Transcript Practice Problem - HCC Southeast Commons

Chapter 6
Introduction
• An alkene
is a hydrocarbon that contains a
carbon-carbon double bond
 It is also called an olefin but alkene is better
 It is abundant in nature
 It is an important industrial product
• It includes many naturally occurring materials
– flavors, fragrances, vitamins
I. Alkenes: An Overview
A.
Industrial Preparation and
Use of Alkenes
B.
Degree of Unsaturation
C.
Naming Alkenes
A.
Industrial Preparation and Use of
Alkenes
Ethylene and propylene are
– the simplest alkenes
– the most important organic chemicals produced
industrially
Compounds derived from ethylene and propylene
Industrial Synthesis of Ethylene, Propylene and Butene
• Ethylene, propylene, and butene are synthesized
industrially by thermal cracking of light (C2-C8)
alkanes:
Industrial Synthesis of ethylene, propylene and butene
• The high-temperature conditions cause spontaneous
homolytic breaking of C-C and C-H bonds, forming
smaller fragments:
• Thermal cracking is an example of a reaction whose
energetics is governed by TDSo rather than DHo
DGo = DHo – TDSo
TDSo > DHo
B.
Degree of Unsaturation
• Alkenes - are unsaturated hydrocarbons.
• They have fewer hydrogens than alkanes with same
number of carbons
• Acyclic Alkanes - have the general formula
CnH2n+2
where n is an integer
• Acyclic Alkenes - have the general formula
CnH2n
where n is an integer
• Degree of Unsaturation is the number of p bonds
and/or rings in the molecule.
• Each ring or multiple bond replaces 2 H's in the alkane
formula CnH2n+2
• It relates molecular formula to possible structures
• Degree of Unsaturation is the number of p bonds
and/or rings in the molecule.
Degree of Unsaturation =
(2n+2) - x
2
where n is the number of carbons
x is the number of hydrogens
Example: A hydrocarbon with a molecular weight
(82 g/mol) C6H10
• Saturated is C6H14
– Therefore 4 H's are not
present
H H
H H
C
C
H3C
C
C
H
• This has two degrees
of unsaturation
–
–
–
–
Two double bonds?
or triple bond?
or two rings?
or ring and double bond?
H
H H
CH3
Degree of Unsaturation With Other Elements
1. Organohalogens (C, H, X where X: F, Cl, Br, I)
• Halogen replaces hydrogen
• Add the number of halogens to the number of
hydrogens
•
C4H6Br2 and C4H8 have one degree of unsaturation
2. Organooxygens (C, H, O)
•
•
Oxygen forms two bonds. It does not affect the total
count of H's
Ignore the number of oxygens
•
C5H8O and C5H8 have two degrees of unsaturation
3. Organonitrogens (C, H, N)
•
•
Nitrogen has three bonds. So if it connects where H
was, it adds a connection point
Subtract the number of nitrogens from the number of
hydrogens
Summary: Degree of Unsaturation
Degree of Unsaturation can be calculated:
• Count pairs of H's below CnH 2n+2
• Add number of halogens to number of H's (X
equivalent to H)
• Ignore or don't count oxygens (O links H)
• Subtract number of nitrogens from H's (N has
two connections)
Degree of Unsaturation and Variation
• Compounds with the same degree of unsaturation
can have many things in common and still be very
different
Practice Problem: Calculate the degree of unsaturation in the
following hydrocarbons:
a. C8H14
b. C5H6
c. C12H20
d. C20H32
e. C40H56 (b-carotene)
Practice Problem: Calculate the degree of unsaturation in the
following formulas, and then draw as many
structures as you can for each:
a. C4H8
b. C4H6
c. C3H4
Practice Problem: Calculate the degree of unsaturation in the
following formulas:
a. C6H5N
b. C6H5NO2
c. C8H9Cl3
d. C9H16Br2
e. C10H12N2O3
f. C20H32ClN
C.
Naming Alkenes
• Like alkanes, alkenes are named according to the
system devised by the International Union of Pure and
Applied Chemistry (IUPAC).
Steps to naming alkenes1
1. Name the parent hydrocarbon
a. Find the longest continuous carbon chain containing the
double bond
b. Name using the suffix -ene
Steps to naming alkenes2
2. Number the carbon atoms in the main chain
a. The double-bond carbons receive the lowest possible
numbers
b. The correct sequence is when the substituents have the
lowest possible number
Steps to naming alkenes3
3. Write the full name
a. Name, number and list the substituents alphabetically
b. Indicate the position of the double-bond
Steps to naming alkenes3
3. Write the full name
c. If more than one double bond is present:
– indicate the position of each and
– Use the suffixes -diene, -triene, …
Naming Cycloalkenes
• Like alkenes, cycloalkenes are also named by the
rules devised by the International Union of Pure and
Applied Chemistry (IUPAC).
Steps to naming cycloalkenes1
1. Name the parent hydrocarbon
a. Number the cycloalkene so that the double bond is
between C1 and C2
b. Name using the prefix cyclo- and the suffix -ene
Steps to naming cycloalkenes2
2. Number the substituents and write the name
•
The correct sequence is when the substituents have the
lowest possible number
Common Names of Alkenes
• Many alkenes are known by their common names.
• These common names are recognized by the
International Union of Pure and Applied Chemistry
(IUPAC).
Practice Problem: Give IUPAC names for the following
compounds:
Practice Problem: Draw structures corresponding to the
following IUPAC names:
a. 2-Methyl-1,5-hexadiene
b. 3-Ethyl-2,2-dimethyl-3-heptene
c. 2,3,3-Trimethyl-1,4,6-octatriene
d. 3,4-Diisopropyl-2,5-dimethyl-3-hexene
e. 4-tert-Butyl-2-methylheptane
Practice Problem: Name the following cycloalkenes:
II. Alkenes: Structure
A.
Electronic Structure
B.
Cis-Trans Isomerism
C.
Sequence Rules: The E, Z
Designation
D.
Stability of Alkenes
A.
Electronic Structure of Alkenes
• Carbon atoms in a double bond are sp2-hybridized
– Three equivalent orbitals are in a plane at 120º angle
– Fourth orbital is atomic p orbital (perpendicular to the plane)
Two sp2-hybridized Carbon atoms form:
•  bond
– Head-on overlap of two sp2 orbitals forms a  bond
• p bond
– Side-to-side overlap of two p orbitals forms a p bond
Molecular Orbitals (MO)
• Additive interaction of p orbitals (combining p orbital
lobes with the same algebraic sign) creates a p
bonding orbital
• Subtractive interaction (combining lobes with
opposite signs) creates a p anti-bonding orbital
• Only bonding MO is occupied.
• Occupied p orbital prevents rotation.
– Rotation prevented by p bond - high barrier, about 268
kJ/mole in ethylene
Rotation of p Bond is Prohibitive
• The p bond does not have circular cross-section.
• The p bond must break for rotation to take place around
a carbon-carbon double bond (unlike a carbon-carbon
single bond).
• It creates possible alternative structures
B.
Cis-Trans Isomerism
• The presence of a carboncarbon double can create
two possible structures
– cis isomer - two similar
groups on same side of the
double bond
– trans isomer - similar
groups on opposite sides
Cis-Trans Isomerism requires that end groups must
differ in pairs
Each carbon must have two different groups for these
isomers to occur
Cis-Trans Isomerism requires that end groups must
differ in pairs
• 180°rotation superposes
the upper pair
• Bottom pair cannot be
superposed without
breaking C=C
X
Practice Problem: Which of the following compounds can exist
as pairs of cis-trans isomers? Draw each cistrans pair, and indicate the geometry of each
isomer.
a. CH3CH=CH2
b. (CH3)2C=CHCH3
c. CH3CH2CH=CHCH3
d. (CH3)2C=C(CH3)CH2CH3
e. ClCH=CHCl
f. BrCH=CHCl
Practice Problem: Name the following alkenes, including the cis
or trans designation:
C.
Sequence Rules: The E, Z Designation
• Neither compound is
clearly “cis” or “trans”
– Substituents on C1 are
different than those on C2
– We need to define “similarity”
in a precise way to
distinguish the two
stereoisomers
• Cis-trans nomenclature
only works for
disubstituted double
bonds
Develop a System for Comparison of Priority
of Substituents
• Assume a valuation system
– If Br has a higher “value” than
Cl
– If CH3 is higher than H
• Then, in A, the higher value
groups are on opposite
sides
• In B, they are on the same
side
– Requires a universally
accepted “valuation”
E,Z Stereochemical Nomenclature
• Priority rules of Cahn, Ingold,
and Prelog are used.
– Compare where higher priority
group is with respect to C=C
bond and designate a prefix
– E - entgegen: opposite sides
– Z - zusammen: together on the
same side
E - entgegen
Z - zusammen
Cahn-Ingold-Prelog Rules
1. Ranking Priorities
a. Look at the atoms directly attached to each double-bond
carbon
b. Rank them according to atomic number. Higher atomic
number gets higher priority.
In this case, the higher priority
groups are opposite:
(E )-1-bromo-1-chloro-propene
Cahn-Ingold-Prelog Rules
2. Extended Comparison
a. If atomic numbers are the same for the first atoms,
compare the second, third, or fourth atoms away from
the double bond carbons
b. Compare until something has higher atomic number
c. Do not combine – always compare
Cahn-Ingold-Prelog Rules
3. Dealing With Multiple Bonds
a. Multiple-bonded atoms are equivalent to the same
number of single-bonded atoms.
• Substituent is drawn with connections shown and no
double or triple bonds
• Added atoms are valued with 0 ligands themselves
Practice Problem: Which member in each of the following sets
has higher priority?
a. -H or -Br
b. -Cl or -Br
b. -CH3 or -CH2CH3
c. -NH2 or -OH
d. -CH2OH or –CH3
e. -CH2OH or –CH=O
Practice Problem: Which member in each of the following sets
has higher priority?
a. -CH3, -OH, -H, -Cl
b. -CH3, -CH2CH3, -CH=CH2, -CH2OH
c. -CO2H, -CH2OH, -C≡N, -CH2NH2
d. -CH2CH3, -C≡CH, -C≡N, -CH2OCH3
Practice Problem: Assign E or Z configuration to the following
alkenes:
Practice Problem: Assign stereochemistry (E or Z) to the
following alkene, and convert the drawing into
a skeletal structure (red = O)
D.
Stability of Alkenes
• Cis alkenes are less stable than trans alkenes.
– This is due to steric (spatial) strain between the two
bulky substituents on the same side of the double bond
<
Determination of Relative Stabilities of Alkenes
The relative stabilities of alkenes can be determined by:
1. establishing a cis-trans equilibrium through treatment
with strong acid
2. comparing the heats of combustion (DHocombustion) for
the two isomers
3. comparing the heats of hydrogenation (DHo hydrogenation)
for the two isomers
Less stable isomer is higher in energy and gives off
more heat; More stable isomer is lower in energy and
gives off less heat.
1. Cis-trans equilibrium through treatment with strong acid
• After treatment of 2-butene with strong acid, at equilibrium,
the trans isomer is more favored than the cis isomer by a
ratio of 76 to 24 :
• Cis-2-butene is less stable than trans-2-butene by 2.8 kJ/mol
DE = - RT ln K
2. Heats of combustion (DHocombustion)
• Comparing the heats of combustion of the two isomers of 2butene indicates that cis-2-butene is more strained than
trans-2-butene by 3.3 kJ/mol:
3. Heats of hydrogenation (DHohydrogenation)
• Alkenes undergo a hydrogenation reaction on treatment
with H2 gas in the presence of a catalyst (Pd or Pt):
• More energy is released in the hydrogenation of the cis
isomer than the trans isomer (DGocis > DGotrans)
because the cis alkene has a higher energy level.
• The energy difference between the 2-butene isomers
as calculated from the heats of hydrogenation
(DHohydro) is ~ 4kJ/mol and is in agreement with the
other two methods.
– Trans-2-butene generates 4 kJ/mol less heat than cis-2-butene
Alkenes become more stable with increasing substitution:
Stabilities of Alkenes
As a general rule, alkenes follow the stability order:
This order of stability is due to:
• hyperconjugation
• bond strength
Hyperconjugation
Hyperconjugation - is a stabilizing interaction between
the unfilled antibonding C=C p bond orbital and a
filled C-H  bond orbital on a neighboring substituent
The more substituents, the more opportunities exist for
hyperconjugation and the more stable the alkene
• Electrons in neighboring filled  orbital stabilize
vacant antibonding p orbital – net positive
interaction
• Alkyl groups are better than H
Bond Strength
• A bond between an sp2 carbon and an sp3 carbon is
stronger than a bond between two sp3 carbons.
sp2-sp3 > sp3-sp3
• More highly substituted alkenes always have a
higher ratio of sp2-sp3 bonds to sp3-sp3 bonds than
less highly substituted alkenes and thus are more
stable
Practice Problem: Name the following alkenes, and tell which
compounds in each of the following pairs are
more stable:
III. Alkenes: Reactivity
A.
Electrophilic Addition of HX
to Alkenes
B.
Orientation of Electrophilic
Addition: Markovnikov’s
Rule
C.
Carbocation Structure and
Stability
III. Alkenes: Reactivity
D.
The Hammond Postulate
E.
Carbocation Rearrangements
A.
Electrophilic Addition of HX to
Alkenes
• General reaction mechanism of electrophilic
addition:
– Attack on electrophile (such as HBr) by a p bond of
alkene (nucleophile)
– This produces carbocation and bromide ion
– Carbocation is an electrophile, reacting with nucleophilic
bromide ion
Mechanism: electrophilic addition of HBr to
2-methylpropene
• Addition of hydrogen
bromide to 2-Methylpropene
• H-Br transfers proton to
C=C
• Forms carbocation
intermediate
– More stable cation
forms
• Bromide adds to
carbocation
Electrophilic Addition Energy Path
• Two step process
• First transition state is high energy point
Energy Diagram for Electrophilic Addition
• Rate determining (slowest)
step has highest energy
transition state
– Independent of direction
– In this case it is the first
step in forward direction
– “rate” is not the same as
“rate constant”
Electrophilic Addition for preparations
• The reaction is successful with HCl and with HI as
well as HBr
• Note that HI is generated from KI and phosphoric acid
Writing Organic Reactions
•
•
•
•
•
No established convention – shorthand
Can be formal kinetic expression
Not necessarily balanced
Reactants can be before or on arrow
Solvent, temperature, and details are on arrow
• Reactants can be before or on arrow
• Solvent, temperature, and details are on arrow
B.
Orientation of Electrophilic
Addition: 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
regiospecific
Markovnikov’s Rule
• In the 19th century, Markovnikov observed that:
In the addition of HX to alkene, the H attaches to
the carbon with the most H’s and X attaches to
the carbon with the most alkyl substituents
– This is Markovnikov’s rule
Examples of Markovnikov’s Rule
• Addition of HCl to 2-methylpropene
• Regiospecific – one product forms where two are
possible
• If both ends have similar substitution (the same
degree of substitution), then not regiospecific
Markovnikov’s Rule (Restated)
• Markovnikov’s rule can be restated:
In the addition of HX to alkene, the more highly
substituted carbocation is formed as the
intermediate rather than the less highly
substituted one.
Energy of Carbocations and Markovnikov’s Rule
• More stable carbocation forms faster
• Tertiary cations and associated transition states are
more stable than primary cations
Mechanistic Source of Regiospecificity in
Addition Reactions
• If addition involves a
carbocation intermediate
– and there are two
possible ways to add
– the route producing the
more alkyl substituted
cationic center is lower
in energy
– alkyl groups stabilize
carbocation
Practice Problem: Predict the products of the following
reactions:
Practice Problem: What alkenes would you start with to prepare
the following alkyl halides?
a. Bromocyclopentane
b. 1-Ethyl-1-iodocyclohexane
C.
Carbocation Structure and Stability
• Carbocations are planar
• The trivalent carbon:
– is sp2 hybridized
– has a vacant p-orbital
Structure
• Carbocations are planar
• The trivalent carbon is sp2
hybridized.
• The tricoordinate carbon is
surrounded by only 6
electrons in sp2 orbitals
• The three substituents are
oriented to the corners of
equilateral triangle
• The fourth orbital on
carbon is a vacant p-orbital
Stability
• The stability of carbocations increases with increasing
substitution:
• More highly substituted carbocations are more stable than
less highly substituted ones.
• Therefore stability of carbocations: 3º > 2º > 1º > +CH3
• The stability of carbocation (R+) can be determined by
measuring energy needed to form it from R-X:
R-X  R+ + :X• 3º alkyl halides dissociate to give R+ more easily than 2º
and 1º
Why are more highly substituted carbocations more
stable than less highly substituted ones?
Carbocations follow the stability order:
This order of stability is due to:
• Inductive effects
• hyperconjugation
Inductive Effects
Inductive Effects – result from shifting of electrons in a 
bond in response to the electronegativity of nearby
atom
–
Electrons from a relatively large and polarizable alkyl group can
shift toward neighboring positive charge more easily than the
electron from H
–
The more alkyl groups attached to positively charged carbon,
the more electron density shift and the more inductive
stabilization
Hyperconjugation
Hyperconjugation - is a stabilizing interaction between a
vacant p orbital and a properly oriented neighboring C-H
 bond.
–
The more alkyl groups there are on the carbocation, the more
possibilities there are for hyperconjugation and the more stable
the carbocation
Practice Problem: Show the structures of the carbocation
intermediates you would expect in the
following reactions:
Practice Problem: Draw a skeletal structure of the following
carbocation. Identify it as primary, secondary,
or tertiary, and identify the hydrogen atoms
that are involved in hyperconjugation in the
conformation shown:
D.
The Hammond Postulate
• Electrophilic addition to an unsymmetrically
substituted alkene gives the more highly
substituted carbocation:
– A more highly substituted carbocation forms faster
and rapidly goes to the final product.
– A more highly subsbtituted carbocation is more
stable than a less substituted one
Why does the stability of the carbocation intermediate
affect the rate of its formation?
– The relative stability of the carbocation intermediate
is determined by DGº
•
It is related to an equilibrium constant
– The reaction rate is determined by DG‡.
•
The relative stability of the transition state (which describes
the size of the rate constant) is the activation energy (DG‡)
– This is not a direct relationship
In general, the faster reaction forms the more stable
intermediate and the slower reaction forms the less stable
intermediate
Typical
Transition State Structures
• A transition state is the highest energy species in a
reaction step
– By definition, its structure is not stable enough to exist
for one vibration
– The transition state is transient and cannot be
examined
– But the structure controls the rate of reaction
– Its properties may be determined in an informed way -- the Hammond Postulate
Hammond’s Postulate
The structure of a transition state resembles the structure
of the nearest stable species
– Transition states for endergonic steps structurally resemble
products
– Transition states for exergonic steps structurally resemble
reactants
Statement of the Hammond Postulate
• A transition state should be similar to an intermediate
that is close in energy
• Sequential states on a reaction path that are close in
energy are likely to be close in structure - G. S.
Hammond
In a reaction involving
a carbocation, the
transition states look
like the intermediate
G
carbocation
Reaction
Competing Reactions and the Hammond Postulate
• Normal Expectation:
Faster reaction gives more
stable intermediate
• Intermediate resembles transition state
“Non-Hammond” Behavior
• More stable intermediate from slower reaction
• Conclusion: Transition state and intermediate must
not be similar in this case – not common
How does the Hammond postulate apply to
electrophilic addition reactions?
Formation of carbocation by protonation of an alkene is an
endergonic step:
– The transition state is similar to carbocation intermediate
– Any factor that makes the carbocation stable makes the
nearby transition state more stable
Transition State for Alkene Protonation
• Resembles carbocation
intermediate
• Close in energy and
adjacent on pathway
• Hammond Postulate
says they should be
similar in structure
Practice Problem: What about the second step in the
electrophilic addition of HCl to an alkene – the
reaction of chloride ion with the carbocation
intermediate? Is this step exergonic or
endergonic? Does the transition state for this
second step resemble the reactant
(carbocation) or product (alkyl chloride)?
Make a rough drawing of what the transitionstate structure might look like.
E.
Carbocation Rearrangements
• Carbocations undergo structural rearrangements.
• A less stable carbocation rearranges to a more stable
ion.
• Carbocations undergo
structural rearrangements
following set patterns:
– 1,2-hydride shift
– 1,2-alkyl shift
• Goes to give more stable
carbocation
• Can go through less
stable ions as
intermediates
• Hydride shift –
is the shift of a hydrogen atom and
its electron pair (a hydride :H-)
between neighboring carbons
• Alkyl shift – is the shift of an alkyl and its electron pair
(an alkyl anion :R-) between neighboring
carbons
Practice Problem: On treatment with HBr, vinylcyclohexane
undergoes addition and rearrangment to yield
1-bromo-1-ethylcyclohexane. Using curved
arrows, propose a mechanism to account for
this result:
Chapter 6