecise. The term is usually taken to be a measure of the affinity of a nucleophile for a carbon atom in the $\mathrm{S}_{\mathrm{N}} 2$ reaction, but the reactivity of a given nucleophile can change from one reaction to the next. The exact nucleophilicity of a species in a given reaction depends on the substrate, the solvent, and even the reactant concentrations. Detailed explanations for the observed nucleophilicities aren't always simple, but some trends can be detected from the data of TABLE 11.1.


An acetylide anion -
- Nucleophilicity roughly parallels basicity when comparing nucleophiles that have the same reacting atom. Thus, $\mathrm{OH}^{-}$is both more basic and more nucleophilic than acetate ion, $\mathrm{CH}_{3} \mathrm{CO}_{2}{ }^{-}$, which in turn is more basic and more nucleophilic than $\mathrm{H}_{2} \mathrm{O}$. Since "nucleophilicity" is usually taken as the affinity of a Lewis base for a carbon atom in the $\mathrm{S}_{\mathrm{N}} 2$ reaction and "basicity" is the affinity of a base for a proton, it's easy to see why there might be a correlation between the two kinds of behavior.
- Nucleophilicity usually increases going down a column of the periodic table. Thus, $\mathrm{HS}^{-}$is more nucleophilic than $\mathrm{HO}^{-}$, and the halide reactivity order is $\mathrm{I}^{-}>\mathrm{Br}^{-}>\mathrm{Cl}^{-}$. Going down the periodic table, elements have their valence electrons in successively larger shells where they are successively farther from the nucleus, less tightly held, and consequently more reactive. This matter is complex, though, and the nucleophilicity order can change depending on the solvent.
- Negatively charged nucleophiles are usually more reactive than neutral ones. As a result, $\mathrm{S}_{\mathrm{N}} 2$ reactions are often carried out under basic conditions rather than neutral or acidic conditions.

PROBLEM What product would you expect from $\mathrm{S}_{\mathrm{N}} 2$ reaction of 1-bromobutane with each of the following?
11-4 (a) NaI
(b) KOH
(c) $\mathrm{H}-\mathrm{C} \equiv \mathrm{C}-\mathrm{Li}$
(d) $\mathrm{NH}_{3}$


An acetylide anion -
PROBLEM Which substance in each of the following pairs is more reactive as a nucleophile? Explain.
11-5 (a) $\left(\mathrm{CH}_{3}\right)_{2} \mathrm{~N}^{-}$or $\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH}$
(b) $\left(\mathrm{CH}_{3}\right)_{3} \mathrm{~B}$ or $\left(\mathrm{CH}_{3}\right)_{3} \mathrm{~N}$
(c) $\mathrm{H}_{2} \mathrm{O}$ or $\mathrm{H}_{2} \mathrm{~S}$


The Leaving Group -
Still another variable that can affect the $S_{N} 2$ reaction is the nature of the group displaced by the incoming
nucleophile, the leaving group. Because the leaving group is expelled with a negative charge in most $\mathrm{S}_{\mathrm{N}} 2$ reactions, the best leaving groups are those that best stabilize the negative charge in the transition state. The greater the extent of charge stabilization by the leaving group, the lower the energy of the transition state and the more rapid the reaction. But as we saw in Section 2.8 , the groups that best stabilize a negative charge are also the weakest bases. Thus, weak bases such as $\mathrm{Cl}^{-}, \mathrm{Br}^{-}$, and tosylate ion make good leaving groups, while strong bases such as $\mathrm{OH}^{-}$and $\mathrm{NH}_{2}{ }^{-}$make poor leaving groups.

It's just as important to know which are poor leaving groups as to know which are good, and the preceding data clearly indicate that $\mathrm{F}^{-}, \mathrm{HO}^{-}, \mathrm{RO}^{-}$, and $\mathrm{H}_{2} \mathrm{~N}^{-}$are not displaced by nucleophiles. In other words, alkyl fluorides, alcohols, ethers, and amines do not typically undergo $\mathrm{S}_{\mathrm{N}} 2$ reactions. To carry out an $\mathrm{S}_{\mathrm{N}} 2$ reaction with an alcohol, it's necessary to convert the ${ }^{-}$OH into a better leaving group. This, in fact, is just what happens when a primary or secondary alcohol is converted into either an alkyl chloride by reaction with $\mathrm{SOCl}_{2}$ or an alkyl bromide by reaction with $\mathrm{PBr}_{3}$ (Section 10.5).

Alternatively, an alcohol can be made more reactive toward nucleophilic substitution by treating it with para-toluenesulfonyl chloride to form a tosylate. As noted previously, tosylates are even more reactive than halides in nucleophilic substitutions. Note that tosylate formation does not change the configuration of the oxygen-bearing carbon because the $\mathrm{C}-\mathrm{O}$ bond is not broken.


The Leaving Group -
A $1^{\circ}$ or $2^{\circ}$ alcohol

A tosylate

The one general exception to the rule that ethers don't typically undergo $\mathrm{S}_{\mathrm{N}} 2$ reactions pertains to epoxides, the three-membered cyclic ethers that we saw in Section 8.7. Because of the angle strain in their three-membered ring, epoxides are much more reactive than other ethers. They react with aqueous acid to give 1,2-diols, as we saw in Section 8.7, and they react readily with many other nucleophiles as well. Propene oxide, for instance, reacts with HCl to give 1-chloro-2-propanol by an $\mathrm{S}_{\mathrm{N}} 2$ backside attack on the less hindered primary carbon atom. We'll look at the process in more detail in Section 18.5.

PROBLEM Rank the following compounds in order of their expected reactivity toward $\mathrm{S}_{\mathrm{N}} 2$ reaction:
11-6
$\mathrm{CH}_{3} \mathrm{Br}, \mathrm{CH}_{3} \mathrm{OTos}$, $\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCl}$, $\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCl}$


The Solvent -
The rates of $\mathrm{S}_{\mathrm{N}} 2$ reactions are strongly affected by the solvent. Protic solvents-those that contain an -OH or -NH group-are generally the worst for $\mathrm{S}_{\mathrm{N}} 2$ reactions, while polar aprotic solvents, which are polar but don't have an - OH or -NH group, are the best.

Protic solvents, such as methanol and ethanol, slow down $\mathrm{S}_{\mathrm{N}} 2$ reactions by solvation of the reactant nucleophile. The solvent molecules hydrogen-bond to the nucleophile and form a cage around it, thereby lowering its energy and reactivity.

enhanced ground-state stability)


The Solvent -
enhanced ground-state stability)

In contrast with protic solvents-which decrease the rates of $S_{N} 2$ reactions by lowering the ground-state energy of the nucleophile-polar aprotic solvents increase the rates of $\mathrm{S}_{\mathrm{N}} 2$ reactions by raising the ground-state energy of the nucleophile. Acetonitrile $\left(\mathrm{CH}_{3} \mathrm{CN}\right)$, dimethylformamide [ $\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NCHO}$, abbreviated DMF], and dimethyl sulfoxide $\left[\left(\mathrm{CH}_{3}\right)_{2} \mathrm{SO}\right.$, abbreviated DMSO] are particularly useful. A solvent known as hexamethylphosphoramide $\left\{\left[\left(\mathrm{CH}_{3}\right)_{2} \mathrm{~N}\right]_{3} \mathrm{PO}\right.$, abbreviated HMPA $\}$ can also be useful but it should only be handled with great care and not be allowed to touch the eyes or skin. These solvents can dissolve many salts because of their high polarity, but they tend to solvate metal cations rather than nucleophilic anions. As a result, the bare, unsolvated anions have a greater nucleophilicity and $\mathrm{S}_{\mathrm{N}} 2$ reactions take place at correspondingly increased rates. For instance, a rate increase of 200,000 has been observed on changing from methanol to HMPA for the reaction of azide ion with 1-bromobutane.

PROBLEM Organic solvents like benzene, ether, and chloroform are neither protic nor strongly polar. What 11-7 effect would you expect these solvents to have on the reactivity of a nucleophile in $\mathrm{S}_{\mathrm{N}} 2$ reactions?


A Summary of $\mathrm{S}_{\mathrm{N}} 2$ Reaction Characteristics -
The effects on $\mathrm{S}_{\mathrm{N}} 2$ reactions of the four variables-substrate structure, nucleophile, leaving group, and solvent-are summarized in the following statements and in the energy diagrams of FIGURE 11.8:

Substrate $\quad$ Steric hindrance raises the energy of the $S_{N} 2$ transition state, increasing $\Delta G^{\ddagger}$ and decreasing the reaction rate (FIGURE 11.8a). As a result, $\mathrm{S}_{\mathrm{N}} 2$ reactions are best for methyl and primary substrates. Secondary substrates react slowly, and tertiary substrates do not react by an $\mathrm{S}_{\mathrm{N}} 2$ mechanism.
Nucleophile Basic, negatively charged nucleophiles are less stable and have a higher ground-state energy than neutral ones, decreasing $\Delta G^{\ddagger}$ and increasing the $\mathrm{S}_{\mathrm{N}} 2$ reaction rate (FIGURE 11.8b).

Leaving Good leaving groups (more stable anions) lower the energy of the transition state,
group
Solvent decreasing $\Delta G^{\ddagger}$ and increasing the $S_{N} 2$ reaction rate (FIGURE 11.8c). Protic solvents solvate the nucleophile, thereby lowering its ground-state energy, increasing $\Delta G^{\ddagger}$, and decreasing the $S_{N} 2$ reaction rate. Polar aprotic solvents surround the accompanying cation but not the nucleophilic anion, thereby raising the ground-state energy of the nucleophile, decreasing $\Delta G^{\ddagger}$, and increasing the reaction rate (FIGURE 11.8d).

FIGURE 11.8 Energy diagrams showing the effects of (a) substrate, (b) nucleophile, (c) leaving group, and (d) solvent on $\mathbf{S}_{\mathbf{N}} \mathbf{2}$ reaction rates. Substrate and leaving group effects are felt primarily in the transition state. Nucleophile and solvent effects are felt primarily in the reactant ground state.