Alkenes
• Contain at least one C=C double bond
• General formula: CnH2n (like cycloalkanes)
• Each carbon atom in a C=C double bond is sp2 hybridized
 bond
 bonds
E
• The double bond consists of a  bond and a  bond
•  bond from head-on overlap of sp2 orbitals
•  bond from side-on overlap of p orbitals
A  bond is stronger than a  bond.
 bond
C
 bond
• Rotating a double bond requires breaking the  bond
• NO FREE ROTATION at room temperature
C
Alkene Nomenclature (Naming)
2-hexene
3-butyl-2,4-hexadiene
1. Parent chain = longest chain that includes the double bond(s)
2. The double bonds have priority and must have the lowest
3.
4.
5.
6.
7.
number(s) possible
The first C atom in the C=C bond indicates the double bond’s
location (or number in naming)
Name, number, & alphabetize substituents as usual
Replace –ane ending with –ene ending
– Two double bonds: -diene; three double bonds: -triene
Put double bond number in front of entire root name
(i.e. 2-pentene indicates the double bond starts on carbon 2)
Cyclic alkenes: number the atoms in the ring starting with the
double bond
Naming Practice
4-ethyl-3,5-dimethyl-2-heptene
3,4-dimethyl-1,3-pentadiene
2,4-hexadiene
4-isopropyl-3,5-dimethyl1,3,5-heptatriene
Cis-trans isomerism in alkenes
• Substituents will stay on the same or opposite sides of
the double bond (no C=C bond rotation)
H3C
CH3
C
H
C
H
cis- 2-butene
(same side)
X
H3C
H
C
H
C
CH3
trans- 2-butene
(opposite sides)
• For cis-trans isomerism, each C in the double bond must have
2 different substituents attached (i.e. a C and a H, etc.)
• Determining cis or trans: follow the parent chain through the
double bond
cis
cis-3-methyl-2-heptene
Cis/Trans Naming Practice
trans-4-ethyl-3,5-dimethyl2-heptene
Both CH3’s
No Cis or Trans
Both H’s
3,4-dimethyl-1,3-pentadiene
trans,trans-2,4-hexadiene
No Cis or Trans
Both H’s
trans,trans-4-isopropyl-3,5dimethyl-1,3,5-heptatriene
Important Common Names
Ethylene
H
Propylene
H
H
C
H
C
C
Allyl (branch)
HC
CH2
i.e. polyvinyl chloride = PVC pipe
CH3
i.e. polypropylene = milk jugs
Vinyl (branch)
R
C
H
H
i.e. polyethylene = plastic bags
HC
H
R
CH2
CH2
i.e. diallyllysergamide
= derivative of LSD
Arranging many double bonds
• Cumulated
C=C double bonds all in a row: C=C=C=C
• Conjugated
Single and double bonds alternate: −C=C−C=C−C=C−
• Isolated
>1 single bond between double bonds: −C=C−C−C=C−C−C=C−
Lycopene
11 conjugated double bonds
2 isolated double bonds
Reactions of Alkenes
• Alkanes – substitution reactions
– R-H + A-B  R-A + H-B
– R = “residue”, a generic alkyl group
• Alkenes – addition reactions

H
H
C
C

 bond is electron-rich
C
C
+ A-B 
C
C
A B
Thermodynamics:
Hrxn = bonds broken – bonds formed
= ( bond +  bond) – ( bond +  bond)
Exothermic reaction
H
H
Alkene Addition Reactions
C
C
+ A-B 
C
C
A
B
Reaction
A-B
Hydrohalogenation
(addition of H-X, X = halogen)
H-F, H-Cl, H-Br, H-I
Hydration
(addition of H2O)
H-OH
Addition of halogens
Br-Br, Cl-Cl, F-F
Hydrogenation
(addition of H2)
H-H
Reactions at the Double Bond
Electrophile:
• “electron-lover”
• Electron deficient reagent (often + charged cations)
• seeks e- in reactions
Nucleophile:
• “nucleus-lover”
• electron-rich reagent (often – charged anions)
• tries to donate e- to an electrophile in reactions, forming bonds
H
H

C
C

H
H
 bond is electron-rich
• Acts as a nucleophile
• Attracts electrophiles to
form bonds
Markovnikov’s Rule
• Consider the reaction
H
H
C
C
H
+
H
H
Cl
CH3
H
H
C
C
Cl
H
CH3
1-chloropropane
1-propene
H
H
H
C
C
H
Cl
CH3
2-chloropropane
• Two products are possible
• Experimentally, only 2-chloropropane is formed
Markovnikov’s Rule:
The alkene carbon with the most H atoms gets the H
• Hydrohalogenation (H-X), hydration (H2O) of alkenes
Vladimir
Markovnikov
Why? Look at the reaction mechanism to find out...
Carbocation Stability
• Relative carbocation stability: 3° > 2° >> 1° > methyl
R
R
C
R
>
R
R
3°
C
>>
H
2°
Most stable
R
C
H
1°
>
H
C
H
H
methyl
Least stable
Why?
Electrons can “drift” into the
empty p orbital from C-H bonds
on neighboring C atoms to help
stabilize
H
Result:
The final product comes from
the most stable carbocation
H
C
H
H
H
C
H
intermediate
Mechanism of Hydrohalogenation
Step 1: Alkene  electrons attack H+, forming a carbocation
H


H
Cl
carbocation
C
H
C
H
H
H
H
C
C
H
CH3
H
H
CH3
or
H
C
C
H
+ Cl
CH3
Step 2: Cl- nucleophile forms bond with carbocation
H
H
C
H
C
H
CH3
Cl
H
H
H
C
C
H
Cl
CH3
H
H
H
C
C
Cl
H
CH3
The carbocation formed in step 1 determines the final product
Hydrohalogenation
Predict the product of the following hydrohalogenation reaction
Br
+ HBr
Anti-Markovnikov Product
Remember:
Markovnikov’s rule says
that the H (from HBr) will
bond to the alkene C with
the most H’s
OR
Br
Markovnikov Product
Hydration
Predict the product of the following hydration reaction
OH
+ H2O
Anti-Markovnikov Product
Remember:
Markovnikov’s rule says
that the H (from H2O) will
bond to the alkene C with
the most H’s
OR
OH
Markovnikov Product
Mechanism of Hydration (R + H2O)
Step 1:  bond attacks electrophilic H+ catalyst forming carbocation
H
H
H
H
C
H
C
H
C
C
H
H
Most stable carbocation
CH3
CH3
Step 2: Nucleophilic H2O attacks carbocation, forming oxonium ion

H
H
C
C
H
H
CH3
H

H
O
H

H
H
C
C
H
H
O
CH3
oxonium ion
H
Step 3: Oxonium ion loses a proton, regenerating H+ catalyst
H
H
H
C
C
H
H
O
CH3
H
H
H
H
C
C
H
OH
CH3
+
H
Markovnikov product
Mechanism of Halogen Addition (R + X2)
Step 1: Br2 is polarized by  electrons, Br+ attaches to alkene
CH3
H3 C
 Br
CH3
H3C
Br
 Br
+
Br
bromonium ion
intermediate
Step 2: Br+ blocks access to one face of the alkene
Br- adds to the other face (anti addition)
Br
CH3
Br
Br
CH3
CH3
H3C
Br
trans isomer is the only product
Halogenation
Br
+ Br2
Br
Trans isomer
Br
+ Br2
The Br’s will add to opposite sides of a RING (anti addition)
The Br’s will be forced into a trans conformation ALWAYS
Br
Mechanism of Hydrogenation (R + H2)
• Occurs in the presence of a metal catalyst (like Pt)
Step 1: H2 adsorbs to catalyst surface
H
H
H−H
Pt
+
Pt
Step 2: Both H atoms add to same face of alkene (syn addition)
H3C
CH3
CH3
H
H
CH3
H
H
Pt
Pt
cis isomer is the
only product
Hydrogenation
• Occurs in the presence of a metal catalyst (like Pt)
+ H2/Pt
cis isomer
+ H2/Pt
Both H’s will add to the same side of a RING (syn addition)
If branches are present, they will be forced into a cis conformation
Hydrogenation of alkenes
+ H2/Pt
vegetable oils
unsaturated
Contains double bonds
saturated
Contains NO double bonds
Hydrohalogenation
Determine any reactant(s) that could yield the given product of
the following hydrohalogenation reaction
Br
+ HBr
3-methyl-1-hexene
Only possible reactant for this product
+ HBr
Br
trans-3-methyl-2-hexene
Major product using M’s rule
Hydrohalogenation
Determine any reactant(s) that could yield the given product of
the following hydrohalogenation reaction
+ HBr
Br
trans-3-methyl-2-hexene
trans-3-methyl-3-hexene
All three reactants could give this
product
2-ethyl-1-pentene
Halogenation
Determine any reactant(s) that could yield the given product of
the following halogenation reaction
Br
+ Br2
trans-3,5-dimethyl-2-heptene
Only one possible reactant in this case
Br
Alkene Reaction Summary
• Hydrohalogenation (+ HX)
Markovnikov’s rule
• Hydration (+ H2O)
• Halogenation (+ X2)
Cis/trans with rings
• Hydrogenation (H2/Pt)
Polymerization of Alkenes
• Polymer: a large molecule made by linking together
small repeat units called monomers
• Polymerization mechanism: radical chain reaction
Monomer
CH2
CH2
Polymer
CH2
CH2
CH2
n
ethene
(ethylene)
H
CH2
polyethylene
H
C
CH3
CH
C
H
propene
(propylene)
CH3
CH2
CH
CH3
polypropylene
CH2
n
Alkynes
• Contain at least one CC triple bond with sp-hybridized C atoms
• Triple bond: one  bond (sp orbitals), two  bonds (p orbitals)
E
• Naming: triple bond indicated by –yne ending
H
C
C
ethyne
(acetylene)
H
HC
C
4-methyl-1-pentyne
• Reactivity: same addition reactions as alkenes
• Use 2 equivalents of addition reagent (i.e. + 2HCl)
• Use Markovnikov’s rule in the same manner
Alkyne Naming Practice
3,4-dimethyl-1-pentyne
2-methyl-3-hexyne
3-methyl-1-pentyne
2,5-dimethyl-3-hexyne
Hydrohalogenation of Alkynes
Predict the product of the following hydrohalogenation reaction
+ 2HBr
?
1-pentyne
Break the reaction into two steps, adding 1 HBr each time to the multiple bond
Br
+ HBr
Br
Br
Br
+ HBr
Final Product
Markovnikov’s rule still applies…
Hydration of Alkynes
Predict the product of the following hydration reaction
+ 2H2O
?
3,3-dimethyl-1-butyne
Break the reaction into two steps, adding H2O each time to the multiple bond
+ H2O
OH
+ H2O
OH
Markovnikov’s rule still applies…
HO
OH
Final Product
Halogenation of Alkynes
Predict the product of the following halogenation reaction
+ 2Br2
?
3,3-dimethyl-1-butyne
Break the reaction into two steps, adding Br2 each time to the multiple bond
+ Br2
Br
Br
Br
Br
Br
+ Br2
Br
Br
Br
Final Product
Hydrogenation of Alkynes
Predict the product of the following hydrogenation reaction
+ 2H2
?
3,3-dimethyl-1-butyne
This reaction will simply turn the alkyne to an alkene, and then to an alkane
+ H2
+ H2
Final Product
Alkyne Reaction Practice
1.
Cl
+ 2HCl →
Cl
2.
HO
+ 2H2O →
Br
3.
5.
Br
+ 2Br2 →
Br
4.
OH
Br
+ 2H2 →
Cl
+ 2HCl →
Cl
Cl
AND
Cl