Download Alkenes Key features sp -hybridized carbons, 120 bond angles

Alkenes
(β-carotene,
an antioxidant
pigment)
CnH2n
(acyclic)
R
R
C
CnH2n-2
(cyclic)
C
R
R
Key features
sp2-hybridized carbons, 120o bond angles
σ + π bonding between C = C
planar geometry around C = C
"unsaturated" hydrocarbons
stereoisomerism is common
Nomenclature: summary of important rules
1. Root of name depends on longest chain of C containing the double bond; ends
in "ene"
H 3C
C
Cl
C
H
H
C
CH 3
CH 3
2. Numbering begins with end closest to C = C
3. Substituents are numbered and named using same rules as for alkanes (lowest
possible numbers, ABC order, etc).
4. Cycloalkene rings are numbered with the
C = C bond taking positions 1 & 2:
5. Cis/trans or E/Z used to distinguish geometric isomers (more on this later)
6. New "R" groups:
vinyl: - CH = CH2
allyl: - CH2 - CH = CH2
Using formulas to predict number of rings & π bonds:
Review: "Degrees of unsaturation" formula:
# of rings + π bonds = 1/2 (2C + 2 - H - X + N)
"Cis-trans" isomerism:
Geometric isomerism may exist based on placement of atoms/groups around
C = C bond.
Free rotation of groups around C = C is eliminated due to π-bonding …
interconversion requires breaking of π bond (64 kcal/mole)
For cis-trans isomers to exist, each of the C = C carbon atoms must be bonded to
two different groups or atoms; usually there is one H atom on each C:
2-butene:
H 3C
C
H
C
H
CH 3
H3C
Cis-2-butene
(same side)
C
(opposite)
H
Trans-2-butene
CH3
C
1-butene:
H
(only one form exists)
C
H
H3C
C
H
H
H2C
C
CH 3
CH3
C
H
H
To visualize cis/trans isomers, draw C = C first, then arrange groups at 120oC angles
Geometric isomerism affects shape, physical properties, packing, molecular recognition.
E/Z nomenclature system:
When there are 3 or more different groups or atoms bonded to the C = C, the E/Z
priority system is used instead of cis/trans designations:
1. Groups around C = C are assigned priority based on the atomic numbers of the two atoms
bonded to each carbon. The atom with highest atomic number gets highest priority.
Br > Cl > S > P > O > N > C > H
2. In case of a "tie", priority is based on the next bonded atom. Double bonds count as two
bonds to the next atom
3. Compare relative positions of the highest priority group on each C of the C=C:
E (entgagen) = opposite:
Higher priority groups opposite
Z (zusammen) = together
Higher priority groups on same side of bond
Cahn-Ingold-Prelog Priority System Hints for assigning E/Z
Extended Comparison for assigning priority groups
• If atomic numbers are the same, compare at next connection
point at same distance
• Compare until something has higher atomic number
• Do not combine – always compare
Alkene Formation and Stability
Alkenes are produced industrially by thermal cracking reactions:
850-900oC
Ex:
H2 + CH4 + CH2=CH2 + CH3CH=CH2 etc.
small alkanes
They also form as products of elimination reactions from alkyl halides or alcohols
Alkene formation depends on stability: the lower the energy of a particular isomer,
the more likely it will form.
Two factors that influence alkene stability:
1. Hyperconjugation:
sp3-sp2 interactions can occur when
an unfilled C=C pi antibonding MO
is adjacent to filled C-H or C-C sigma
bonding orbitals: increases stability
Alkenes which are more highly substituted
with alkyl groups around C=C have more of
these stabilizing interactions.
2. Steric strain:
cis > trans
Close proximity of large groups in a cis-arrangement creates steric repulsion:
H3C
CH3
C
H
H
H3C
C
C
H
H
C
CH 3
Compare heats of hydrogenation in Table 6.2:
Energy is released (ΔHo ) when H2 is added to C=C bond to make an alkane
The more stable the alkane, the lower the energy
Stability-wise:
Trans > cis
More substituted > less substituted
Electrophilic Addition Reactions: Predicting products based on
stability of intermediate and transition state
In this polar reaction, only one possible structure forms from the intermediate:
H
H 3C
C
C
H
H
2-butene
CH3
Br
H3C
H
C
H
Br C
CH3
H 3C
H
H
C
H
Br
C CH3
H
2-bromobutane
In the reaction of 2-methylpropene, two possible pathways:
Why does only one product actually form? From energy diagrams, we know that the
more stable the intermediate, the faster it will be produced…if more than one intermediate is
possible, the more stable one predominates (the smaller the ΔGt, the faster the reaction).
Trend in relative stabilities of carbocations:
Why? The flow of electrons toward a positively charged C stabilizes the
charge; Alkyl groups provide greater electron density than a hydrogen atom.
What does the transition state look like?
--Partially formed and broken bonds; may resemble either reactants or products
Hammond postulate: the transition state will be closer in structure to the
species that is closest in energy - either the reactant or product.
In an electrophilic addition, the rate-determining step is carbocation
formation. It is an exergonic process, in which the transition state
will more closely resemble the carbocation intermediate.
Relative stability of carbocation intermediates therefore determines product
Regioselectivity of reactions: Which side does the nucleophile end up?
Some reactions favor reaction to occur with a particular orientation, where a given
region of the molecule is more likely to participate in bond formation.
This regioselectivity occurs with hydrogen halide addition to alkenes, so products
have predictable structures based on the pattern of reactivity.
One major product forms (one or more minor products may form as well)
Ex: 1-methylcyclohexene + HBr
Looking back at the methyl propene reaction:
The halide ends up on the most substituted carbon.
The H+ must have added to the less-substituted carbon of the C = C bond.
Markovnikov’s Rule: The electrophile adds to C = C in a way that produces the
most stable carbocation intermediate
Markovnikov's Rule Restated: "The rich get richer rule"
For electrophilic addition to alkenes, the hydrogen goes to the carbon atom
which is already bonded to the greatest number of hydrogens (also results in
more stable carbocation structure)
Thus reaction products from electrophilic can be predicted in 2 ways:
1.) Apply the “rich get richer” rule: the C with more H’s gains another H
2.) Figure out which carbocation structure would be more stable…that positivelycharged carbon will then bond to the nucleophile
Both ways should lead to the same answer, but method (2) helps predict whether
rearrangement will occur.
Carbocations and Rearrangement:
How can you explain these outcomes?
CH3
H3C
C
H
H
C
H
CH2
mostly
CH3
H3C
C
CH3
H
C
H
CH3
Br
H 3C
C
Br
CH 3
Br
CH 2
mostly
H3C
H2
C CH3
C
H
C
Cl
CH 3
CH3
Look at the carbocation intermediate in each case and think about how its stability
could be improved upon…by rearranging it.
3 common types of rearrangement occur in reactions with a carbocation
intermediate:
1) 1,2-hydride shift:
CH3
In first reaction:
H3C
C
H
H
C
CH3
CH3
H 3C
C
H2
C CH3
2o carbocation becomes a 3o carbocation
2) 1,2-methyl shift:
CH3
In the second rxn:
H3C
C
H
C
CH3
CH3
H 3C
CH3
C
H
C
CH3
CH3
2o carbocation becomes a 3o carbocation
3) Ring-expansion:
CH3
H+
CH 3
CH3
CH 3
C
H
C
H
CH 3
C
H
C
H2
CH 3
C
H2
What are the intermediates?