CHAPTER 6-ALKENES;STRUCTURE AND REACTIVITY Alkenes are ...
CHAPTER! 6!-A ! LKENES!;S ! TRUCTURE! AN!D! REACTIVITY! ! A ! lkenes! are! hydrocarbons! that! c! ontain! a! carbon-carbon! double!b! ond.! = ! ! o! lefin!.! ! V ! ery! c! ommon! function! group! in! nature!.! ! H3C H
H C
H
CH3
C H CH3
Ethylene (Ethene)
a-Pinene
!Ethylene! (a! ripening! agent! in! fruit)! and! a-pinene!(! turpentine)! a! re! e! xamples!.! b-Carotene
!
!b-Carotene!(! pigment, vitamin A precursor and "antioxidant") is an example of a multi-alkene. Ethylene and propylene, simpletst alkenes are industrially important - all kinds of chemicals are made from them. See text p. 190 Made from cracking of natural gas. Ethylene (Ethene) CH2=CH2
Propylene (Propene) CH3CH=CH2
Calculating Degree of Unsaturation Note general formula for a mono alkene is CnH2n. Remember that the general formula for an alkane was CnH2n+2. Because of C=C, alkene has fewer hydrogens. For each double bond in a hydrocarbon, there are two fewer hydrogens than in the alkane with the same number of carbons. The alkane is referred to as saturated - has all its hydrogens
H
H
H
C
C
H
H
H
Ethane
The alkene is referred to as unsaturated. H
H C
C
H
H Ethylene (Ethene)
For each ring in a hydrocarbon, the compound has two fewer hydrogens. CH2 H
CH2
CH2 H
CH2
In general, each ring or double bond in a molecule corresponds o a loss of two hydrogens from the alkane formula CnH2n+2. Example: What are the possible structures for a hydrocarbon with the molecular formula, C5H8? Answer: A saturated hydrocarbon of 5 carbons will have the formula C5H12. The compound in question has only 8 hydrogens. H12 - H8 = H4. Since one unsaturation per every two hydrogens, divide H4 by 2. Therefore H4/2 = H2 and there are two degrees of unsaturation in the moleule. This means there are a combination of two double bond, two rings (unusual for so few carbons), a ring and a double bond, or a triple bond.
Some examples -
CH3CH=CHCH=CH2
CH3C
CCH2CH3
For compounds with elements other than carbon and hydrogen. 1.Halohydrocarbons: C, H,and X Both H and the halogens are monovalent and a hydrogen is simply a replacement for an H. So count the halogen(s) as hydrogen(s). C7H8Br2 is equivalent to C7H10 in a calculation of unsat'n. So C7H16 ia alkane. H16 - H10 = H6, Some examples CHBr2
6/2 = 3. Three degrees on unsat'n.
Br CH3CH=CHC
CCH2CHBr2
Br
2. Organooxygen compounds: C, H, O Oxygen is divalent and does not affect the formula. Examine: CH3-H, vs. CH3-O-H CH3CH2-CH2CH3
vs. CH3CH2-O-CH2CH3
So C5H8O is equivalent to C5H8 when calculating deg. of unsat'n. Some examples OH
O CH2=CHCH=CHCH2 OH
[Since S is also divalent, the same rule applies as with O]. 3. Organonitrogen compounds: C, H, N
Nitrogen is trivalent. A compound containing N has one more H than a corresponding hydrocarbon. Examine: CH3-H vs. CH3-NH2 CH3-CH3
vs. CH3-NH-CH3
So subtract on H from molecular formula of the compounds to compare to hydrocarbon. C5H9N is equivalent to C5H8. Summary: 1. Add the number of halogens to the number of hydrogens. 2. Ignore the number of oxygens or sulfurs. 3. Subtract the number of nitrogens from the number of hydrogens. Nomencalture 1. Find the longest carbon chain containing the double bond. Name the compound with same stems from alkanes but use suffix -ENE.
CH3CH2 C
CH3CH2
CH2
C
CH3CH2CH2
CH2
CH3CH2CH2
Six-carbon chain does NOT contain the double bond
Five-carbond chain contains the double bond - so a PENTENE
2. Number the chain beginning are the nearer the double bond (C=C has lowest possible number). CH3 CH3CH2CH2 CH=CHCH3 6
5
4
3
2
1
CH3 CH CH=CHCH2CH3 1 6
2 5
3 4
4 3
5 2
6 1
In both cases, is a 3-hexene, but top numbers give CH3 lower number.
3. Write full name numbering substituents as to position and listing alphabetically. Indicate by the first number the position of the double bond. If more than one double bond present, indicate position of each by number of first carbon and the suffix -DIENE, -TRIENE, etc.
CH3CH2
CH3 C
CH2
CH3CH2CH2 CH=CHCH3
CH3 CH CH=CHCH2CH3
CH3CH2CH2
2-ethyl-1-pentene
2-hexene
2-methyl-3-hexene CH3
CH3
CH3
Problem CH3 6.6 (b) in text.
CH3
1-methylcyclohexene 1,4-cyclohexadiene 1,5-dimethylcyclopentene
4,4-dimethylcycloheptene
Some common names H2C=CH2 is commonly known as ethylene (systematic name = ethene) CH3CH=CH2 is propylene (systematic name = propene) Sometimes an unsaturated group has to be named as a substituent. H2C
CH2=CH
CH2=CHCH2
Electronic Structure of Alkenes The carbons in the double bond are sp2 hybridized. The carbon - carbon bond consists of one sigma bond from the overlap of an sp2 from each carbon. The second bond in the double bond is a pi bond which results from overlap of the unhybridized p orbitals. In M.O. Theory, two p orbitals give two molecular orbitals - one bonding (lower energy) and one antibonding (higher energy). The two electrons in the bond are in the bonding orbital.
p Antibonding molecular orbital Combine
p Bonding molecular orbital
Two p atomic orbitals
The two p orbitals must be parallel to get "maximum overlap". If they are not aligned there is no overlap and therefore there is no bonding. This means there is no free rotation around a C=C.
Cis-Trans Isomerism The two compounds below cannot be superimposed and therefore are different. H3C
CH3 C
H3C
C
H C
C
3C HH3C
H
CH3
CC CC HH H CH3 Cis-trans isomerism occurs when each of the carbons in the double bond are attached to two different things. If one carbon in the double bond has two identical groups on it, there is no cis-trans isomerism. H
H
H
CH3
Cis-trans nomenclature fails with tri- or tetra-substituted alkenes. So a system of sequene rules was devised which eliminates the problem. Sequence Rules: The E-Z Designation
Each carbon in the double bond is considered separately. The sequence rules are applied to each carbon and a priority is assigned to the substituent on that carbon. After determining the priority on the second carbon, the relative positions of the two high priority groups are compared. If the two high priority groups are on the same side of the double bond, use the designation, Z- (zusammen, German for together). High
High
Low
Low
High
Low
Low
High E double bond (entgegen)
Z double bond (zusammen)
If the two high priority groups are on opposite sides of the double bond, use the designation, E- (entgegen, German for opposite).
Cahn-Ingold-Prelog Sequence Ruules Remember to consider each carbon separately. 1. Look at the atom DIRECTLY attached to the carbon. An atom with a higher atomic number receives a higher priority than a lower atomic number. Example: H H3C
Cl
CH3
(E)-2-chloro-2-butene
H H3C
CH3 Cl
(Z)-2-chloro-2-butene
2. If a decision cannot be reached by ranking the atom directly attached to the C=, go out to the second atoms out. Then the third, fourth, etc until a difference is determined so that a decision can be made.
H
H H lower than
C H H
C C H
O H
H H
H lower than
C CH3 H
H
lower than
O C H H
CH3
CH3
C CH3
C NH2
H
H
H lower than
C Cl H
3. Multiple bonds are equivalent to the same number of single-bonded atoms. This is just a bookkeeping method for determining priorities. In the examples below only the atoms in bold print are real. The equivalent structures are needed when and if you have to go out to further atoms to make a decision. H C
is equivalent to
O
H
H
O C O
H is C equivalent H to
C
C
H
C C
C C H
H
C
C
C
is equivalent to
H
C C
C C
H
C
Examples H H
C C
H3C
CH2
CH(CH3 )2
H
C
C CH3
(E)-3-Methyl-1,3-pentadiene
H3C
H
C
C CH2OH
Z configuration
C
H3C E configuration
CN
Problem 6.11 (d)
CH2NH 2
Alkene Stability When the two isomers of butene are hydrogenated, different amount of energy are given off per mole. Since both go to the same product, the one giving off less energy must have been of lower eneergy to start with. H3C
CH3 C
H2/Pd
C
H
H
H3C
H2/Pd
CH3CH2CH2 CH3
DH = -120kJ/mol
H C
C
DH = - 116kJ/mol H
CH3
There is steric strain in the cis isomer that is not present in the trans isomer. This can also be calculated from equilibrium data. When several alkenes are examined it is found that the more alkyl substituents on the alkene carbons the more stable the molecule. When several alkene are studies in this manner, it is found that the greater the number of alkyl substituents on the alkene carbons, the more stable the alkene. So tetrasubstituted alkenes are more stable than trisubstituted alkenes, etc. R C R
R
R >
C R
R
R C
>
C
H
R
R
H C
C
R
= H
R C
H
C
R >
H
H C
H
Two explanations have been postulated about this. One explanation is called hyperconjugation. If electrons can "delocalize" (spread out) over several atoms instead of localizing on one - such as in resonance - than that is a stabilizing effect. It is theorized that the possibility of interaction of the electrons of the adjacent alkyl substituent can interact with the unfilled antibonding pi orbital. Antibonding C-C pi orbital (unfilled)
H C
C
Bonding C-C sigmai orbital (filled)
C
Another argument is based on bond strength and compares the bonds in the alkenes. A bond between and sp3 and an sp2 is a little stronger than a bond between two sp3 bonds.
C H
3
2
2
sp -sp Ø
3
3
sp -sp Ø
CH3-CH=CH-CH3
3
3
2
sp -sp , sp -sp Ø Ø
vs
CH3-CH2-CH=CH2
Electrophilic Additions Reactions to Alkenes Remember the reaction we saw as an example of a polar reaction in Chapter 5. Br
H+ Br-
H
H
H
H C
C
-
H H
H
C
C H
Br H
H
H C
C
H
H
H
Review the energy profile for this reaction ( a two step reaction) that we saw in Chapter 5. This reaction is called electrophilic addition to the double bond because the electrophile, H+, adds first. This general reaction allows us to prepare a variety of haloalkanes from alkenes. Cl
H3C C
CH2
+ HCl
Ether
H3C
H3C
CH3 CH3
CH3
CH3 + HBr
Ether
Br
(HI is usually not added directly but is generated in the reaction vessel using KI and H3PO4). REVIEW HOW REACTIONS ARE WRITTEN AND WHAT GOES ABOVE/BELOW THE ARROW, ETC. Notice the first reaction above. There are two possible ways that the H-Cl can add across the double bond. The two products are shown below. But notice that one product is not formed.
H3C H3C
H Cl C
CH2
H C
Cl Cl
CH2
H3C
H3C
CH3 CH3
H3C
H
H H3C C
CH2
Cl
H3C
CH3
CH2Cl
NOT FORMED
CH3
When HX is added to an unsymmetrical alkene one product is formed either in much greater yield than or exclusively over the other possible addition product. A Russian chemist, Vladimir Markovnikov examined many such reactions and from the results postulated his "rule". Markovnikov's Rule: In the addition of HX to an alkene, the H attaches to the carbon with fewer alkyl substituents and the X attaches to the carbon with more alkyl substituents. (Re=-examine the two reactions above). When both carbons in the double bond have the same degree of substitution, a mixture of products is formed. Show example (p202). This type of reaction is called a REGIOSPECIFIC reaction. Let's go back and look at this mechanism (steps). The H+ can become attached to either of the carbons of the double bonds. Two different carbocations are formed. Obviously one is formed over the other. H3C H3C
H Cl C
CH2
H C
Cl Cl
CH2
H3C
H3C
CH3 CH3
H3C
H
H H3C C CH3
CH2
Cl
H3C
CH2Cl CH3
Structurally, carbocations are planar - sp2 hybridized - with a vacant p orbital. Similar to boron trifluoride.
NOT FORMED
sp2
R 'R
C
F
R"
B
F
F
Experimental data shows that 3˚ carbocations are easier to form than 2˚, etc. Thus the stability of carbocations is as shown below. R R
R +
C R
>
R
R +
C
>
H
H
C+ H
H >
H
C+ H
The reasons for this are in part inductive effects and in part hyperconjugation. Inductive effects - electrons in bonds in the alkyl groups shift in response to the positive charge on the carbocation. Alkyl groups are polarizable. R
C R
R
Hyperconjugation - Interaction of the vacant p orbital with the C-H orbital of the adjacent alkyl groups is a stabilizing effect. H C
C
HAMMOND POSTULATE How does stability of the intermediate carbocation affect the rate at which it is formed? Hammond postulate states a reasonable explanation of experimental facts. When comparing two similar reactions, the more stable intermediate forms faster than the less stable one.
E
E Slower reaction
Less stable intermediate
Slower reaction Less stable intermediate
More stable intermediate
More stable intermediate
Reaction progress
Reaction progress
Hammond Postulate states: The structure of a transition state resembles the structure of the nearest stable species. Thus what stabilizes the intermediate also stabilized the transition state and the energy diagram on the left is the more likely scenario. The more stable intermediate forms faster because the transition state is of lower energy. Draw other energy profiles to show endothermic reaction, transition state resembles product exothermic reaction, transition state resembles starting materials
Carbocation Rearrangements It is known that with certain alkenes, in reactions with the addtion of HX, structural rearrangements occur during the reaction. If H-X added in one concerted step this could not happen. This discovery was supporting evidence for the stepwise theory and the formation of carbocations. Carbocation rearrangements occur to form more stable cations with what is called a hydride shift or an alkyl shift. In both cases, a group (:H- or :CH3 -) moves to a positively charged carbon taking its electron pair with it. ! H H3C
C
H3C
H
H C
+
C
H-Cl
H
H3C
C
H3C
H
H H3C
C C
H3C
C
C
C
+
H Cl
H3C
H
H
H
C
C
H
+
H
H
H3C H3C
C C H
H C
C
H
.!!
H
C
C H
H
H
H C H
Show other examples with alkyl shift CH3
H
H3C H3C C
H C
H3C H3C
H
Cl
H
CH3
H
Cl
H
C H
H
H C
H
CH3
C H CH3
Ethylene (Ethene)
a-Pinene
!Ethylene! (a! ripening! agent! in! fruit)! and! a-pinene!(! turpentine)! a! re! e! xamples!.! b-Carotene
!
!b-Carotene!(! pigment, vitamin A precursor and "antioxidant") is an example of a multi-alkene. Ethylene and propylene, simpletst alkenes are industrially important - all kinds of chemicals are made from them. See text p. 190 Made from cracking of natural gas. Ethylene (Ethene) CH2=CH2
Propylene (Propene) CH3CH=CH2
Calculating Degree of Unsaturation Note general formula for a mono alkene is CnH2n. Remember that the general formula for an alkane was CnH2n+2. Because of C=C, alkene has fewer hydrogens. For each double bond in a hydrocarbon, there are two fewer hydrogens than in the alkane with the same number of carbons. The alkane is referred to as saturated - has all its hydrogens
H
H
H
C
C
H
H
H
Ethane
The alkene is referred to as unsaturated. H
H C
C
H
H Ethylene (Ethene)
For each ring in a hydrocarbon, the compound has two fewer hydrogens. CH2 H
CH2
CH2 H
CH2
In general, each ring or double bond in a molecule corresponds o a loss of two hydrogens from the alkane formula CnH2n+2. Example: What are the possible structures for a hydrocarbon with the molecular formula, C5H8? Answer: A saturated hydrocarbon of 5 carbons will have the formula C5H12. The compound in question has only 8 hydrogens. H12 - H8 = H4. Since one unsaturation per every two hydrogens, divide H4 by 2. Therefore H4/2 = H2 and there are two degrees of unsaturation in the moleule. This means there are a combination of two double bond, two rings (unusual for so few carbons), a ring and a double bond, or a triple bond.
Some examples -
CH3CH=CHCH=CH2
CH3C
CCH2CH3
For compounds with elements other than carbon and hydrogen. 1.Halohydrocarbons: C, H,and X Both H and the halogens are monovalent and a hydrogen is simply a replacement for an H. So count the halogen(s) as hydrogen(s). C7H8Br2 is equivalent to C7H10 in a calculation of unsat'n. So C7H16 ia alkane. H16 - H10 = H6, Some examples CHBr2
6/2 = 3. Three degrees on unsat'n.
Br CH3CH=CHC
CCH2CHBr2
Br
2. Organooxygen compounds: C, H, O Oxygen is divalent and does not affect the formula. Examine: CH3-H, vs. CH3-O-H CH3CH2-CH2CH3
vs. CH3CH2-O-CH2CH3
So C5H8O is equivalent to C5H8 when calculating deg. of unsat'n. Some examples OH
O CH2=CHCH=CHCH2 OH
[Since S is also divalent, the same rule applies as with O]. 3. Organonitrogen compounds: C, H, N
Nitrogen is trivalent. A compound containing N has one more H than a corresponding hydrocarbon. Examine: CH3-H vs. CH3-NH2 CH3-CH3
vs. CH3-NH-CH3
So subtract on H from molecular formula of the compounds to compare to hydrocarbon. C5H9N is equivalent to C5H8. Summary: 1. Add the number of halogens to the number of hydrogens. 2. Ignore the number of oxygens or sulfurs. 3. Subtract the number of nitrogens from the number of hydrogens. Nomencalture 1. Find the longest carbon chain containing the double bond. Name the compound with same stems from alkanes but use suffix -ENE.
CH3CH2 C
CH3CH2
CH2
C
CH3CH2CH2
CH2
CH3CH2CH2
Six-carbon chain does NOT contain the double bond
Five-carbond chain contains the double bond - so a PENTENE
2. Number the chain beginning are the nearer the double bond (C=C has lowest possible number). CH3 CH3CH2CH2 CH=CHCH3 6
5
4
3
2
1
CH3 CH CH=CHCH2CH3 1 6
2 5
3 4
4 3
5 2
6 1
In both cases, is a 3-hexene, but top numbers give CH3 lower number.
3. Write full name numbering substituents as to position and listing alphabetically. Indicate by the first number the position of the double bond. If more than one double bond present, indicate position of each by number of first carbon and the suffix -DIENE, -TRIENE, etc.
CH3CH2
CH3 C
CH2
CH3CH2CH2 CH=CHCH3
CH3 CH CH=CHCH2CH3
CH3CH2CH2
2-ethyl-1-pentene
2-hexene
2-methyl-3-hexene CH3
CH3
CH3
Problem CH3 6.6 (b) in text.
CH3
1-methylcyclohexene 1,4-cyclohexadiene 1,5-dimethylcyclopentene
4,4-dimethylcycloheptene
Some common names H2C=CH2 is commonly known as ethylene (systematic name = ethene) CH3CH=CH2 is propylene (systematic name = propene) Sometimes an unsaturated group has to be named as a substituent. H2C
CH2=CH
CH2=CHCH2
Electronic Structure of Alkenes The carbons in the double bond are sp2 hybridized. The carbon - carbon bond consists of one sigma bond from the overlap of an sp2 from each carbon. The second bond in the double bond is a pi bond which results from overlap of the unhybridized p orbitals. In M.O. Theory, two p orbitals give two molecular orbitals - one bonding (lower energy) and one antibonding (higher energy). The two electrons in the bond are in the bonding orbital.
p Antibonding molecular orbital Combine
p Bonding molecular orbital
Two p atomic orbitals
The two p orbitals must be parallel to get "maximum overlap". If they are not aligned there is no overlap and therefore there is no bonding. This means there is no free rotation around a C=C.
Cis-Trans Isomerism The two compounds below cannot be superimposed and therefore are different. H3C
CH3 C
H3C
C
H C
C
3C HH3C
H
CH3
CC CC HH H CH3 Cis-trans isomerism occurs when each of the carbons in the double bond are attached to two different things. If one carbon in the double bond has two identical groups on it, there is no cis-trans isomerism. H
H
H
CH3
Cis-trans nomenclature fails with tri- or tetra-substituted alkenes. So a system of sequene rules was devised which eliminates the problem. Sequence Rules: The E-Z Designation
Each carbon in the double bond is considered separately. The sequence rules are applied to each carbon and a priority is assigned to the substituent on that carbon. After determining the priority on the second carbon, the relative positions of the two high priority groups are compared. If the two high priority groups are on the same side of the double bond, use the designation, Z- (zusammen, German for together). High
High
Low
Low
High
Low
Low
High E double bond (entgegen)
Z double bond (zusammen)
If the two high priority groups are on opposite sides of the double bond, use the designation, E- (entgegen, German for opposite).
Cahn-Ingold-Prelog Sequence Ruules Remember to consider each carbon separately. 1. Look at the atom DIRECTLY attached to the carbon. An atom with a higher atomic number receives a higher priority than a lower atomic number. Example: H H3C
Cl
CH3
(E)-2-chloro-2-butene
H H3C
CH3 Cl
(Z)-2-chloro-2-butene
2. If a decision cannot be reached by ranking the atom directly attached to the C=, go out to the second atoms out. Then the third, fourth, etc until a difference is determined so that a decision can be made.
H
H H lower than
C H H
C C H
O H
H H
H lower than
C CH3 H
H
lower than
O C H H
CH3
CH3
C CH3
C NH2
H
H
H lower than
C Cl H
3. Multiple bonds are equivalent to the same number of single-bonded atoms. This is just a bookkeeping method for determining priorities. In the examples below only the atoms in bold print are real. The equivalent structures are needed when and if you have to go out to further atoms to make a decision. H C
is equivalent to
O
H
H
O C O
H is C equivalent H to
C
C
H
C C
C C H
H
C
C
C
is equivalent to
H
C C
C C
H
C
Examples H H
C C
H3C
CH2
CH(CH3 )2
H
C
C CH3
(E)-3-Methyl-1,3-pentadiene
H3C
H
C
C CH2OH
Z configuration
C
H3C E configuration
CN
Problem 6.11 (d)
CH2NH 2
Alkene Stability When the two isomers of butene are hydrogenated, different amount of energy are given off per mole. Since both go to the same product, the one giving off less energy must have been of lower eneergy to start with. H3C
CH3 C
H2/Pd
C
H
H
H3C
H2/Pd
CH3CH2CH2 CH3
DH = -120kJ/mol
H C
C
DH = - 116kJ/mol H
CH3
There is steric strain in the cis isomer that is not present in the trans isomer. This can also be calculated from equilibrium data. When several alkenes are examined it is found that the more alkyl substituents on the alkene carbons the more stable the molecule. When several alkene are studies in this manner, it is found that the greater the number of alkyl substituents on the alkene carbons, the more stable the alkene. So tetrasubstituted alkenes are more stable than trisubstituted alkenes, etc. R C R
R
R >
C R
R
R C
>
C
H
R
R
H C
C
R
= H
R C
H
C
R >
H
H C
H
Two explanations have been postulated about this. One explanation is called hyperconjugation. If electrons can "delocalize" (spread out) over several atoms instead of localizing on one - such as in resonance - than that is a stabilizing effect. It is theorized that the possibility of interaction of the electrons of the adjacent alkyl substituent can interact with the unfilled antibonding pi orbital. Antibonding C-C pi orbital (unfilled)
H C
C
Bonding C-C sigmai orbital (filled)
C
Another argument is based on bond strength and compares the bonds in the alkenes. A bond between and sp3 and an sp2 is a little stronger than a bond between two sp3 bonds.
C H
3
2
2
sp -sp Ø
3
3
sp -sp Ø
CH3-CH=CH-CH3
3
3
2
sp -sp , sp -sp Ø Ø
vs
CH3-CH2-CH=CH2
Electrophilic Additions Reactions to Alkenes Remember the reaction we saw as an example of a polar reaction in Chapter 5. Br
H+ Br-
H
H
H
H C
C
-
H H
H
C
C H
Br H
H
H C
C
H
H
H
Review the energy profile for this reaction ( a two step reaction) that we saw in Chapter 5. This reaction is called electrophilic addition to the double bond because the electrophile, H+, adds first. This general reaction allows us to prepare a variety of haloalkanes from alkenes. Cl
H3C C
CH2
+ HCl
Ether
H3C
H3C
CH3 CH3
CH3
CH3 + HBr
Ether
Br
(HI is usually not added directly but is generated in the reaction vessel using KI and H3PO4). REVIEW HOW REACTIONS ARE WRITTEN AND WHAT GOES ABOVE/BELOW THE ARROW, ETC. Notice the first reaction above. There are two possible ways that the H-Cl can add across the double bond. The two products are shown below. But notice that one product is not formed.
H3C H3C
H Cl C
CH2
H C
Cl Cl
CH2
H3C
H3C
CH3 CH3
H3C
H
H H3C C
CH2
Cl
H3C
CH3
CH2Cl
NOT FORMED
CH3
When HX is added to an unsymmetrical alkene one product is formed either in much greater yield than or exclusively over the other possible addition product. A Russian chemist, Vladimir Markovnikov examined many such reactions and from the results postulated his "rule". Markovnikov's Rule: In the addition of HX to an alkene, the H attaches to the carbon with fewer alkyl substituents and the X attaches to the carbon with more alkyl substituents. (Re=-examine the two reactions above). When both carbons in the double bond have the same degree of substitution, a mixture of products is formed. Show example (p202). This type of reaction is called a REGIOSPECIFIC reaction. Let's go back and look at this mechanism (steps). The H+ can become attached to either of the carbons of the double bonds. Two different carbocations are formed. Obviously one is formed over the other. H3C H3C
H Cl C
CH2
H C
Cl Cl
CH2
H3C
H3C
CH3 CH3
H3C
H
H H3C C CH3
CH2
Cl
H3C
CH2Cl CH3
Structurally, carbocations are planar - sp2 hybridized - with a vacant p orbital. Similar to boron trifluoride.
NOT FORMED
sp2
R 'R
C
F
R"
B
F
F
Experimental data shows that 3˚ carbocations are easier to form than 2˚, etc. Thus the stability of carbocations is as shown below. R R
R +
C R
>
R
R +
C
>
H
H
C+ H
H >
H
C+ H
The reasons for this are in part inductive effects and in part hyperconjugation. Inductive effects - electrons in bonds in the alkyl groups shift in response to the positive charge on the carbocation. Alkyl groups are polarizable. R
C R
R
Hyperconjugation - Interaction of the vacant p orbital with the C-H orbital of the adjacent alkyl groups is a stabilizing effect. H C
C
HAMMOND POSTULATE How does stability of the intermediate carbocation affect the rate at which it is formed? Hammond postulate states a reasonable explanation of experimental facts. When comparing two similar reactions, the more stable intermediate forms faster than the less stable one.
E
E Slower reaction
Less stable intermediate
Slower reaction Less stable intermediate
More stable intermediate
More stable intermediate
Reaction progress
Reaction progress
Hammond Postulate states: The structure of a transition state resembles the structure of the nearest stable species. Thus what stabilizes the intermediate also stabilized the transition state and the energy diagram on the left is the more likely scenario. The more stable intermediate forms faster because the transition state is of lower energy. Draw other energy profiles to show endothermic reaction, transition state resembles product exothermic reaction, transition state resembles starting materials
Carbocation Rearrangements It is known that with certain alkenes, in reactions with the addtion of HX, structural rearrangements occur during the reaction. If H-X added in one concerted step this could not happen. This discovery was supporting evidence for the stepwise theory and the formation of carbocations. Carbocation rearrangements occur to form more stable cations with what is called a hydride shift or an alkyl shift. In both cases, a group (:H- or :CH3 -) moves to a positively charged carbon taking its electron pair with it. ! H H3C
C
H3C
H
H C
+
C
H-Cl
H
H3C
C
H3C
H
H H3C
C C
H3C
C
C
C
+
H Cl
H3C
H
H
H
C
C
H
+
H
H
H3C H3C
C C H
H C
C
H
.!!
H
C
C H
H
H
H C H
Show other examples with alkyl shift CH3
H
H3C H3C C
H C
H3C H3C
H
Cl
H
CH3
H
Cl
H
C H
H
