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Alkenes

Building Bridges to Knowledge

The Carbon-Carbon Double Bond

The simplest molecule with a double bond is ethene or ethylene, C2H4. Following the octet rule, the structure for C2H4 would be

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A sigma, σ, and a pi, π, bonds comprise the double bond.

In order to understand the nature of the double bond, let’s revisit the electron configuration of carbon. The electron configuration of carbon is 1s2 2s2 2p2.

An electron in the 2s atomic orbital of carbon can be promoted to the 2p atomic orbital of carbon. Diagram 3.1 is an illustration of this process.

Diagram 3.1 promotion of an electron from the 2s atomic orbital of carbon to the 2p atomic orbital of carbon

Two “2p” atomic orbitals (each containing one electron) linearly combine with the 2s atomic orbital that contains one electron to form three degenerate 2sp2 hybridized atomic orbitals. Diagram 3.2 is an illustration of this process.

1s

2s

2px 2pz2py 2py 2pz2px

2s

1s

3

Diagram 3.2 the formation of three degenerate 2sp2 hybridized atomic orbitals

A 2sp2 hybridized atomic orbital of carbon linearly combines with a 1s atomic orbital of hydrogen to form a C-H sigma bond. Four of these C-H bonds are formed in the ethylene molecule. Diagram 3.3 describes each C-H single bond.

Diagram 3.3: linear combination of a 1s atomic orbital of hydrogen with a 2sp2 atomic orbital of carbon to form the bonding and anti-bonding sigma molecular orbitals

The third 2sp2 hybridized atomic orbital linearly combines with the

1s

2sp2 2sp2 2sp2

2p

1s

2sp2

σ

σ

*

(2sp2 + 1s)

(2sp2 - 1s)

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2sp2 hybridized atomic orbital of the second carbon to form a (2sp2+ 2sp2) σ bond. The remaining electron in the 2p atomic orbital of carbon linearly combines with the remaining electron in the 2p atomic orbital of the second carbon to form a (2p +2p) π bond. These two types of linear combinations of atomic orbitals comprise the C-C double bond. Diagram 3.4 illustrates the linear combination of atomic orbitals that form the molecular orbitals of the C-C double bond.

Note that the two 2p atomic orbitals linearly combine vertically and the two 2sp2 hybridized atomic orbitals linearly combine horizontally.

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Diagram 3.4 the sigma and pi bonds that comprise the C-C double bond

The overall pictorial representation of the ethylene molecule is illustrated in Diagram 3.5.

π (2p -2p)*

σ(2sp2 -2sp2 )*

σ(2sp2 +2sp2 )

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Diagram 3.5 the bonding in ethylene

The pi bond can be described as the vertical overlap of two “2p” atomic orbitals and the sigma bond can be described as the horizontal overlap of two 2sp2 hybridized atomic orbitals. Together, the pi and sigma bonds form the C-C double bond. Each C-H single bond can be described as the horizontal overlap of a 1s atomic orbital of a hydrogen atom with a 2sp2 hybridized atomic orbital of a carbon atom.

The H-C-H bond angle is 120o and the H-C-C bond angle is 120o. The hybridization around any carbon atom forming a double bond of an alkene can be described in an analogous manner, i.e., the hybridization around any carbon atom of a double bond of an alkene before it participates in the formation of a molecular orbital would be 2sp2. The pi bond is made by the linear combination of a 2p atomic orbital of one carbon atom combining with a 2p atomic orbital of another carbon atom.

C CH

H H

H

σ(2sp2 + 2sp2)

π(2p +2p)

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Nomenclature

Alkenes end in the suffix “ene.” “Ene” represents unsaturation, i.e., a double bond within the carbon structure. Identifying the longest continuous chain that contains the double bond designates the position of the double bond. After locating the carbon backbone containing the double bond, number this continuous chain so that the double bond will receive the lowest number. Once the double bond has been located, name the parent chain as an alkene and indicate the location of the double bond with a number in front of the alkene.

For example, in the following structure,

the carbon backbone is numbered in such a manner that the double bond would be designated as 2, and the structure would be 2-pentene.

Even if there is a group on the longest continuous chain, the numbering would start from the carbon backbone that would designate the double bond with the lowest possible number. For example, the name of the following compound

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would be 5,5,7-trimethyl-2-octene not 2,4,4-trimethyl-6-octene.

It is important to understand that the longest continuous chain may not necessarily be in a straight line; however, the longest continuous chain must contain the double bond.

For example, the name of the following alkene

would be 5-ethyl-5,6-dimethyl-1-nonene.

Unlike sigma, σ, bonds, there is no rotation about the double bond. The inability of the double bond to rotate leads to cis and trans isomers in double bond systems. Cis-trans isomers differ in that two groups can be fixed on adjacent carbons of the double bond on

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opposite side which gives rise to the trans isomer or on the same side which gives rise to the cis isomer.

3-Hexene is an illustration of cis-trans isomerism.

Cis and trans 3-hexene differ only in their location about the double bond and are referred to as geometric isomers.

Table 3.1 lists the small differences between the physical properties of the two isomers.

Compound melting

point boiling point

density, g/mL at 20oC

refractive index

trans-3-hexene 160 K 340 K 0.67797 1.3940 cis-3-hexene N/A 339 K 0.68720 1.3942

Table 3.1 physical properties of 3-hexene

Terminal double bonds, i.e., double bonds at the end of a carbon

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chain, do not exhibit cis-trans isomerism. For example, compounds I a and I b, 2-methyl-1-butene, are identical, i.e., cis-trans isomerism doesn’t exist in this system.

Double bonds with two identical groups attached do not exhibit cis-trans isomerism. For example, compounds II a and II b, 4-ethyl-3-methyl-3-hexene or 3-ethyl-4-methyl-3-hexene are identical.

Now, try drawing geometrical isomers, if possible, for the following structures:

(a)

(b)

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(c)

(d)

Answers

(a) does not exhibit cis-trans isomerism

(b) does not exhibit cis-trans isomerism

(c) Following are the geometric isomers for (c):

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(i) Following are the geometric isomers for (d):

None of the groups in problem (d) are the same. Consequently, how would one differentiate the cis isomer from the trans isomer? Structures III a and III b represent different compounds.

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IIIa

IIIb

Is “III a” the cis isomer or is “III b” the cis isomer? Is “III a” the trans isomer or is “III b” the trans isomer? Is it getting confusing? Clearly the cis and trans notation is only useful when two groups attached to the double bond are identical.

Alkenes of the type delineated in “III a” and “III b” are differentiated using “Z” and “E” notations. “Z” means on the same side, and has its origin in the German word “zusammen” which means together. “E” means on opposite sides, and has its origin in the German word “entgegen” which means opposite. To make this notation applicable, a priority must be assigned to the groups attached to each carbon atom associated with the double bond. Each carbon atom of the double bond has two groups attached. The atoms attached to the carbon atoms of the double bond are given priorities of one or two. If the two atoms with priority 1 attached to the carbon

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atoms of the double bond are on the same side, then the molecule is specified as “Z.” If the two groups with priority 1 attached to the carbon atoms of the double bond are on opposite sides, then the molecule is specified as “E.” Priorities of atoms attached to the double bond are established using the Cahn, Ingold, Prelog prioritization rules (http://en.wikipedia.org/wiki/Cahn-Ingold-Prelog_priority_rule). These rules simply state that a priority must be established for the two atoms attached to each carbon atom of the double bond. If the atoms with priority 1 are on the same side, then the molecule is specified as “Z”. If the atoms with priority 1 are on the opposite side, then the molecule is specified as “E”. The priority is based on atomic numbers. If the two atoms connected to the carbon atom exhibiting the double bond have the same atomic number, then move to the next atom. If these two have the same atomic number, then continue this process until atoms are reached that have different atomic numbers. For example, in structures III a and III b, the Br atom has the higher atomic number compared to the carbon atom of the ethyl group. The second carbon atom of the double bond is attached to two carbon atoms; therefore, the atomic number of the next atom must be considered and the higher atomic number would be the group containing three carbon atoms, “CH3CH2CH2,” since this group contains three carbon atoms and the other group has less than three carbon atoms. Consequently, the propyl group would have priority over the methyl group, and structure “III a” would be (E)-3-bromo-4-methyl-3-heptene and “III b” would be (Z)-3-bromo-4-methyl-3-heptene.

The following site may be helpful in understanding this concept.

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http://en.wikipedia.org/wiki/Cis-trans_isomerism

Try naming compounds “IV a” and “IV b.”

Answers

Compound IV a

(Z)-2-bromo-1-chloropropene

Compound IV b

(E)-2-bromo-1-chloropropene

The IUPAC nomenclature for cyclic alkenes follows analogous rules as alkenes except that the prefix “cyclo” precedes the name of the alkene. For example the carbon skeleton of compound V would be numbered in the following manner.

V

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Since the molecule is cyclic, the carbon atoms containing the center of unsaturation are understood to be C1 and C2 and the methyl groups are numbered in such a fashion that they would receive the smallest numbers with the double bond receiving preference; therefore, the IUPAC name for compound V would be 1,6-dimethylcyclohexene and not 2,3-dimethylcyclohexene.

It would be incorrect to number the system in the following manner.

This numbering system does not give the lowest numbers for the first methyl groups attached to the cyclohexene ring structure.

Suggest an IUPAC name for the following compound.

Answer

1,3-dimethylcyclohexene

The IUPAC name for Compound VI would be (E)-5-cyclobutyl-2-pentene.

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VI

Try drawing structures for (E)-2,2,5,5-tetramethyl-3-heptene, (Z)-6-ethyl-4-nonene, (2E,7E)-3,7-dimethyl-2,7-decadiene, (2E,7Z)-3,7-dimethyl-2,7-decadiene, and (2Z,7Z)-3,7-dimethyl-2,7-decadiene.

Answers

(E)-2,2,5,5-tetramethyl-3-heptene

(Z)-6-ethyl-4-nonene

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(2E,7Z)-3,7-dimethyl-2,7-decadiene

(2Z,7Z)-3,7-dimethyl-2,7-decadiene

The Nature of the Carbon-Carbon Double Bond

Alkenes are characterized by C-C double bonds, and a double bond consists of a sigma, σ, bond and a pi, π, bond. In the double bond of an alkene, the σ bond results from the linear combination of a 2sp2 hybridized atomic orbital of one carbon atom with a 2sp2 hybridized atomic orbital of an adjacent carbon; and the π bond results from the linear combination of an electron in a non-hybridized 2p atomic orbital of a carbon atom with the non-hybridized 2p atomic orbital of the adjacent carbon atom. The resulting π bond is weaker than the σ bond. The 2p atomic orbitals that give rise to the π bond have electron densities above and below the plane and a node at the plane defined by the two carbon atoms and the four hydrogen atoms

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pictured in Diagram 3.6.

Diagram 3.6: sigma and pi bonds in ethene or ethylene

The strength of the π bond is approximately 297 kJ/mol while the strength of the σ bond is between 377 kJ/mol and 387.5 kJ/mol. Therefore, the pi bond is more easily broken than the σ bond.

A double bond is stronger than a single bond because there are two linear combinations of atomic orbitals holding a double bond (the σ bond and the π bond) together compared with one linear combination of atomic orbitals holding a single bond, σ bond, together.

Reactions of Double Bonds

As indicated earlier, the π bond has lower bond energy than a σ bond. This factor along with the accessibility of the π electrons makes these bonds susceptible to rupture. When this occurs, two stronger σ bonds are generally formed. Such reactions are typical of alkenes, and the reactions are exothermic. A variety of reagents will add to the double bond. These reactions are referred to as addition reactions. Diagram 3.7 is a representation of some reagents that will add to the double bond. The stereochemistry, i.e., the spatial

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arrangements of the resulting products will be addressed later in this paper.

Diagram 3.7 some addition reactions of alkenes

Alkenes are more reactive than alkanes. Diagram 3.7 illustrates the reactions that many of the alkenes can undergo, but the manner in which the reagents add depends on the pathway or mechanism of the reaction. A rule that is important to this process is referred to as Markovnikov’s rule. http://wps.prenhall.com/esm_organic_wade_5/5/1361/348631.cw/index.html is an animation that explains Markovnikov’s rule. Click on any of the listed animations to visualize 3-D addition reactions of alkenes. The Website animates:

(a) The addition of HX to alkenes; (b) The addition of water in the presence of H3O+ to alkenes;

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(d) The addition of diborane to alkenes; and

(e) The addition of Br2 to alkenes

Markovnikov’s rule has applications to hydrogen halides, as well as with other electrophilic reagents. The addition of HX to alkenes takes place in such a manner that the hydrogen of the hydrogen halide will attach itself to the carbon atom of the double bond having the greater number of hydrogen atoms attached, and the halide will attach to the carbon of the double bond having the fewer number of hydrogen atoms attached. For example in the following reaction,

Carbon 1 has more hydrogen atoms attached to it than carbon 2. Carbon 1 has two hydrogen atoms attached, and carbon 2 has one hydrogen atom attached. Therefore, when the reaction is complete, the hydrogen of the hydrogen halide will be attached to carbon atom 1 not carbon atom 2:

The products formed from reacting 1-methylcyclohexene (VII) with HBr are 1-bromo-1-methylcyclohexane (VIII a) and 1-bromo-1-methylcyclohexane (VIII b). 1-Bromo-2-methylcyclohexane (IX) is not

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produced in this reaction.

The difference between VIIIa and VIIIb will be explained in a future paper.

The anti-Markovnokov product 1-bromo-2-methylcyclohexane (compounds IXa and IXb) is obtained in the presence of peroxide.

VII IXa IXb

where x + y = 1.00

Compounds analogous to IXa and IXb will be discussed in a future paper.

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These observations can be understood when the mechanism or pathway, the series of elementary steps that explain the formation of product or products, is considered. In the case of the Markovnikov observation, the mechanism of the reaction involves the formation of a carbocation intermediate; and in the case of the anti-Markovnikov product, the mechanism involves the formation of a free radical intermediate.

Therefore, the addition of HX to an alkene, in the absence of peroxide, involves an electrophilic attack by a proton in a slow rate step followed by the nucleophilic attack by a halide in a fast step. The slow step would be the rate determining step. An electrophilic reagent is a reagent that is attacked by a negatively charge entity and a nucleophilic reagent is a reagent that attacks a positive center. The two steps for the addition of HX to a double bond can be represented by the following equations.

When steps 1 and 2 are added, the result would be the following overall reaction.

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The slow step is the rate determining step; consequently, the rate of the reaction would depend upon the concentrations of both the alkene and the hydrogen halide.

or

Therefore, when compound VII reacts with HBr, the mechanism of the reaction would be:

(1)

a tertiary carbocation

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(2)

where x + y = 1.00

The electrophilic attack of the proton on the double bond in step 1 occurs according to Markovnikov’s addition rule where the proton is attached to the carbon of the double bond having the greater number of hydrogen atoms. The result would be the more stable carbocation, the tertiary carbocation. If the hydrogen attached to the carbon of the double bond with the fewer number of hydrogen atoms, then the resulting carbocation would be the secondary

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carbocation, and secondary carbocations are not as stable as tertiary carbocations. The more stable carbocation dominates because the transition state leading to the tertiary carbocation has lower potential energy than the transition state leading to the formation of the less stable secondary carbocation.

A tertiary carbocation is more stable than a secondary carbocation, and a secondary carbocation is more stable than a primary carbocation. Therefore, the relative reactivity of alkenes can be predicted on the basis of the expected stability of the intermediate carbocations formed. This is the underlining principle of Markovnikov’s rule.

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The observed major product produced when HBr is added to an alkene in the presence of peroxide, ROOR, is just the opposite of the major product expected in the absence of peroxide. This observation is referred to as the anti-Markovnikov product. The apparent anti-Markovnikov effect can be explained by the following mechanism.

step 1 R-O-O-R → 2RO∙

step 2 RO∙ + HBr → ROH + Br∙

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step 3

step 4

Step 3 involves the formation of a tertiary free radical. Therefore, the stability of the intermediate is product controlling. Consequently, the addition of the bromine radical in step 3 occurs at the carbon of the double bond that has the greater number of hydrogen atoms. Addition of the bromine radical to this carbon site results in the formation of the more stable tertiary free radical. This mechanism explains the formation of compound IX, 1-bromo-2-methylcyclohexane when compound VII, 1-methylcyclohexene, is treated with HBr in the presence of peroxide, ROOR.

For the reaction

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The mechanism is

(1) Initiation Step

(2) Propagation Step

(3) Propagation Step

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(4) Propagation Step

(5) Propagation Step

We can select x or y since either will lead to the same two products. Let’s select fraction x (that will become apparent in a future paper).

IXa IXb

Where x + y = 1.00

The steps not shown in the mechanism would be a series of termination steps.

The anti-Markovnikov reaction with HX only occurs with HBr not with HCl or HI, because the propagation step that involves the generation of the halo free radical, i.e., step 2 of the mechanism is not thermodynamically favorable.

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Later, you will be able to ascertain the multiple products that are possible when compound VII reacts with HBr in the presence of peroxides. There is a difference between IXa and IXb, and the compounds are not produced in equal amounts. Compound IXb would be the predominant (major) product. This will be resolve in a future paper.

Predict the major products obtained from the addition of HBr in the presence and absence of peroxide to camphene (X); 4-isopropyl-1-methylcyclohexene (XI); and 4-cyclohexyl-2-methylcyclohexene (XII).

Answers:

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http://www.organic-chemistry.org/namedreactions/markovnikovs-rule.shtm provides an overview of the Markovnikov and anti-Markovnikov rules.

Often the addition of hydrogen halides to alkenes is accompanied by the rearrangement of the intermediate carbocation to a more stable carbocation. (Note that usually a rearranged product implies a carbocation mechanism.) The rearrangement involves the initial formation of a carbocation that rearranges to a more stable carbocation. The rearrangement may involve a 1,2-hydride shift or a 1,2-methide shift.

(a) 1,2-Hydride shift

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(b) 1, 2-Methide shift

What major product would you predict from the reaction of 3,3-dimethyl-1-butene with HBr?

Answer

2-bromo-2,3-dimethylbutane

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The Addition of X2 to the Double Bond

The addition of halogens, Br2, Cl2 and I2 to the double bond follows a different path than that described by Markovnikov rule. Incidentally, F

2 does not work well for this process. Studies have shown that the

addition of X2 to the double bond involves a cyclic halonium ion that makes the reaction stereospecific, i.e., different arrangements of atoms in space will lead to different stereoisomers that give rise to stereochemically different products. For example, the addition of Br2 to a trans-alkene would lead to different products than the addition of Br2 to a cis alkene.

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Addition of Br2 to trans-2-butene

meso-2,3-dibromobutane

In the above rotational conformation of meso-2,3-dibromobutane, each atom or constituent species is equal distance from its twin atom or constituent species, i.e., the bromine atom on carbon atom 2 is equal distance from the bromine atom on carbon atom 3; the methyl group attached to carbon atom 2 is equal distance from the methyl group attached to carbon atom 3; and the hydrogen atom on carbon atom 2 is equal distance from the hydrogen atom on carbon atom 3.

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In the above rotational conformation of meso-2,3-dibromobutane, the molecule exhibits a plane of symmetry that reflects both sides of the molecule, i.e., the imaginary plane divides carbon atoms 2 and 3, and creates a mirror reflection for the right half of the molecule as well as the left half of the molecule. This plane is referred to as a plane of symmetry.

Addition of Br2 to cis-2-butene

The products are composed of a racemic mixture (frequently referred to as a racemic modification) where neither molecule exhibits a plane of symmetry or a point of symmetry. These molecules have identical physical properties except they rotate plane polarized light in opposite directions (a discussion in another paper).

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The dihalo products observed from the addition of X2 to the double bond of the cis or trans alkenes are characterized by an anti or trans (opposite side) addition rather than a syn (the same side) addition. In the anti-addition, the positive ion, i.e., the electrophile, adds to one side of the double bond while the negative ion, i.e., the nucleophile, adds to the opposite side. The following mechanisms describe this phenomenon.

(a) Bromination of the cis-alkene leads to what is referred to as the enantiomeric pair or racemic mixture (topics that will be discussed in a future paper). Bromination of the cis-alkene could be represented by the following four equations:

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(1)

(2)

2.3-dibromobutane

(Compound A)

(A more descriptive name will be assigned to compound A in a future paper)

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2,3-dibromobutane

(Compound B)

(A more descriptive name will be assigned to compound B in a future paper)

(3)

2,3-dibromobutane

(Compound B)

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(4)

2,3-dibrombutane

(Compound A)

Bromination of trans-2-butene follows a similar mechanistic pathway, but leads to what is referred to as the meso product. The meso product has the same name; however, a more descriptive name will be assigned to compound A in a future paper.

(ii) The following three chemical equations rationalize the formation of the meso compound.

(1)

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(2)

2,3-dibrombutane

(Compound C)

(A more descriptive name will be assigned to compound C in a future paper)

2,3-dibrombutane

(Compound C)

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(3)

2,3-dibrombutane

(Compound C)

2,3-dibrombutane

(Compound C)

The product of these elementary steps is Compound C. Compound C is referred to as a meso compound.

Let’s get a glimpsed of why compounds A, B, and C resulting from the bromination of trans-2-butene and cis-2-butene are not identical.

The meso compound, Compound C, can rotated around carbon

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atoms 2 and 3 resulting in the formation of an imaginery plane that divides the molecule in such a manner that the two halves of the molecule are mirror reflections of each other. This mirror reflection is referred to as a plane of symmetry. In this case, a plane of symmetry is an indication that the molecule is a meso compound, and the mirror reflection of the meso compound creates a second compound where the spatial arrangements of all the atoms are identical. This means that all forms of Compound C can be superimposed (where there is a match or superposition of each atom for each atom) on one another so that the atoms can match.

For example, treating trans-2-butene with bromine in carbon tetrachloride results in the formation of two possible compounds. These compounds can be divided in such a manner that two halves of the molecules are mirror reflections. Consequently, the resulting two compounds can be superimposed, and are, therefore, identical.

These two compounds are mirror images of each other; therefore, they are identical.

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This is not the case with cis-2-butene. This can be visualized by reacting cis-2-butene with bromine, and drawing the resulting three- dimensional structure of 2,3-dibrombutane. Once the possible structures resulting from the reaction of cIs-2-butene with bromine are drawn, it is obvious that a plane of symmetry cannot be found to produce a mirror reflection of the molecules. Consequently, the two compounds, Compound A and Compound B, resulting from the reaction of cis-2-butene with bromine are not identical.

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The proposed cyclic halonium ion intermediate for the formation of compounds A, B, and C is consistent with the experimental results from the reactions of alkenes with halogens. However, let’s suppose that a carbocation intermediate is proposed in the mechanistic sequence. Would the dibromo alkanes be consistent with predicted and experimental expectations? The answer is no, since carbocation intermediates would give the following results assuming that the carbocation can rotate before a nucleophilic reaction can occur.

(a) cis-2-butene

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Only the racemic mixture is formed when cis-2-butene is treated with bromine in carbon tetrachloride. This implies that the pathway above doe not occur, i.e., a carbocation mechanism does not occur. However, a cyclic halonium ion would support the observed products.

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(b) trans-2-butene

However, only the meso compound, Compound C is observed when trans-2-butene is treated with bromine in carbon tetrachloride. This implies that the pathway drawn above does not occur.

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Table 3.2 summarizes the expectations when Br2 is added to cis and

trans 2-butene

halogen alkene mechanism predicted

products experimentally

observed products

Br2 trans-alkene carbocation meso and racemic mixture

meso

Br2 cis-alkene carbocation meso and racemic mixture

racemic mixture

Br2 trans-alkene cyclic halonium ion

meso meso

Br2 cis-alkene cyclic halonium ion

racemic mixture

racemic mixture

Table 3.2: bromination of cis and trans 2-butene

The products predicted from the cyclic halonium ion mechanism are consistent with the experimentally observed results. This is not the case for the carbocation mechanism. The predicted products from the carbocation mechanism differ from what is experimentally observed; consequently, the halonium ion mechanism must be the accepted mechanism in the halogenations of alkenes.

There is additional evidence in support of the halonium ion mechanism. The bromination of cyclohexene produces only the trans-product. If the mechanism was by way of the formation of a carbocation, then the cis and trans 1,2-dibromocyclohexane would be produced.

If the mechanism for the formation of 1,2-dibromocyclohexane is via the formation of a cyclic bromonium ion, then only the trans product

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would be obtained; however, the trans product can be formed by two different pathways.

If the mechanism for the formation of 1,2-dibromocyclohexane is via the formation of the carbocation mechanism, then both the cis and trans isomers would be expected to be obtained since the carbocation is flat and the nucleophile can attack from either side.

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The cis product is not observed; therefore, halogenation of alkenes must not follow a carbocation mechanism, but must take place through a cyclic halonium ion since only the trans product is observed.

Other Reactions of the C-C Double Bond

The reactions of alkenes usually involve additions to the double bond. These reactions include ozonolysis, carbene insertion, catalytic hydrogenation, addition of sulfuric acid, hydration in the presence of a mineral acid, addition of hypohalous acids, addition of free radicals, addition of peroxy acids, oxidation with dilute potassium permanganate, oxidation with osmium tetroxide, and

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addition of diborane.

Addition of ozone to the double bond

Compounds with more than one double bond will also add ozone to each double bond. The oxidation products will depend on the groups attached to the carbon atoms that form the double bond. If a hydrogen atom is attached to the carbon atom of the double bond, then an aldehyde will be produced. If two hydrogen atoms are attached to the carbon of the double bond, then formaldehyde will be produced. If two alkyl groups are attached to the carbon of the double bond, then a ketone will be produced.

Following are some examples of how ozonolysis works with alkenes:

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Ozonolysis of 2-methyl-2,4-hexadiene

Ozonolysis of 2-methyl-2,4-hexadiene leads to the formation of acetone (XIII), glyoxal or ethanedial (XIV) and acetaldehyde (XV).

Ozonolysis of alkenes can be used for structural identification since the products give insight into the location of the double bond in the alkene. For example, ozonolysis of 1-phenylpropene would produce acetaldehyde (XV) and benzaldehyde (XVI).

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On the other hand, ozonolysis of 3-phenylpropene would produce formaldehyde (XVII) and phenylethanal (XVIII).

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The mechanism of ozonolysis proceeds through the formation of a molozonide first. The molozonide then rearranges to form the ozonide.

(1)

(2)

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(3)

Cleavage of the ozonide can be accomplished with Zn in water to produce formaldehyde, ketones, and/or aldehydes depending on the attachments to the double bonds.

(1)

(2)

The sum of the steps of the mechanism gives the stoichiometry of the reactants and products of the reaction.

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A dynamic visualization of the formation of the molozonide then rearrangement to the ozonide, and cleavage with zinc can be observed at http://bcs.whfreeman.com/vollhardtschore4e/cat_010/ch12/12010-04.htm.

The R groups can be an alkyl group, an aryl group, or hydrogen atoms.

Suggest products for the ozonolysis of the following compounds.

(a)

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(b)

Answers

(a)

(b)

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Even though ozonolysis will produce carbonyl compounds from alkenes, it is a less desirable pathway for the syntheses of aldehydes and ketones. There are much better synthetic methods for making carbonyl compounds.

For more on ozonolysis, please look at the following website:

http://www.organic-chemistry.org/namedreactions/ozonolysis-criegee-mechanism.shtm

Carbenes

Carbenes have the general formula R2C:. If the R groups are

hydrogen, H2C:, the species is referred to as methylene. As

expected, carbenes are very reactive chemical species. They will add to double bonds to produce cyclopropane or cyclopropane derivatives.

Equations (1) and (2) below are examples of carbenes reacting with alkenes to form cyclopropane or cyclopropane derivatives.

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(1)

(2)

Theoretically, the cyclopropane product, bicycle[4.1.0]heptane, in equation 2 can assume both a cis and trans form, but the reaction product is always cis; therefore, the reaction is stereospecific.

Could you suggest a rationale for the dominance of the cis isomer?

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Answer

The trans isomer would be too strained.

Carbenes are also of the type X2C:, where X represents Cl, I or Br.

Carbenes are highly reactive; therefore, they are generated in situ in the presence of the alkene. A couple of common carbene generators are:

1.

(a) RCHI2 + Zn-Cu → [IRCHZnI] + Cu

(b) IRCHZnI → ZnI2 + :CHR

or

2. K+ -OC(CH3)

3 + CHX

3 → X

2C: + KCl + HOC(CH

3)

3

In reaction 1, the diiodoalkane reacts with ZnCu alloy to give the unstable IRCHZnI [step (a)] which decomposes to the carbene [step (b)].

Reaction 2 uses the powerful base potassium tert-butoxide to generate the carbene from trihalomethanes.

Catalytic Hydrogenation

Alkanes can be reduced to alkanes by catalytic hydrogenation.

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For example cyclohexene can be converted to cyclohexane by hydrogen and finely divided nickel.

Catalytic hydrogenation occurs on the surface of the catalyst. Hydrogen is adsorbed on the surface of the catalyst, and the hydrogen adds “syn” to the double bond.

Figure 3.1 is an illustration of the Activated Complex during catalytic hydrogenation of alkenes.

Figure 3.1 Activated Complex during Catalytic hydrogenation of Alkenes

63

Addition of Sulfuric Acid

Sulfuric Acid adds to the double bond producing alkyl sulfonic acids. Following is an illustration of the addition of sulfuric acid to a double bond. The reaction follows Markovnikov’s rule.

Addition of Water

Water, H2O, adds to double bonds in the presents of mineral acids to form alcohols. The reaction follows the Markovnikov’s rule, and is traditionally referred to as a hydration reaction.

64

Addition of Hypohalous Acid

Hypohalous acid, HOX, adds to double bonds, and the reaction follows Markovnikov’s rule. The reaction is carried out in the presence of the diatomic halogen molecule in water. The halogen reacts with water to form hypohalous acid which in turn reacts with the center of unsaturation, the double bond, to produce the halohydrin. The reaction is referred to as halohydrin formation which is illustrated by the following chemical equations.

X2 + HOH → HOX + HX where X = Cl

2 and Br

2

65

XVII

Notice that the formation of chorohydrin product XVII, 2-chloro-1-methylcyclohexanol, results in the OH attached to the carbon atom containing the fewer number of hydrogen atoms. This means that, in the addition reaction, the positive charge is on the halogen atom, and the negative charge is on the oxygen atom. This polarity is in agreement with the fact that the electronegativity of oxygen is greater than the electronegativity of chlorine.

Addition of Free Radicals

Free radicals add to the double bond to form products that are not too surprising if you understand the mechanism of the reaction. For example, bromotrichloromethane can react with 1-octene in the presence of hydrogen peroxide to form 3-bromo-1,1,1-trichlorononane, compound XXIII.

66

XXIII

The formation of compound XXIII from bromotrichloromethane in the presence of hydrogen peroxide follows a free radical mechanism. The mechanism involves a series of steps that include the formation of stable free radical intermediates. The first step of the mechanism involves the homolytic cleavage of the oxygen-oxygen bond of hydrogen peroxide.

Step 1

Step 2 is a propagation step that produces the more stable trichloromethyl radical. This radical is stabilized by resonance.

HO

OH

..

.. ....

..

..O

H.

2

67

Take a few minutes and draw the resonance structures for the trichloromethyl radical.

Answer

Step 3 of the mechanism is another propagation step that produces the more stable alkyl radical, i.e., a secondary radical versus a primary radical.

..

..

HO . + C

Cl

Cl

ClBr

..

..

..

..

:

:

:

: :

OH

....

Br

..

....

..:

:+

::

:

:

:

..

..

..

..

Cl

Cl

Cl

C.

68

The secondary alkyl radical is more stable than the primary alkyl radical; therefore, the secondary radical would be produced preferentially over the primary radical.

secondary alkyl radical

primary alkyl radical

Step 4 of the mechanism is another propagation step that leads to the formation of the product and regenerates another trichloromethyl radical.

69

3-bromo-1,1,1-trichlorononane

The four-step mechanism explains why 3-bromo-1,1,1-trichlorononane is the primary product rather than 1-bromo-2-trichloromethyloctane.

The trichlormethyl radical generated can react with another 1-octene molecule until all the starting material is converted into the product. In addition, there are several possible free radical quenching steps that do not lead to the formation of primary product, but could explain side products that may form.

Oxidation of alkenes to diols can occur by:

(a) The Addition of Cold Potassium Permanganate to Alkenes

(b) The Addition of Peroxyacid Followed by Acid Hydrolysis to Alkenes

(c) The Addition of Osmium Tetroxide to Alkenes Followed by

70

Treatment with t-Butyl Hydroperoxide

Cold dilute potassium permanganate or peroxyacid followed by acid hydrolysis or osmium tetroxide followed by treatment with t-butyl hydroperoxide or hydrogen sulfide react with carbon-carbon double bonds to form vicinal (vic)-diols. These reactions are referred to as glycol formation reactions.

The addition of cold dilute potassium permanganate to alkenes to produce vicinal diols in the cis configuration can be represented by the following balanced chemical equation:

The addition of peroxyacid to alkenes followed by acid hydrolysis to produce vicinal diols in the trans configuration can be represented by the following chemical equations.

71

The addition of osmium tetroxide to alkenes followed by oxidation with tert-butyl hydroperoxide to produce vicinal diols in the cis configuration can be represented by the following chemical equations:

(a)

(b)

72

Following are examples of the formation of vic (vicinal) glycols by reacting cyclohexene with cold dilute potassium permanganate, peroxyacid followed by acid hydrolysis, and osmium tetroxide followed by treatment with t-butyl hydroperoxide

Cyclohexene with Cold Dilute Potassium Permanganate:

cis glycol

Cyclohexene with Peroxyacid Followed by Acid Hydrolysis:

trans glycol

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Cyclohexene with (a) osmium tetroxide followed by (b) treatment with t-butyl hydroperoxide

(a)

(b)

cis glycol

This paper will examine the details of glycol formation by way of dilute potassium permanganate and osmium tetroxide. Also, the formation of vicinol glycols from alkenes with peroxyacids followed by acid hydrolysis will be discussed in this paper.

Following is the mechanism for the formation of a glycol from the addition of dilute potassium permanganate to an alkene. The series of elementary steps rationalize the formation of products. In addition, a mechanism should be supported by kinetic data; therefore, a definitive mechanism must be supported by a set of

74

careful experiments. The mechanisms suggested in this paper and in future papers are scientific guesses based on knowledge and intuition. The definitive proof of a mechanism should be based on a planned set of reproducible experiments. Nevertheless, in the absence of experimental data, a suggested mechanism should follow a series of elementary steps that result in the production of stoichiometric quantities of reactants and products. This process is demonstrated in the elementary steps that explain the formation of vicinal diols resulting from the reaction of dilute potassium permanganate with alkenes.

To understand the reaction and gain insight into the multiple steps that lead to the formation of the vicinal glycol, the reduction-oxidation (REDOX) balanced equation between an alkene and dilute potassium permanganate in acidic medium should be considered.

The oxidation half-reaction of the REDOX reaction is

and the reduction half-reaction of the REDOX reaction is

The number of electrons lost must equal the number of electrons gain; therefore, equation (1) must be multiplied by 3 and equation (2) must be multiplied by 2 to give:

75

Oxidation

Reduction

Adding equations (1) and (2) would give the following results:

or

This is the net ionic equation for the REDOX reaction. This reaction can also be written in its molecular form.

76

Note that only two moles of permanganate will react with three moles of alkene to produce 3 moles of the diol. This information gives insight into a possible mechanism of the reaction.

The following fourteen (14) steps could represent a suggested mechanism for the formation of vicinal glycols from alkenes and dilute potassium permanganate.

(1)

Mn has oxidation state equal to +7 Mn has oxidation state equal to +5

77

(2)

Mn has oxidation state equal to +5

(3)

Mn has oxidation state equal to +5

78

(4)

Mn has oxidation state equal to +5

(5)

Mn has oxidation state equal to +5

(6)

79

(7)

(8)

(9)

80

(10)

(11)

Mn has oxidation state equal to +3

(12)

81

(13)

Mn has oxidation state equal to +3

(14)

Mn has oxidation state equal to +4

Following are steps that may explain the formation of manganese (IV) oxide from the interaction of HMnO

3 and HMnO

2 to produce

manganese (IV) oxide.

(a)

82

(b)

(c)

manganese (IV) oxide

Mn has oxidation state equal to +4

Manganese can have an oxidation state of +7, +6, +5, +4, +3, and +2.

In reality, the structure of manganese (IV) oxide is more complicated than the simple structure in 14 (c). Electron Spin Resonance (ESR) spectrometry suggests that the complex exhibits unpaired electrons, and the complex is paramagnetic. A more representative picture of manganese (IV) oxide is a crystal lattice structure that suggests eight oxygen atoms surrounding each manganese atom.

83

The structure above is a partial crystalline lattice structure of manganese (IV) oxide showing that each manganese atom is surrounded by eight oxygen atoms.

Adding steps 1 through 14 would give the net ionic equation for the REDOX reaction:

84

The mechanism for the formation of a glycol with osmium tetroxide would follow a series of elementary steps that clearly makes an inextricable connection with inorganic chemistry. Before we begin this discussion, let’s get some insight into the structural formula of osmium tetroxide.

Valance Shell Electron Pair Repulsion (VSEPR) Theory suggests that the geometry of OsO4 is tetrahedral, AL44

,

VSEPR Theory Tetrahedral Geometry

Bond Angle Equal to 109.5o

The electron configuration for Os (76 electrons) is [Xe]6s24f145d6; therefore, Os has 8 (6s25d6) electrons in its outer shell.

85

To understand the bonding in osmium tetroxide, six (6) electrons in the 5d atomic orbitals must be promoted to the 6p and 6d atomic orbitals. One electron in the 6s atomic orbital is also promoted to a higher energy level. This may be visualized using the following diagram.

86

87

Four (4) degenerate 6sd3 atomic orbitals linearly combine with four (4) separate 2p atomic orbitals of four oxygen atoms. The 2p atomic orbitals of each oxygen atom has one electron, and the four degenerate 6sd3 atomic orbitals have one electron each. The four (4) Os-O sigma, σ, bonding molecular orbitals formed can be described as the linear combination of 6sd3 atomic orbital of Os with 2p atomic

orbital of oxygen to form four σ (6sd3 + 2p) bonds or molecular

orbitals. The O-Os-O bond angle is approximately 109.5o. Three (3) 6p atomic orbitals of osmium linearly combine with three 2p atomic orbitals of three oxygen atoms to form three Os-O π bonds

88

described as π (6p + 2p) bonds or bonding molecular orbitals. Each atomic orbital contains one electron, and the resulting bonding molecular orbitals contain two paired electrons each. The fourth π bond results from the linear combination of a 6d atomic orbital of Os with a 2p atomic orbital of oxygen. Each atomic orbital contains one electron, and the resulting bonding molecular orbital contains two paired electrons. Incidentally, “d” atomic orbitals can linearly combine with “p” atomic orbitals to form “pi” bonds. The most probable “d” atomic orbital for the formation of a “pi” bond is the 6dz

2 atomic orbital since it resembles a 6p atomic orbital.

Following is the most likely structure for the Osmium tetroxide. One bond was arbitrarily selected to exhibit the “d” atomic orbital participation in the double bond. Any of the four double bonds could have been selected to show “d” atomic orbital sharing.

X-ray analyses of crystalline osmium tetroxide suggest that the

89

compound is tetrahedral; that the O-Os-O bond angle is approximately 109o ; and that each Os-O bond length is 166 pm (1.66 x 10-10 m) . These data strongly contribute to the understanding that osmium tetroxide is a tetrahedral molecule.

The formation of glycols from alkenes and osmium tetroxide occurs in two steps. The first step is the formation of a cyclic osmate ester, and the second step is the cleavage of the cyclic osmate ester to form the cis-glycol.

1. Formation of A Cyclic Osmate Ester

cyclic osmate ester

2. Cleavage of the Osmate Ester

cis-glycol

90

The following elementary steps provide insight into the formation of cis glycols by way of osmate esters followed by cleavage of the osmate ester with t-butyl hydroperoxide.

The first part of the process, i.e., the formation of the cyclic osmate ester, is a one step bimolecular reaction.

cyclic osmate ester

Cyclic osmates are stable, but can be cleaved by a variety of reagents. One such reagent is t-butyl hydroperoxide, and another is hydrogen sulfide.

Following are a series of elementary steps that explain the formation of the cis glycol when the cyclic osmate ester is cleaved with t-butyl hydroperoxide in the presence of a base like sodium hydroxide and t-butyl alcohol as a solvent.

91

1.

(The reaction is written twice, because two moles of the t-butyl hydroperoxide are required)

+CH3 C

CH3

CH3

OOH -OH

OO-

CH3

CH3

CCH3 + H2O

+CH3 C

CH3

CH3

OOH -OH

OO-

CH3

CH3

CCH3 + H2O

92

2.

3.

4.

93

5.

6.

94

7.

8.

The sum of the steps of the proposed mechanism will give the stoichiometric quantities of the reactants and the products. The sum of steps 1 through 8 will give the following reaction.

95

Potassium permanganate and osmium tetroxide will react with alkenes to produce cis glycols. Peroxyacids will react with alkenes followed by acid hydrolysis to produce trans glycols.

(1)

This bimolecular step occurs in the following manner.

96

activated complex

Once the epoxide is formed, it reacts with dilute acid to form the trans-glycol by the following mechanism..

(1)

(2)

97

(3)

Hydroboration-Oxidation

Diborane adds to carbon-carbon double bonds followed by treatment with hydrogen peroxide to form alcohols that have an apparent anti-Markovnikov’s arrangement. This reaction is referred to as a hydroboration-oxidation reaction. Dr. Herbert C. Brown (05/22/1912-12/19/2004) pioneered the work in organoboranes at Purdue University. He received the 1979 Nobel Prize in chemistry for his revolutionary work in the syntheses of organoboranes from alkenes. Organoboranes can be used to synthesize leading medicinals including antidepressants and cholesterol-lowering drugs.

Following is a general reaction that represents the addition of diborane to alkenes to form alkylboranes.

98

Once the alkylboranes have been formed, they can be oxidized with hydrogen peroxide in basic media to form apparent anti-Markovnikov alcohols. A general equation for the oxidation of organoboranes can be represented by the following chemical equation.

99

Following is an illustration of H. C. Brown’s hydroboration-oxidation reaction for the synthesis of 2-methylcyclohexanol from 1-methylcyclohexene.

1-methylcyclohexene

2-methylcyclohexanol

The product, 2-methylcyclohexanol, has the OH group on the carbon atom with the fewer number of carbon atoms.

The following mechanism represents a pathway that explains hydroboration-oxidation reactions.

100

(1) HOOH + -OH → HOO- + HOH

(2)

101

(3)

(4)

(5)

HOOH + -OH → HOO- + HOH

102

(6)

(7)

103

(8)

(9)

HOOH + -OH → HOO- + HOH

(10)

104

(11)

(12)

The sum of elementary steps (1) -(12) gives the reactants and products with their stoichiometric quantities. Hydroxide, -OH, is the catalyst for the reaction; therefore, it does not appear as a reactant or a product.

105

Preparation of Alkenes

Alkenes are commonly prepared by elimination reactions. These reactions may involve the loss of water from an alcohol (dehydration reaction); the loss of hydrogen halide from an alkyl halide (dehydrohalogenation reaction); the loss of halogen, X

2, from alkyl

dihalides (dehalogenation reaction); and addition of hydrogen to acetylenic compounds.

Dehydration of Alcohols

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phenylethene or styrene

Dehydrohalogenation of Alkyl Halides

XXIV

There is another possible structure in addition to compound XXIV, 1-methylcyclohexene, but it is not the major product of the reaction. The mechanism for this reaction will be discussed in a future paper.

Alkenes can be synthesized from the dehalogenation of alkyl dihalides. The reaction scheme takes place in the presence of a metal, and Zn is a popular metal used for this transformation.

107

A plausible mechanism for the reaction could include the following series of elementary steps. Remember that an actual mechanism involves an experimental process that includes acquiring kinetic data, examining the stereospecificity, if any, of the products, and designing cleaver experimental processes for trapping intermediates. If experimental data are not available to support the proposed mechanism, then the mechanism is merely a scientific guess. The mechanisms in this paper and future papers are, for the most part, a combination of scientific guesses and data from established experiments.

The first step in the mechanism would probably result in the formation of a radical anion formed from the transfer of an electron from zinc to the dihalide as represented in the following equations. It doesn’t make a different which halide atom receives the extra electron from the Zn.

(1)

The second step in the mechanism would be a transfer of the

108

electron remaining on the outer shell of Zn+to form Zn++ and the carbanion that will eliminate the halide ion to form the more stable alkene described in the final step of the mechanism.

(2)

(3)

Step (3), the final step in the mechanism, may be thought of as progressing in a manner that would lead to the more stable trans isomer if cis-trans isomerism is possible, and if the molecule can assume a position that would lead to the trans isomer.

109

Cracking

Alkenes can be prepared by the cracking or dehydrogenation of alkanes; however, this is not a very good method for synthesizing alkenes since it would result in mixtures of products.

Platinum or palladium catalysts along with high temperatures are required for the dehydrogenation of alkanes to alkenes. The reaction is difficult to control as to the location of the double bond; therefore, it is not a reaction of choice for laboratory synthesis of alkenes.

X equals the fraction of trans-2-butene, y equals the fraction of cis-

110

2-butene, and z equals the fraction of 1-butene. The sum of x, y, and z is 1.00 (x + y + z = 1.00). X is greater than y, and y is greater than z. According to Diagram 3.7, the trans alkene would be more stable than the cis-alkene, and the cis-alkene is more stable than the alkene with one alkyl group attached.

Addition of Hydrogen to Alkynes

Alkenes can be prepared by adding H2 to Alkynes. The following reactions illustrate catalytic addition of hydrogen to alkynes (alkyne molecules contain triple bonds, a topic that will be discussed in a future paper).

(1)

H2/Pd/CaCO

3 is referred to as the Lindlar Catalyst.

111

(2)

Pd with a poisoned surface can be used with the hydrogenation of alkynes to alkenes. Catalytic hydrogenation of alkynes leads to the formation of the cis alkene. The reaction follows the same pathway that was described in the paper on “Alkanes” where hydrogen is adsorbed on the surface of the catalyst and then adds to the alkyne. Since the surface of the catalyst is poisoned, the reaction stops at the alkene.

If one desires to prepare the trans alkene from an alkyne, then chemical reduction is used. The trans isomer is accomplished by treating the alkyne with sodium or lithium in liquid ammonia. The following example, reaction (3), illustrates the chemical reduction of an alkyne to an alkene.

112

(3)

The mechanism of this reaction will be discussed in a future paper.

Elimination Reactions

When two or more alkenes are possible from an elimination reaction, the alkene containing the largest number of alkyl groups on the carbon-carbon double bond predominates since alkyl groups stabilize the double bond. Consequently, the major product from the dehydration of 2,3-dimethyl-2-butanol, compound XXV, is 2,3-dimethyl-2-butene, compound XXVI, not 3-methyl-1-butene, compound XXVII.

113

Where

X% + (100-X)% = 100%

and

X% > (100-X)%

If compound XXVIII, 2-halo-2,3-dimethylbutane, were treated with base in the presence of an alcohol, the major elimination product would be compound XXVI, 2,3-dimethyl-2-butene. 2,3-Dimethyl-2-butene is the same major product produced when compound XXV, 2,3-dimethyl-2-butanol, is treated with sulfuric acid.

114

Where

X% + (100-X)% = 100%

and

X% > (100-X)%

For dehydration and dehydrohalogenation reactions, the order of removal of H is 3o > 2o > 1o. Therefore, the stability of alkenes decreases in a manner that is illustrated in Diagram 3.7.

115

Diagram 3.7: Order of Stability of Alkenes

116

Dehydration of Alcohols

Alkenes can be made by the dehydration of alcohols. The oxygen of alcohols is highly susceptible to protonation in a mineral acid. Once the alcohol is protonated, the facile loss of water produces the more stable carbocation, and the stable carbocation leads to the formation of the more stable alkene. When possible, carbocations can rearrange by way of 1,2-hydride shifts or 1,2-methide shifts to form more stable carbocations.

2-Methyl-2-butanol will dehydrate to 2-methyl-2-butene in the presence of water, sulfuric acid or hydrochloric acid and heat following a three step sequence schema.

Step 1

2-methyl-2-butanol

Step 2

117

Step 3

2-methyl-2-butene

The same alkene, 2-methyl-2-butene, would be the major product when 3-methyl-2-butanol is treated with water in the presence of a mineral acid. The mechanism would be similar except that the secondary carbocation formed in step (2) of the mechanism undergoes a 1,2-hydride shift to form the more stable tertiary carbocation (step 3) which ultimately loses a hydrogen ion to form 2-methyl-2-butene (step 4). Following are illustrations of these reactions.

Step 1

3-methyl-2-butanol

118

Step 2

Step 3: 1,2-hydride shift to produce the more stable tertiary carbocation

A secondary carbocation A tertiary carbocation

Step 4

2-methyl-2-butene

A competing alkene would be 2-methyl-1-butene (which has two

119

alkyl groups around the double bond); however, 2-methyl-1-butene is less stable than 2-methyl-2-butene (which has three alkyl groups around the double bond).

2-methyl-1-butene

A second illustration of the ability of carbocations to rearrange can be observed by the conversion of 2,2-dimethylpropanol in water and mineral acid to produce the same alkene, 2-methyl-2-butene, which was formed by the dehydration of 2-methyl-2-butanol and 3-methyl-2-butanol. 2,2-Dimethylpropanol forms a primary carbocation in step (2) of the mechanism. The primary carbocation undergoes a 1,2-methide shift to form the more stable tertiary carbocation (step 3) which ultimately loses a hydrogen ion to form 2-methyl-2-butene (step 4).

Step 1

2,2-dimethylpropanol

120

Step 2

Step 3: 1,2-methide shift to produce the more stable tertiary carbocation

a primary carbocation a tertiary carbocation

step 4

2-methyl-2-butene

121

Alkenes

Problems

1. Draw geometrical isomers, if possible, for the following structures:

2. Name the following compounds using the IUPAC system of nomenclature:

CH2 CHCH3

CH3

CH3CH3

H C C

CH3CH2CH2 C

CH3 H

CH2CH2CH3C

(a)

(b)

(c)

122

3. Write structures for the following alkenes:

(a) trans-2,3-dimethyl-3-heptene

(b) trans-4-octene

(c) 1,2-propadiene (allene)

CH3CH2CH2 C

H

CH2CH2CH3

(a)

(b)

(c)

CH2CH2CH CHCH3

CH2CH3

CH CH

(d)

CH3

CH3

CH3 CH3

CH3

CH2CH3CH CHC C

123

(d) 2,4,4-trimethyl-1-pentene

(e) cis-1,3-pentadiene

(f) 2,3-dichloropropene

(g) 1,2-dimethylcyclopentene

(h) 2,3-dimethylcyclopentene

4. Predict the major product obtained in the following reactions:

124

5. Give a probable mechanism for the following reaction:

6. Considering the stability of the carbocation intermediate, predict the relative ease of dehydration of the following alcohols:

(a) CH3CH2CH2OH

(b) (CH3)3COH

(c) CH3CH(OH)CH2CH3

(d) CH3CH=CHCH2OH

125

7. Predict the dehydration products that would be obtained in (a); (b); (c); and (d) in problem 6.

8. Explain, using compatible mechanisms, why trans-2-butene reacts with Br2 to give meso-2,3-dibromobutane and no racemic-2,3-dibromebutane, but cis-2-butene reacts with

Br2 to

give racemic-2,3-dibromobutane and no meso-2,3-dibromebutane.

9. Use the following information to predict the structure of the alkenes that exhibit the ozonolysis products in a, b, c, d and e.

(a)

(b)

(c)

(d)

A C10H20 alkene that gives

126

as the only product

(e)

A diene with the molecular formula C8H12 that gives

as the only product.

10. Suggest a mechanism for following observation:

11. Give the IUPAC name for the compound produced in problem 10.

12. Suggest syntheses for the following alkenes from any starting

127

materials:

(a)

CH3CH2 C CH2

H

(b)

(c)C CCH3 CH2CH3

HCH3

(d)

CH3CH2CH2

C CH

CH3

H