#5: Alkenes

Question 1

Alkene Nomenclature Rules

Answer 1

Name of the alkene is obtained by replacing the suffix -ane or the corresponding alkane into -ene.

The longest parent chain is chosen so that it includes the double bond.

Double bond should get the least possible numbering. The locant (or numerical position) of only one of the doubly bonded carbons is specified in the name; it’s understood that the other doubly bonded carbon must follow in sequence.

Double bond gets precedence over substituent alkyl groups and halogens.

Hydroxyl group gets precedence over the double bond. Both -en and -ol suffixes are used when both double bond and the hydroxyl group are present in the same compound.

Question 2

Vinyl

Answer 2

Alkenyl group. H2C-(DB)-CH-

Question 3

Allyl

Answer 3

Alkenyl group. H2C-(DB)-CHCH2-

Question 4

Isopropenyl

Answer 4

H2C-(DB)-C
|
CH3

Alkenyl group.

Question 5

Structure and Bonding in Alkenes

Answer 5

Planar, each carbon is sp2-hybridized. The double bond has a sigma and pi component.

Sigma component arises from overlap of sp2 hybrid orbitals along a line connecting the two carbons.

Pi component arises via a “side-by-side” overlap of two p orbitals.

Question 6

Isomerism in Alkenes

Answer 6

Although ethylene is the only two-carbon alkene, and propene the only three-carbon alkene, there are four isomeric alkenes of molecular formula C4H8.

1-Butene has an unbranched carbon chain with a double bond between C-1 and C-2. It is a constitutional isomer of the other three. Similarly, 2-methylpropene, with a branched carbon chain, is a constitutional isomer of the other three.

The pair of isomers designated cis- and trans-2-butene have the same constitution; both have an unbranched carbon chain with a double bond connecting C-2 and C-3. They differ from each other int hat the cis isomer has both of its methyl groups on the same side of the double bond, but the methyl groups in the trans isomer are on opposite sides of the double bond.

Cis-trans stereoisomerism in alkenes is not possible when one of the doubly bonded carbons bears two identical substituents. Thus, neither 1-butene nor 2-methylpropene can have steroisomers.

Question 7

Naming Stereoisomeric Alkenes by the E-Z Notational System

Answer 7

When the groups on either end of a double bond are the same or are structurally similar to each other, it’s a simple matter to describe the configuration of the double bond as cis or trans.

The terms cis and trans are ambiguous, however, when it’s not obvious which substituent on one carbon is similar or analogous to a reference substituent on the other. A completely unambiguous system for specifying double-bond stereochemistry has been adopted by the IUPAC based on an atomic number criterion for ranking substituents on the doubly bonded carbons. When atoms of higher atomic number are on the same side of the double bond, we say that the double bond has the Z configuration, where Z stands for the German word zysammen, meaning “together”. When atoms of higher atomic number are on opposite sides of the double bond, the configuration is E, standing for the German word entgegen, meaning “opposite.”

Question 8

Cahn-Ingold-Prelog Priority Rules

Answer 8

1) Higher atomic number takes precedence over lower.
2) When two atoms directly attached to the same carbon of the double bond are identical, compare the atoms attached to these two on the basis of their atomic numbers. Precendence is determined at the first point of difference.
3) Work outward from the point of attachment, comparing all the atoms to a particular atom before proceeding further along the chain.
4) When working outward from the point of attachment, always evaluate substituent atoms one by one, never as a group.
5) An atom that is multiply bonded to another atom is considered to be replicated as a substituent on that atom.

Question 9

Degree of Substitution

Answer 9

We classify double bonds as monosubstituted, disubstituted, trisubstituted, or tetra substituted according to the number of carbon atoms directly attached to the C–C structural unit. Mono is one carbon, di two, etc.

Alkenes with more highly substituted double bonds are more stable than isomers with less substituted double bonds.

Like the sp2-hybridized carbons or carbocations and dree radicals, the sp2-hybridized carbons of double bonds are electron attracting, and alkenes are stabilized by substituents that release electrons to these carbons. Alkyl groups are better electron-releasing substituents than hydrogen and are, therefore, better able to stabilize an alkene.

Question 10

van der Waals Strain

Answer 10

Alkenes are more stable when large substituents are trans to each other than when they are cis. Less strain in trans.

For cyclic alkenes however, the reverse is true. Cis is more stable than trans.

Question 11

Dipole Moments of Alkenes

Answer 11

Dipole moments can cancel each other out if they’re both coming from the same kind of atom and they’re both directed towards each other. If these two atoms are different however, even though they are still facing each other, the one with the stronger dipole moment will push the direction a bit in their favor, but the effect is usually very weak.

The strongest effects usually come when both atoms have a dipole moment facing the same direction.

Question 12

Preparation of Alkenes: Dehydration of Alkanes

Answer 12

X-C.a-C.b-Y –> C–C + X-Y

Alkene formation requires that X and Y be substituents on adjacent carbon atoms. By making X the reference atom and identifying the carbon attached to it as the a(alpha) carbon, we see that atom Y is a substituent on the b(beta) carbon. Only B elimination reactions will be discussed. (Beta (b) elimination reactions are also known as 1,2 eliminations.)

Ethylene and propene are prepared on an industrial scale by the high temperature dehydrogenation of ethane and propane. Both reactions involve b elimination of H2.

CH3CH3 -(heat)-> H2C–CH2 + H2
CH3CH2CH3 -(heat)-> CH3CH–CH2 + H2

Many reactions classified as dehydrogenations occur within the cells of living systems at 25 degrees Celsius. H2 is not one of the products however. Instead, the hydrogens are lost in separate steps of an enzyme-catalyzed process.

Dehydration of alkanes is not a practical lab synthesis for the vast majority of alkenes.

Question 13

Preparation of Alkenes: Dehydration of Alcohols

Answer 13

In the dehydration of alcohols, the H and OH are lost from adjacent carbons. An acid catalyst is necessary.

H-C-C-OH -H+-> C–C + H2O

Before dehydrogenation of ethane became the dominant method, ethyl alcohol was prepared by heating ethyl alcohol with sulfuric acid.

CH3CH2OH -H2SO4-> H2C–CH2 + H2O

Other alcohols behave similarly. Secondary alcohols undergo elimination at lower temperatures than primary alcohols. Tertiary even lower.

Reaction conditions, such as the acid used and the temperature, are chosen to max the formation of alkene by elimination. Sulfuric acid (H2SO4) and phosphoric acid (H3PO4) are the acids most used. Potassium sulfate (KHSO4) is also often used.

Question 14

Regioselective Reaction

Answer 14

Reaction that can proceed in more than one direction but in which one direction is preferred over the other.

Question 15

Zaitsev’s Rule

Answer 15

Generalization describing the regioselectivity of beta eliminations.

States that the alkene formed in greatest amount is the one that corresponds to removal of the hydrogen from the beta carbon having the fewest hydrogens.

Rule as applied to the acid-catalyzed dehydration of alcohols is now more often expressed in a different way: beta elimination reactions of alcohols yield the most highly substituted alkene as the major product. Because the most highly substituted alkene is also normally the most stable one, Zaitsev’s rule is sometimes expressed as a preference for predominant formation of the most stable alkene that could arise by beta elimination.

Question 16

Stereoselective Reaction

Answer 16

Reactions in which one of the stereoisomers is preferred.

One in which a single starting material can yield two or more stereoisomeric products, but gives one of them in greater amounts than any other.

Alcohol dehydrations tend to produce the more stable stereoisomer of an alkene. Dehydration of 3-pentanol, for example, yields a mix of trans-2-pentene and cis-2-pentene, in which the more stable trans stereoisomer predominates.

Question 17

E1 and E2 Mechanisms of Alcohol Dehydration

Answer 17

The dehydration of alcohols resembles the reaction of alcohols with hydrogen halides in two ways.

1) Both reactions are promoted by acids.
2) The relative reactivity of alcohols increases in the other primary < seconday < tertiary.

These common features suggest carbocations are key intermediates in alcohol dehydrations, just as they were in reaction of alcohols withhydrogen halides.

Question 18

E1 Mechanism

Answer 18

(CH3)3COH -(H2SO4, heat)-> (CH3)2C–CH2 + H2O

Steps 1 and 2 describe the generation of tert-butyl cation by a process similar to that which led to its formation as an intermediate in the reaction of tert-butyl alcohol with hydrogen chloride.

1) Protonation of tert-butyl alcohol:
(CH3)3C-O-H + H-O-H2 (CH3)3C-O+-H2 + H2O

2) Dissociation of tert-butyloxonium ion
(CH3)3C-O+-H2 (CH3)3C+ + H2O

Step 2 is rate-determining and is unimolecular.

Step 3 is an acid-base reaction in which the carbocation acts as a Bronsted acid, transferring a proton to a Bronsted base (water). This is the property of carbocations that is of the most significance to elimination reactions. Carbocations are strong acids; they are the conjugate acids of alkenes and readily lose a proton to form alkenes. Even weak bases such as water are sufficiently basic to abstract a proton from a carbocation.

3) Deprotonation of tert-butyl cation:
(CH3)2-C+-CH2-H + H2O (CH3)2-C–CH2 + H3O+

Used by secondary and tertiary alcohols.

Question 19

E2 Mechanism

Answer 19

Primary carbocations are too high in energy to be intermediates in most chemical reactions. If primary alcohols don’t form primary carbocations, then how do they undergo elimination? A modification of the E1 mechanism offers the answer. For primary alcohols it’s believed that a proton is lost from the alkyloxonium ion in the same step in which carbon-oxygen bond cleavage takes place.

For example, the rate-determining step in the sulfuric acid-cataylyzed dehydration of ethanol may be represented as:

H2O + H-CH2-CH2-H2O+[ethyloxonium ion] –> H2O+-H + H2C–CH2(Ethylene) + H2O

Two molecules, alkyloxonium ion and water, involved in this step. Bimolecular elimination, E2.

Question 20

Rearrangement

Answer 20

Some alcohols undergo dehydration to yield alkenes having carbon skeletons different from the starting alcohols. Not only has elimination taken place, but the rearrangement of atoms in the alkene is different from that int he alcohol. A rearrangement has occurred.

For example, take 3,3-Dimethyl-2-butanol:
(CH3)3-C-CH(OH)-CH3 –> (CH3)C-CH–CH2[3,3-Dimethyl-1-butene (3%)] + (CH3)2C–C(CH3)2[2,3-Dimethyl-2-butene(64%)] + (CH2)(CH3)C-CH(CH3)2[2,3-Dimethyl-1-butene(33%)]

The carbon skeleton rearrangement occurs in a separate step following carbocation formation. Once the alcohol was converted to the corresponding carbocation, that carbocation could either lose a proton to give an alkene having the same carbon skeleton or rearrange to a different carbocation. The rearranged alkenes arise by loss of a proton from the rearranged carbocation.

Carbocations rearrange because they want to be in an energetically favorable state. For example, a secondary carbocation may want to rearrange to be a teritary carbocation. It’s also the more stable state.

Look in notes to see how carbocation actually rearranges.

When carbocation rearrangement occurs, alpha and beta assignments change as well.

The structure with the most substitutes usually gets the most yield in product.

Anytime you form an alkene from an alcohol, think about rearrangements. Write ALL products on exam.

Question 21

Dehydrohalogenation of Alkyl Halides

Answer 21

Dehydrohalogenation is the loss of a hydrogen and a halogen from an alkyl halide. It is one of the most useful methods for preparing alkenes by B elimination.

When applied to the preparation of alkenes, the reaction is carried out in presence of a strong base.

The regioselectively follows the Zaitsev rule; B elimination predominates in the direction that leads to the more highly substituted alkene.

Also a stereoselective reaction and favors formation of the more stable sterioisomer. Favors trans for acyclic products, cis for cyclic when there’s less than 10 carbons.

Question 22

Strong Bases Used In Dehydrogenation of Alkyl Halides

Answer 22

Sodium Ethoxide (NaOCH2CH3) in ethanol (CH3CH2OH).

Sodium Methoxide (NaOCH3) in Methanol (CH3OH).

Potassium tert-butoxide [KOC(CH3)3] is the preferred base when the alkyl halide is primary. Used in either terybutanol or dimethyl sulfoxide as solvent.

Question 23

Dehydrohalogenation of Alkyl Halides: E2 Mechanism

Answer 23

In the 1920s, Sir Christopher Ingold proposed a mechanism for dehydrohaloganation of alkyl halides that’s accepted as the best description of how these reactions occur. Based on these facts

1) The reaction exhibits second-order kinetics; it’s first-order in alkyl halide and first-order in base.
Rate = k[alkyl halide][base]
- Doubling the concentration of either the alkyl halide or the base doubles the reaction rate. Doubling the concentration of both reactants increases the rate by a factor of 4.

2) The rate of elimination depends on the halogen, the reactivity of alkyl halides increasing with decreasing strength of the carbon-halogen bond. (RF - slowest rate of elimination, strongest carbon-halogen bond < RCl < RBr < RI - fastest rate of elimination, weakest carbon-halogen bond)
- Iodine is the best leaving group in a dehydrogenation reaction, fluoride is the poorest.

What are the implications of second-order kinetics? Ingold reasoned that second-order kinetics suggests a biomolecular rate-determining step involving both a molecule of the alkyl halide and a molecule of base. He concluded that proton removal from the B carbon by the base occurs during the rate-determining step rather than in a separate step following the rate-determining step.

What are the implications of the effects of the various halide leaving groups? Because the halogen with the weakest bond to carbon reacts fastest, Ingold concluded that the carbon-halogen bond breaks in the rate-determining step. The weaker the carbon-halogen bond, the easier it breaks.

Ingold proposed the E2 mechanism the four key elements:
1) B-H bond making
2) C-H bond breaking
3) C–H pi bond formation
4) C-X bond breaking
are all taking place at the same transition state in a concerted process. The carbon-hydrogen and carbon-halogen bonds are in process of being broken, the base is becoming bond to the hydrogen, a pi bond is being formed, and the hybridization is being changed from sp3 to sp2.

The E2 mechanism is followed whenever an alkyl halide-be it primary, secondary, or teritary-undergoes elimination in presence of a strong base.

The regioselectivity of elimination is accommodated int he E2 mechanism by noting that a partial double bond develops at the transition state. Because alkyl groups stabilize double bonds, they also stabilize a partially formed pi bond in the transition state. The more stable alkene therefore requires a lower energy of activation for its formation and predominates in the product mixture because it is formed faster than a less stable one.

Question 24

Dehydrohalogenation of Alkyl Halides - Anti Elimination in E2 Reactions: Stereoelectronic Effects

Answer 24

Comparing the rate of elimination of the cis and trans isomers of 4-tert-butylcyclohexyl bromide, both of them yield 4-tert-bytylcyclohexene as the only alkene, but they do so at very different rates. The cis isomer reacts over 500 times faster than the trans.

The difference in reaction rate results from different degrees of pi bond development in the E2 transition state. Since pi overlap of p orbitals requires their axes to be parallel, pi bond formation is best achieved when the four atoms of H-C-C-X unit lie in the same plane at the transition state. The two conformations that permit this are termed syn coplanar (orbitals aligned but bonds are eclipsed) and anti coplanar (orbitals aligned and bonds are staggered).

Because adjacent bonds are ecliped when the H-C-C-S unit is syn coplanar, a transition state with this geometry is less stable than one that has an anti coplanar relationship between the proton and the leaving group.

Bromine is axial and anti coplanar to two axial hydrogens int he most stable conformation of cis-3-tery-butylcyclohexyl bromide and has the proper geometry for ready E2 elimination. The transition state is reached with little icnrease in strain, and elimination occurs readily.

In its most stable conformation, the trans sterioisomer has no B hydrogens anti to Br; all four are gauche. Strain increases significantly in going to the E2 transition state, and the rate of elimination is slower than for the cis stereoisomer.

Effects that arise because of one spatial arrangement of electrons (or orbitals or bonds) is more stable than another are called stereoelectronic effects. There’s a stereoelectronic preference for the anti coplanar arrangement of proton and leaving group in E2 reactions. Although coplanarity of the p orbitals is the best geometry for the E2 process, modest deviations from it can be tolerated.

Stereoelectronic effects are also important in the dehydrogenation of acyclic alkyl halides by an E2 pathway. Again,t he most favorable arrangement for the hydrogen and the halide being lost is anti coplanar.

Question 25

Dehydrohalogenation of Alkyl Halides - Isotope Effects and the E2 Mechanism

Answer 25

The E2 mechanism receives support from studies of alkyl halides that contain deuterium (D), H with two neutrons at the B carbon. The fundamental kinds of reactions a substance undergoes are the same regardless of which isotope is present, but the reaction RATES can be different.

A C-D bond is 12 kj/mol stronger than a C-H bond, making the activation energy for breaking a C-D bond slightly greater. Consequently, the rate constant k for an elementary step in which a C-D bond breaks is smaller than for a C-H bond. This difference in rate is expressed as a ratio of the respective rate constants (KH/KD) and is a type of kinetic isotope effect. Because it compares D to H, it is also referred to as a deuterium isotope effect.

Typical deuterium isotope effects for reactions in which C-H bond breaking is rate-determining lie in the range KH/KD = 3-8. If the C-H bond breaks after the rate-determining step, the overall reaction rate is affected only slightly and KH/KD = 1-2. Thus, measuring the deuterium isotope effect can tell us if a C-H bond breaks in the rate-determining step.

According to the E2 mechanism for dehydrogenation, a base removes a proton from the B carbon in the same step as the halide is lost. This step, indeed it is the only step in the mechanism, is rate-determining. Therefore, elimination by the E2 mechanism should exhibit a deuterium isotope effect.

The size of an isotope effect depends on the ratio of the atomic masses of the isotopes; thus, those that result from replacing H by D or T(H with 3 neutrons, tritium) are easiest to measure. This, plus additional facts that most organic compounds contain hydrogen and many reactions involving breaking C-H bonds, have made rate studies involving hydrogen isotopes much more common that those of other elements.

Question 26

The E1 Mechanism of Dehydrogenation of Alkyl Halides

Answer 26

The E2 mechanism is a concerted process in which the carbon-hydrogen and carbon-halogen bonds both break in the same elementary step. What if these bonds break in separate steps?

One possibility is a mechanism in which the carbon-halogen bond breaks first to give a carbocation intermediate, followed by deprotonation of the carbocation in a second step.

The alkyl halide ionizes to a carbocation and a halide anion by a heterolytic cleavage of the carbon-halogen bond. Like the dissociation of an alkyloxonium ion to a carbocation, this step is rate-determining. Study the figure on Page 215 to get familiar with this mechanism.

Typically, elimination by the E1 is observed only for tertiary and some secondary alkyl halides, and then only when the base is weak or in low concentration. Unlike eliminations that follow an E2 pathway and exhibit second-order kinetic behavior, those that follow the E1 mechanism obey a first-order rate law.

Rate = k[alkyl halide]

The reactivity parallels the ease of carbocation formation.
- RCH2X(Primary alkyl halide, slowest rate of E1 elimination) < R2CHX < R3CX (Tertiary alkyl halide, fastest rate of E1 elimination)

Because the carbon-halogen bond breaks in the slow step, the rate of the reaction depends on the leaving group. Alkyl iodides have the weakest carbon-halogen bond and are the most reactive; alkyl fluorides have the strongest carbon-halogen bond and are the least reactive.

This mechanism and the E1 mechanism for alcohol dehydration is very similar. The main difference between them is the source of the carbocation. When the alcohol is the substrate, it is the corresponding alkyloxonium ion that dissociates to form the carbocation. The alkyl halide ionizes directly to the carbocation.

Like alcohol dehydrations, E1 reactions of alkyl halides can be accompanied by carbocation reaarangements. Eliminations by the E2 mechanism, on the other hand, normally proceed without rearrangement. Consequently, if one wishes to prepare an alkene from an alkyl halide, conditions favorable to E2 elimination should be chosen. In practice, this simply means carrying out the reaction in presence of a strong base.