Transcript Alkenes

Alkenes
CnH2n
© E.V. Blackburn, 2011
Alkenes
• contain carbon - carbon double bonds
• called unsaturated hydrocarbons
• also known as olefins
(oleum, latin, oil; facere, latin, make)
• CnH2n
• CnH2n + H2  CnH2n+2 - one degree of unsaturation
© E.V. Blackburn, 2011
Degree of unsaturation
Degree of unsaturation = (2NC - NX + NN – NH + 2)/2
NC = number of carbons
NX = number of halogens
NN = number of nitrogens
NH = number of hydrogens
NH 2
H
Cl
O
© E.V. Blackburn, 2011
Nomenclature – the E/Z system
1. To name alkenes, select the longest carbon
chain which includes the carbons of the double
bond. Remove the -ane suffix from the name of the
alkane which corresponds to this chain. Add the
suffix -ene.
a derivative of octene not nonane
© E.V. Blackburn, 2011
Nomenclature – the E/Z system
2. Number this chain so that the first carbon of the
double bond has the lowest number possible.
4
2
3
1
4-butyl-2-octene
© E.V. Blackburn, 2011
Nomenclature – the E/Z system
H3C
H
C C
H
CH3
trans
H3C
CH3
C C
H
H
cis
diastereoisomers
geometric isomers
© E.V. Blackburn, 2011
cis/trans problems
H 3C
Br
C C
H
Cl
This molecule is a 1-bromo-1-chloropropene
but is it cis or trans!
© E.V. Blackburn, 2011
Nomenclature – the E/Z system
H 3C
Br
C C
H
Cl
(Z)-1-bromo-1-chloropropene
• use the Cahn-Ingold-Prelog system to assign priorities to
the two groups on each carbon of the double bond.
• then compare the relative positions of the groups of
higher priority on these two carbons.
• if the two groups are on the same side, the compound
has the Z configuration (zusammen, German, together).
• if the two groups are on opposite sides, the compound
has the E configuration (entgegen, German, across).
© E.V. Blackburn, 2011
E-Z designations
Br
H3CH2C
OCH 3
CH3
CH2Br
CH2CH2CH3
H3C
H3CH2C
CH(CH 3)2
C(CH3)3
© E.V. Blackburn, 2011
Relative stabilities of alkenes
• Cis isomers are generally less stable than trans
isomers due to strain caused by crowding of the two
alkyl groups on the same side of the double bond
• Stabilities can be compared by measuring heats of
hydrogenation of alkenes.
H3C
CH3
H
H
H
CH3
H3C
H
H2
catalyst
H2
catalyst
CH3CH2CH2CH3
H = -119.7 kJ/mol
CH3CH2CH2CH3
H = -115.5 kJ/mol
© E.V. Blackburn, 2011
Overall relative stabilities of
alkenes
H3C
CH3
>
H3C
CH3
H3C
H3C
C2H5
H
>
H
C2H5
C2H5
H
H
CH2CH2CH2CH3
H
H
>
>
C 2H 5
H
C 2H 5
H
© E.V. Blackburn, 2011
Synthesis of alkenes by
elimination reactions
X
base
- HX
OH
H+/
dehydrohalogenation:
H
dehydration:
H
- H2O
© E.V. Blackburn, 2011
Dehydrohalogenation of alkyl
halides
H
C C
X
+ KOH
Reactivity: RX
C2H5OH
+ KX + H2O
3o > 2o > 1o
a 1,2 elimination reaction
© E.V. Blackburn, 2011
Dehydrohalogenation
H
C C
X
Br
+ KOH
C2H5OH
C2H5ONa
o
C2H5OH, 55 C
Br
+ H2O
+ NaBr + C 2H5OH
79%
C2H5ONa
C2H5OH, 55 oC
+ KX
+ NaBr + C 2H5OH
91%
© E.V. Blackburn, 2011
Alkoxide ions – bases used in
dehydrohalogenation
CH3CH2OH
Na
CH3CH2O- Na +
sodium ethoxide
(CH3)2CHOH
Al
((CH3)2CHO -)3 Al 3+
aluminum isopropoxide
(CH3)3COH
K
(CH3)3CO- K+
potassium t-butoxide
© E.V. Blackburn, 2011
Dehydrohalogenation of alkyl
halides
CH3CH2CH2Cl
CH3CH2CH2CH2Cl
KOH
C2H5OH
KOH
C2H5OH
- no rearrangement
KOH
CH3CH2CHCH 3
C2H5OH
Cl
CH3CH=CH 2
CH3CH2CH=CH 2
CH3CH=CHCH 3 +
80%
CH3CH2CH=CH 2
20%
© E.V. Blackburn, 2011
The mechanism
In the presence of a strong base, the reaction follows
second order kinetics:
rate = k[RX][B-]
However, with weak bases at low concentrations and as
we move from a primary halide to a secondary and a
tertiary, the reaction becomes first order.
There are two mechanisms for this elimination: E1 and
E2.
© E.V. Blackburn, 2011
E2 mechanism
X
C C
H
:B
X
C C
H
B
-
+ HB + X -
© E.V. Blackburn, 2011
E1 mechanism
X
C C
H
+
C C
H :B
+
C C
H
+ X
-
slow
+ HB
fast
© E.V. Blackburn, 2011
Evidence for the E1
mechanism
• Follows first order kinetics
• Same structural effects on reactivity as for SN1
reactions - 3 > 2 > 1
• Rearrangements can occur indicative of the
formation of carbocations
© E.V. Blackburn, 2011
Evidence for the E2
mechanism
• The reaction follows second order kinetics
• There are no rearrangements
• There is a large deuterium isotope effect
• There is an anti periplanar geometry requirement
© E.V. Blackburn, 2011
Isotope effects
A difference in rate due to a difference in the isotope
present in the reaction system is called an isotope
effect.
© E.V. Blackburn, 2011
Isotope effects
If an atom is less strongly bonded in the transition state
than in the starting material, the reaction involving the
heavier isotope will proceed more slowly.
H
C H + Z
k
C H Z
C + HZ
The isotopes of hydrogen have the greatest mass
differences. Deuterium has twice and tritium three times
the mass of protium. Therefore deuterium and tritium
isotope effects are the largest and easiest to determine.
© E.V. Blackburn, 2011
Primary isotope effects
These effects are due to breaking the bond to the
isotope.
C H + Z
C D + Z
kH
kD
C H Z
C + HZ
C D Z
C + DZ
kH
D = 5-8
k
Thus the reaction with protium is 5 to 8 times faster than
the reaction with deuterium.
© E.V. Blackburn, 2011
Evidence for the E2
mechanism - a large isotope
effect
kH
CH3CHCH 3
NaOEt
Br
kD
CD3CHCD 3
NaOEt
Br
CH3CH=CH 2
CD3CH=CD 2
kH/kD = 7
© E.V. Blackburn, 2011
Further evidence for the E2
mechanism
X
C C
H
:B
X
C C
H
B
-
+ HB + X -
RI > RBr > RCl > RF
© E.V. Blackburn, 2011
Orientation and reactivity
KOH
CH3CH2CHCH3
C2H5OH
Cl
CH3CH=CHCH3 + CH3CH2CH=CH 2
80%
20%
The ease of alkene formation follows the sequence:R2C=CR2 > R2C=CHR > R2C=CH2, RHC=CHR > RHC=CH2
This is also the order of alkene stability. Therefore the
more stable the alkene formed, the faster it is formed.
Why?
© E.V. Blackburn, 2011
Orientation and reactivity
Let’s look at the transition state for the reaction:
X
C C
H
:B
X
C C
H
+ HB + X B
-
The double bond is partially formed in the transition state
and therefore the transition state resembles an alkene.
Thus the factors which stabilize alkenes will stabilize this
nascent alkene.
A Zaitsev elimination.
© E.V. Blackburn, 2011
anti elimination
:B
X
H
B:
H
X
© E.V. Blackburn, 2011
anti elimination
anti
H3C
H
Cl
H
H
anti
CH(CH 3)2
KOH
?
H
H
neomenthyl
chloride
+
© E.V. Blackburn, 2011
anti elimination
H
H 3C
Cl
H
H
H
KOH
H
?
CH(CH 3)2
menthyl
chloride
© E.V. Blackburn, 2011
Formation of the less substituted
alkene
Dehydrohalogenation using a bulky base favours the
formation of the less substituted alkene:
CH3
o
75
C
(CH3)3CO + CH3CH2-C-Br
(CH3)3COH
CH3
CH3
H3CHC C
CH3
+ H3CH2CC
CH3
27.5%
72.5%
CH2
© E.V. Blackburn, 2011
Substitution vs elimination
SN2 v E2
X
H
Nu:
substitution
X
X
H
Nu:
SN2
elimination
H
:Nu
E2
© E.V. Blackburn, 2011
Substitution vs elimination
SN1 v E1
X
C C
H
+
C C
H
-
+ X
Nu
SN1
Nu: +
H
H
:Nu
E1
© E.V. Blackburn, 2011
Substitution vs elimination
RX + -OR'
1o
RX + 'R C C
1o
RX + CN
1o
R-O-R' + X'R C C R + X-
SN2
RCN + X-
elimination
RX = 1o
2o
3o
substitution
© E.V. Blackburn, 2011
Dehydration of alcohols
H
C C
OH
acid
+ H 2O

H2SO4, Al 2O3
or H3PO4
CH3
H3C C OH
CH3
H3C
H
H3C
H
+ H2O
© E.V. Blackburn, 2011
Dehydration of alcohols - the
mechanism
1. (CH3)3C-OH + H
+
2. (CH3)3C-O-H
H
H3C
3.
H3C
+
H
C H
H
+
(CH3)3C-O-H
H
+
(CH3)3C+ + H2O
H3C
H
H3C
H
+
+ ROH2
ROH
© E.V. Blackburn, 2011
Dehydration of alcohols orientation
CH3CH2CH2CH2OH
CH3
CH3CH2CHCH 2OH
H+
+
H
CH3CH=CHCH 3
CH3
CH3CH=CCH 3
© E.V. Blackburn, 2011
The Zaitsev product
predominates
© E.V. Blackburn, 2011
The Zaitsev product
predominates
The transition state explains the orientation:
© E.V. Blackburn, 2011
Dehalogenation of vicinal
dihalides
C C
X X
Zn

+ ZnX 2
© E.V. Blackburn, 2011
Hydrogenation of alkynes
R
H2
H
H
Lindlar's
catalyst
R
R
Na or Li
H
R
NH 3
R
H
R
Lindlar catalyst:
Pd/CaCO 3/Pb(OAc) 2/quinoline
© E.V. Blackburn, 2011
Synthesis of alkynes by
elimination reactions
H H
C C
X X
H NaNH 2
KOH
alcohol
CH3CH=CH 2
X
Br2
CH3CHBrCH 2Br
C C
CH3CHBrCH 2Br
NaNH 2
CH3-C C H
© E.V. Blackburn, 2011
1.
H2, Pd/CaCO3
quinoline
Na, NH3, -78oC
H H
H
H
© E.V. Blackburn, 2011