1. Alkenes: Structure and Reactivity
Chapter 7
Alkenes: Structure and Reactivity
Suggested Problems -
1-21,26,29-31,38-9,45-49

2. Alkene - Hydrocarbon With Carbon-Carbon Double Bond
Also called an olefin but alkene is better
Hydrocarbon that contains a C-C double bond
Includes many naturally occurring materials
Flavors, fragrances, vitamins

3. Industrial Preparation and Use of Alkenes
Ethylene and propylene are important organic chemicals produced

4. Thermal Cracking
Thermal cracking may take place w/o catalyst at very high temperatures
Since considerable heat is required to effect cracking the reaction is endothermic, this reaction is driven by a very large entropy term involving the formation of four molecules from one molecule. Since delta S is large and a high temperature is used, the T(delta S) term is larger than the delta H term and delta G is negative (exergonic).

5. Saturated and Unsaturated Hydrocarbons
Alkanes have the maximum number of C-H bonds – they are “saturated” with hydrogens and are therefore termed saturated hydrocarbons.
In contrast, alkenes are called unsaturated hydrocarbons because they have fewer than the maximum number of hydrogens.
Saturated hydrocarbons have no double bonds.
Unsaturated hydrocarbons have one or more double bonds.

6. Molecular Formula and the Degree of Unsaturation
molecular formula = CnH2n+2 – 2 hydrogens for every π bond or ring
degree of unsaturation = the total numbers of π bonds and rings
Alkenes are hydrocarbons that contain a carbon-carbon double bond.
Like alkanes, they come in acyclic and cyclic varieties.
The general molecular formula for an acyclic alkene is CnH2n, because the double bond means that the alkene has two fewer hydrogens than an alkane with the same number of carbons.
Note that an acyclic alkene has the same number of hydrogens as a cyclic alkane.
Two hydrogens are subtracted from the alkane molecular formula for every ring and every π bond in the molecule.
Thus, the general molecular formula for a hydrocarbon is CnH2n+2 minus two hydrogens for every π bond or ring in the molecule.
The total number of π bonds and rings is called the compound’s degree of unsaturation. Cyclopentene above has two degrees of unsaturation – a ring and a π bond.

7. Example: C6H10 Saturated is C6H14 therefore 4 H's are not present
This has two degrees of unsaturation
Two double bonds?
or triple bond?
or two rings?
or ring and double bond?

8. Another Example: C8H14 Saturated: CnH2n+2 = C8H18
Therefore, C8H14 has two degrees of unsaturation.
compounds with six hydrogens and two degrees of unsaturation
Here is another example illustrating two degrees of unsaturation in a C-8 hydrocarbon.
A C-8 alkane would have the formula – C8H18 (from CnH2n+2 where n = 8).
The molecules here have four fewer hydrogens. Any combination of π bonds or rings equal to 2 satisfies this condition.

9. Degree of Unsaturation With Other Elements
Organohalogens (X: F, Cl, Br, I)
Halogen replaces hydrogen
C4H6Br2 and C4H8 have one degree of unsaturation
Simply rewrite the formula counting each halogen as a hydrogen, then determine degree of unsaturation

10. Degree of Unsaturation (Continued)
Organooxygen compounds (C,H,O) – Oxygen forms 2 bonds
these don't affect the formula of equivalent hydrocarbons
May be ignored in calculating degrees of unsaturation

11. Organonitrogen Compounds
Nitrogen has three bonds
So if it connects where H was, it adds one connection point
Subtract one H for equivalent degree of unsaturation in hydrocarbon

12. Summary - Degree of Unsaturation
Count pairs of H's below CnH2n+2
Add number of halogens to number of H's (X equivalent to H)
Ignore oxygens (oxygen links H)
Subtract number of N’s from number of H’s

13. Nomenclature of Alkenes
Replace “ane” of alkane with “ene.”
The double bond is the functional group of an alkene. This functional group is designated by the the suffix “ene”. A number is used to designate the position of the double bond.
The IUPAC rules for naming are:
1) The longest continuous chain containing the functional group (in this case the double bond) is numbered in the direction that gives the functional group suffix the lowest possible number. A chain is therefore numbered assigning the lowest number to the first carbon of the alkene. The molecule on the right above is 2-hexene and not 4-hexene. Note the naming of the alkene at the bottom of this slide. The alkene is included in the longest parent chain.
The functional group gets the lowest possible number. 13

14. Nomenclature of Dienes
For a compound with two double bonds, the “ne” ending of the corresponding alkane is replaced with “diene”. Note the parent chain is numbered in the direction which gives the first double bond encountered the lowest number.
Thus, 1,3-pentadiene is not 2,4-pentadiene (which would be obtained if numbering commenced with the left most carbon).
two double bonds = diene

15. Nomenclature of Alkenes
3) The name of a substituent is given together with a number designating its position before the name of the longest continuous chain that contains the functional group. The functional group’s suffix (in this case, that corresponding to the alkene) rather than the substituent gets the lowest possible number.
Number in the direction so that the functional group gets the lowest number.

16. Nomenclature of Alkenes
4) If a chain has more than one substituent, the substituents are stated in alphabetical order. Then the appropriate number is assigned to each substituent.
In the case of the molecule on the right, the name is not 4-chloro-5-bromo-1-heptene.
Substituents are stated in alphabetical order.

17. Nomenclature of Alkenes
5) If counting in either direction results in the same number for the alkene functional group suffix, the correct name is the one containing the lowest substituent number.

18. Nomenclature of Cyclic Alkenes
A double bond in a ring is numbered such that the double bond is between C1 and C2. This being the case, you do not need to indicate the position of the double bond in its name (it is between C1 and C2!)
The ring is otherwise numbered to give substituents the lowest possible number.
A number is not needed to denote the position of the functional group;
it is always between C1 and C2.

19. Nomenclature of Alkenes
Again, the lowest substituent number is used. If a single substituent is on the double bond, the carbon to which it is attached receives the number 1.

20. Nomenclature of Alkenes
7) Where numbering is the same in both directions for the alkene functional group and first substituent, number the chain or ring assigning the lowest possible number to the next substituent.

21. Many Alkenes Are Known by Common Names

22. Vinylic and Allylic Carbons
The sp2 carbons of an alkene are called vinylic carbons. An sp3 carbon adjacent to a vinylic carbon is called an allylic carbon.
A hydrogen bonded to an allylic carbon is called an allylic hydrogen. One bonded to a vinylic carbon is called an allylic hydrogen.
vinylic carbon: the sp2 carbon of an alkene
allylic carbon: a carbon adjacent to a vinylic carbon

23. Six Atoms of an Alkene are in the Same Plane
Because three points determine a plane, each sp2 carbon and the two atoms singly bonded to it lie in a plane. In order to achieve maximum orbital-orbital overlap, the two p orbitals must be parallel to each other. For the two p orbitals to be parallel, all six atoms of the double bond system must be in the same plane.

24. Double Bonds Have Restricted Rotation
Each double-bonded carbon of an alkene has three sp2 orbitals. Each of these orbitals overlaps an orbital of another atom to form σ bond, one of which is one of the bonds in the double bond.
Thus, the σ bond of the double bond is formed by the overlap of an sp2 orbital of one carbon with an sp2 orbital of the other carbon, and the other bond of the double bond is a π bond formed from side-to-side overlap of the remaining p orbital on each of the sp2 carbons.
Rotation about a double bond breaks the π bond. 24

25. Alkenes Have Cis–Trans Isomers
Cis: The hydrogens are on the same side of the ring.
Trans: The hydrogens are on opposite sides of the ring.
As long as each of the sp2 carbons of an alkene is bonded to one hydrogen, we can use the terms cis and trans to designate the geometric isomers. If the hydrogens are on the same side of the double bond, it is the cis isomer; if the hydrogens are on opposite sides of the double bond, it is the trans isomer.
They have different configurations; they can be separated. 25

26. Which Is Cis and Which Is Trans?

27. Which Is Cis and Which Is Trans?
But what if each carbon does not bear a hydrogen? How are geometrical isomers named in this instance?

28. Alkene Stereochemistry and the E,Z Designation
Cis-Trans naming system discussed thus far only works with disubstituted alkenes
Tri- and Tetra substituted double bonds require more general method
Method referred to as the E,Z system

29. The E,Z System of Nomenclature
The E,Z system of nomenclature was devised for alkenes that do not have a hydrogen attached to each of the sp2 carbons.
To name an isomer by the E,Z system, we must first determine the relative priorities of the two groups bonded on each of the two sp2 carbons, distinguishing a high priority and a low priority substituent on each carbon.
Z = Zusammen (together)
E = Entgegen (opposite)
Pneumonic – zusammen = “zi zame zide”

30. Relative Priorities
If the two high priority groups (one from each carbon) are on the same side of the double bond, the isomer is the Z isomer (Z is for zusammen, German for “together”).
If the two high priority groups are on opposite sides of the double bond, the isomer is the E isomer (E is for entgegen, German for “opposite”).
The relative priorities depend on the atomic numbers of the atoms bonded directly to the sp2 carbon. The greater the atomic number, the higher the priority.
Thus, the molecule on the left, with the two high priority groups on the same side of the double bond is the Z isomer; the molecule on the right, with the two high priority groups on opposite sides of the double bond is the E isomer.
This system is identical to that used in assigning R and S configurations that we saw earlier.
The relative priorities of the two groups depends on the atomic numbers of the atoms attached to the sp2 carbon. 30

31. The E and Z Isomers
If the two atoms attached to an sp2 carbon are the same (there is a tie), then consider the atomic numbers of the atoms that are attached to the ”tied” atoms.
In the isomer on the left, the lower right carbon is attached to an O, H, and H. The upper right carbon is attached to a C, C, H. Oxygen wins out the the CH2OH group has a higher priority than the isopropyl group.
In the isomer on the left, the lower left carbon is attached to a Cl, H, and H, whereas the upper left carbon is attached to a C, H, and H. Chlorine beats carbon, and the lower left group has a higher priority than the group above it.

32. E and Z
If an atom is doubly bonded to another atom, the priority system treats it as if it were singly bonded to two of those atoms. If triply bonded, it is treated as if singly bonded to three of those atoms.
Thus the two carbons attached to the double bond on the left are considered to be attached to HHO in the top case, and CCC in the bottom case. Oxygen is of higher priority than a carbon, so the upper left group receives the highest priority.

33. E and Z
If two isotopes are compared, the mass number is used to determine the relative priorities. Deuterium has a higher priority than H.
Note the case of the two groups on the right side of the double bond in the left structure. The first carbon at top is bonded to C, C, and H. The first carbon at the bottom is bonded to C, C, H. This constitutes another tie. One then moves out to the next carbon. In the group at upper right, the second carbon out is bonded to H, H, and H. The second carbon out on the lower right is bonded to C, H, and H (using the double bond consisting of two singly bonded carbons). Thus the group at lower right has the higher priority.

34. Isomers for Compounds with
Two Double Bonds
For isomers with two double bonds, the chain is numbered in the direction giving the lowest numbered positions for the two double bonds. In the molecule at upper left, this is a 2,4-heptadiene versus a 3,5-heptadiene using numbering from the right.
The double bonds are assigned Z and E configurations based on the priorities of the two groups attached to them. For the molecule at top left, the 2 double bond is in the Z configuration and so is the double bond in the 4 position. Thus the molecule is a (2Z,4Z)-2,4-heptadiene. The chlorine attachment is included last.
Note with the two double bonds there are four geometric isomers.

35. Stability of Alkenes Cis alkenes are less stable than trans alkenes
Compare heat given off on hydrogenation: Ho
Less stable isomer is higher in energy
And gives off more heat
tetrasubstituted > trisubstituted > disubstituted > monosubstituted
hyperconjugation stabilizes

36. Stability of Alkenes (Continued): Hyperconjugation
Electrons in neighboring filled  orbital stabilize vacant antibonding  orbital – net positive interaction
Alkyl groups are better than H

37. Stability of Alkenes (Continued): Comparing Stabilities of Alkenes
Evaluate heat given off when C=C is converted to C-C
More stable alkene gives off less heat
trans-Butene generates 5 kJ less heat than cis-butene

38. Relative Stabilities of Alkenes
The heat released in a hydrogenation reaction is called the heat of hydrogenation.
It is customary to give it a positive value.
Hydrogenation reactions, however, are exothermic (they have negative ΔHo values), so the heat of hydrogenation is the value of ΔHo without the negative sign.
Note in this slide that different heats of hydrogenation are obtained from three different alkenes as they form the same product.
This suggests that each of the alkenes has a different energy than the others, since the products all are identical.

39. The Most Stable Alkene Has the Smallest Heat of Hydrogenation
Here is another way to look at the energy of the three alkenes based on their heats of hydrogenation.
The most stable alkene (the 2-alkene) has the smallest heat of hydrogenation.

40. Relative Stabilities of Alkenes
The stability of an alkene increases as the number of alkyl substituents bonded to its sp2 carbons increases.
Note that an alkene with four substituents on the double bond is more stable than one with three, etc…

41. Trans is More Stable Than Cis
Both trans-2-butene and cis-2-butene have two alkyl substituents bonded to their sp2 carbons, but trans-2-butene has a smaller heat of hydrogenation.
This means that the trans isomer, in which the large substituents are farther apart, is more stable than the cis isomer.

42. The Cis Isomer Has Steric Strain
When large substituents are on the same side of the double bond, as in a cis isomer, their electron clouds can interfere with each other, causing steric strain in the molecule.
Steric strain makes a compound less stable.
When the large substituents are on opposite sides of the double bond, as in a trans isomer, their electron clouds cannot interact, so there is no destabilizing steric strain.

43. The Relative Stabilities of Disubstituted Alkenes
relative stabilities of dialkyl-substituted alkenes
The heat of hydrogenation of cis-2-butene, in which the two alkyl substituents are on the same side of the double bond, is similar to that of 2-methylpropene, in which the two alkyl substituents are on the same carbon.
The three dialkyl-substituted alkenes are all less stable than a trialkyl-substituted alkene, and they are all more stable than a monoalkyl-substituted alkene.

44. Electrophilic Addition of Alkenes
General reaction mechanism of electrophilic addition
Attack on electrophile (such as HBr) by  bond of alkene
Produces carbocation and bromide ion
Carbocation is an electrophile, reacting with nucleophilic bromide ion

45. The Mechanism of the Reaction
The π bond of a double bond is weak, so it is easily broken. This allows alkenes to undergo addition reactions.
An alkene is a nucleophile, so the first species it reacts with is an electrophile.
Therefore, we can say that alkenes undergo electrophilic addition reactions.
When an alkene undergoes an electrophilic addition reaction with HBr, the first step is a relatively slow addition of a proton (an electrophile) to the alkene ( a nucleophile).
A carbocation intermediate (an electrophile) is formed, which then reacts rapidly with a bromide ion (a nucleophile) to form an alkyl halide.
Each step involves the addition of an electrophile with a nucleophile.
The overall reaction is the addition of an electrophile to one of the sp2 carbons of the alkene and the addition of a nucleophile to the other sp2 carbon.

46. Electrophilic Addition of Alkenes: Energy Path
Two step process
First transition state is high energy point
First step is slower than second

47. Electrophilic Addition of Alkenes (Continued)
The reaction is successful with HCl and with HI as well as HBr
HI is generated from KI and phosphoric acid

48. Addition of Hydrogen Halides
If the electrophilic reagent that adds to an alkene is a hydrogen halide (HF, HCl, HBr, HI), the product of the reaction will be an alkyl halide.
Because the alkenes in these examples have the same substituents on both sp2 centers, it is easy to predict the product of the reaction: the electrophile (H+) adds to either one of the sp2 carbons and the nucleophile (X-) adds to the other sp2 carbon.
It does not matter which sp2 carbon the electrophile adds to because the same product will be obtained in either case.

49. Orientation of Electrophilic Additions: Markovnikov’s Rule
In an unsymmetrical alkene, HX reagents can add in two different ways, but one way may be preferred over the other
If one orientation predominates, the reaction is regioselective
Markovnikov observed in the 19th century that in the addition of HX to alkene, the H attaches to the carbon with more H’s and X attaches to the other end (to the one with more alkyl substituents)
This is Markovnikov’s rule

50. A Regioselective Reaction
A regioselective reaction is one in which two constitutional isomer can be obtained as product, but more of one is obtained than the other.
In other words, a regioselective reaction selects for a particular constitutional isomer.
Recall that a reaction can be moderately, highly, or completely regioselective depending on the relative amounts of the constitutional isomers formed in the reaction.

51. Not Regioselective
Here is an example of a reaction that is not regioselective.
Because the addition of H+ to either of the sp2 carbons produces a secondary carbocation, both carbocations are formed at about the same rate.
Therefore, approximately equal amounts of the two alkyl halides are obtained.
Vladimir Markovnikov was the first to recognize that the major product obtained when a hydrogen halide adds to an alkene is a result of the addition of the H+ to the sp2 carbon bonded to the most hydrogens.
His law is often stated as “them that has git” inferring that the electrophile adds to the sp2 carbon bonded to the most hydrogens (the less substituted sp2 carbon).
The rule is simply a quick way to determine the major product of an electrophilic addition reaction.
What products are favored if 1-pentene is used in this reaction?
If one wanted to synthesize 2-bromopentane from a pentene, which pentene would one use?

52. Which sp2 Carbon Gets the H+?
But what happens if the alkene does not have the same substituents on both sp2 carbons have the same substituents? Which sp2 carbon gets the hydrogen?
Does the above reaction form tert-butyl chloride or isobutyl chloride?
Markovnikov’s Rule pneumonic – “them that has gets”

53. Example of Markovnikov’s Rule
Addition of HCl to 2-methylpropene
Regiospecific – one product forms where two are possible
If both ends have similar substitution, then not regiospecific
The reaction at the bottom could still be regioselective if more of one product is formed than the other. If only one product were to form the reaction would be referred to as regiospecific.

54. The Electrophile Adds to the sp2 Carbon Bonded to the Most Hydrogens
A regioselective reaction forms more of one constitutional isomer than another.
Highly regioselective
Completely regioselective - regiospecific
Moderately regioselective
Two products are formed in each of these reactions, but the major product is the one that results from reaction of the nucleophile with the faster formed tertiary carbocation.
The two products formed in these reactions are constitutional isomers. They have the same molecular formula, but they differ in how the atoms are connected.
A reaction in which two or more constitutional isomers could be obtained as products, but one of them predominates, is called a regioselective reaction.
There are degrees of regioselectivity: a reaction can be moderately regioselective, highly regioselective, or completely regioselective, depending on the amounts of the products formed.

55. Why is the First Reaction More Highly Regioselective?
In this case, the addition of HCl to 2-methylpropene (where the two possible carbocations are tertiary and primary) is more highly regioselective than the addition of HCl to 2-methyl-2-butene (where the two possible carbocations are tertiary and secondary), because the two carbocations formed in the latter are closer in stability.
In a completely regioselective reaction, only one of the possible products is formed.

56. The Mechanism Carbocation formation is the rate-limiting step.
We find experimentally that only tert-butyl chloride is formed.
By learning the mechanism of the reaction, it is possible to predict which isomer will be formed in other examples.
Mechanistically, the first step of the reaction – the addition of H+ to an sp2 carbon to form the tert-butyl cation or the isobutyl cation - is the slow rate-determining step of the reaction.
If there is any difference in the rate of formation of these two carbocations, then the one that is formed faster will be the predominant product of the first step.
Moreover, the particular carbocation formed in the first step determines the final product of the reaction.
That is, if the tert-butyl cation is formed, it will react rapidly with Cl- to form tert-butyl chloride.
On the other hand, if the isobutyl cation is formed, it will react rapidly with Cl- to form isobutyl chloride.
Since we know that the only product of the reaction is tert-butyl chloride, the tert-butyl cation must be formed much faster than the isobutyl cation.
The question is, why is the tert-butyl cation formed faster?
To answer this question, we need to look at two things: (1) the factors that affect the stability of a carbocation, and (2) how its stability affects the rate at which it is formed.
Carbocation formation is the rate-limiting step.

57. What Product Will Be Formed?
Now that we know that the major product of an electrophilic addition reaction is the one obtained by adding the electrophile to the sp2 carbon that results in formation of the more stable carbocation, we can predict the major product of the reaction of an unsymmetrical alkene.
In the addition reaction above, the proton can add to C-1 to form a secondary carbocation or it can add to C-2 to form a primary carbocation.
Because the secondary carbocation is more stable, it is formed more rapidly.
(Primary carbocations are so unstable that they form with great difficulty).
As a result, the only product is 2-chloropropane.

58. Markovnikov’s Rule (restated)
More highly substituted carbocation forms as intermediate rather than less highly substituted one
Tertiary cations and associated transition states are more stable than primary cations

59. Markovnikov’s Rule (restated)

60. Carbocation Structure and Stability
Carbocations are planar and the tricoordinate carbon is surrounded by only 6 electrons in sp2 orbitals
the fourth orbital on carbon is a vacant p-orbital
the stability of the carbocation (measured by energy needed to form it from R-X) is increased by the presence of alkyl substituents

61. Relative Stabilities of Carbocations
Carbocations are classified based on the carbon that carries the positive charge: a primary carbocation has a positive charge on a primary carbon, a secondary carbocation has a positive charge on a secondary carbon, and a tertiary carbocation has a positive charge on a tertiary carbon.
It turns out the stability of a carbocation increases as the number of alkyl substituents attached to the positively charged carbon increases - 3o carbocations are more stable than 2o carbocations which are more stable than 1o carbocations.

62. Carbocation Structure and Stability (Continued)
An inductive stabilized cation species

63. Hyperconjugation Stabilizes a Carbocation
Alkyl groups stabilize carbocations because they decrease the concentration of the positive charge on the carbon.
This is because the orbital of an adjacent C-H σ bond can overlap the empty p orbital of the positively charged carbon.
The movement of electrons from a σ bond orbital toward the vacant p orbital decreases the charge on the sp2 carbon and causes a partial positive charge to develop on the two atoms bonded by the overlapping σ bond orbital (the H and C in the figure below).
With three atoms sharing the positive charge the carbocation is stabilized because the charge is dispersed over three atoms.
In contrast, the positive charge in the methyl cation is concentrated solely on one atom.

64. Hyperconjugation
Hyperconjugation can only occur if the orbital of the σ bond and the empty p orbital have the proper orientation.
The proper orientation is readily achieved in this case because there is free rotation about the C-C σ bond.
Note that the σ bond orbitals that can overlap the empty p orbital are those attached to an atom that is attached to the positively charged carbon.
In the case of the tert-butyl cation, nine σ bond orbitals can potentially overlap the empty p orbital of the positively charged carbon.
The isopropyl cation has six such orbitals, whereas the ethyl and propyl cations each have three.
Thus, hyperconjugation stabilizes the 3o carbocation more than the 2o carbocation which in turn is stabilized more than a 1o carbocation.
Note that both C-H and C-C σ bond orbitals can overlap with the empty p orbital.

65. The Hammond Postulate
Why does the stability of the carbocation intermediate affect the rate at which it is formed?
the relative stability of the intermediate is related to an equilibrium constant (DGº)
the relative stability of the transition state (which describes the size of the rate constant) is the activation energy (DG‡)
DGº and DG‡ are not directly related
the transition state is transient and cannot be examined

66. The Transition State Looks Like
What it is Closer To
The Hammond postulate says that the transition state is more similar in structure to the species to which it is more similar in energy.
The question is answered by the Hammond postulate, which says that the transition state is more similar in structure to the species to which it is more similar in energy.
Thus, in an exergonic reaction (ΔGo < 0), the energy of the transition state is more similar to the energy of the reactants (Curve I), so its structure will be more similar to that of the reactants.
In an endergonic reaction (ΔGo > 0), the energy of the transition state is more similar to the energy of the products (Curve III), so its structure will be more similar to that of the products.
Only when the reactants and products have identical energies (Curve II) would we expect the structure of the transition state to be exactly halfway between that of the reactants and that of the products.

67. Competing Reactions and the Hammond Postulate
Normal Expectation: Faster reaction gives more stable intermediate
Intermediate resembles transition state

68. Why the Difference in Rate?
The more stable carbocation is formed more rapidly.
Because the formation of a carbocation is an endergonic reaction, the structure of the transition state resembles the structure of the carbocation product.
This means that the transition state has a significant amount of positive charge on a carbon. The same factors that stabilize the positively charged carbocation stabilize the partially positively charged transition state.
Therefore, the transition state leading to the tert-butyl cation ( a partially charged tertiary cation) is more stable (lower in energy) than the transition state leading to the isobutyl cation (a partially charged primary cation).
Thus, the tert-butyl cation, with a smaller energy of activation, is formed faster than the isobutyl cation.
In an electrophilic addition reaction, the more stable carbocation is formed more rapidly.

69. The Difference in Carbocation Stability Determines the Products
The relative rates of formation of the two carbocations determine the relative amounts of products formed, because formation of the carbocation is the rate-limiting step of the reaction.
If the difference in rates is small, both products will be formed, but the major product will be the one formed from reaction of the nucleophile with the faster formed carbocation.
If the difference in the rates is sufficiently large, the product formed from reaction of the nucleophile with the faster formed carbocation will be the only product.
In the case of the reaction of 2-methylpropene with HCl, the rates of formation of the two possible carbocation intermediates – one primary and the other tertiary – are sufficiently different to cause tert-butyl chloride to be the only product of the reaction.

70. The Major Product is a Surprise
Some electrophilic addition reactions give products that are not what you would get by adding an electrophile to the sp2 carbon bonded to the most hydrogens and a nucleophile to the other sp2 carbon.
In this reaction, we have come to expect that 2-bromo-3-methylbutane would be the major product. But it is the minor product. How come? And where did 2-bromo-2-methylbutane come from?

71. The Major Product is a Surprise
Similarly, shouldn’t 3-chloro-2,2-dimethylbutane be the major product of this reaction? And again, where did the product on the right come from?

72. Evidence for the Mechanism of Electrophilic Addition: Carbocation Rearrangments
Carbocations undergo structural rearrangements following set patterns
1,2-H and 1,2-alkyl shifts can occur
Most stable carbocation is formed
Can go through less stable ions as intermediates

73. Carbocation Rearrangement
(a 1,2-hydride shift)
In each of the previous reactions, the unexpected product results from a rearrangement of the carbocation intermediate.
Not all carbocations rearrange. Carbocations rearrange only if they become more stable as a result of the rearrangement.
Shown on this slide is the reaction we saw two slides ago. In the first step, a secondary carbocation is formed initially.
However, the secondary carbocation has a hydrogen that can shift with its pair of electrons to the adjacent positively charged carbon, creating a more stable tertiary carbocation.
Because a hydrogen shifts with its pair of electcrons, the rearrangement is called a hydride shift. More specifically, it is called a 1,2-hydride shift because the hydride ion moves from one carbon to an adjacent carbon.
As a result of the carbocation rearrangement, two alkyl halides are formed, one from adding the nucleophile to the unrearranged carbocation and one from adding the nucleophile to the rearranged carbocation.

74. Carbocation Rearrangement
(a 1,2-methyl shift)
In the second reaction we saw previously, again a secondary carbocation is formed initially. Then one of the methyl groups, with its pair of electrons, shifts to the adjacent positively charged carbon to form a more stable tertiary carbocation.
This kind of rearrangement is called a 1,2-methyl shift – the methyl group moves with its electrons from one carbon to an adjacent carbon.
Again, the major product is the one formed by adding the nucleophile to the rearranged carbocation.

75. The Carbocation Does Not Rearrange
(No Improvement in Carbocation Stability)
If a rearrangement does not lead to a more stable carbocation, then it typically does not occur.
In this reaction, a secondary carbocation is formed initially. A 1,2-hydride shift would form a different secondary carbocation.
Since both carbocations are equally stable, there is no energetic advantage to the rearrangement.
Consequently, the rearrangement does not occur, and only one alkyl halide is formed.
It is important to note, that it doesn’t matter what the nucleophile is that is trapping the carbocation. Here it is a bromide, but the same thing could occur if water or an alcohol, etc… were the nucleophile

76. Hydride shifts in biological molecules

77. Let’s Work a Problem
Addition of HCl to 1-isopropylcyclohexene yields a rearranged product. Propose a mechanism, showing the structures of the intermediates and using curved arrows indicate electron flow in each step.
The hydrogen is approached by the nucleophilic double bond. A new C-H bond is formed, along with a 3˚ carbocation. At the same time, Cl takes on 1more electron, making it a Cl anion.
There is a “Methyl” shift, creating a new C-H bond with the 1st 3˚ carbocation to give an isomeric 3˚ carbocation that is less hindered.
The Cl-ion then approaches the new carbocation. A Cl-C bond is formed which gives charged neutral product.

78. Worked Example
On treatment with HBr, vinylcyclohexane undergoes addition and rearrangement to yield 1-bromo-1-ethylcyclohexane
Using curved arrows, propose a mechanism to account for this result
Evidence for the mechanism of electrophilic additions: Carbocation rearrangements

79. Worked Example Solution:
Step 1 - Electrophilic addition of H+ to double bond
Step 2 - Hydride shift that forms a more stable tertiary carbocation
Step 3 - Reaction of carbocation with Br-
Evidence for the mechanism of electrophilic additions: Carbocation rearrangements