Alkenes CnH - P.G.G.C.G.-11, E-Content Management...
Transcript of Alkenes CnH - P.G.G.C.G.-11, E-Content Management...
Alkenes CnH2n ―unsaturated‖ hydrocarbons
C2H4 ethylene
Functional group = carbon-carbon double bond
sp2 hybridization => flat, 120o bond angles
σ bond & π bond => H2C=CH2
No rotation about double bond!
C C
H
H H
H
C3H6 propylene CH3CH=CH2
C4H8 butylenes CH3CH2CH=CH2
α-butylene
1-butene
CH3
CH3CH=CHCH3 CH3C=CH2
β-butylene isobutylene
2-butene 2-methylpropene
there are two 2-butenes:
cis-2-butene trans-2-butene
―geometric isomers‖ (diastereomers)
C C C CH3C
H H
CH3 H3C
H CH3
H
C=C are called ―vinyl‖ carbons
If either vinyl carbon is bonded to two equivalent groups,
then no geometric isomerism exists.
CH3CH=CHCH3 CH3CH2CH=CH2
yes no
CH3
(CH3)2C=CHCH3 CH3CH=CCH2CH3
no yes
Confusion about the use of cis- and trans-. According to
IUPAC rules it refers to the parent chain.
―cis-‖
????????
C C
H
H3C CH2CH3
CH3
C C
H3C
H Br
Cl
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‖ (E)- ―entgegen‖
together opposite
C C
H
H3C CH2CH3
CH3
C C
H3C
H Br
Cl
*
* *
*
(Z)-
(E)-
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.
C C
H
H3C CH2CH3
CH3
C C
H3C
H Br
Cl
*
* *
*
(Z)-3-methyl-2-pentene
(3-methyl-cis-2-pentene)
(E)-1-bromo-1-chloropropene
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 3-buten-2-ol
OH
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
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— C — C — + Zn — C = C — + ZnX2
| |
X X
eg. CH3CH2CHCH2 + Zn CH3CH2CH=CH2 + ZnBr2
Br Br
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.) — C = C — + KX + H2O
| |
H X
a) RX: 3o > 2o > 1o
b) no rearragement
c) may yield mixtures
d) Saytzeff orientation
e) element effect
f) isotope effect
g) rate = k [RX] [KOH]
h) Mechanism = E2
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
One step! ―Concerted‖ reaction.
base:
C
W
C
H
C C + H:base + :WRDS
CH3CHCH3 + KOH(alc) CH3CH=CH2
Br isopropyl bromide propylene
CH3CH2CH2CH2-Br + KOH(alc) CH3CH2CH=CH2 n-butyl bromide 1-butene
CH3CH2CHCH3 + KOH(alc) CH3CH2CH=CH2
Br 1-butene 19%
sec-butyl bromide +
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?
CH3 CH3
isobutylene KOH(alc) + CH3CCH3 or CH3CHCH2-X
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!
2-methyl-2-butene
H2C CHCHCH3
CH3
+ CH3CH CCH3
CH3
CH3CH CCH3
CH3KOH(alc)
????????? PURE!
KOH(alc)
CH3CHCHCH3
X
CH3
CH3CH2CCH3
CH3
X
KOH(alc)
CH3CH2C CH2
CH3
+ CH3CH CCH3
CH3
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 + KOH(alc) CH3CH2CH=CH2 RCH=CH2
Br 1-butene 19%
sec-butyl bromide +
CH3CH=CHCH3 RCH=CHR
2-butene 81%
KOH (alc)
CH3CH2CH2CHBrCH3 CH3CH2CH=CHCH3 + CH3CH2CH2CH=CH2
71% 29%
CH3 CH3 CH3
CH3CH2CCH3 + KOH(alc) CH3CH=CCH3 + CH3CH2C=CH2
Br 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 CH2=CH2 3 adj. H’s
CH3CHCH3 CH3CH=CH2 6 adj. H’s & more stable
X alkene
CH3 CH3
CH3CCH3 CH=CCH3 9 adj. H’s & most stable
X alkene
C C
H W
RDSC C
H
1) + :W
2) C C
H
- HC C
Elimination unimolecular E1
Elimination, unimolecular E1
a) RX: 3o > 2o > 1o
b) rearragement possible
c) may yield mixtures
d) Saytzeff orientation
e) element effect
f) no isotope effect
g) 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 E1
a) RX: 3o > 2o > 1o carbocation
b) rearragement possible ―
c) may yield mixtures
d) Saytzeff orientation
e) element effect C—W broken in RDS
f) no isotope effect C—H not broken in RDS
g) rate = k [RW] only R-W in RDS
alkyl halide + base substitution or elimination?
C C
H
X
:Z
SN2
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 E2 CH3
CH3CCH3 + NaOCH3 CH3C=CH2 + HOCH3
Br
3o
E2
CH3CH2CH2-Br + KOH(alc) CH3CH=CH2
CH3 CH3
CH3CHCHCH3 + dilute OH- CH3CCH2CH3 SN1
Br OH
CH3
+ CH3C=CHCH2 E1
CH3
CH3CHCHCH3 CH3
+ CH2=CCH2CH3 E1
[1,2-H]
CH3
CH3CCH2CH3
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
C C
H OH
+ H C C
H OH2
1)
2) C C
H OH2
RDSC C
H
+ H2O
C C
H
3) C C + H
CH3CH2-OH + 95% H2SO4, 170oC CH2=CH2
CH3 CH3
CH3CCH3 + 20% H2SO4, 85-90oC CH3C=CH2
OH
CH3CH2CHCH3 + 60% H2SO4, 100oC CH3CH=CHCH3
OH
+ CH3CH2CH=CH2
CH3CH2CH2CH2-OH + H+, 140oC CH3CH2CH=CH2
rearrangement! + 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)
R-OH
R-X Alkene
vicinal
dihalide
H+
Zn
KOH
(alc.)