Four new mechanisms to learn: SN2 vs E2 and SN1 vs E1psbeauchamp/pdf/314_SN_E_2013.pdf ·...

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Nucleophilic Substitution & Elimination Chemistry 1

y:\files\classes\314\314 Special Handouts\314 SN & E Spring 2013.doc

Four new mechanisms to learn: SN2 vs E2 and SN1 vs E1

S = substitution = a leaving group (X) is lost from a carbon atom (R) and replaced by nucleophile (Nu:)N = nucleophilic = nucleophiles (Nu:) donate two electrons to carbon in a manner similar to bases donating to a proton (B:)E = elimination = two vicinal groups (adjacent) disappear from the skeleton and are replaced by a pi bond1 = unimolecular kinetics = only one concentration term appears in the rate law expression, Rate = k[RX]2 = bimolecular kinetics = two concentration terms appear in the rate law expression, Rate = k[RX] [Nu: or B:]

SN2 competes with E2SN1 competes with E1

These electrons always leave with X.SN2

E2

SN1

E1

XRBNu BHNuHCompeting

ReactionsCompeting Reactions

Carbon Group

LeavingGroup

Nu: / B: = is an electron pair donor to carbon (= nucleophile) or to hydrogen (= base). It can be strong (SN2/E2) or weak (SN1/E1).

(strong)(concerted)

(weak)(multi-step)

R = methyl, primary, secondary, tertiary, allylic, benzylic

X = -Cl, -Br, -I, -OSO2R (possible leaving groups in neutral, basic or acidic solutions)

X = -OH2 (only possible in acidic solutions)

CH

C X

B

Nu

E2 approach (usually from the backside in an "anti" conformation,"syn" is possible, but not common)

SN2 approach (alwaysfrom the backside, resulting in inversion of configuration) B Nu

strong base strong nucleophile

One step, concerted reactions, from the backside. CNu

C

H

SN2 product (a nucleophile substitutes for a leaving group)

E2 product (a pi bond forms)

C

C X

Terms

S = substitution E = elimination Nu: = strong nucleophile B: = strong base X = leaving group

2 = bimolecular kinetics (second order, rate in slow step depends on RX and Nu: /B: )

R-X = R-Cl, R-Br, R-I, R-OTs, ROH2+

stableleaving group

E or Z forms, depending on conformationR or S forms,

depending on configuration

SN2 reactions = substitution nucleophilic bimolecularRate = kSN2[RX][Nu: ] bimolecular kinetics, both RX and Nu: participate in the slow step of the reaction; kSN2 = Ax10

- Ea(SN2)

2.3xRxT

E2 reactions = elimination bimolecularRate = kE2[RX][B: ] bimolecular kinetics, both RX and B: participate in the slow step of the reaction; kE2 = Ax10

SN1 reactions = substitution nucleophilic unimolecularRate = kSN1[RX] unimolecular kinetics, only RX participates in the slow step of the reaction; kSN1 = Ax10

E1 reactions = elimination unimolecularRate = kE1[RX] unimolecular kinetics, only RX participates in the slow step of the reaction; kE1 = Ax10

- Ea(E2)

2.3xRxT

- Ea(SN1)

2.3xRxT

- Ea(E1)

2.3xRxT

SN and E reactions (A is a pre-exponential term and can be considered to be the maximum possible rate constant when Ea = 0)



XH3C

H2CR X

HCR X

R

CR X

R

R

CH

H2CH2C X

H2C X

methyl (Me) primary (1o) secondary (2o) tertiary (3o) allylic benzylic

C C X

C-beta carbon (0-3 of these)

C-alpha carbon (only 1 of these)

Relative rates of SN2 reactions - steric hindrance at the C carbon slows down the rate of SN2 reactions.

XC

H

H

Hk 30

methyl (unique)

XC

H

H3C

Hk 1

ethyl (primary)reference compound

XC

CH3

H3C

Hk 0.025

isopropyl (secondary)

XC

CH3

H3C

CH3

k 30t-butyl (tertiary)

(very low)140

C

H

HH

X

methyl RX

Methyl has three easy paths of approach by the nucleophile. It is the least sterically hindered carbon in SN2 reactions, but it is unique.

C

H

RH

X

primary RX

Primary substitution allows two easy paths of approach by the nucleophile. It is the least sterically hindered "general" substitution pattern for SN2 reactions.

C

R

RH

X

secondary RX

Secondary substitution allows one easy path of approach by the nucleophile. It reacts the slowest of the possible SN2 substitution reactions.

tertiary RX

C

R

RR

X = C

C

X

C

H

HH

H

HH

CH

H

H

Nu Nu

Tertiary substitution has no easy path of approach by the nucleophile from the backside. We do not propose any SN2 reaction at tertiary RX centers.

When the C carbon is completely substituted the nucleophile cannot get close enough to make a bond with the C carbon. Even C-H sigma bonds block the nucleophiles approach.

H2CC

H

H

Hk 1ethyl

reference compound

H2CC

H

H3C

Hk 0.4propyl

H2CC

CH3

H3C

Hk 0.03

2-methylpropyl

H2CC

CH3

H3C

CH3

k 0.000012,2-dimethylpropyl

(1o neopentyl)

X X XX

All of these are primary R-X structures at C, but substituted differently at C.

C

C

X

H

C

CH3CH3

HH

H

H

Nu

A completely substituted C carbon atom also blocks the Nu: approach to the backside of the C-X bond. A large group is always in the way at the backside of the C-X bond. C

C

X

H

H

CH3CH3

HNu

If even one bond at C has a hydrogen then approach byNu: to the backside of the C-X bond is possible and an SN2 reaction is possible.

SN2 is the most important mechanism



SN2 versus E2 overview

CH

CH3H

C Br

HH

1-bromopropane

Example - primary RX, requires strong nucleophile/base, SN2 > E2, except when potassium t-butoxide or sodium amide.

Nu

CH

CH3H

C Br

HH

1-bromopropane

O

C

CNu

HH

H

CH3

H

E2 > SN2

(when t-butoxide) C

C

H

H

H3C

H

alkene

E2

Nu B=

strong = anything with negative charge,and neutral sulfur, phosphorous or nitrogen

SN2 > E2

(unless t-butoxide)

(2R)-2-bromobutane

NuC

CNu

C

HH

H

H

HD

CH3less basic = SN2 > E2 at secondary RX

Some examples

OH Na

OR Na

R

O

O

Na

O KCH

HH

C Br

C

HH

CH3

H

B SN2 and E2depends on steric size andbasicity of donor

C

C

H

H

CH2

H

1-butene 2E-butene

R

CH

HH

C Br

CHa

CH3

Hb

CH3 C

C

CH3

H

Hb

H3C

SN2

more basic = E2 > SN2 at secondary RX

HCC

NC

E2 reaction at 3o RX with strong base-nucleophile

2-methylbut-1-ene

C

C

CH3

H3C

H

CH3

C

C

H

H

H3C

CH2

CH3

only first mechanismis shown

C

H3C

H3C

H3C

B

H

(2R)-2-bromobutane

SN2 and E2depends on steric size andbasicity of donor

2Z-butene

C

C

CH3

H

CH3

Ha

BR

CH

HH

C Br

CHa

CH3

Hb

B

CH2

(2R)-2-bromobutane

HB

2-methylbut-2-ene

Example - secondary RX, requires strong nucleophile/base, SN2 or E2, depends on strength and size of nucleophile / base

Example - tertiary RX, requires strong nucleophile/base, only E2, generally favors more stable alkene = more substituted alkene

(also R2N = E2)



SN1 / E1 overview: Always SN1 > E1 in our course, except when ROH / H2SO4 /

E1 approach comes from parallel C-H with either lobe of 2p orbital.

SN1 approach is from either the top face or the bottom face.

Bweak base

SN1 and E1 reactions are multistep reactions.

E1 products(pi bond forms)

X stableleaving group

Terms

S = substitution E = elimination H-Nu: = weak nucleophile H-B: = weak base X = leaving group

1 = unimolecular kinetics (first order reaction, the rate in the slow step depends only on RX)

R-X = R-Cl, R-Br, R-I, R-OTs, ROH2+

SN1 and E1 reactions begin exactly the same way. The leaving group, X, leaves on its own, forming a carbocation.

CC

X

H

The R-X bond ionizes with help from the polar, protic solvent, which is alsousually the weak nucleophile/base.

CC

H

X

solvated carbocation

addition of a weak nucleophile (SN1 type reaction)

Loss of a beta hydrogen atom, (C-H in most E1 reactions that we study).

rearrangement to a new carbocation of similar or greater stability

add Nu: (SN1)

lose beta H (E1)

rearrange

H

Nu H

weak nucleophile

CC

H

C

Nu

C

H

CC

H

Nu H

H

Nu H

Nu H

C C

CC

H

Nu

C

Nu

C

H

solvated carbocation

Bweak base

H Nu H weak nucleophile

=

SN1

E1

Often a final proton transfer is necessary.

SN1 product (a nucleophile substitutes for a leaving group)

BHNu H = usually a polar, protic solvent (or mixture) of H2O, ROH or RCO2H=

NuH

BH

R

SN1 and E1 relative reactivities of R-X compounds with weak base/nucleophiles (H-B: / H-Nu:)

R-X R SN1 > E1(usually)

XH3C XH2CH3C XC

HH3C XCH3C

CH3 CH3

CH3relative

rates = 10-510-4

These rates are probably by SN2 reaction.

Reference compound

1.0 106 = 1,000,000

Relative SN1/E1 reactivity: methyl << primary << secondary << tertiary.

one exception where E1 > SN1X = "OH", ROH + H2SO4 /



Relative Stability of Carbocations, R+

Some of these values are esitimated from different sets of data.

We will not proposethese carbocations insolution in this book.

CH3

CH2

CH2

CH2

CH3CH2

methyl & primary

-277

-315

-270

-265

-267

tertiary "C" resonance "X" resonance other / misc.

C

CH3

H3C

CH3

CHH2C CH2

CH2

H

CH C

CH

HC H

-232

-227

-227

-256

-239

-229

-220

-210

CH2H2N

OH CH2 -248

-218

-230

-386

-287

ethynyl carbocation(see problem below)

CH

HC H

phenyl carbocation(see problem below)

-300

vinyl carbocation(see problem below)

secondary

CHH3C

H3C

-249

H -246

Inductive effectsResonance effects

adjacent pi bond adjacent lone pair

CO CH3

Hydride Affinity

H = energy released =

HR

RH

Potential EnergyA larger hydride

affinity means a less stable carbocation.

empty sp orbital

empty 2p orbital

empty sp2

orbital



These structures represent your hydrocarbon starting points to synthesize target molecules (TM) that are specified. You will need to propose a step-by-step synthesis for each target molecule from these given structures. Every step needs to show a reaction arrow with the appropriate reagent(s) above each arrow and the major product of each step. This is often accomplished by using retrosynthetic thinking. You start at the target molecule (the end) and work your way backwards (towards the beginning), one step at a time until you reach an allowed starting material. The starting material of each step becomes the target molecule for the next step until you reach the beginning. Allowed starting structures – our main sources of carbon – 1. free radical substitution of alkane sp3 C-H bonds to form sp3 C-Br bonds at the weakest C-H position and 2. Anti-Markovnikov addition to alkenes makes 1o R-Br

CH4

Br Br

BrWe need to make these 1o RBr from anti-Markovnikov free radical addition of H-Br (ROOR) to alkenes.

Br2 / h Br2 / h Br2 / h Br2 / h Br2 / h Br2 / h Br2 / h

H3CBr

Br

BrBr Br

Br

Br

HBrH2O2 / h

HBrH2O2 / h

HBrH2O2 / h

E2 reaction

O

K

E2 reaction

O

K

E2 reaction

O

K

E2 reaction

O

K

E2 reaction

O

K

Br2 / h

Br2 / h

Br2 / h

Br

Br

Br

Examples of allylic RBrcompounds: This is just free radical substitution

at allylic sp3 C-H positionof an alkene.

Br

benzylic RBr, f rom above

Br



1. Mechanism for free radical substitution of alkane sp3 C-H bonds to form sp3 C-Br bonds at weakest C-H position

H3C

H2C

CH3 H3CCH

CH3

Br

BrBr

overall reactionh

BrH

1. initiation

BrBrh BrBr H = 46 kcal/mole

weakest bond ruptures first

2b propagation

H3CC

CH3

H

BrBrH3C

CCH3

H Br

BrH = -22 kcal/mole (overall)

BE = +46 kcal/moleBE = -68 kcal/mole

2a propagation

H3CC

CH3

H H

Br BrH

H3CC

CH3

H

H = +7 kcal/mole (overall)


H = -15

both steps

3. termination = combination of two free radicals - relatively rare because free radicals are at low concentrations

BrH3C

CCH3

H

H3CC

CH3

H Br

H3CC

CH3

H

CH3

CH3C

H

CH

CH3

H3C CHCH3

CH3

H = -68 kcal/mole

H = -80 kcal/mole

very minor product

2. Free radical addition mechanism of H-Br alkene pi bonds (alkenes can be made from E2 or E1 reactions at this

point in course) (anti-Markovnikov addition to alkenes)

H3C

H2C

CH2

Br

H3C

HC

CH2

HBrR2O2 (cat.)

h

overall reaction

1. initiation (two steps)

RO

OR h R

OO

R

BrHR

O RO

H Br

H = 40 kcal/mole

H = -23 kcal/mole


(cat.)

reagent

2a propagation

H3C

HC

CH2Br

H3CC

CH2

Br

H

H = -5 kcal/mole

BE = +63 kcal/mole BE = -68 kcal/mole



H = -15

both steps(2a + 2b)

2b propagation

H3CC

CH2

Br

H

BrH H3C

H2C

CH2

BrBr

H = -10 kcal/mole

BE = +88 kcal/mole BE = -98 kcal/mole

3. termination = combination of two free radicals - relatively rare because free radicals are at low concentrations

H3CC

CH2

Br

H

Br

H3CCH

CH2

Br

Br

H = -68 kcal/mole

very minor product

Reagent list for our course (Chem 314)

What kind of mechanisms are possible? What is the major mechanism occuring? Be able to write in ALL mechanism details (lone pairs, formal charge, curved arrows, etc.) for each step of each reaction studied. Mechanism choices at this point in organic chemistry: SN2, E2, SN1, E1,acid/base reaction, free radical substitution, free radical addition to alkenes and no reaction.

Nuclephile/Bases (and other reagents) to choose from include:

Br2 h

Na HNasodium amide

OH Na

O

OH

CN NaSH Na

AlD

D

D

D

Li

NR2

sodiumhydroxide

water

ethanoic acid(acetic acid)

sodium hydride(only basic)

lithium aluminium hydride

(nucleophilic)

BD

D

D

D

Na

sodium borohydride(nucleophilic)

sodiumcyanide

monohydrogensulfide

K H potassium hydride

(only basic)

Na

sodiumazide

N3

purchase / given - You can invoke these whenever you need them. Some of these will disappear when we learn how to make them.

AlH

H

H

H

Li

lithium aluminium hydride

(nucleophilic)

BH

H

H

H

Na

sodium borohydride(nucleophilic)

O

propanone(acetone)

HO

H

ethanol

OH

HN

diisopropylamine

H3C

H2C

CH2

CH2

Lin-butyl lithium

t-butyl alcohol

OH

S

S

PhS

Ph

dithianediphenyl sulfide

(useful for making epoxides)triphenylphosphine

(used for making Wittig reagents)

PhP

PhPh

Cl2 h

strong acids: HCl, HBr, HI, H2SO4 (buy all of these = given)strong bases: NaOH, NaH, KH, n-BuLi, NaNR2, (buy all of these = given) t-BuOK, LDA (make these) strong nucleophiles: NaOH, LiAlH4, LiAlD4, NaBH4, NaBD4, NaCN, Ph2S, Ph3P, (buy all of these = given) NaOR, NaO2CR, NaCCR, Li-enolates, Li-dithiane (make these)weak nucleophiles: H2O, ROH, RCO2H (buy all of these = given)common electrophiles: RX (= RCl, RBr, RI) (make these)free radical reactions: alkane C-H substitution (Br2/h), anti-Markovnikov addition to alkenes (HBr/ROOR (cat)/h)

These two are exceptions to our strong nucleophiles = anions. They are good neutral necleophiles. Both are used to make ylids (one sulfur and one phosphorous)

Some of our reagents have to be made (often by acid/base reactions)

Make alkoxides and use as nucleophiles only at Me-X and 1o RCH2-X in SN2 reactions.

OR

Na

alkoxides

OR H

alcohols

Na

H

BrR

O

SN2 ethers



Make potassium t-butoxide and use as sterically large, strong base at 1o RCH2-X, 2o R2CH-X and 3o R3C-X in E2 reactions.

Make carboxylates and use as nucleophiles at Me-X, 1o RCH2-X and 2o R2CH-X in SN2 reactions.

O

O

Na

ethanoate(acetate)

O H

Na

O

Oethanoic acid(acetic acid)

HBr

O

O

estersSN2

Make terminal acetylides and use as nucleophiles only at Me-X and 1o RCH2-X in SN2 reactions.

CCR

Na

terminalacetylides

CCR

terminalalkynes

HNa

NR2

Br

SN2

R

alkynes

Make thiolates and use as nucleophiles at Me-X, 1o RCH2-X and 2o R2CH-X in SN2 reactions.

SR

Na

alkylthiolatesNa

SR H

thiols

O HBr

SN2R

S

sulfides

Synthesis of lithium diisopropyl amide, LDA, sterically bulky, very strong base used to remove Cα-H proton of carbonyl groups. (acid / base reaction) to make carbonyl enolates.

Keq =Ka1

Ka2

10-37

10-50= = 10+13

CH2 Li

N

H

NLi

Think - sterically bulky, very basic that goes after weakly acidic protons.

LDA = lithium diisopropyl amide

given

given

React ketone enolate (nucleophile) with R-Br electrophile (Me, 1o and 2o RX compounds)

O

HN

Li

N

H

CH2

O

Li

Keq =Ka1

Ka2

10-20

10-37= = 10+17

ketones (and other carbonyl compounds)

Make enolate (ketones)

LDA

ketone enolates(resonance stabilized)

given

-78oC



H2C

O Li


React ketone enolate with RX compounds (methyl, 1o and 2o RX)

Br

O

1o RX key bondR

S Larger ketone made from smaller ketone.

SN2reactions

-78oC

React ester enolate (nucleophile) with R-Br electrophile (Me, 1o and 2o RX compounds)

O

OH

N

Li

N

H

CH2

O

O

Li

Keq =Ka1

Ka2

10-25

10-37= = 10+12

ketones (and other carbonyl compounds)

Make enolate (esters)

LDA


given

R R-78oC

H2C

O

O

Li

ester enolates(resonance stabilized)

React ester enolate with RX compounds (methyl, 1o and 2o RX)

Br

O

O

1o RX key bondR

S Larger ester made from smaller ester.

R RSN2reactions

-78oC



R-X patterns - typical reaction patterns in our course (bold = atypical result)

Reagents methyl RX1o RX

primary2o RX

secondary3o RXtertiary

1o neopentyl RX

allylic RX benzylic RX

Na OH

O2CR

O-C(CH3)3t-butoxide

C N

C C R

N3

SH

Li AlH3H

BH3H

AlD3D

BD3D

Na

Na

K

Na

Na

Na

Na

Na

Na

Na

Li

SN2 > E2only SN2

only SN2

only SN2

only SN2

only SN2

only SN2

only SN2

only SN2

only SN2

only SN2

only SN2

only SN2

only SN2

SN2 > E2

SN2 > E2

E2 > SN2

SN2 > E2

SN2 > E2

SN2 > E2

SN2 > E2

SN2 > E2

SN2 > E2

SN2 > E2

SN2 > E2

SN2 > E2

SN2 > E2

only E2

SN2 > E2

SN2 > E2

SN2 > E2

SN2 > E2

SN2 > E2

SN2 > E2

SN2 > E2

SN2 > E2

E2 > SN2

E2 > SN2

OLi

ketone enolates

only SN2 SN2 > E2 SN2 > E2

E2 > SN2

only E2

only E2

only E2

only E2

only E2

only E2

only E2

only E2

only E2

only E2

only E2

only E2

only E2

only E2

no reaction

no reaction

no reaction

no reaction

no reaction

no reaction

no reaction

no reaction

no reaction

no reaction

no reaction

no reaction

no reaction

no reaction

very fast SN2 very fast SN2



NA











OR

SR

1. NaN3 (SN2)2. LiAlH4 (SN2)3. workup (acid/base)

makes1o amines

makes1o amines

makes1o amines

makes1o amines

makes1o amines

no reactiononly E2

NA

O

O Li

ester enolates

only SN2 SN2 > E2 SN2 > E2 only E2 no reaction very fast SN2 very fast SN2R

The following are synthetic approaches to some of the target molecules above. They are presented as examples of how you might approach synthesis problems (these and others like them). You need to not only look at structures that I provide above but think about other possibilities using all of the structures available to you. I do not list every possibility (there are too many), but you need to be able to consider every possibility. The best strategy is to work backwards from the target molecule to an allowed starting structure (cover up everything but the target structure and write out an answer for how you could make that molecule, then start all over on target molecule minus one, which is called ‘retrosynthetic analysis’). That way you are only thinking about “one” backward step at a time, instead of an “unknown” number of steps forward from a starting structure you are not sure of. This is the way I presented the approaches below. You should also know the mechanism for every reaction below. I can write these pages until I am blue in the face, but they can’t help you learn, unless you do the work of trying them out. My job is to give you an opportunity to learn, and your job is to take advantage of that opportunity. These schemes were made quickly, so be on the lookout for mistakes. (I hope not too many.)



1. Synthesis of RBr compounds - Free radical substitution at sp3 C-H and free radical addition to alkenes (anti-Markovnikov addition). Making these represents our most common starting points.

BrBr

Br

1o RX 1o RX 1o RX

These are our target molecules (TM).

BrBr

Br

allylic RXallylic RX allylic RX

H of each step is controlled by relative bond energies

Sample R-Br from R-H using Br2 / h - you have to make these

BrH3C Br

BrBr

Br

BrBr

methyl RX 1o RX 2o RX 3o RX

2o RX 1o benzylic RX 2o benzylic RX

Free radical substitution mechanism: (R-H + Br2 + light)

1. initiation 2a and 2b. propagation 3. termination.

H of each step is controlled by relative bond energiesFree radical addition mechanism: (alkene + HBr/ROOR + light)

1. initiation 2a and 2b. propagation 3. termination.

O

K

E2potassiumt-butoxide

free radical substitution2o RX

Br2 h

Br

propane

Br2H2O2 (cat.)

h

free radical addition

Br

1o RXwe need these

TMTM - 2TM - 1 allowed

starting material

alkene

Br2H2O2 (cat.)

h

free radical substitution

H3o RX

Br2 h

BrO

K



Br

1o RXwe need these

2-methylpropanealkene

TM TM - 1 TM - 2 allowed starting material

(isobutane)

Br Br2H2O2 (cat.)

h


O

K


TM TM - 1 TM - 2Br


Br2 h

allowed starting material

ethylbenzene

allylic RBr compounds - (NBS is more likely reagent, but we only know Br2). You will need to make the starting alkene (E2 reactions)

Br

Br2 h


1o allylic RX

O

K



H

3o RX

Br2 h

Br 2-methylpropane(isobutane)

TM TM - 1 TM - 2 allowed starting material



Br2 h


Br

1o allylic RX

O

K


free radical substitution3o RX

Br2 hBr

propane

TM TM - 1 TM - 2allowed

starting material

Br Br2 h


2o allylic RX

Br

potassiumt-butoxide

Br2 h


E2

TM TM - 1 TM - 2

allowed starting material

cyclohexane

2. alcohol synthesis – starting from RX (later there will be many other approaches)

Sample alcohols - SN2 reactions using (R-X + HO ) at Me, 1o, 2o, allylic RX or SN1 reactions using (R-X + H2O) at 2o, 3o RX

OHH3C OHOH

OH

OH

OH OH

OH

OH

OH

OHOH

OH

These are our target molecules (TM).

SN2 SN2 SN2 SN2

SN2 SN2 SN2 SN2

SN1SN1

SN2 or SN1 SN2 or SN1

OH Na

OH

Br

alcohols - various approaches, SN2 at methyl and primary RX using NaOH and SN1 at secondary and tertiary RX using H2O

see above, from propene

1o alcohol SN2 > E2

Hydroxide and alkoxides are OK nucleophiles at methyl and 1o RX, but E2 > SN2 at 2o RX and only E2 at 3o RX.

TM TM - 1

1o RBr

OH Br Br2 h


allowed starting structure

H2O

SN1 > E1

Generally, SN1 > E1 at 2o and 3o RXif we don't have to worry about rearrangements. No SN1/E1 at methyl and 1o RX.

TM

2o alcohol 1o RBr

TM - 1

Generally, SN1 > E1 at 2o and 3o RXif we don't have to worry about rearrangements. No SN1/E1 at methyl and 1o RX.

Br2 h

H2O

BrOH

SN1 > E1free radical substitution


TM

3o alcohol 3o RBr

TM - 1

SN2 mechanisms using aqueous hydroxide introduces a new alcohol functional group (R-OH)

BrH3C

nucleophile = ?electrophile = ?

SN2 OCH3

key bond

BrOHNa

H Na



BrH2C

H3Cnucleophile = ?electrophile = ?

SN2 OBr

key bond

Na

OH

NaH

BrH2C

CH2H3Cnucleophile = ?electrophile = ?

SN2O

key bond

Br

Na

OH

NaH

SN1 mechanisms (if no rearrangement) using water introduces a new alcohol functional group (R-OH)

Br

H2O

SN1 > E1

H

O H

H

H

O

H

H

O H

H

H

O

H

water is a nucleophile water is a

base

SN1 > E1 at 2o and 3o RX and we don't have to worry about rearrangements here.

Br

H2O

SN1 > E1

O H

H

O

H

H

O H

H

O

H

water is a nucleophile

water is a base

3. ether synthesis - starting from RX (SN2 at Me and 1o RX with alkoxide, SN1 at 2o and 3o RX and ROH)

Sample ethers - SN2 reactions using (R-X + RO ) at Me, 1o, 2o, allylic RX or SN1 reactions using (R-X + ROH) at 2o, 3o RX There are many more possibilities.

OO O

O

(SN1 or SN2)(SN1 or SN2) SN2 SN2

TM

Otarget molecule SN1 > E1

Br2 h


Br

OH

1. Na H

2.

SN2 > E2

acid / base Br

OHGenerally, SN1 > E1 at 2o and 3o RX. No SN1/E1at methyl and 1o RX.

ethers - various approaches, SN2 at methyl and primary RX using NaOR and SN1 at secondary and tertiary RX using ROH

BrNa OH

SN2

Br2 h


HO

H

Br

TM - 1 TM - 2



TM - 1 TM - 2



Na OH

O

1. Na H

2.

SN2 > E2

acid / base

HOPh Br

Br2 h

Ph = phenylBr Ph OHSN1 > E1

SN2

Ph Br

already made

already made

free radical substitutiontarget molecule

given

TM TM - 1

TM - 1 TM - 1

TM - 2

SN2 mechanisms using aqueous alkoxide introduces a new ether functional group (R-O-R’)

BrH3C


OCH3

key bond

BrOCNa

C NaSN2

H3C

H3C

H3C

CH3

H3C

H3C

BrH2C


SN2 O

Br

key bond

NaOH3C

Na H3C

BrH2C


SN2O

key bond

Br

Na

OH3C

NaH3C

SN1 mechanisms (if no rearrangement) using alcohol introduces a new ether functional group (R-O-R’)



4. ester synthesis (using the oxygen side and using the carbon enolate side, also ketone examples using an enolate)

Sample alcohols - SN2 reactions using (R-X + RCO2 ) at Me, 1o, 2o, allylic RX or SN1 reactions using (R-X + RCO2H) at 2o, 3o RX

These are our target molecules (TM) - using the oxygen side (carboxylates).

O

O

CH3

O

O

O

O

O

O

O

O

O

O

O

O

O

O

O

O

O

O

O

O

O

O

O

O

SN2 SN2 SN2 SN2 SN1 SN2

SN1 SN2 SN1

SN2 SN1

SN2 SN2

SN2 SN2 SN2 SN1

esters - various approaches, SN2 at methyl, primary and secondary RX using RCO2Na. SN1 at secondary and tertiary RX using RCO2H but possible rearrangement.

O

O

O

OHtarget molecule

Br

1. Na OH

2.

SN2 > E2

acid / basealready made

O

OH

Br

SN1 > E1

Carboxylates are OK nucleophiles at methyl, 1o RX and 2o RX but only E2 at 3o RX.

given

given

O

O

Br

already made

TM

TM - 1

TM - 1TM - 1

TM

SN1 > E1

SN2 mechanism using carboxylate at RX center introduces a new ester functional group (RCO2R’)

BrH3C


key bond

Br

Na

NaO

O

O

O

CH3SN2



SN1 mechanisms (if no rearrangement) using carboxylic acid introduces a new ester functional group (RCO2-R’)

Br

SN1 > E1

HH

acid is a nucleophile

acid is a base

SN1 > E1 at 2o and 3o RX and we don't have to worry about rearrangements here.

Br

SN1 > E1

OO

O

OH

O

OH

resonance

O

OH

H

O

O

O

OH

O

OH

acid is a nucleophile

O HO

OH

O

acid is a baseresonance

Sample alcohols - SN2 reactions using (R-X + ketone enolate) at Me, 1o, 2o, allylic RX

These are our target molecules (TM) - using the carbon side (ketone enolates from only propanone for now).

O

SN2newbond

O

SN2newbond

O

SN2newbond

O

SN2newbond

O

SN2newbond

O

SN2newbond

O

SN2newbond

etc, etc, etc

O

SN2newbond



O

target molecule

key bond

HN

n-butyl lithiumacid / base

1.N

R R

Li

LDA(makes enolate)

Br2.

SN2 > E2

acid / base

O

diisopropylamine

via ester enolates

Enolates are OK nucleophiles at methyl, 1o RX and 2o RX but only E2 at 3o RX.

-78oC

TM

TM - 1

already made


SN2 mechanism using ester enolate at RX center makes a larger ester on the carbon side (RCO2R’)

BrH3C


SN2

H2C

Li

C

O

O

OC

CH2

O

CH3

(3C = 2C + 1C)

key bondLi(alkylation)

-78oC

R

RBr

BrH2C

H3C

Br


H2C

Li

C

O

OR

OC

CH2

O

H2C

(4C = 2C + 2C)

key bond

RCH3

LiSN2

(alkylation)

-78oC

BrH2C


BrLiO

CCH2

O

H2C

(5C = 2C + 3C)

key bond

RCH2

CH3

H2C

Li

C

O

OR

-78oC

SN2

(alkylation)

ketone enolates (good nucleophiles at epoxides, carbonyls and CH3X, 1oRX and 2oRX) bigger ketones

BrBrH3C


SN2

H2C

Li

C

O

H3C

H3CC

CH2

O

CH3

(4C = 3C + 1C)

key bondLi(alkylation)

-78oC

BrH2C


H2CC

O

H3C

Li

O key bond

LiBr

SN2

(alkylation)

-78oC



BrH2C


H2CC

O

H3C

Li O key bond

(6C = 3C + 3C)

LiBr

SN2

(alkylation)

-78oC

5. azide and primary amine synthesis (azides can be made into primary amines via 1. LiAlH4 2. workup)

Proposed mechanisms

RN

NN

H3Al H

RN

NN

H OH2H

2. workup

RN

H

H

lithiumaluminium

hydride alkyl azide nitrogen 1o amines

SN2 at nitrogen, N2 leaving group

Br

NN

N

NN

N

-2

2-azidopropane

SN2= N3



6. alkene synthesis (via one E2 reaction with RBr and t-BuOK, we will see other ways to make alkenes later)

Sample alkenes using E2 reactions: R-X + t-butoxide; alkenes can make allylic R-X compounds

Sample allylic bromides from alkenes (Br2 / h)

BrBr

Br

alkenes

Br

Br

1. Na OHacid / base

already made

inexpensive base is OK because RX is tertiary

O OHK already made

K Hacid / base

OK

already made

Br

already made

E2 > SN2 (need strong, bulky base to favor E2 at 1o RX)

E2

Use steric bulk and/or extreme basicity to drive E2 > SN2.

Br

already made

OK

already made

E2


relative alkene stabilities = tetrasub. > trisub. > trans-disub > cis-disub gem-disub > monosub > unsub.


TM

TM

TM

TM - 1

TM - 1

TM - 1

7. alkyne synthesis (via two E2 reactions with RBr2 and 2xNaNR2, two times)

Sample alkynes using double E2 reactions of 1. RBr2 + NaNR2 2. workup

Make starting alkynes using two leaving groups and very basic NR2

Na sodium amide(very strong base)

Br Br

Br

Brand

Br2 h


(2 equivalents)(2 equivalents)

NR2

E2 twice

1.

2. workup


Br2 h


(2 equivalents)Br Br


(2 equivalents)

NR2

E2 twice

1.

2. workup



Br2 h


(2 equivalents)

Br Br


(2 equivalents)

NR2

E2 twice

1.

2. workup

Sample alkynes using SN2 reactions of 1. terminal alkyne + NaNR2 2. SN2 only at methyl or primary R-X (can do twice with ethyne)

(1 equivalent)


1.

2. BrH3C

acid / base

SN2 > E2

NR2newbond

(1 equivalent)


1.

2.

acid / base

SN2 > E2Br

NR2

Terminal acetylides are OK nucleophiles at methyl and 1o RX, but E2 > SN2 at 2o RX and only E2 at 3o RX.

newbond

Possible steps in mechanism (E2 twice then 2. acid/base or 2. RX electrophile)

BrBr

HN

R R

Na

Br

H

H

NR R

H

NR R

Na

Na2. workup

H

H2C

R

X

H2C

R

HO

H

H

2.

a

b

a

b

SN2



1. RNH Na RNH2

1 equivalent

NaBr

2.

(4C = 2C + 2C)terminal alkyne

SN2

Br

2.

Na

1. RNH Na RNH2

1 equivalent

(5C = 3C + 2C)internal alkyne

SN2

1. RNH Na RNH2

1 equivalent

NaBr

2.


SN2

Br

2.

Na

1. RNH Na RNH2

1 equivalent

(5C = 3C + 2C)

internal alkyne

(zipper reaction)

1. excess NaNRH2. workup (neutralize)

SN2acid/base

like tautomers

1. RNH Na RNH2

1 equivalent

Na

2.

2.

Na


1. RNH Na RNH2

1 equivalent

(6C = 3C + 3C)

internal alkyne

(zipper reaction)

1. excess NaNRH2. workup (neutralize)

Br

Br

like tautomers

SN2

SN2

acid/base

acid/base

terminal acetylides (SN2 only at methyl RX and 1o RX) larger alkynes

BrH3CCCH3C

Na SN2CCH3C CH3


Na

Br

BrH2CCCH3C

Na SN2CCH3C CH2

H3C CH3nucleophile = ?electrophile = ?

Na

Br



BrH2CCCH3C

Na SN2CCH3C CH2

CH2H2CH3C CH3nucleophile = ?

electrophile = ?

Na

Br

8. nitrile synthesis (via SN2 reaction at Me, 1o, 2o RX)

CN

CN

CN

H3CC

NCN

CN

Sample nitriles using SN2 reactions of cyanide + methyl, primary or secondary R-X

CN

CN

CN

nitriles

CN

BrSN2 > E2

Na

NC

already made

allowedCyanides are OK nucleophiles at methyl, 1o RX and 2o RX but only E2 at 3o RX.

CN

SN2 > E2

Na

NC

Br

already made

allowed

cyanide nitriles

BrH3CNa SN2

CN CH3


Na

Br

CN

BrH2C

Na SN2CN CH2

H3C CH3nucleophile = ?electrophile = ?

Na

Br

CN

BrH2CNa SN2

CN CH2

CH2H2CH3C CH3nucleophile = ?

electrophile = ?

Na

Br

CN



9. thiols (via SN2 reaction with HS-- at Me, 1o, 2o RX) and sulfides synthesis (via SN2 reaction with RS-- at Me, 1o, 2o RX)

Sample thiols using SN2 reactions of HS + methyl, primary, secondary, allylic, benzylic R-X

SHH3CSH

SH SHSH

SHSH

SH

SH

SHSH

SH

Sample sulifides using SN2 reactions of RS + methyl, primary or secondary R-X (many possibilities)

SS

SS

thiols and sulfides

SN2 > E2

SBr

already made

Na SH

SH

thiol(pKa 8)

1. Na OHacid / base

Br2.

sulfideSN2 > E2

Sulfides are OK nucleophiles at methyl, 1o RX and 2o RX but only E2 at 3o RX.

Possible mechanisms

BrH3C


SN2 SCH3

key bond

BrSHNa

H Na

BrH2C


SN2 S

key bond

NaNaH

Br

SH

BrH2C


SN2S

key bond

Na

NaH Br

SH



BrH3C


SN2 SCH3

key bondNa

H3C NaBr

SH3C

BrH2C


SN2 S

key bond

NaNa H3C

Br

SH3C

BrH2C


SN2S

key bond

Na

NaH3C Br

SH3C

10. introducing nucleophilic hydride or deuteride into organic molecules (via SN2 reaction) Sample deuterium added using SN2 reactions of LiAlD4 or NaBD4 + methyl, primary, secondary, allylic, benzylic R-X

DH3CD

D

D

D

DD

D

D

D D

D

lithium aluminium hydride = LAH = (Li AlH4) and sodium borohydride (NaBH4) are both nucleophilic hydride. For now we will consider them to be equivalent reagents.

Br

AlD

D

D

D

BH

H

H

H

NaLiD

AlH

H

H

H

Li

SN2

lithium aluminium hydride (LiAlH4) and sodium borohydride (NaBH4) = nucleophilic hydride (using “deuteride” shows where the “hydrogen” goes)

BrH3C


SN2 DCH3

key bondLi

AlD

D

D

D

Li Br

BrH2C


SN2Li H

AlD

D

D

DLi

ethane

Br

key bond



BrH2C


SN2H

key bond

propane

AlH

H

H

HLi Li Br

Mechanisms Worksheet

You need to be able to write your own mechanism from starting structures. I suggest you should practice using the nucleophiles from the list with the various RX compounds listed above. Writing mechanisms for each possibility would give you more than enough practice to learn the mechanism. Common mistakes include lone pairs, formal charge, curved arrows, correct Lewis structures, correct products predicted. This stuff takes practice – AND – correcting your mistakes. I have set up the following worksheet to give you templates to fill in so that you can use them multiple times by copying the basic framework and then filling in the details. You can do this IF YOU DO THE WORK!

SN2 then E2 examples first, followed by SN1 and E1 reactions

Example 1 - CH3-X (deuterium = D and tritium = T are isotopes of hydrogen that can be distinguished from H) The only possible choice at methyl is SN2. There is no C-H for E2 reactions, and methyl R+ is too high energy for solution chemistry, so no SN1/R1.

C Br

T

HD

Nu

B

choose from

our list

?

OR H

Examples you might consider:

R

O

OH

OH H

only SN2

weak nucleophile/bases = no reaction - in our course

these are SN1 and E1 conditions (weak)

S-bromodeuteriotritiomethane

strong = anything with negative charge,and neutral sulfur or nitrogen

Easier methyl example - CH3-X

C Br

H

HH

Nu

B

only SN2 (with strongnucleophiles)

CNu

H

HH

bromomethane

Nu B=

strong = anything with negative charge,and neutral sulfur or nitrogenWe cannot see inversion of configuration,

but we assume it occurs if the kinetics is bimolecular.

choose from

our list

choose from above,

Harder methyl example - CHDT-X

C Br

T

HD

Nu

B

only SN2 (with strongnucleophiles)

CNu

T

HD

SR

S-bromodeuteriotritiomethane

Nu B=

strong = anything with negative charge,and neutral sulfur or nitrogen

We can see inversion of configuration, because there is a chiral center and the kinetics is bimolecular.



CbH

HH

Ca Br

HH

Ca

Cb

H

Br

HH

HH

bromoethane

Example 2 - primary RCH2-X (deuterium = D and tritium = T are isotopes of hydrogen that can be distinguished from H) SN2 is the main reaction unless the strong electron pair donor is bulky and basic, then E2 is the main reaction (only potassium t-butoxide in our course).

Nu

B

NuB

Easiest primary example - CH3-X

choose from

our list

a = E2

b = SN2

a

a = E2b = SN2

b

bromoethane

a b

Cb

CaNuCa

Cb

Nu

HH

Cb

Ca

H

H

H

H

verticalview

horizontalview

CC

H

H

H

Ha = E2

a = E2b = SN2 b = SN2H

H

H

H

HH

H H

CbH

CH3D

Ca Br

DH

Ca

Cb

H

Br

HD

DH3C

(1S,2S)- 1-bromo-1,2-dideuteriopropane

?

(1S,2R)- 1-bromo-1,2-dideuteriopropane

Nu

B

Nu

B

?

Harder primary example - RCH2-X

choose from

our list

choose from

our list

CH

CH3D

C Br

DH

(1S,2S)- 1,2-dideuterio-1-bromopropane

Example 2 - primary RX

Nu

CH

CH3D

C Br

DH

(1S,2S)- 1,2-dideuterio-1-bromopropane

B

C

CNu

DH

H

CH3

D

SN2 > E2

(unless t-butoxide)C

C

D

D

H3C

H

E-alkene

C

C

CH3

D

H

HZ-alkene

E2

H or D can beanti to Ca-Br

Nu B=

strong = anythingwith negative charge,and neutral sulfur ornitrogen

SN2 > E2

(unless t-butoxide)



CH

TD

C Br

DH

(1S,2S)-1,2-dideuterio-1-bromoethane

Example 3 - primary RX

Nu

CH

TD

C Br

DH

(1S,2S)-1-bromo-1,2-dideuterio-1-tritioethane

B

C

CNu

DH

H

T

D

SN2 > E2

(unless t-butoxide) C

C

D

D

T

H

E-alkene(if H is anti)

C

C

T

D

H

H

E2

H or D can beanti to Ca-Br

Nu B=

strong = anythingwith negative charge,and neutral sulfur ornitrogen

(1R,2S)-1,2-dideuterio-1-nucleophile-ethane

C

C

H

D

D

H

SN2 mechanism

E2 mechanism

Z-alkene(if D is anti)

E-alkene(if T is anti)

SN2 > E2

(unless t-butoxide)

CD

TH

C Br

HD

(1R,2R)-1-bromo-1,2-dideuterio-1-tritioethane

B

Nu C

CNu

HD

D

T

H C

C

T

H

D

D


C

C

H

H

T

D

E2

C

C

D

H

H

DZ-alkene

(if D is anti)E-alkene

(if T is anti)(1S,2R)

CH

TD

C Br

HD

(1R,2S)-1-bromo-1,2-dideuterio-1-tritioethane

B

Nu C

CNu

HD

H

T

D C

C

D

H

T

D

Z-alkene(if H is anti)

C

C

T

H

H

D

E2

C

C

H

H

D

DE-alkene

(if D is anti)Z-alkene

(if T is anti)(1S,2S)

CD

TH

C Br

DH

(1S,2R)-1-bromo-1,2-dideuterio-1-tritioethane

B

Nu C

CNu

DH

D

T

H C

C

T

D

D

H


C

C

H

D

T

H

E2

C

C

D

D

H

HZ-alkene

(if D is anti)E-alkene

(if T is anti)(1R,2R)

H3CC

CH2

Br

H H

other primary RX to consider.

H3CC

CH2

Br

H3C CH3

Br

Br

Br

Mechanism predictions with weak nucleophile/bases:

OR H

R

O

OH

OH H ?

No SN1/E1 is possible at primary RX in solution chemistry. The primary R+ is too high in energy.

look at C - fully substitutedno SN2 or E2 at 1o neopentyl RX,E2 is possible at 2o neopentyl RX

benzylic = fast SN2



CH

DCH3

C Br

C

HC

C

D

Br

CH

HH3C

H

CH2CH3CH3

H

CH3

CH2CH3

(2S,3R,4S)-2-deuterio-4-methyl-3-bromohexane

Example 4 - secondary RX (deuterium is an isotope of hydrogen that can be distinguished from H)

(2S,3S,4S)-2-deuterio-4-methyl-3-bromohexane

Nu

B choose from above

?

Nu

B choose from above

?

(2R,3R,4S)-2-deuterio-3-bromo-4-methylhexane

NuSN2 and E2depends on basicity of donor

C

CNu

C

HH

H

H

HD

CH3less basic = SN2 > E2 at secondary RX

Some examples

OH Na

OR Na

R

O

O

Na

O KCH

DCH3

C Br

C

HH

CH3

CH2CH3

BSN2 and E2depends on basicity of donor

C

C

CH3

D

C

H

2Z-alkene 2E-alkene 3Z-alkene

C

C

H

C

CH3

D

H

Rstill R

CH

DCH3

C Br

C

HH

CH3

CH2CH3

(2R,3R,4S)-2-deuterio-3-bromo-4-methylhexane

H3C

CH2CH3H CH3

CH2CH3still S

C

C

H

H3C

C

H H CH3

CH2CH3still S

E2

SN2

more basic = E2 > SN2 at secondary RX

HCC

NC

BSN2 and E2depends on basicity of donor

C

C

CH3

D

CH3

H

2Z-alkene

C

C

H

H3C

CH3

H

2E-alkene 1-butene, neither E or Z

H2C

H

C

CH3

D

H

R

still R

(2S,3R)-3-deuterio-2-bromobutane

CH

DCH3

C Br

C

HH

H

H

SN2 and E2depends on basicity of donor

C

CNu

C

HH

H

H

HD

CH3

more basic = E2 > SN2Some examples

OH Na

OR Na

R

O

O

Na

OKSN2

Nu

CH

DCH3

C Br

C

HH

H

H

(2S,3R)-3-deuterio-2-bromobutane

2Rless basic = SN2 > E2



H3CC

CC

H H

other secondary RX to consider.

H3CC

CCH3

H3C CH3

Br

Mechanism predictions with:

OR H

R

O

OH

OH H ?

Br H

HH

H

Br HBr H

Br

Br

Br

Br

Usually SN1 > E1 (except ROH + H2SO4/.)

2o neopentyl allylic

(2R,3R,4S)-2-deuterio-4-methyl-3-bromohexane

Example 5 - tertiary RX (deuterium is an isotope of hydrogen that can be distinguished)

(2S,3S,4S)-2-deuterio-4-methyl-3-bromohexane

Nu

B ?

Nu

B?

CH

DCH3

C Br

C

CH2

H

CH3

CH2CH3

H C

C

D

Br

C

CH2

HH3C

H

CH2CH3CH3

H

choose from

our list

choose from

our list

only E2 possible at 3o RX and strong base-nucleophile, need Cb-H though.

Some examples

OH Na

OR Na R

O

O

Na

OKNu No SN2 reaction

at 3o RX, backside is too sterically hindered

(2S,3R,4S)-2-deuterio-4-methyl-3-bromohexane

CH

DCH3

C Br

C

CH2

H

CH3

CH2CH3

H

E2 reaction at 3o RX with strong base-nucleophile

(2R,3R,4S)-2-deuterio-4-methyl-3-bromohexane

B

2Z-alkene(with D)

2E-alkene(without D) 1-butene, neither E or Z

C

C

H3C

C

CH3

D

H

Rstill R

H3C

H2C CH3

3Z-alkene from "S"

H2CC

C

C

CH3

H

D

H2C

CH3

still R

still SS

H CH3

CH

DCH3

C Br

C

CH2

H

CH3

CH2CH3

H

(3E-alkene if "R")

C

C

H

H3C

H3C

C

H2C

CH3

H CH3

still S

C

C

CH3

D

H3C

C

H2C

CH3

H CH3

still Sonly first mechanismis shown



Usually SN1 > E1 (except ROH + H2SO4/D.)

other tertiary RX to consider.

H3CC

C

H2C

H3C CH3


OR H

R

O

OH

OH H ?

Br Br CH3

Br

Br H

CH3

Br

Br Br Br

Example 5 - cyclohexane structures to consider (X must be axial to react by SN2 and E2) (X can be axial or equatorial to react by SN1 and E1). Essential details can be filled in on the following templates.

cyclohexanewithout branches

cyclohexanewith a branch


OR H

R

O

OH

OH H

Many substitution patterns are possible.

XX X

X

X X X

X

X

a b c d e

fg

h ij

Some examples

OH Na OR Na R

O

O

Na

O K

X

weaknucleophiles

strong nucleophile/bases + other anions shown above



Simple Examples – general patterns (this is as easy as it gets, more complicated examples follow)

BrH3CBr

Br

Br

Br

OH Na

OHH3C

Br

OH

OH

No Reaction(neopentyl)

nucleophile / base

only SN2 SN2 > E2 E2 > SN2 only E2 only E2

No SN2 Reaction(neopentyl)

BrH3CBr

Br

Br

Br

NaOR

OH3C

Br

O

OR

nucleophile / base

only SN2 SN2 > E2 E2 > SN2 only E2 only E2

R

R

No Reaction(neopentyl)


BrH3CBr

Br

Br

Br

O

O

Na

Br

O

O

nucleophile / base

only SN2 SN2 > E2 only E2 only E2

O

O

O

O

SN2 > E2No Reaction(neopentyl)




BrH3CBr

Br

Br

Br

O K

Br

O

nucleophile / base

only SN2only E2

only E2

O

E2 > SN2 only E2 No Reaction(neopentyl)

No SN2 Reaction(neopentyl

and t-butoxide)

BrH3CBr

Br

Br

Br

OH H

Br

nucleophile / base

No Reaction(1o RX)

SN1 > E1 SN1 > E1SN1 > E1

with rearrangement(main product?)

OH

R R

R& S

OH

OHOH

R& S

No Reaction(1o neopentyl)No Reaction

(1o RX)



BrH3CBr

Br

Br

Br

OR H

Br

nucleophile / base

SN1 > E1 SN1 > E1SN1 > E1

OR

R R

R& S

OR

OROR

R& S

No Reaction(1o RX)

No Reaction(1o RX)

No Reaction(1o neopentyl)

with rearrangement(main product?)

BrH3CBr

Br

Br

Br

R

O

OH

Br

nucleophile / base

SN1 > E1 SN1 > E1SN1 > E1

with rearrangement

O

R R

R& S

O

OO

R& S

O O

O

O

No Reaction(1o RX)

No Reaction(1o RX)

No Reaction(1o neopentyl)



SN1 / E1 possibilities –extra complications at Cβ positions, 2o RX, rearrangements NOT considered (H2O,ROH,RCO2H)

OR HR

O

OOH H

Examples of weak nucleophile/bases in our course

a b

c

H

water alcohols carboxylic acids

E1 product from left C carbon atom

H3CC

CC

H2C

CH3

H H

H

H

Br

H

H3CC

CC

H2C

CH3

H H HH

H

Redrawn

(achiral)

(achiral)

CCH

H2C

H

H3CH

CH2

CH3

H2O

CCH

C

H

CH2H

CH3H

H3CH

H2O

same first step for SN1/E1

add nuclephile (top/bottom) = SN1

lose -H (top/bottom) = E1

rearrangement to similar or more stable carbocation (R+)(start all over)

"H" parallel to empty 2p orbital on top

"H" parallel to empty 2p orbital on bottom

C C

H3C H

H CH2

H2C

CH3

C C

H H

H3C CH2

H2C

CH3

2E-alkene

2Z-alkene

H2O

CH2C

H

C

H

H3C

CH2H

CH3

CCH2

H

C

H3C


"H" parallel to empty 2p orbital on bottom 3Z-alkene

3E-alkene

OH2

H

H

OH2

C

C CH2

H2C

H

H

CH3

H3C

C

C H

H2C

H2C

HH3C

CH3

OH2

CH2C

H

H2CH3C

CH2

CH3

OH2

H2O

a

b

SN1 product (a. add from top and b. add from bottom)

CCH2 H2C

H

O

HH

H3C H2C

CH3

C

H2C

H H2C

O

HH

H3CCH2

CH3

a

b

OH2

OH2

CCH2 C

H2

H

O

H

H3C H2C

CH3

C

H2C

H H2C

O

H

H3CCH2

CH3

(achiral)

(3S)

enantiomers

acid/base proton transfer


E1 product from right C carbon atom

H2C

H3C

(3R)

(3R)

carbocation choices



SN1 / E1 possibilities –extra complications at Cβ positions, 2o RX, rearrangements NOT considered, with deuterium (makes it a little harder)

OR HR

O

OOH H

Examples of weak nucleophile/bases

a b

c

H

water alcohols carboxylic acids

E1 product from left C carbon atom

H3CC

CC

H2C

CH3

D H

H

D

Br

H

H3CC

CC

H2C

CH3

D H DH

H

Redrawn

(2S,3R,4S) (2S,4S)

(2S,4S)

CCH

C

D

H

H3CH

CH2D

CH3

H2O

CCH

C

H

CH2D

CH3D

H3CH

H2O




CCH

C

H

H

DH3C

CH2D

CH3

H2O

CCH

C

H

CH2D

CH3H

DH3C

H2O

"D" parallel to empty 2p orbital on top

"D" parallel to empty 2p orbital on bottom



C C

H3C H

H CHD

H2C

CH3

C C

H H

H3C CHD

H2C

CH3

C C

D H

H3C CHD

H2C

CH3

C C

H3C H

D CHD

H2C

CH3

4S,2E-alkene without "D"

4S,2Z-alkene without "D"

4S,2Z-alkene with "D"

4S,2E-alkene with "D"

H2O

(2S,4S)

(2S,4S)

(2S,4S)

(2S,4S)

(4S)

(4S)

(4S)

(4S)


carbocation choices



Redrawn from above (2S,4S)

CCH

C

D

H

H3CH

CH2D

CH3

CCH

CD

H3CH

"D" parallel to empty 2p orbital on top

"D" parallel to empty 2p orbital on bottom



2S,3E-alkene without "D"

2S,3Z-alkene without "D"

2S,3Z-alkene with "D"

2S,3E-alkene with "D"

OH2

H

DH2C

OH2

CCH

C

D

D

H3CH

HCH2

CCH

CD

H3CH

OH2

D

H2C

H

OH2

H3C

H3C

CH3

C

CH2C

HDC

D

H

CH3

H3C

C

C D

CDH

H2C

HH3C

C

CH2C

CDH

H

H

CH3

H3C

CH3

C

C H

CDH

H2C

HH3C

CH3

OH2

CCH

C

D

H

H3CH

CH2D

CH3

OH2

H2O

a

b

SN1 product (a. add from top and b. add from bottom)

CCDH HDC

H

O

HH

H3C H2C

CH3

C

DHC

HDHC

O

HH

H3CCH2

CH3

a

b

OH2

OH2

CCDH HDC

H

O

H

H3C H2C

CH3

C

DHC

HDHC

O

H

H3CCH2

CH3

(2S,4S) (2S,3R,4S)

(2S,3S,4S)

diastereomers



E1 product from right C carbon atom

(2S,4S)(2S)

(2S)

(2S)

(2S,4S)

(2S,4S)

(2S,4S)

(2S)



SN1 / E1 possibilities –extra complications at Cβ positions of 2o RX, rearrangement to more stable 3o R+ considered

E1 product from left C carbon atom (without rearrangement)

H3CC

CC

H2C

CH3

H H

H

CH3

Br

H

H3CC

CC

H2C

CH3

H H CH3H

H

Redrawn

(4S)

(achiral)

CCH

C

H

H3CH

H2O

CCH

CH

H3CH

H2O






C C

H3C H

H C

H2C

CH3

C C

H H

H3C C

H2C

CH3

4S,2E-alkene

4S,2Z-alkene

H2O

CH2C

H

C

H

H3C

CH2H3C

CH3

CCH2

H

C

H3C


"H" parallel to empty 2p orbital on bottom 3Z-alkene

3E-alkene

OH2

H

CH3

OH2

C

C CH2

CH2

H3C

H

CH3

H3C

C

C CH3

H2C

H2C

HH3C

CH3

OH2

CH2C

H

CH3C

OH2

H2O

a

b

SN1 product (a. add from top and b. add from bottom), (without rearrangement)

CCH2 C

H

O

HH

H3C H2C

CH3

C

H2C

H

C

O

HH

H3C

a

b

OH2

OH2

CCH2 C

H

O

H

H3C H2C

CH3

C

H2C

H

C

O

H

H3C

(3S,4S)

diastereomers



E1 product from right C carbon atom (without rearrangement)

H2C

H3C

(3R,4S)

(3R,4S)

H2C

CH3

CH3H CH3H

H2C

CH3

CH3H CH3

H

H2C

CH3

H CH3 CH3H2C

CH3

CH3H

H H

H

H2C

CH3

CH3

CH3

(4S)

(4S)

(4S)

(4S)

(4S)

diastereomers

diastereomers

3R stereochemistry is lost

(3R,4S)

(3S,4S)


carbocation choices



After rearrangement to 3o carbocation (R+) – We will skip rearrangements in Chem 314

E1 products from left C carbon atom (top and bottom, after rearrangement)

Redrawn (achiral)

H2O

CC

CH3

H2C

HH3C

2E-alkene2Z-alkene

3Z-alkene3E-alkene

OH2

C

C CH2

H2C

H3C

H

CH3

H3CC

CH2C

H

H3C

CH2

OH2

H2O

a

b

SN1 product (a. add from top and b. add from bottom), (after rearrangement)

C

O

HH

C

O

HH

a

b

OH2

OH2

(achiral) acid/base proton transfer

E1 product from right C carbon atom (top and bottom, after rearrangement)

(3R)

diastereomers

CH2C

H

C

H

H3C

CH2H3C

CH3

(4S)

initial 2o carbocation rearranges via hydride shift to a 3o carbocation

C

C

H3C

H

CH2

H3CH

C

CH3

4S stereochemistry is lost in new carbocation



rearrangement to similar or more stable carbocation (R+)(none is possible here)

R+ reaction possibilities start all over again with new carbocation

H

H

C

C

H3C

H

CH2

H3CH

C

CH3

H

H

C

C

H3C

H

CH2

H3CH

C

CH3

H

H

"H" parallel to empty 2p orbital on top and bottom of right C position

CH2

H3C

CC

HH2C

CH3H3C

CH2

H3C

diastereomers

C

C

H3C

H

CH2

H3CH

C

CH3

H

H

H2C

CH3

CH2

H3C

CH2

H3C

CH2

CH3CH2H3C

H2C

H3C


C

O

H

CH2

CH3CH2H3C

H2C

H3C

C

O

H

H2C

CH3

CH2

H3C

CH2

H3C

(3S)

enantiomers

CH3

H3C

CH2

C

CH3

H2C

H2C 1-alkene(does not have E/Z stereochemistry)

OH2

E1 product from methyl C carbon atom (top and bottom, after rearrangement, only one product from the methyl)

C

C

H2C

H

CH2

H3CH

CH2

CH3CH2

H3C

H

(3R)

(3S)





Homework problems: The number of each type of product (SN1, E1, SN2, E2) is listed after a reaction arrow for each starting structure (assuming I analyzed the possibilities accurately in my head, while sitting at the computer). See if you can generate those products using a valid mechanism for each one.

Br H

H3C H

OH

O

OH

HO

H

SN1 E1 SN2 E2

O

O

O

HO

2 41 3

1 3

1 3

a.

2 5b.

2 4a.

2 5b.

2 4a.

2 5b.b. after rearrangement

a. initial carbocation

Br H

H3C H

Br H

H3C H

OH

O

OH

SN1 E1 SN2 E2

O

O

O

HO

2 61 3

1 3

1 3

a.

2 5b.

2 6a.

2 5b.

2 6a.

2 5b.b. after rearrangement

a. initial carbocation

Br H

H3C H

HO

H

D H D H

HO

H

2 3

chair 1 chair 2

SN1

XX X

a.

b.

SN2 E2

1 1

E1

1 1

1

SN1 SN2 E2

1 2

E1

2 2

SN1 SN2 E2

0 1

E1

5 6SN1

XX X

SN2 E2

1 2

E1

2 2

4

SN1 SN2 E2

1 2

E1

2 2

SN1 SN2 E2

1 2

E1

2 2

8 9

SN1

XX

SN2 E2

1 2

E1

2 2

7

SN1 SN2 E2

0 3

E1

2 3

SN1 SN2 E2

0 1

E1

2 1

1 2

Only consider carbocation rearrangements that immediately go to a more stable carbocation (not equally stable carbocations = too many possibilities)

a.

b.2 2

1 2

X

a.

b.c.

d.

consider any 3o R+ possibilityand consider stereoisomers

4 51 21 2

HO Na

condition 1 condition 2



C

C

C

Br

CH

Hb

Ha H

H HH H

C

C

Ha

Br

CH

H3CHb H

H

Rotation of C-C1 brings Ha or Hb anti to C-Br, which allows SN2 and two different E2 possibilities: Ha (Z) and Hb (E). Since C2 is a simple methyl, there is no C2 substituent to inhibit either of these reactions.

C

C

Hb

Br

CH

Ha

H3C H

H

When methyl on C1 is anti to C-Br, no SN2 is possible and no E2 is possible from C1.

SN2 possibleE2 from C1 (2Z- butene)E2 from C2 (1-butene)

SN2 possibleE2 from C1 (2E- butene)E2 from C2 (1-butene)

No SN2 possible and noE2 from C1, but E2 from C2 (1-butene) is possible.

E2 possible here

Use these ideas to understand cyclohexane reactivity.

H H

H

H

H HBr

H

H

H

H

H

HH

H

H

HH

BrH

H

H

HH

No SN2 is possible (1,3 diaxial positions block approach of nucleophile), and no E2 is possible because ring carbons are anti.

No SN2 or E2 when "X" is in equatorial position.

SN2 possible if C is not tertiary and there is no anti C "R" group.

E2 possible with anti C-H.

Both SN2 and E2 are possible in this conformation with leaving group in axial position.

H H

H

H

H HBr

H

C

H

H

H

HH

H

H

HH

BrH

H

H3C

HH

No SN2 or E2 when "X" is in equatorial position.

No SN2 possible if there is an anti C "R" group.

E2 possible with anti C-H.

Only E2 is possible in this conformation. Leaving group is in axial position.

H

H

H

full rotation atC-C is possible in chain

only partial rotation is possible in ring

only partial rotation is possible in ring

H H H

equatorial leaving group

axial leaving group

No SN2 is possible (1,3 diaxial positions block approach of nucleophile), and no E2 is possible because ring carbons are anti.

No E2 possible, no anti C-H.

full rotation atC-C is possible in chain

alkene stabilities tetrasubstituted > trisubstituted > trans-disubstituted > gem-disubstituted cis-disubstituted > monosubstituted



Examples - group A

C

H3C

H3C

H3C

H

H

CCH3H3C

H

H

severe steric repulsion

19,999

Keq »t-butyl substituent locks in chair conformation with equatorial t-butyl

X X

XX

XX

X

1. Is SN2 possible? Requires an open approach at C and C.2. Is E2 possible? Requires anti C-H.3. How many possible products are there?4. What is the relationship among the starting structures?5. What is the relationship among the products?6. Are any of the starting structures chiral?7. Are any of the product structures chiral?

X X

X X

1 2

3

4

5

6

7

8

9

10

11

Cyclohexane structures have two chair conformations possible.

X = Cl Br I OTs

H3C

1. Which conformation is reactive?2. Is SN2 possible? Requires an open approach at C and C.3. Is E2 possible? Requires anti C-H.4. How many possible products are there?5. What is the relationship among the starting structures?6. What is the relationship among the products?7. Are any of the starting structures chiral?8. Are any of the product structures chiral?

Examples - group B

1

TsO OTs

OTsTsO

TsO OTs

OTs

TsO OTsTsO OTs2

3

4

5

6

7

8

9

10

11

chair 1 chair 2



strong base/nucleophile conditions (E+ = electrophile, Nu: = nucleophile)

n-butyl lithium (powerful nucleophile at epoxides and carbonyls (C=O), and the “most powerful base” at other times)

BrH3Cnucleophile = ?electrophile = ?

H3C

H2C

CH2

H2C

LiShould be SN2, but does not work well. Too many side reactions. Another approach is necessary,using cuprates (R2Cu Li ).

BrH2C


H3C

H2C

CH2

H2C

Li

BrH2C


H3C

H2C

CH2

H2C

Li

Should be SN2, but does not work well. Too many side reactions. Another approach is necessary,using cuprates (R2Cu Li ).

Should be SN2, but does not work well. Too many side reactions. Another approach is necessary,using cuprates (R2Cu Li ).

Four new mechanisms to learn: SN2 vs E2 and SN1 vs E1psbeauchamp/pdf/314_SN_E_2013.pdf ·...

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Transcript of Four new mechanisms to learn: SN2 vs E2 and SN1 vs E1psbeauchamp/pdf/314_SN_E_2013.pdf ·...