Alkenes Bonding and Structure - School of...
Transcript of Alkenes Bonding and Structure - School of...
Alkenes
Bonding and Structure:• Carbons in the double bond of butene are sp2 hybridized.
• Side on p-p orbital overlap creates a π-bond.
• Angles around the carbons in the double bond are ~ 120º. Thus, all of theatoms bonded to the sp2 hybridized carbon lie in a plane.Carbon-Carbon double bond length is ~ 1.34 Å (single bonds in alkaneare ~ 1.54 Å.
H3C H
H3CCH3
H
H3C
H
HCH3
H
H
CH390° 90°
• In order to interconvert between the isomers with the methyl groups onthe same, and opposite sides, the double bond must be broken.
• As a result the two isomers do not interconvert at ordinary temperatures.
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Introduction to Stereoisomers
• For 2 butene there are two isomers with the same connectivity, one withthe methyl groups on the same side of the double bond and the otherwith the methyl groups on the opposite sides.
H3C H
H3CCH3
H
H
H
CH3
cis-2-butenetrans-2-butene(Z)-2-butene(E)-2-butene
• Thus, while the compounds have the same connectivity, they have theiratoms fixed in different regions of space. Such isomers are calledstereoisomers. The carbons are called stereo centers or stereogenicatoms.
• The cis case, both methyl groups are on the same side of the double bondand in the trans case, opposite sides of the double bonds.
• For more complicated olefins, a system based on the priorities of thegroups attached to the double bond is used. This is called the E and Zpriority system where E is used for the isomer with the high prioritygroups on the opposite side of the double bond and Z when they are onthe same side—more later.
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Nomenclature (IUPAC)
i. The same prefixes are used for alkenes as alkanes but the ane is replacedby ene.
The principle chain is the longest chain containing the greatest numberof double bonds. (It is not necessarily the longest chain in the molecule).
Example:
2-propyl-1-hexenenot an octane (longest hydrocarbon chain)not 2-butyl-1-pentene
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34
5 6
345
ii. The chain is numbered such that the double bond(s) receive(s) thelowest number(s). (First point of difference rule applies.)
Example:
1 2 3 4 5 6 7 1234567
3-heptene...correct 4-heptene...incorrect
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First point of difference:
Example:
2,3-diethyl-4-methyl-1,3-pentadiene
not3,4-diethyl-2-methyl-2,4-pentadiene
First point of difference with numbering of double bonds
2,3-diethyl-4-methyl-1,3-pentadiene
# of double bondsposition of double bonds
position of substituents
1
2
34
51
23
4
5
5
For cycloalkenes, begin numbering at the double bond and proceed throughthe bond in the direction to generate the lowest number at the first point ofdifference (FPOD). (Thus a double bond should always be carbons 1 and2.)
One of the most common mistakes in naming cycloalkenes is to generatethe lowest number sequence around the ring, disregarding this rule. Onceagain, the numbering must begin at the double bond and proceed throughthe bond in the direction to generate the lowest number sequence.
1,5-diethylcyclopentenenot 2,3-diethylcyclopentenenot 1,2-diethylcyclo-2-pentene
Note # of ethyl determined by FPOD
Note that the alkene is assumed to be in position 1and thus is not stated explicitly
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2
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Some common names of molecules containing double bonds:
H2C CH2
ethylene propylene styrene isoprene
Some common names of groups containing double bonds:
HC CH2
R
R
R
vinyl allyl isopropenyl
Examples:
BrBr
allylbromide isopropenylbromide3-bromo-1-propene 2-bromo-1-propene
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Nomenclature for stereoisomers: The Z (zusammen, together in German)and (E entgegen, across in German) nomenclature.
Ca
High
Low
Cb
Low
High
Ca
High
Low
Cb
High
LowZ-isomer E-isomer
Z is similar to cis E is similar to trans
Cis and trans typically used only when the low priority groups are hydrogens.1) Consider each carbon of double bond separately.2) Rank the priority of the substituent on the carbon as follows:
The atom with the higher molecular weight takes top priority. If there are twoisotopes of the same atom, the isotope with the higher mass takes priority.
Ca
Cl
H
Cb
H3C
Br
Low
High
Low
High
Ca
Cl
H
Cb
H3C
H3C
Low
High
Ca
H
D
Cb
H3C
Br
Low
High Low
High
Z-isomer E-isomer no stereocenter on Cb
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If the atom directly attached to the alkene does not distinguish, then movedown the chain of the substituent assigning priorities down the chain untilyou reach the first point of difference to assign the priority.
CH2CH2Cl
CH2CH2CH2CH2IH
H3C
FPOD
FPOD
Cl > C
High
LowLow
High
Z-isomer
Multiple bonds count as multiples of that same atom.
Ca
X
X
Cb
H
alkene
treated as Ca
X
XCbH
alkene
Ca Cb
Example:
Ca
HC
CH2OH
Cb
H3C
H O
CaCb
H3C
H O
C
H
O
counts as
[O,H,H]
[O,O,H]
Low
LowHigh
High
H
H OH
E-isomer
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Calculating Degrees of Unsaturation
Alkanes, (CnH2n+2), are fully saturated.
Cycloalkanes, (CnH2n), have two fewer hydrogens than the parent hydrocarbonsince two of the hydrogens on the terminal methyl groups are lost in order toform the carbon-carbon bond that closes the ring: cyclopentane is C5H10, whilepentane is C5H12.
Alkenes, have one, (CnH2n), or more double bonds. An alkene with one doublebond has two fewer hydrogens than the saturated hydrocarbon of the samelength. For example, butene (H2C=CHCH2CH3) has the molecular formulaC4H8 and butane (CH3CH2CH2CH3) has the formula C4H10.
Alkynes have one, (CnH2n-2), or more triple bonds. An alkyne with one triplebond has four fewer hydrogens than the saturated hydrocarbon of the samelength. For example, butyne (HC≡CCH2CH3) has the molecular formula C4H6
and butane (CH3CH2CH2CH3) has the formula C4H10.
Knowing this relationship, it is possible to take a molecular formula andcalculate the degree of unsaturation; that is, the total number of multiple bonds orrings in a molecule.
For hydrocarbons, the process is simple: take the parent hydrocarbon andcalculate the number of hydrogens using the 2n + 2 rule, every two hydrogensthat are "missing" in the analysis of the unknown represents one degree ofunsaturation.
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For compounds containing elements other than carbon and hydrogen, degrees ofunsaturation can be calculated as follows:
Halogen containing compounds: since a halogen is simply a replacement fora hydrogen in an organic molecule (a valence of one), you simply add thetotal number of halogens to the carbon-hydrogen analysis, and calculate theunsaturation number as described above.
Oxygen containing compounds: since oxygen is divalent, it has no effect onthe calculation for the degree of unsaturation, and can simply be ignored.Considering methanol (CH3OH) removing the oxygen yields methane(CH4). For a carbonyl, (i.e., acetone, CH3COCH3), ignoring the oxygengives C3H6, two hydrogens short of (2n + 2), and one degree of unsaturation.The carbonyl is therefore equivalent to one degree of unsaturation.
Nitrogen containing compounds: since nitrogen is trivalent, a nitrogencontaining compound has one more hydrogen than an equivalenthydrocarbon has. Therefore you should subtract the number of nitrogensfrom the total number of hydrogens and calculate as described above.
The unsaturation number, U is the sum of the number of multiple bondsand/or the rings in a compound. Double bond adds one, triple bond addstwo, (these bonds need not be between the same type of atoms), ring addsone.
U = 0.5 × [2C + 2 + N - ( H + X)]C = # of CarbonsN = # of NitrogensH = # of HydrogensX = # of Halogens
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Examples: C8H8 U = 0.5 x [2(8) + 2 + 0 - ( 8)] = 5
CH3
H
CC
CC
CC
C
H
H
H
H
H
CH3
1 ring4 double bonds0 triple bonds
1 ring2 double bonds1 triple bonds
2 rings3 double bonds0 triple bonds
0 rings3 double bonds1 triple bonds
Examples: C9H16ClNO2 U = 0.5 x [2(9) + 2 +1 - ( 16 + 1)] = 2
O
O
Cl
H2N
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Relationship between Equilibrium Constant and Free Energy
∆Gº = ∆Hº-T∆Sº∆Gº is the Gibbs free energy change at equilibrium for standard states ofreactants and products(for gases 1 atm, for solution [1 M]).∆Hº is the standard enthalpy of the reaction.∆Sº is the standard entropy of the reaction.T is the temperature in Kelvin.
So, ∆Gº is defined as the free energy change, ∆G, of an ideal solution in whicheach of the reactants and products are present at 1 Molar. For such a system∆G= ∆Gº.
However, at equilibrium ∆G = 0, and concentration of all of the componentsneed not be 1 Molar. At this point:
∆G = ∆Gº + RT ln Keq
So ∆Gº = -RT ln Keq
Where:R is the universal gas constant 1.987 x 10-3 kcal/deg mol, and
Keq =[Products]eq
[Reactants]eq
Or ∆Gº = -RT2.3 log Keq
∆Gº, is the driving force of a reaction. As we can see from the equation above, itis related to equilibrium constant:
logKeq =-ΔG0
2.3RT
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Reactions tend to go in the direction to create more stable compounds.
Gproducts
Gproducts
GreactantsGreactants
ΔG0 > 0,Keq < 1
ΔG0 < 0,Keq > 1
REACTANTS PRODUCTS
Reaction goes to the left Reaction goes to the right
FREE
EN
ERG
Y
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Reaction Coordinate Diagram
Reactants Products
ΔGo
ΔG‡
Transition state
Stan
dard
Fre
e En
ergy
Reaction Coordinate
(activation energy)
• The height of the barrier for the transition state (ΔG‡) determines the rateof the reaction.
• The transition state is an activated complex that is the highest energypoint on a simple reaction coordinate.
• The transition state is unstable with respect to both the reactants andproducts, and, therefore, is not present at any appreciable concentration andexists for extremely short periods of time.
• The direction of the reaction is determined by standard free energy of thereaction (ΔG°).
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Examples:
A B A BBA
E
reaction doesn’t occur reaction slow reaction fastΔG‡ is extremely high ΔG‡ is high ΔG‡ is low
A AA
AB
BB
B
fast fast slow slow(reactants favored) (products favored) (reactants favored) (products favored)
ΔG‡ is low ΔG‡ is low ΔG‡ is high ΔG‡ is highΔG° is positive ΔG° is negative ΔG° is positive ΔG° is negative
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More Complex Reaction Coordinate Diagrams
Intermediate
Transition state(t.s.) #1
Transition state(t.s.) #2
ReactantsProducts
ΔGo
ΔG‡(1) ΔG‡(2)
Reaction Coordinate
Stan
dard
Fre
e En
ergy
Highest energy processis slowest step, so thisis the rate determiningstep, RDS.
• A reaction mechanism is a detailed series of steps that track relatively lowenergy bond breaking and making processes, which taken in sequence,provides a chemically reasonable path from the reactants to the products.
• Such a mechanism may typically involve the formation of intermediates,which are metastable species that are higher in energy than either thereactant or the product but lower in energy than the transition state.
• Intermediates may or may not have significant lifetimes.
• The rate determining step is the slowest step in a multistep process. This actsas a bottleneck for the progress of a reaction that involves intermediates.
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Hammond Postulate
There is no fundamental relationship between the energy of the transitionstate and the stability of the reactants and products (intermediates). However,there are trends that often occur and are worth noting.
Slower ReactionLess stableproduct
Less stableproduct
More stableproduct
More stableproduct
Slower ReactionFaster reaction
Faster reactionEn
ergy
Ener
gy
Reaction coordinate Reaction coordinate
The reaction with the lowerenergy transition state (fasterreaction) gives the more stableproduct (intermediate).
The reaction with the lowerenergy transition state(faster reaction) gives lessstable product (intermediate).
In general, the situation on the left is what is observed and NOT the right-hand case. Thus, typically the more stable intermediate gives the fasterreaction, when there is a choice.
George Hammond summarized these observations in what is now known asthe Hammond Postulate.
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The diagrams below help to illustrate the point. On the left, the reaction is
endergonic and the transition state is similar in energy as the product. On
the right, the reaction in exergonic and the transition state is similar in
energy to the reactant.
Reactant
ReactantProduct
Product
T.S. T.S.
Ener
gy
Reaction Coordinate
Ener
gy
Reaction Coordinate
Hammond Postulate: The structure of a transition state resembles that of the
nearest stable species (in energy). The structure of the transition state on the
left resembles that of the product and the structure of the transition state on
the right resembles that of the reactant.
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Solvents
Although we draw chemical reactions with reagents on the left of an arrowand the products on the right, the details of how the reaction is carried out inthe lab are absolutely critical for a successful outcome.
Various aspects of conditions that must be considered (not in any particularorder of importance) are:
1. order of addition of reagents 2. rate at which one reagent is added to the other 3. temperature at which the reagents are mixed 4. temperature at which the reaction is allowed to proceed 5. time of the reaction 6. details of stoichiometry--is one reagent used in excess 7. concentration of the reagents 8. is an equilibrium forced to completion by removing one of the products 9. adequate stirring10. use of catalyst11. inherent incompatibility of reagents12. choice of solvent
A chemist must consider each of these factors carefully when planning areaction.Lack of consideration of these factors will often result in: mixtures ofproducts, no desired product, decomposition of the desired product, oraccidents -- explosions, liberation of noxious gases, etc.For now we will focus a bit on properties of solvents. In the next chapters itwill become clear why choices of particular solvents are favorable for variousreactions.
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Choosing Solvents
Choice of solvent is dictated by many things including boiling point,solubility of the reactants and products, whether the solvent is inert towardreactants and products, ability of the solvent to stabilize transition states byvirtue of its polarizability, its polarity, or its ability to form hydrogen bonds.
While there are many ways to classify solvents, organic chemists typically usethree classifications:1. Polar or apolar (really a continuous gradation)2. Protic and aprotic (referring to its ability to act as a hydrogen bond donor)3. Donor or nondonor (referring to its Lewis basity)
There are two functional definitions of solvent polarity: • One is simply the dipole moment of the solvent • The other refers to the solvent ability to (by whatever mechanism)
stabilize charge or decrease the attractive or repulsive interactionbetween ions
This latter property goes back to our discussion of ions, dipoles and theenergy of charges interacting with each other back in the chapter on acidsand bases. As promised, the energy of interaction of charges rears its uglyhead once again.
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• One simple definition of a solvent polarity in the latter sense of the word is themagnitude of a solvent’s dielectric constant.
The potential energy of two charges separated by some distance is given by:
Kq1q2εr12
PE =
The ε in the denominator is the dielectric constant of the medium between thetwo charges.
The larger the dielectric constant, the lower the interaction energy between ions.
• A solvent, which lowers the interaction between charges, is said to screen orshield the charges from each other.
Organic chemists (somewhat arbitrarily) define a polar solvent as one with adielectric constant greater than 15. Solvents lower the interaction energybetween ions by aligning their dipoles around the ion in such a manner as tostabilize the ion as shown below.
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Also, specific interactions between the solvent and the ions can have aprofound effect on a solvent's ability to screen ions.
In particular, the ability of a solvent to act as either a hydrogen bond donoror acceptor plays an extremely important role in solvent stabilization.
Protic (i.e., hydrogen bond donor) solvents can form hydrogen bonds withLewis bases because they preferentially stabilize anions.
OH H
O HH
OH
HOH
H
OH
HOH H
OH
HOH
H
OH
HO H
H
OH
H
OH
HOH H
OH
HOH H
OH
HO HH
OHH
Real molecules are more complex than arrows representing dipoles.Consider DMSO (below), its dipole moment arises from a significantcontribution of the right hand, charge-separated resonance structure.However, the positive end is sterically congested relative to the oxygen. Asa result, the oxygen can have a closer approach to a positively (or partiallypositively) charged species than the anion. Thus, positively (or partiallypositively) charged species will be preferentially stabilized by solvents inwhich the negative end of the dipole is exposed and the positive end issterically crowded.
H3C CH3SOr
r' dimethylsulfoxideDMSO
H3C CH3SO
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Effects of Solvation on Reaction Rates
ΔG
Reference Solvent Solvent thatstabilizes T.S.
Reference Solvent Solvent thatstabilizes reactants
ΔGΔG
ΔG
ΔΔG is negativereaction is faster
ΔΔG is positivereaction is slower
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Classification of Solvents
SolventDielectricConstant
ε
DipoleMomentµ (D)
Polarizabilityα (cm3 x 10–24)
Apolar ProticCH3COOH 6.15 1.68 5.16(CH3)3COH 12.47 1.66 8.82
Hexanol 13.3 1.55 12.46Apolar Aprotic
CCl4 2.24 0 10.49Hexane 1.88 0.085 11.87
Diethylether 4.34 1.15 8.92Tetrahydrofuran (THF) 7.58 1.75 7.92
Polar ProticWater 78.5 1.84 1.48
HCOOH 58.5 1.82 3.39Methanol 32.70 2.87 3.26Ethanol 24.55 1.66 5.13
Polar AproticCH3COCH3 (Acetone) 20.70 2.69 6.41(CH3)2NCHO (DMF) 36.71 3.86 7.0CH3CN (Acetonitrile) 37.5 3.44 4.95
DMSO 46.68 3.9 7.99