Transcript Wade slides Chapter 7

Structure and Synthesis
of Alkenes
Introduction
• Alkenes are hydrocarbon with carbon-carbon
double bonds.
• Alkenes are also called olefins, meaning “oilforming gas”.
• The functional group of alkenes is the carboncarbon double bond, which is reactive.
Chapter 7
2
Sigma Bonds of Ethylene
Chapter 7
3
Orbital Description
• Sigma bonds around the double-bonded
carbon are sp2 hybridized.
• Angles are approximately 120º and the
molecular geometry is trigonal planar.
• Unhybridized p orbitals with one
electron will overlap forming the double
bond (pi bond) .
Chapter 7
4
Bond Lengths and Angles
• sp2 hybrid orbitals have more s character
than the sp3 hybrid orbitals.
• Pi overlap brings carbon atoms closer
shortening the C—C bond from 1.54 Å in
alkanes down to 1.33
Å in alkenes.
Chapter 7
5
Pi Bonding in Ethylene
• The pi bond in ethylene is formed by overlap of the
unhybridized p orbitals of the sp2 hybrid carbon
atoms.
• Each carbon has one unpaired electron in the p
orbital.
• This overlap requires the two ends of the molecule to
be coplanar.
Chapter 7
6
Cis-Trans Interconversion
• Cis and trans isomers cannot be interconverted.
• No rotation around the carbon—carbon bond is
possible without breaking the pi bond (264
kJ/mole).
Chapter 7
7
Elements of Unsaturation
• Unsaturation: A structural element that
decreases the number of hydrogens in the
molecule by two.
• Double bonds and rings are elements of
unsaturation.
• To calculate: Find number of hydrogens if
they were saturated, subtract the actual
number of hydrogens, then divide by 2.
Chapter 7
8
Example: Calculate the Unsaturations
for a Compound with Formula C5H8.
• First calculate the number of hydrogen atoms for a
saturated compound with five carbons:
(2 x C) + 2
(2 x 5) + 2 = 12
• Now subtract from this number the actual number of
hydrogen atoms in the formula and divide by 2:
12 – 8 = 4 = 2 unsaturations
2
2
• The compound has two unsaturations, they can be
two double bonds, two rings, or one double bond and
one ring.
Chapter 7
9
Elements of Unsaturation:
Heteroatoms
• Halogens replace hydrogen atoms in
hydrocarbons, so when calculating
unsaturations, count halides as hydrogen
atoms.
• Oxygen does not change the C:H ratio, so
ignore oxygen in the formula.
• Nitrogen is trivalent, so it acts like half a
carbon. Add the number of nitrogen atoms
when calculating unsaturations.
Chapter 7
10
Elements or Degrees of
Unsaturation: p-Bond or Ring
Halogens replace Hydrogens
C5H10Cl 2 is saturated
C6H9Br has 2 degrees of unsaturation
Br
Br
Cl
Cl
Br
Br
Saturated compounds with
Oxygen and Nitrogen
H
H
C
H
H
C
H
H
C
H
C2H6
CnH2n+2
O
H H
C
H
H
H
H
H
H
C2H6O
CnH2n+2 O
H
C
N
H H
C
H
C2H7N
CnH2n+3 N
Determine the # elements of
Unsaturation
C8H10Br2O
C8H18 is saturated
compound is deficient by 6 "H's"
3 degrees of unsaturation
CH2OH
Br
e.g.
O
O
H
Br
Br
Br
Br
Br
CH3
Example: Calculate the Unsaturations for
a Compound with Formula C4H7Br.
• First calculate the number of hydrogens for a
saturated compound with four carbons:
(2 x C) + 2 + N
(2 x 4) + 2 = 10
• Now subtract from this number the actual number of
hydrogens in the formula and divide by 2.
Remember to count halides as hydrogens:
10 – 8 = 2 = 1 unsaturation
2
2
Chapter 7
16
Example: Calculate the Unsaturations
for a Compound with Formula C6H7N.
• First calculate the number of hydrogens for a
saturated compound with six carbons. Add the
number of nitrogens:
(2 x C) + 2 + N
(2 x 6) + 2 + 1 = 15
• Now subtract from this number the actual number of
hydrogens in the formula and divide by 2:
15 – 7 = 8 = 4 unsaturations
2
2
Chapter 7
17
IUPAC Nomenclature
• Find the longest continuous carbon chain
that includes the double-bonded carbons.
• -ane changes to -ene.
• Number the chain so that the double
bond has the lowest possible number.
• In a ring, the double bond is assumed to
be between Carbon 1 and Carbon 2.
Chapter 7
18
IUPAC and New IUPAC
Chapter 7
19
Alkene Nomenclature
Ring Nomenclature
In a ring, the double bond is assumed to
be between Carbon 1 and Carbon 2.
CH3
3
1
CH3
2
2
1
1-methylcyclopentene
3-methylcyclopentene
Chapter 7
22
Multiple Double Bonds
• Give the double bonds the lowest numbers
possible.
• Use di-, tri-, tetra- before the ending “-ene” to
specify how many double bonds are present.
Chapter 7
23
Cis-Trans Isomers
• Similar groups on same side of double bond, alkene
is cis.
• Similar groups on opposite sides of double bond,
alkene is trans.
• Not all alkenes show cis-trans isomerism.
Chapter 7
24
Cyclic Compounds
• Trans cycloalkenes are not stable unless the
ring has at least eight carbons.
• Cycloalkenes are assumed to be cis unless
otherwise specifically named trans.
Chapter 7
25
Cyclic alkenes
Alkyl Groups with p-Bonds
CH=CH2
a vinyl group
(or ethenyl)
CH2CH=CH2
an allyl group
(or 2-propenyl)
Br
cis 1-bromo-3-vinyl cyclohexane
a phenyl group
3-phenyl -1-nonene
4-allyl cyclopentene
Alkylidene Groups
Double Bonds Fused to Rings
CH2
methylene cyclohexane
CHCH3
ethylidene cyclopentane
Polyenes
Name these Alkenes
cis-trans Isomers
Cis and Trans
no rotation about p-bond
Solved Problem 1
Which of the following alkenes are stable?
Solution
Compound (a) is stable. Although the double bond is at a bridgehead, it is not a bridged bicyclic
system. The trans double bond is in a ten membered ring. Compound (b) is a Bredt’s rule violation and
is not stable. The largest ring contains six carbon atoms, and the trans double bond cannot be stable in
this bridgehead position.
Compound (c) (norbornene) is stable. The (cis) double bond is not at a bridgehead carbon.
Compound (d) is stable. Although the double bond is at the bridgehead of a bridged bicyclic system,
there is an eight-membered ring to accommodate the trans double bond.
Chapter 7
33
E-Z Nomenclature
• Use the Cahn–Ingold–Prelog rules to
assign priorities to groups attached to
each carbon in the double bond.
• If high priority groups are on the same
side, the name is Z (for zusammen).
• If high priority groups are on opposite
sides, the name is E (for entgegen).
Chapter 7
34
Example
1
2
1
2
E-1-bromo-1-chloropropene
• Assign priority to the
substituents according
to their atomic number
(1 is highest priority).
• If the highest priority
groups are on opposite
sides, the isomer is E.
• If the highest priority
groups are on the same
side, the isomer is Z.
Chapter 7
35
E/Z System
Establish Priority of
2
Substituents on Each sp
Carbon
What’s My Name?
(Z)-4-ethyl-5-isopropyl-4-nonene
Lo
Lo
Hi
Hi
Priorities with Multiple Bonds
E or Z?
E or Z?
Name These
3,3-dimethyl-1,4-cyclohexadiene
2-butyl-1,5-hexadiene
(E) 1-ethylidene-2-methylcyclopentane 3-allylcyclohexene
Hydrogenation Data Helps to
Determine Stability
DHhydrogenation of Alkenes
Enthalpy Change Shows
Relative Energy of Alkene
Both cis and trans 2-Butene are
Hydrogenated to Butane
“E” is More Stable than “Z”
by 2.3 KJ/mol
Relative Stabilities of Alkenes
Hyperconjugation
p bond associates with adjacent C-H s bond
1-butene
trans 2-butene
C
C
C
C
mono-substituted
disubstituted
Commercial Uses of Ethylene
Chapter 7
55
Commercial Uses of
Propylene
=>
Chapter 7
56
Addition Polymers
Chapter 7
57
Heat of Hydrogenation
• Combustion of an alkene and hydrogenation of an
alkene can provide valuable data as to the stability of
the double bond.
• The more substituted the double bond, the lower its
heat of hydrogenation.
Chapter 7
58
Relative Stabilities
Chapter 7
59
Substituent Effects
• Among constitutional isomers, more substituted
double bonds are usually more stable.
• Wider separation between the groups means less
steric interaction and increased stability.
Chapter 7
60
Disubstituted Isomers
• Stability: cis < geminal < trans isomer
• The less stable isomer has a higher exothermic
heat of hydrogenation.
cis-2-butene
CH3
C C
H
iso-butene
trans-2-butene
CH3
H
(CH3)2C=CH2
H
-120 kJ
C C
CH3
Chapter 7
CH3
-117 kJ
-116 kJ
H
61
Stability of Cycloalkene
• Cis isomer more stable than trans in small cycloalkenes.
• Small rings have additional ring strain.
• Must have at least eight carbons to form a stable trans double
bond.
• For cyclodecene (and larger), the trans double bond is almost
as stable as the cis.
Chapter 7
62
Bredt’s Rule
• A bridged bicyclic compound cannot have a
double bond at a bridgehead position unless
one of the rings contains at least eight carbon
atoms.
Chapter 7
63
Physical Properties of Alkenes
•
•
•
•
Low boiling points, increasing with mass.
Branched alkenes have lower boiling points.
Less dense than water.
Slightly polar:
 Pi bond is polarizable, so instantaneous dipole–
dipole interactions occur.
 Alkyl groups are electron-donating toward the pi
bond, so may have a small dipole moment.
Chapter 7
64
Polarity and Dipole Moments of
Alkenes
• Cis alkenes have a greater dipole moment than trans
alkenes, so they will be slightly polar.
• The boiling point of cis alkenes will be higher than the
trans alkenes.
Chapter 7
65
Alkene Synthesis
Overview
•
•
•
•
E2 dehydrohalogenation (-HX)
E1 dehydrohalogenation (-HX)
Dehalogenation of vicinal dibromides (-X2)
Dehydration of alcohols (-H2O)
Chapter 7
66
Dehydrohalogenation by the E2
Mechanism
• Strong base abstracts H+ as double bond forms and
X- leaves from the adjacent carbon.
• Tertiary and hindered secondary alkyl halides give
good yields.
Chapter 7
67
Bulky Bases for E2 Reactions
• If the substrate is prone to substitution, a bulky base
can minimize the amount of substitution.
• Large alkyl groups on a bulky base hinder its
approach to attack a carbon atom (substitution), yet it
can easily abstract a proton (elimination).
Chapter 7
68
Hofmann Products
Bulky bases, such as potassium tert-butoxide,
abstract the least hindered H+ giving the less
substituted alkene as the major product (Hofmann
product).
Chapter 7
69
Stereochemistry of E2 Elimination
• Depending on the stereochemistry of the alkyl halide,
the E2 elimination may produce only the cis or only
the trans isomer.
• The geometry of the product will depend on the anticoplanar relationship between the proton and the
leaving group.
Chapter 7
70
Solved Problem 2
Show that the dehalogenation of 2,3-dibromobutane by iodide ion is stereospecific by showing that the
two diastereomers of the starting material give different diastereomers of the product.
Solution
Rotating meso-2,3-dibromobutane into a conformation where the bromine atoms are anti and coplanar,
we find that the product will be trans-2-butene. A similar conformation of either enantiomer of the (±)
diastereomer shows that the product will be cis-2-butene. (Hint: Your models will be helpful.)
Chapter 7
71
E2 Reactions on Cyclohexanes
• An anti-coplanar conformation (180°) can only be
achieved when both the hydrogen and the halogen
occupy axial positions.
• The chair must flip to the conformation with the axial
halide in order for the elimination to take place.
Chapter 7
72
Solved Problem 3
Explain why the following deuterated 1-bromo-2-methylcyclohexane undergoes dehydrohalogenation
by the E2 mechanism, to give only the indicated product. Two other alkenes are not observed.
Solution
In an E2 elimination, the hydrogen atom and the leaving group must have a trans-diaxial relationship.
In this compound, only one hydrogen atom—the deuterium—is trans to the bromine atom. When the
bromine atom is axial, the adjacent deuterium is also axial, providing a trans-diaxial arrangement.
Chapter 7
73
Dehalogenation of Vicinal
Dibromides
• Remove Br2 from adjacent carbons.
• Bromines must be anti-coplanar (E2).
• Use NaI in acetone, or Zn in acetic acid.
Chapter 7
74
E1 Elimination Mechanism
• Tertiary and secondary alkyl halides:
3º > 2º
• A carbocation is the intermediate.
• Rearrangements are possible.
• Weak nucleophiles such as water or alcohols.
• Usually have substitution products too.
Chapter 7
75
Dehydration of Alcohols
• Use concentrated sulfuric or phosphoric acid,
remove low-boiling alkene as it forms to shift
the equilibrium, and increase the yield of the
reaction.
• Carbocation intermediate: 3º alcohols react
faster than 2º. Primary alcohols are the least
reactive.
• Rearrangements are common.
• Reaction obeys Zaitsev’s rule.
Chapter 7
76
Dehydration Mechanism: E1
Step 1
Step 2
Step 3
Chapter 7
77
Solved Problem 4
Propose a mechanism for the sulfuric acid–catalyzed dehydration of t-butyl alcohol.
Solution
The first step is protonation of the hydroxyl group, which converts it to a good leaving group.
The second step is ionization of the protonated alcohol to give a carbocation.
Abstraction of a proton completes the mechanism.
Chapter 7
78
Catalytic Cracking of Alkanes
• Long-chain alkane is heated with a catalyst to
produce an alkene and shorter alkane.
• Complex mixtures are produced.
Chapter 7
79
Alkenes
Properties
Nomenclature
Stability
Addition Reactions
The Addition Reaction
HBr Addition
Markovnikov’s Rule
The addition of H-X across a
double bond results in the more
highly substituted alkyl halide as
the major product.
Depicting a Reaction
Addition of HBr or HCl
Markovnikov Addition
H
CH3
C
CH3
C
H
Br H
HBr
H
CH3
C
C
CH3
H
Markovnikov
CH3
H
Br
C
C
CH3
H
H
not formed
Regiochemistry Determined by
Stability of Intermediate
Br H
HBr
H-Br
Br
H
o
3 carbocation
Hyperconjugation
Carbocation Stability
more highly substituted, lower energy
3o Carbocation forms
Preferentially
What Alkenes are Needed to
form Theses Alkyl Halides?
Definitions
• Regioisomers – two constitutional isomers
that could result from an addition reaction.
• Regiospecific – only one regiosisomer
forms at the expense of the other.
• Regioselective – both regioisomers are
formed, but one is formed in preference.
Determine the major product:
a)
b)
c)
HCl
HBr
2 mol HCl
a)
Cl
HCl
H
b)
c)
HBr
2 mol HCl
Br
H
H
Cl
Cl
H
cis and trans
Rearrangements
CH3
CH3
HCl
CH3
Cl
H
H Cl
CH3
CH3
CH3
methide shift
CH3
H
2
o
o
3 CH3
Cl
Propose a Mechanism
2 Possible pathways to the
Carbocation Intermediates
Endo- transition state looks like product
Exo- transition state looks like Reactant
Hammond’s Postulate
• Related species that are close in energy are close
in structure.
• In an endothermic reaction, the transition state is
similar to the product in structure and stability.
• In an exothermic reaction, the transition state is
similar to the reactant in structure and stability.
• i.e. the structure of the transition state resembles
the structure of the most stable species.