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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
Nucleophilic Substitution & Elimination Chemistry Beauchamp 2
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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
Nucleophilic Substitution & Elimination Chemistry Beauchamp 3
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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
Nucleophilic Substitution & Elimination Chemistry Beauchamp 4
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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)
Nucleophilic Substitution & Elimination Chemistry Beauchamp 5
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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
Nucleophilic Substitution & Elimination Chemistry Beauchamp 6
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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
Nucleophilic Substitution & Elimination Chemistry Beauchamp 7
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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)
Nucleophilic Substitution & Elimination Chemistry Beauchamp 8
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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
Nucleophilic Substitution & Elimination Chemistry Beauchamp 9
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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
Nucleophilic Substitution & Elimination Chemistry Beauchamp 10
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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
Nucleophilic Substitution & Elimination Chemistry Beauchamp 11
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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
Nucleophilic Substitution & Elimination Chemistry Beauchamp 12
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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
Nucleophilic Substitution & Elimination Chemistry Beauchamp 13
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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
Nucleophilic Substitution & Elimination Chemistry Beauchamp 14
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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
Nucleophilic Substitution & Elimination Chemistry Beauchamp 15
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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
Nucleophilic Substitution & Elimination Chemistry Beauchamp 16
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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
Nucleophilic Substitution & Elimination Chemistry Beauchamp 17
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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
Nucleophilic Substitution & Elimination Chemistry Beauchamp 18
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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
Nucleophilic Substitution & Elimination Chemistry Beauchamp 19
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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
Nucleophilic Substitution & Elimination Chemistry Beauchamp 20
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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
Nucleophilic Substitution & Elimination Chemistry Beauchamp 21
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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
Nucleophilic Substitution & Elimination Chemistry Beauchamp 22
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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
Nucleophilic Substitution & Elimination Chemistry Beauchamp 23
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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
(resonance stabilized)
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
Nucleophilic Substitution & Elimination Chemistry Beauchamp 24
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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
(resonance stabilized)
-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.
Nucleophilic Substitution & Elimination Chemistry Beauchamp 25
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?
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
Nucleophilic Substitution & Elimination Chemistry Beauchamp 26
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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
Nucleophilic Substitution & Elimination Chemistry Beauchamp 27
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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"?
Nucleophilic Substitution & Elimination Chemistry Beauchamp 28
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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
Nucleophilic Substitution & Elimination Chemistry Beauchamp 29
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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..
Nucleophilic Substitution & Elimination Chemistry Beauchamp 30
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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
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
SN1 and E1 are possible here
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
SN1 and E1 are possible here
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.
SN1 and E1 are possible here
CH3
CH3CH3
CH3
H3C
H3C
H3C H3C
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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.
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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)
Nucleophilic Substitution & Elimination Chemistry Beauchamp 33
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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
conjugate acid pKa = 16
conjugate acid pKa = 5
conjugate acid pKa = 19
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
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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
Nucleophilic Substitution & Elimination Chemistry Beauchamp 35
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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
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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
Nucleophilic Substitution & Elimination Chemistry Beauchamp 37
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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
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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
Nucleophilic Substitution & Elimination Chemistry Beauchamp 39
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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 group
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
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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.
water is a good leaving group
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
water is a good leaving group
Nucleophilic Substitution & Elimination Chemistry Beauchamp 41
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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.
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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
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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
Nucleophilic Substitution & Elimination Chemistry Beauchamp 44
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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
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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
Nucleophilic Substitution & Elimination Chemistry Beauchamp 46
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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)
exception(too basic)
exception(too 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)
Nucleophilic Substitution & Elimination Chemistry Beauchamp 47
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
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 = _____?
oxidation state of carbon = _____?
oxidation state of carbon = _____?
oxidation state of carbon = _____?
oxidation state of carbon = _____?
oxidation state of carbon = _____?
oxidation state of carbon = _____?
oxidation state of carbon = _____?
oxidation state of carbon = _____?
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 =
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 =
oxidation state of Cr =
oxidation state of C =
oxidation state of Cr =
Nucleophilic Substitution & Elimination Chemistry Beauchamp 48
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
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
Nucleophilic Substitution & Elimination Chemistry Beauchamp 49
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
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
Nucleophilic Substitution & Elimination Chemistry Beauchamp 50
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
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)
Nucleophilic Substitution & Elimination Chemistry Beauchamp 51
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
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)
BE = +95 kcal/moleBE = -88 kcal/mole
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
BE = +88 kcal/moleBE = -111 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
Nucleophilic Substitution & Elimination Chemistry Beauchamp 52
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
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)
Nucleophilic Substitution & Elimination Chemistry Beauchamp 53
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
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)
Nucleophilic Substitution & Elimination Chemistry Beauchamp 54
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
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
Nucleophilic Substitution & Elimination Chemistry Beauchamp 55
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
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
Nucleophilic Substitution & Elimination Chemistry Beauchamp 56
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
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
Nucleophilic Substitution & Elimination Chemistry Beauchamp 57
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
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