Alkenes, Syntheses

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Alkenes
C2H4 ethylene
CnH2n
“unsaturated” hydrocarbons
H
H
C C
H
H
Functional group = carbon-carbon double bond
sp2 hybridization => flat, 120o bond angles
σ bond & π bond => H2C=CH2
No rotation about double bond!
C3H6 propylene
C4H8 butylenes
CH3CH=CH2
CH3CH2CH=CH2
α-butylene
1-butene
CH3CH=CHCH3
β-butylene
2-butene
CH3
CH3C=CH2
isobutylene
2-methylpropene
there are two 2-butenes:
H
H
C C
H3C
CH3
cis-2-butene
“geometric isomers”
H
CH3
C C
H3C
H
trans-2-butene
(diastereomers)
C=C are called “vinyl” carbons
If either vinyl carbon is bonded to two equivalent groups,
then no geometric isomerism exists.
CH3CH=CHCH3
yes
(CH3)2C=CHCH3
no
CH3CH2CH=CH2
no
CH3
CH3CH=CCH2CH3
yes
Confusion about the use of cis- and trans-. According to
IUPAC rules it refers to the parent chain.
H3 C
CH2CH3
C C
H
CH3
H3 C
Cl
????????
C C
H
“cis-”
Br
E/Z system is now recommended by IUPAC for the
designation of geometric isomerism.
1. Use the sequence rules to assign the higher priority * to
the two groups attached to each vinyl carbon.
2.
*
*
*
*
(Z)- “zusammen”
together
(E)- “entgegen”
opposite
*
H3 C
*
CH2CH3
C C
H
CH3
*
H3 C
Cl
(E)-
C C
H
(Z)-
Br
*
Nomenclature, alkenes:
1. Parent chain = longest continuous carbon chain that
contains the C=C.
alkane => change –ane to –ene
prefix a locant for the carbon-carbon double bond using
the principle of lower number.
2. Etc.
3. If a geometric isomer, use E/Z (or cis/trans) to indicate
which isomer it is.
*
H3 C
*
CH2CH3
C C
H
CH3
*
H3 C
(3-methyl-cis-2-pentene)
Cl
C C
H
(Z)-3-methyl-2-pentene
(E)-1-bromo-1-chloropropene
Br
*
CH3
CH3CH2
CHCH2CH3
\
/
C=C
3-ethyl-5-methyl-3-heptene
/
\
CH3CH2
H
(not a geometric isomer)
-ol takes precedence over –ene
CH2=CHCH2-OH
2-propen-1-ol
CH3CHCH=CH2
OH
3-buten-2-ol
Physical properties:
non-polar or weakly polar
no hydrogen bonding
relatively low mp/bp
~ alkanes
water insoluble
Importance:
common group in biological molecules
starting material for synthesis of many plastics
Syntheses, alkenes:
1. dehydrohalogenation of alkyl halides
2. dehydration of alcohols
3. dehalogenation of vicinal dihalide
4. (later)
3. dehalogenation of vicinal dihalides
|
|
—C—C—
|
|
X
X
+ Zn 
| |
— C = C — + ZnX2
eg.
CH3CH2CHCH2 + Zn
Br Br
 CH3CH2CH=CH2 + ZnBr2
Not generally useful as vicinal dihalides are usually made
from alkenes. May be used to “protect” a carbon-carbon
double bond.
1. dehydrohalogenation of alkyl halides
|
|
— C — C — + KOH(alc.) 
|
|
H X
a)
b)
c)
d)
e)
f)
g)
h)
RX: 3o > 2o > 1o
no rearragement 
may yield mixtures 
Saytzeff orientation
element effect
isotope effect
rate = k [RX] [KOH]
Mechanism = E2
| |
—C=C—
+ KX + H2O
rate = k [RX] [KOH]
=> both RX & KOH in RDS
R-I > R-Br > R-Cl “element effect”
=> C—X broken in RDS
R-H > R-D “isotope effect”
=> C—H broken in RDS
 Concerted reaction: both the C—X and C—H bonds are
broken in the rate determining step.
Mechanism = elimination, bimolecular E2
W
RDS
C
C
C
H
base:
One step! “Concerted” reaction.
C
+ H:base + :W
CH3CHCH3
Br
+ KOH(alc) 
CH3CH=CH2
isopropyl bromide
propylene
CH3CH2CH2CH2-Br
+ KOH(alc)
n-butyl bromide
CH3CH2CHCH3
Br
sec-butyl bromide

CH3CH2CH=CH2
1-butene
+ KOH(alc) 
CH3CH2CH=CH2
1-butene 19%
+
CH3CH=CHCH3
2-butene 81%
Problem 8.6. What akyl halide (if any) would yield each of the following
pure alkenes upon dehydrohalogenation by strong base?
isobutylene 
CH3
KOH(alc) + CH3CCH3
X
or
CH3
CH3CHCH2-X
1-pentene  KOH(alc) + CH3CH2CH2CH2CH2-X
note: CH3CH2CH2CHCH3 would yield a mixture! 
X
2-pentene  KOH(alc) + CH3CH2CHCH2CH3
X
2-methyl-2-butene  KOH(alc) + NONE!
KOH(alc)
?????????
CH3
CH3CH CCH3
PURE!
2-methyl-2-butene
CH3
CH3CHCHCH3
X
KOH(alc)
CH3
CH3CH2CCH3
X
KOH(alc)
CH3
H2C CHCHCH3
CH3
CH3CH2C CH2
+
CH3
CH3CH CCH3
+
CH3
CH3CH CCH3
No alkyl halide will yield 2-methyl-2-butene as the only product of dehydrohalogenation
Saytzeff orientation:
Ease of formation of alkenes:
R2C=CR2 > R2C=CHR > R2C=CH2, RCH=CHR > RCH=CH2 > CH2=CH2
Stability of alkenes:
R2C=CR2 > R2C=CHR > R2C=CH2, RCH=CHR > RCH=CH2 > CH2=CH2
CH3CH2CHCH3 +
Br
sec-butyl bromide
KOH(alc) 
CH3CH2CH=CH2
1-butene 19%
+
CH3CH=CHCH3
2-butene
81%
RCH=CH2
RCH=CHR
KOH (alc)
CH3CH2CH2CHBrCH3  CH3CH2CH=CHCH3 + CH3CH2CH2CH=CH2
71%
29%
CH3
CH3CH2CCH3 + KOH(alc) 
Br
CH3
CH3
CH3CH=CCH3 + CH3CH2C=CH2
71%
29%
CH3
CH3
CH3
CH3CHCHCH3 + KOH(alc)  CH2=CHCHCH3 + CH3CH=CCH3
Br
major product
Order of reactivity in E2: 3o > 2o > 1o
CH3CH2-X

CH3CHCH3 
X
CH3
CH3CCH3 
X
CH2=CH2
3 adj. H’s
CH3CH=CH2
6 adj. H’s & more stable
alkene
CH3
CH=CCH3
9 adj. H’s & most stable
alkene
Elimination unimolecular E1
1)
2)
C C
H W
C C
H
RDS
-H
C C
H
C C
+ :W
Elimination, unimolecular
a)
b)
c)
d)
e)
f)
g)
E1
RX: 3o > 2o > 1o
rearragement possible 
may yield mixtures

Saytzeff orientation
element effect
no isotope effect
rate = k [RW]
E1:
Rate = k [RW] => only RW involved in RDS
R-I > R-Br > R-Cl “element effect” =>
C—X is broken in RDS
R-H  R-D
no “isotope effect” =>
C—H is not broken in the RDS
Elimination, unimolecular
a)
b)
c)
d)
e)
f)
g)
RX: 3o > 2o > 1o
rearragement possible
may yield mixtures
Saytzeff orientation
element effect
no isotope effect
rate = k [RW]
E1
carbocation
“
C—W broken in RDS
C—H not broken in RDS
only R-W in RDS
alkyl halide + base  substitution or elimination?
X
C C
H
SN2
:Z
E2
R-X +
base
 ????????
1) If strong, conc. base:
CH3 > 1o => SN2  R-Z
3o > 2o
=> E2

 alkene(s)
2) If weak, dilute base:
3o > 2o > 1o => SN1 and E1  R-Z + alkene(s) 
3) If KOH(alc.)
3o > 2o > 1o
=> E2  alkene(s) 
SN2
CH3CH2CH2-Br
+
NaOCH3

CH3CH2CH2-O-CH3
1o
CH3
CH3CCH3
Br
3o
+ NaOCH3
CH3CH2CH2-Br
+
E2
CH3
 CH3C=CH2
+ HOCH3
E2
KOH(alc)  CH3CH=CH2
CH3
CH3CHCHCH3 +
Br
dilute OH-


CH3
CH3CHCHCH3

 [1,2-H]

CH3
CH3CCH2CH3

CH3
CH3CCH2CH3
OH
SN1
+
CH3
CH3C=CHCH2
E1
+
CH3
CH2=CCH2CH3
E1
2. dehydration of alcohols:
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|
| |
— C — C — acid, heat  — C = C — + H2O
|
|
H
OH
a) ROH: 3o > 2o > 1o
b) acid is a catalyst
c) rearrangements are possible 
d) mixtures are possible 
e) Saytzeff
f) mechanism is E1
note: reaction #3 for alcohols!
Mechanism for dehydration of an alcohol = E1
1)
C C
H OH
+ H
C C
H OH2
RDS
2)
C C
H OH2
C C
H
+ H2O
3)
C C
H
C C
+ H
CH3CH2-OH
CH3
CH3CCH3
OH
+ 95% H2SO4, 170oC  CH2=CH2
+ 20% H2SO4, 85-90oC 
CH3
CH3C=CH2
CH3CH2CHCH3 + 60% H2SO4, 100oC  CH3CH=CHCH3
OH
+ CH3CH2CH=CH2
CH3CH2CH2CH2-OH + H+, 140oC 
rearrangement!

CH3CH2CH=CH2
+ CH3CH=CHCH3
Synthesis of 1-butene from 1-butanol:
CH3CH2CH2CH2-OH + HBr  CH3CH2CH2CH2-Br
SN2
E2  KOH(alc)
CH3CH2CH=CH2
only!
To avoid the rearrangement in the dehydration of the alcohol
the alcohol is first converted into an alkyl halide.
Syntheses, alkenes:
1. dehydrohalogenation of alkyl halides
E2
2. dehydration of alcohols
E1
3. dehalogenation of vicinal dihalide
4. (later)
H+
R-OH
R-X
KOH
(alc.)
Alkene
Zn
vicinal
dihalide
Alkyl halides:
nomenclature
syntheses:
1. from alcohols
a) HX
b) PX3
2. halogenation of certain alkanes
3.
4.
5. halide exchange for iodide
reactions:
1. nucleophilic substitution
2. dehydrohalgenation
3. formation of Grignard reagent
4. reduction
Alcohols:
nomenclature
syntheses
later
reactions
1. HX
2. PX3
3. dehydration
4. as acids
5. ester formation
6. oxidation
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