alkyl halides substitution and elimination 1 nomenclature ... · alkyl halides : page 1 alkyl...
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Alkyl Halides : page 1
Alkyl Halides Substitution and Elimination
1 Nomenclature • Look for the longest chain that contains the maximum number of functional groups, in this case the halogen is the functional group and so even though the cyclohexane has more carbon atoms, the main chain is the two carbon ethane chain, the structure is named as a substituted alkyl bromide
2 Second Order Nucleophilic Substitution (SN2) Reaction Substitution by making a new bond at the same time as breaking the old bond • Substitution requires a bond to be broken and a new bond to be formed • The LOWEST energy way of doing this (unless precluded by steric or other effects, see later) is to make the new bond (getting some energy "back") at the same time as breaking the old bond, this is SN2 Example:
• Reactions in which all bonds are made and broken at the same time are called CONCERTED • This is fundamentally just a Lewis acid/base reaction of a kind we have seen previously, the Lewis base has the high energy chemically reactive electrons, which are used to make a new bond to the Lewis acid, and a stronger bond is formed (C-O in the example above) and a weaker bond is broken (C-Br above) • HO– is the Lewis Base and Nucleophile • the halide is the Lewis acid/electrophile • the Br– anion is the Leaving Group • This reaction "goes" because…. 1) A weaker bond is converted into a stronger bond
C–Br: B.D.E. ~ 65 kcal/mol (weaker bond) HO–C: B.D.E. ~ 90 kcal/mol (stronger bond)
2) A stronger base (–OH) is converted into a weaker base (Br–) 3) HIGHER energy electrons are converted into LOWER energy electrons. • The reaction also proceeds with inversion (i.e. BACKSIDE ATTACK, think about an umbrella turning inside out in the wind), called a Walden inversion.
Often this will lead to a change in absolute configuration, i.e. R to S or vice versa, but not necessarily!
Br
(1R)-bromo-1-cyclohexylethanenamed as a substiuted alkyl halidetherefore, cyclohexane is a substituent!
12 *H
H
CBr
MeEt
H–O
H
MeEt
BrHO C
‡sp2H
CHO Me
Et
LB
LA
+ Br
partial bonds
"backside attack!"
* *
CONCERTED
leaving group
nucleophile
electrophile
(R) (S)
H–O + Br+
WALDEN INVERSION
H
MeEtBrHO C
‡H
CBr
MeEt
H
CHO
MeEt
Alkyl Halides : page 2
• Even though the transition state apparently has one more bond than either the reactants or the products, the partial bonds are very LONG and thus WEAK, partial bonds thus have very high energy electrons which is why the transition state is higher in energy than either reactants or products (2 weak partial bonds add up to less than one real bond) Examples of SN2 Reactions: Give the major organic product in the following reactions • we understand these SN2 reactions a simple Lewis acid/base processes • identify the Lewis base/NUCLEOPHILE as the reactant with the high energy electrons • the Lewis acid/nucleophile must react with the Lewis acid/ELECTROPHILE
Important: SN2 reactions are one of the most important ways of making new bonds, i.e. of transforming one organic molecule into another one • Here we made a NEW FUNCTIONAL GROUP (nitrile), we made a new C-C bond (larger molecule), we also made a ring structure, we will use SN2 a lot! 2.1 Nucleophilicity versus Basicity and SN2 Why the Name Second Order Nucleophilic Substitution (SN2)? S - Substitution reaction N - Nucleophile does the substitution (like a Lewis base, but see below) 2- kinetically 2nd order, TWO molecules are involved in the rate determining step (the only step) • The halide AND the nucleophile (2 molecules) are involved in the rate determining step and so the reaction rate depends upon the concentration of them both, the reactions is kinetically SECOND (2nd) order
reaction coordinate
Energy
CHO
H
MeEt
Br–
HO–
CBr
H
MeEt
‡
Ea
H
MeEt
BrHO C
‡partial bonds are LONGpartial bonds are WEAK
CONCERTED
Br C N
LB/NucLA/Elec
CN
BrC CCH3
LB/NucLA/Elec
CCCH3
new C–C bond
inversion
Br O Na O
Nuc "end"Elec "end"
DMF(solvent)
CH3CN(solvent)
acetone(solvent)
Na+
Na+
"intramolecular" reaction
new C–C bond
new C–O bond
makes a nitrilefunctional group
makes a ring
makes a larger moleucle
rate = k [nucleophile] [halide]
rate constant
Alkyl Halides : page 3
• an increase in the concentration of EITHER or BOTH the nucleophile and halide results in a proportional increase in reaction rate, the rate depends upon the concentration of BOTH REACTANTS What is a Nucleophile and How is it Different from/Same as a Base? The definition of a base is based on thermodynamics (Keq)
• Lewis/Brønsted base strength measured by size of Keq (thermodynamic definition) • stronger base means stronger new bond means more exothermic reaction larger Keq • weaker base means weaker new bond means less exothermic (or more endothermic) smaller Keq The definition of a nucleophile is based on kinetics (k)
• Nucleophile strength measured by size of rate constant k (kinetic definition) • stronger nucleophile means smaller Ea (stronger partial bonds), larger k, faster reaction rate • weaker nucleophile means larger Ea (weaker partial bonds), smaller k, slower reaction rate All nucleophiles are Lewis bases, the Hammond postulate says that strong bases should also be strong nucleophiles, and this is generally true, although we will meet a few important exceptions later…..
Lewis Base / Nucleophile (nucleus loving), donates electrons Lewis Acid / Electrophile (electron loving), accepts electrons
Why are we concerned with kinetics now when we used to be only concerned with the acid/base understanding of reactivity? We have already seen that when there are competing reactions, the fastest one "wins" (e.g. the most stable intermediate is formed fastest), in other words MOST organic reactions are controlled by kinetics, their reactions are kinetically controlled. For this reason, it makes sense to start talking about nucleophiles and electrophiles, because their definition is based on kinetics. Of course, most strong nucleophiles react fast because they are also strong bases and have very exothermic reactions (although there are some exceptions). We will use the terms nucleophile/electrophile and Lewis base/acid interchangeably. 2.2 Understanding SN2 Reactivity: Molecular Orbitals and Steric Effects • Here we will use a more detailed form of Lewis acid/base theory that considers the important molecular orbitals, i.e. Frontier Molecular Orbital theory (FMO Theory) • Making bonds between atoms requires overlap of atomic orbitals in phase to generate a new bonding molecular orbital • Here, we need to make a bond between two molecules • Making a bond between molecules requires overlapping molecular orbitals in phase to generate a new bonding molecular orbital • FMO theory looks at the overlap between molecular orbital with the highest energy electrons (the HOMO) of nucleophile, with the lowest energy molecular orbital of the electrophile (the LUMO) • The HOMO is where the reactive Lewis basic electrons "are", the anti-bonding LUMO is the only orbital that the electrons can "go to" in the electrophile, all of the bonding orbitals are full of electrons
H3C–O OH
HH
LB/BB LA/BA
KeqH3C–O + H2O
Hnew bond
+
H3C–O XH3C
Nucleophile(and LB)
Electrophile(and LA)
k + X+ H3C O
CH3
Alkyl Halides : page 4
• This provides the best explanation for "backside attack", HOMO/LUMO overlap best at the carbon "end" of the halide LUMO • For this reason, reaction suffers "steric hindrance" when R1, R2, R3 are large
• SN2 reactions get slower with increasing steric hindrance at the backside of the carbon of the electrophile • To the extent that there is no SN2 reaction at a tertiary halide • SN2 reactions at methyl and allylic carbons are particularly facile (see below) • There is no SN2 at a tertiary or vinyl carbons because the nucleophile cannot get close enough to form reasonable partial bonds in the transition state due to a steric effect at the other alkyl substituents on the C atoms
• The vinyl C(sp2)-X sigma* orbital is smaller than a C(sp3)-X sigma* orbital, weakens any potential partial bond • The vinyl C(sp2)-X bond is stronger than a corresponding C(sp3)-X bond SN2 at the allylic position is FASTEST because the transition state is resonance stabilized, lowering the energy of the electrons in the transition state • A lower energy transition state means a smaller activation energy which results in a faster reaction……
R3
R1R2
XNu C
‡
sp2
R3CNu
R1R2
X
NuR3C X
R1R2
σ∗ LUMO larger on Carbon
R3C X
R1R2
Nu
X antibonding
σ∗ LUMO Xno reaction
backside attackbonding
n HOMO
n HOMO
makingbond
bonding
breakingbond
ANTI-bonding
XH3C > H3CH2C > HCH3C
H3C> C
H3C
H3CH3CX X X
decreasing reactivity in SN2
1° 2° 3°
no SN2 reaction at 3° or vinyl (sp2) carbons
>
allylic fastest
CC
CH2
X~ C
CX
XNu
XX
X
steric hinderance no
reactionNu
H3CC X
H3CH3C
H
C
H
H
Nu
C
X
Nu
3°
vinylstrong C(sp2)-X bond
Alkyl Halides : page 5
2.3 Factors Controlling SN2 Reactivity: Nucleophilicity Good nucleophiles are almost always good Lewis bases (but see further below….) Comparing same atom (charged versus non-charged): Anions make stronger nucleophiles than neutrals
• Nucleophilicity is the same as basicity here! • higher energy electrons on negatively charged oxygen both react faster and have more exothermic reactions, are good Lewis bases AND good nucleophiles, this is true in all solvents (see later) Comparing similarly sized atoms (across the periodic table): Electronegative atoms make poor nucleophiles
• Nucleophilicity is the same as basicity here! • lower energy electrons on electronegative fluorine react slower and have less exothermic reactions, are weaker Lewis bases and less good nucleophiles, this is true in all solvents Comparing differently sized atoms (down the periodic table) • A new Concept: Larger atoms have more polarizable electrons, they do not have to "get so close" to make a bond, can make "longer" bonds, and thus can make stronger partial bonds in the transition state
‡
H
transition state "resonance stabilized"
CHH
X
Nu
CC
H
H
CHH
X
Nu
C
C
H
HH
‡
H
CHH
X
Nu
C
C
H
H
~H
CHH
X
Nu
CC
H
H
H3C–O BrH3C
LB/Nuc LA/Elec
H3CCH3
O + Br
BrH3C
LB/Nuc LA/Elec
H3CH
OH3CH
OCH3
ANION STRONG nucleophilefaster reaction
NEUTRAL WEAK nucleophileslower reaction
+ Br
H–O BrH3C
LB/Nuc LA/Elec
HCH3
O
BrH3C
LB/Nuc LA/Elec
FF CH3
stronger nucleophilefaster reaction
weaker nucleophileslower reaction
+ Br
+ Br
H
HH
BrHS C
H
HH
BrHO C‡
‡
H–O BrH3C HCH3
O
BrH3CH–S HCH3
S
+ Br
+ Br
MORE basic, BUT less polarizable electrons, weaker partial bond
LESS basic, BUT polarizable electrons, stronger partial bond
Alkyl Halides : page 6
• So, which "wins", basicity or nucleophilicity because of the partial-bond strength? Unfortunately, which one wins depends upon solvent (see later) • Here we will look at the case that is straightforward, where nucleophilicity and basicity are the same Example: Compare these two reactions
• The larger S makes stronger partial bonds, but the stronger bonds in the product with O simply wins! • Nucleophilicity is the same as basicity here!
2.4 Factors Controlling SN2 Reactivity: Leaving Group Ability Good leaving groups: • Are stable/less reactive as an anion • Are generally weak bases (i.e. they have WEAK bonds to, for example, H atoms) • Polarize the C–LG bond (and are polarizable to make strong partial bonds in transition state) For Example:
• Leaving Group ability increases going down the periodic, as anion stability increases • The iodide anion is stable and unreactive because it is a weak base, it makes weak bonds to H • The chloride anion is less stable and is more reactive because it is somewhat stronger base, it makes somewhat stronger bonds to H Reaction Energy Diagrams: Two Equally correct ways of illustrating this effect • Which of the following energy diagrams best illustrates why iodide is a better leaving group than bromide etc.?
H–O BrH3C HCH3
O BrH3CH–S HCH3
S+ Br + Bracetone
SLOWERFASTERstronger base
stronger nucleophile
(aproticsolvent)
acetone(aproticsolvent)
reaction coordinate
RelativeEnergy
–Br
‡BrH3C
CH3HS
stronger bond,higher exothermicity
smaller Ea
‡
CH3HO
normalized here
HO–
HS–HS–
HO–
H3C Cl OH H3C OHslowest + Cl
+ Br
+ I
SN2+
H3C Br OH H3C OH+
H3C I OH H3C OH+
increasingreaction
rate fastest
more stable anion
better leaving groupWEAKER
C-X BOND
CH3–ICH3–BrCH3–Cl
reaction coordinate
RelativeEnergy
OH+
EaI
EaCl
CH3–OH + X–
normalizedhere
higher energyelectrons in bond
CH3–ICH3–BrCH3–Cl
reaction coordinate
RelativeEnergy
OH+Ea
I EaCl
CH3–OH + Cl–
normalizedhere
least stablebase anion
CH3–OH + Br–
CH3–OH + I–
Alkyl Halides : page 7
• Both of these diagrams are equally correct • The diagram on the LEFT, the diagram shows the difference in the energies of the electrons on the bonds, the weaker (higher electron energy) bond is the most reactive (smallest Ea, largest reaction rate) • The diagram on the RIGHT shows the difference in the stabilities of the anion leaving groups, the more stable the anion the smaller the Ea, the faster the reaction • REMEMBER, the iodide anion is the most stable not because of particularly low electron energy in this case, but because if it reacts it can only make weak bonds because it is so large 2.5 Factors Controlling SN2 Reactivity: Solvent Effects • Nucleophile strength depends upon solvent, this is something new for us • Solvent effects on reactions can be dramatic • But, A complication: solvent effects are different for nucleophiles of different sizes (see below) Polar protic (hydrogen-bonding) Solvents: Mostly alcohols and water • Polar protic solvents have strong intermolecular ion-dipole forces with dissolved anions in particular • These ion-dipole interactions solvent anions strongly (think about ionic compounds dissolved in water) • The stronger the IMF, the more solvated the ion in the solvent since the interactions with the solvent lower the total electron energy more than for comparable weaker intermolecular forces
• Stronger solvation lowers electron energy and chemical reactivity, anions in particular tend to be less reactive in polar protic solvents • Polar protic solvents solvate small anions strongly, but solvate larger anions less strongly, transition states are large, and are thus much less solvated • there is often a large solvation energy difference between reactants and the transition state, particularly when using small anionic nucleophiles in a polar protic solvent • Intermolecular force is actually an unfortunate term because force and energy are not the same thing, but it is not hard to understand that if a nIMF is strong it is more likely to lower the energy of the relevant electrons more Polar Aprotic (Non-hydrogen-bonding) Solvents • There are Several, you need to know these • Polar protic solvents have weaker intermolecular ion-dipole forces with dissolved ions • The ion-dipole interactions certainly solvent and stabilize ions, but there is no H-bonding!
methanol
water
ethanol
i-propanol –
large anion MUCH less solvated/stabilized
charge is "diluted", weaker IMF
H ROδ–
δ+
H
R
Oδ+
δ–
H HO
Me HO
Et HO
i-Pr HO
weaker IMF
example polar protic solvents
H ROδ–
δ+
H
R
Oδ+
δ– small anion highly solvated/stabilized
ion-dipole IMF
dimethylformamide (DMF)
acetonitrile
hexamethylphosphoramide (HMPA)
dimethylsulfoxide (DMSO)
example polar aprotic solvents
acetone
C NMe
N HCOMe
Me
Me2N NMe2PO
NMe2
Me MeSO
Me MeCO
CH3
H3C
Cδ+δ–
O
CH3
CH3
C Oδ+ δ–
small anionBUT, "buried" + charge
much weaker IMF
learn these by using them and working practice problems
+
+
+
+
+
Alkyl Halides : page 8
• There is a usually a smaller energy difference between reactants and the transition state when SN2 reactions are performed in polar aprotic solvents, SN2 reactions in polar aprotic solvents are usually faster than in polar protic solvents (but see further below) • You need to know the polar aprotic solvents, this is not easy because they are different structures, learn them by working with them Medium Polarity or Non-Polar Solvents • These solvents should be very good for SN1 and SN2 reactions since they will solvate the anions so poorly, but that is the problem, they solvate ions so poorly that the reactants often simply don't dissolve in them! • We do see these solvents quite often in organic reactions, in fact we have seen carbon tetrachloride, CCl4) already, as an inert (non-reactive) solvent in more than one reaction • Some common low polarity solvents are summarized here but we will not use them much for Sn1/Sn2
• Medium and in particular non-polar solvents are commonly used in organic chemistry, but not so much for SN2 reactions, since these often involve ionic reactants that will simply not dissolve in non-polar solvents Example Solvent Effects
• Explain the difference in reaction rates using a reaction energy diagram
• Here we have two reactions on one diagram. The absolute energies of the two systems are very different, thus we need to normalize the energies and plot relative energy. • Where to normalize? In general, we will want to emphasize the places where the energies are different and where they are similar. In this case, the energies at the start and end are different due to large differences in anionic solvation, and the energies at the transition states are more similar due to small differences in solvation, thus we normalize at the transition state
chloroform
hexane
benzene
ethyl acetate
diethyl ether
carbon tetrachloride
cyclohexane
Et EtO
Me OCO
Et
CCl4CHCl3
CH3(CH2)4CH3
example medium and nonpolar solvents
H–O BrH3C
BrH3C
faster reaction in polar aproticslower reaction in polar protic
H3C OH
H3C OHH–O MeOH
CH3CN(acetonitrile)
(methanol)
+ Br
+ Br
anion highly solvatedlower in energy reaction coordinate
Br–
‡
H
HH
BrHO C
HO–
BrH3C
OHH3C
CH3CN
MeOH
"big anion" not solvated well in either solvent
normalized here
RelativeEnergy
‡anion less solvatedhigher in energy
Alkyl Halides : page 9
2.6 Differences Between Nucleophilicity and Basicity Bulky bases are strong bases but weak nucleophiles • Compare methoxide and t-butoxide
• Both are equally strong Bronsted bases, both make equally strong bonds to a proton (H+) • But, t-butoxide is a much weaker nucleophile than methoxide, due to electron repulsion/steric effects, it makes a weak O-C partial bond in the transition state in an SN2 reaction
• This steric effect results in the following nucleophilicity trend, note that all are equally strong bases
• SN2 is not possible using the t-butoxide anion • We will use bulky bases to our advantage in controlling the competition between SN2 and E2 reactions frequently, see later Nucleophile strength can depend upon solvent, this is something new for us • Reminder: Larger atoms have more polarizable electrons, they do not have to "get so close" to make a bond, they can make "longer" bonds, and thus can make stronger partial bonds in the transition state • But, the oxygen anion makes stronger bonds, it is smaller, and so which wins? • We already saw a comparison between these two reactions in a polar aprotic solvent • In the polar aprotic solvent the smaller anion making the stronger bond in the product wins, the stronger base is the stronger nucleophile.
CH3O C–O
CH3
CH3
H3Cmethoxide t-butoxide
equally strong bases, both make equally strong bonds to H+
H
HH
XCH3O C
‡H
CCH3O
H HX
backsideattack
bonding
HC X
HH
n HOMO σ* LUMO
CH3O
stronger partial bond
H
HH
XC–O C
‡H
CC–O
H HX
backsideattack
bonding
HC X
HH
C–O
weaker partial bond
CH3
H3C
CH3
CH3
H3C
CH3
CH3
CH3
H3C
electron repulsion
decreasing rate of reaction with an electrophile
OH3C > H3CH2C > HCH3C
H3C> C
H3C
H3C
H3CO O O
strong base andstrong nucleophile
strong, "bulky" base butweak nucleophile, no SN2
= t-BuO–
t-butoxide anionMeO–
methoxide anion
HCH3
O
HCH3
S
+ Br
+ Br
acetone(polar aprotic)
FASTER
slower
stronger baseSTRONGER nucleophile H–O BrH3C
BrH3CH–S
Alkyl Halides : page 10
• However, in a polar protic solvent the larger anion making stronger bonds in the transition state wins, the weaker base is the stronger nucleophile.
• The larger more polarizable sulfur anion forms stronger partial bonds in the transition state • And, the larger sulfur anion is not solvated well (not stabilized) by the polar protic solvent, therefore, it is more reactive • Here, nucleophilicity is opposite to basicity, solvation and polarization effects "win" over exothermicity • Reaction energy diagram curves emphasize the difference in solvation of the nucleophilic ions In a polar Aprotic solvent:
In a polar protic solvent:
Summary: • Bulky bases such as t-butoxide are weak nucleophiles (No SN2 with t-butoxide) • Nucleophilicity is always the same as basicity, except that larger anions are more nucleophilic than smaller anions in protic solvents (but not in polar aprotic solvents) Going down the periodic table often makes things complicated! Even worse, when NUETRAL nucleophiles are compared with different sized atoms, for example Me2O versus Me2S, then nucleophilicity is the same is all solvents, and Me2O is weaker than Me2S because neutral structures are not as affected by solvent and the large size and polarizability of the electrons on the larger atom wins out over electronegativity. However, I don’t think it is really fair to ask you to know this detail, and so you should assume that nucleophilicity and basicity are the same EXCEPT for ions going down the periodic table in polar protic solvents, where it is reversed, this is the only exception you need to know. Well, there is one more exception that we will met soon, which are the bulky bases, but they are actually quite easy!
HCH3
O
HCH3
S
+ Br
+ Br
MeOH(PROTIC solvent)
FASTER
slower
weaker baseSTRONGER nucleophile
H–O BrH3C
BrH3CH–SSLOWER
reaction coordinate
RelativeEnergy
–Br
‡BrH3C
CH3HS
stronger bond,higher exothermicity
smaller Ea
‡
CH3HO
normalized here
HO–
HS–HS–
HO–
polar Aprotic solvente.g. acetone
reaction coordinate
RelativeEnergy
–Br
BrH3C
CH3HS
‡
CH3HO
small solvation difference
normalized here
HO–
HS–HS–
HO–
larger anionlower solvation
H
HH
BrNu C‡
difference in solvation is small in
the very large ‡
polar protic solvente.g. MeOH
Alkyl Halides : page 11
3 First Order Nucleophilic Substitution (SN1) Reaction • What happens if we try to do an SN2 reaction with a very weak (e.g. neutral) nucleophile Lewis base?
• here we have a nucleophilic substitution reaction BUT...... • we have a 3° halide which is a weak electrophile (backside attack is not possible), can't do SN2 • H3COH is a weak nucleophile (no negative charge on the oxygen), shouldn't do SN2 • H3COH is also a PROTIC solvent, which should be slow for SN2 • here the solvent "helps" to break the C–Br bond, the reaction is a solvolysis reaction (lysis - bond breaking) We need a new substitution MECHANISM to account for this: The SN1 Mechanism
• although the alcohol is a weak LB/Nucleophile, the first cation intermediate is a STRONG LA/Electrophile, and so nucleophilic addition at this step in the mechanism is fast • the SN1 reaction requires a polar protic solvent to stabilize the ionic (cation and halide) intermediates • usually requires heat (energy) to break the C–X bond unimolecularly • ONLY the halide (not the nucleophile) involved in the R.D.S., thus SN1 (1 means only 1 reactant in the R.D.S.) • requires a stable intermediate cation, NO SN1 for methyl or primary halides
• No SN1 (OR SN2) at sp2 hybridized carbons, the C-X bond is too strong and the cations are too unstable • In general, SN1 will always occur in preference to SN1 since this makes a bond at the same time the bond is broken, unless SN2 is impossible (e.g. at a 3° carbon)
H3C–OH (solvent)OCH3C
H3C
H3CH3C
substitutionand SOLVOLYSIS
Δ (means heat)BrC
H3C
H3CH3C
3° halideweak Electrophile
NO SN2
weak LB/Nuc, not good for SN2
protic solvent, slow SN2
BrC
H3C
H3CH3C C
CH3
CH3
H3C Br
H CH3O
OC
H3C
H3CH3C
H
CH3
WEAK LB/Nuc H CH3O
OCH3C
H3C
H3CH3C
LB/BB
LA/BA
rate determining
step
ions solvated (stabilized) by polar protic solvent
STRONG LA/Elec
XH3C> H3CH2C >HCH3C
H3C>XC
H3C
H3CH3C XX
decreasing reactivity in SN1
3°
>CH2
X
CH
CH
H X X2° 1°
allylic position, next to C=C, fastest SN1, most stable cation
X
X
methyl
vinyl halideNO substitution at
sp2 hybridized carbon atoms
Alkyl Halides : page 12
3.1 Stereochemistry of SN1 Reactions: Racemization (?) Example:
• We expect racemization, or at least some loss of stereochemistry for SN1 compared to SN2 • depending upon conditions/reactants, attack on the same side as the leaving group may be hindered, resulting in a slight excess of the inversion product • in reality, however, it is not easy to predict exactly how much stereochemistry will be lost, and so we will use the "rule" in this course that if the reaction goes via SN1 we will assume that racemization always occurs 3.2 Cation Rearrangements in SN1 Reactions Example:
• Once the cation is made it will react the same as any other cation, and so this cation will rearrange 3.3 Distinguishing SN1 and SN2 Reactions
• NOTE: the factors above favor the reactions by making them go faster, e.g. SN2 is FASTER at a primary carbon, SN1 is faster at a tertiary carbon, SN1 is faster in polar protic solvents etc. • However, weak nucleophiles do not favor SN1 because they make SN1 reactions faster, they don't, but they do make competing SN2 reactions SLOWER • SN2 reactions are not precluded by polar protic solvents, they are just faster in aprotic solvents
H
CBr
MeEt
H
CCH3O Me
Et(R)– (S)–H3C–O Na/DMF
H3C–OH/heatSN1
H
C
MeEtOH3C
H OCH3
H
–HH
C OCH3EtMe(±)
racemic mixture
inversion
SN2
BrH
C OEtMe H
CH3
OH3C H
(±)
LB
LB
LA
LA/BA
LB/BB
single enantiomer
single enantiomer
Br
CH3C
H
C
CH3
CH3
CH3EtOH
boil
H3C
CH3C
H
C
OEt
CH3
CH3
CH3C
H
C
CH3
CH3
CH3 CH3C
H
C
CH3
CH3
H3C
EtO
H
2° cation 3° cation (more stable)
"alkyl-shift" H3C
CH3C
H
C
O
CH3
CH3
Et H
EtO
H
LA
LB/BB
LA/BA
LB
SN2 favored by: • 1° > 2° > 3°
• strong nucleophile• polar aprotic solvent
SN1 favored by: • 3° > 2° > 1°
• weak nucleophile• polar protic solvent AND heat
Alkyl Halides : page 13
Examples: Assign the mechanism of the following reactions to SN1 or SN2
Example Problems: Give the major organic product of reactions
• polar aprotic solvent, strong nucleophile, SN2, Br- better leaving group • 1 equivalent means exactly the same number of nucleophiles as organic reactants, which in this context means that there is only enough nucleophile to substitute one of the halide leaving groups
• polar protic solvent and heat, no strong nucleophile and allylic halide, must be SN1. Need to draw the mechanism to be sure of the product!
• polar aprotic solvent, strong nucleophile, SN2, allylic position more reactive • 1 EQUIVALENT will ONLY REACT at the carbon where SN2 will be fastest
H3C BrNa CN
DMFH3C C N + Na Br
1° halide, strong Nu, polar aprotic - SN2
Br CH3OH
heat
MeO
2° halide, weak Nu, polar protic/heat - SN1 with rearrangement
Br Na SPh
CH3CN
2° halide, strong Nu, polar aprotic - SN2 (so NO rearrangement and inversion)
SPh
K SCH3CH3CN
Cl Br1 Equivalent
Cl SCH3
BrEtOHboil
OEt
(±)
OH
Et
O EtH
OH
Et
LA
LB
LB/BB
LA/BA(±)
note that the intermediate is racemic
Br Br1 Equiv. K OCH3
DMF
H3CO Br
Alkyl Halides : page 14
4 E2 Elimination Reaction HERE is a reaction that we now know, straightforward SN2 at a primary bromide......
• what about this reaction? SN2 is not possible here (3° bromide), yet there certainly is a reaction
• does E2 instead! • breaks weak C–Br bond and strong C–H bond, makes strong O-H bond and C=C pi bond • not as exothermic as SN2, but the reaction converts 2 molecules into 3, favored by entropy AND "converts" strong -OH base into weak -Br base, this lowers the energy of these electrons, which also helps • Just like SN2, all bonds made and broken at same time (all four!) • the –OH acts as a Brønsted base, a strong base is required for E2! What we are doing here is using the chemical potential energy (reactive electrons) in the strong base to "drive" this reaction, to make it "go" • The reaction is concerted (all four bonds are made and broken at the same time) 4.1 Product Selectivity in E2: Saytzeff Rule (or Zaitsev, etc.) The Saytzeff Rule: Most substituted alkene is formed in an E2 reaction, if possible Example
• When more than one alkene isomer can be formed in an E2 elimination, the more substituted, more stable alkene isomer is usually formed (see an exception below), this more substituted alkene is called the Sayetzeff alkene, after the Russian chemist of the same name • Sayetzeff’s name was translated differently from the Cyrillic into Roman alphabet in different countries, hence Sayetzeff can be spelled multiple ways, Zaitsev is another common spelling • consider how these two E2 products are formed
Br
CC
H H
H
H
HCC
H
HHHH
OHSN2
DMF
H O
+ Br1° bromide substitution
Br
CC
H CH3
H
CH3
H
E2
X
+ BrCCH CH3
CH3HHO–H +
BrCC
H CH3
H
CH3
H
HO‡
DMF
H O
3° bromide
makes strong bondstrong nucleophile and
strong basehigh energy electrons
breaks weak bond 2 molecules become 3 molecules,
favored by entropy
elimination
Br
HO +Na
DMFMAJORminor
SayetzeffE2
Alkyl Halides : page 15
Recall, the most substituted alkene is always the most stable alkene isomer • the Sayetzeff/Zaitsev alkene is the most substituted of all possible structurally isomeric alkenes that can be formed in an elimination reaction • E2 eliminations will always form the Sayetzeff/Zaitsev alkene unless there are steric inhibitions, see below 4.2 Reactivity Order for E2 •Decreasing reactivity order: 3° > 2° > 1° halide Example
• Elimination from a 3° halide tends to give a more substituted alkene product, tends to be faster • This is one way that we can easily distinguish the possibilities of SN2 versus E2, there is simply no SN2 at a tertiary halide, but E2 eliminations tend to be facile (assuming a strong enough base) 4.3 Stereochemistry of E2 Reaction • Again, Molecular Orbital theory provide a very informative picture • We need to make and break ALL FOUR bonds at the same time, the reaction is concerted • We need to make the new bonds by overlapping the HOMO and the LUMO IN PHASE to make new bonding molecular orbital
• The 2 sigma M.O.s on the central carbons (associated with the breaking C-H and C-Br bonds) become the pi M.O.s (bonding and antibonding) • The 2 sigma M.O.s therefore must be parallel in order to be able to make the new pi-bond • The H and leaving group (Br) must be coplanar (periplanar), and preferably "anti" • The electrons in the breaking C-H sigma bond are used to make the new pi-bond, overlap occurs with the anti-bonding M.O. associated with the breaking C-Br bond best as shown, i.e. analogous to "backside attack" in the SN2 reaction, this is the origin of the requirement for ANTI- in addition to co-planar • Only One conformation will tend to be reactive in an E2 Reaction
Br
Hdisubstituted
Saytzeff productTRIsubstituted alkene, MAJOR
Br
H
HO
OH
H
H
H
BrNa OMe
DMF
BrNa O-t-Bu
DMF
BrNa O-t-Bu
DMF
3°
2°
1°
isomericalkyl halides,
increasingrate of E2
isomericalkenes,
increasingstability
DIsubstituted
TRIsubstituted
TETRAsubstituted
HOMO sp3
making π-bond(like backside attack)
C CH H
H H
BHσ-bond
sp3
C
Br
C
H
H
H
H
H
B
BrLUMOσ*
making σ-bond
breaking σ-bond
π-bond
Alkyl Halides : page 16
• The reactive conformation must be attained BEFORE reaction can occur Example: give the major products of the E2 reactions of:
• The reaction is stereospecific, different isomeric halides give different isomeric alkenes because the reaction is concerted Example: give the major product of the E2 reaction of (1R)-iodo-(2S)-methylcyclohexane
C
Et
C
H
Et
Br
Me
Me
C
Br
C
H
Et
Me
Me
EtEt
Br
Me
Me Et
H
= anti-coplanar is the"reactive conformation"
rotate todifferent
conformation
C-H and C-Br bonds NOT anti-coplanar
C
Br
CMe
Ph
H
H
Ph
Ph
Br
Ph
=
CMe H
CPhPh
Ph's on same side
cis-diphenyl alkene formed
C
Br
C
Me
H
H
Ph
Ph
C
Br
C
H
Ph
H
Me
Phanti-coplanar conformations
CPh H
CPhMe
trans-diphenyl alkene formed
Base:
unreactive conformations
C
Br
CH
Me
H
Ph
Ph
Ph
Br
Ph
=
R
R
R
S
Base:H
Br
Ph
Me Ph
Heliminated
H
Br
Ph
Ph Me
Heliminated
Ph's on oppositesides
(1R)-bromo-(1,2R)-diphenylpropane AND (1R)-bromo-(1,2S)-diphenylpropane
E2
anti-coplanar anti-coplanar
I
Me= I
Me
"unreactive" conformation, H and I NOT anti-coplanar
HH
I
H
HMe
B
Me
H and I anti-coplanar
chair interconversion
Alkyl Halides : page 17
Example: give the major product of the E2 reaction of the following compound
• The cyclohexane is "locked" into the chair that has the LARGE t-butyl substituent in the equatorial position, i.e., the energy difference between the two chairs is so large that the one with the axial t-butyl is present at such low concentration that it can be ignored • In this conformation, the C-H bond that is anti-coplanar to the C-Br bond MUST give the Anti-Sayetzeff alkene Example: give the major product of the following reaction......
• Again, the Br is "locked" into an equitorial position because of the LARGE t-butyl group • In this conformation (chair) that are NO anti-coplanar C-H bonds, therefore E2 is NOT POSSIBLE and SN2 is the only reasonable reaction
5 E1 Elimination Reaction • elimination initiated by 1st-order heterolysis Example: a 3° halide with a poor nucleophile, poor Brønsted base and in a polar protic solvent
• the intermediate cation is a strong electrophile that can react with the weak nucleophile MeOH via SN1 • the intermediate cation is ALSO a VERY STRONG Brønsted acid, stronger than hydrochloric acid (pKa < -10) • The intermediate cation can, therefore, react with the weak Bronsted base MeOH in a Bronsted acid/base reaction to give an alkene ELIMINATION product • E1 elimination because one molecule is involved in the rate determining step (kinetically first order) • SN1 and E1 are often competitive, they have the SAME rate determining step, the reactions "partition" at the cation intermediate • It is difficult to select conditions that favor E1 (high temperature can help due to the temperature dependence of entropy), i.e. not useful "synthesis" reaction - see later
Me
Anti-Saytzeff product can not be avoided!!
Br
t-Bu
Me
H
Br
= t-Bu
HO t-Bu
Mebulky t-Bu groups "locks" cyclohexane in this chair
conformation
Saytzeffnot formed
t-Bu
Me
MeH
H
=t-Bu
t-Bu
Me
Br
Et OOEtBr
t-Bu
MeEtO– +Na
acetone
bulky t-Bu groups "locks" cyclohexane in this chair
conformation
E2 not possible, no C-H bond ANTI-Coplanar to the C-Br bond
SN2
substitution
H3C C
CH3
CH3
BrMeOH
heat
H3C CCH3
CH3
HOMe
H3C C
CH3
CH3
OMeSN1
AND/OR
H3C C
CH3
H2C H
HO
MeBB
H3C CCH3
CH2
E1
+
H3C CCH3
CH3
OH
Me
H OMe
weak Nu/Base
weak
weak LBstrong LA
strong BA
3°
Alkyl Halides : page 18
Example: Give the expected ELIMINATION products (ignore substitution) under the following conditions
6 Distinguishing E1, E2, SN1 and SN2 Reactions • Reality - the mechanisms are often mixed : However, favored conditions are........ SN2 - 1° halide, aprotic solvent, strong nucleophile, weak base SN1 - 3° halide, protic solvent, weak nucleophile, weak base E2 - 3° halide, aprotic solvent, weak nucleophile, strong base E1 - 3° halide, protic solvent, but in reality they are difficult to favor! Examples: What was the mechanism that resulted in the PROVIDED organic product? (there may be other reaction products, but the questions ask about the provided ones only)
Example: give the major products of the following reactions and identify the reaction mechanisms
• There is no REQUIREMENT for an SN2 reaction to be in a polar aprotic solvent, they are faster in aprotic solvents but in reality many are actually performed in protic solvents for convenience
Ph
ClMe
H
DMF
Na+ –O-t-Bu
(E2 conditions)
Ph
Anti-Saytzeff
EtOH/heat E1 Conditions
Ph
Me
HPh
Me
HH
H
Et O H
Ph
Saytzeff
hydride shift
Br DMF
Na+ –CNCN
strong Nu and base, polar aprotic, but 1°, therefore must be SN2
substitution
Br heat
CH3OH
OCH3
3° halide, weak Nu and weak base, polar protic and heat, must be SN1
+
E1 would also occur
substitution
Br DMF
Na+ –OH
3° halide, strong Nu and base, polar aprotic, must be E2
elimination
EtOH
Na+ –OEt
I
"allylic", strong base/Nu, polar protic, E2 not possible, must be SN2
OEt
SN2 substitution
no H atoms no H atomselimination
notpossible
Alkyl Halides : page 19
• Secondary halides don't favor any mechanism in particular and often undergo more than one reaction
• It is usually a good idea to draw out at least a partial mechanism when carbocation intermediates are involved to avoid missing any rearrangements • SN1 and E1 are often competitive, unless elimination is not possible because there are no adjacent hydrogen atoms
• Elimination is not possible in this case
• Note the use of the t-butoxide anion bulky base to force E2 elimination • For E2 eliminations where there is stereochemistry in the reactant, you will usually have to setup the correct conformation for elimination (anti-coplanar) in order to get the correct stereochemistry in the alkene product
acetone
Na+ –OEtBr
2° halide (tough!), strong base and strong Nu, polar aprotic - mixture!
OEt
+E2
andSN2
heat
EtOHBr
2° halide (tough!), weak base/Nu, polar aprotic - mixture!
+
H
OH
EtHO
H
Etweak base weak nucleophilestrong
acid/electrophile
–H+
SN1E1 and OEt
CPh
Ph Br
3° halide, weak base/Nu, polar protic - No elimination possible - SN1
CPh
PhOEt
no H
heat
EtOH
CPh
Ph
Ph
PhEtOH
–H+
SN1
Ph
acetone
Na+ –O-t-Bu
3° halide strong base but weak nucleophile so can't do SN2, polar aprotic, must be E2
E2
Br
Ph CC MeBr
Et
H
H
get the stereochemistry
of the alkene product correct! Ph
CC
Et
H H
Br Me=
Br- and -H anti-coplanarPh and Me on the "same side"
H Et
Ph Me
redraw as 3D structure
same as above
E2
Alkyl Halides : page 20
7 Elimination Using Bulky (Sterically Hindered) Bases • The products of E2 eliminations can be different for 2 versus 3° halides with or without bulky bases Examples:
Looking at the Transition States can explain these product distributions • the fstest reaction occurs, the reaction is kinetically controlled
• The transition state for formation of the MAJOR product has CH3-H electron repulsion/steric effects, which costs LESS energy that CH3-CH3 electron repulsions/steric effects below, the transition state is thus lower in energy, the reaction is faster and this reaction forms the MAJOR product
• The transition state for formation of the minor product has CH3-Ch3 electron repulsion/steric effects, which costs MORE energy that CH3-H electron repulsions/steric effects in the reaction that forms the major product, the transition state is thus higher in energy, the reaction is slower and this reaction forms the minor (Hoffman) product In Summary • A non-bulky with a 3° halide forms the most substituted alkene (normal Saytzeff product) • But, a bulky base with a 3° halide forms the least substituted alkene (Hofmann product) for steric reasons
no substitutionSaytzeff productBr
3°halide
t-BuO– +K
Na+–OCH3
no substitutionBUT
Anti-Saytzeff (Hoffman) product
Hbulky baseabstracts least hindered proton
Br
t-BuO– +K
Na+–OCH3
OCH3+
2°halide
(E2 + SN2)
(E2 only)
H
NOSN2
Br
HO CH3
CH3
CH3
‡
Me-H steric interactions
transition state for MAJOR product
Br
K+ –O-t-Bu
major
H
HHlower energy
faster
CH3
BrH3C
‡
HOH3C
CH3
CH3
transition state for MINOR product
Me-Me steric interactions
Br
K+ –O-t-Bu
minor
H
Hhigher energy slower
Alkyl Halides : page 21
8 Water as a Leaving Groups • Substitution and elimination reactions are NOT LIMITED to alkyl halides with halide anions as leaving groups • It is a good idea to look at some other leaving groups now, not to create more stuff to learn, but to facilitate learning of these new reaction types in different contexts • Good leaving groups are stable as anions • Leaving groups that are stable NEUTRAL molecules therefore should be expected to be EVEN BETTER! • A classic example is water, consider the following reaction that converts an alcohol into an alkyl bromide
• This reaction doesn't work! In fact, goes in reverse –OH will substitute for X– (think about a standard SN2 reaction that has -OH as the NUCLEOPHILE and Br- as the Leaving group), this reaction has the OPPOSITE! •–OH is too poor a leaving group, need to make a better leaving group Consider the following reaction instead
• The first step is a standard Bronsted acid/base reaction, with H-Br as the strong Bronsted acid • NOW we have a very good potential leaving group, H2O, the next step is standard SN2 and it works well, even though the bromide anion is a poor nucleophile Example Problem: • Give the mechanism of the following substitution reaction
• FIRST, standard Bronsted acid base reaction between the -OH and the strong Bronsted acid H-Br, this converts the -OH from a poor leaving group into a very good leaving group • After that it is standard SN1 mechanism with an expected rearrangement of the cation intermediate
X + OH
BrSN2
doesn't work!
LA/Elec
OHNa+ –Br
DMFBr
goes the opposite way!
VERY POORleaving group
+
BrBA
OH HBr
BrH
BBO
H
H
Br
O
H
Hwater
VERY GOODleaving group
SN2
Br
BA/LA
OH HBr
HBB/LB
Br
O
H
H
Br
strong LA/Elec
weakLB/Nuc
+ H2O
+ H2Overy goodleaving group
SN1
SN1
HH
convert-OH intoa good
leaving group
standard SN1 from here
Alkyl Halides : page 22
Elimination of Water • E1 and E2 eliminations are also observed where water can be a good leaving group • Again, the reactions start by protonating the -OH of an alcohol to form a good leaving group, and then standard E1 and/or E2 mechanisms after that Example Problem: Give the product AND Mechanism for the following elimination reaction.
• Sulfuric acid protonates the -OH in a Bronsted acid/base reaction to convert the -OH into a good leaving group • Water is such a good leaving group that the elimination is almost always E1 with 3° and 2° alcohols • Water is such a good leaving group that E1 is occurs even at a secondary carbon to make a secondary cation • carbocation intermediates mean rearrangements! (that hasn't changed, of course) • the conjugate base anion of the sulfuric acid, the bisulfate anion, is the most likely base to deprotonate the carbocation intermediate, thus regenerating the acid catalyst • The alkene formed will be the Sayetzeff (Zaitsev), there are no stereochemical constraints in the E1 mechanism and the most stable alkene will form. Why does the alcohol make an alkene + water when previously we learned that water + alkene gives an alcohol? • THIS is what we learned previously
• The addition reaction "goes" because the weaker pi-bond is converted into a stronger sigma-bond • The reagents/conditions have a LARGE quantity of water and a SMALL quantity of sulfuric acid • THIS is what we now learned
• The reagents have ZERO water and a HIGH concentration of sulfuric acid (opposite of previous reaction) • The elimination reaction "goes" because the water is highly solvated in the concentrated sulfuric acid • note a special kind of SOLVENT EFFECT here! In an aqueous medium, acid catalyzes water ADDITION to the alkene to make an alcohol. In conc. sulfuric acid medium, the acid helps to REMOVE water from an alcohol to make an alkene (the sulfuric acid DEHYDRATES the alcohol)
OHconc. H2SO4
heat
OH2 H
Sayetzeff major
LA/BALB/BB
LA/BA LB/BB
standard E1 elimination (including rearrangement)
no water!!
E1
H O S O HO
O
O S O HO
O
FIRSTconvert poor leaving group
to good leaving group
+ H2O very goodleaving group
C CH2SO4 (cat.)/heat
addition of H2OH2O H
C COH
WATER
SMALL quantity of acid
HC C
OHC C
conc. H2SO4
heat+
NO water!
elimination
HIGH acid concentration (100%)
STRONG IMFH-bonding
OH
HH O S O H
O
O
H O S O HO
O
Alkyl Halides : page 23
• Alternate reagents and conditions are H2SO4/P2O5, and others…. Example: Primary (1°) Alcohols: E2 elimination (with rearrangement…)
• With a primary alcohol the mechanism must be E2, formation of a primary carbocation CAN'T occur • BUT, even though the elimination does not involve a rearrangement, the final alkene product is usually the same one that would have been formed via an E1 reaction due to protonation followed by deprotonation (isomerization) of the primary alkene into a final more stable product Look again at the second part of the mechanism, the rearrangement
• This effectively converts a less stable less substituted alkene into a more stable more substituted alkene, this is why this isomerization reaction "goes" • To solve the mechanism problem, add some hydrogen atoms back to the line-angle structure, the H atoms tell you exactly where you need to protonate and deprotonate • in the presence of acid, PROTONATION will occur first, followed by deprotonation • a less substituted/less stable alkene is converted into a more substituted/more stable alkene • this is a rearrangement, the acid is only the catalyst (no atoms are overall added or subtracted) • In a strong acid, especially with heat, protonation and deprotonation can OFTEN occur, and if this can result in formation of a more stable alkene, then the more stable alkene will form, and you should always include this step when doing acid catalyzed dehydrations of alcohols The final product is the same most substituted, whether the mechanism is E1 followed by cation rearrangement (2° and 3° alcohols) or E2 followed by protonation/deprotonation (1° alcohols)
OH
OH2
HLB/BB
LB/BB
LA/BA
LA/BAHH
LB/BB
LB/BB
conc. H2SO4
heat
E2 elimination
LA/BA
rearranged stable alkene formed
same as from E1!
protonation/deprotonation
LA/BAH O S O H
O
O
O S O HO
O H O S O HO
O
O S O HO
O
E2 elimination for 1° alcohols, NO E1!
FIRSTconvert poor leaving group
to good leaving group
HH
rearranged stable alkene formed
H O S O HO
O
OSOHO
OH
H
H HH
H
H H HH
H
Alkyl Halides : page 24
10 Reaction Summary Do NOT start studying by trying to memorize these reactions! Work as many problems as you can with this list of reactions in front of you, if necessary, so that you can get through as many problems as you can without getting stuck on the reagents/conditions. AFTER you have worked all of the problems, just before an exam, then do the following: • Cover the entire page of reagents/conditions with a long vertical strip of paper, see if you can write down the reagents/conditions for each reaction, check to see which you get correct, if COMPLETELY correct, circle Y, if incorrect or even slightly incorrect, circle N. In this way you keep track of what you know and what you don't know. • Keep coming back to this list and so the same thing only for those reactions you circled N, until all are circled Y. • Knowing the reagents/conditions on this page is INSUFFICIENT to do well on an exam since you will ALSO need to recognize how to use and solve reaction problems in different contexts, this page ONLY helps you to learn the reagents/conditions that you have not YET learned by working problems. ALSO, SN2 reactions in particular can occur in MANY DIFFERENT CONTEXTS, knowing the reactions summarized here is INSUFFICIENT for you to solve substitution and elimination reaction problems
Br
t-BuO– +K
Br
Na+ –OMe
Br
t-BuO– +K
Sayetzeff
2°
3°
3°
bulky base avoids SN2
nonbulky base
Sayetzeff
Anti-Sayetzeffbulky base
Br Na OH OH
and many other SN2 reactions of 1° halides
Y / N
Br
MeOH3° heat E1
E2
E2
E2
SN2
any alcoholY / N
Y / N
Y / N
Y / N
Y / NSN1
Br OMe
MeOHheat rearranged
Sayetzeff
OH conc. H2SO4
heat
rearranged
OHconc. H2SO4
heatY / N
Y / N
most stable alkene
E1
E2
HBrOH BrY / N
HBrOHY / N
SN1
BrSN2
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