CHE-240
Unit 3
Physical and Chemical Properties and
Reactions of Alkenes and Alkynes
CHAPTER SEVEN
TERRENCE P. SHERLOCK
BURLINGTON COUNTY COLLEGE
2004
Orbital Description
•
•
•
•
•
Sigma bonds around C are sp2 hybridized.
Angles are approximately 120 degrees.
No nonbonding electrons.
Molecule is planar around the double bond.
Pi bond is formed by the sideways overlap of
parallel p orbitals perpendicular to the plane
of the molecule.
=>
Chapter 7
2
Pi Bond
• Sideways overlap of parallel p orbitals.
• No rotation is possible without breaking
the pi bond (63 kcal/mole).
• Cis isomer cannot become trans without
a chemical reaction occurring.
=>
Chapter 7
3
Elements of
Unsaturation
• A saturated hydrocarbon: CnH2n+2
• Each pi bond (and each ring) decreases the
number of H’s by two.
• Each of these is an element of unsaturation.
• To calculate: find number of H’s if it were
saturated, subtract the actual number of H’s,
then divide by 2.
=>
Chapter 7
4
Propose a Structure:
for C5H8
• First calculate the number of elements of
unsaturation.
• Remember:
 A double bond is one element of unsaturation.
 A ring is one element of unsaturation.
 A triple bond is two elements of unsaturation. =>
Chapter 7
5
Heteroatoms
• Halogens take the place of hydrogens, so
add their number to the number of H’s.
• Oxygen doesn’t change the C:H ratio, so
ignore oxygen in the formula.
• Nitrogen is trivalent, so it acts like half a
carbon.
H H
H
C C N C
H H H H
Chapter 7
=>
6
Structure for C6H7N?
• Since nitrogen counts as half a carbon,
the number of H’s if saturated is
2(6.5) + 2 = 15.
• Number of missing H’s is 15 – 7 = 8.
• Elements of unsaturation is 8 ÷ 2 = 4.
=>
Chapter 7
7
IUPAC Nomenclature
• Parent is longest chain containing the
double bond.
• -ane changes to -ene. (or -diene, -triene)
• Number the chain so that the double
bond has the lowest possible number.
• In a ring, the double bond is assumed to
be between carbon 1 and carbon 2.
=>
Chapter 7
8
Name These Alkenes
CH2
CH CH2
CH3
1-butene
CHCH2CH3
CH3
C CH CH3
CH3
H3C
2-sec-butyl-1,3-cyclohexadiene
2-methyl-2-butene
CH3
3-n-propyl-1-heptene
3-methylcyclopentene
=>
Chapter 7
9
Name these:
H
CH3
Br
C C
CH3CH2
Br
C C
H
H
trans-2-pentene
H
cis-1,2-dibromoethene
=>
Chapter 7
10
E-Z Nomenclature
• Use the Cahn-Ingold-Prelog rules to
assign priorities to groups attached to
each carbon in the double bond.
• If high priority groups are on the same
side, the name is Z (for zusammen).
• If high priority groups are on opposite
sides, the name is E (for entgegen).
=>
Chapter 7
11
Example, E-Z
1
1
H3C
Cl
C C
H
2Z
1
H
CH2
2
2
Cl
2
CH CH3
C C
H
2
1
5E
(2Z, 5E)-3,7-dichloro-2,5-octadiene
=>
Chapter 7
12
Commercial Uses:
Ethylene
=>
Chapter 7
13
Commercial Uses:
Propylene
=>
Chapter 7
14
Other Polymers
=>
Chapter 7
15
Substituent Effects
• More substituted alkenes are more stable.
H2C=CH2 < R-CH=CH2 < R-CH=CH-R < R-CH=CR2 < R2C=CR2
unsub. < monosub. < disub.
< trisub. < tetra sub.
• Alkyl group stabilizes the double bond.
• Alkene less sterically hindered.
=>
Chapter 7
16
Cycloalkene Stability
• Cis isomer more stable than trans.
• Small rings have additional ring strain.
• Must have at least 8 carbons to form a
stable trans double bond.
• For cyclodecene (and larger) trans
double bond is almost as stable as the
cis.
=>
Chapter 7
17
Alkene Synthesis
Overview
•
•
•
•
E2 dehydrohalogenation (-HX)
E1 dehydrohalogenation (-HX)
Dehalogenation of vicinal dibromides (-X2)
Dehydration of alcohols (-H2O)
=>
Chapter 7
18
Removing HX via E2
• Strong base abstracts H+ as X- leaves
from the adjacent carbon.
• Tertiary and hindered secondary alkyl
halides give good yields.
• Use a bulky base if the alkyl halide
usually forms substitution products.
=>
Chapter 7
19
Some Bulky Bases
CH3
H3C
_
CH(CH3)2
C O
N
CH3
H
tert-bu toxi de
CH(CH3)2
H3C
di i sopropyl am i n e
N
CH3
2,6-di m e th yl pyri di n e
(CH3CH2)3N :
triethylamine
=>
Chapter 7
20
Hofmann Product
• Bulky bases abstract the least hindered H+
• Least substituted alkene is major product.
H CH3
CH3 C C CH2
H Br H
_
CH3CH2O
CH3 CH3CH2
H3C
C C
C C
CH3CH2OH
CH3
H
CH3 C C CH2
H Br H
_
(C H3)3C O
CH3CH2OH
H
H3C
71%
H CH3
H
29%
CH3 CH3CH2
H3C
C C
C C
H
CH3
28%
Chapter 7
H
H3C
=>
H
72%
21
E2: Cyclohexanes
Leaving groups must be trans diaxial. =>
Chapter 7
22
E2: Vicinal Dibromides
• Remove Br2 from adjacent carbons.
• Bromines must be anti-coplanar (E2).
• Use NaI in acetone, or Zn in acetic acid.
I
-
Br
CH3
H
CH3
Br
H
H3C
H
Chapter 7
C C
CH3
H
=>
23
Removing HX via E1
•
•
•
•
Secondary or tertiary halides
Formation of carbocation intermediate
Weak nucleophile
Usually have substitution products too
=>
Chapter 7
24
Dehydration of
Alcohols
• Reversible reaction
• Use concentrated sulfuric or phosphoric
acid, remove low-boiling alkene as it
forms.
• Protonation of OH converts it to a good
leaving group, HOH
• Carbocation intermediate, like E1
• Protic solvent removes adjacent H+
=>25
Chapter 7
Dehydration
Mechanism
H
H O H
C C
H O H
O
H O S
C C
O H
_
HSO4
O
H
H O H
H
C C
C C
Chapter 7
H2O:
C C
=>
+
H3O
26
POWER POINT IMAGES FROM
“ORGANIC CHEMISTRY, 5TH EDITION”
L.G. WADE
ALL MATERIALS USED WITH PERMISSION OF AUTHOR
PRESENTATION ADAPTED FOR BURLINGTON COUNTY COLLEGE
ORGANIC CHEMISTRY COURSE
BY:
ANNALICIA POEHLER STEFANIE LAYMAN
CALY MARTIN
Chapter 7
27