Download - Alkyl Halides Substitution and Elimination 1 Nomenclature ... · Alkyl Halides : page 1 Alkyl Halides Substitution and Elimination 1 Nomenclature • Look for the longest chain that

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


• 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


• 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


• 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


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


• 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–


• 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

+

+

+

+

+


• 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


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


• 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


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


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


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


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


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


• 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


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°


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


• 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


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


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


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


• 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


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