3-Methyl-3-hexanol - 8.2 Halogenation of Alkenes: Addition of $X_{2}$
Bromine and chlorine add rapidly to alkenes to yield 1,2-dihalides, a process called halogenation. For example, nearly 50 million tons of 1,2-dichloroethane (ethylene dichloride) are synthesized worldwide each year, much of it by addition of $\mathrm{Cl}_{2}$ to ethylene. The product is used both as a solvent and as starting material for the manufacture of poly(vinyl chloride), PVC, the third most widely synthesized polymer in the world afterpolyethelyne and polypropolyne. Fluorine is too reactive and difficult to control for most laboratory applications, and iodine does not react with most alkenes.

Ethylene
1,2-Dichloroethane (ethylene dichloride)
Based on what we've seen thus far, a possible mechanism for the reaction of bromine with alkenes might involve electrophilic addition of $\mathrm{Br}^{+}$to the alkene, giving a carbocation intermediate that could undergo further reaction with $\mathrm{Br}^{-}$to yield the dibromo addition product.


Possible -
mechanism?

Although this mechanism seems plausible, it's not fully consistent with known facts. In particular, it doesn't explain the stereochemistry of the addition reaction. That is, the mechanism doesn't account for which product stereoisomer is formed.

When the halogenation reaction is carried out on a cycloalkene, such as cyclopentene, only the trans stereoisomer of the dihalide addition product is formed, rather than the mixture of cis and trans isomers that might have been expected if a planar carbocation intermediate were involved. We say that the reaction occurs with anti stereochemistry, meaning that the two bromine atoms come from opposite faces of the double bond-one from the top face and one from the bottom face.

An explanation for the observed stereochemistry of addition was suggested in 1937 by George Kimball and Irving Roberts, who proposed that the reaction intermediate is not a carbocation but is instead a bromonium ion, $\mathbf{R}_{\mathbf{2}} \mathbf{B r}^{+}$, formed by electrophilic addition of $\mathrm{Br}^{+}$to the alkene. (Similarly, a chloronium ion contains a positively charged, divalent chlorine, $\mathrm{R}_{2} \mathrm{Cl}^{+}$.) The bromonium ion is formed in a single step by interaction of the alkene with $\mathrm{Br}_{2}$ and the simultaneous loss of $\mathrm{Br}^{-}$.

How does the formation of a bromonium ion account for the observed anti stereochemistry of addition to cyclopentene? If a bromonium ion is formed as an intermediate, we can imagine that the large bromine atom might "shield" one side of the molecule. Reaction with $\mathrm{Br}^{-}$ion in the second step could then occur only from the opposite, unshielded side to give the trans product.


Possible -
The bromonium ion postulate, made more than 85 years ago to explain the stereochemistry of halogen addition to alkenes, is a remarkable example of deductive logic in chemistry. Arguing from experimental results, chemists were able to make a hypothesis about the intimate mechanistic details of alkene electrophilic reactions. Subsequently, strong evidence supporting the mechanism came from the work of George Olah at the University of Southern California, who prepared and studied stable solutions of cyclic bromonium ions in liquid $\mathrm{SO}_{2}$. There's no question that bromonium ions exist.

Alkene halogenation reactions occur in nature just as they do in the laboratory but are limited primarily to marine organisms living in halide-rich environments. These biological halogenation reactions are carried out by enzymes called haloperoxidases, which use $\mathrm{H}_{2} \mathrm{O}_{2}$ to oxidize $\mathrm{Br}^{-}$or $\mathrm{Cl}^{-}$ions to a biological equivalent of $\mathrm{Br}^{+}$or $\mathrm{Cl}^{+}$. Electrophilic addition to the double bond of a substrate molecule then yields a bromonium or chloronium ion intermediate just as in the laboratory, and reaction with another halide ion completes the process. Halomon, for example, an antitumor pentahalide isolated from red alga, is thought to arise by a route that involves twofold addition of BrCl through the corresponding bromonium ions.

Halomon
PROBLEM What product would you expect to obtain from addition of $\mathrm{Cl}_{2}$ to 1,2 -dimethylcyclohexene? Show 8-3 the stereochemistry of the product.

PROBLEM Addition of HCl to 1,2-dimethylcyclohexene yields a mixture of two products. Show the 8-4 stereochemistry of each, and explain why a mixture is formed.


Possible - 8.3 Halohydrins from Alkenes: Addition of HO-X
Another example of an electrophilic addition is the reaction an alkene with either $\mathrm{Br}_{2}$ or $\mathrm{Cl}_{2}$ in the presence of
water to yield a 1,2-halo alcohol, called a halohydrin.

An alkene

$+\mathrm{HX}$

A halohydrin

We saw in the previous section that when $\mathrm{Br}_{2}$ reacts with an alkene, the cyclic bromonium ion intermediate reacts with the only nucleophile present, $\mathrm{Br}^{-}$ion. If the reaction is carried out in the presence of an additional nucleophile, however, the intermediate bromonium ion can be intercepted by the added nucleophile and diverted to a different product. In the presence of a high concentration of water, for instance, water competes with $\mathrm{Br}^{-}$ion as a nucleophile and reacts with the bromonium ion intermediate to yield a bromohydrin. The net effect is addition of $\mathrm{HO}-\mathrm{Br}$ to the alkene by the pathway shown in FIGURE 8.3.

FIGURE 8.3 MECHANISM

Bromohydrin formation by reaction of an alkene with $\mathrm{Br}_{2}$ in the presence of water.
Water acts as a nucleophile in step 2 to react with the intermediate bromonium ion.
(1) Reaction of the alkene with $\mathrm{Br}_{2}$ yields a bromonium ion intermediate, as previously discussed.
(2) Water acts as a nucleophile, using a lone pair of electrons to open the bromonium ion ring and form a bond to carbon. Since oxygen donates its electrons in this step, it now has the positive charge.

(1) $\mathrm{Br}_{2}$

2

Loss of a proton $\left(\mathrm{H}^{+}\right)$from oxygen then gives $\mathrm{H}_{3} \mathrm{O}^{+}$and the neutral bromohydrin addition product.


(3) -
3-Bromo-2-butanol

In practice, few alkenes are soluble in water, and bromohydrin formation is often carried out in a solvent such as aqueous dimethyl sulfoxide, $\mathrm{CH}_{3} \mathrm{SOCH}_{3}$ (DMSO), using a reagent called N -bromosuccinimide (NBS) as a source of $\mathrm{Br}_{2}$. NBS is a stable, easily handled compound that slowly decomposes in water to yield $\mathrm{Br}_{2}$ at a controlled rate. Bromine itself can also be used in the addition reaction, but it is more dangerous and more difficult to handle than NBS.

Notice that the aromatic ring in the above example does not react with $\mathrm{Br}_{2}$, even though it appears to have three carbon-carbon double bonds. As we'll see in Section 15.2, aromatic rings are a good deal more stable and less reactive than might be expected.

There are a number of biological examples of halohydrin formation, particularly in marine organisms. As with halogenation (Section 8.2), halohydrin formation is carried out by haloperoxidases. For example:

PROBLEM What product would you expect from the reaction of cyclopentene with NBS and water? Show the 8-5 stereochemistry.

PROBLEM When an unsymmetrical alkene such as propene is treated with $N$-bromosuccinimide in aqueous
8-6 dimethyl sulfoxide, the major product has the bromine atom bonded to the less highly substituted carbon atom. Is this Markovnikov or non-Markovnikov orientation? (Section 7.8) Explain.


(3) - 8.4 Hydration of Alkenes: Addition of $\mathrm{H}_{2} \mathrm{O}$ by Oxymercuration
Water adds to alkenes to yield alcohols, a process called hydration. The reaction takes place on treatment of the alkene with water and a strong acid catalyst, such as $\mathrm{H}_{2} \mathrm{SO}_{4}$, by a mechanism similar to that of HX addition. Thus, as shown in FIGURE 8.4, protonation of an alkene double bond yields a carbocation intermediate, which reacts with water to yield a protonated alcohol product, $\mathrm{ROH}_{2}{ }^{+}$. Loss of $\mathrm{H}^{+}$from this protonated alcohol gives the neutral alcohol and regenerates the acid catalyst.


FIGURE 8.4 MECHANISM -
Mechanism of the acid-catalyzed hydration of an alkene to yield an alcohol. Protonation of the alkene gives a carbocation intermediate, which reacts with water. The initial product is then deprotonated.

A hydrogen atom on the electrophile $\mathrm{H}_{3} \mathrm{O}^{+}$ is attacked by $\pi$ electrons from the nucleophilic double bond, forming a new $\mathrm{C}-\mathrm{H}$ bond. This leaves the other carbon atom with $\mathrm{a}+$ charge and a vacant $p$ orbital. Simultaneously, two electrons from the $\mathrm{H}-\mathrm{O}$ bond move onto oxygen, giving neutral water.

2-Methylpropene
(1)

2 The nucleophile $\mathrm{H}_{2} \mathrm{O}$ donates an electron pair to the positively charged carbon atom, forming a $\mathrm{C}-\mathrm{O}$ bond and leaving a positive charge on oxygen in the protonated alcohol addition product.

Protonated alcohol
(3) Water acts as a base to remove $\mathrm{H}^{+}$, regenerating $\mathrm{H}_{3} \mathrm{O}^{+}$and yielding the neutral alcohol addition product.

(3)

2-Methyl-2-propanol

Most ethanol throughout the world is now made by fermentation of biological precursors, such as corn and sugar, but acid-catalyzed alkene hydration is particularly suited to large-scale industrial procedures, and approximately 90,000 tons of ethanol is manufactured each year in the United States by hydration of ethylene. The reaction is of little value in the laboratory, however, because it requires high temperatures- $250{ }^{\circ} \mathrm{C}$ in the case of ethylene-and strongly acidic conditions.


Ethylene -
Acid-catalyzed hydration of double bonds is also uncommon in biological pathways. Instead, biological
hydrations usually require that the double bond be adjacent to a carbonyl group for reaction to proceed. Fumarate, for instance, is hydrated to give malate as one step in the citric acid cycle of food metabolism. Note that the requirement for an adjacent carbonyl group in the addition of water is the same as in Section 8.1 for the elimination of water. We'll see the reason for this requirement in Section 19.13, but will note for now that the reaction is not an electrophilic addition but instead occurs through a mechanism that involves formation of an anion intermediate followed by protonation by an acid HA.

When it comes to circumventing problems like those with acid-catalyzed alkene hydrations, laboratory chemists have a great advantage over the cellular "chemists" in living organisms. Laboratory chemists are not constrained to carry out their reactions in water solution; they can choose from any of a large number of solvents. Laboratory reactions don't need to be carried out at a fixed temperature; they can take place over a wide range of temperatures. And laboratory reagents aren't limited to containing carbon, oxygen, nitrogen, and a few other elements; they can contain any element in the periodic table.

In the laboratory, alkenes are often hydrated by the oxymercuration-demercuration procedure, which involves electrophilic addition of $\mathrm{Hg}^{2+}$ to the alkene on reaction with mercury(II) acetate $\left\left(\mathrm{CH}_{3} \mathrm{CO}_{2}\right)_{2} \mathrm{Hg}\right.$. The intermediate organomercury compound is then treated with sodium borohydride, $\mathrm{NaBH}_{4}$, and demercuration occurs to produce an alcohol. For example:
![


1-Methylcyclopentene -
1-Methylcyclopentanol
(92\%)

Alkene oxymercuration is closely analogous to halohydrin formation. The reaction is initiated by electrophilic addition of $\mathrm{Hg}^{2+}$ (mercuric) ion to the alkene to give an intermediate mercurinium ion, whose structure resembles that of a bromonium ion (FIGURE 8.5). Nucleophilic addition of water as in halohydrin formation, followed by the loss of a proton, then yields a stable organomercury product. The final step, demercuration of the organomercury compound by reaction with sodium borohydride, is complex and involves radicals. Note that the regiochemistry of the reaction corresponds to Markovnikov addition of water; that is, the - OH group attaches to the more highly substituted carbon atom, and the -H attaches to the less highly substituted carbon. The hydrogen that replaces mercury in the demercuration step can attach from either side of the molecule depending on the exact circumstances.

FIGURE 8.5 Mechanism of the oxymercuration of an alkene to yield an alcohol. (1) Electrophilic addition of $\mathrm{Hg}^{2+}$ gives a mercurinium ion, which (2) reacts with water as in halohydrin formation. Loss of a proton gives an organomercury product, and (3) reaction with $\mathrm{NaBH}_{4}$ removes the mercury. The product of the reaction is a more highly substituted alcohol, corresponding to Markovnikov regiochemistry.

PROBLEM What products would you expect from oxymercuration-demercuration of the following alkenes?
8-7 (a)
$\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}=\mathrm{CH}_{2}$
(b)

PROBLEM From what alkenes might the following alcohols have been prepared?
8-8 (a)

(b)


1-Methylcyclopentene - 8.5 Hydration of Alkenes: Addition of $\mathrm{H}_{2} \mathrm{O}$ by Hydroboration
In addition to the oxymercuration-demercuration method, which yields the Markovnikov product, a complementary method that yields the non-Markovnikov product is also useful. Discovered in 1959 by H.C. Brown at Purdue University and called hydroboration, the reaction involves addition of a B-H bond of borane, $\mathrm{BH}_{3}$, to an alkene to yield an organoborane intermediate, $\mathrm{RBH}_{2}$. Oxidation of the organoborane by reaction with basic hydrogen peroxide, $\mathrm{H}_{2} \mathrm{O}_{2}$, then gives an alcohol. For example:

Borane is very reactive as a Lewis acid because the boron atom has only six electrons in its valence shell rather than an octet. In tetrahydrofuran solution, $\mathrm{BH}_{3}$ accepts an electron pair from a solvent molecule in a Lewis acid-base reaction to complete its octet and form a stable $\mathrm{BH}_{3}-$ THF complex.

When an alkene reacts with $\mathrm{BH}_{3}$ in THF solution, rapid addition to the double bond occurs three times and a trialkylborane, $\mathrm{R}_{3} \mathrm{~B}$, is formed. For example, 1 molar equivalent of $\mathrm{BH}_{3}$ adds to 3 molar equivalents of cyclohexene to yield tricyclohexylborane. When tricyclohexylborane is then treated with aqueous hydrogen $\mathrm{H}_{2} \mathrm{O}_{2}$ in basic solution, an oxidation takes place. The three $\mathrm{C}-\mathrm{B}$ bonds are broken, -OH groups bond to the three carbons, and 3 equivalents of cyclohexanol are produced. The net effect of the two-step hydroboration-oxidation sequence is hydration of the alkene double bond.


1-Methylcyclopentene - 8.5 Hydration of Alkenes: Addition of $\mathrm{H}_{2} \mathrm{O}$ by Hydroboration
One of the features that makes the hydroboration reaction so useful is the regiochemistry that results when an unsymmetrical alkene is hydroborated. For example, hydroboration-oxidation of 1-methylcyclopentene yields trans-2-methylcyclopentanol. In this process, boron and hydrogen add to the alkene from the same face of the double bond-that is, with syn stereochemistry, the opposite of anti-with boron attaching to the less highly substituted carbon. During the oxidation step, the boron is replaced by an -OH with the same stereochemistry, resulting in an overall syn non-Markovnikov addition of water. This stereochemical result is particularly useful because it is complementary to the Markovnikov regiochemistry observed for oxymercuration-demercuration.

Why does alkene hydroboration take place with syn, non-Markovnikov regiochemistry to yield the less highly substituted alcohol? Hydroboration differs from many other alkene addition reactions in that it occurs in a single step without a carbocation intermediate (FIGURE 8.6). Because the $\mathrm{C}-\mathrm{H}$ and $\mathrm{C}-\mathrm{B}$ bonds form at the same time and from the same face of the alkene, syn stereochemistry results. Non-Markovnikov regiochemistry occurs because attachment of boron is favored at the less sterically crowded carbon atom of the alkene.

FIGURE 8.6 Mechanism of alkene hydroboration. The reaction occurs in a single step in which the $\mathrm{C}-\mathrm{H}$ and $\mathrm{C}-\mathrm{B}$ bonds form at the same time and on the same face of the double bond. The lower energy, more rapidly formed transition state is the one with less steric crowding, leading to non-Markovnikov regiochemistry.


Predicting the Products of a Hydration Reaction -
What products would you obtain from reaction of 2-methyl-2-pentene with:
(a) $\mathrm{BH}_{3}$, followed by $\mathrm{H}_{2} \mathrm{O}_{2}, \mathrm{OH}^{-}$
(b) $\mathrm{Hg}(\mathrm{OAc})_{2}$, followed by $\mathrm{NaBH}_{4}$


Strategy -
When predicting the product of a reaction, you have to recall what you know about the kind of reaction being carried out and apply that knowledge to the specific case you're dealing with. In the present instance, recall that the two methods of hydration-hydroboration-oxidation and oxymercuration-demercuration-give complementary products. Hydroboration-oxidation occurs with syn stereochemistry and gives the nonMarkovnikov addition product; oxymercuration-demercuration gives the Markovnikov product.


Solution -
(a) 2-Methyl-2-pentene

2-Methyl-3-pentanol

PROBLEM What alkenes might be used to prepare the following alcohols by hydroboration-oxidation?
8-10 (a)

(b)

(c)

PROBLEM The following cycloalkene gives a mixture of two alcohols on hydroboration followed by oxidation.
8-11 Draw the structures of both, and explain the result.


Solution - 8.6 Reduction of Alkenes: Hydrogenation
Alkenes react with $\mathrm{H}_{2}$ in the presence of a metal catalyst such as palladium or platinum to yield the corresponding saturated alkanes. We describe the result by saying that the double bond has been hydrogenated, or reduced. Note that the word reduction is used somewhat differently in organic chemistry from what you might have learned previously. In general chemistry, a reduction is defined as the gain of one or more electrons by an atom. In organic chemistry, however, a reduction is a reaction that results in a gain of electron density for carbon, caused either by bond formation between carbon and a less electronegative atom-usually hydrogen-or by bond-breaking between carbon and a more electronegative atom-usually oxygen, nitrogen, or a halogen. We'll explore this topic in more detail in Section 10.8.

Reduction Increases electron density on carbon by:
-forming this: $\mathrm{C}-\mathrm{H}$

- or breaking one of these: $\mathrm{C}-\mathrm{O} \quad \mathrm{C}-\mathrm{N} \quad \mathrm{C}-\mathrm{X}$


A reduction: -
An alkene
An alkane
Platinum and palladium are the most common laboratory catalysts for alkene hydrogenations. Palladium is normally used as a very fine powder "supported" on an inert material such as charcoal ( $\mathrm{Pd} / \mathrm{C}$ ) to maximize surface area. Platinum is normally used as $\mathrm{PtO}_{2}$, a reagent known as Adams' catalyst after its discoverer, Roger Adams at the University of Illinois.

Catalytic hydrogenation, unlike most other organic reactions, is a heterogeneous process rather than a homogeneous one. That is, the hydrogenation reaction does not occur in a homogeneous solution but instead takes place on the surface of solid catalyst particles. Hydrogenation usually occurs with syn stereochemistry: both hydrogens add to the double bond from the same face.

As shown in FIGURE 8.7, hydrogenation begins with adsorption of $\mathrm{H}_{2}$ onto the catalyst surface. Complexation between catalyst and alkene then occurs as a vacant orbital on the metal interacts with the filled alkene $\pi$ orbital
on the alkene. In the final steps, hydrogen is inserted into the double bond and the saturated product diffuses away from the catalyst. The stereochemistry of hydrogenation is syn because both hydrogens add to the double bond from the same catalyst surface.

FIGURE 8.7 MECHANISM

Mechanism of alkene hydrogenation. The reaction takes place with syn stereochemistry on the surface of insoluble catalyst particles.
catalyst surface and dissociates into hydrogen atoms.

(2) The alkene adsorbs to the catalyst surface, using its $\pi$ bond to complex to the metal atoms.
(3) A hydrogen atom is transferred from the metal to one of the alkene carbon atoms, forming a partially reduced intermediate with a $\mathrm{C}-\mathrm{H}$ bond and carbon-metal $\sigma$ bond.
(4) A second hydrogen is transferred from the metal to the second carbon, giving the alkane product and regenerating the catalyst. Because both hydrogens are transferred to the same face of the alkene, the reduction has syn stereochemistry.

> Alkane plus regenerated catalyst


A reduction: -
> Alkane plus regenerated catalyst

An interesting feature of catalytic hydrogenation is that the reaction is extremely sensitive to the steric environment around the double bond. As a result, the catalyst usually approaches the more accessible face of an alkene, giving rise to a single product. In $\alpha$-pinene, for example, one of the methyl groups attached to the fourmembered ring hangs over the top face of the double bond and blocks approach of the hydrogenation catalyst from that side. Reduction therefore occurs exclusively from the bottom face to yield the product shown.

Alkenes are much more reactive toward catalytic hydrogenation than most other unsaturated functional groups, and the reaction is therefore quite selective. Other functional groups, such as aldehydes, ketones, esters, and nitriles, often survive alkene hydrogenation conditions unchanged, although reaction with these groups does occur under more vigorous conditions. Note that, particularly in the hydrogenation of methyl 3-phenylpropenoate shown below, the aromatic ring is not reduced by hydrogen and palladium even though it contains apparent double bonds.

Cyclohex-2-enone
Cyclohexanone
(ketone not reduced)

Cyclohexylideneacetonitrile

Cyclohexylacetonitrile
(nitrile not reduced)

In addition to its usefulness in the laboratory, catalytic hydrogenation is also important in the food industry, where unsaturated vegetable oils are reduced on a large scale to produce the saturated fats used in margarine and cooking products (FIGURE 8.8). As we'll see in Section 27.1, vegetable oils are triesters of glycerol, $\mathrm{HOCH}_{2} \mathrm{CH}(\mathrm{OH}) \mathrm{CH}_{2} \mathrm{OH}$, with three long-chain carboxylic acids called fatty acids. The fatty acids are generally polyunsaturated, and their double bonds have cis stereochemistry. Complete hydrogenation yields the corresponding saturated fatty acids, but incomplete hydrogenation often results in partial cis-trans isomerization of a remaining double bond. When eaten and digested, the free trans fatty acids are released, raising blood cholesterol levels and potentially contributing to coronary problems.

A polyunsaturated
fatty acid in
vegetable oil

A vegetable oil


A saturated fatty -
acid in margarine

A trans fatty acid

FIGURE 8.8 Catalytic hydrogenation of polyunsaturated fats leads primarily to saturated products, along with a small amount of isomerized trans fats.

Double-bond reductions are very common in biological pathways, although the mechanism is completely different from that of laboratory catalytic hydrogenation over palladium. As with biological hydrations (Section 8.4), biological reductions usually occur in two steps and require that the double bond be adjacent to a carbonyl group. In the first step, the biological reducing agent NADPH (reduced nicotinamide adenine dinucleotide phosphate), adds a hydride ion ( $\mathrm{H}:^{-}$) to the double bond to give an anion. In the second, the anion is protonated by acid HA, leading to overall addition of $\mathrm{H}_{2}$. An example is the reduction of trans-crotonyl ACP to yield butyryl ACP, a step involved in the biosynthesis of fatty acids (FIGURE 8.9).

NADPH
FIGURE 8.9 Reduction of the carbon-carbon double bond in trans-crotonyl ACP, a step in the biosynthesis of fatty acids. One hydrogen is delivered from NADPH as a hydride ion, $\mathrm{H}^{-}$; the other hydrogen is delivered by protonation of the anion intermediate with an acid, HA.

PROBLEM What product would you obtain from catalytic hydrogenation of the following alkenes?
8-12 (a)

(b)

(c)

=


A saturated fatty - 8.7 Oxidation of Alkenes: Epoxidation and Hydroxylation
Like the word reduction used in the previous section for the addition of hydrogen to a double bond, the word oxidation has a slightly different meaning in organic chemistry than what you might have previously learned. In general chemistry, an oxidation is defined as the loss of one or more electrons by an atom. In organic chemistry, however, an oxidation is a reaction that results in a loss of electron density for carbon, caused either by bond formation between carbon and a more electronegative atom-usually oxygen, nitrogen, or a halogen-or by bond-breaking between carbon and a less electronegative atom-usually hydrogen. Note that an oxidation often adds oxygen, while a reduction often adds hydrogen.

Oxidation $\quad \begin{array}{lll}\text { Decreases electron density on carbon by: } \\ & - \text { forming one of these: } \mathrm{C}-\mathrm{O} \quad \mathrm{C}-\mathrm{N} & \mathrm{C}-\mathrm{X} \\ & - \text { or breaking this: } \mathrm{C}-\mathrm{H} & \end{array}$
In the laboratory, alkenes are oxidized to give epoxides on treatment with a peroxyacid, $\mathrm{RCO}_{3} \mathrm{H}$, such as meta-chloroperoxybenzoic acid. An epoxide, also called an oxirane, is a cyclic ether with an oxygen atom in a three-membered ring. For example:

Peroxyacids transfer an oxygen atom to the alkene with syn stereochemistry-both C-O bonds form on the same face of the double bond-through a one-step mechanism without intermediates. The oxygen atom farthest from the carbonyl group is the one transferred.

Another method for the synthesis of epoxides involves the use of halohydrins, prepared by electrophilic addition of HO-X to alkenes (Section 8.3). When a halohydrin is treated with base, HX is eliminated and an epoxide is produced.


A saturated fatty - 8.7 Oxidation of Alkenes: Epoxidation and Hydroxylation
Epoxides undergo an acid-catalyzed ring-opening reaction with water (a hydrolysis) to give the corresponding 1,2-dialcohol, or diol, also called a glycol. Thus, the net result of the two-step alkene epoxidation/hydrolysis is hydroxylation-the addition of an - OH group to each of the two double-bond carbons. In fact, approximately 204 million tons of ethylene glycol, $\mathrm{HOCH}_{2} \mathrm{CH}_{2} \mathrm{OH}$, most of it used for automobile antifreeze, are produced worldwide each year by the epoxidation of ethylene and subsequent hydrolysis.

Acid-catalyzed epoxide opening begins with protonation of the epoxide to increase its reactivity, followed by nucleophilic addition of water. This nucleophilic addition is analogous to the final step of alkene bromination, in which a cyclic bromonium ion is opened by a nucleophile (Section 8.2). That is, a trans-1,2-diol results when an epoxycycloalkane is opened by aqueous acid, just as a trans-1,2-dibromide results when a cycloalkene is brominated. We'll look at epoxide chemistry in more detail in Section 18.6.

Hydroxylation can also be carried out directly (without going through an intermediate epoxide) by treating an alkene with osmium tetroxide, $\mathrm{OsO}_{4}$. The reaction occurs with syn stereochemistry and does not involve a carbocation intermediate. Instead, it takes place through an intermediate cyclic osmate, which is formed in a single step by addition of $\mathrm{OsO}_{4}$ to the alkene. This cyclic osmate is then cleaved using aqueous sodium bisulfite, $\mathrm{NaHSO}_{3}$.

1,2-Dimethylcyclopentene

A cyclic osmate intermediate
cis-1,2-Dimethyl-1,2-cyclopentanediol (87\%)


A saturated fatty - 8.7 Oxidation of Alkenes: Epoxidation and Hydroxylation
1,2-Dimethylcyclopentene

A cyclic osmate intermediate
cis-1,2-Dimethyl-1,2-cyclopentanediol (87\%)

Because $\mathrm{OSO}_{4}$ is both very expensive and very toxic, the reaction is usually carried out using only a small, catalytic amount of $\mathrm{OsO}_{4}$ in the presence of a stoichiometric amount of a safe and inexpensive co-oxidant such as $N$-methylmorpholine $N$-oxide, abbreviated NMO. The initially formed osmate intermediate reacts rapidly with NMO to yield the product diol plus $N$-methylmorpholine and reoxidized $\mathrm{OsO}_{4}$, which reacts with more alkene in a catalytic cycle.

PROBLEM What product would you expect from reaction of cis-2-butene with meta-chloroperoxybenzoic 8-13 acid? Show the stereochemistry.

PROBLEM Starting with an alkene, how would you prepare each of the following compounds?
8-14 (a)

(b)

(c)


A saturated fatty - 8.8 Oxidation of Alkenes: Cleavage to Carbonyl Compounds
In all the alkene addition reactions we've seen thus far, the carbon-carbon double bond has been converted into a single bond but the carbon skeleton has been unchanged. There are, however, powerful oxidizing reagents that will cleave $\mathrm{C}=\mathrm{C}$ bonds and produce two carbonyl-containing fragments.

Ozone $\left(\mathrm{O}_{3}\right)$ is perhaps the most useful double-bond cleavage reagent. Prepared by passing a stream of oxygen through a high-voltage electrical discharge, ozone adds rapidly to a $\mathrm{C}=\mathrm{C}$ bond at low temperature to give a cyclic intermediate called a molozonide. Once formed, the molozonide spontaneously rearranges to form an ozonide. Although we won't study the mechanism of this rearrangement in detail, it involves the molozonide coming apart into two fragments that then recombine in a different way.

$$
3 \mathrm{O}_{2} \xrightarrow[\text { discharge }]{\text { Electric }} 2 \mathrm{O}_{3}
$$

Low-molecular-weight ozonides are explosive and therefore not isolated. Instead, the ozonide is immediately treated with a reducing agent, such as zinc metal in acetic acid, to produce carbonyl compounds. The net result of the ozonolysis/reduction sequence is that the $\mathrm{C}=\mathrm{C}$ bond is cleaved and an oxygen atom becomes doubly bonded to each of the original alkene carbons. If an alkene with a tetrasubstituted double bond is ozonized, two ketone fragments result; if an alkene with a trisubstituted double bond is ozonized, one ketone and one aldehyde result; and so on.

Isopropylidenecyclohexane
(tetrasubstituted)


Cyclohexanone Acetone -
84\%; two ketones

Several oxidizing reagents other than ozone also cause double-bond cleavage, although such reactions are not often used. For example, potassium permanganate $\left(\mathrm{KMnO}_{4}\right)$ in neutral or acidic solution cleaves alkenes to give carbonyl-containing products. If hydrogens are present on the double bond, carboxylic acids are produced; if two hydrogens are present on one carbon, $\mathrm{CO}_{2}$ is formed.

3,7-Dimethyl-1-octene 2,6-Dimethylheptanoic acid (45\%)
In addition to direct cleavage with ozone or $\mathrm{KMnO}_{4}$, an alkene can also be cleaved in a two-step process by initial hydroxylation to a 1,2-diol, as discussed in the previous section, followed by treatment of the diol with periodic acid, $\mathrm{HIO}_{4}$. If the two -OH groups are in an open chain, two carbonyl compounds result. If the two -OH groups are on a ring, a single, open-chain dicarbonyl compound is formed. As indicated in the following examples, the cleavage reaction takes place through a cyclic periodate intermediate.


Predicting the Reactant in an Ozonolysis Reaction -
What alkene would yield a mixture of cyclopentanone and propanal on treatment with ozone followed by reduction with zinc?


Strategy -
Reaction of an alkene with ozone, followed by reduction with zinc, cleaves the $\mathrm{C}=\mathrm{C}$ bond and becomes two $\mathrm{C}=\mathrm{O}$ bonds. Working backward from the carbonyl-containing products, the alkene precursor can be found by removing the oxygen from each product and joining the two carbon atoms.

Solution

PROBLEM What products would you expect from reaction of 1-methylcyclohexene with the following 8-15 reagents?
(a) Aqueous acidic $\mathrm{KMnO}_{4}$
(b) $\mathrm{O}_{3}$, followed by $\mathrm{Zn}, \mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}$

PROBLEM Propose structures for alkenes that yield the following products on reaction with ozone followed by 8-16 treatment with Zn :
(a) $\left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}=\mathrm{O}+\mathrm{H}_{2} \mathrm{C}=\mathrm{O}$
(b) 2 equiv $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}=\mathrm{O}$


Strategy - 8.9 Addition of Carbenes to Alkenes: Cyclopropane Synthesis
Yet another kind of alkene addition is the reaction with a carbene to yield a cyclopropane. A carbene, $\mathbf{R}_{\mathbf{2}} \mathbf{C}$ :, is a neutral molecule containing a divalent carbon with only six electrons in its valence shell. It is therefore highly reactive and generated only as a reaction intermediate, rather than as an isolable molecule. Because they're electron-deficient, carbenes behave as electrophiles and react with nucleophilic $\mathrm{C}=\mathrm{C}$ bonds. The reaction occurs in a single step without intermediates.


An alkene A carbene A cyclopropane -
One of the simplest methods for generating a substituted carbene is by treatment of chloroform, $\mathrm{CHCl}_{3}$, with a strong base such as KOH . As shown in FIGURE 8.10, the loss of a proton from $\mathrm{CHCl}_{3}$ gives trichloromethanide anion, ${ }^{-}: \mathrm{CCl}_{2}$, which spontaneously expels a $\mathrm{Cl}^{-}$ion to yield dichlorocarbene, $: \mathrm{CCl}_{2}$.


FIGURE 8.10 MECHANISM -
Mechanism of the formation of dichlorocarbene by reaction of chloroform with strong base. Deprotonation of $\mathrm{CHCl}_{3}$ gives the trichloromethanide anion, ${ }^{-}: \mathrm{CCl}_{3}$, which spontaneously expels a $\mathrm{Cl}^{-}$ion.

1. Base abstracts the hydrogen from chloroform, leaving behind the electron pair from the $\mathrm{C}-\mathrm{H}$ bond and forming the trichloromethanide anion.

Chloroform
(1)

richloromethanide
anion
Spontaneous loss of chloride ion then yields the neutral dichlorocarbene.

๑]

Dichlorocarbene

The carbon atom in dichlorocarbene is $s p^{2}$-hybridized, with a vacant $p$ orbital extending above and below the plane of the three atoms and with an unshared pair of electrons occupying the third $s p^{2}$ lobe. Note that this electronic description of dichlorocarbene is similar to that of a carbocation (Section 7.9) with respect to both the $s p^{2}$ hybridization of carbon and the vacant $p$ orbital. Electrostatic potential maps further illustrate the similarity (FIGURE 8.11).

FIGURE 8.11 The structure of dichlorocarbene. Electrostatic potential maps show how the positive region coincides with the empty $p$ orbital in both dichlorocarbene and a carbocation $\left(\mathrm{CH}_{3}{ }^{+}\right)$. The negative region in the dichlorocarbene map coincides with the lone-pair electrons.

If dichlorocarbene is generated in the presence of an alkene, addition to the double bond occurs and a dichlorocyclopropane is formed. As the reaction of dichlorocarbene with cis-2-pentene demonstrates, the addition is stereospecific, meaning that only a single stereoisomer is formed as product. Starting from a cis alkene, for instance, only cis-disubstituted cyclopropane is produced; starting from a trans alkene, only transdisubstituted cyclopropane is produced.


FIGURE 8.10 MECHANISM -
The best method for preparing nonhalogenated cyclopropanes is by a process called the Simmons-Smith reaction. First investigated at the DuPont company, this reaction does not involve a free carbene. Rather, it utilizes a carbenoid-a metal-complexed reagent with carbene-like reactivity. When diiodomethane is treated with a specially prepared zinc-copper mix, (iodomethyl)zinc iodide, $\mathrm{ICH}_{2} \mathrm{ZnI}$, is formed. In the presence of an alkene, $\mathrm{ICH}_{2} \mathrm{ZnI}$ transfers a $\mathrm{CH}_{2}$ group to the double bond to yield cyclopropane. For example, cyclohexene reacts cleanly and with good yield to give the corresponding cyclopropane. Although we won't discuss the mechanistic details, carbene addition to an alkene is one of a general class of reactions called cycloadditions, which we'll study more carefully in Chapter 30.

$$
\mathrm{CH}_{2} \mathrm{I}_{2}+\mathrm{Zn}(\mathrm{Cu}) \longrightarrow \mathrm{ICH} 2-\mathrm{ZnI} \quad\left[": \mathrm{CH}_{2}^{\prime \prime \prime}\right]
$$

Diiodomethane
(Iodomethyl)zinc iodide
(a carbenoid)

Cyclohexene

PROBLEM What products would you expect from the following reactions?

8-17 (a)

(b)


FIGURE 8.10 MECHANISM - 8.10 Radical Additions to Alkenes: Chain-Growth Polymers
In our brief introduction to radical reactions in Section 6.6 , we said that radicals can add to $\mathrm{C}=\mathrm{C}$ bonds, taking one electron from the double bond and leaving one behind to yield a new radical. Let's now look at the process in more detail, focusing on the industrial synthesis of alkene polymers. A polymer is a large-sometimes very large-molecule, built up by repetitive joining together of many smaller molecules, called monomers.

Nature makes wide use of biological polymers. Cellulose, for instance, is a polymer built of repeating glucose monomer units; proteins are polymers built of repeating amino acid monomers; and nucleic acids are polymers built of repeating nucleotide monomers.


Cellulose-a glucose polymer -
Protein-an amino acid polymer

An amino acid
Nucleic acid-a nucleotide polymer

A nucleic acid
Synthetic polymers, such as polyethylene, are much simpler chemically than biopolymers, but there is still a great diversity to their structures and properties, depending on the identity of the monomers and on the reaction conditions used for polymerization. The simplest synthetic polymers are those that result when an alkene is treated with a small amount of a suitable catalyst. Ethylene, for example, yields polyethylene, an enormous alkane that may have a molecular weight up to 6 million $u$ and may contain as many as 200,000 monomer units. Worldwide production of polyethylene is approximately 88 million tons per year.


Polyethylene-a synthetic alkene polymer -
Polyethylene and other simple alkene polymers are called chain-growth polymers because they are formed in a chain-reaction process in which an initiator adds to a carbon-carbon double bond to yield a reactive intermediate. The intermediate then reacts with a second molecule of monomer to yield a new intermediate, which reacts with a third monomer unit, and so on.

Historically, ethylene polymerization was carried out at high pressure (1000-3000 atm) and high temperature $\left(100-250{ }^{\circ} \mathrm{C}\right)$ in the presence of a radical initiator such as benzoyl peroxide. Like many radical reactions, the mechanism of ethylene polymerization occurs in three steps: initiation, propagation, and termination:

- Initiation The polymerization reaction is initiated when a few radicals are generated on heating a small
amount of benzoyl peroxide catalyst to break the weak $\mathrm{O}-\mathrm{O}$ bond. The initially formed benzoyloxy radical loses $\mathrm{CO}_{2}$ and gives a phenyl radical ( $\mathrm{Ph} \cdot$ ), which adds to the $\mathrm{C}=\mathrm{C}$ bond of ethylene to start the polymerization process. One electron from the ethylene double bond pairs up with the odd electron on the phenyl radical to form a new $\mathrm{C}-\mathrm{C}$ bond, and the other electron remains on carbon.

- Propagation Polymerization occurs when the carbon radical formed in the initiation step adds to another ethylene molecule to yield another radical. Repetition of the process for hundreds or thousands of times builds the polymer chain.

- Termination The chain process is eventually ended by a reaction that consumes the radical. The combination of two growing chains is one possible chain-terminating reaction.

$$
2 \mathrm{R}-\mathrm{CH}_{2} \mathrm{CH}_{2} \cdot \longrightarrow \mathrm{R}-\mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2}-\mathrm{R}
$$


Polyethylene-a synthetic alkene polymer -
Ethylene is not unique in its ability to form a polymer. Many substituted ethylenes, called vinyl monomers, also undergo polymerization to yield polymers with substituent groups regularly spaced on alternating carbon atoms along the chain. Propylene, for example, yields polypropylene, and styrene yields polystyrene.

When an unsymmetrically substituted vinyl monomer such as propylene or styrene is polymerized, the radical addition steps can take place at either end of the double bond to yield either a primary radical intermediate $\left(\mathrm{RCH}_{2} \cdot\right)$ or a secondary radical $\left(\mathrm{R}_{2} \mathrm{CH} \cdot\right)$. Just as in electrophilic addition reactions, however, we find that only the more highly substituted, secondary radical is formed.

TABLE 8.1 shows some commercially important alkene polymers, their uses, and the monomers from which they are made.

TABLE 8.1 Some Alkene Polymers and Their Uses

| Monomer |  | Trade or common name of
 polymer |  |
| :--- | :--- | :--- | :--- |
| Ethylene | $\mathrm{H}_{2} \mathrm{C}=\mathrm{CH}_{2}$ | Polyethylene | Uses |
| Propene (propylene) | $\mathrm{H}_{2} \mathrm{C}=\mathrm{CHCH}_{3}$ | Polypropylene | Moldings, rope,
 carpets |
| Chloroethylene (vinyl
 chloride) | $\mathrm{H}_{2} \mathrm{C}=\mathrm{CHCl}$ | Poly(vinyl chloride) | Insulation, films, pipes |
| Styrene | $\mathrm{H}_{2} \mathrm{C}=\mathrm{CHC}_{6} \mathrm{H}_{5}$ | Polystyrene | Foam, moldings |
| Tetrafluoroethylene | $\mathrm{F}_{2} \mathrm{C}=\mathrm{CF}_{2}$ | Teflon | Gaskets, nonstick |
| coatings |  |  |  |


Predicting the Structure of a Polymer -
Show the structure of poly(vinyl chloride), a polymer made from $\mathrm{H}_{2} \mathrm{C}=\mathrm{CHCl}$, by drawing several repeating units.


Strategy -
Mentally break the carbon-carbon double bond in the monomer unit, and form single bonds by connecting numerous units together.


Solution -
The general structure of poly(vinyl chloride) is

PROBLEM Show the monomer units you would use to prepare the following polymers:


8-18 (a) -
(b)

PROBLEM One of the chain-termination steps that sometimes occurs to interrupt polymerization is the
8-19 following reaction between two radicals. Propose a mechanism for the reaction, using fishhook arrows to indicate electron flow.

2

$$
\geqslant \mathrm{CH}_{2} \dot{\mathrm{C}} \mathrm{H}_{2} \longrightarrow \quad \geqslant \mathrm{CH}_{2} \mathrm{CH}_{3}+\frac{>}{<} \mathrm{CH}=\mathrm{CH}_{2}
$$


8-18 (a) - 8.11 Biological Additions of Radicals to Alkenes
The same high reactivity of radicals that enables the alkene polymerization we saw in the previous section also makes it difficult to carry out controlled radical reactions on complex molecules. As a result, there are severe limitations on the usefulness of radical addition reactions in the laboratory. In contrast to an electrophilic addition, where reaction occurs once and the reactive cation intermediate is rapidly quenched by a nucleophile, the reactive intermediate in a radical reaction is not usually quenched. Instead, it reacts again and again in a largely uncontrollable way.


Radical addition -
(Intermediate is not quenched, so reaction does not stop.)

In biological reactions, the situation is different from that in the laboratory. Only one substrate molecule at a time is present in the active site of an enzyme, and that molecule is held in a precise position, with other necessary reacting groups nearby. As a result, biological radical reactions are more controlled and more common than laboratory or industrial radical reactions. A particularly impressive example occurs in the biosynthesis of prostaglandins from arachidonic acid, where a sequence of four radical additions take place. Its reaction mechanism was discussed briefly in Section 6.6.

As shown in FIGURE 8.12, prostaglandin biosynthesis begins with abstraction of a hydrogen atom from C13 of arachidonic acid by an iron-oxy radical to give a carbon radical that reacts with $\mathrm{O}_{2}$ at C 11 through a resonance form. The oxygen radical that results adds to the $\mathrm{C} 8-\mathrm{C} 9$ double bond to give a carbon radical at C8, which adds to the $\mathrm{C} 12-\mathrm{C} 13$ double bond and gives a carbon radical at C 13 . A resonance form of this carbon radical adds at C 15 to a second $\mathrm{O}_{2}$ molecule, completing the prostaglandin skeleton. Reduction of the $\mathrm{O}-\mathrm{O}$ bond then gives prostaglandin $\mathrm{H}_{2}$, called $\mathrm{PGH}_{2}$. The pathway looks complicated, but the entire process is catalyzed with exquisite control by a single enzyme.

Arachidonic acid

FIGURE 8.12 Pathway for the biosynthesis of prostaglandins from arachidonic acid. Steps 2 and 5 are radical addition reactions to $\mathrm{O}_{2}$; steps 3 and 4 are radical additions to carbon-carbon double bonds.


Radical addition - 8.12 Reaction Stereochemistry: Addition of $\mathrm{H}_{2} \mathrm{O}$ to an Achiral Alkene
Most of the biochemical reactions that take place in the body, as well as many organic reactions in the laboratory, yield products with chirality centers. For example, acid-catalyzed addition of $\mathrm{H}_{2} \mathrm{O}$ to 1-butene in the laboratory yields 2 -butanol, a chiral alcohol. What is the stereochemistry of this chiral product? If a single enantiomer is formed, is it $R$ or $S$ ? If a mixture of enantiomers is formed, how much of each? In fact, the 2 -butanol produced is a racemic mixture of $R$ and $S$ enantiomers. Let's see why.

To understand why a racemic product results from the reaction of $\mathrm{H}_{2} \mathrm{O}$ with 1-butene, think about the reaction
mechanism. 1-Butene is first protonated to yield an intermediate secondary carbocation. Because the trivalent carbon is $s p^{2}$-hybridized and planar, the cation has a plane of symmetry and is achiral. As a result, it can react with $\mathrm{H}_{2} \mathrm{O}$ equally well from either the top or the bottom. Reaction from the top leads to ( $S$ )-2-butanol through transition state 1 (TS 1) in FIGURE 8.13, and reaction from the bottom leads to ( $R$ )-2-butanol through TS 2. But the two transition states are mirror images, so they have identical energies, form at identical rates, and are equally likely to occur.

FIGURE 8.13 Reaction of $\mathbf{H}_{2} \mathbf{O}$ with the carbocation resulting from protonation of 1-butene. Reaction from the top leads to $S$ product and is the mirror image of reaction from the bottom, which leads to $R$ product. Because they are energetically identical, they are equally likely and lead to a racemic mixture of products. The dotted $\mathrm{C} \cdots \mathrm{O}$ bond in the transition state indicates partial bond formation.


Radical addition - 8.12 Reaction Stereochemistry: Addition of $\mathrm{H}_{2} \mathrm{O}$ to an Achiral Alkene
As a general rule, the formation of a new chirality center by achiral reactants always leads to a racemic mixture of enantiomeric products. Put another way, optical activity can't appear from nowhere; an optically active product can only result by starting with an optically active reactant or chiral environment (Section 5.12).

In contrast to laboratory reactions, enzyme-catalyzed biological reactions often give a single enantiomer of a chiral product, even when the substrate is achiral. One step in the citric acid cycle of food metabolism, for instance, is the aconitase-catalyzed addition of water to (Z)-aconitate (usually called cis-aconitate) to give isocitrate.

Even though cis-aconitate is achiral, only the ( $2 R, 3 S$ ) enantiomer of the product is formed. As discussed in Section 5.11 and Section 5.12, cis-aconitate is a prochiral molecule, which is held in a chiral environment by the aconitase enzyme during the reaction. In this environment, the two faces of the double bond are chemically distinct, and addition occurs on only the Re face at C2.


Radical addition - 8.13 Reaction Stereochemistry: Addition of $\mathrm{H}_{2} \mathrm{O}$ to a Chiral Alkene
The reaction discussed in the previous section involves an addition to an achiral reactant and forms an optically inactive, racemic mixture of two enantiomeric products. What would happen, though, if we were to carry out the reaction on a single enantiomer of a chiral reactant? For example, what stereochemical result would be obtained from addition of $\mathrm{H}_{2} \mathrm{O}$ to a chiral alkene, such as $(R)$-4-methyl-1-hexene? The product of the reaction, 4-methyl-2-hexanol, has two chirality centers and so has four possible stereoisomers.

Let's think about the two chirality centers separately. What about the configuration at C4, the methyl-bearing carbon atom? Since C 4 has the $R$ configuration in the starting material and this chirality center is unaffected by the reaction, its configuration is unchanged. Thus, the configuration at C 4 in the product remains $R$ (assuming that the relative rankings of the four attached groups are not changed by the reaction).

What about the configuration at C2, the newly formed chirality center? As shown in FIGURE 8.14, the stereochemistry at C 2 is established by reaction of $\mathrm{H}_{2} \mathrm{O}$ with a carbocation intermediate in the usual manner. But this carbocation doesn't have a plane of symmetry; it is chiral because of the chirality center at C4. Because the carbocation is chiral and has no plane of symmetry, it doesn't react equally well from the top and bottom faces. One of the two faces is likely, for steric reasons, to be a bit more accessible than the other, leading to a mixture of $R$ and $S$ products in some ratio other than $50: 50$. Thus, two diastereomeric products, ( $2 R, 4 R$ )-4-methyl-2-hexanol and ( $2 S, 4 R$ )-4-methyl-2-hexanol, are formed in unequal amounts, and the mixture is optically active.


Radical addition - 8.13 Reaction Stereochemistry: Addition of $\mathrm{H}_{2} \mathrm{O}$ to a Chiral Alkene
(2S,4R)-4-Methyl-2-hexanol
(2R,4R)-4-Methyl-2-hexanol
FIGURE 8.14 Stereochemistry of the acid-catalyzed addition of $\mathbf{H}_{\mathbf{2}} \mathbf{O}$ to the chiral alkene, ( $R$ )-4-methyl-1-hexene. A mixture of diastereomeric $2 R, 4 R$ and $2 S, 4 R$ products is formed in unequal amounts because reaction of the chiral carbocation intermediate is not equally likely from top and bottom. The product mixture is optically active.

As a general rule, the formation of a new chirality center by a chiral reactant leads to unequal amounts of diastereomeric products. If the chiral reactant is optically active because only one enantiomer is used rather than a racemic mixture, then the products are also optically active.

PROBLEM What products are formed from acid-catalyzed hydration of racemic ( $\pm$ )-4-methyl-1-hexene? What 8-20 can you say about the relative amounts of the products? Is the product mixture optically active?

PROBLEM What products are formed from hydration of 4-methylcyclopentene? What can you say about the 8-21 relative amounts of the products?


Terpenes: Naturally Occurring Alkenes -
Ever since its discovery in Persia around 1000 A.D., it has been known that steam distillation, the distillation of plant materials together with water, produces a fragrant mixture of liquids called essential oils. The resulting oils have long been used as medicines, spices, and perfumes, and their investigation played a major role in the emergence of organic chemistry as a science during the 19th century.

FIGURE 8.15 The wonderful fragrance of leaves from the California bay laurel is due primarily to myrcene, a simple terpene. (credit: "California Bay Umbellularia californica" by Don Loaire/Flickr, CC BY 2.0)

Chemically, plant essential oils consist largely of mixtures of compounds called terpenoids-small organic molecules with an immense diversity of structure. More than 60,000 different terpenoids are known. Some are open-chain molecules, and others contain rings; some are hydrocarbons, and others contain oxygen. Hydrocarbon terpenoids, in particular, are known as terpenes, and all contain double bonds. For example:

Myrcene (oil of bay)

$\alpha$-Pinene (turpentine)

Humulene (oil of hops)

$\beta$-Santalene (sandalwood oil)
the isoprene rule, they can be thought of as arising from head-to-tail joining of 5 -carbon isoprene units ( 2 -methyl-1,3-butadiene). Carbon 1 is the head of the isoprene unit, and carbon 4 is the tail. For example, myrcene contains two isoprene units joined head to tail, forming an 8-carbon chain with two 1-carbon branches. $\alpha$-Pinene similarly contains two isoprene units assembled into a more complex cyclic structure, and humulene contains three isoprene units. See if you can identify the isoprene units in $\alpha$-pinene, humulene, and $\beta$-santalene.


Terpenes: Naturally Occurring Alkenes -
Terpenes (and terpenoids) are further classified according to the number of 5 -carbon units they contain. Thus, monoterpenes are 10-carbon substances derived from two isoprene units, sesquiterpenes are 15-carbon molecules derived from three isoprene units, diterpenes are 20-carbon substances derived from four isoprene units, and so on. Monoterpenes and sesquiterpenes are found primarily in plants, but the higher terpenoids occur in both plants and animals, and many have important biological roles. The triterpenoid lanosterol, for instance, is the biological precursor from which all steroid hormones are made.

Lanosterol
(a triterpene, $\mathrm{C}_{30}$ )
Isoprene itself is not the true biological precursor of terpenoids. Nature instead uses two "isoprene equivalents"-isopentenyl diphosphate and dimethylallyl diphosphate-which are themselves made by two different routes depending on the organism. Lanosterol, in particular, is biosynthesized from acetic acid by a complex pathway that has been worked out in great detail. We'll look at the subject more closely in Sections 27.5 and 27.7 .

Isopentenyl diphosphate

Dimethylallyl diphosphate


Key Terms -
- anti stereochemistry
- bromonium ion
- carbene, $\mathrm{R}_{2} \mathrm{C}$
- chain-growth polymer
- epoxide
- glycol
- halohydrin
- hydroboration
- hydrogenated
- hydroxylation
- monomer
- oxidation
- oxirane
- oxymercuration-demercuration
- ozonide
- polymer
- reduction
- Simmons-Smith reaction
- stereospecific
- syn stereochemistry


Summary -
With the background needed to understand organic reactions now covered, this chapter has begun the systematic description of major functional groups.
Alkenes are generally prepared by an elimination reaction, such as dehydrohalogenation, the elimination of HX from an alkyl halide, or dehydration, the elimination of water from an alcohol. The converse of this elimination reaction is the addition of various substances to the alkene double bond to give saturated products.
$\mathrm{HCl}, \mathrm{HBr}$, and HI add to alkenes by a two-step electrophilic addition mechanism. Initial reaction of the nucleophilic double bond with $\mathrm{H}^{+}$gives a carbocation intermediate, which then reacts with halide ion. Bromine and chlorine add to alkenes via three-membered-ring bromonium ion or chloronium ion intermediates to give addition products having anti stereochemistry. If water is present during the halogen addition reaction, a halohydrin is formed.

Hydration of an alkene-the addition of water-is carried out by either of two procedures, depending on the product desired. Oxymercuration-demercuration involves electrophilic addition of $\mathrm{Hg}^{2+}$ to an alkene, followed by trapping of the cation intermediate with water and subsequent treatment with $\mathrm{NaBH}_{4}$. Hydroboration involves addition of borane $\left(\mathrm{BH}_{3}\right)$ followed by oxidation of the intermediate organoborane with alkaline $\mathrm{H}_{2} \mathrm{O}_{2}$. The two hydration methods are complementary: oxymercuration-demercuration gives the product of Markovnikov addition, whereas hydroboration-oxidation gives the product with non-Markovnikov syn stereochemistry.


Summary -
Alkenes are reduced by addition of $\mathrm{H}_{2}$ in the presence of a catalyst such as platinum or palladium to yield alkanes, a process called catalytic hydrogenation. Alkenes are also oxidized by reaction with a peroxyacid to give epoxides, which can be converted into trans-1,2-diols by acid-catalyzed hydrolysis. The corresponding cis-1,2-diols can be made directly from alkenes by hydroxylation with $\mathrm{OsO}_{4}$. Alkenes can also be cleaved to produce carbonyl compounds by reaction with ozone, followed by reduction with zinc metal. In addition, alkenes react with divalent substances called carbenes, $\mathbf{R}_{\mathbf{2}} \mathbf{C}$ :, to give cyclopropanes. Nonhalogenated cyclopropanes are best prepared by treatment of the alkene with $\mathrm{CH}_{2} \mathrm{I}_{2}$ and zinc-copper, a process called the Simmons-Smith reaction.

Alkene polymers-large molecules resulting from repetitive bonding of many hundreds or thousands of small monomer units-are formed by chain-reaction polymerization of simple alkenes. Polyethylene, polypropylene, and polystyrene are examples. As a general rule, radical addition reactions are not common in the laboratory but occur frequently in biological pathways.

Many reactions give chiral products. If the reactants are optically inactive, the products are also optically inactive. If one or both of the reactants is optically active, the products can also be optically active.


LEARNING REACTIONS -
What's seven times nine? Sixty-three, of course. You didn't have to stop and figure it out; you knew the answer immediately because you long ago learned the multiplication tables. Learning the reactions of organic chemistry requires the same approach: reactions have to be learned for quick recall if they are to be useful.

Different people take different approaches to learning reactions. Some people make flashcards; others find studying with friends to be helpful. To help guide your study, most chapters in this book end with a summary of the reactions just presented. In addition, the accompanying Study Guide and Student Solutions Manual has several appendixes that organize organic reactions from other perspectives. Fundamentally, though, there are no shortcuts. Learning organic chemistry does take effort.


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

1. Addition reactions of alkenes
a. Addition of $\mathrm{HCl}, \mathrm{HBr}$, and HI (Section 7.7 and Section 7.8)

Markovnikov regiochemistry occurs, with H adding to the less highly substituted alkene carbon and halogen adding to the more highly substituted carbon.

b. Addition of halogens $\mathrm{Cl}_{2}$ and $\mathrm{Br}_{2}$ (Section 8.2)

Anti addition is observed through a halonium ion intermediate.

c. Halohydrin formation (Section 8.3)

Markovnikov regiochemistry and anti stereochemistry occur.

d. Addition of water by oxymercuration-demercuration (Section 8.4) Markovnikov regiochemistry occurs.

e. Addition of water by hydroboration-oxidation (Section 8.5) Non-Markovnikov syn addition occurs.

f. Catalytic hydrogenation (Section 8.6)

Syn addition occurs.

g. Epoxidation with a peroxyacid (Section 8.7)

Syn addition occurs.

h. Hydroxylation with $\mathrm{OsO}_{4}$ (Section 8.7)

Syn addition occurs.

i. Addition of carbenes to yield cyclopropanes (Section 8.9)
(1) Dichlorocarbene addition

(2) Simmons-Smith reaction

2. Hydroxylation by acid-catalyzed epoxide hydrolysis (Section 8.7)

Anti stereochemistry occurs.

3. Oxidative cleavage of alkenes (Section 8.8)
a. Reaction with ozone followed by zinc in acetic acid

b. Reaction with $\mathrm{KMnO}_{4}$ in acidic solution

4. Cleavage of 1,2 -diols (Section 8.8)


Visualizing Chemistry -
PROBLEM Name the following alkenes, and predict the products of their reaction with (1)
8-22 meta-chloroperoxybenzoic acid, (2) $\mathrm{KMnO}_{4}$ in aqueous acid, (3) $\mathrm{O}_{3}$, followed by Zn in acetic acid:
(a)

(b)

PROBLEM Draw the structures of alkenes that would yield the following alcohols on hydration (red = 0). Tell in
8-23 each case whether you would use hydroboration-oxidation or oxymercuration-demercuration.
(a)

(b)

PROBLEM The following alkene undergoes hydroboration-oxidation to yield a single product rather than a
8-24 mixture. Explain the result, and draw the product showing its stereochemistry.

PROBLEM From what alkene was the following 1,2-diol made, and what method was used, epoxide hydrolysis $\mathbf{8 - 2 5}$ or $\mathrm{OsO}_{4}$ ?


Mechanism Problems -
PROBLEM Predict the products for the following reactions, showing the complete mechanism and appropriate 8-26 stereochemistry:
(a)

(b)

(c)

PROBLEM Draw the structures of the organoboranes formed when borane reacts with the following alkenes,
8-27 including the regiochemistry and stereochemistry as appropriate. Propose a mechanism for each reaction.
(a)

(b)

(c)

PROBLEM meta-Chlorobenzoic acid is not the only peroxyacid capable of epoxide formation. For each reaction 8-28 below, predict the products and show the mechanism.

(a)

$+\mathrm{CF}_{3} \mathrm{CO}_{3} \mathrm{H} \longrightarrow$
(b)

PROBLEM Give the mechanism and products for the following acid-catalyzed epoxide-opening reactions,
8-29 including appropriate stereochemistry.
(a)

(b)

(c)

$\qquad$ ?

PROBLEM Which of the reactions below would result in a product mixture that would rotate plane-polarized 8-30 light?
(a)

(b)

(c)

PROBLEM Reaction of 2-methylpropene with $\mathrm{CH}_{3} \mathrm{OH}$ in the presence of $\mathrm{H}_{2} \mathrm{SO}_{4}$ catalyst yields methyl tert-butyl
8-31 ether, $\mathrm{CH}_{3} \mathrm{OC}\left(\mathrm{CH}_{3}\right)_{3}$, by a mechanism analogous to that of acid-catalyzed alkene hydration. Write the mechanism, using curved arrows for each step.

PROBLEM Iodine azide, $\mathrm{IN}_{3}$, adds to alkenes by an electrophilic mechanism similar to that of bromine. If a
8-32 monosubstituted alkene such as 1-butene is used, only one product results:


Mechanism Problems -
(a) Add lone-pair electrons to the structure shown for $\mathrm{IN}_{3}$, and draw a second resonance form for the molecule.
(b) Calculate formal charges for the atoms in both resonance structures you drew for $\mathrm{IN}_{3}$ in part (a).
(c) In light of the result observed when $\mathrm{IN}_{3}$ adds to 1-butene, what is the polarity of the $\mathrm{I}-\mathrm{N}_{3}$ bond? Propose a mechanism for the reaction using curved arrows to show the electron flow in each step.

PROBLEM 10-Bromo- $\alpha$-chamigrene, a compound isolated from marine algae, is thought to be biosynthesized
8-33 from $\gamma$-bisabolene by the following route:

Draw the structures of the intermediate bromonium and cyclic carbocation, and propose mechanisms for all three steps.

PROBLEM Isolated from marine algae, prelaureatin is thought to be biosynthesized from laurediol by the
8-34 following route. Propose a mechanism.

PROBLEM Dichlorocarbene can be generated by heating sodium trichloroacetate. Propose a mechanism for
8-35 the reaction, and use curved arrows to indicate the movement of electrons in each step. What relationship does your mechanism bear to the base-induced elimination of HCl from chloroform?

PROBLEM Reaction of cyclohexene with mercury(II) acetate in $\mathrm{CH}_{3} \mathrm{OH}$ rather than $\mathrm{H}_{2} \mathrm{O}$, followed by treatment 8-36 with $\mathrm{NaBH}_{4}$, yields cyclohexyl methyl ether rather than cyclohexanol. Suggest a mechanism.

Cyclohexene
Cyclohexyl methyl ether

PROBLEM Use your general knowledge of alkene chemistry to suggest a mechanism for the following reaction.
8-37

PROBLEM Treatment of 4-penten-1-ol with aqueous $\mathrm{Br}_{2}$ yields a cyclic bromo ether rather than the expected
8-38 bromohydrin. Suggest a mechanism, using curved arrows to show electron movement.


Mechanism Problems -
4-Penten-1-ol 2-(Bromomethyl)tetrahydrofuran
PROBLEM Hydroboration of 2-methyl-2-pentene at $25{ }^{\circ} \mathrm{C}$, followed by oxidation with alkaline $\mathrm{H}_{2} \mathrm{O}_{2}$, yields
8-39 2-methyl-3-pentanol, but hydroboration at $160{ }^{\circ} \mathrm{C}$ followed by oxidation yields 4-methyl-1-pentanol. Suggest a mechanism.


Reactions of Alkenes -
PROBLEM Predict the products of the following reactions (the aromatic ring is unreactive in all cases). Indicate
8-40 regiochemistry when relevant.

PROBLEM Suggest structures for alkenes that give the following reaction products. There may be more than 8-41 one answer for some cases.
(a)
? $\xrightarrow{\mathrm{H}_{2} / \mathrm{Pd}}$

(b)

(c)
$? \xrightarrow{\mathrm{Br}_{2}}$

(d)

(e)

(f)

PROBLEM Predict the products of the following reactions, showing both regiochemistry and stereochemistry
8-42 where appropriate:
(a)

(b)

(c)

(d)

PROBLEM Which reaction would you expect to be faster, addition of HBr to cyclohexene or to
8-43 1-methylcyclohexene? Explain.
PROBLEM What product will result from hydroboration-oxidation of 1-methylcyclopentene with deuterated
8-44 borane, $\mathrm{BD}_{3}$ ? Show both the stereochemistry (spatial arrangement) and the regiochemistry (orientation) of the product.

PROBLEM The cis and trans isomers of 2-butene give different cyclopropane products in the Simmons-Smith
8-45 reaction. Show the structures of both, and explain the difference.

PROBLEM Predict the products of the following reactions. Don't worry about the size of the molecule;
8-46 concentrate on the functional groups.

PROBLEM Addition of HCl to 1-methoxycyclohexene yields 1-chloro-1-methoxycyclohexane as a sole product.
8-47 Use resonance structures of the carbocation intermediate to explain why none of the alternate regioisomer is formed.

$\xrightarrow{\mathrm{HCl}}$

1-Methoxycyclohexene

1-Chloro-1-methoxy-
cyclohexane

Synthesis Using Alkenes
PROBLEM How would you carry out the following transformations? What reagents would you use in each case?

8-48 (a)

(b)

(c)

(d)


Reactions of Alkenes -
Synthesis Using Alkenes
PROBLEM How would you carry out the following transformations? What reagents would you use in each case?

8-48 (a)

(b)

(c)

(d)

PROBLEM Draw the structure of an alkene that yields only acetone, $\left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}=\mathrm{O}$, on ozonolysis followed by
8-49 treatment with Zn .
PROBLEM Show the structures of alkenes that give the following products on oxidative cleavage with $\mathrm{KMnO}_{4}$
8-50 in acidic solution:
(a) $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CO}_{2} \mathrm{H}+\mathrm{CO}_{2}$
(b) $\left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}=\mathrm{O}+\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CO}_{2} \mathrm{H}$
(c)

(d)

PROBLEM In planning the synthesis of one compound from another, it's just as important to know what not to
8-51 do as to know what to do. The following reactions all have serious drawbacks to them. Explain the potential problems of each.
(a)

(b)

(c)

(d)

PROBLEM Which of the following alcohols could not be made selectively by hydroboration-oxidation of an 8-52 alkene? Explain.
(a)

(b)

(c)

(d)


Polymers -
PROBLEM Plexiglas, a clear plastic used to make many molded articles, is made by polymerization of methyl
8-53 methacrylate. Draw a representative segment of Plexiglas.

Methyl methacrylate

PROBLEM Poly(vinyl pyrrolidone), prepared from $N$-vinylpyrrolidone, is used both in cosmetics and as a
8-54 component of a synthetic substitute for blood. Draw a representative segment of the polymer.

$N$-Vinylpyrrolidone

PROBLEM When a single alkene monomer, such as ethylene, is polymerized, the product is a homopolymer.
8-55 If a mixture of two alkene monomers is polymerized, however, a copolymer often results. The following structure represents a segment of a copolymer called Saran. What two monomers were copolymerized to make Saran?

Saran


General Problems -
PROBLEM Compound A has the formula $\mathrm{C}_{10} \mathrm{H}_{16}$. On catalytic hydrogenation over palladium, it reacts with
8-56 only 1 molar equivalent of $\mathrm{H}_{2}$. Compound $\mathbf{A}$ also undergoes reaction with ozone, followed by zinc treatment, to yield a symmetrical diketone, $\mathbf{B}\left(\mathrm{C}_{10} \mathrm{H}_{16} \mathrm{O}_{2}\right)$.
(a) How many rings does $\mathbf{A}$ have? (b) What are the structures of $\mathbf{A}$ and $\mathbf{B}$ ?
(c) Write the reactions.

PROBLEM An unknown hydrocarbon $\mathbf{A}$ with the formula $\mathrm{C}_{6} \mathrm{H}_{12}$ reacts with 1 molar equivalent of $\mathrm{H}_{2}$ over a
8-57 palladium catalyst. Hydrocarbon $\mathbf{A}$ also reacts with $\mathrm{OsO}_{4}$ to give diol $\mathbf{B}$. When oxidized with $\mathrm{KMnO}_{4}$
in acidic solution, A gives two fragments. One fragment is propanoic acid, $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CO}_{2} \mathrm{H}$, and the other fragment is ketone $\mathbf{C}$. What are the structures of $\mathbf{A}, \mathbf{B}$, and $\mathbf{C}$ ? Write all reactions.

PROBLEM Using an oxidative cleavage reaction, explain how you would distinguish between the following two
8-58 isomeric dienes:


General Problems -
PROBLEM Using an oxidative cleavage reaction, explain how you would distinguish between the following two
8-58 isomeric dienes:

PROBLEM Compound $\mathbf{A}, \mathrm{C}_{10} \mathrm{H}_{18} \mathrm{O}$, undergoes reaction with dilute $\mathrm{H}_{2} \mathrm{SO}_{4}$ at $50{ }^{\circ} \mathrm{C}$ to yield a mixture of two
8-59 alkenes, $\mathrm{C}_{10} \mathrm{H}_{16}$. The major alkene product, $\mathbf{B}$, gives only cyclopentanone after ozone treatment followed by reduction with zinc in acetic acid. Identify $\mathbf{A}$ and $\mathbf{B}$, and write the reactions.


Cyclopentanone -
PROBLEM Draw the structure of a hydrocarbon that absorbs 2 molar equivalents of $\mathrm{H}_{2}$ on catalytic 8-60 hydrogenation and gives only butanedial on ozonolysis.

Butanedial

PROBLEM Simmons-Smith reaction of cyclohexene with diiodomethane gives a single cyclopropane product,
8-61 but the analogous reaction of cyclohexene with 1,1-diiodoethane gives (in low yield) a mixture of two isomeric methylcyclopropane products. What are the two products, and how do they differ?

PROBLEM The sex attractant of the common housefly is a hydrocarbon with the formula $\mathrm{C}_{23} \mathrm{H}_{46}$. On treatment
8-62 with aqueous acidic $\mathrm{KMnO}_{4}$, two products are obtained, $\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{12} \mathrm{CO}_{2} \mathrm{H}$ and $\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{7} \mathrm{CO}_{2} \mathrm{H}$. Propose a structure.

PROBLEM Compound A has the formula $\mathrm{C}_{8} \mathrm{H}_{8}$. It reacts rapidly with $\mathrm{KMnO}_{4}$ to give $\mathrm{CO}_{2}$ and a carboxylic
8-63 acid, B $\left(\mathrm{C}_{7} \mathrm{H}_{6} \mathrm{O}_{2}\right)$, but reacts with only 1 molar equivalent of $\mathrm{H}_{2}$ on catalytic hydrogenation over a palladium catalyst. On hydrogenation under conditions that reduce aromatic rings, 4 equivalents of $\mathrm{H}_{2}$ are taken up and hydrocarbon $\mathbf{C}\left(\mathrm{C}_{8} \mathrm{H}_{16}\right)$ is produced. What are the structures of $\mathbf{A}, \mathbf{B}$, and $\mathbf{C}$ ? Write the reactions.


Cyclopentanone -
PROBLEM How would you distinguish between the following pairs of compounds using simple chemical tests?
8-64 Tell what you would do and what you would see.
(a) Cyclopentene and cyclopentane
(b) 2-Hexene and benzene

PROBLEM $\alpha$-Terpinene, $\mathrm{C}_{10} \mathrm{H}_{16}$, is a pleasant-smelling hydrocarbon that has been isolated from oil of
8-65 marjoram. On hydrogenation over a palladium catalyst, $\alpha$-terpinene reacts with 2 molar equivalents of $\mathrm{H}_{2}$ to yield a hydrocarbon, $\mathrm{C}_{10} \mathrm{H}_{20}$. On ozonolysis, followed by reduction with zinc and acetic acid, $\alpha$-terpinene yields two products, glyoxal and 6-methyl-2,5-heptanedione.

Glyoxal

6-Methyl-2,5-heptanedione
(a) How many degrees of unsaturation does $\alpha$-terpinene have?
(b) How many double bonds and how many rings does it have?
(c) Propose a structure for $\alpha$-terpinene.

PROBLEM Evidence that cleavage of 1,2-diols by $\mathrm{HIO}_{4}$ occurs through a five-membered cyclic periodate
8-66 intermediate is based on the measurement of reaction rates. When diols $\mathbf{A}$ and $\mathbf{B}$ were prepared and the rates of their reaction with $\mathrm{HIO}_{4}$ were measured, it was found that diol A cleaved approximately 1 million times faster than diol B. Make molecular models of $\mathbf{A}$ and $\mathbf{B}$ and of potential cyclic
periodate intermediates, and then explain the results.

A
(cis diol)

B
(trans diol)


Cyclopentanone -
A
(cis diol)

B
(trans diol)

PROBLEM Reaction of HBr with 3-methylcyclohexene yields a mixture of four products: cis- and
8-67 trans-1-bromo-3-methylcyclohexane and cis- and trans-1-bromo-2-methylcyclohexane. The analogous reaction of HBr with 3-bromocyclohexene yields trans-1,2-dibromocyclohexane as the sole product. Draw structures of the possible intermediates, and then explain why only a single product is formed in this reaction.

PROBLEM We'll see in the next chapter that alkynes undergo many of the same reactions that alkenes do. What
8-68 product might you expect from each of the following reactions?

$\begin{cases}\text { (a) } \xrightarrow{\text { 1 equiv } \mathrm{Br}_{2}} & \text { ? } \\ \text { (b) } \xrightarrow[\text { equiv } \mathrm{H}_{2}, \mathrm{Pd} / \mathrm{C}]{ } & \text { ? } \\ \text { (c) } \xrightarrow{\text { 1 equiv } \mathrm{HBr}} & \text { ? }\end{cases}$
PROBLEM Hydroxylation of cis-2-butene with $\mathrm{OsO}_{4}$ yields a different product than hydroxylation of
8-69 trans-2-butene. Draw the structure, show the stereochemistry of each product, and explain the difference between them.


Cyclopentanone -
PROBLEM Compound $\mathbf{A}, \mathrm{C}_{11} \mathrm{H}_{16} \mathrm{O}$, was found to be an optically active alcohol. Despite its apparent
8-70 unsaturation, no hydrogen was absorbed on catalytic reduction over a palladium catalyst. On treatment of $\mathbf{A}$ with dilute sulfuric acid, dehydration occurred and an optically inactive alkene $\mathbf{B}, \mathrm{C}_{11} \mathrm{H}_{14}$, was the major product. Alkene $\mathbf{B}$, on ozonolysis, gave two products. One product was identified as propanal, $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CHO}$. Compound $\mathbf{C}$, the other product, was shown to be a ketone, $\mathrm{C}_{8} \mathrm{H}_{8} \mathrm{O}$. How many degrees of unsaturation does $\mathbf{A}$ have? Write the reactions, and identify $\mathbf{A}, \mathbf{B}$, and C.


Alkynes: An Introduction to Organic Synthesis -
FIGURE 9.1 Synthesizing organic compounds is like conducting a musical group. When in tune, chemists can create highly complex organic compounds. (credit: modification of work "Jazz great visits Navy" by U.S. Navy, Michael Worner/Wikimedia Commons, Public Domain)


CHAPTER CONTENTS - 9.1 Naming Alkynes
9.2 Preparation of Alkynes: Elimination Reactions of Dihalides
9.3 Reactions of Alkynes: Addition of HX and $\mathrm{X}_{2}$
9.4 Hydration of Alkynes
9.5 Reduction of Alkynes
9.6 Oxidative Cleavage of Alkynes
9.7 Alkyne Acidity: Formation of Acetylide Anions
9.8 Alkylation of Acetylide Anions
9.9 An Introduction to Organic Synthesis

WHY THIS CHAPTER? Alkynes are less common than alkenes, both in the laboratory and in living organisms, so we won't cover them in great detail. The real importance of this chapter is that we'll 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. Without the ability to design and synthesize new molecules in the laboratory, many of the medicines we take for granted would not exist and few new ones would be made.

An alkyne is a hydrocarbon that contains a carbon-carbon triple bond. Acetylene, $\mathrm{H}-\mathrm{C} \equiv \mathrm{C}-\mathrm{H}$, the simplest alkyne, was once widely used in industry as a starting material for the preparation of acetaldehyde, acetic acid, vinyl chloride, and other high-volume chemicals, but more efficient routes to these substances using ethylene as starting material are now available. Acetylene is still used in the preparation of acrylic polymers, such as

Plexiglas and Lucite, but is probably best known as the gas burned in high-temperature oxy-acetylene welding torches.

In addition to simple alkynes with one triple bond, research is also being carried out on polyynes-linear carbon chains of alternating single and triple bonds. Polyynes with up to eight triple bonds are thought to be present in interstellar space, and evidence has been presented for the existence of carbyne, an allotrope of carbon consisting of alternating single and triple bonds in long chains of indefinite length. The electronic properties of polyynes are being explored for potential use in nanotechnology applications.

A polyyne detected in interstellar space


CHAPTER CONTENTS - 9.1 Naming Alkynes
Alkyne nomenclature follows the general rules for hydrocarbons discussed in Section 3.4 and Section 7.3. The suffix -yne is used, and the position of the triple bond is indicated by giving the number of the first alkyne carbon in the chain. Numbering the main chain begins at the end nearer the triple bond so that the triple bond receives as low a number as possible.

Begin numbering at the end nearer the triple bond.

> 6-Methyl-3-octyne
> (New: 6-Methyloct-3-yne)

Compounds with more than one triple bond are called diynes, triynes, and so forth; compounds containing both double and triple bonds are called enynes (not ynenes). Numbering of an enyne chain starts from the end nearer the first multiple bond, whether double or triple. When there is a choice in numbering, double bonds receive lower numbers than triple bonds. For example:

1-Hepten-6-yne
(New: Hept-1-en-6-yne)

4-Methyl-7-nonen-1-yne
(New: 4-Methylnon-7-en-1-yne)

As with alkyl and alkenyl substituents derived from alkanes and alkenes, respectively, alkynyl groups are also possible.

1-Butynyl
(an alkynyl group)
(New: But-1-ynyl)
PROBLEM Name the following alkynes:
9-1 (a)

(b)

(c)

(d)

(e)
$\mathrm{CH}_{3} \mathrm{CH}=\mathrm{CHCH}=\mathrm{CHC} \equiv \mathrm{CCH}_{3}$

PROBLEM There are seven isomeric alkynes with the formula $\mathrm{C}_{6} \mathrm{H}_{10}$. Draw and name them.
9-2


CHAPTER CONTENTS - 9.2 Preparation of Alkynes: Elimination Reactions of Dihalides
Alkynes can be prepared by the elimination of HX from alkyl halides in a similar manner as alkenes (Section 8.1). Treatment of a 1,2-dihaloalkane (called a vicinal dihalide) with an excess amount of a strong base such as KOH or $\mathrm{NaNH}_{2}$ results in a twofold elimination of HX and formation of an alkyne. As with the elimination of HX to form an alkene, we'll defer a full discussion of this topic and the relevant reaction mechanisms to Chapter 11.

The starting vicinal dihalides are themselves readily available by addition of $\mathrm{Br}_{2}$ or $\mathrm{Cl}_{2}$ to alkenes. Thus, the overall halogenation/dehydrohalogenation sequence makes it possible to go from an alkene to an alkyne. For example, diphenylethylene is converted into diphenylacetylene by reaction with $\mathrm{Br}_{2}$ and subsequent base treatment.

The twofold dehydrohalogenation takes place through a vinylic halide intermediate, which suggests that vinylic halides themselves should give alkynes when treated with strong base. (A vinylic substituent is one that is attached to a double-bond.) This is indeed the case. For example:


CHAPTER CONTENTS - 9.3 Reactions of Alkynes: Addition of HX and $\mathrm{X}_{2}$
You might recall from Section 1.9 that a carbon-carbon triple bond results from the interaction of two $s p$-hybridized carbon atoms. The two $s p$ hybrid orbitals of carbon lie at an angle of $180^{\circ}$ to each other along an axis perpendicular to the axes of the two unhybridized $2 p_{\mathrm{y}}$ and $2 p_{\mathrm{z}}$ orbitals. When two $s p$-hybridized carbons approach each other, one $s p-s p \sigma$ bond and two $p-p \pi$ bonds are formed. The two remaining $s p$ orbitals form bonds to other atoms at an angle of $180^{\circ}$ from the carbon-carbon bond. Thus, acetylene is a linear molecule with $\mathrm{H}-\mathrm{C} \equiv \mathrm{C}$ bond angles of $180^{\circ}$ (FIGURE 9.2). The length of the $\mathrm{C} \equiv \mathrm{C}$ bond is 120 pm , and its strength is approximately $965 \mathrm{~kJ} / \mathrm{mol}(231 \mathrm{kcal} / \mathrm{mol})$, making it the shortest and strongest known carbon-carbon bond.

FIGURE 9.2 The structure of acetylene, $\mathbf{H}-\mathbf{C} \equiv \mathbf{C}-\mathbf{H}$. The $\mathbf{H}-\mathbf{C} \equiv \mathbf{C}$ bond angles are $\mathbf{1 8 0}^{\circ}$, and the $\mathbf{C} \equiv \mathbf{C}$ bond length is $\mathbf{1 2 0}$ pm. The electrostatic potential map shows that the $\pi$ bonds create a negative belt around the molecule.


CHAPTER CONTENTS - 9.3 Reactions of Alkynes: Addition of HX and $\mathrm{X}_{2}$
As a general rule, electrophiles undergo addition reactions with alkynes much as they do with alkenes. Take the reaction of alkynes with HX, for instance. The reaction often can be stopped with the addition of 1 equivalent of HX, but reaction with an excess of HX leads to a dihalide product. For example, reaction of 1-hexyne with 2 equivalents of HBr yields 2,2-dibromohexane. As the following examples indicate, the regiochemistry of addition follows Markovnikov's rule, with halogen adding to the more highly substituted side of the alkyne bond
and hydrogen adding to the less highly substituted side. Trans stereochemistry of H and X normally, although not always, occurs in the product.

HBr addition

1-Hexyne
2-Bromo-1-hexene
2,2-Dibromohexane
HCl addition

Bromine and chlorine also add to alkynes to give addition products, and trans stereochemistry again results.
$\mathrm{Br}_{2}$ addition

1-Butyne (E)-1,2-Dibromo-1-butene 1,1,2,2-Tetrabromobutane
The mechanism of alkyne addition is similar but not identical to that of alkene addition. When an electrophile such as HBr adds to an alkene, the reaction takes place in two steps and involves an alkyl carbocation intermediate (Section 7.7 and Section 7.8). If HBr were to add by the same mechanism to an alkyne, an analogous vinylic carbocation would be formed as the intermediate.

A vinylic carbocation has an $s p$-hybridized carbon and generally forms less readily than an alkyl carbocation (FIGURE 9.3). As a rule, a secondary vinylic carbocation forms about as readily as a primary alkyl carbocation, but a primary vinylic carbocation is so difficult to form that there is no clear evidence it even exists. Thus, many alkyne additions occur through more complex mechanistic pathways.

A $2^{\circ}$ vinylic carbocation

A $\mathbf{2}^{\circ}$ alkyl carbocation


CHAPTER CONTENTS - 9.3 Reactions of Alkynes: Addition of HX and $\mathrm{X}_{2}$
A $2^{\circ}$ vinylic carbocation

A $\mathbf{2}^{\circ}$ alkyl carbocation

FIGURE 9.3 The structure of a secondary vinylic carbocation. The cationic carbon atom is sp-hybridized and has a vacant $p$ orbital perpendicular to the plane of the $\pi$ bond orbitals. Only one R group is attached to the positively charged carbon rather than two, as in a secondary alkyl carbocation. The electrostatic potential map shows that the most positive regions coincide with lobes of the vacant $p$ orbital and are perpendicular to the most negative regions associated with the $\pi$ bond.

PROBLEM What products would you expect from the following reactions?


CHAPTER CONTENTS - 9.4 Hydration of Alkynes
Like alkenes (Section 8.4 and Section 8.5), alkynes can be hydrated by either of two methods. Direct addition of water catalyzed by mercury(II) ion yields the Markovnikov product, and indirect addition of water by a hydroboration-oxidation sequence yields the non-Markovnikov product.


Mercury (II)-Catalyzed Hydration of Alkynes -
Alkynes don't react directly with aqueous acid but will undergo hydration readily in the presence of mercury(II) sulfate as a Lewis acid catalyst. The reaction occurs with Markovnikov regiochemistry, so the - OH group adds to the more highly substituted carbon and the -H attaches to the less highly substituted one.


An enol -
2-Hexanone (78\%)
Interestingly, the actual product isolated from alkyne hydration is not a vinylic alcohol, or enol (ene $+o l$ ), but is instead a ketone. Although the enol is an intermediate in the reaction, it immediately rearranges into a ketone by a process called keto-enol tautomerism. The individual keto and enol forms are said to be tautomers, a word used to describe two isomers that undergo spontaneous interconversion accompanied by the change in position of a hydrogen. With few exceptions, the keto-enol tautomeric equilibrium lies on the side of the ketone; enols are almost never isolated. We'll look more closely at this equilibrium in Section 22.1.

As shown in FIGURE 9.4, the mechanism of the mercury(II)-catalyzed alkyne hydration reaction is analogous to the oxymercuration reaction of alkenes (Section 8.4). Electrophilic addition of mercury(II) ion to the alkyne gives a vinylic cation, which reacts with water and loses a proton to yield a mercury-containing enol intermediate. In contrast with alkene oxymercuration, however, no treatment with $\mathrm{NaBH}_{4}$ is necessary to remove the mercury. The acidic reaction conditions alone are sufficient to effect replacement of mercury by
hydrogen. Tautomerization then gives the ketone.
FIGURE 9.4 MECHANISM
Mechanism of the mercury(II)-catalyzed hydration of an alkyne to yield a
ketone. The reaction occurs through initial formation of an intermediate enol, which tautomerizes to the ketone.
(1) The alkyne uses a pair of electrons to attack the electrophilic mercury(II) ion, yielding a mercury-containing

vinylic carbocation intermediate.
(2) Nucleophilic attack of water on the carbocation forms a $\mathrm{C}-\mathrm{O}$ bond and yields a protonated mercurycontaining enol.

$\bigcirc$

-

(4) Replacement of $\mathrm{Hg}^{2+}$ by $\mathrm{H}^{+}$ occurs to give a neutral enol.


An enol -
$\bigcirc$

-

(4) Replacement of $\mathrm{Hg}^{2+}$ by $\mathrm{H}^{+}$ occurs to give a neutral enol.

$$
\text { (4) } \mathrm{H}_{3} \mathrm{O}^{+}
$$

(5)

A mixture of both possible ketones results when an unsymmetrically substituted internal alkyne ( $\mathrm{RC} \equiv \mathrm{CR}^{\prime}$ ) is hydrated. The reaction is therefore most useful when applied to a terminal alkyne ( $\mathrm{RC} \equiv \mathrm{CH}$ ) because only a methyl ketone is formed.


A terminal alkyne -
A methyl ketone
PROBLEM What products would you obtain by mercury-catalyzed hydration of the following alkynes?

PROBLEM What alkynes would you start with to prepare the following ketones?

(b)


Hydroboration-Oxidation of Alkynes -
Borane adds rapidly to an alkyne just as it does to an alkene, and the resulting vinylic borane can be oxidized by $\mathrm{H}_{2} \mathrm{O}_{2}$ to give an enol, which tautomerizes to either a ketone or an aldehyde, depending on the alkyne. Hydroboration-oxidation of an internal alkyne such as 3-hexyne is straightforward and gives a ketone, but hydroboration-oxidation of a terminal alkyne is more complex because two molecules of borane often add to the triple bond, complicating the situation. To prevent this double addition, a bulky, sterically encumbered borane such as bis(1,2-dimethylpropyl)borane, known commonly as disiamylborane is used in place of $\mathrm{BH}_{3}$. When a terminal alkyne such as 1-butene reacts with disiamylborane, addition to the triple bond occurs normally, but a second addition is hindered by the bulk of the dialkylborane. Oxidation with $\mathrm{H}_{2} \mathrm{O}_{2}$ then gives an enol, which tautomerizes to the aldehyde.

The hydroboration-oxidation sequence is complementary to the direct, mercury(II)-catalyzed hydration reaction of a terminal alkyne because different products result. Direct hydration with aqueous acid and mercury(II) sulfate leads to a methyl ketone, whereas hydroboration-oxidation of the same terminal alkyne
leads to an aldehyde.

PROBLEM What alkyne would you start with to prepare each of the following compounds by a 9-6 hydroboration-oxidation reaction?
(a)

(b)

PROBLEM How would you prepare the following carbonyl compounds starting from an alkyne (reddish brown
9-7 = Br)?


Hydroboration-Oxidation of Alkynes - 9.5 Reduction of Alkynes
Alkynes are reduced to alkanes by addition of $\mathrm{H}_{2}$ over a metal catalyst. The reaction occurs in two steps through an alkene intermediate, and measurements show that the first step in the reaction is more exothermic than the second.

Complete reduction to the alkane occurs when palladium on carbon ( $\mathrm{Pd} / \mathrm{C}$ ) is used as catalyst, but hydrogenation can be stopped at the alkene stage if the less active Lindlar catalyst is used. The Lindlar catalyst is a finely divided palladium metal that has been precipitated onto a calcium carbonate support and then deactivated by treatment with lead acetate and quinoline, an aromatic amine. The hydrogenation occurs with syn stereochemistry (Section 8.5), giving a cis alkene product.

The alkyne hydrogenation reaction has been explored extensively by the Hoffmann-LaRoche pharmaceutical company, where it is used in the commercial synthesis of vitamin A. The cis isomer of vitamin A produced initially on hydrogenation is converted to the trans isomer by heating.

7-cis-Retinol
(7-cis-vitamin A; vitamin A has
a trans double bond at C7)
An alternative method for the conversion of an alkyne to an alkene uses sodium or lithium metal as the reducing agent in liquid ammonia as solvent. This method is complementary to the Lindlar reduction because it produces trans rather than cis alkenes. For example, 5-decyne gives trans-5-decene on treatment with lithium in liquid ammonia. The mechanism is explained below.


5-Decyne -
trans-5-Decene (78\%)
Alkali metals dissolve in liquid ammonia at $-33^{\circ} \mathrm{C}$ to produce a deep blue solution containing the metal cation and ammonia-solvated electrons. When an alkyne is then added to the solution, reduction occurs by the mechanism shown in FIGURE 9.5. An electron first adds to the triple bond to yield an intermediate anion radical-a species that is both an anion (has a negative charge) and a radical (has an odd number of electrons). This anion radical is a strong base, able to remove $\mathrm{H}^{+}$from ammonia to give a vinylic radical. Addition of a second electron to the vinylic radical gives a vinylic anion, which abstracts a second $\mathrm{H}^{+}$from ammonia to give trans alkene product.

FIGURE 9.5 MECHANISM
Mechanism of the lithium/ammonia reduction of an alkyne to produce a trans alkene.
(1) Lithium metal donates an electron to the alkyne to give an anion radical . . .
2. . . which abstracts a proton from ammonia solvent to yield a vinylic radical.

The vinylic radical accepts another electron from a second lithium atom to produce a vinylic anion . . .
4. . . which abstracts another proton from ammonia solvent to yield the final trans alkene product.

A trans alkene

Trans stereochemistry of the alkene product is established during the second reduction step (3) when the less-hindered trans vinylic anion is formed from the vinylic radical. Vinylic radicals undergo rapid cis-trans equilibration, but vinylic anions equilibrate much less rapidly. Thus, the more stable trans vinylic anion is formed rather than the less stable cis anion and is then protonated without equilibration.

PROBLEM Using any alkyne needed, how would you prepare the following alkenes?
9-8 (a) trans-2-Octene
(b) cis-3-Heptene
(c) 3-Methyl-1-pentene


5-Decyne - 9.6 Oxidative Cleavage of Alkynes
Alkynes, like alkenes, can be cleaved by reaction with powerful oxidizing agents such as ozone or $\mathrm{KMnO}_{4}$, although the reaction is of little value and it is mentioned only for completeness. A triple bond is generally less reactive than a double bond, and yields of cleavage products can be low. The products obtained from cleavage of an internal alkyne are carboxylic acids; from a terminal alkyne, $\mathrm{CO}_{2}$ is formed as one product.


A terminal alkyne - 9.7 Alkyne Acidity: Formation of Acetylide Anions
The most striking difference between alkenes and alkynes is that terminal alkynes are relatively acidic. When a terminal alkyne is treated with a strong base, such as sodium amide, $\mathrm{Na}^{+}{ }^{-} \mathrm{NH}_{2}$, the terminal hydrogen is removed and the corresponding acetylide anion is formed.


Acetylide anion -
According to the Brønsted-Lowry definition (Section 2.7), an acid is a substance that donates $\mathrm{H}^{+}$. Although we usually think of oxyacids $\left(\mathrm{H}_{2} \mathrm{SO}_{4}, \mathrm{HNO}_{3}\right)$ or halogen acids $(\mathrm{HCl}, \mathrm{HBr})$ in this context, any compound containing a hydrogen atom can be an acid under the right circumstances. By measuring dissociation constants of different acids and expressing the results as $\mathrm{p} K_{\mathrm{a}}$ values, an acidity order can be established. Recall from Section 2.8 that a lower $\mathrm{p} K_{\mathrm{a}}$ corresponds to a stronger acid and a higher $\mathrm{p} K_{\mathrm{a}}$ corresponds to a weaker one.

Where do hydrocarbons lie on the acidity scale? As the data in TABLE 9.1 show, both methane ( $\mathrm{p} K_{\mathrm{a}} \approx 60$ ) and ethylene $\left(\mathrm{p} K_{\mathrm{a}}=44\right)$ are very weak acids and thus do not react with any of the common bases. Acetylene, however, has $\mathrm{p} K_{\mathrm{a}}=25$ and can be deprotonated by the conjugate base of any acid whose $\mathrm{p} K_{\mathrm{a}}$ is greater than 25 . Amide ion $\left(\mathrm{NH}_{2}{ }^{-}\right.$), for example, the conjugate base of ammonia ( $\mathrm{p} K_{\mathrm{a}}=35$ ), is often used to deprotonate terminal alkynes.

TABLE 9.1 Acidity of Simple Hydrocarbons


Acetylide anion -
TABLE 9.1 Acidity of Simple Hydrocarbons

| Family | Example | $K_{\mathrm{a}}$ |  | $\mathrm{p} K_{\mathrm{a}}$ |
| :--- | :--- | :--- | :--- | :---: |
| Alkyne | $\mathrm{HC} \equiv \mathrm{CH}$ | $10^{-25}$ | 25 | Stronger
 acid |
| Alkene | $\mathrm{H}_{2} \mathrm{C}=\mathrm{CH}_{2}$ | $10^{-44}$ | 44 |  |
| Alkane | $\mathrm{CH}_{4}$ | $10^{-60}$ | 60 | Weaker
 acid |

Why are terminal alkynes more acidic than alkenes or alkanes? In other words, why are acetylide anions more stable than vinylic or alkyl anions? The simplest explanation involves the hybridization of the negatively charged carbon atom. An acetylide anion has an sp-hybridized carbon, so the negative charge resides in an orbital that has $50 \%$ " $s$ character." A vinylic anion has a $s p^{2}$-hybridized carbon with $33 \% s$ character, and an alkyl anion $\left(s p^{3}\right)$ has only $25 \% s$ character. Because $s$ orbitals are nearer the positive nucleus and lower in energy than $p$ orbitals, the negative charge is stabilized to a greater extent in an orbital with higher $s$ character (FIGURE 9.6).

FIGURE 9.6 A comparison of alkyl, vinylic, and acetylide anions. The acetylide anion, with $s p$ hybridization, has more $s$ character and is more stable. Electrostatic potential maps show that placing the negative charge closer to the carbon nucleus makes carbon appear less negative (red).


Acetylide anion -
PROBLEM The $\mathrm{p} K_{\mathrm{a}}$ of acetone, $\mathrm{CH}_{3} \mathrm{COCH}_{3}$, is 19.3. Which of the following bases is strong enough to 9-9 deprotonate acetone?
(a) $\mathrm{KOH}\left(\mathrm{p} K_{\mathrm{a}}\right.$ of $\left.\mathrm{H}_{2} \mathrm{O}=15.7\right)$
(b) $\mathrm{Na}^{+-} \mathrm{C} \equiv \mathrm{CH}\left(\mathrm{p} K_{\mathrm{a}}\right.$ of $\left.\mathrm{C}_{2} \mathrm{H}_{2}=25\right)$
(c) $\mathrm{NaHCO}_{3}\left(\mathrm{p} K_{\mathrm{a}}\right.$ of $\left.\mathrm{H}_{2} \mathrm{CO}_{3}=6.4\right)$
(d) $\mathrm{NaOCH}_{3}\left(\mathrm{p} K_{\mathrm{a}}\right.$ of $\left.\mathrm{CH}_{3} \mathrm{OH}=15.6\right)$


Acetylide anion - 9.8 Alkylation of Acetylide Anions
The negative charge and unshared electron pair on carbon make an acetylide anion strongly nucleophilic. As a result, an acetylide anion can react with electrophiles, such as alkyl halides, in a process that replaces the halide and yields a new alkyne product.

We won't study the details of this substitution reaction until Chapter 11, but for now you can picture it as happening by the pathway shown in FIGURE 9.7. The nucleophilic acetylide ion uses an electron pair to form a bond to the positively polarized, electrophilic carbon atom of bromomethane. As the new $\mathrm{C}-\mathrm{C}$ bond forms, $\mathrm{Br}^{-}$ departs, taking with it the electron pair from the former $\mathrm{C}-\mathrm{Br}$ bond and yielding propyne as product. We call such a reaction an alkylation because a new alkyl group has become attached to the starting alkyne.


A mechanism for the alkylation reaction of acetylide anion with bromomethane to give propyne. -
(1) The nucleophilic acetylide anion uses its electron lone pair to form a bond to the positively polarized, electrophilic carbon atom of bromomethane. As the new $C-C$ bond begins to form, the $\mathrm{C}-\mathrm{Br}$ bond begins to break in the transition state.

(1)

Transition state
(2) The new $\mathrm{C}-\mathrm{C}$ bond is fully formed and the old $\mathrm{C}-\mathrm{Br}$ bond is fully broken at the end of the reaction.
(2)

Alkyne alkylation is not limited to acetylene itself. Any terminal alkyne can be converted into its corresponding anion and then allowed to react with an alkyl halide to give an internal alkyne product. Hex-1-yne, for instance, gives dec-5-yne when treated first with $\mathrm{NaNH}_{2}$ and then with 1-bromobutane.

Because of its generality, acetylide alkylation is a good method for preparing substituted alkynes from simpler precursors. A terminal alkyne can be prepared by alkylation of acetylene itself, and an internal alkyne can be prepared by further alkylation of a terminal alkyne.

Acetylene
A terminal alkyne


A terminal alkyne -
An internal alkyne
The only limit to the alkylation reaction is that it can only use primary alkyl bromides and alkyl iodides because acetylide ions are sufficiently strong bases to cause elimination instead of substitution when they react with secondary and tertiary alkyl halides. For example, reaction of bromocyclohexane with propyne anion yields the elimination product cyclohexene rather than the substitution product 1-propynylcyclohexane.

PROBLEM Show the terminal alkyne and alkyl halide from which the following products can be obtained. If 9-10 two routes look feasible, list both.
(a)
$\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{C} \equiv \mathrm{CCH}_{3}$
(b)
$\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHC} \equiv \mathrm{CCH}_{2} \mathrm{CH}_{3}$
(c)

PROBLEM How would you prepare cis-2-butene starting from propyne, an alkyl halide, and any other reagents
9-11 needed? This problem can't be worked in a single step. You'll have to carry out more than one reaction.