315 sn e syn & chem catalog
TRANSCRIPT
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Nucleophilic Substitution & Elimination Chemistry Beauchamp 1
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Four mechanisms to learn: SN2 vs E2 and SN1 vs E1
S = substitution = a leaving group (X) is lost from a carbon atom (R) and replaced by nucleophile (Nu:)
N = nucleophilic = nucleophiles {Nu:) donate two electrons in a manner similar to bases (B:)
E = elimination = two vicinal groups (adjacent) disappear from the skeleton and are replaced by a pi bond
1 = unimolecular kinetics = only one concentration term appears in the rate law expression, Rate = k[RX]
2 = bimolecular kinetics = two concentration terms appear in the rate law expression, Rate = k[RX] [Nu: or B:]
SN2 competes with E2
SN1 competes with E1
These electronsalways leave with X.SN2
E2
SN1
E1
XRBNu BHNuHCompetingReactions
CompetingReactions
CarbonGroup
LeavingGroup
Nu: / B: = is an electron pair donor to carbon (= nucleophile) orto 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 coming from anions of all types and neutral nitrogen, sulfur and phosphorous atoms. Weak electron pair
donors will typically be neutral solvent molecules, usually water (H2O), alcohols (ROH), mixtures of the two,
or simple, liquid carboxylic acids (RCO2H).
2. What is the substitution pattern of the R-X substrate at the Ccarbon attached to the leaving group, X? Is it a
methyl, primary, secondary, tertiary, allylic, or benzylic carbon? What about any Ccarbon atoms? How
many additional carbon atoms are attached at a Cposition (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.
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Strong electron pair donation in our course (SN2 / E2).
Na
O H
Na
S HO R
Na K
O
Na
O
O
hydroxide alkoxides potassium t-butoxide carboxylates(make, acetate)
hydrogen sulfide (thiolate)(buy) (make) (make, strong base) (buy)
Nu
B
=
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)
Na
S R
alkyl sulfide(thiolate)
(make)
Na
C NN
O
Ophthalimidate(an imidate)
C C H
Na CH2
OLi
ketone enolates
Li
Na
cyanideacetylides
ester enolates
H2C
O
O
(make)
(make)(make)(buy) (make)
Na
N
N
N
(buy)
azide
BH
H
H
H
Na
AlH
H
H
H
Li
sodiumborohydride
lithiumaluminium
hydride
S
S
H
Li
dithiane anion
(make)
BD
D
D
D
Na
sodiumborodeuteride
AlD
D
D
D
Li
lithiumaluminiumdeuteride
CH2
Li
n-butyl lithium(very strong baseor nucleophile, useanytime)
Na Cl
Br
INa
Na
good nucleophileswith tosylates and
in strong acid(buy) (buy) (buy) (buy) (buy) (buy)
S
Ph
Ph
CH2
sulfurylids
(MgBr)
Grignard reagents(very strong basesand nucleophiles)
R Li
alkyl lithium(very strong basesand nucleophiles)
R R2Cu
organocuprates(good carbonnucleophiles)
Li
Na H
N
lithiumdiisopropylamide(very strong base)
sodium hydride(very strong base)
Na
sodium amide(very strong base)
SCl
O
O= Ts-Cl (tosyl chloride)makes ROH into tosylates
N
pyridine =proton sponge
KH
potassium hydride(very strong base)
R = C or H
NR2
= py
BrCu
cuprousbromide
Li
(buy) (buy) (buy)
(make)
(make) (make)(make) (buy)
(buy)
Important additions for us.
SPh Ph
Ph = phenylP
Ph Ph
diphenylsulfide,used with C=O
to make epoxides
triphenylphosphineused with C=Oto make alkenes
Ph
CrO3/ pyridine
pyridinium chlorochromatePCC
oxidizing E2 reaction
CrO3/ H2O
Jones reagentoxidizing E2 reaction
(buy)
(buy)
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These two reactions look similar, but there are important differences.
Br O
majorminorH
O
H
O
H
H
Br
Br
H
O
H
O
majorminor
H
Br
H
O
H
Rate = kSN2[HO ]
1
[RBr]
1
Rate = kE2[HO ]1
[RBr]1
Rate = kE1[RBr]1
Rate = kSN1[RBr]1
SN2 versus E2 overview (essential features)
Example: 1o
RX, requires strong nucleophile/base, SN2 > E2, exceptions: potassium t-butoxide or sodium amide.
CH
CH3H
C Br
HH
1-bromopropane
Nu
CH
CH3H
C Br
H
H
1-bromopropane
O
C
CNu
HH
H
CH3
H
E2 > SN2
(when t-butoxide) C
C
H
H
H3C
H
alkene
E2
Nu B=
strong = anythingwith negative charge,and neutral sulfur,
phosphorous or nitrogen
SN2 > E2
(unless t-butoxide)
C
H3C
H3C
H3C
(also R2N = E2)
SN2 always
backside attack
E2 alwaysanti C-H C-X
SN2 reactions are the most important reactions always backside attack at C-X carbon
XH3CH2CR X
HCR X
R
CR X
R
R
CH
H2CH2C X
H2C X
methyl (Me) primary (1o) secondary (2o) tertiary (3o)allylic
(1o, 2o, 3oversions)
C C X
C-beta carbon(0-3 of these)
C-alpha carbon(only 1 of these)
benzylic
(1o, 2o, 3oversions)
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Relative rates of SN2 reactions - steric hindrance at the Ccarbon slows down the rate of SN2 reactions.
XC
H
H
H
k 30methyl (unique)
XC
H
H3C
H
k 1ethyl (primary)
reference compound
XC
CH3
H3C
H
k 0.025isopropyl (secondary)
XC
CH3
H3C
CH3
k 0t-butyl (tertiary)
(very low)140
C
H
HH
X
methyl RX
Methyl has three easy paths ofapproach by the nucleophile. It isthe least sterically hindered carbonin SN2 reactions, but it is unique.
C
H
RH
X
primary RX
Primary substitution allows two easy pathsof approach by the nucleophile. It is theleast sterically hindered "general" substitutionpattern for SN2 reactions.
C
R
RH
X
secondary RX
Secondary substitution allows one easy pathof approach by the nucleophile. It reacts theslowest of the possible SN2 substitutionreactions.
tertiary RX
C
R
RR
X = C
C
X
C
H
H
H
H
HH
CH
H
H
Nu Nu
Tertiary substitution has no easy path of approachby the nucleophile from the backside. We do notpropose any SN2 reaction at tertiary RX centers.
When the Ccarbon is completelysubstituted the nucleophile cannotget close enough to make a bond
with the Ccarbon. Even C-Hsigma bonds block the nucleophilesapproach.
All of these are primary R-X structures at C, but substituted differently at C.
H2CC
H
H
H
k 1
ethylreference compound
H2CC
H
H3C
H
k 0.4
propyl
H2CC
CH3
H3C
H
k 0.03
2-methylpropyl
H2CC
CH3
H3C
CH3k 0.00001 0
2,2-dimethylpropyl(1oneopentyl)
X X XX
C
C
X
H
C
CH3CH3
H
H
H
H
Nu
A completely substituted Ccarbon atom also blocks theNu: approach to the backsideof the C-X bond. A large groupis always in the way at thebackside of the C-X bond.
C
C
X
H
H
CH3CH3
HNu
If even one bond at Chas a
hydrogen then approach byNu: to the backsideof the C-X bond is possibleand an SN2 reaction ispossible.
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sp2carbon transition state as carboninverts configuration forms a high PEcarbon with 10 electrons at carbon.This is a concerted, one-step reaction.
"Transition State"
C
H
H
H
BrOH
BrC
H
H H
HO
C
H
H
H
HO Br
reactants
products
Ea- this energy difference determines
how fast the reaction occurs: "kinetics".
Ea
GG - this energy difference determines theextent of the reaction; the ratio of products
versus reactants at equilibrium (whenkinetics allows the reaction to proceed.Thermodynamics is determined mostly by thestrengths of the bonds and solvation energies ofthe reactant and product speces: "thermodynamics".
POR = progress of reaction
PERate = kSN2[RX][Nu] = bimolecular reaction
kSN2= 10
Ea= -2.3RT log(kSN2)
-Ea
2.3RT
Go= -2.3RT log(Keq)
Keq= 10
-Go
2.3RT
Problem 1 - How can you tell whether the SN2 reaction occurs with front side attack, backside attack or front and
backside attack? Use the two molecules to explain you answer. Follow the curved arrow formalism to show
electron movement for how the reaction actually works.
HH3C
H Br
C Br
H
H3C
CH3CH2
I
CH3C
O
O
a b
Problem 2 - Why might C3and C4rings react so slowly in SN2 reactions? (Hint-think about bond angles in the
transition state versus bond angles in the starting ring structure.)
krelative
(SN2)
Br BrBr Br Br
BrBr
1.0 0.00001 0.008 1.6 0.01 0.97 0.22
Problem 3 - Why might C6rings react slower in SN2 reactions? What are the possible conformations from which a
reaction is expected? Trace the path of approach for backside attack across the cyclohexane ring to see what positions
block this approach. Which chair conformation would have the leaving group in a more reactive position (axial or
equatorial)? Is this part of the difficulty (which conformation is preferred)?
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H H
H
HH
H
H
H
H
HH
H
H
H
H
H
H
H
H
H
H
H
These two chairconformations
interconvert in avery fast equilibrium.
X
X
Problem 4 - In each of the following pairs of nucleophiles one is a much better nucleophile than its closely related
partner. Propose a possible explanation.
N N O O CC
C
C
O
HH
HH
H
HH
HH
b. c.a. relative rates 250/1
OC
H
H
H
methoxide
t-butoxide
Problem 5 Write out the expected SN2 product for each possible combination (4x8=32 possibilities).
AlD
D
D
D Li
Nu
X
Nu
Nu
Nu =H
O
H3C
O
O
O N
CH
S
H3C
S
N
N
N
?
?
X = Cl , Br , I ...a good leaving group
1 3 4 6 7
X
X
X
?
?
Nu
A
B
C
D
25
8
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Problem 6 Using R-Br compounds from page 2 and reagents from page 3 to propose starting materials to make each
of the following compounds. One example is provided. TM-1 is an E2 product (see page 15), all the others are SN2
products.
N
NN
?
Br
problemsolution
NaN3
N3
2-azidopropane
TM = target molecule
N
O
O
O
TM-1 TM-2 TM-3
O
R
TM-4 TM-5 TM-6O
S
Ph Ph
CH3C
O
TM-7TM-8 TM-9 TM-10 TM-11
P
PhPh
Ph
Br
N
Br
O
SH
TM-12 TM-13 TM-14 TM-15
HO
O
O
S
R
D
TM-16TM-17
S
S
H
Acid/base reactions important to our course (some reagents have to be made, often by acid/base reactions) and
subsequent reactions (mostly SN2)
Make alkoxides and use as nucleophiles only at Me-X and 1oRCH2-X in SN2 reactions.
alkoxides are good nucleophiles at methyl,primary, allyl and benzyl RX, mostly E2 atsecondary and only E2 tertiary RX, they arealso used as moderately strong bases
OR
Na
alkoxides
OR H
alcohols Na
H
Br R
O
SN2
ethers
Keq=Ka1
Ka2
10-16
10-37= = 10+21
H
Hacid/base
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Make potassium t-butoxide and use as sterically large, strong base at 1oRCH2-X, 2oR2CH-X and 3
oR3C-X in E2
reactions.
OK
K
O
H
t-butylalcohol potassiumt-butoxide
HBr
H
R E2
R
alkenes
H
H
Keq=Ka1
Ka2
10-19
10-37= = 10+18
potassium t-butoxide, sterically bulky basethat mostly does E2 reactions with RX
compounds (except SN2 with CH3-X).
acid/base
Make thiols using NaSH as the nucleophile at Me-X, 1oRCH2-X and 2oR2CH-X in SN2 reactions.
SH
Na
hydrogen sulfide
Br H
S
thiolSN2
Make thiolates and use as nucleophiles at Me-X, 1
o
RCH2-X and 2
o
R2CH-X in SN2 reactions.
acid/base
SR
Na
alkylthiolates
NaSR H
thiols
O HBr
SN2R
S
sulfides
Keq=Ka1
Ka2
10-9
10-16= = 10+7
Synthesis of lithium dithiane anion (acid/base) 2. SN2 with RX 3. Hydrolyze to make aldehydes or ketones)
A good nucleophile thatcan be made into aldehydesand ketones
S
S
dithiane anion
LiS
S
dithiane
aldehydes
andketones(later)
Br
SN2
H2CH
HLi
S
S
H
Keq=Ka1
Ka2
10-50
10-35= = 10+15
acid/base
Make terminal acetylides and use as nucleophiles only at Me-X and 1oRCH2-X in SN2 reactions.
CCR
Na
terminalacetylides
CCR
terminalalkynes
H
NaNR2
H3C
Br
SN2
R
CH3alkynes
Keq=Ka1
Ka2
10-25
10-37= = 10+12
terminal acetylides are good nucleophilesat methyl, primary, allyl and benzyl RX, butmostly E2 at secondary and only E2 tertiary RX
acid/base
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Zipper reaction moves triple bond to end of linear chain to form the most stable anionic charge, further chemistry
is possible. This reaction is almost identical to tautomers, without any heteroatoms.
Synthesis of lithium diisopropyl amide, LDA, sterically bulky, very strong base used to remove C-H proton of
carbonyl groups. (acid / base reaction) to make carbonyl enolates (next).
Keq =Ka1
Ka2
10-37
10-50= = 10+13
CH2 Li
N
H
N
Li
Think - sterically bulky, very basic
that goes after weakly acidic protons.
LDA = lithium
diisopropyl amide
given
given
acid/base
React ketone enolate (nucleophile) with R-Br electrophile (Me, 1oand 2oRX compounds)
H2C
O Li
ketone enolates(resonance stabilized)
React ketone enolate with RX compounds (methyl, 1oand 2oRX)
Br
O
1oRX key bond
RS Larger ketone made
from smaller ketone.
SN2reactions
-78oC
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React ester enolate (nucleophile) with R-Br electrophile (Me, 1oand 2oRX compounds)
H2C
O
O
Li
ester enolates(resonance stabilized)
React ester enolate with RX compounds (methyl, 1oand 2oRX)
Br
O
O
1oRX key bondR
S Larger ester madefrom smaller ester.
R RSN2
reactions
-78oC
diphenylmethylsulfonium
bromide salt
S
Ph
Ph
CH2
sulfur ylid
to make epoxides
S
Ph
Ph
CH2make epoxides
from aldehydes
and ketones
H2C
Li
Keq =Ka1
Ka2
10-50
10-35= = 10+15
H
S
Ph
Ph
H3C
Br
diphenyl
sulfide
SN2
acid/base
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Clarification of Hydride electron pair donors: In our course sodium hydride and potassium hydride are
always basic, and lithium aluminum hydride and sodium borohydride are always nucleophilic hydride.
Basic hydride
In our course, sodium hydride (NaH) and potassium hydride (KH) are alwaysbasic (= electron pair donation by
hydride to a proton), nevera nucleophile. The conjugate acid of hydride is hydrogen gas (with a pKa= 37, H2can
hardly be considered an acid).
Sodium hydride and potassium hydride (KH)
Na
Aldrich, 2012$39 / 100 grams
60% oil dispersionMW = 24.0 g/mol
H KH
Aldrich, 2012$128 / 75 grams
30% oil dispersionMW = 40.1 g/mol
Problem 8 Write an arrow pushing mechanism for each of the following reactions.
O
H
O
H
H
Na
K
H
pKa
ROH 16-19
H-H 37pKa= 17
pKa= 19
Nucleophilic hydride Formation of C-H bonds using nucleophilic lithium aluminum hydride and sodium
borohydride.
In our course, sodium borohydride (NaBH4) and lithium aluminum hydride (LiAlH4= LAH) are inorganic saltscontaining nucleophilic hydride, very unusual nucleophiles. Both reagents supply nucleophilic hydride that can
displace X in SN2-like reactions with RX compounds. Sodium borohydride and lithium aluminum hydride are used
in many other reactions. They also react with carbonyl compounds (C=O) and epoxides (both to be discussed more
later). We will often use the deuterium version of borohydride and aluminum hydride so we can identify where a
reaction occurred. In reality, this is not very common because of the expense. But, for purposes of probing your
understanding of SN2 reactions, using them shows if you understand what is happening. The deuterium isotope of
hydrogen reacts similarly to the proton isotope, but there are experimental methods which allow us to observe
where a reaction took place (e.g. NMR).
Sodium borohydride and lithium aluminum hydride (LAH) - 4 equivalents of hydride per anion
B H
H
H
H Na Li
Al H
H
H
H
B D
D
D
D Na
Aldrich, 2012$312 / 100 gramsMW = 37.8 g/mol
Aldrich, 2012$120 / 100 gramsMW = 38 g/mol
Aldrich, 2012$254 / 5 grams
MW = 41.8 g/mol
Aldrich, 2012$125 / 5 gramsMW = 42 g/mol
Li
Al D
D
D
D
The deuteride salts areway more expensive.
The hydride salts are
way more common.
All these reagents will undergo SN2 reactions at RX centers. Because there is a greater difference in
electronegativity between aluminum and hydrogen than boron and hydrogen, LAH is more reactive than sodium
borohydride. This wont be important for us to know until we study carbonyl reactions.
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Problem 9 Write an arrow pushing mechanism for the following reaction.
Li
Al D
D
D
D
C
Cl
CH3
H
+ ?
Na
B H
H
H
H
C
Cl
DCH3
+ ?
a. b.
Problem 10 It is hard to tell where the hydride was introduced since there are usually so many other hydrogens in
organic molecules. Where could have X have been in the reactant molecule? There are no obvious clues. Which
position(s) for X would likely be more reactive with the hydride reagent? Could we tell where X was if we used
LiAlD4?
LAH
Where was "X"?
H2C
CH2
H2C
CH3
E2 Reactions Compete with SN2 Reactions
E2 reactions also occur at the backside relative to the leaving group, but at C-H instead of C-X. The C-H
proton has to be anti to the C-X to allow for parallel overlap of the 2p orbitals that form the new pi bond. This
allows elimination to occur in a concerted manner. The syn conformation also has parallel overlap of 2p orbitals,
but is present in less than 1% due to an eclipsed conformation (staggered conformations > 99%).
E2 Potential Energy vs. Progress of Reaction Diagram (= concerted, energy picture looks very similar to SN2)
transition state
reactants
products
PEpotential
energy
POR = progress of reaction - shows how PE changes as reaction proceeds
higher(lessstable)
lower(morestable)
Ea= -2.3RTlog(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> R2and R3> R4then this would be Zconfiguration, which is fixed by therequirement for anti C-H / C-Xelimination.
E2 mechanism depends on
steric factors and basicity ofthe electron pair donor. Moresteric hindrance and more basicfavors E2 over SN2
= negative(exergonic)
Keeping Track of the C Hydrogens in E2 reactions
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At least one Cposition, with a hydrogen atom, is necessary for an E2 reaction to occur. In more complicated
systems there may be several different types of hydrogen atoms on different Cpositions. In E2 reactions there can
be anywhere from one to three Ccarbons, and each Ccarbon can have zero to three hydrogen atoms.
C
H
H H
X
C
H
C H
X
C
C
C H
X
C
C
C C
X
zero Cpositions
unique methyl,
only SN2 possible
one Cposition
1oRX, usually SN2 > E2
(except with t-butoxide)
two Cpositions
(2oRX, SN2 / E2
both competitive)
three Cpositions
(3oRX, only E2)
0-3 H at C0-6 H at C 0-9 H at C
With proper rotations, each C-H may potentially be able to assume an anti conformation necessary for an E2
reaction to occur. There are a lot of details to keep track of and you must be systematic in your approach to
consider all possibilities. Using one of these two perspectives may help your analysis of E2 reactions Lets
consider a moderately complicated example (next problem).
B
Either sketch will work for everypossibility above, IF you fill inthe blank positions correctly.
C
C
H
X
CH
C X?
B
?
horizontal perspectiveC-H / C-X
vertical perspectiveC-H / C-X
Problem 11 - How many total hydrogen atoms are on Ccarbons in the given RX compound? How many different
types of hydrogen atoms are on Ccarbons (a little tricky)? How many different products are possible? Hint - Be
careful of the simple CH2. The two hydrogen atoms appear equivalent, but E/Z (cis/trans) possibilities are often
present. (See below for relative expected amounts of the E2 products.)
C
CH2
CH3
CH3
X
CH
H3C
CH2H3C
Redraw structure to explicitely show allof the Chydrogen atoms. Notice the
base/nucleophile is strong, so SN2 or E2reaction are considered. The RX pattern istertiary, so no SN2 is possible. There arethree Ccarbon atoms and all of themhave hydrogen atoms on them. That leavesE2 reaction as the only possible choice.
OR OHR
Na
C
C
CH3
C
X
C1
H3C
CH2H3C
HHH
H
H
H
There are two chiral centers, Cand
C1, that make this structure morecomplicated than we are treating it.Possible stereoisomers are RR, RS,SR and SS.
?
Stability of pi bonds
In elimination reactions, more substituted alkenes are normally formed in greater amounts. Greater substitution
of carbon groups in place of hydrogen atoms at alkene carbons (and alkyne carbon atoms too) translates into greater
stability (lower potential energy). Alkene substitution patterns are shown below. There are three types of
disubstituted alkenes and their relative stabilities are usually as follows: geminal cis trans.
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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
R1 > R2 > R3
S
R
Problem 24 Draw in all of the mechanistic steps in an SN1 reaction of 2R-bromobutane with a. water, b. methanol
and c. ethanoic acid. Add in necessary details (3D stereochemistry, curved arrows, lone pairs, formal charge).
What are the final products?
Br
a.
H
O
H
b.
H3CO
H
c.
H3C
C
O
O
H
Donate the carbonyl (C=O)electrons to the carbocation.
SN1 products will generally outcompete E1 products, in our course. The only exception for us (presented
later) will be when alcohols are mixed with concentrated sulfuric acid at high temperatures to form alkenes (the E1
product).
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Keeping Track of the C Hydrogens in E1 Reactions(more possibilities than E2 reactions)
E1 products arise from the same carbocation intermediate formed in SN1 reactions. Just as in E2 reactions, we
have to examine each type of C-H. In more complicated R-X molecules there may be several different types of
hydrogen atoms on Cpositions. After all, there can be either two or three Ccarbons with zero to three hydrogen
atoms on each. We will only consider secondary and tertiary RX compounds below, since methyl and primary
carbocations do not form (in our course). That still leaves a lot of possibilities.
CC
C
H
X
CC
C
C
X
2o R-X has six C positions3o R-X has nine C positions
CC
C
H
CC
C
C
XH3CR
H2C
XNo Reaction
in our course.
NuH
NuHNuH
methyl primary
HNu: = H2O, ROH, RCO2H (weak, in our course)
E1 Mechanism (unimolecular kinetics) loss of proton from any adjacent C-H position, top or bottom
The high reactivity (low stability) of carbocations forces some very quick choices to try and stabilize the
situation. The carbocation needs two electrons to complete its octet (in a hurry!). There are three ways it typically
does this. We have studied the two ultimate pathways, SN1 and E1 reactions.The third possibility, rearrangements, is discussed next. 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.
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SN1 / E1 possibilitiesextra complications at Cpositions, In this problem 2oRX, rearrangements are NOT considered
(H2O,ROH,RCO2H)
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form by rearrangement. In the end, SN1 and E1 possibilities are the ultimate fates of even the most stable
carbocation that can form. Our goal at this point is to understand how rearrangements occur and what SN1 and E1
products are possible.
All groups on any Ccarbon can potentially migrate to the adjacent carbocation carbon (also called a 1,2
shifts), if a similar or more stable carbocation can form. If hydrogen with its two electrons is the group migrating,
the rearrangement is called a 1,2 hydride shift. If a carbon group migrates with its two electrons, the rearrangement
if called a 1,2 alkyl shift. Hydride and alkyl shifts can occur from further away than a Cposition or even between
two positions in completely different molecules. However, these we will not emphasize such possibilities in ourcourse.
Transition state of a carbocation rearrangement
C
CH2
H3C
H
HC
X
CH2
SN1
E1
rearrangement
2ocarbocation
SN1 and E1 arepossible here
XH3C
1,2-hydride shift
CC
H
CH2CH2
H
transition state(no finite existance)
migrating group is positionedbetween two vicinal carbon atoms
(vicinity = neighbors, Latin)
CC
H
CH2
HH2C
H3C
3ocarbocation
CC
H2C
H3C
H
HH2C
R-X
TS1 TS2TS3
Ea
G
2oR3oR
SN1 & E1products
PE
Progress of Reaction (POR)
H: = hydride shift
R: = alkyl shift
The migrating gropu is always attached tothe carbon skeleton; it is never a free anion.
SN1 and E1 arepossible here
CH3
CH3CH3
CH3
H3C
H3C
H3C H3C
Two main rules will help guide you in evaluating possible rearrangements.
1. Rearrangements usually occur so that the migrating groups moves from a Catom to the Catom (the
carbocation center). These are the 1,2 hydride or 1,2 alkyl shifts mentioned above. The Catom that gives up
the migrating group becomes the new electron deficient carbocation center, often because it is a more stable
carbocation site.
2. If a 1,2 shift of a hydrogen atom or an alkyl group can form a similar or more stable carbocation, then such a
rearrangement is likely to be competitive with other reaction choices (SN1 and E1). When interpretinga
reaction mechanism involving rearrangements, you may have to consider both equal (3o3oR+) and more
stable (2o3oR+) carbocation possibilities. However in this book when you are asked topredictwhat mightbe possible, you usually only need to consider more stable carbocation possibilities (2o3oR+).
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Problem 27 - Consider all possible rearrangements from ionization of the following RX reactants. Which
are reasonable? What are the possible SN1 and E1 products from the reasonable carbocation possibilities?
d. What would happen to the complexity of the above problems with a small change of an ethyl for a
methyl? Use the key of b and c as a guide. This problem is a lot more messy than those above,
(which is the point of asking it). There are too many possibilities to consider listing every answer.
There are other features that must also be considered in carbocation rearrangements in addition to
the relative stabilities of 1o, 2
oand 3
ocarbocations. One such feature that modifies the relative potential
energies of the possible choices is strain energy. Consider the possible rearrangement choices available
to the following tertiary carbocation in a polar ionizing solvent.
Br
slow
stepR.D.S.
R
O
H
CH2
HH
ab
c
CH2
H
H
CH3
H
CH3
H
H
CH3
CH3
a
b
c
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 tertiarycarbocation), 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/mole6 kcal/mole), resulting in an overall potential energy change of -5 kcal/mole.
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What is actually observed? (Only E1 reactions are shown, SN1 possibilities are not included.)
Rearrangement c occurs, followed by another rearrangement from a secondary to tertiary carbocation.
H
CH3
CH3
H CH3
CH3
major(85%)
minor(14%)
H CH3
CH3
HH
Path c leads toring expansion
rearrangement.2ocarbocation
3ocarbocation
E1
H CH3
CH2
veryminor(1%)
E1 products
Problem 28 Lanosterol is the first steroid skeletal structure on the way to cholesterol and other steroids in our
bodies. It is formed in a spectacular cyclization of protonated squalene oxide. The initially formed 3ocarbocation
rearranges 4 times before it undergoes an E1 reaction to form lanosterol. Add in the arrows and formal charge to
show the rearrangements and the final E1 reaction.
R
CH3
HH
CH3
CH3
H
CH3
HO
H
R
CH3
H
CH3
CH3
CH3
HO
H
H
lanosterol precursor
lanosterol
B H
R
O protonatedsqualene
acidcatalysis
rearrangement1
R
CH3H
CH3
CH3
H
CH3
HO
H
H
rearrangement2
R
CH3
CH3
H
CH3
HO
H
H
H
rearrangement4
CH3
R
CH3
H
CH3
HO
H
H
HCH3
CH3
R
CH3
CH3
CH3
H
CH3
HO
H
H
H
rearrangement3
E1 reaction
19 moresteps
H3CH
CH3
CH3
HO
H
H H
H
cholesterol
other
bodysteroids
H
squalene
Requires 5arrows toshow thereaction.
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SN1 / E1 possibilitiesextra complications at Cpositions of 2oRX, rearrangement to more stable 3oR+considered
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Alcohols in strong acid = Protonated Alcohols - Water as a Good Leaving Group
We can make the hydroxyl group of an alcohol, OH, into a good leaving group in strong acid conditions (HCl,
HBr, HI and H2SO4). Strong protic acids are used to extensively protonate the alcohol OH. When the alcohol OH
is protonated, the leaving group is water, not hydroxide. Waters conjugate acid is H3O+, (pKa= -2), while
hydroxides conjugate acid is H2O, (pKa= 16). If substitution is the desired goal, then the strong halide acids are
normally used, HCl, HBr or HI. If elimination is the desired goal, then concentrated sulfuric acid (H2SO4) is used
at an elevated temperature ().Using the hydrohalic acids (HCl, HBr or HI), very polar, strongly acidic conditions encourage SN1 reactions,
and these are assumed to be operating at all tertiary and secondary alcohol (ROH) centers. Rearrangements are
frequently observed under these conditions. The large energy expense of a methyl or primary carbocation prevents
the escape of water on its own. The H2O at methyl and primary ROH2+is assumed to be pushed off by the halide
(SN2) of the strong acid to form a methyl or primary haloalkane without rearrangement.
a. 1o, 2oand 3oROH reacted with HX acids (HCl, HBr, HI) - usually SNchemistry
i. methyl alcohols (SN2 emphasized, no rearrangement)
H3C
O HBrH
methanol
H3C
O H
H
Br
H3C
O
H
H
Br
SN2
BrH
O
H
H H
bromomethane
Br
water is a goodleaving group
pKa= -9
ii. primary alcohols (SN2 emphasized, no rearrangement)
H3C CH2
O H
BrH
primary alcohol
H3
C CH2
O H
H
Br
H3C CH2
O
H
H
BrSN2
BrH
O
H
H
primarybromoalkane
Br
pKa= -9
water is a goodleaving group
iii. secondary alcohols (SN1 emphasized, rearrangements possible)
water is a goodleaving group
ClH
secondary alcohol (trans OH)
H2C H3C H
SN1
Cl
H3C
H
Cl
H3C
Cl
H
trans Cl
cis Cl
top and bottomattack
H3C
H
O
H
H
O
H
H
secondarychloroalkane
pKa= -7
O
H
H O
H
H H
ClH Cl
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iv. tertiary alcohols (SN1 emphasized, rearrangements possible)
IH
tertiary alcohol (trans OH)
H3C H3C CH3
O
H
H
SN1I
O
H
H H
I
I
trans I
cis I
top and bottomattack
H3CCH3
O
H
CH3
O
H
H
IH
pKa= -10
Problem 29 - Propose a synthesis of monodeuterated cyclohexane from cyclohexanol.
O H
H H
D
b. 1o, 2oand 3oROH reacted with H2SO
4and high temperature = 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, an exception to our rule about no primary carbocations the conditions
are very harsh). Rearrangements are possible and observed.
i. primary alcohols (with high temperature E1-? - emphasized, rearrangements possible)
water is a goodleaving group
primary alcohol
O
H
H O
H
H H
O
H OH SO3H
(heat)
O
H
H
C
H
H
H H
rearrangementH
H
O SO3H
E1bp = -4
distillsbp = +82oC
very difficult(high temperature)
OH SO3H O SO3H
alcohol alkene
Tbp= 129oC
pKa= -10
ii. secondary alcohols (E1 emphasized, rearrangements possible)
secondary alcohol
O
H
H O
H
H H
O
H OH SO3H
(heat)
O
H
H
O SO3H
E1
bp = +83oC
distills outbp = +161oC
OH SO3HO SO3H
H
H
H
alcohol alkene
Tbp= 78oC
pKa= -10 water is a goodleaving group
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iii. tertiary alcohols (E1 emphasized, rearrangements possible)
tertiary alcohol
O
H
H O
H
H H
OH SO3H
(heat)
O SO3H
bp = +33oC
distills out
bp = +102oC
OH SO3HO SO3H
O
H O
H
H
H
H H
E1
bp = +39o
distills ou
90% < alke(more substitu
minor alkene(less substituted)b
a
a b
alcohol alkene
Tbp63oC
pKa= -10water is a goodleaving group
We have some, limited control to direct the alcohol functionality toward SNor E choices. The conditions to
effect these different pathways are important, so you must be aware of the details mentioned above (halide acids =
SNreactions and H2SO4/= E1 reactions). Heat is a crucial aspect of the E1 reaction, since it allows the lower
boiling alkene to escape from the reaction mixture by distillation, while the higher boiling alcohol or inorganic
esters remains, in the reaction pot to reestablish equilibrium by forming more alkene, which distills......etc. The
alkene boils much lower than the alcohol it comes from because it does not have an OH to form hydrogen bonds.
Examples of Boiling Point Differences Between Alcohols and Possible Alkene Products
boiling points
of alcohols (oC)
boiling point
of alkenes (oC)
OH H2C CH2
OH
OH
OH
(79 oC)
(82 oC)
(-104 oC)
(-47 oC)
(97 oC)
(100 oC)
OH (161oC)
(-47 oC)
(-6 oC)
(1 oC)
(4 oC)
(83 oC)
DTbp
(183 oC)
(129 oC)
(144 oC)
(106 oC)
(99 oC)
(96 oC)
(78 oC)
There are important other ways to change OH into Br. The OH group can be an alcohol or a carboxylic acid.
Some possibilities are shown below.
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1. Formation of tosylates from ROH + TsCl (toluenesulfonyl chloride = tosyl chloride), SN/E chemistry is possible
without rearrangements.
ORORH
S
Cl
H
SCl
O
O
O
O
N
OS
ClO
O
N H
OS
O
O
Br
Na
separate
step
Br
HBr
rearrangement
and
R,S (racemic)
S
1. TsCl/pyridine
2. NaCl
(prevents R+ formation
and any rearrangement)
alkyl tosylate = RX compound
compare - a diff erent result
R
R RR
RR
RBr
Br
SN1
toluene sulfonylchloride
(tosyl chloride)
SN2
Problem 30 We can now make the following molecules. Propose a synthesis for each. (Tosylates formed fromalcohols and tosyl chloride/pyridine via acyl substitution reaction, convert OH from poor leaving group into a very
good leaving group, similar to iodide)
2. 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, amides
and anhydrides.
Thionyl chloride with methyl, 1oROH = acyl-like substitution at SOCl2, then SN2 at methyl and primary RX.
R O
H S
Cl
O
Cl
H2C
R O
S
O Cl
Cl
OS
O
H
H2C
R O
SCl
O
CH2R
synthesis of an alkyl chloride from an alcohol + thionyl chloride (SOCl2) [can also make RBr from SOBr2]
SN2 at methyl andprimary alcohols
No rearrangement because no R+.
H
ClH
Cl
O
SCl
O
Hproduct
O
S
O
H
acyl-likesubstitution
Cl
Cl
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Thionyl chloride with 2oand 3oROH = acyl substitution, then SN1 (there are various ways you can write this
mechanism)
O H O
S
O Br
Br
OS
O
H
O
SBr
O
HH
Br
SN1 at secondary andtertiary alcohols
synthesis of an alkyl chloride from an alcohol + thionyl chloride (SOCl2) [can also make RBr from SOBr2]
O
SBr
O
H
BrH
RR
R
RR/S
acyl-likesubstitution
S
Br
O
Br
Br
Br
Synthesis of acid chlorides from acids + thionyl chloride (SOCl2), use the carbonyl oxygen instead of the OH.
C
R
O
O
HC
R
O
O
H
S
O Cl
Cl
resonance
C
R
O
O
H
S
O
Cl
Cl
C
R
O
O
H
S
O
Cl
Cl
Base
C
R
O
O
S
O Cl
C
BaseHC
R
O
O
SCl
O
CR OCR O
resonance
C
R
Cl
O acylium ion
S
Cl
O
Cl
OS
O
Cl
Cl
resonance
Formation of esters from ROH + acid chlorides, amides from RNH2or R2NH + acid chlorides and anhydrides from
RCO2H + acid chlorides
OR
O
R
H
Cl
OCl 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
R
O
H
R
O
H
H
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Problem 31 Look back at the table of R-Br structures on page 2. Include stereoisomers together. Be able to list
any relevant structures under each criteria below.
1. Isomers that can react fastest in SN2 reactions
2. Isomers that give E2 reaction but not SN2 with sodium methoxide
3. Isomers that react fastest in SN1 reactions
4. Isomers that can react by all four mechanisms, SN2, E2, SN1 and E1 (What are the necessary
conditions?)5. Isomers that might rearrange to more stable carbocation in reactions with methanol.
6. Isomers that are completely unreactive with methoxide/methanol
7. Isomers that are completely unreactive with methanol, alone.
The number of each type of product (SN1, E1, SN2, E2) is listed after a reaction arrow for each starting structure
(assuming I analyzed the possibilities accurately in my head, while sitting at the computer). Do you agree with
these numbers? Can you draw a valid mechanism for each one?
Br H
H3C H
OH
O
OH
H
O
H
SN1 E1 SN2 E2
O
O
O
H
O
2 41 3
1 3
1 3
a.
2 5b.
2 4a.
2 5b.
2 4a.
2 5b.b. after rearrangement
a. initial carbocation
Br H
H3C H
Br H
H3C H
OH
O
OH
SN1 E1 SN2 E2
O
O
O
H
O
2 6
1 3
1 3
1 3
a.2 5b.
2 6a.
2 5b.
2 6a.
2 5b.b. after rearrangement
a. initial carbocation
Br H
H3C H
H
O
H
D H D H
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Similar patterns in cyclohexane RX structures. The leaving group has to be axial in SN2 and E2 reactions.
C
C
C
Br
C
H
Hb
Ha H
H H
H H
C
C
Ha
Br
C
H
H3C
Hb H
H
Rotation of C-C1brings Haor Hbanti to C-Br, which allows SN2 andtwo different E2 possibilities: Ha(Z) and Hb(E). Since C2is a simplemethyl, there is no C2substituent to inhibit either of these reactions.
C
C
Hb
Br
C
H
Ha
H3C H
H
When methyl on C1 is antito C-Br, no SN2 is possible andno E2 is possible from C1.
SN2 possibleE2 from C1(2Z- butene)
E2 from C2(1-butene)
SN2 possible
E2 from C1(2E- butene)
E2 from C2(1-butene)
No SN2 possible and no
E2 from C1, but E2 from
C2(1-butene) is possible.
E2 possible here
Use these ideas to understand cyclohexane reactivity.
H H
H
H
H HBr
H
H
H
H
H
HH
H
H
H
HBr
H
H
H
H
H
No SN2 is possible (1,3 diaxialpositions block approach ofnucleophile), and no E2 is
possible because ring carbonsare anti.
No SN2 or E2 when "X"is in equatorial position.
SN2 possible if Cisnot tertiary and thereis no anti C"R" group.
E2 possible
with anti C-H.
Both SN2 and E2 are possiblein this conformation with leavinggroup in axial position.
H H
H
H
H HBr
H
C
H
H
H
HH
H
H
H
HBr
H
H
H3C
HH
No SN2 or E2 when "X"is in equatorial position.
No SN2 possible if thereis an anti C"R" group. E2 possible
with anti C-H.
Only E2 is possible in thisconformation. Leaving groupis in axial position.
H
H
H
full rotation at
C-Cis possiblein chain
only partialrotation is
possible in ring
only partialrotation is
possible in ring
H H H
equatorial leaving group
axialleavinggroup
No SN2 is possible (1,3 diaxialpositions block approach ofnucleophile), and no E2 is
possible because ring carbonsare anti.
No E2 possible,no anti C-H.
full rotation atC-Cis possiblein chain
alkene stabilitiestetrasubstituted > trisubstituted > trans-disubstituted > gem-disubstituted cis-disubstituted > monosubstitute
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The number of each type of product (SN1, E1, SN2, E2) is listed after a reaction arrow for each starting structure
(assuming I analyzed the possibilities accurately in my head, while sitting at the computer). Only consider
rearrangements in part 9. Do you agree with these numbers? Can you draw a valid mechanism for each one?
H
O
H
2 3
chair 1 chair 2
SN1
XX X
a.
b.
SN2 E2
1 1
E1
1 1
1
SN1 SN2 E2
1 2
E1
2 2
SN1 SN2 E2
0 1
E1
5 6
SN1
X
X X
SN2 E2
1 2
E1
2 2
4
SN1 SN2 E2
1 2
E1
2 2
SN1 SN2 E2
1 2
E1
2 2
8 9
SN1
XX
SN2 E2
1 2
E1
2 2
7
SN1 SN2 E2
0 3
E1
2 3
SN1 SN2 E2
0 1
E1
2 1
1 2
Only consider carbocationrearrangements thatimmediately go to a more
stable carbocation (notequally stable carbocations= too many possibilities)
a.
b.
2 2
1 2
X
a.
b.
c.
d.
consider any 3oR+possibility
and consider stereoisomers
4 5
1 2
1 2
H
ONa
condition 1 condition 2
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Available chemicals from the catalog
Sources of carbon - you can invoke these whenever needed:
CH4
Br
Br2 Cl2
Na
C N
Na
O H
CH2
Li n-butyl lithium(very strong baseor nucleophile, useanytime)
Na H
N
Hdiisopropylamine
N
O
O
H
Na
S H
O
OH
H2O
O
sodium hydride(very strong base)
ketone(a carbonyl)
pre alds / ketsphthalimide(an imide)
ethanoic acida carboxylic acid)
pre alds / kets
Na
sodium amide(very strong base)
HCl HBr HI H2SO4
Commercially available chemicals and reagents - you can invoke these whenever you need them.
NOO
Br
NBS = N-bromosuccinimide(supplies free radical brominefor allylic & benzylic substitution)
SClO
O= Ts-Cl (tosyl chloride)makes ROH into tosylate
N
pyridine =proton sponge
lithium aluminumhydride = LAH(very strong nucleophilic hydride)(LiAlD4 too)
sodium borohydride(nucleophylic hydride)
S
S
dithiane Br2
Pd / H2
quinoline(Lindlar's cat)
Pd / H2
H3C Li
methyl lithium(very strong base)
CLi
phenyl lithium(very strong base)
Na
N N N
sodium azide(excellent nucleophile)
H3PO4 HNO3 H2O2
KH
OH
palladium &hydrogen
phosphoricacid
nitric acid
sodium nitrite
hydrogenperoxide
diboraneuse w/ alkenes
dialkylboraneuse w/ alkynes
t-butyl alcohol(use to maket-butoxide)
pyridineH2O / H3O
+
CO2carbon dioxidebromine chlorine
hydrogenchloride
hydrogenbromide
hydrogeniodide
sulfuricacid
water
sodiumhydroxide
sodiumhydrogen
sulfide
sodiumcyanide
NH3ammonia
potassium hydride(very strong base)
R = C or H
NHR2
HCCl3
HCBr3
CH2I2Zn / Cu
chloroform
bromoform
SimmonsSmith
reagent
potassium permanganateosmium tetroxide
Na
sodiummetal
3 ozonereactions
1. O3, -78oC
2. CH3SCH3
1. O3, -78oC
2. NaBH4
1. O3, -78oC
2. H2O2, HO
1. Hg(OAc)2/H2O
2. NaBH4
1. Hg(OAc)2/ROH2. NaBH4
O
O O
H
Cl
meta chloroperbenzoic acid(mCPBA)
Na / NH3
in ammonia(Birch reagent)
= py
(PCC) (Jones)
N
O O
Na
Mn
O O
OO
Os
O O
OOK
O
O O Cr
O
OO
Cr
O
OO
B
H
H
H
H AlH
H
H
H
NaLi
B
H
H
H
B
R
R
H
pre aldehydes/ketones
P
Ph
Ph
Phtriphenylphosphine
(to make Wittig salts)
Na Cl
Br
INa
NaBrCu
Salts / ionic substances
1.
2. H2O2/HO 2. Br2/CH3O
1.
2. H2O2/HO
diboraneuse w/ alkenes
B
H
H
H1.
Mgmagnesium
metal
Li
lithiummetal
Zn
zincmetal
Lewis aciAlBr3FeBr3BF3
SOCl2PBr3SO3etc.
Various metals
HgX2
(mecuric salts)
HOOH
H2
NNH2 NH
ethyleneglycol
(protect C=O)
hydrazine(Wolff-Kishner)
pyrrolidine(enamines)
O
THP(protect O-H) SHO
O
O
toluene sulfonic acid = TsOH(very strong organic acid)
sodium cyanoborohydride(nucleophylic hydride)(NaBD4 too)
BH
H
H
CNNa
cuprousbromide
For now, the structures below represent your hydrocarbon starting points to synthesize target molecules (TM) that are
specified. We will only study two free radical reactions 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.
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Allowed starting structures our main sources of carbon 1. Free radical substitution of sp3C-H bonds to form sp
C-Br bonds at the weakest C-H position and 2. Anti-Markovnikov addition to alkenes makes 1oR-Br. From these two
reactions we can make 13 R-Br molecules below.
CH4
Br Br
BrWe need to make these 1oRBr
from anti-Markovnikov free
radical addition of H-Br (ROOR)
to alkenes.
Br2/ h Br2/ h Br2/ h Br2/ h Br2/ h Br2/ h Br2/ h
H3C
BrBr
Br
Br BrBr
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 RBr
compounds: This is just
free radical substitution
at allylic sp3C-H position
of an alkene.
Br
benzylic RBr,f rom above
Br
PhBr2/ h
2 eqs.
Ph
Br
Br
Br Br Br Br
E2 reaction(twice)
NaR2N
1. 3 eqs.
2. workup
Ph
a b c a ab
b
c
c
Br
bromobenzeneis given untilaromatic chemistryis covered in 316
You will need to propose a step-by-step synthesis for each target molecule from these given structures. Every step
needs to show a reaction arrow with the appropriate reagent(s) above each arrow and the major product of each step.
This is often accomplished by using retrosynthetic thinking. You start at the target molecule (the end) and work your
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way backwards (towards the beginning), one step at a time until you reach an allowed starting material. The starting
material of each step becomes the target molecule for the next step until you reach the beginning.
1. Mechanism for free radical substitution of alkane sp3C-H bonds to form sp3C-Br bonds at weakest C-H
position
H3C
H2
CCH3 H3C
CHCH3
Br
BrBr
overall reactionh
BrH
1. initiation
BrBrh
BrBr H = 46 kcal/mole
weakest bond ruptures first
2b propagation
H3C
C
CH3
H
BrBr
H3C
C
CH3
H Br
BrH = -22 kcal/mole (overall)
BE = +46 kcal/moleBE = -68 kcal/mole
2a propagation
H3C
C
CH3
H H
Br BrH
H3C
C
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
Br
H3C
C
CH3
H
H3C
C
CH3
H Br
H3CC
CH3
H
CH3
C
H3C
H
CH
CH3
H3C CHCH3
CH3
H = -68 kcal/mole
H = -80 kcal/mole
very minor product
2. Free radical addition mechanism of H-Br alkene pi bonds(alkenes can be made from E2 or E1 reactions at this
point in course) (anti-Markovnikov addition to alkenes)
H3C
H2C
CH2
Br
H3C
HC
CH2
HBrR2O2(cat.)
h
overall reaction
1. initiation (two steps)R
O
O
R hR
O
O
R
BrH
R
O R
O
H Br
H = 40 kcal/mole
H = -23 kcal/mole
BE = +88 kcal/moleBE = -111 kcal/mole
(cat.)
reagent
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2a propagation
H3C
HC
CH2 Br
H3C
C
CH2
Br
H
H = -5 kcal/mole
BE = +63 kcal/moleBE = -68 kcal/mole
H = -15
both steps(2a + 2b)
2b propagation
H3C
C
CH2
Br
H
BrH H3C
H2C
CH2
Br Br
H = -10 kcal/mole
BE = +88 kcal/moleBE = -98 kcal/mole
Miscellaneous E2 mechanism not included above; An E2 reaction that makes carbonyl compounds (C=O)
1. PCC = pyridinium chlorochromate, (CrO3/pyridine), CrO3oxidations of alcohols (methyl, 1oand 2oROH)
without water. Steps are: 1. Cr=O addition, 2. acid/base and 3. E2 to form C=O (aldehydes and ketones).
H3C
CH2
C
H
O
H
H
Cr
O
OO H3C
CH2
C
H
O
H
H
Cr
O
O
O
H3C
CH2
C
H
O
H
Cr
O
O
O
NN H
N
H3C
CH2
C
H
O
Cr
O
O
ON H
PCC = pyridinium chlorochromate oxidationof primary alcohol to an aldehyde (no water tohydrate the carbonyl group)
primary alcohols
aldehydes
E2
CrO3oxidations of alcohols (methyl, 1oand 2oROH) without water = PCC, Cr=O addition, acid/base and E2 to
form C=O (aldehydes and ketones)
H3C
CH2
C
CH3
O
H
H
Cr
O
OO H3C
CH2
C
CH3
O
H
H
Cr
O
O
O
H3C
CH2
C
CH3
O
H
Cr
O
O
ON N H
N
H3C
CH2
C
CH3
O
Cr
O
O
O
N H
PCC = pyridinium chlorochromate oxidationof primary alcohol to an aldehyde (no water tohydrate the carbonyl group)
primary alcohols E2
ketones
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2. Jones reagent = CrO3/water/acid, CrO3oxidations of alcohols (methyl, 1oand 2oROH) with water. Steps are:
1. Cr=O addition, 2. acid/base and 3. E2 to form C=O (aldehydes and ketones) 4. hydration of C=O and repeat
reactions when the starting alcohol is a 1oalcohol (forms carboxylic acids from primary alcohols and ketones
from secondary alcohols).
H3C
CH2
C
H
O
H
H
H3C
CH2
C
H
O
Jones = CrO3/ H2O / acidprimary alcohols oxidize to carboxylic acids
(water hydrates the carbonyl group,
which oxidizes a second time )
primary alcohols
aldehydes (cont. in water)
H
O
HH
OH
H
H3C
CH2
C
H
O
H
H3C
CH2
C
H
O
H
H
O
H
H3C
CH2
C
O
H
O
H
H
H
H
O
HH3C
CH2
C
O
O
H H
H
H
O
H
H
O
H
H3C
CH2
C
O
OH
hydration ofthe aldehyde
second oxidation of the carbonyl hydrate
H
O
H
resonance
carboxylic acids
Cr
O
OOH3C
CH2
C
H
O
H
H
Cr
O
O
O
H3C
CH2
C
H
O
H
Cr
O
O
O
Cr
O
O
O
H
O
H
H
H
O
H
H
Cr
O
OO H3C
CH2
C
O
O
H
Cr
O
O
O
H
H
H3C
CH2
C
O
O
H
Cr
O
O
O
H
H
O
H
H
Cr
O
O
O
H
O
H
H
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