2
3
A.
Effects of a neighboring double bond on a CH
3 group
1) Weakens the C—H bond H
2
C
H a) Allyl C—H DH o = 87 kcal/mol b) Primary C—H DH o = 98 kcal/mol
CH
3
CH
2
H
CH
2 c) d)
Secondary C—H DH
Tertiary C—H DH o o = 94.5 kcal/mol
= 93 kcal/mol
(CH
3
)
2
CH H
(CH
3
)
3
C H
H
H
2
C
H
CH
2
H
2
C
H
CH
2
Cl
2) S
N
1 solvolysis rate is increased
H
MeOH
MeOH
-Cl
-
H
2
C H
2
C
CH
2
H
CH
2
OMe
3) pKa is greatly lowered (more acidic)
H H
H
2
C H
2
C
CH
2
H CH
2
pKa = 40
CH
3
CH
2
Cl
MeOH
No Reaction
CH
3
CH
2
H pKa = 50

B.
Delocalization explains these observations
1.
The three observations generate primary C
.
, C+, and C- respectively.
How can they be stabilized?
2.
Delocalization = resonance structures spread the p
-electrons over several
C’s. This helps stabilize the radical, carbocation, and carbanion.
H
2
C CH CH
2
H
2
C CH CH
2
H
2
C CH CH
2
H
2
C CH CH
2
H
2
C CH CH
2
H
2
C CH CH
3.
Molecular Orbital (MO) Representation of p
-orbitals a.
H
2
C=CH—CH
2 has each C sp 2 hybridized b.
One unhybridized p-orbital for each carbon c.
C—C bonds are equivalent d.
Treat s
-bonds as in Lewis structures
2

a.
MO theory needed to describe the delocalized p
-bond i.
3 p atomic orbitals must give us 3 MO’s ii.
We get a bonding (O nodes), a nonbonding (1 node), and an antibonding (2 nodes) MO iii.
Fill up these orbitals with the appropriate number of p
-electrons cation has 2 e-, radical has 3 e-, anion has 4 e- (B.O. = 1 for all)

H
2
C
0.5+
4) Resonance Indicates that the charge or radical is found only on terminal C’s a.
MO theory shows differences only in p nb eb.
That orbital has a node at C
2 c.
MO theory agrees with the resonance hybrids
H H H
C C C
CH
2
0.5+
H
2
0.5
C CH
0.5
2
H
2
C
0.5-
CH
0.5-
2
A.
Radical Allylic Halogenation
1) We would expect normal chlorination of propene (and we get it)
H Cl Cl
Cl
2
H
2
C C CH
2
CHCH
3
CH
3
2) But, at low Cl
2
H concentrations, a radical reaction occurs for allylic groups
H
Cl
2
H
2
C C H
2
C C dilute
CH
3
CH
2
Cl

O
O
N Br
3) NBS Radical Bromination is the most common way to do this reaction a) NBS = N-Bromosuccinimide
CCl
4
HBr b) NBS gives a very low concentration of Br
2
O in CCl
NBS
4
CCl
4
N H + Br
2
HBr
+ HBr
Br
O c) Light (hv) or a radical initiator (RO
.
) needed to start the reaction
R R d) Allyl Radical resonance forms make two products possible.
Br i.
Symmetric allyl groups give only one product
R R R R
Br
2 R R
H
CH
3
CH
2
H H
H ii.
Assymetric allyl groups give product mixtures
Br CH
3
CH
2
H CH
3
CH
2
H
Br
H
H
Br
2
CH
3
CH
2
Br
70%
H
H
+
CH
3
CH
2
30%
Br
H
H

B.
Nucleophilic Substitution of Allylic Halides
1.
S
N
1 is possible due to resonance stability of carbocation
CH
3
CH=CHCH
2
Cl or
CH
3
CHCH=CH
2
Cl
-Cl
CH
3
CH=CHCH
2
CH
3
CHCH=CH
2
H
2
O
CH
3
CH=CHCH
2
OH minor
+
CH
3
CHCH=CH
2 major
OH
2.
At room temperature, major product is the less stable alkene a) Kinetic Control = less stable isomer is formed faster through a lower energy transition state b) Thermodynamic Control = most stable isomer is formed due to overall lower energy of products
3.
Even though the less stable isomer is formed first under room temperature conditions (kinetic control), we can get the most stable product by heating the reaction for long periods (thermodynamic control).
a) H + conditions mean the reaction is reversible b) Allylic Alcohol reverts back to carbocation c) Eventually, we will form the most stable product

f) Unstable isomer has lower Ea because its transition state is the most stable carbocation

4.
S
N
2 Reactions of allylic halides are faster than for alkyl halides
1.
Overlap of the p
-bond with the p-orbital in the transition state stabilizes the T.S. for allylic reactants
2.
Example:
H
2
C=CHCH
2
Cl + I
-
CH
3
CH
2
CH
2
Cl + I
acetone acetone
H
2
C=CHCH
2
I + Cl
-
rate = 73
CH
3
CH
2
CH
2
I + Cl
-
rate = 1

C.
Allylic Organometallic Reagents
1.
Synthesis of Allylic Organometallic Reagents a) Reaction with an Alkylithium reagent i.
Alkyl anion is more basic (less acidic) than allylic group ii.
Alkyl anion deprotonates allylic group
TMEDA
H
2
C=CHCH
3
+ CH
3
CH
2
CH
2
CH
2
Li H
2
C=CHCH
2
Li + CH
3
CH
2
CH
2
CH
3
N N iii.
TMEDA (like HMPA) is a good polar aprotic solvent b) Normal Grignard Formation
THF
H
2
C=CHCH
2
Br + Mg H
2
C=CHCH
2
MgBr
2.
Allylic Organometallic Reagents work as good 3C nucleophiles
H
2
C=CHCH
2
MgBr +
O
H
2
C=CHCH
2
OH
C
CH
3
CH
3

A.
Naming Dienes (compounds with two double bonds)
1) Allenes have both double bonds on the same C. H
2
C=C=CH
2 a) Two p
-bonds on the same C must be perpendicular to each other b) The central C is sp hybridized c) Nonconjugated
2) Conjugated = two double bonds separated by a C—C single bond a) All C’s are sp 2 hybridized with a single p-orbital left b) The 4 p-orbitals can overlap
H
2
C=CH—CH=CH
2

3) Naming is similar to alkenes, except diene at the end
H
3
C H
H
H
H H trans -1,3-pentadiene
H CH
2
CH
3
H
3
C
H
H H cis -2trans -4-heptadiene
H
3
C H
H
Br
H H
(Z)-4-bromo-1,3-pentadiene trans -1,4-heptadiene
1,3-cyclohexadiene
1,4-cycloheptadiene
B.
Stability of Dienes
1) Conjugated dienes are more stable than nonconjugated dienes:
D
H o
CH
3
CH
2
CH=CH
2
H
2
, Pt
CH
3
CH
2
CH
2
CH
3
D
H o = -30 kcal/mol
H
2
C=CHCH
2
CH
2
CH=CH
2
H
2
, Pt
H
2
C=CH CH=CH
2
H
2
, Pt
CH
3
(CH
2
)
4
CH
3
CH
2
CH
2
CH
3
CH
3
D
H o = -60 kcal/mol
D
H o = -57 kcal/mol

2) Conjugated Dienes are stabilized by resonance
3) Resonance Energy = 3.5 kcal/mol

C.
Overlap of p
-bonds Creates Conjugation
1) Four p-orbitals overlap over the entire molecule
2) Rotational barrier around the single bond is raised. Planarity is needed for conjugation.
a) The scis molecule has its p
-bonds on the same side of C—C b) The strans molecule has its p
-bonds on the opposite sides of C—C c)
The “s” tells us there is a single bond between the C=C’s d) The strans molecule is 3 kcal/mol more stable due to steric interactions in the scis molecule

a) 4 p-orbitals give 4 MO’s b) Number of nodes increases up c) 4 p electrons fill p
1 and p
2 d) p
-Bond Order = 2

D.
Electrophilic Attack on Conjugated Dienes
1.
Conjugated dienes are more reactive that alkenes
CH
2
CH
2 or
CH=CH
2
H
2
C=CH CH=CH
2
HCl
CH
3
CH CH=CH
2
CH
3
Cl
CH CH=CH
2
+
80%
CH
3
CH=CH CH
2
Cl 20%
CH
3
CH=CH CH
2 a) 3-chloro-1-butene is the normal Markovnikov product (1,2 addition) b) 1-chloro-2-butene is the result of 1,4 Addition to a butadiene
2) It is common for dienes to give both 1,2 and 1,4 addition products
H
2
C=CH CH=CH
2
Br
2
CCl
4
BrCH
2
Br
CH CH=CH
2
+ BrCH
2
CH=CH CH
2
Br