2 r-x is the electrophile sn1psbeauchamp/pdf/315_sn_e_slides_fall_2015.pdfy:\files\classes\0 organic...

$: 2 R-X is the electrophile SN1psbeauchamp/pdf/315_SN_E_slides_fall_2015.pdfy:\files\classes\0 Organic Topics - latest\315 topics\20 315 lecture notes, 7-9-15\2c 315 SN and E & chem$
Nucleophilic Substitution & Elimination Chemistry Beauchamp 1

y:\files\classes\0 Organic Topics - latest\315 topics\20 315 lecture notes, 7-9-15\2c 315 SN and E & chem lecture slides 2.doc

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

SN2 competes with E2

SN1 competes with E1

These electrons always leave with X.

R-X is the electrophileSN2

E2

SN1

E1

XRBNu BHNuHCompeting Reactions

Competing 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) (weak)

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)

Important details to be determined in deciding the correct mechanisms of a reaction.

1. Is the nucleophile/base considered to be strong or weak? We simplistically view strong electron pair donation as anions of all types and certain neutral nitrogen, sulfur and phosphorous atoms. Weak electron pair donors will be neutral solvent molecules, usually water (H2O), alcohols (ROH), or simple, carboxylic acids (RCO2H).

2. What is the substitution pattern of the R-X substrate at the C carbon attached to the leaving group, X? Is it a methyl, primary, secondary, tertiary, allylic, or benzylic carbon? What about any C carbon atoms? How many additional carbon atoms are attached at a C position (none, one, two or three)?

Answers to these questions will determine SN2, E2, SN1 and E1 reactivities and alkene substitution patterns and relative stabilities in E2 and E1 reactions.

SN2 versus E2 overview (Essential features: strength of the nucleophile/base (as judged by its conjugate acid pKa), and steric factors (size) of the nucleophile/base) These are competing reactions.

CH

CH3H

C Br

HH

1-bromopropane

CH

CH3H

C Br

HH

1-bromopropane

O

C

CO

HH

H

CH3

H

E2 > SN2

(when t-butoxide)C

C

H

H

H3C

Halkene

E2 = major

Nu B=

strong = anything with negative charge,and neutral sulfur, phosphorous or nitrogenin our course.

SN2 > E2

(unless t-butoxide)

C

H3C

H3C

H3C

(also R2N = E2)

SN2 alwaysbackside attack

E2 alwaysanti C-H, C-X(in our course)

OH3C

H3C

SN2 = major

most anions react by SN2 at 1o RX

Br

Br



reactants

products

PEpotential energy

POR = progress of reaction - shows how PE changes as reaction proceeds

higher(is lessstable)

lower(is morestable)

G

OH C

H

HH

Br

As carbon inverts configuration, its sp2 transition state forms a high PE carbon with 10 electrons at carbon. This is a concerted, one-step reaction.

Br= negative(exergonic)

C

H

HH

HO

CHO

Transition State

H

H H

Br

Ea - this energy difference determines how fast the reaction proceeds = 'kinetics'

Rate = kSN2[RX][Nu] = bimolecular reaction

kSN2 = 10

-Ea2.3RT

Ea = -2.3RT log(kSN2)

G = -2.3RT log(Keq)

G - this energy difference determines the extent of reaction, the ratio of products vs. reactants at equilibrium (when kinetics allows the reaction to proceed). Thermodynamics is determined by the strengths of the bonds and solvation energies ofthe reactant and product species.

KSN2 = 10

-Go

2.3RT

The stability of the bromide anion helps drive the equilibrium to the right.

SN2 PE vs. POR Diagram (= concerted energy picture that looks very similar to E2 reactions)

Transition State

reactants

products

PEpotential energy

POR = progress of reaction - shows how PE changes as reaction proceeds

higher(is lessstable)

lower(is morestable)

Ea = -2.3RT log(kE2)

G = -2.3RT log(Keq)

OH C

C

R1R2

Br

C

C

R1 R2

Transition state - requires parallel overlapof the two 2p orbitals forming the pi bond. This is easiest when C-H is anti to C-X.

Br

HO

R3R4

H

R3 R4

H O

H

Br

C

C

R1 R2

R3 R4

If stereochemical priority is R1 > R2 and R3 > R4 then this would be Z configuration, which is fixed by the requirement for anti C-H / C-X elimination. If syn elimination occurred the stereochemistry would be E ("syn" is not typically observed).

E2 mechanism depends on steric factors and basicity of the electron pair donor. More steric hindrance and more basic favors E2 over SN2

= negative(exergonic)

H

staggered eclipsed

E2 PE vs. POR Diagram (= concerted energy picture that looks very similar to SN2 reactions)

Rate = kE2[RX][Nu] = bimolecular reaction



SN2 reactions are arguably the most important reactions in organic chemistry. They always occur by backside attack at C-X carbon. These are the common R-X patterns we encounter. Steric factors are very important.

XH3CH2CR X

HCR X

R

CR X

R

RH2C X

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

(1o, 2o, 3o versions)

C C X

C-beta carbon (0-3 of these)

C-alpha carbon (only 1 of these)

benzylic(1o, 2o, 3o versions)

These are the importantcarbon atoms to recognize.

H2C

HC

CH2

X

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 0t-butyl (tertiary)

(very low)140

As SN2 slows down E2 gets more competitive.C carbon patterns

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

XNu

Tertiary substitution has no easy path of approach by the nucleophile from the backside. The C carbon is completely substituted, so the nucleophile cannot get close enough to make a bond with the C carbon. We do not propose any SN2 reaction at tertiary RX centers. E2 becomes the dominant mechanism here.

H2C

HC

CH2

X

k 100allyl (primary)

resonance sp2 in T.S.

CHO

H

C

Br

C

H H

H

Negative charge is stabilized by delocalizaton into adjacent pi bond. Lowers T.S. energy, thus faster rate.

stabilized "allylic" transition state (benzylic too)

Nu

excellent

Nu

good

Nu

OK

verypoor

H



Relative rates of SN2 reactions - Steric hindrance at the Cβ carbon slows down the rate of SN2 reactions. All of these are primary R-X structures at C, but substituted differently at C.

C carbon patterns

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 0.00001 02,2-dimethylpropyl

(1o neopentyl)

X X XX

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 even when 1o RX. SN2 is not possible.

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.

Steric size of the nucleophile/base also affects the SN2/E2 distribution of products.

C

H

C

Br

HH

CH2

HH3C

O

H3C

H3C

H2C

CH

CH2H3C

CH2

CH2

CH2

OH3C

SN2

E2

SN2 = 90%E2 = 10%

C

H

C

Br

HH

CH2

H

CO

H3C

H3C

H2C

CH

CH2

SN2

E2

SN2 = 15%E2 = 85%

H3C

H3CCH3

H3CCH2

CH2

CH2

OH3C

1-bromobutane (1o RBr)

1-bromobutane (1o RBr)

methoxide

t-butoxide

(pKa = 16)

(pKa = 19,sterically large)

The first clue to recognize is strong nucleophile/bases (most anions, neutral sulfur and phosphorous, some nitrogen) vs. weak nucleophile/bases (neutral solvents = H2O, ROH, RCO2H)



Nucleophile / Bases: Strong electron pair donation (SN2 / E2) or Weak electron pair donation (SN1 / E1)

Na

O H

Na

S HO R

Na

Na

O

O

hydroxide

alkoxides

hydrogen sulfide (thiolate)

Na

S R

alkyl sulfide(thiolate)

Na

C N N

O

Ophthalimidate (an imidate)

C C H CH2

OLi

ketone enolates

Li

Na

cyanide

acetylides ester enolates

H2C

O

OMe

Na

NN

N

azide

BH

H

H

H

Na

AlH

H

H

H

Li

sodiumborohydride

lithiumaluminium

hydride

S

S

H

Li

dithiane anion

BD

D

D

D

Na

sodiumborodeuteride

AlD

D

D

D

Li

lithiumaluminiumdeuteride

Na H

N

LDA (lithiumdiisopropylamide)(sterically large, very strong base)

sodium hydride (very strong base)

Na

sodium amide (very strong base)

KH

potassium hydride (very strong base)

R = C or HNR2

Li

SPh Ph

Ph = phenyl

PPh Ph

diphenylsulfide,used with C=O

to make epoxides

triphenylphosphine,used with C=O to make alkenes

Ph

CH2

CLi

nitrile enolates

CH2

O

N

Li

3o amide enolates

N Li

acid dianions

H2C

O

O

almost always a base in our course

CH2

Li

n-butyl lithium (very strong base

used to make other strong bases/nucleophiles)

Na Cl

Br

INa

Na

good nucleophileswith tosylates and

in strong acid

SCl

O

O= Ts-Cl (tosyl chloride)makes ROH into tosylates

N

pyridine = proton sponge

= py BrCu

cuprous bromide(makes cuprates)

CrO3 /

pyridinium chlorochromatePCC

oxidizing E2 reaction,makes aldehydes and ketones

CrO3 / H2O

Jones reagentoxidizing E2 reaction,

makes carboxylic acids and ketones

oxidizing reagents at ROH and RCHO miscellaneous reagents

You can used any of these whenever you need (buy) You have to make these from given compounds (acid/base chem)(use reagents and example functional groups, for now)

organocuprates(good carbon

nucleophiles at RX,made from R-Li)

LiCuR R

(MgBr)

Grignard reagents(very strong basesand nucleophiles)

R Li

alkyl lithium (very strong basesand nucleophiles)

R

poor nucleophiles at RBr, but good nucleophiles at C=O and epoxides

N

S

S

H

H

H

free radical chemistry

Br2h

Br2 / ROORh

ylid chemistry reagents

dithiane chemistryenolate chemistry

used to make LDA, which is the base used

to make enolates of all types

used to make aldehydes

and ketones

carboxylates

CH3C O

CH3

H3C K

potassium t-butoxide(sterically large,

strong base)

Na

almost always a base in our course

neutral functional group examples to make enolates (we can make our own nitriles)

O

O

O

MeN

O

Me

O

O

H

propanone methyl ethanoate N,N-dimethylethanamide ethanoic acid

HO

H

weak nucleophiles (in our course)

RO

H R

O

OH

(usually SN1 > E1)

water alcohols carboxylic acids

E1 exception: ROH + H2SO4 / Me

miscellaneous reagents

Me

Me

SN2 / E2 concerted reactions (one step)

SN2 = always backside attack at C (inversion of configuration)

E2 = anti C-H / C-X elimination (forms pi bonds)

Strong nucleophile and/or base Weak nucleophile and/or baseSN1 / E1 multistep reactions,Carbocation formation requires 2o, 3o, allylic or benzylic RX and a polar hydrogen bonding solvent.

rearrangements are likely to form similar or more stable carbocations

nucleophiles can add to both sides of R+ and beta protons can be lost from both faces of R+, generally not as useful due to many possibilities.

usually SN1 > E1, except for dehydration of ROH using H2SO4 and heat (distills out the E1 alkene)

N

O

Ophthalimide

H

RBr

MgR

BrLi



Special requirements for cyclohexane rings (SN2 and E2)

SN2 Inversion of Configuration in Cyclohexane Rings – Axial leaving group is preferred.

H H

H

HH

H

H

H

H

HH H

H

H

H

H

H

H

H

H

H

HThese two chair conformations

interconvert in a very fast equilibrium.

Equatorial X is preferred, but axial X is reactive for both SN2 and E2.

X

X

Nu ? Nu?

H

H

H

H

H

H

HX

Nu

cyclohexyl ring versus isopropyl RX as reference compound.Backside attack IS possible.

Backside approach is not possible when leaving group is equatorial.

Backside approach is possible when leaving group is equatorial.

An axial leaving group is also required for E2 reactions because it is the only way to have the required anti Cβ-H / CαX orientation

X

X

H

H

H

H

H

H

No anti C-H / C-X, so no E2 reaction(the ring is anti)

anti C-H / C-X, so E2 reaction is possible

(these are trans in the ring)H

H

Br Br

Br

Br

Br

Br

Br

Predict products. Explain choices. In pairs, predict the faster reacting stereoisomer (why?).

More stable conformation,

but less reactive.

Less stable conformation,

but more reactive.

Br

axial Br, then fill in the blanks



Stability of pi bonds

Greater substitution of carbon groups in place of hydrogen atoms at alkene carbons translates into greater stability (lower potential energy). There are three types of disubstituted alkenes and their relative stabilities are usually as follows: geminal cis trans. This is also true in alkynes.

C

H

H

C

H

H

C

R

H

C

H

H

C

R

H

C

R

H

C

R

R

C

H

H

C

R

H

C

H

R

C

R

R

C

R

H

C

R

R

C

R

R

<

The more substituted alkenes are usually more stable than less substitued alkenes. Substitution here, means an R group for a hydrogen atom at one of the four bonding positions of the alkene.

1 2 3 4 5 6 7

Relative stabilities of substitued alkenes.

1 = unsubstituted alkene (ethene is unique)2 = monosubstituted alkene3 = cis disubstituted alkene4 = geminal disubstituted alkene5 = trans disubstituted alkene6 = trisubstituted alkene7 = tetrasubstituted alkene

Saytzeff's rule: More stable alkenes tend to form faster (because of lower Ea) in dehydrohalogenation reactions (E2 and E1). They tend to be the major alkene product,though typically a little of every alkene product possible is obtained.

"unsubstituted" "mono" "di-cis" "di-geminal" "di-trans" "tri" "tetra"

< < < <

Possible explanations for greater stability with greater substitution of the pi bond

A fairly simple-minded explanation (the one we will use) for the relative alkene stabilities is provided by considering the greater electronegativity of an sp2 orbital over an sp3 orbital. An alkyl group (R) inductively donates electron density better than a simple hydrogen. The more R groups at the four sp2 positions of a double bond, the better. Hyperconjugation is also used to explain electron donation into the pi bond. However, be aware that other features, such as steric effects or resonance effects, can reverse expected orders of stability.

C C C C

HR

...inductively donating "R" substituent is more stable at sp2 C than unsubstituted "H"...

hyperconjugation reason (MO argument) (sigma resonance?)

C C

H

H

H

C

H C C

H

H

H

C

H

inductive donation to more electronegative sp2 orbital (than sp3 orbital)



Summary chart of similar looking nucleophiles versus bases (2o RX centers are the most ambiguous)

carbonnucleophile / bases

nitrogennucleophile / bases

Ka = 10-25

SN2 > E2

only E2

only SN2

R-X patterns

CC

NC

Ka = 10-9

SN2 > E2

only SN2

E2 > SN2 SN2 > E2

only E2

morebasic

lessbasic

RN

R

Ka = 10-37

NA

E2 > SN2

only E2

only E2

NN

N

Ka = 10-5

lessbasic

N

O

O

Ka = 10-9

lessbasic

only SN2 only SN2

SN2 > E2 SN2 > E2

SN2 > E2 SN2 > E2

only E2 only E2

morebasic &sterics

oxygennucleophile / bases hydrogen

nucleophile / bases("D" shows rxn site)

OH

ORR

O

O

Ka = 10-5

lessbasic

Ka = 10-16

morebasic

only SN2 only SN2

SN2 > E2 SN2 > E2

E2 > SN2 SN2 > E2

only E2 only E2

O

only SN2

E2 > SN2

only E2

only E2

Ka = 10-19

Ka = 10-37

more basic Ka = ?less basic

H

B H

H

H

H

NA

only E2

only E2

only E2

only SN2

SN2 > E2

SN2 > E2

only E2

morebasic &sterics

H3CX

CH2

XR

CHXR

R

CXR

RR

R-X patterns

methyl

primary

secondary

tertiary

H3CX

CH2

XR

CHXR

R

CXR

RR

methyl

primary

secondary

tertiary

R



Simple reactions to practice on: 1. hydroxide nucleophile

HO

H3CBr

Br

Br

Br

Br

Br

Br

Br

HO

HO

HO

HO

HO

HO

HO

2. alkoxide nucleophiles

H3CO

H3CBr

Br

Br

Br

Br

Br

Br

Br

H3CO

H3CO

H3CO

H3CO

H3CO

H3CO

H3CO



3. carboxylate nucleophiles

OH3C

Br

Br

Br

Br

Br

Br

Br

Br

O

O

O

O

O

O

O

O

O

O

O

O

O

O

O

4. t-butoxide base

H3CBr

Br

Br

Br

Br

Br

Br

BrO

O

O

O

O

O

O

O

K

K

K

K K

K

K

K



5. imidate nucleophile

H3CBr

Br

Br

Br

Br

Br

Br

BrN

O

O

N

O

O

N

O

O

N

O

O

N

O

O

N

O

O

N

O

O

N

O

O

6. azide nucleophile

H3CBr

Br

Br

Br

Br

Br

Br

Br

NN

NNa

NN

NNa

NN

NNa

NN

NNa

NN

NNa

NN

NNa

NN

NNa

NN

NNa



7. dialkylamide anion

H3CBr

Br

Br

Br

Br

Br

Br

BrR

N

R

Na

RN

R

Na

RN

R

Na

RN

R

Na

RN

R

Na

RN

R

Na

RN

R

Na

RN

R

Na

8. hydrogen sulfide nucleophile

HS

H3CBr

Br

Br

Br

Br

Br

Br

Br

HS

HS

HS

HS

HS

HS

HS



9. thiolate nucleophiles

H3CS

H3CBr

Br

Br

Br

Br

Br

Br

Br

H3CS

H3CS

H3CS

H3CS

H3CS

H3CS

H3CS

10. cyanide nucleophile

H3CBr

Br

Br

Br

Br

Br

Br

Br

NC

Na

NC

Na

NC

Na

NC

Na

NC

Na

NC

Na

NC

Na

NC

Na



11. terminal acetylide nucleophiles

H3CBr

Br

Br

Br

Br

Br

Br

Br

CC

R

Na

CC

R

Na

CC

R

Na

CC

R

Na

CC

R

Na

CC

R

Na

CC

R

Na

CC

R

Na

12. ketone enolate nucleophiles

H3CBr

Br

Br

Br

Br

Br

Br

Br

CCH2H3C

O Li

CCH2H3C

O Li

CCH2H3C

O Li

CCH2H3C

O Li

CCH2H3C

O Li

CCH2H3C

O Li

CCH2H3C

O Li

CCH2H3C

O Li



13. ester enolate nucleophiles

H3CBr

Br

Br

Br

Br

Br

Br

Br

CCH2O

OLi

R

CCH2O

OLi

R

CCH2O

OLi

R

CCH2O

OLi

R

CCH2O

OLi

R

CCH2O

OLi

R

CCH2O

OLi

R

CCH2O

OLi

R

14. 3o amide nucleophiles

H3CBr

Br

Br

Br

Br

Br

Br

Br

CCH2N

OLi

R

R

CCH2N

OLi

R

R

CCH2N

OLi

R

R

CCH2N

OLi

R

R

CCH2N

OLi

R

R

CCH2N

OLi

R

R

CCH2N

OLi

R

R

CCH2N

OLi

R

R



15. nitrile enolate nucleophiles

H3CBr

Br

Br

Br

Br

Br

Br

Br

NC

CH2Li

NC

CH2Li

NC

CH2Li

NC

CH2Li

NC

CH2Li

NC

CH2Li

NC

CH2Li

NC

CH2Li

16. carboxylate dianion enolate nucleophiles

H3CBr

Br

Br

Br

Br

Br

Br

Br

CCH2O

OLiLi

2. workup

2. workup

2. workup

2. workup

2. workup

2. workup

2. workup

2. workup

CCH2O

OLiLi

CCH2O

OLiLi

CCH2O

OLiLi

CCH2O

OLiLi

CCH2O

OLiLi

CCH2O

OLiLi

CCH2O

OLiLi



17. dithiane nucleophiles

H3CBr

Br

Br

Br

Br

Br

Br

Br

S

S

Li

S

S

Li

S

S

Li

S

S

Li

S

S

Li

S

S

Li

S

S

Li

S

S

Li

18. aluminum hydride (Li AlH4 or LiAlD4) / borohydride nucleophiles (NaBH4 or NaBD4)

H3CBr

Br

Br

Br

Br

Br

Br

Br

BH

H

H

H

Na

BD

D

D

D

Na

BH

H

H

H

Na

BD

D

D

D

Na

AlD

D

D

D

Li

AlH

H

H

H

Li

AlD

D

D

D

Li

AlH

H

H

H

Li



19. diphenylsulfide nucleophile (sulfonium salt is used to make epoxides with aldehydes and ketones)

H3CBr

Br

Br

Br

Br

Br

Br

Br

S

Ph

Ph

S

Ph

Ph

S

Ph

Ph

S

Ph

Ph

S

Ph

Ph

S

Ph

Ph

S

Ph

Ph

S

Ph

Ph

20. triphenylphosphine nucleophile (phosphonium salt is used to make E or Z alkenes with aldehydes and ketones)

H3CBr

Br

Br

Br

Br

Br

Br

Br

P

Ph

Ph

Ph

P

Ph

Ph

Ph

P

Ph

Ph

Ph

P

Ph

Ph

Ph

P

Ph

Ph

Ph

P

Ph

Ph

Ph

P

Ph

Ph

Ph

P

Ph

Ph

Ph



Other reactions we need to know.

‘acyl substitution,’ many carboxyl groups react this way, and it can occur in aqueous base or aqueous acid.

O

O

O H

NaO

O

O H OH

OO O

O

OH

Me, 1o, 2o alcoholsSN2 > E2tetrahedral intermediate (TI)

NaNa

2

acid/base

The elements of water have been added to the ester to make a carboxylic acid and an alcohol. Hydrolysis means 'addition of water' (can also be called hydration) and is a very common reaction in organic chemistry and biochemistry. We will see similar reactions many times, as well as the opposite reactions (dehydration = removal of water).

O

O

OH

O

OH

1. NaOH / H2O

O

O

OH

1. H3O+ / H2O(neutralization)

H2O is added.

OH H

H

ester carboxylic acid alcohol

organic ambiguitya. attack C=Ob. attack C-Oc. attack C-H

Problem 7 – Show the acyl substitution mechanism for each functional group below with hydroxide.

Na

O HC

R

O

X

X = possible leaving group

CR

O

X

OH

tetrahedralintermediate

CR

O

OH

X

CR

O

OXH

CR

O

Cl

Functional Groups to use - What is the leaving group?

CR

O

OC

R

O

O

CR

O

NC

O

RR

R

Racid chloride anhydride ester 3o amide

acid/base



Make imidate and use as nucleophile at Me-X, 1o RCH2-X and 2o R2CH-X in SN2 RX centers to convert to primary amines. Gabriel 1o amine synthesis = 1. Make alkyl imides by SN2 2. Hydrolyze in base to make primary amines (acyl substitution) 3. Workup (neutralize base conditions) For an alternative approach see the azide primary amine synthesis, next: 1. SN2 with NaN3 2. SN2 with LiAlH4 at nitrogen 3. workup)

N

O

O

HNa

O H

N

O

O

Na SN2rxns

N

O

Oimide(buy)

imidatealkyl imide

Keq =

Na

O H

N

O

O OH

N

O

O

OH

N

O

O

O

H

acyl substitution #1reaction leads to a primary amineO H

N

O

O

O

H

O

H

N

O

O

H

O

OH

O

O

O

O

N

H

H

primary aminealso made by1. NaN32. LiAlH43. workup

throw away

resonance stabilized makes imidate less basic and a better behaved nucleophile

1

acid/base rxn

2

3

Look at similarities with ester hydrolysis, just above.

Br

acyl substitution #2

acid/base rxn

acid/base

Keq =

Keq =

pKa 9

HO

H

Alternative azide strategy to make primary amines (SN2 and acid/base reactions) at methyl, 1o and 2o RBr.

(pKa 9)

AlH

H

H

HLi

NN

N

Br

N

N

N

resonanceN

N

N

alkyl azide

N

N

N

N

N

N

H

gas

OH H

H

2. workup

NH

1o amine

H

step 1 - make alkyl azide

step 2 - make primary amine

SN2

SN2

acid/base

Na

1. LAH



Alkyne synthesis (via two E2 reactions with RBr2 and NaNR2, two times)

To make starting alkynes from our simple given alkanes requires substituting on two leaving groups (Br for H), followed by 3 equivalents of very basic R2N

--. The 3rd equivalent of R2N-- is necessary because of the acidity

of the sp C-H bond. This would have to be added back on in a final workup step.

Na sodium amide(very strong base)

Br2 h

free radical substitution

(2 equivalents)

(3 equivalents)

NR2

E2 twice

1.

2. workup

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

Br Br

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

BrBr

HN

R R

Na

Br

H

H

NR R

HN

R R

Na

Na2. workup

H

H2C

R

X

H2C

R

HO

H

H

2.

a

b

a

b

SN2

2 eqs. Br2h

3rd equivalent

most stable anion in mixture

H

H E2

E2

acid/base

R H

RH2C H

H2N H

pKa

25

37

37

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

CCR

Na

terminalacetylides

CCR

terminalalkynes

H

NaNR2

Br

SN2

R

alkynes

Keq =

acid/base

possible zipper rxn

1. excess NaNR22. workup

R



The Zipper reaction moves a triple bond from an intermal position along an unbranched chain to the end of linear chain to form the most stable anionic charge (sp carbanion). Further nucleophile chemistry is possible using the terminal acetylide carbanion. This reaction is similar to tautomers, without any heteroatoms (it just uses a much stronger base).

R H

RH2C H

H2N H

pKa

25

37

37

CPh C C

CCH2C

Nanonterminal

alkyne

H2N H

H2N

H

H

H

CPh C C

H

H

resonance

Na

CPh C C

H

H

CPh C C

H

H

H

H2N

Na

CPh C C

HH

resonance

CPh C C

H

H

H2N H

Ph CCH2CPh H

H2N

manypossibilities

react with RX

react with C=O

react with epoxide

workup

acid/base

acid/base

acid/base

acid/base

Most stable anion in the mixture, reaction stops here.

many possibile alkyne reactions

1. R2N Na R2NH

excess

acid/base

2. workup(H3O+)

Ph

Ph

Ph

Ph

Ph

H

terminal acetylide

terminal alkyne

reaction is over in seconds

2. CH3Br

2. H2C=O3. workup

2.

3. workup

O

Ph

CH3

Ph

H2C

OH

Ph

H2C

CH2

OHPh

1 eq.

R2N Na R2NH

acid/base

Na



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 of ketones, esters, nitriles and others (next).

pKa = 37Keq =

CH2 Li

N

H

NLi

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

LDA = lithium diisopropyl amide

givengiven

acid/base

pKa = 50

React ketone enolate (nucleophile) with R-Br electrophile (SN2 reaction at Me, 1o and 2o RX compounds)

O

HN

Li

N

H

CH2

O

Li

Keq =ketones (and other carbonyl compounds)

Make enolate (ketones) (organic ambituity - attack C or attack C-H?)

LDA ketone enolates (resonance stabilized)

given -78oC

resonanceacid/base

H2C

OLi

ketone enolates(resonance stabilized)

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

Br

O

1o RX key bond

RS

Larger ketone made from smaller ketone.

SN2reactions

-78oC

pKa = 20

pKa = 37

React ester enolate (nucleophile) with R-Br electrophile (SN2 reaction at Me, 1o and 2o RX compounds)

O

OH

N

Li

N

H

CH2

O

O

Li

Keq =

Make enolate (esters)

LDAester enolates

(resonance stabilized)

R R-78oC

resonanceacid/base

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

givenfor now pKa = 37

pKa = 25

React carboxylate dianion enolate (nucleophile) with R-Br electrophile (SN2 reaction at Me, 1o and 2o RX compounds)

O

OH

N

Li

N

H

CH2

O

O

Li

Keq =

Make dianion of carboxylic acids

LDAacid dianion


H

-78oC

resonanceacid/base

twice

H2C

O

O

Li

acid dianions(resonance stabilized)

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

Br

O

O

1o RX key bondR

S

Larger acid made from smaller ester.

HSN2reactions

-78oC

givenfor now pKa = 37

pKa = 25pKa = 5

2 2

2. workup



React nitrile ‘enolate’ with R-Br electrophile (SN2 reaction at Me, 1o and 2o RX compounds)

CH2

CHN

Li

N

H

Li

Keq =nitriles

Make nitrile enolate (nitriles)

LDAnitrile enolates


-78oC

resonanceacid/base

N

H2CC

N

Li

nitrile enolates(resonance stabilized)

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

Br

C

1o RX key bondR

S

Larger nitrile made from smaller ester.

SN2reactions

-78oC

CH2

CN

N

pKa = 37

pKa = 30

Problem 11 – Predict the major product of each set of conditions below and write a plausible mechanism for how the reaction(s) work.

a.

NaOH

Brb.

?

OH 1. NaH2.

Br

?

c.

O

1. NaOH2.

alcoholproductOH

Br

3. NaOH

N

O

O

H

1. NaOH2.

3. NaOH

Br

?

d. e. f.

Br

1. NaN32. LiAlH43. workup

? Br

NaSH?

1. NaOH2.

?

g. h. i.

Br

1. n-butyl lithium2. CH3Br3. workup ?

?

SHBr

S

SC

CH3C

H

1. NaNR22.

1. LDA, -78oC2.

?

j. k. l.

BrO O

OR

1. LDA, -78oC2.

?

Br NC

H3C

1. LDA, -78oC2.

?

Br

?

m. n.

Br

?

Br

Ph PhS Ph Ph

P

Ph

1.

2. n-butyl lithium

1.

2. n-butyl lithium

BrLiAlD4

?

o.



?

p. q.

? ?

r.

Br2 / h t-butoxide

2 eqs.Br2 / h

Br

?

s. t.

?

u.

1. excess NaNR22. workup

Br Br

1. LDA, -78oC2. CH3Br3. workup

O

NR2 ?

1. LDA, -78oC (2 eqs.)2. CH3Br3. workup

O

OH

Special case of intramolecular SN2 reaction - synthesis of epoxides (oxiranes) under base conditions

Problem 23 – Predict SN2 products and propose mechanisms for the following reactions. What constitutes a ‘strong’ nucleophile in our course? How could this form under the reaction conditions? What is the necessary stereochemistry for an SN2 reaction? What conformation in a cyclohexane ring allows this approach?

O

Br

HO

H

Br

CH3

H3CH

H OH

H3CH

Br

HCH3

mildNaOH

O

How?

mildNaOH

How?

mildNaOH

How?? ?

Example Mechanisms shown below.

BrH

D

12

34

56

3R-bromo-2S-deuteriohexane

Na

O H

H secondary RX (2o)

C

C

HBr

H

CH3

D

CHa

Hb

CH2CH3

OHC

C

HHO

C

HCH3

D

Ha CH2CH3

Hb

a

bc

(2S,3S)

C

C

H3C

H

D

CH2CH2CH3

C

C

H

H3CH2C

CHDCH3

Hb

a, b, c

C

C

HBr

D

H

CH3

CHb

CH2CH3

Ha

OH

deC

C

H

H

CH3

CH2CH2CH3

C

C

H

Ha

CHDCH3

CH2CH3

d, e

sigma bond rotations

One SN2 product and four E2 products.

b

a

c

d

e

only one SN2 productfour possible E2 products

(2S,3R)(2E, with "D")

(3E, lost Ha)

(3Z, lost Hb)(2Z, without "D")

1. strong Nu: /B:2. 2o R-X



SN1 and E1 Competition – Multistep Reactions Arising From Carbocation Chemistry

SN1 versus E1 overview (essential feature: stability of the carbocation) These are competing reactions.

Example: requires 2o or 3o RX and a weak nucleophile/base. SN1 generally out competes E1.

CH

HH

C Br

CH

2-bromopropaneE1 = minorSN1 = major

H

H

H

ionization,leaving

group leaves C

CH H

H

HCH

H

H

sp2 carbocation(secondary R+)

HO

CH3

weaknucleophile/base

attacks from both faces

CH

HH

C O

CH

H

H

H

CH3

C

C

H H

H CH3

rearrangementsare also possible,

but not in this example

HO

H3C

R-X Substitution Pattern and rates of SN1 reactions (backwards from SN2 reactions)

SN1 (and E1) relative reactivities of R-X compounds:

R

relativerates =

106 = 1,000,00010-5 0

1.0reference

compoundthese two rates are probably by SN2 reaction

H3CX

H3C

H2C

X H3CCH

X

CH3

H3CC

X

CH3

H3C

methyl primary (1o) secondary (2o) tertiary (3o)

carbocationintermediate

10-4 0

SN1 (R-Nu)

when HNu / HB are weak nucleophiles

H2OROH

RCO2H

Xleavinggroup

H-Nu:

E1 (alkenes)

rearrangement (new carbocation)

H-B:

XR

polarprotic

solvent

ion formation requires assistance from the polar solvent

R'O

H startover

The order of stability at the electron deficient carbocation carbon is methyl primary secondary << tertiary. This is explained by either inductive effect or hyperconjugation or both. Hyperconjugation can be considered as sigma resonance.

C

X

CHH

H

X

CC

H

HH

H

H

X

Sigma electrons are pulled toward the carbocation carbon. Part of the + is distributed on to the hydrogen atoms, but not typically shown with formal charge.

CHH

H

+ +

+

+

Additional sigma bonds of alkyl substituent(s) allow further polarizations of electrons from more bonds(inductive donating effect), which spreads out + charge through sigma bond polarizations and helpsstabilize the electron deficient carbocation carbon.

CC

H

HH

H

H

+ +

++

+

+

+

inductive donating effects help stabilize C



CC

H

H

H

H

H

H

two electron delocalization

CC

H

HH

H

hyperconjugation in a carbocation

resonance

Resonance effects also make carbocations more stable, from an adjacent pi bond or lone pair.

C C

C H

HH

H

H

C C

C H

HH

H

H

C C

C H

HH

H

H

C C

C H

HH

H

H

resonance from adjacent pi bond

2D

3D

O C

R

RHresonance from adjacent lone pair

CO

H R

R

CO

H R

R

O C

R

RH

Resonance effects help stabilize carbocations (from pi bonds and from lone pairs).

2D

3D

C C

C H

HH

X

H H

CO

R

R

A H

strongacid

resonance resonance

resonance

resonance

Gas phase Stabilities as Indicated by Hydride Affinities

Hydride affinity is the energy released when a hydride is added to a carbocation (gas phase reaction). The energy of reaction, H, is very negative (favorable). How much do inductive and resonance effects help a carbocation center? The following gas phase data below show the differences in carbocation stability are enormous. In fact, differences are so large that we will almost never propose methyl or primary carbocation possibilities as reaction pathways in solutions in our course. We will consider these two patterns (CH3-X and RCH2-X) as unreactive in SN1 and E1 chemistry, and that should make your life a little bit easier.

Problem 25 - Explain the differences in stability among the following carbocations (hydride affinities). All relative energy values in kcal/mole versus a primary carbocation. A positive value is less stable and a negative value is more stable relative to the reference primary carbocation.

CH3CH2

CH2

CH

CH3H3C

CH3

CH3C CH3 H2C

HC

CH2

CH2

HOCH2 H2N

CH2

CO CH3

315

H3C

270reference

249 232 256 239 230248 218

= +45 = -21 = -38 = -14 = -31 = -22 = -52 = -40

Too unstable to propose in our course.

2o R+ 3o R+ 1o allylic R+ 1o benzylic R+

lone pair resonance stabilization of R+1o R+

methyl R+

pi bond resonance stabilization of R+

= 0

A more negative (compared to 1o R+) is a more stable carbocation.

Inductive effect stabilization of carbocations (3o > 2o R+)

What is the hybridization of "O", "N", "O"?



Problem 27 – The bond energy depends on charge effects in the anions too. Can you explain the differences in bond energies below? (Hint: Where is the charge more delocalized?) We won’t emphasize these differences.

CH3C

CH3

CH3

X

X = Gas Phase B.E.

Cl +157Br +149I +140H + 230

The activation energies for ionization in solvents are on the order of 20-30 kcal/mole (SN1 and E1 reactions) It is clear from the difference in the gas phase energies of ionization (> 200 kcal/mole) that the solvent is the most stabilizing factor in ion formation. Because solvent structure is so complex we ignore it, but we do so at our own peril.

solvation is the most important energy factor

CH3C CH3

CH3

RO

H

HO

R H

O R

R

OH

X

HO

R

HO

R

HO

R

OH

R Many small solvent / ion interactions make up for a single large covalent bond (heterolytic cleavage). A typical hydrogen bond is about 5-7 kcal/mole and typical covalent bonds are about 50-100 kcal/mole. In a sense the polar protic solvent helps to pull the C-X bond apart. The "polarized" protons tug on the "X" end and the lone pairs of the solvent molecules tug on the "C" end. If the carbocation is stable enough, the bond will be broken.

reactants E1 products

Progress of reaction (POR)

(10)

-Ea (E)

2.3RT

Rate SN1 = kSN1[RBr]1

Rate E1 = kE1[RBr]1 =(10)

-Ea (SN)

2.3RT

(10)

-Ea

2.3RT

Say Ea(SN) = 1.3 kcal/mol and Ea(E) = 2.6 kcal/mol

=

(10)

-Ea

2.3RT= (10)

- (1.3 - 2.6)

1.3= (10)

1.3

1.3= = 101 =

Rate SN1Rate E1

101

SN1 products

E1 rate (slower)

SN1 rate (faster)R

Ea

overall reaction rate

PE

Weak Nucleophile/Bases are used in SN1/E1 Reactions R+ has to be secondary, tertiary or resonance stabilized carbon.

Nu

B=

H

H

HO

H RO

H

O

OH

water alcohols carboxylic acids

SN1/E1 reactions - form carbocation (R+) in first step,

R+ has three common choices1. rearrange to similar or more stable R+

2. add nucleophile to top/bottom (R/S)3. lose any beta proton from top/bottom (E/Z) (forms pi bonds)

these are our weak nucleophile / bases



We will view the attack on an sp2 carbocation as equally accessible from either face (top or bottom). In reality, the leaving group often shields (blocks) the face it is leaving from, giving a preference for inversion over retention.

CR1

R3

R2

achiral carbocation carbon

mirrorplane

CR1

R3

R2

HNu

top face attack

bottom face attack

top

bottom

C

Nu

R3

R2

R1

C

Nu

R2

R3

R1

R / S assumes priorities are Nu > R1 > R2 > R3

S

Rachiral carbocation carbon with no other chiral centers in R1, R2 or R3.

If all 3 attached groups at a carbocation carbon are different from one another and the attacking nucleophile,then a racemic mixture of enantiomers will form. If R2 = R3, then the C carbon is achiral.

CC

R3

R2

HNu

top face attack

bottom face attack

top

bottom

C

Nu

R3

R2

C

Nu

R2

R3

The new stereogenic center forms both R and S absolute configurations. If another chiral center is present, that does not change in the reaction then diastereomers will form (RR) vs. (RS). These would likely form in different amounts.

S

R

* = chiral branch in carbocation

H3C

DH

*R

C

D

HH3C

C

D

HH3C

R

R

There is, perhaps, a slower rate of attack from the face where the methyl is positioned, but remember, it rotates through a full 360o..



Rearrangements of Carbocations – searching for a more stable carbocation (a common complication)

Consider the migration of every group on a C position, whether H or C. To keep our choices simpler (than they really are) we will only consider rearrangements of 2o to 3o and 3o to 3o carbocations. What are the likely SN1 and E1 products of the initial carbocation and the rearranged carbocations from “a”, “b” and “c”?

C

CH2

H3C

H

HC

X

CH3

The RX compound must be 2o, 3o, allylic or benzylic to form the initial carbocation.

SN1

E1redrawn

All potentially migrating bonds drawn with bold lines.

Consider all C-groups (H and C).

SN1 and E1 are possible here

2o carbocation

An adjacent C group migrates with its electrons to the C carbocation position. There are four possibilities in this problem.

C

CH2

H3C

H

C

H

CH3

C

CH2

H3C

H

C

H

CH

HH

d

a

b

dd

a H: hydride migrates

H3C: methyl migrates

cb

C

CH2

H3C C

H

CH3 C

H

C

CH3

CH3C

CH2

H3C

H

C

H

CH2

HH H

1o carbocationlooks very poor

2o carbocationlooks OK

3o carbocationlooks very good

We are not likely to observe this

choice.SN1 and E1 are possible here


a

c

H3C H3C

c

d

d

H3C

b

H3CH2C: ethyl migrates

CH3C

H

C

X

CH3

CH2

2o carbocationlooks OK


XX

CH2

H3C H3C H3CH3C

possible rearrangements

H: hydride migrates

H-Nu/H-B

Transition state of a carbocation rearrangement

C

CH2

H3C

H

HC

X

CH2

SN1

E1

rearrangement

2o carbocation


XH3C

1,2-hydride shift

CC

H

CH2

CH2

H

transition state (no finite existance)

migrating group is positioned between two vicinal carbon atoms

(vicinity = neighbors, Latin)

CC

H

CH2

HH2C

H3C

3o carbocation

CC

H2C

H3C

H

HH2C

R-X

TS1 TS2TS3

Ea

G

2o R3o R

SN1 & E1products

PE

Progress of Reaction (POR)

H: = hydride shift

R: = alkyl shift

The migrating group is always attached to the carbon skeleton; it is never a free anion.


CH3

CH3CH3

CH3

H3C

H3C

H3C H3C



The three fates of carbocations are to add a nucleophile (from either face), to lose any Cβ-H from either face and to rearrange. Rearrangements are a temporary solution for an unstable carbocation. Rearrangements transfer the unstable carbocation site to a new position having a similar energy or, better yet, to a site where the positive charge is more stable. If such possibilities exist, this will very likely be one of the observed reaction pathways. However, even with a rearrangement a carbocation will not gain the two needed electrons. The electron deficiency is merely moved to a new position. This process can occur a number of times before a carbocation encounters its ultimate fates, discussed above, SN1 and E1. Usually SN1 will outcompete E1.

CC

CBr

H3C

CH3

H3C H H

H

H3CO

H

?

Strain energy is another factor that must be considered in carbocation rearrangements (in addition to the relative stabilities of 1o, 2o and 3o carbocations). Consider the possible rearrangement choices available to the following tertiary carbocation in a polar ionizing solvent.

Br

slowstep

R.D.S.

RO

H

CH2

HH

ab

c

CH2

HH

CH3

H

CH3

H

H

CH3

CH3

a

b

c

Primary R+ is completely unstable.

Tertiary R+ looks good, but the angle strain (90o vs. 120o) has gone way up.

Secondary R+ doesn't look so good (vs 3o R+), but the ring strain (cyclobutane vs. cyclopentane) has gotten more stable by 20 kcal/mole, and can rearrange again to 3o R+. =

ring strain 26 kcal/mole(90o vs. 109o)

ring strain 6 kcal/mole

a. A hydride migration makes a primary carbocation from a tertiary carbocation. This reaction would increase the potential energy by about 35 kcal/mole and is not a realistic option.

b. At first this option (hydride shift) seems very reasonable (tertiary carbocation to tertiary carbocation), but there would be much additional ring strain energy because of bond angle changes in the small cyclobutane ring (109o = sp3 to 120o = sp2), while geometric shape in the ring is trying to be 90o. This would, therefore, not be a favorable option.

c. At first this looks like a very poor reaction (tertiary carbocation to secondary carbocation vial alkyl migration of a ring carbon) and would be uphill by about 15 kcal/mole based on carbocation stabilities. However, the reduction in ring strain would be downhill by about 20 kcal/mole (26 kcal/mole 6 kcal/mole), resulting in an overall potential energy change of -5 kcal/mole.



Example SN1 / E1 Mechanisms with rearrangement (2oR+ to 3oR+ rearrangement)

C

C

HH

H

H CCH3

CH2CH2CH3

H

carbocation continued with rearrangement)

C

CH2

CH2O

H3C

H

CH3

rearrangement from 2oR+ to 3oR+,

redrawn with new C.

H3CO

H

CH3

OH

C

C

H3CH2C CH3

Same product forms from loss of methyl proton from either face (no E or Z).

H3C

O

H

CH3

O

H

CH2CH3H

Two possible products from loss of proton from left face or right face on propyl branch.

C

C CH3

CH2CH2CH3

CH3

H

C

C

H

H

H3CH2C

CH2CH2CH3

Two possible products from loss of proton from left face or right face on ethyl branch.

a

b

c

d e

a

b

c

c

a

H3CO

H

d

e

CH3

OH

H3C

C

CH

CH2CH3

H

CC H

HHH3C

H

H

redraw R+ to show SN1 reaction

a, b, c = E1 products(protons lost from

either face)

SN1 products

3E alkene

3Z alkene

2Z alkene2E alkene

(3S)(3R)

C2

H2C

H2C CH3

H3C

CH2CH3

C

H2C

H2C

O

CH3

H

H3CCH3

CH2CH3 C

H2C

H2C

O

CH3

H3CCH3

CH2CH3

H3CH2C

C

CH2

CH2O

H3C

CH3H3C

H3CH2C

2o R+ from previous page

C

C CH3

CH2CH2CH3

H

H3C

C

C

H3CH2C CH3

HH3CH2C

3o R+

3o R+

CH3H

H

12

34

56

2R-bromo-3R-methylhexane

H3CO

H

Br secondary RX (2o)

C

C

HBr

H

H

H

CHCH3

CH2CH2CH3

C

C

HO

C

HCH3

D

Ha CH2CH3

Hb

C

C

H

CH3

CH3

CH2CH2CH3

1. The first 2o R+ forms two SN1 products and three E1 products2. The rearranged 3o R+ forms two SN1 products and five E1 products (next page)

C

C

HH

H

H CCH3

CH2CH2CH3

H

a, b, c show attack on left face of carbocation,attack also occurs from the right side of R+.

H3CO

H

H3CO

H

b

a

c

C

C

H

H

H

C

CH3

H CH2CH2CH3

a

b c

E1 product after loss of beta protonfrom methyl (CH3)

E1 product after loss of beta protonfrom methine (CH) from either face

C

C

HO

C

HCH3

D

Ha CH2CH3

Hb

SN1 product from attack of left face

H

H3C

H3CO

H

C

C

HO

H

H

H

CHCH3

CH2CH2CH3

H

CH3

H3CO

H

C

C

HO

H

H

H

CHCH3

CH2CH2CH3

CH3

SN1 product from attack of right face

b, c

rearrangement(below)

d

d

C

C

H

CH2CH2CH3

CH3

CH3

c

H3C

d

a

1. weak H-Nu:/H-B:2. 2o R-X

R+ reactions1. add H-Nu: (SN1)2. lose C-H (E1)3. rearrange (start over)



Reaction Templates - sideways and vertical perspectives (either one will work)

SN2/E2 (Nu: / B: ) always backside for SN2 and usually anti C-H/C-X attack for E2

SN1/E1 (H-Nu: / H-B:) - form R+, attack from either face for both reactions (usually SN1 > E1)

easier (all C = H) harder (C = varies)iotopes of hydrogenH = protium (proton)D = deuteriumT = tritium

simplistic view

Nu:BO

O

O

O=OH

O

O

Examples of Strong base / nucleophiles that can be used below = SN2 / E2 (many others are possible)

conjugate acid pKa = 16




Examples of Weak base / nucleophiles that can be used below = SN1 / E1 (most common for us)

HO

H RO

HH

water liquid alcohols liquid carboxylic acids

C

D

HT

X

methyl (Me)

Nu:strong

side views

C

DH

T

X

vertical views

Nu:strong

simplistic views

Nu: H3CBr



primary (1o)

C

C

HD

X

H

R

D

Nu:

B

side views

C

CH

D

X

H

RD

vertical views

Nu

B

B

CH2

CHH3C

Br

HNu

simplistic views

secondary (1o)

C

C

HX

H

R1

D

CHD

R2

Nu:

B

side views

priorities: R1 > R2

C

CH

X

H

R1D

C

H

D

R2

vertical views

NuB



B

CHCH

H3C

Br

HNu

simplistic views

CH2

H

tertiary (1o)

C

C

X

H

R1

D

CHD

R2

CH

D

priority R1 > R2 > R3

Nu:

B

R3

side views

C

C

X

H

R1D

C

H

D

R2

C

H

DR3

vertical views

NuB

B

CC

CH2

Br

HNu

simplistic views

CH2

CHH

H

H3C

CH3

CH3



C

D

HT

X

methyl (Me)

C

DH

T

X

weak

side views

vertical views

simplistic views

H3CBr

NuH

weak

NuH

weak

NuH

BH

NuH

primary (1o)

C

C

HD

X

H

R

D

side views

C

C

H

D

X

H

RD

vertical views

BH

Nu H

CH2

CHH3C

Br

H

simplistic views

BH

Nu H



secondary (1o)

C

C

HX

H

R1

D

CHD

R2

side views

BH

NuH

priorities: R1 > R2

C

CH

X

H

R1D

C

H

D

R2

vertical views

BHNu H

CHCH

H3C

Br

H

simplistic views

CH2

HBH

Nu H

tertiary (1o)

C

C

X

H

R1

D

CHD

R2

CH

D

priority R1 > R2 > R3

R3

side views

BH

NuH



C

C

X

H

R1D

C

H

D

R2

C

H

DR3

vertical views

B

BH

Nu H

CC

CH2

Br

H

simplistic views

CH2

CHH

H

H3C

CH3

CH3

BHNu H

Example: 3-bromo-4-deuterio-2-methoxyhexane (RRR), (SSS), (RRS), (SSR), (RSR), (SRS), (SRR), (RSS) ? It might help to draw a 2D structure first.

BrH

12

34

56 C

C

HBr

H

CH

C

C

H

Br

C

HH

template side view vertical view

3R

HBr

12

34

56

3S



Alcohols in strong acid = Protonated Alcohols - Water as a Good Leaving Group

a. methyl, 1o, 2o and 3o ROH reacted with HX acids (HCl, HBr, HI) - usually SN2 or SN1chemistry

H3C

O HBrH

methanol

H3C

O H

H

Br

H3C

O

H

H

Br

SN2BrH

O

H

H H

bromomethane

BrpKa = -9

water is a good leaving group

H3C CH2

O H

BrH

primary alcohol

H3C CH2

O H

H

Br

H3C CH2

O

H

H

Br

SN2

primary bromoalkane

pKa = -9


water is a good leaving groupClH

secondary alcohol (trans OH)

H2C

H3C H

SN1

Cl

H3CH

Cl

H3CCl

H

trans Cl

cis Cltop and bottom

attack

H3CH

O

H

H

O

H

H

2o chloroalkane

pKa = -7

O

H

H O

H

H H

ClH

water is a good leaving groupClH

secondary alcohol (trans OH)

H2C

H3C HSN1 Cl

H3CH

Cl

H3CCl

H

trans Cl

cis Cltop and bottom

attack

H3CH

O

H

H

O

H

H

2o chloroalkane

pKa = -7

O

H

H O

H

H H

ClH

IH

tertiary alcohol (trans OH)

H3C

H3C CH3

O

H

H

SN1I

O

H

H H

I

I

trans I

cis Itop and bottom

attack

H3CCH3

O

H

CH3

O

H

H

IH

pKa = -10



b. 1o, 2o and 3o ROH reacted with H2SO4 and high temperature ( = heat) = E1 chemistry

Using strongly acidic sulfuric acid, H2SO4, at elevated temperatures favors E1 reactions because lower boiling alkenes distill out and continually shift the equilibrium to make more alkene, which continues to distill out, until there is no more alcohol left in the reaction pot. We will assume that an E1 mechanism is operating in all of the reactions below (even the primary alcohol). Rearrangements are possible and observed.


primary alcohol

O

H

HO

H

H H

OH OH SO3H

(heat)

OH

H

CH

H

H H

rearrangementH

H

O SO3H

E1

bp = -47oCdistills out

bp = +82oC

very difficult(high temperature)

OH SO3H O SO3H

alcohol alkene Tbp = 129oC

pKa = - 5

secondary alcohol

O

H

H O

H

H H

OH OH SO3H

(heat)

OH

H

O SO3H

E1

bp = +83oCdistills outbp = +161oC

OH SO3HO SO3H

H

H

H

alcohol alkene Tbp = 78oC

pKa = - 5water is a good leaving group

tertiary alcohol

O

H

H O

H

H H

OH SO3H

(heat)

O SO3H bp = +33oCdistills out

bp = +102oC

OH SO3HO SO3H

OH O

H

H

H

H H

E1

bp = +39oCdistills out

90% < alkene(more substituted)

minor alkene(less substituted)

b

a

a b

alcohol alkene Tbp 63oC

pKa = - 5




Sulfonated Alcohols as Good Leaving Groups

Both Sulfuric acid and Sulfonic Acids are very strong acids, (they have very stable conjugate bases)

SO

O

O

O

H

S

O

O

O

H

H

toluenesulfonic acid

sulfuric acid

O H

H

O H

H

SO

O

O

O

H

bisulfate(very stable anion)

Sulfonates are stable anions, which make excellent leaving groups when bonded to carbon.

S

O

O

O

O H

H

H

O H

H

H

Ka pKa

103 -3

Ka pKa

103 -3

Formation of an inorganic sulfonate ester (mechanism = acyl-like substitution)

S

O

O

Cl

toluenesulfonyl chloride

O

H

S

O

OCl

O

H

N

S

O

OCl

ONH

S

O

O

O

Sulfonates esters have an excellent leaving groups and are useful in SN2 chemistry.

acid/base

Cl

The pyridinium ion is a stable form of the otherwise very acidic proton.

alcohol

pyridine

sulfur valency = 4sulfur valency = 5

sulfur valency = 4

Formation of an analogous organic alkanoate ester (mechanism = acyl substitution)

carbon valency = 3

RC

O

Cl

O

HR

C

OO

Clacid chloride

alcohol

RC

OO

Cl

HNR3

acid/base RC

O

O

organic ester

NR3H Cl discardtetrahedral intermediatecarbon valency = 4

carbon valency = 3

Problem 34 – Write a detailed arrow-pushing mechanism for each of the following transformations.



NNH

S

Cl

SCl

O

O

O

O

N

NS

Cl O

O

N H

NS

O

O

toluene sulfonamide

Br

toluene sulfonylchloride(tosyl chloride)

HH H H

H

NH

Cl

H

C

O

Cl CH3

C

O

Cl CH3

N

H

H

RN

RR

C

O

Cl CH3

N

RN

RR

H

H

C

O

N CH3

H

SN2 at 2o RBr without rearrangement. 1. make tosylates from ROH + TsCl (toluenesulfonyl chloride = tosyl chloride) and 2. NaBr, SN/E chemistry is possible without rearrangements (SN2).

OROR

H S

Cl

H

SCl

O

O

O

O

N

OS

Cl O

O

N H

OS

O

O

Br

Na

separate step

Br

HBr

rearrangementand

R,S (racemic)

S

1. TsCl/pyridine2. NaCl(prevents R+ formation and any rearrangement)

alkyl tosylate = RX compound

compare a different resultif HBr is used

R RR

R

RR

RBr

Br

SN1

toluene sulfonylchloride(tosyl chloride)

SN2



Other acyl-like transformations include thionyl chloride (SOCl2) or thionyl bromide (SOBr2) with alcohols (makes R-Cl and R-Br) or carboxylic acids (makes acid chlorides, RCOCl). Acid chlorides formed can make esters, thioesters, amides and anhydrides.

R OH S

Cl

O

Cl

H2C

R O

S

O Cl

Cl

OS

O

H

H2C

R O

SCl

O

CH2

R

SN2 at methyl and primary alcohols

No rearrangement because no R+.

H

ClH

Cl

O

SCl

O

H

product

O

S

O

H

acyl-like substitution

Cl

Cl

Thionyl chloride with methyl, 1o ROH = acyl-like substitution at SOCl2, then SN2 at methyl and primary RX.

Thionyl chloride with 2o and 3o ROH = acyl substitution, then SN1 (there are various ways you can write this mechanism)

OH O

S

O Br

Br

OS

O

H

O

SBr

O

HH

Br

SN1 at secondary and tertiary alcohols

O

SBr

O

H

BrH

RR

R

RR/S

acyl-like substitution

SBr

O

Br

Br

Br

Synthesis of acid chlorides from acids + thionyl chloride (SOCl2), use the carbonyl oxygen instead of the OH.

CR

O

OH

CR

O

OH

S

O Cl

Cl

resonance

CR

O

OH

S

O Cl

Cl

CR

O

OH

S

O Cl

Cl

Base

CR

O

O

S

O Cl

Cl

BaseH

CR

O

O

SCl

O

CR OCR O

resonance

CR

Cl

Oacylium ion

SCl

O

Cl

OS

O

Cl

Cl

resonance



Formation of esters from ROH + acid chlorides, amides from RNH2 or R2NH + acid chlorides and anhydrides from RCO2H + acid chlorides

OR

O

R

H

Cl

O Cl O

H

O

Cl

H

O

O

O

R

R

There are many variations of ROH and RCO2H joined together by oxygen.

ester synthesis from acid chloride and alcohols

RO

H

RO

H

H

N

RN

R

H

Cl

O Cl O

N

O

N

O

R

RThere are many variations of RNH2 or R2NHand RCO2H joined together by nitrogen.

H

H

H H H H

amide synthesis from acid chloride and amines

RH2N

RH3N

Cl

O

O

Cl

O Cl O

Cl

There are many variations of R1CO2H and R2CO2H joined together by oxygen.

anhydride synthesis from acid chloride and carboxylic acids

O

OH

H

O

O

H

O

O

O

O ClHresonance



Phosphorous trichloride (PCl3) = SN2 of alcohol at phosphorous, then SN2 (at methyl and primary R(OH)PCl2+)

OH

OP

H

Cl

ClCl

Cl

OP

H

Cl

Cl

reacts twice more

P

Cl

Cl

Cl

SN2 SN2

Phosphorous tribromide (PBr3) = SN2 of ROH at phosphorous, then SN1 (at secondary, tertiary, allylic and benzylic R(OH)PCl2

+)

OH

S

Br

H

R,S

OH

R

P

Br

Br

Br

SN2

OP

H

Br

Br

Br

SN2

BrO

P

H

Br

Br

reacts twice more

P

Br

Br

BrO

P

H

Br

Br

Br

OP

H

Br

Br

reacts twice more

BrSN2 SN1



Chart of SN and E Chemistry (note exceptions)

typical strong basenucleophiles are:(for our course)

HO

RO

R

O

O

HO

RO

O

OH HH

NC C

C

RHS

RS

NN

N

H3CX

CH2

XR

CHXR

R

CXR

RR

methyl

primary

secondary

tertiary

only SN2 only SN2 only SN2 only SN2 only SN2 only SN2 only SN2 only SN2

SN2 > E2 SN2 > E2 SN2 > E2 SN2 > E2 SN2 > E2 SN2 > E2 SN2 > E2 SN2 > E2

E2 > SN2 E2 > SN2 E2 > SN2SN2 > E2 SN2 > E2 SN2 > E2 SN2 > E2 SN2 > E2

only E2 only E2 only E2 only E2 only E2 only E2 only E2 only E2

exception(too basic)

O

t-butoxide

only SN2

E2 > SN2exception

(bulky & basic)

E2 >> SN2

only E2

typical weak basenucleophiles are:(for our course)

H3CX

CH2

XR

CHXR

R

CXR

RR

methyl

primary

secondary

noreaction

noreaction

noreaction

noreaction

noreaction

noreaction

SN1 > E1

SN1 > E1

SN1 > E1

SN1 > E1

SN1 > E1

SN1 > E1

BH

H

H

H

only SN2

SN2 > E2

SN2 > E2

NA

exception(bulky & basic)



tertiary

AlD

D

D

D

alcohol reactions in strong acid:

(for our course)

H3COH

CH2

OHR

CHOHR

R

COHR

RR

methyl

primary

secondary

tertiary

HX

(X = Cl, Br or I)

SN1

SN1

E1

notdiscussed

H2SO4

E1

E1

SN2

SN2

SN1

SN1

SN2

SN2

SOCl2SOBr2

PCl3PBr3

1. TsCl/py2. NaBr

NA

SN2

SN2

SN2

enolatesdithianes

(too)



Oxidation of alcohols – E2 elimination reactions can form carbonyl functional groups (C=O), including aldehydes, ketones and carboxylic acids.

Recall that in freshman oxidation/reduction electron counting rules, all electron credit in bonds goes to the more electronegative atom. Oxygen almost always is –2. Hydrogen atoms are usually +1. Formal charge on the other hand, views all bonded electrons as shared evenly between bonded atoms (single, double or triple bonds).

Problem 38 – What are the oxidation states of each carbon atom below? All of the atoms in these examples have a formal charge of zero.

OHH3CCH4 C O C O

HOH

CO O

H3C

H2C

OHH3C

CO

H

H3CC

O

OH

H3C CH3

H H

C O

HO

HO

H2O

oxidation state of carbon = _____?










Problem 39 – What are the oxidation states below on the carbon atom and the chromium atom as the reaction proceeds? Which step does the oxidation/reduction occur? (PCC, B: = pyridine and Jones, B: = water)

C

H

R

R

O

H

Cr

O

O

O C

H

R

R

O

H

Cr

O

O

O

B

C

H

R

R

OCr

O

O

O

oxidation state of C =

oxidation state of Cr =



BC

R

R

OCr

O

O

O

HB

HB R.D.S. = slow step

E2 reaction

partial negative of oxygen bonds to partial positive of chromium

(step 1)

acid/base (step 2)

(step 3)oxidation state of C =






PCC = pyridinium chlorochromate, (CrO3/pyridine), CrO3 oxidations of alcohols (methyl, 1o and 2o ROH) without water. Steps are: 1. Cr=O addition, 2. acid/base and 3. E2 to form C=O (aldehydes and ketones).

H3CCH2

CH

OH

H

Cr

O

OO H3CCH2

CH

OH

H

Cr

O

O

O

H3CCH2

CH

O

H

Cr

O

O

ON

N H

N

H3CCH2

CH

O

Cr

O

O

ON H

PCC = pyridinium chlorochromate oxidation of primary alcohol to an aldehyde (no water to hydrate the carbonyl group)

primary alcohols

aldehydes

E2

CrO3 oxidations of alcohols (methyl, 1o and 2o ROH) without water = PCC, Cr=O addition, acid/base and E2 to form C=O (aldehydes and ketones)

H3CCH2

CCH3

OH

H

Cr

O

OO H3CCH2

CCH3

OH

H

Cr

O

O

O

H3CCH2

CCH3

O

H

Cr

O

O

ON N H

N

H3CCH2

CCH3

O

Cr

O

O

O

N H

PCC = pyridinium chlorochromate oxidation of primary alcohol to an aldehyde (no water to hydrate the carbonyl group)

sedondary alcohols

E2

ketones



Problem 40 – Supply all of the mechanistic details in the sequences below showing 1. the oxidation of a primary alcohol, 2. hydration of the carbonyl group and 3. oxidation of the carbonyl hydrate (Jones conditions).

H3CCH2

CH

OH

H

Cr

O

OOH3C

CH2

CH

OH

H

Cr

O

O

O

H3CCH2

CH

O

H

Cr

O

O

O

H3CCH2

CH

OCr

O

O

O

Jones = CrO3 / H2O / acid primary alcohols oxidize to carboxylic acids

(water hydrates the carbonyl group, which oxidizes a second time )

primary alcohols

aldehydes

HO

HH

OH

H

HO

H

H

H3CCH2

CH

OH

H3CCH2

CH

OH

HO

H

hydration of the aldehyde

Under aqueous conditions, a hydrate of a carbonyl group has two OH groups which allow a second oxidation, if another C-H bond is present. This is only possible if the starting carbonyl group was an aldehyde (true when starting with methyl and primary alcohols).

chromicanhydride

inorganicester

nucleophile addition at Cr

acid/base

H

O

H

resonance

nucleophile addition at C

E2 reaction(oxidation here)

H3CCH2

C

OH

O

H

H

H

HO

HH3C

CH2

C

O

O

H H

HO

H

H

Cr

O

OO

H

H3CCH2

C

O

O

H H

H Cr

O

O

OHO

H

H3CCH2

C

O

O

H H

Cr

O

O

OH

OH

H

HO

H

H3CCH2

C

O

OH

HO

H

H

Cr

O

O

O

second oxidation of the carbonyl hydrate

carboxylic acid from an aldehyde or a primary alcohol

acid/base

nucleophile addition at Cr

acid/baseinorganic

ester

E2 reaction(oxidation here)

Your next step is to write out the above mechanism completely on your own, using the following equation.

H3CCH2

CH2

O

Cr

O

OO

H3CCH2

C

O

OH Cr

O

O

O

HH

OH

carbonyl hydrate



Typical oxidation possibilities are shown below for common alcohol patterns. Currently we can make the alcohols from RBr compounds using SN chemistry, and that’s a lot of alcohols. The alcohol carbon with oxygen is the key. Does it have any hydrogen atoms (is it methyl, primary or secondary)? How many hydrogen atoms does it have? Potentially all of the hydrogen atoms can be oxidized off, or only one of them, depending on the conditions you choose (Jones or PCC).

1. Primary alcohols (or methanol), without any water in the reaction mixture (PCC), oxidize only to aldehydes. No carbonyl hydrates can form without water, so there is no way to oxidize a second time. Pyridine is the base.

HC

H

OH

H

PCCCrO3/pyridine

HC

H

O

RC

H

OH

H

PCCCrO3/pyridine

RC

H

O

aldehydesprimary alcoholsmethanol methanal

2. Primary alcohols (or methanol) with water in the reaction mixture can oxidize twice (Jones). Once the aldehyde is formed, it can hydrate (add H2O) and form a chromium ester a second time, which oxidizes off a second hydrogen atom. Water is the base.

RC

H

OH

H

RC

OH

O

aldehydesprimary alcohols

CrO3H2SO4H2O

Jones conditions

RC

H

O H3O+

H2OR

CH

HO OH

carbonylhydrates

CrO3H2SO4H2O

Jones conditionscarboxylic

acidagain

3. Secondary alcohols can only oxidize once in either aqueous or nonaqueous conditions. Either reagent produces a ketone product.

RC

H

OH

R'ketonessecondary alcohols

either method

RC

R'

O There are no additional C-H bonds at the original alcohol carbon, so there is no additional oxidation possible at the ketone carbon.(Jones or PCC)

4. Tertiary alcohols can form chromium esters, but there is no hydrogen atom to eliminate at the alcohol carbon. Tertiary alcohols are unreactive with either reagent. At higher temperatures C-C bonds can be cleaved (usually making a mess).

RC

R"

OH

R'tertiary alcohols

either method No productive reaction

(no removable C-H at the alcohol carbon in tertiary alcohols)



Allowed starting structures – our main sources of carbon – 1. Free radical substitution of 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

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

(faster at 3o > 2o > 1o C-H positions)

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

2. Free radical addition mechanism of H-Br to 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



For now, the structures below represent your hydrocarbon starting points to synthesize target molecules (TM) that are specified. We will only use the two free radical reactions, above, in our course, but they are very important reactions because they make versatile functionalized starting molecules for synthesis of all the other functional groups studied in this course. From these two free radical reactions and E2 reactions with potassium t-butoxide (to make alkenes) we can make 13 R-Br molecules below. We can use double E2 reactions with sodium dialkylamides to make 3 terminal alkynes.

CH4

Br Br

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

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

PhBr2 / h2 eqs.

Ph

Br

Br

Br Br Br Br

E2 reaction(twice)

NaR2N

1. 3 eqs.

2. workup

Ph

a b ca

ab

b

c

c

Br

bromobenzene is given until aromatic chemistry is covered in 316

Br2 / h

Br

messyE2 reaction(3 products)



Br Br

Br

H3CBr

Br

Br

BrBr

Br

Br

BrBr

Br

allylic RBrbenzylic RBr,

Ph

Br

Br

Br Br Br Br

Ph

Br

Br

RBr, RBr2, alkene and alkyne compounds we can currently make.

OH

O

ketone(propanone)

carboxylic acid(ethanoic acid)

Simple examples of functional groups we can currently make.

Additional organic compounds that are available as examples until we can make them

O

H

O

O

O

NHR

O

OH

BrHO Br

HO Br

OH

O

O

O

anhydride(ethanoic anhydride)

ester(ethyl ethanoate)

O

Cl

acid chloride(ethanoyl chloride)

2o amide(ethanamide)

aldehyde(ethanal)

alcohol(ethanol)

SH

thiol(ethanethiol)

NH2

amine(ethanamine)

O

ether(ethoxyethane)(diethyl ether)

H3CC

N

nitrile(ethanenitrile)

S

sulfide(ethylthioethane)(diethyl sulfide)

N

azide(azidoethane)

OH

Br

NN

BrBr

alkenes

alkynesdibromohydrocarbons

Br

bromoalkane(bromoethane)

alkene(propene)

alkyne(propyne)



Problem 41 – We can now make the following molecules. Propose a synthesis for each from our starting materials.

C

O

H H

C

O

H3C H

C

O

H3C CH3

C

O

CH2

H

C

O

CH3

H3C

O

C

O

CH HH3C

CH3

C

O

H C

O

CH2

H

C

O

H OH

C

O

H OCH3

C

O

H3C OH

C

O

CH2

OHH3C

C

O

CH OHH3C

CH3

C

O

OH C

O

CH2

OH

C

O

H3C O

C

O

CH2

OH3C

C

O

CH OH3C

CH3

CH3CH3

CH3 C

O

OC

O

CH2

O

CH3

CH3

aldehydes

ketones

C

O

CH

HH2C C

O

CH

OHH2C C

O

CH

OH2C CH3

O

C

O

C HH2C

CH3

C

O

C OHH2C

CH3

C

O

C OH2C

CH3

CH3

conjugated aldehydes, carboxylic acids and esters

C

O

H3C Cl

C

O

CH2

ClH3C

C

O

CH ClH3C

CH3

C

O

ClC

O

CH2

Cl

C

O

H NH

C

O

H3C NH

C

O

CH2

NH

H3CC

O

CH NH

H3C

CH3

C

O

NH C

O

CH2

NH

(not stable)

CH3 CH3 CH3

CH3CH3

CH3

anhydrides

C

O

H3C OC

O

HC

O

H3C OC

O

CH3

C

O

H3C OC

O

CH2

CH3C

O

H3C OC

O

CHCH3

CH3

C

O

H3C OC

O

carboxylic acids

acid chlorides

amides

esters

nitriles

H3CC

NC

NC

N CN

CN

CN

CN

starting sources of carbon

CH4

Possible RBr compounds from these starting hydrocarbons.

Br

Br

BrBr

Br Br Br

BrH3CBr

Possible alkenes and alkynes from these starting hydrocarbons.

BrBr

Br

Br Br

Ph

Br Br

dibromoalkanes can make alkynes



R-Br examples C1 C7 (For now, these are given. Soon, we will have to make them.) You need to recognize reactive substitution patterns as: methyl (Me), primary (1o), secondary (2o), tertiary (3o), allylic, benzylic and unreactive substitution patterns as: primary neopentyl, vinyl and phenyl.

Br

Br

Br

Br

Br

BrBr

H3CBr

1C 2C 3C 4C

a b ab

c d

* = chiral centers

*2o1o1o 1o 1o

3o2omethyl

methyl1o = primary RX2o = secondary RX3o = tertiary RX

1o = primary neopentyl RX2o = secondary neopentyl RX3o = tertiary neopentyl RX

categories of RX compounds:allyl RXbenzyl RXvinyl RX phenyl RX

Br Br Br

Br

Br

Br

Br

Br

5C a b c d e f g h

*

* *

1o neopentyl

1o

2o3o 1o1o2o2o1o

Br

Br

Br

Br Br Br

6Ca

b c d e f

*

*

**

1o 2o 2o 1o3o 2o

Br

BrBr

Br Br

Br Br

6C g h i j k lm

** *

3o 1o 1o neopentyl2o 2o1o 1o

Br

Br

Br

Br

6C no p q

* *

2o neopentyl 1o 1o 3o

Br

Br

Br

Br

Br

7C ab c d e

*

*

*

2o 2o1o2o 1o

BrBr

Br

Br

Brfg h i j

* * *

3o 1o2o2o

2o

7C

7C

Br

BrBr

Br

Br

k l m n o*

* *****

3o2o1o

2o2o

Br

Br Br Br

Br

Br

p q r s t u

***

*

1o 1o1o neopentyl 2o

2o 1o

7C



Br

Br

BrBr

BrBr Br

7C

v w x y z aa bb

* **

* **

2o 1o1o1o

3o 3o1o

BrBr

Br

Br

Br Br

7C

cc dd ee ffgg hh

*

2o 1o neopentyl 1o1o3o 2o

BrBr

BrBr

Brkk ll mmii jj

*

3o2o

1o neopentyl 3o 1o

7C

Extra patterns to know (allylic and benzylic RX are very fast SN2 patterns) (1o neopentyl, vinyl and phenyl RX patterns are unreactive).

Br BrBr

Br BrBr

BrBr

allylic

benzylic

1o neopentyl vinyl X phenyl X

a b c a b

allylic allylic

benzylic



O

Na Li K

HO

N

O

O

H3CBr

Br

Br

Br

Br

Br

Br

Br

N

O

O

O

O

ORN

RN

NN

NC

CC

R

CCH2O

OLi

CCH2H3C

O LiR C

CH2N

OLi

R

R

CCH2O

OLiLi

2. workup

2. workup

2. workup

2. workup

2. workup

2. workup

2. workup

2. workup

NC

CH2Li S

S

Li

BH

H

H

H

AlH

H

H

HNaLi S

Ph

Ph

P

Ph

Ph

Ph

P

Ph

Ph

Ph

P

Ph

Ph

Ph

P

Ph

Ph

Ph

P

Ph

Ph

Ph

P

Ph

Ph

Ph

P

Ph

Ph

Ph

P

Ph

Ph

Ph

P

Ph

Ph

Ph

BrBr

Ph PhP

Ph

1. 2. n-butyl lithium 2. n-butyl lithium Br

LiAlD4

Br2 / hROOR (cat.)

Br2 / h

2 r-x is the electrophile sn1psbeauchamp/pdf/315_sn_e_slides_fall_2015.pdfy:\files\classes\0 organic...

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