Configuration
Electron
configuration
Electron
configuration of
electron
The
Atomic
element describes
an
distributed
are
its
in
orbitals.
atomic
Symbol
mass
number
electrons
how
T
A
"symbol of element
Z
atomic !
Atomic number
Mass number
The number of
How much it
number
protons
including protons electrons
This
element has
.
an
,
identify
used to
is
weighs
and neutrons
,
.
the element.
He
hydrogen positively charged,
is
just
it is
meaning
hydrogen does not
Ht
proton
=
proton
a
have
since
neutrons
.
Ho
hydrogen
H :
hydride
(negatively
#
hydronium
(positively charged hydrogen)
hydrogen)
charged
leverything
(base)
with electron pairs and
a
minus
is a
(acid)
①
=
Isotopes
A form of an element that has the same number of protons
therefore
different numbers of neutrons ,
but
weigh differently
.
the
Examples
Reactivity
C
C
carbon
Carbon-13
protons :
6
p:
6
6
n :
9
electrons : 6
e :
6
neutrons
:
Aufbau
14
·
p
g
: 6
352
i
5
6
3 p
Principal
.
.
V E.
.
Bonds
Orb
one
must
fill lower energy levels
are
the
electrons
orbital
every
in
a
sublevel
must
10)
f Cholds 14e/
,
di
-
total
f :
,
C1
17 e
can
.
many bonds
the
occupy
singly occupied
be
it needs
same
before
to
get
orbital
and
double
to
eight
their group
in
periodic table
the
their current
valence
electrons
.
the
same
orbital
must have
2p6392
3p
from
electrons
the
to
in
different
.
.
spin
.
filling
it
:
configuration
.
Ge
E C
.
.
4
V E
4
Bonds
.
11
1
1
.
1s
as
can
1525242
IV
the most outermost
in
the
Hund's rule
EC
holds
before moving to higher energy levels.
shell of an atom
The valence electrons correspond
create
ability to create bonds. One
Parli exclusion principal no more than two electrons
=
I
completely
*
C
P!
,
in
79
Valence electrons
Electron
pholds Get
,
T periods
6d"
P
*
Example
Si
2)
period in the periodic table .
for each
one
It
Aufbau
*
&Cholds
,
electron
an
Spin
*
*
to describe
shell numbers
n :
Orbitals
p
541
6
needed
Subshells
14
75
Valency is
things
shells
21&
2
different.
are
6
:
:
n
2s
6
g
same
Carbon-14
4)
5s2
the
is
of isotopes
Physical properties
C
1s
4
isotopes
(metodo de la lluvial
diagram
4 g
of
25
24
Valence electrons
Orb
.
.
:
15 25
2 p. 352
zp5
EC
.
T
V E
1
Bonds
10
Is
Br
=
13
252
IV
25
.
TV To
24
10
.
12
35
·
In
101
39
Orb
>
45 3610 4p5
T
.
.
35 e
.
1
1v
-
.
14
15
25
IV
111
44
-
Tr In TV
24
-
12
35
.
12
in
39
in
In
4S
.
In
er
en
3d
to
er
base
Dot
Lewis
How to
create
Find
the
Determine
Form
dot
lewis
a
Structure
total
number
the
central atom
.
bonds
single
of
elections.
valence
between
Move electrons around to
Examples
structure
(Usually
the
the
one
that
atoms
and
then
to
octet rule
.
adhere
the
can
place
4e
H
1 x 4
:
H
:
remaining electrons
on
the
atoms
lone pairs.
as
NHz
H
Ha
the
bonds
most
:
CH4
C=
the
create
H
C
He
Or
C
H :
:
H
N =
5 e
H
1x3
=
=
#
Se
H
or
-
8 e
Total:
H
H
Total
H28
8 e
=
HzCO
:
H
=
1x2
0
=
6
H
ze
=
H
Ge
:
-
0
-
H
1x 2
=
= 1x4
C
:
=
Zei
0
8 e
=
=
1x6
Total
CCl
2:
Cl
1 x 4
4
:
x
Ye
=
4
=
28
If
12e
:
·
32e-
:
:
Ge
=
.
electrons
valence
:
8
-H
4e
H
Total
HiNH
H
HNC
Valence electrons
H
=
1 x 1
N
=
1 x5e
2
=
1 x4
=
1
HC
5
=
=
N
:
4
10e-
CO2
Valence electrons
2
=
0
:
1
x
4
2x6e
=
=
:
4
12
:
Co
16e
The number
S
area
P
area
work
d
area
the
f
area
for
p
of valence electrons only
elements
.
area
in
the
sarea
and
Molecular shapes
Orbital
hybridisation
Electron
density
Molecular
is
region
the three
the
is
domain
electron
or
shape
.
concept of mixing atomic orbitals
the
electron
or
Electron geometry
is
are
measure
based
on
of
the
being present at
the total number of electron pairs
electron
around
space
atom
.
central
a
and
Aluminum (Al)
.
arrangement of atoms in
dimensional
Boron (B)
given point.
a
follow
the octant
rule ; can
Visual
explanation
6
·
⑧
H : Sp
H:
·
Sp
do not
valence
survive
with
electrons
Central a tom
Bonds
⑧
H:
⑨
Sp
Electron density
Hisp
6
Hybridisation (number of orbitals)
H
S
We must keep electron densities
Electron
geometry
and
Representation
Electron
density
as
their
far from each other
possible
Hybridisation
1 p 36 5 f :
:
:
possible .
molecular shapes.
Electron
2
SI3
3
S4
I rigonal
3
Sia
4
O
as
=
geometry
Angle
Molecular shaples
1800
Linear
Linear
1200
Trigono planar
Trigonal planar
774 7200
Bent
Spe
Tetrahedral
109 50
Tetrahedral
4
Spe
Tetrahedral
less than 109 50
4
Sp3
Tetrahedral
less less than 109 5.
⑨
2
C
planar
-
(loan
⑧
1/100
front
-
·
III
:
.
Trigonal pyramidal
.
back
: 1118
Bent
.
·
CH4
H
4
C
H
Sp3
tetrahedral
109 50
tetrahedral
.
+
H
Bettz
4e
H
2
SP
linear
1800
linear
3
Sp2
trigonal planar
1280
trigonal glanar
4
spe
tetrahedral
less than
4
Sp
tetrahedral
less less than 109 5
H
Be
BFg
24e-
:
F
B
:
:
Fi
NHy
Se
j
H
H
trigonal pyramidal
109 50
.
H28"
Se
·
g
.
bent
:
H
H
CHat
Ge
3
Sp2
trigonal planar
Se
4
sp3
tetrahedral
+
HC
H
trigonal planar
1200
H
-
:
H
Che
- H
H
less than 109 50
.
trigonal pyramidal
I
pairs do not
count)
Molecular
Orbital
Theory
Molecular orbital theory
is
for
method
a
the electronic structure of molecules
describing
sp
3
Sp2
hybrid
hybrid
4)
sp3
hydrid orbitals
Molecular
orbital energy
orbitals
diagram
1s
CH4
H
H
1
h
Hybridisation Sp
:
*
-
IV
T
hybridisation
Since
the
is
sp"
we
will
one
s
and
three p's.
Outer core
electrons
This
2S
---
4
L
sp3
Sp3
L
electron
15
Atomic
orbitals
24
3534
3d
4544
4448
545f
S
5
an
IV
15
H
i"i'i I
is
2s
G
H
outer core
-
1V
mix
G
4e-
I
-
1
2P
H
Hy
Ge
Carbon :
C
mechanics
.
orbitals
2.
H
quantum
using
Molecular orbitals
G S
5
75
7
d
P
of central atom
do not bond
Here
do not
we
follow aufbar principle
N
NHy
A
HH
1V
--
Hybridisation Sp3
:
:
is
34
Sigma bonds (0)
are
single bonds
sigma
bonds (1)
another
on
+
1
ii"
In
E
1V
15
15
00
40
L
3
4sp
Molecular orbitals
bonds and Pi bonds
Sigma
P:
-
3 e-
=
Atomic orbitals
Electrons from N : T
All
Hy
1
24
H
Ye
=
atom
.
are
are
The
the
strongest type of covalent chemical
bond.
They are formed by
density ofSigma
bonds
is
between
each of which
lobes
of
an
.
bonds
Electron
covalent chemical bonds
overlap
occurs
,
in
two
-O
C
T
orbital
H20
G
L
-
Pi bond
+
·
Nucleus
⑧
Electron
Signa bond
of
⑨
S
orbitals
Shape of
↑ orbitals
S orbital
Shape
of
Gas
4 sp
+
Gas8
↑
I
8
8
p
orbitals
8
0
o
is
not
overlap
to
easy
with two
break.
lobes of
an
H
representation
Overlap
o
8
- -
18
G
⑳
atom
one
on
it
atomic orbitals.
C
M
representation
t
&
and
between
Hy CH
H
M
T
bound
the nuclei
overlapping
laterally
.
Ho C
Sigma
head-on
M
.
face to
----X
-
-
8
o
⑧
-
----
~
=
O
O
o
face
overlap
=
M
bond
orbital
Identify
sigma bonds
↳
C
C
it
H
bonds
&
H
1
0
o
Single bonds
are
saturated.
bonds
are
unsaturated
Double
the
Do
and pi bonds
:
I
o
1
L
reactive
.
the following
.
Oxygen
M
o
o
L
1
1V
1
L
>
unhybridised
11
Sp
Se
:
!1
11
p orbital
24
o
-
W
11 IV
v
1
,
unhybridised
porbital
3 sp
25
V
M
loan pairs
In
3 sp2
25
-
more
L
24
energy
are
diagram of
energy
Carbon : Ge
C
HH
sp2
orbital
molecular
and
,
IV
IS
IS
molecular
Atomic orbitals
Atomic
orbitals
Molecular
orbital
orbital
HC
N:
Carbon : Ge
.
1
of
1
24
energy
4
SD
11
2S
↓
L
1
1
1r
1
.
Nitrogen
>
unhybridised
↑
:
-
11
orbitals
24
I
2 Sp
11
Te
I
SP
loan
e
pairs
~
i
M
L
IV1
2SP
IV
1S
15
↑
Atomic
Molecular
orbitals
orbitals
unhybridised
W
Atomic
Molecular
orbitals
orbitals
3
P orbitals
formal
Charge
Formal charge= valence electrons
everything you
touch
can
Bonds
counts
Loan pairs
Find
the formal charge .
N
:
Val e-touch
.
Oxygen f
c
.
jo
H
6-7 :
:
Nitrogen
H
N8
6
5
.
=
individually
counted
are
-1
8
:
·
H
:
f C .:
one
as
H
HC
H
C:
HC
H
XC :
X
f C
Carbon
4-4 : 0
=
.
:
neutral
f c
Carbon
4-5
=
.
=
1
4-3 :
=
.
H
C
+
f C
Carbon
It
H
H
H
H C C:
j:
:
C" is
a
carbocation
C
a
carbanion
..
%
H00H
6
-
8
=
-
6
2
1
=
2-
.
H
+
H
is
5
-
H
or
H
# "hydronium
1 0
=
1
HH
N
He
hydrogen
1 1
=
0
H
H:
hydride
1 2=
H
5
-
4
=
1
+
Electronegativity
Flourine
is
from left to
right
will increase
the
electronegative
most
element
-
-
-
and
from
down-up
in
the
periodic table.
.
Difference
electronegativity
in
Ionic
+1
1 %
Polar covalent
1
.8
F
3 5
O
3 0
N
4
.
.
2 5
C
.1
2
H
.
.
0 5
.
0 5
covalent
Nonpolar
7
.
.
0 8
.
In ionic
2
Joscuas
this
Memorize
C
,
Metal
rest is
Flores
No
positive ,
is
the
negative
Claras
,
Ging
Classify
the
electronegativity.
C C
2C
-
2 5-3 0
:
.
Which is delta
·
j
2-
:
:
H2
at H
slightly polar
delta +?
and
-
0
I
+
2
2+
2 5
.
:
8
Nonpolar
ones
that
8+
will
be
8-
will be the most
The
2 5-3 5
2t
.
HH
-
-
C-8
8-
:
.5
2
05
.
.
:
.
1
Polar
are
the
will
polar
have
a
2
-
or
electronegative
least
electronegative
C
If
a
molecule is
symmetrical it will be nonpolar
If it is not symmetrical ,
or
nitrogen
,
and it has
an
hallogen
,
HC
.
Es
3
mu
=
polar
1 nitrogen
OHCH2CHzCHzOH
:
1 5 74
.
polar
60 : 388
244
polar
H
N and O will
Ot
Symmetrical
Nonpolar
always be bent
. 5- 2 1 = 1 4
3
it has
.
.
↑
CH3OH
Polar
2HyCHz OH
Polar
CHy CHzCHz OH
Polar
CHy CHz CHzCHzOH
Polar
polar
nonpolar
Overall
polar
Cha
two of itself
=
:
gi
C
.
.
OH
j
Che
unless
the
same
.
Nonpolar
OH
OH
has
four hallogens will be nonpolar
CO2
:
3) : 20
H
Every carbon that
Symmetrical
Nonpolar
ratio
Pcarbons : 1 oxygen
C C
-
-
c
-
H
mmm
Polarity
C
H
oxygen,
likely will be polar
it most
↑
Chr
,
CH3CHz CH2 CH2CHz OH
=
nonpolar
2 +
Resonance Structure
set
of two
or
single polyatomic
species
including fractional
has
lowest numbers
structures
Resonance
bonding of
a
The
best
Do
not
Electrons
structures
resonance
do
the
from
more
g
that
ones
Y
C
e
If
>
they
to
where
:
↑N
H
A
negative and
>
&
1
o
---
%
8:
a
-
·
H
C
-
C
This
H
is
more
stable
C
C
H
H
c
C
H
+
-
H
3
i
C
H
.
H
H
H
H
·
O
.
H
Y
:
:
(%
>
&
C
H
wrong
&
S
C
H
-
are
·
j
:
H
next to each other
cannot be
positive
These
...
0 :
D
H
lacking
.
:
:
:
C
of formal charges.
are
=
L
xamples :
-ir
charges.
2
---
C
fractional
and
H
H
[
C
H
H
2
H
C+
H
H
H
HH
HH
H
j
H
H
H
C
i
&
H
H
H
H
S
↳
the
describe
plus charges.
not the
j:
:
8
bonds
collectively
H
O:
C
c
+
H
H
j
:..
,
:: -
H
that
structures
:
:
lif
:
more
high density
Charge separation example
:
the
Lewis
more
charge separation .
are
Electrons
a
are
H
S!
H
H
H
C
H
S
H
-
·
-
C
c
H
H
H
H
.
electronic
Alkanes
Alkanes
formula
General
All
hydrocarbones
are
alkanes
Subfix
:
Different
the carbon atoms
are
together by single bonds
held
.
Alkanes
saturated
are
compounds.
CnHente
:
made of
are
molecules
These
which
in
.
points
nonpolar molecules. They have low boiling
are
force they feel is the London dispersion forces
intermolecular
The strongest
hydrogens
and
carbons
and
.
insoluble in
are
water.
ane
structures :
u-pentane
Bond line formula
formula
Structural
-
H
H
ChzCH2CHzCHzCHe
HHHH
↑
-
C
-
C
-
-
c
-
c
-
formula
Condensed
c
I
111
I
H
HHH
H
-
H
Prefixes
1. meth-
6-hexa-
eth-
7-hepta-
3-prop-
8- Octa-
2-
4
-
5-
but-
9
pent-
10
Example
-
-
nona-
deca-
:
CH4
CHyCHy
Chy CHzCHe
Methane
Ethane
Propane
The
Butane
more
added
Pentane
,
the
carbons
higher the
boiling point is
,
because
of the molecular
weight.
Substitutions
Find
the
longest chain
Alkyl group is
subfix -y/
alkane
an
all
,
the remainings
missing
are
called
substitutions
.
hydrogen
one
.
:
Example
3
Bromo
:
Br
.
4
-
2
·
I
4
5
2
3-methyl pentance
i
ethyl
I
5
methyl butane
3-ethyl pentane
2 bromo
1
I
Br
-
-
2
1
3
1
-
-
1.
)
4
2
-
Br
G
4
23
-
5
3
↑
2
3
I
&
2
.
2-dimethyl propane
2
. 3
Br
,
4-trimethyl
1-bromo propane
hexane
2
3
2
Isobromo propane
OH
OH
&
"
1- propano
1-bromo ethane
~
,
Naming rules
Find
the
longest
Find
all
the substitutions
.
Start
counting
.
chain
where
there is
State the number of the
Write it in
If there is
alphabetical
a
more
carbon
electronegativity.
where
the substitution
is
order.
repeated substitution
,
we
put them
together
.
EJ :
1,
C-bromo propane
2-dimethyl
Isopropyl alcohol
2-propanol
Special alkyls
to remember
Iso
when
used
is
-
a
hydrogen
from
taken
is
the second
to
carbon
attach
.
group
a
CH3CHCHy
I
V
n-butyl
isobutyl
isopropyl
10
20
10
The attachment
It
first
to
is
the
on
is
attatch
one
carbon
sec-butyl
fertbutyl
20
zo
It is attached
2 because it
It is attached
to two carbons
has two carbon
to 3 carbons
attachment
carbon
Examples
:
isopropyl
6
-
L
2
"
↑
L
n-butyl
-
C
-
g
g
4
5
3
2
2
I
I
terbuty I
7
5
3
T
:
Seebutyl
propyl
~
5
3
&
. sec-butyl
I
7
5
3
I
·
4-propyl heptane
heptane
4
G
Cycloalkanes
(
cyclobutane
Cyclopropane
These
are
M088
They
strain
.
cyclohexane
cyclopentane
-
the most
Stable
109 5 %
is
feel
109 50
.
unstable because their
angle
.
a
strong angle
Specially cyclopropane
.
Example :
1
-
5
43
1
.
heptane
2-dimethyl cyclopentance
strain :
bond angel
(
6
!
-Br
(Br
1 3-cibromo
,
cyclohexane
heptane
&
Deviation of
90
1600
isopropyl
T
isobutyl
Angle
Formula: CnHan
C
5
3
4-
"
4 - isobutyl
4
&
4-ferbutyl heptane
4
2
&
7
5
-
6
4
2
3
2
2
~
6
Y
I
4-butyl octane
4-isopropyl octane
isoprops
I
Cycles
109 50
.
.
are
preffered
1109 5%
.
109 50
.
and
sps after
Newman Projections
Newman projections
molecule
.
Normally done
Examples:
Look
C
-
drawings used to help visualize
are
CH3CHzCH
front
is
This
H
side
is
View
view
C,
on
front
22
on
back
OH
OH
OH
H
680
>
Che
H
600
680
3
H
CHy
H
Che
7
H
H
600
CHz
>
H
HH
H
Gauche
H
H
CHa
H
H
H
Eclipsed
staggered
Torsional strain
Gauche
staggered
Anti
H
H
HHH
Eclipsed
Staggered
OH
OH
CH3
600
H
H
organic
22
This
H
an
for alkanes
.
-
OH
3-dimensional structure of
the
Eclipsed
Stability chart
staggered garche
staggered anti
Staggered
anti
the most
is
confirmation
stable
,
the
since
two
Staggered gauche is the second most stable confirmation The two
Eclipsed is when the groups of the back carbons
right behind
.
larger groups
larger groups
the two
when
Torsional strain
Steric strain
is
or
larger groups
repulsion
the
Look
atoms
between two
hindrance
steric
right behind each other
are
is
two
when
while
bulky
front
the
are
unstable
Eclipsedtorsional strain
Eclipsed Eclipsed
staggered garche
,
around
of
molecule
a
the
furthest possible from each other (180 %
are
600
apart
carbon
.
they feel the
because
rotating
groups
are
a
.
Eclipsed is
very unstable , but
mosttorsional
it
is
even
more
strain.
sigma bond.
come
closer together
.
22-23
CHyCHzCH2CHzCHz
i
CHy
CHy
CHy
H
H
H
CHzCHa
go
600
H
CH2CHy
H
H CHuCHy
chuchy
Staggered
H
H
H
600
H
H
CHy
Goo
CH2CHs
Eclipsed
H
H
H
n
PucHy
H
H
Staggered
Anti
*
H
H
H
H
#
CHy
CHy
680
*
Gauche
Eclipsed
staggered
Least stable
Garche
Eclipsed
Most stable
the most
Draw
and least stable .
Most stable
Chy
Look
Chachuc
22-3
H
CH3CHaCHy
H
H
H
Che
CHy
HH
CHzCHa
CH3
CH3
Staggered
Br
Br
Br CHzCHz Br
"
Look
C
,
-
Eclipsed
Br
Br
(2
Br
H
H
-
c
H
H
H
HH
H
Br
Staggered anti
Most stable
Eclipsed
Least stable
Least Stable
Look C3-C will be
the same
Name the
compound
based
on
the
newman
projection
.
Chy
3-bromo hexane
Br
Br
CH2CH2CHy
H
4
CHz CHz CHCHuCHzCHy
H
Br
CH3
Br
↑
H
CH2CHy
Br
H
CH-CCHzCH
2-methyl pentane
Chy
CH3
1-bromo 1-methyl
H
H
cyclohexane
Br
H
H
H
I
H
2 bromo
H
H
Br
Br
Br
1. bromo
CH3
Bra
C
-
I
2-doro
3-iodo
1-methyl cyclopentance
Confirmation
Chair
Cyclohexane
upper bond
Axial
lower bond
that
The peaks
so
Equatorial (1095 apart)
they
so
close
are
hindrance
.
They feel
feel steric
interactions
.
pointing up
are
in
space,
13 diaxial
.
We do not like big groups to feel steric hindrance
,
we avoid putting
them on axial positions.
flip
chair
A
equatorial
axial will become
Upper bonds will
will remain
taking it
is
remain
inside out. The bonds that
and
viceversa
upper bonds
lower bonds
after the flip.
It
is
hindrance there
It
very
is
very
were
.
and
lower
bonds
Boat confirmation
H
H))HH
Every time
cyclohexane goes through
boat
chair flips it goes through
confirmation
a
a
a
high energy meaning
,
unstable
·
.
Cis
means
Trans
No organism
the same side.
means
different sides
makes
transfats transfat
,
are
.
human-made
.
Example
1. 4-dimethyl cyclohexane
:
Trans
Dis
Cha
CHy
This
Che
is
the most
they are
both
This carbon
is
on
stable
because
equatorial positions
.
CHz
Both
are
on
lowerbonds
:
cis
One is
.
a
Which is
cis
and
which
is
an
upperbond and the other in
lower bond : trans.
trans ?
CHy
CH3
in
Cha
CHy
Cis
Trans
Write
1
,
4-dimethyl cyclohexane
as
CHy
Cis
a
chair flip.
Chy
This carbon
CHy
is
this carbon
this carbon
CHz
CHy
Trans
CH3
&
CHy
CHy
This
is
the most stable
The carbons
will more clockwise
Do the
1
,
4-dimethyl
cyclohexane
rotation.
Trans
Che
Cha
CHy
1-bromo
2-coro
cyclohexane
written
as
flip.
chair
Cis
Br
Bromine
because it
C
C
1-bromo
2-cloro
cyclohexane
Br
written
as
rotation .
Cis
Br
c
C
Least
Stable
Cis 1-bromo
2-cloro
cyclohexane
2-cloro
cyclohexane
Br
.
Cl
Most
stable
Cis 1-bromo
Br
C
.
Br
should be
is
bigger
on
equatorial
than
chlorine.
>unctional
Groups
-
Functional
groups
Name
Representation
Representation
:
Che
R
Alkane
C
R
:
2
Alkenes
C
0
0:
C
:
OH
Alcohol
R
NHz
Amine
·
0R
g
R
O
Ester
:
C
NH2
Amide
Ether
:
O
:
R
.0 : /R
Peroxide
R
C
Acyl
R
SH
Thio
R
X
Alkyl halide
R
g
R
:
is
:
Disulfide
R
R
carbon
X
halogen
What
chain
(F Cl Br
,
,
functional groups
:
CH> C- OCH2CHz
Ester
halide
Sulfide
R
:
acid
O:
C
R
R
R
Carboxylic
OH
R
:
R
O:
C
Aromatics
or
Ketone
R
Alkynes
C
Aldehyde
H
R
C
Name
:
,
1
are
,
At Ts
,
C
present ?
·
0
2
:
CHy-2- CHy
1
+ SH 3
-
Ketone
4
1
Aromatic
2
Alkene
3
Thirt
Ketone
5
I Alkyne
6
Alkane
Constitutional Isomers
Constitutional isomers
Examples
have
the
same
molecular
formula
,
but
different
connectivity
of
the
atoms.
:
25Hiz
n-pentane
2-methyl butane
2
,
2-dimethyl
propane
CH14
n-hexane
2.
3-methyl pentane
methyl pentane
2,
C Ho O
OH
OH
1
propano
2-propand
24 H100
OH
OH
1-butand
-
OH
OH
2-butand
2-metyl
1
propano
2-methyl
C-propano
3-dimethyl butane
2,
2-dimethyl butane
Acids and Bases
definition
Arrhenius'
Acids
are
Bases
are
compounds that
compounds that
Bronsted-Lowry definition
Acids
Bases
are
Hoion
concentration
increases
the
OH-ion
.
concentration
Acids
Everything
accepters.
that
H Br
HNOz
HClOs
HSO4
HClOp
HCI
acid if
an
Strong
lone pair is
one
accepter
an
.
It is
positively charged
It is
a
CalOHIc
we
NaOH
SrCOHIz
from
KOH
Ba (OH) z
metal
.
transition metal
+
If it has
element , it is
.
H20
add
:
to
a
lone pair
The
a
conjugate
conjugate
base
acid
C :"
con
acid
H20
>
acid
Ka
con
pKa
pKa
H28
stronger
&
and
9
=
which
conjugate
OH
Pka
CHyCOO
+
HoOt
conjugate
base
is
-
CH,
:
CO:
base :
+
[HA]
acid
base
CH4
NH2-1
cou based
50
con
.
is
will go
stronger
acid
away
.
The reaction
is
CH3-
stronger base
the
in
.
reverse
from
5
16
H20
Alcohols
17
Amines
36
Alkanes
50
Alkenes
48
Alkynes
25
The
stronger
from the
because the pha
is lower
acid
the
t
CH3COO
is
&
=
,
NH2
Pha 36
stronger acid.
CHzCOOH
The
is
NH2
is
the
is
.
in
the stronger base
con
+
.
acid
CHyCOOH
pra
stronger acid.
reaction
comes
acid
CHyCOOH
base
=
amine
weaker
NHe't
cor
NHy
+
36
an
,
acid
-
base
8:
WHy
-
Base
NHz
t
pka
(NHy is
Nhe
Pha
acid
stronger
is
CHyCOO
acid
=
reaction
remember
to
o
+
is
[A] [H ]
stronger ?
CH3COOH
Is the reaction
?
going toward or in reverse
NH2CHy
CHzNHz
The
.
acid
,
=
Carboxylic
Nat NO-
+
conjugate base
4
:
acid.
the stronger the
-log (hal
=
base
.
&
CHyCOOH
acid
pka the stronger the
the
Products
,
reacti
&
L
t
the Ka
higher
:
base
t
is
.
weaker
base
&
H20
t
Which acid
com
HyOt
+
-
CHy COOH
base
Y
NaOH
CHOCOOH
proton
a
Functional group
base
t
stronger
Phas
acid
we
the result
base
.
H30t
t
base
NaOH
t
%
>
<
when
from the
The lower
z
W
HNOz
get
negatively charged
zu
H20
HNO 3
a
a
i
acid
+
we
The
u
proton
a
acid.
an
is
:
t
remove
we
OH-
Beryllium (Bel
a
when
Conjugate acid
are
on
Hig
get
the result
is
.
,
Examples
Conjugate base
LiOH
I is an incomplete octet :
Boron (B)
aluminium (AI) and
=
least
bases
Most bases
:
,
Pha
at
has
.
acids
9
:
OH --
acceptors.
pair
electron-pair donors
are
Stron
HI
lone
are
Bases
Hi
Ht
.
definition
Lewis
It is
the
(H'l
donors
.
proton
proton
are
increases
reverse
strongest
.
base
=
5
would
Which
be
a
better
Acid
be
better
pull off
to
H
H
C
S
H
A) CHyCOO-
CHyCOO-
B) NHz
NH2
from
comes
:
from
comes
NHy
CHyCOOH pla 5
NHy
pra
36
=
Pra
It must
than
be
18
an
CHyCOO
-
B) NH2
NH2
C) CHy-
CHe-
D) CH3OH
E) HCI
CHyOH
lakohol)
/n
=
-
CHzCOOH
>
-
- -
NHy
>
the
on
same
and
H20
is
elements that
We
is
conjugate
electronegative than
HF
acidic
or
CH4
N is
more
electronegative than C
NHy
is
HzS
is
H
C
gets
stable
more
the
the group,
on
.
acidic
or
HaS
is
than
more
CHzOH
acidic
the acidic
proton
in
the element ?
0
therefore
F
,
therefore
,
stable
.
Br
is
5
is
more
5
is
more
stable
I
is
more
stable
more
stable
.
acidic
↑
H
O
-
c
it
-
C
is
H
g
is
The
,
therefore
acidic
Cl
than
,
therefore
acidic
more
more
oxygen
an
element have , the
HClOg
or
HCIO
HClOy
because
it has
HNOy
HNO2
because it has more oxygen
HNOz
·
HI
bigger
is
O
more
or
is
HI
than
bigger
HC
I
CHySH
or
CH > SH
:
H
...
H
HH
~
electronegative than P
more
greater than
that
HBr
bigger
H2S
H
C
are
elements
for
base
As it goes down
is
.
more
H2O
acidic
S
Which is
the conjugate
bigger than
is
HBr is
Phy
more
conjugate base
element , the
the
base
or
S is
S is
Both
C
.
NHy
or
CHy-
.
stronger bases
group
.
same
the bigger it
Br
HeS
the
The bigger
the most
H20
more
the
stability of
use
on
are
The right most
more
are
,
stability of
.
more
NH2- and
36
=
19
.
electronegative
or
5
=
period.
&
CHy
pKa
=
acid acidic ?
electronegativity for
We use
pra
50
CH4
pha
HCl are neutral so they are discarted.
>
-
Electronegativity
is
stronger base is
stronger
A) CHy COO
What makes
O
,
Bases
CHyCHzOH
are
acid
.
Acid
Pha
the weaker
NHz7
Ig
=
is
the
So
H
H
H
proton ?
a
Bases
O
H
would
base
more
oxygen.
or
.
H2SOy
or
HeSO
H2SO4
because
it has
H2SO4
is
acidic
more
more
.
oxygen
more
acidic it
is
.
Inductive
effect
It is
electronegativity over distance
Not
the
the
Ho
It
somewhere
on
to it.
.
is
the
or
element attached
else
.
BI
A)
O
F
H
/
OH
HS
↓
This
this
A
is
ClOH
over
distance
more
structures
resonance
the
means
,
the
more
H-c -c
B's
:
H
B)
H
or
conjugate base is
is
also
B
is
have
more
acidic
more
-
C-
#
acidic
j
.
M
.
0.
·
j:
HH
2 resonance structures
acidic
or
F
electronegative
is
more
stable
is
CBryCOOH
more
acidic
,
3
resonance
↑
H
i
H
Or
H
more
C-
.
·
i
CFCOOH
H
.
oxygens
B)
is
or
structures
stable
.
more
i
Al
B
more
H
↑
y
-
it
i
structure
1
H
is
structures
H-EH
↑
acidic.
more
2
A)
F
CFz COOH
acidic
more
therefore
H
O
B)
H
.
electronegative
more
The
4
-
but
BrOH
or
is
,
I
c
-
acidic
more
is
-
A
proton
it more acidic
makes
ClOH
CI
acidic
the
is
HH
OH
-
H
or
H
C
f
-C
A)
↑
-
H
·
Arrow Formalism
Do not draw the
The
will go
arrow
.
from the negative charge .
arrow
something that is positive (H)
to
Nucleophiles
Nu
of the electron pair
from the middle
Drawn
Electrophiles
:
t
Nu :
St
or
positive (8+).
delta
I
:
The
least
electronegative
.
H
H
=
Et
Nucleophiles attack
electrophiles
Examples
:
HN
I
t
g-
o
S·
u↳
H
·:
Nat
t
H
u
bond
This
This
are
spectator
ions
broke
·
%
S
o
↳
CH3Cj·-H
t
G
:
Nat
H
+:
B
8-
H
t
ChyCjH
⑧
o
Hs
:
8-
+
G
+:
H CHy
(1 :
- L
:
Nat
H
H
H
Nau
t
Che
·:
CNCHa
↳
o
a
H
·
C
g-
↓
:
H
Acid
gt
Il C
j
:
4
H
t
H
N
-
-
+::
Na
t
H
Base
H
Y
CHy COO-
t
H
N
H
H
Enantiomers
Enantiomers
Chiral center
rotates
.
non-superposable mirror
images
stereocenter of an atom is the atom
are
or
light. They
polarized
plane
rotate at the
-
=0x
but
bonded
the
to
to
chemical species
.
four different
.
side
opposite
+
:
these
Tell if
amount
same
is
Chiral center
Light everywhere
Examples
,
that
molecule
a
the
time
same
in
carbons
are
chiral or not.
H
H
NO.
Cll 2
:
CHyl1
OH
C
.
I
Highlight
the
all
C
yes.
NO
.
C
chiral
carbons
.
Br
Cy
OH
This
-
chiral
is
one
these
because
them
makes
different.
NHz
If it has
We must check the
CHy
cyclohexane to
entire
if they
see
or
triple
double
bond
cannot be
are
a
,
it
chiral .
different.
How to
We will
The
which side
know
priority based on atomic number
priority must be in the back
assign
lowest
not
is
four
in
and
R
the front
on
Br 2
h
4HI"C
is
clockwise
the back ,
OH
back.
the
back.
more
priority it has
is
to
it
more
If
I
is
the front
on
C3
4
S
G
I1
H11 C
4
f
Br2
H4
N
1
It
because
on
4
is
2
the front
It
3
,
because
is
the
on
2
front
3 CH3
R
P
It
CH2CHz
because
,
is
on
the
front
2
G
2CHz CH2
H
because
S,
H
OH 1
is
on
4't
4
H
Che
V
CH3
OH
Br
H4y
H
3
OH 1
Ja
↑
H
1
R
C2
OH
because
R,
3
R
S
Br
2
OH1
CH2CHy
2
would be
C
G
H
4
CH2CHy
This
R
is
on
front
·
flip it.
S
2
CH2CHzCHz 1
1
S
we
/"& CHuCHa
S
Cly
OH
,
CHz3
Br 2
3 CH3
1
the
H
when
we
1
H
in
the
it
Br 2
-
Cl
number,
.
to read
order
or
ge
R
C3
In
S
the
I1
↑ H//1)
I
bigger
.
ignore the number
S is
counterclockwise
I
The
.
We
If
rotates
chiral carbon
the
1
A s
front
H
OH
Si
1
S
3
This
Diastereomers
Diastercomers
nonsuperimposable
Examples
R
compounds which have
are
,
When
.
images
non-mirror
the
chiral center
one
changes
R
R
R
S
R
S
R
S
S
S
7
R
horizontal
Straight
lines
wedges
are
meaning
,
H
H
OH
H
OH
R
R
R
S
S
R
in
H
OH
H
OH
of
each chiralcenter
.
H
2
H
R , becase
G
OH 1
2
CH20H
3
-
C
-
O
G
"
↳He
,
oh
H4
3
diasteromers
-
H
of
S
CH20H
this
H
H
OH
H
OH
OH
H
H
OH
OH
H
H
OH
CH20H
CH20H
CH20H
them
enantiomers
in
diastereomers
.
or
H
OH
H
OH
H
OH
OH
H
"
OH
H
O
OH
H
H
H8
OH
of
H
OH
H
H
H8
CH2OH
CH2OH
T
#
CH2OH
CH2OH
T
#
-
Enantiomers
IV
CHLOH
S
H
and
3
-
2
H
III
Co-
H
OH
Diastercomers
1
1
and
11
and 111
and
IV
11 and
IV
1
11
and
is
Oxygen
this .
CH20H
-
because
H
2-
4
OH
Classify
,
1-
OH 1
H
enantiomer of
the
R
OH
3 CH20H
OH
Draw
2
H
H
H
H
0
=
C
-
4-
the
front
on
is
1-oxygen
OH
H
Enantiomers
:
the rotation
Draw
i
front.
H
H
3
CH20H
Example
3
S
H
CH20H
·
.
projection
Fisher
H
stays the same
Diastereomers
Enantiomers
Diastereomer
4
or
of bonded
sequence
:
7
State
formula and
molecular
same
III
-
H
on
front.
elements
but
are
Meso
compounds
Meso
compounds
have
chiral centers
but
chiral.
they are not
They
have
a
plane
of
50 %
will be
symmetry
enantiomers
.
Examples
:
!
CH2OH
It
!
of
------------
H
OH
OH
CH2OH
OH
Racemic mixture
When
reaction
do the
we
Enantiomer
When
S
Examples
R
or
is
30 %
R and S
EE
,
than
85 %
S
=
=
=
R
15 %
80 %
S
20
50 %
will be
40 %
R
and
S
=
R
and
50 %.
=
15 %
S = 15%
40%
R
60 %
R
Total
20 %
80
chiralcenter
EE
Total
-
more
R
R
=
a
:
70 %
R
get
we
excess
90 %
EE
and
=
60 %
60 %
R
R
R
R
=
10 %
S
=
30 %
95 %
S
5 %
R
95
5
=
-
EE
=
=
90 %
90 % S
S
R
=
30 %
5
=
30 %
.
S
,
and
they
do not
have
Alkenes
Alkenes
CHy
H
unsaturated
H
H
hydrocarbons containing
CHa
CHy
2
H
H
alkene
CH3
CH3
C
2
C
CHz
CH3
CHy
H
Mono substituted
double bond.
a
CHy
CHz
C
2
C
2
of
series
any
are
Disubstituted
Trisubstituted
Tetrasubstituted
alkene
alkene
alkene
Most stable
E-trans
Examples
Z-cis
:
Are these
substituted ?
Trisubstituted
(cis)(z)
Disubstituted
Priority
1
~
Br
2
i
H
I
(trans)
E
the
Cl
priorities
~
2
i
kis)
same
1
Z
opposing
kcis) because the
priorities
side
CH32
Br
N
~
Z
because
are
(trans) (E)
W
H2
2
other.
each
Disubstituted
are
on
the
Z
side
same
.
H
H
This
is
because
priority
neither
z
you
not
pick
same
element.
can
the
on
E
or
(Left side
Naming
4
2
v
S
3
I
-
4
2
5
3
I
.....↑
2
3
6
6
L
5
I
~
4
2
-
(E) 2-pentene
Br
v
1-pentene
(z) 3-bromo 2-methyl
3-hexene
1-methyl
"T-cyclohexene
Alkene
Reactions
An organic reactant will
give you
A reagent is
substance or compound
organic
an
added
a
*
Reagents
go
top of
on
the
to
product.
system
to
a
cause
a
chemical
Ni
or
Pt
reaction
,
or
test if
also
reagents.
it
.
occurs
arrow
.
Hydrogenation
reagent
the
add
We
He
,
and
metal
a
syn addition
Type of reaction
alkane
Obtained functional group
reaction
:
catalyst
such as
,
4d
,
,
,
which
are
.
:
Addition reactions
HH
Hz
H
H
H
C
C
H
4d ,
H
This
of
syn because
is
can
be
-
C
-
C
Syn
H
-
1
Ni , or Pt
the mechanism
Hz
added to
side
same
Anti-different
H
H
=
The
anti.
or
syn
is
being
the
alkene
.
side
.
Mechanism
H
H
C
H
C
bottom
,
i
trigonoplanar it
it's
Since
and
will
plane and it will be attacked
.
from the same side
is
,
attacked
be
by the catalyst from
the
top or
H
HH
Ni
Examples
:
He
H
L
Pt
H
H
H
Hz
Since
H
Ni
we
H
have chiral carbons,
we
have to show
The
two
both
in
in
the
hydrogens
the back
front
---
L Ill
---
stereochem
must be
or
-Che
Stereochem
H
Pd
Li
H
....
H
---- -/CHy
CHe
both
H
These
mess
are
.
CH2CHs
Hz
H
---
H
-H
(
-// cHzCH3
t
11 H
H
III Che
CHy
H
Enantiomers
D
CHz
D2
of
D2
D
Ni
D
is
an
stereochem
Y
(
isotope
Y CHy
t
ill D
Ill
H
D
H
Enan tiomers
hydrogen
CH2CH3
Chucky
Hz
H
H
Stereochem
=
>
CHy
t
Po
Enantiomers
He
H
Hydration
H20
weaddthereagent H AlsoHoH
alcohol
Obtained functional group :
H
H
C
C
H20
H
>
H
H
tracto acids
[
H2S84 cat
~
,
H
Nor
H
OH
C
C
H
H
syn
anti
or
H
Mechanism
H
H
H
C
C
H
-
H
Pibond
A water molecule
that
will
from
other
H
the
water molecule
attack
will
swimming around
an
%
.
is
H
pull
H
H
H
H
H
an
+
C
C
H
---
O
.
H
pair
will attack
the
carboncation
.
H
H
lone
The
&
·
+
H
H
!
C
H
H
+
H
S
of
H
H
H
!
C
O
H
H
G
&
·
H
H
H
t
H
G
&
·
.
Carbonation
H
H
3020To
H
C
C
H
: 0
doe to
Markovnikov
stability chart
H
methyl
hyperconjugation
"
goes
carbon
,
the
addition
to
the
one
with
LESS
more
says
that
substituted
hydrogens.
H
Recimic
H
mixture
50/50
·
Examples :
H2O
OH
H
H2SOy cat
CH3
OH
H20
H2SO4 cat
OH
H
Stereochem
>
OH
t
Enantiomers
CHy
Hydrohalogenation
We
reagents
the
add
with
Type of
reaction :
Obtained
functional group
hydrogen
Markornikov
addition
alkane
:
halogen
and
reaction
with
,
HBr
as
,
HCl
,
but
HI ,
never
HF
.
.
.
halogen
a
such
H
Br
C
C
H
7
H
H
C
HBr
C
H
H
H
H
Mechanism
-H
.
Electron pair from
pibond will
attack the H
will go
the carbon
and it
with
more
CHy
H
H
t
oo
H
H
to
8
+
-
t
H
hydrogen
.
--
This
orbital with
empty
orbital
from
top
and
+
Chy
"+"
a
that
is
an
attacked
be
can
Br
bottom
Br
Not
CH3
chiral
CHy
H
Examples
:
HC)
a
H
HI
CHy
Stereochem
F
CHy
t
H
#
H
I
Enantiomers
Bottom
Cl
Hi :
~
Chiral
Stereochem
H
Cli
4
Gr
2
or
S
H
CHy 3
L
Racemic
Mixture
Branching
the
boiling
lowers
point.
top attack
>
R
4
A
2C
,
3
CHz
H
:
Halogenation
halogen molecules
add
We
Mechanism
Type of reaction
Obtained
:
functional
as
addition
Anti
group
reagents
:
alkane
,
such
Bra
as
Cl
,
and F2
.
,
-
reaction
vicinal dihalide
.
with
on
are
the test :
Halogenation
Hydrohalogenation
-
-
that
Bromination
Mechanism
.
1
L
·
:
17
t
:)
:
We don't
Bri
see
·
B.
(
T:
these
: Br :
Br :
6
structures
electron pair will attack
This
one
This
of the
This
bond
bromine
electron
pair
is
Bromonium
(Br).
it
created
charge
a
is
Vicinal
o
dihalide
the
Ion
.
cyclic
is
-S
*
·
The
on
This
and
anti addition
is
because of the bromonium
positive
the inside
.
ion
,
3
member
ring
.
.
Geminal dihalide
on
Examples
the
same
is
when
two
halide
carbon
:
Br
Brz
them
We need
Br
be
to
opposite to each
other.
CH2CHy
D2
C
Stereochem
]
Cl
=
C
Ch
t
CHy
CH2 CHy
Cl
C
CHy
Enantiomers
Brz
Br
Br
Stereochem
CH3
Br
Yety
t
ill Br
H
Enantiomers
-
- Br
Br
has to be
Ill H
opposite because
Br
this
is
anti addition
.
Carbonation
Every
time
Methyl
shift
The
carbon
rearrangement
a
carbonation
next to
shift occurs
The
the
because
is
fromed
rearrangement will occur
a
,
double bond
you
attached
must be
from
going
are
to
primary
a
.
Just
three
or
as
more
in
hydration and hydrohalogenation
carbons
.
secondary
carbon
At least
to
one
a
of
tertiary
.
them
carbon
has to
,
be
which is
a
methyl group.
more
stable
2H3
CH3
zo
-i
H
Che
T
Example
-
a
Hydride
The
>
T
:
H-
The
I
↓
20
:
with shift
Cl
=
shift
carbon next
H
is
the
to the
one
that
double bond must have
will be moving
two
carbons
.
.
H
H
I
-
HBr
H
↑20
-
↑
zo
H
Br
HBr
t
Br
i
:
Br
Halo hydrin formation
We add
a
halogen
and
water
or
anti addition
Type of reaction
Obtained functional group : alkane
:
something
with OH
as
reagents
.
reaction
.
with
a
halogen
and
an
alcohol.
H
:
↑ :Bri
It
*j
Brz
H20
is
of the bromonium ion.
Br
H
H
O will attack the
most
substituted carbon and
then it
Example
will be
deprotonated
CHz
:
OH
OH
CHy
OH
Brz
stereochem
H20
CHy
t
H
Br
Br
Br
H
L
nan tiomers
=
CH2CHzCHy
Brz
stereochem
OH
H20
Br
Br
H
Enantiomers
H
Br
Br
CHy O
CHz OH
CH2CHzCHa
t
H
Br
Brz
OH
OH
Stereochem
H
t
CHzO
CHzOH lost an
because
H
it got
deprotonated
Br
CH2CH 3
CHzCHz
CHz0
Enan tiomers
anti because
Oxymercuration
reduction
Hg(OAc
We add
Type of reaction
,
water
and
Markovnikov
:
functional group
Obtained
,
Racemic
:
a
.
agent
reduction
anti addition
mixture
hydration reaction
with
alkane
,
The mechanism
formation of the
.
0]
Reduction
oxidizing agents
Oz Oz KMn04 HzOz
CrOg
HCrOp
Some
,
are
,
,
:
,
H
g
Some
reduction agents
are :
Hz
NaBH4
,
,
through
the
.
ion
mercurinium
alcohol.
an
There
Oxidation
is
is
carbonation rearrangent.
no
LiAlH4
,
OH
1) Hg(OAcla
Hz0
,
2) NaBHy
Mechanism
It
-
won't be
the test
on
OAc
t
1 Hg(0Acz
2)
,
OAc
-
-Hy
Hz8
-
NaBHy
HgOAc
Ha
NaBH4
-o
&
t
Mercurium
lon
Examples
H
:
H
H
It
OH
O
OH
H
:
OH
1) Hg(A0cI2
2)
,
H28
t
OH
Stereochem
NaBH4
H
Enantiomers
OH
OH
1) Hg(A0cIz , H2O
Stereochem
t
2) NaBH4
=
nantiomers
L
CHz
OH
1) Hg(AOck Hz8
,
2) H2O2
,
OH-
H
H
OH
OH
CHz
Stereochem
t
Enan tiomers
OH
oxidation
Hydroboration
BHy and
We add
Type of reaction
an
oxidating agent
Nonmarkovnikov
:
reagents
as
.
hydration addition
syn
No
.
formation
carbonation
functional group : alcohol
Obtained
R
H
1) BH 3
O
Since
it
#
to
W
2) H2O2 , OH
Trialkyl
OH
goes
nonmarkovnikov ,
is
the
B
R
R
with
carbon
less
H's
.
Mechanism
H
H
=
%
H
·
B
B
H
This
3
H
·
O
B
-:
...
H
H
total
~
H
OR
Trialnyl borate
B
H
&
-
H
%
OR
OR
This
happens
two
times
more
Example
...
O
·H
t
O
-
B
O
H
H
happens twice again
in
H
#H
O
:
H
BHz
1)
CH3
cHz
H
L
H
OH
Stereochem
2) H2O2 , OH-
W
t
H
OH
Enantiomers
dihydroxilation
Syn
We add
osmium
Type of reaction :
tetroxide
Examples
and
NaHSOn
reagents.
as
addition
Syn
Obtained functional group
(OsOy)
:
with vicinal did
alkane
,
meaning
two alcohol
next to each other.
:
H
OH
OH
1) OsO4
21
NaHSO3
Stereochem
OH
L
--
H
---
+
-
H
·
L
- -
OH
--
OH
H
"Meso
H
OH
1) Os04
2) NaHSO3
OM
stereochem
H
OH
OH
&
Ol
CH3
CH3
OH
Enantiomers
Cha
H
Borane
mCPB A
Anti
dihydroxylation
We add
mCPBA
or
Type of reaction
:
O
ROyH
Anti addition
functional group
Obtained
Two
:
Ho0t
and
,
alcohols
.
of the epoxide
because
g
reagents
as
g
formation.
C
opposing
.
1) m CPBA
CHy
H20
2)
--
H
O
OM
Che
+
OH
t
Epoxide
H
with
Enan tiomers
CHzOH
-
&
streochem
H
2) HzOt
1) mCPBA
t
ph
Cty
PH
of
t
-
OH
CHzCHz
HH
H
Markovnikov
Br
HBr
Br
H
Non mar Kornikov
peroxide
H
Ozonolysis
and
dimethyl sulfide (DMS)
as
reagents
.
On bonds cleavage
,
meaning
it
breaks
the bonds.
Type of reaction : organic redox reaction
Obtained functional group : Ketones and alhydes
O
1) Oz
2) BMS
Li
O
O
↑
1)
=
Oz
8
>
2) DMS
O
g
1) Oz
2) DMS
00
>
H2
city
nonmar Kornikov
HBr
Op
Ph
H
Nonmarkovnikov
We add
.
oxygen
CHzOH
Enantiomers
Peroxides give
three member ring
feny1
OH
stereochem
OH
2) HzOt
ph
CHzCH2
-
OH
H
a
Enantiomer
OH
CHy
an
is
H
OH
OH
form epoxides
.
or
OH
OH
-
radical initiators
,
These are
CH3
H
OH
1) mCPBA
RO3H
H
O
t
C
H
+
OH
H
PH
Cold
potassium permanganate
We add
cold
Type of reaction
KMu0y
:
syn
as
a
.
reagent
dihydroxylation
Obtained functional group :
two
alcohols.
OH
Cold
OH
KMn04
OH
Stereochem
H
H
OH
OH
of
chzChy
Enantiomers
OH
CH2OH
streochem
OH
cold
OH
t
CHICHz
t
=
FMn04
-
OH
CH20H
Enantiomers
Hot potassium
permanganate
We add KMnO4
reagent
a
as
Type of reaction
Obtained
Oxydaxion
:
functional group
:
,
reduction
carboxilic
Hot
KMnO4
and this will
cause
bond cleavage
.
Ketones
.
Carboxylic acid
?
and
acids
formed if it is on the
is
first
or
last carbon.
g
OH
Li
OH
g
-
Hot
L
-
Hydrogens
H
a
KMn04
·
this
Since
H
CO2
,
carbon
it will become
CO2
O
G
Hot
+
KMn04
Hot
KMm04
O
:
O
Alcohol production
OH
H20
"
An alrene
of
cat
·
H
·
H2S04
with
these
Trace acid
any
reagent
alcohol.
will produce
an
·
has two
CO2
O
t
CO2
Alkynes
Alkynes
are
Alkynes
are
unsaturated hydrocarbons
they
and
linear
containing at least
can't be
cis
carbon
to
Internal
alkyne
one
carbon
triple bond.
trans
.
or
-
Terminal
alkyne
Naming
5
5
-
3
S
I
C
4
2-hexyne
1-pentyre
4-bromo
2
.
3
er
S
2
3
4-methyl
I
2-
hexyne
Reactions
Hydrogenation
reagent
the
add
We
He
and
,
syn addition
Type of reaction
alkane
Obtained functional group
reaction
:
catalyst
metal
a
,
4d
such as
Ni
,
,
or
Pt
,
which
are
also
reagents.
.
:
We
are
twice
doing this reaction
Hz
CH
4d,
to
alkyne
it has two
because
No
pibonds.
stereochemistry
.
excess
2x
HC
,
4t
,
or
Most
CHyCHz
Ni
It will
alkane
reaction
do not
need to be balanced.
always happen
twice ,
you
organic
Examples :
Hz
Pt
#
S
Y
2
.
3
5
hexane
2-hexyne
H
H
Hz
Linlard's
catalyst
alkyne
Since it
is
it will give
Lindlard's
the
syn addition ,
a
cis
altere
catalyst stops
reaction
can't
H2
Lindlard's
tell
cat
H
-
He
Lindlard's
Na
and
We add
cat
NHy
Na
and
Type of reaction
:
NHy
anti
Obtained functional group :
as
reagents
to the
alkyne
.
addition
trans alkene
H
Na
H
NHz
H
Na
NHz
H
cis
or
trans
on
this one
can
never
stop it
at the
alkene level.
Hydrohalogenation
We add
the reagents
with
Type of
reaction :
Obtained
functional group
hydrogen
Markornikov
addition
reaction
with
alkane
:
halogen
and
a
,
such
HBr
as
,
HCl
but
HI ,
,
never
HF
.
.
geminal dihalide
.
H
O
Cl
2X
HCl
HC
H
H
O
H
H
Intermediate
H
H
H
excess
H Br
Br
intermediate
Br
Br
c
C
HC
Halogenation
We
halogen molecules
add
Type of reaction
Obtained
as
addition
Anti
:
functional
reagents
group
such
Bra
as
,
Cl
,
and F2
.
with
Br
2X
·
Markovnikov addition
cannot be
applied
to internal alkynes because it has
reaction
alkane
:
,
vicinal
Marnovnikov addition
tetrahalide .
can
be
to terminal alnynes
because
If
alkyne
H.
no
applied
it has
H
an
Br
Brz
Br
Br
Br
Br
Intermediate
Oxymercuration
We add
Hg(OAc ,
and
a
reduction
S
Type of reaction
Obtained
anti addition
Markovnikov
:
functional group
alkane
:
.
agent
with
a
hydration
reaction .
O
H
H2504
OH
End
Intermediate
undergo
Keto-end
tautomerization
Tautomerization
> Ht
this is
-
.
Stable
.H
OH
·
End tautomer
Keto
tautomer
Examples :
O
&
HgS04
H2S04
O
Hg S04
H2504
Ol
OH
O
oxymercuration
Ketone,
ketone.
Hg SO4
This will
a
more
,
goes
through
it will become
wether is internal
or
a
ferminal .
oxidation
Hydroboration
BHy and
We add
Type of reaction
oxidating agent
an
Nonmarkovnikov
:
functional group :
Obtained
.
hydration addition
syn
aldehyde if it
reagents
as
is
.
Ketone if it
terminal.
internal
is
If
a
terminal alsyne
hybroboration
O
we
,
goes thro
will get
ugh
an
aldehyde
1) BHe
H
2) H202 , OH-
I
OH
can
be
BHy
or
(Sialz BH
Intermediate
Examples :
(Sial zBH
1)
2) H202
,
OH
O
Double elimination
When you
NaNtz
use
Type of reaction
on
geminal dihalide
we
elimination and
get double
an
alkyne
.
:
functional group :
Obtained
a
alkyne
Br
CHy
Na NHz
CHz
C
Double
CHzC
elimination
H
C
Be
Example
:
Cl
Cl
XS
Na NH2
C
O
O
CH
o
2x
NaNHz
Br
CCHy
CHzCHzC
Br
Chain elongation
-a
We add
Nattz
Type of reaction
:
.
alnythalide
and
:
Obtained functional group :
alkyne
H
4na
=
36
H
CH
CH3C
NH2
Examples
H2
=
:
CHzC
CHyCl
proton pha
Nucleophile
25
which
:
1) Na NHz
CH
2) ChzCHz Br
HC
CCHICHz
:
+
Ch
1) Na NHz
-
o
2) CH3 Br
3) NaNHz
4)
Br
g
+
Not
CHzC
CCHe
O
CHyC
CC
-
alkylhalide
will attack an
Organic product
HC
:
C
-O
:
H
Nat
important
H
H
+
Write
the
HC
CH
reagents
in
order
that
are
needed
for each
to
reaction
work.
HC
1) Na NHz
2) CHzCH2CHz Br
CHyCHaCHnG
S
Ch
1) NaNHz
CHC
2
4) CHyCHz Br
4) Che Br
Then
,
5) Hz
Lindlard's
2) CH Br
3)
Br
HC
1)
2
3)
O
CH
1) NaNHz
2
2) CH3CHzBr]
HgS043
4)
3)
Hz SOn
NaNHy
5) HBr ?
catalyst
1) NaNHz
1) Na NHz
NaNHs]
4) Chy Br
6) Brz
HC
Ch
2) CHz Br
NaNH
Lindlard's
catalyst
Br
4)) CHy Br
5) Hz
2
Br
CH
1) Na NHz
2) CHzBr
or
3)
eHz
Nant
3)
HC
&
&
2) ChyCHz Br
3) NaNHz
Br
CHyCHz
&
NaNHnf
4) CHz Br
HS
g
or
2) ChyCHz Br/
3)
Br
Br
2
HBr
Y
Br
Br
Radicals
Stability of radicals
Allylic
3020
benzilic
methyl
To
Homolytic bond cleavage
Bond
bonding
breaking in which the
follow
Carbon radicals
↓
stability of
the
electron pair is
split evenly
products
between
.
carbonations
.
&
R
0
R
2x
R
.
radical
CHy
⑧
·
Co
Cha
10
radical
Radical
3° radical
CHy
3° radical
cal
20
halogenation
We add
halogen
a
molecule
light (hrl
and
peroxide
,
,
or
heat) I
as
reagents
.
Type of reaction :
Obtained functional group
Brz
CH4
:
Chy Br
↳ r
or
peroxide
Or
Mechanism
Monohalogenated product
1) Initiation
H
~
Brz
H
H
C
Radicals are formed
:
·
45
·
.
H
2) Propagation
Radicals react
:
and form other
H
H
C
↑
:
i
ir :
L
.
All radicals
:
it
Br:
H
Br~
:
H
↑
Br
C
H
-
-
H
H
H
H
C
H
~u
I
&
H
H
H
H
H
H
S
C
H
H
Halogens go
Cl
radicals
302
5
:
4
:
ratio
Br
radical
ratio
10
zo
zo
go
1
1608
80
I
Thermal cracking
H
where
It grew
there
is
S
H
will be combined
H
:
H
H
3) Termination
.
H
:
H
H
:
B
H
H
-
H
radicals
hydrogen
Br
+
Examples
:
the reaction
Do
and
percentage
show
.
C
Clz
C
t
ht
43 %
57 %
20
Primary hydrogens : 6 H
Multiply the number of
Secundary hydrogens : 2H
radical ratio
10
10
18
:
20
x
2H
x4
:
1
with its
corresponding
Get
then add them
.
,
GH
H
=
=
10
percen
tage
6
=
X
100
=
43 :
14
6
2
g
G
=
100
x
=
57
:
14
14
Br
Brz
Br
t
ur
4%
96 %
G
2
6
10
10
10
=
=
2
GH
10
2
2H
GHx1
=
10 :
6
=
166
* 100
=
108
=
4 %
160
2HX80
=
:
160
20 ·
X
160
96 %
166
Br
Brz
t
Br
↳ r
99 4 %
0 6%
.
.
10
1 > 2
30
1 :
10
10
yo
30
=
=
30
80 :
10 :
9H x
30 :
1600
1H
1600
x
10 :
9
1 =
=
9H
zo
1600
6
x 100
1609
1600
:
1609
0 6 %
=
.
X 100
99 4 %
=
.
1609
1H
Major
Brz
Br
nu
product
and stereochemistry
t
Br
Br
10 : GH
6
: 4H
2
1 % 20330
10
80 :
20
1
:
1608
:
GH x 1
:
4HX80
10 :
6
=
2 :320
328
:
326
x100
=
100
=
1 8 %
Br
.
x
H
%
98 2
.
326
Positions
to
know
vinylic position
Allylic position
:
double bond
On the
most
:
On the carbon next to the
Benzilic
stable
double bond
.
position
On the carbon
C
O
Cl
Cl
NBS
:
source
of Br
Examples :
NBS
Br
ha
Br
NBS
O
-
peroxide
O
NBS
nu
Br
:
next to
a
.
benzene ring
Br
Substitution
reaction
this
Rate law
R
substitution
bimolecular
bond
and
nucleophilic
SN2
In
Reactions
one
,
is
broken
another
one
formed
is
in
a
concerted way
.
:
k[nucleophile] (alry) halide]
=
Charged nucleophiles are best for these reactions.
Examples :
Which nucleophile
best ?
is
:
:
...
O
:
H
or
HzS
H
H
H
CHzC
HS
or
Cha C
or
.0.
O:
0:
.
Mechanism
Sp3 has
-:
Nu
a
front and
a
back.
:
+
-
↳
im
leaving group
NU :
The attack
Backsided
from the
is
back.
explanation
attack
.
H
H
7
Nr :
The back
:
C
the
is
leaving
Nucleophiles
opposite to
can
.
big alryl halide
if it is methyl
group.
if it is
not get in
a
It
best
is
H
Mechanism
SN 2
Nu
is
step
one
.
T
:
i
Nu :
Pentavalent
Nu
transition
state
Rate of attack
Methyl
Examples
Which
zo
20
10
~
So
is
slow it
does not
I
doesn't happen
because
so
even
happen
it is
so
we
use
It
.
big
is
hindered backside attack
.
.
:
alkyl halide
is
better in SN2 ?
Br
or
Br
Because it is primary
Solvents
Solvents
can
hydrogen
bonds.
Best
solvents :
help you determine
which reaction it is.
For
SN2
,
polar aprotic solvents.
Best
)(
Acetonitrile
Examples
HMPA
DMSO
(CHOCN)
I
with no
These do not
protons.
make
leaving groups :
g
Acetone
Polar
Br
C
The
DMF
bigger the leaving group
fact
Due to
the
more
stable it is
that
,
the
the better.
bigger it is the
by itself.
:
Cl
Na CN
Nat
Br
Su
Ne
N3
O
Br
O
Br
CH3C
ONa
CHaSK
CH3S-Kt
will
Ris
CN-
Na No
Nat
SN2
O
o
SCHy
cHy
have
inversion
STR
of
configuration
,
nucleophilic substitution
SN1
Nucleophiles
unimolecular
usually neutral.
are
product will be
The
a
racemic
This mechanism
Rate law :
[alky) halide]
R: k
Slightly more inversion
mixture .
than
retention.
not concerted.
is
Concerted happening at
:
the
same
time .
Mechanism
steps
2
1) Formation of
:
2I
carbonation
a
trigonoplanar
G:
=
+
carbon cation
Rate
of attack
go
20
10
10
The
attack
from the top or bottom
No-
NU
always be SN2 reaction
will
-
attack
Nu
be
can
Nucleophilic
30
is
SN1
always
Solvents
help you determine
Solvents
can
Best
solvents
H28
which reaction it is
For
.
SN1
,
we
polar protic solvents. They make hydrogen bonds.
use
:
R
OH
Why
COOH
R
(not so much
Example :
Na OH
20 and acetone
=
SN2
No enantiomer
acetone
because
OH
Br
H28
2) and H2O
SN1
=
t
Br
OH
OH
2
Cho OH
=
and
CH3OH
t
SN1
OCHy
I
CH30 Na
2
CH3Cy
CH3CN
OCH3
and sol.
=
SN2
OCHy
I
SH
SH
Cl
H2S
2 and H2S
:
t
SN1
Cl
HSNa
20 and HSNa
SN2
SH
of
inversion
of
product
Elimination
We will get
E2
In
E2
will
we
,
E2
.
The
.
and a
base
of carbonation
often
used
are
in
E2
Primary alryl halide
.
better for
E2 , and
only do
will
always
go
through
.
E2
opposite
I must be
and the
is
Hindered bases
.
leaving group
I
Heat )
.
.
Calkyl halide] (base]
law :
Hindered
alkyne
formation
No
anti
is
an
alkyl halide
use
concerted.
is
E2 elimination
Rate
or
elimination
bimolecular
E2
.
alene
an
bases :
Alkoxide
T-butoxide
R:
:
CHzCHz :
Ch
=
Example :
~Bri
Sometimes
+:
H
↑
: Base
Br
:
bases act
as
nucleophiles.
You can only form
double bonds where
Mechanism
1
there
step
coplanar
or
hydrogen
Br
elimination must be
This
Br
is
periplanar.
Must be on the same plane
H
H
Examples
:
Br
CHO
M
NO H
unimolecular elimination
El
forms
Et
Hoffman
Zaitsev Rule :
alkene
The most stable
is
that will be
Rate law
k
:
least substituted alkene
is
El .
We take out the
leaving group
do the double bond
to where there is
hydrogen
the
.
It could
give you two
and
.
answers
.
formed.
Calky) halide]
Step 1 Formation of
:
Step 2 :
through
elimination :
The least stable alrene
the
most substituted alkene. This is the
one
So go
carbonation . It is not concerted. The base won't accelerate the process
.
a
The base
a
carbonation
.
pulls of a proton so it forms
a
double bond
.
Illustration :
:
Br :
:
irj
t
+:
nig
H
H
H
: Base
Br :
Example :
Elimination
t
Elimination
&
Primary
-
only
E2
t
Major-Zaitser
can't be trans
Zaitser
stable
.
.
Trans is
or
cis
Hoffman
stable
more
because it
Steric
Elimination
Br
Major group because it is
more
Hoffman
Br
Br
T
feels less
hindrance
.
When
do an elimination
we
Step 1 :
Draw
chair confirmation
Step 2 :
Draw
the
Step 3 :
The
H
Example
most stable
must be
in
cyclohexane
a
on
leaving
group in
axial position
chair confirmation
you can
get
with the
position as
axial
well
in
order to
,
order to
where
know
bond will go.
the double
.
check if it
form
in
cis
is
or
trans.
double bond.
a
:
Elimination
1-chloro-2-t-butyl cyclohexane
Cis
must do the chair confirmation
we
,
Hoffman
Cl
H
H
up and
are
H
up because
they
cis
I is
can
zaitser- major
opposite to H ,
be
t
formed to
.
either side
t-butyl
H
both sides
the double bond
H
It
On
H
H
H
H
Trans
Limination
1-chloro-2-t-butyl Cyclohexane
C
H
H
elim
H
H
n
H
H
Only
.
is
H
the Hoffman
formed
.
E-butyl
H
CH3
Ph
These
will switch
-~
CHy
Ph
Br
Che
&
this must be rotated
-
Ph
H
Ph
Ph
Br
CHz
Ey
Ph
-
CHy
CH
Alcohols
boiling points than its corresponding alnane
They have higher
R
Their
.
pia
is
17
.
create
They
hydrogen
bonds .
OH
Subfix :
d
Corresponding alkane
We take out the
Examples
add
and
OH
CH3
.
an
:
CHyOH
CH3CHzOH
CHy CHzCHzOH
CHy CHy
CHyCHzCHy
CHz CHz CHz OH
Naming
OH
OH
OH
OH
OH
2-butand
Ethand
1-
cyclohexano
2-propano
propanol
Reactions
We
can
convert
alcohol to
an
aldehyde using three
swern
oxylation
an
DMP
PCC
methods :
They
must be
10
alcohols
Examples
Oxidation
PCC
OH
>
8
H
OH
DMP
O
H
On
oxydation
swern
OH
alcohol
secondary
will
Secundary
Ketone
Tertiary
They don't
metone
PCC
-
HzCr04
O
OH
A cohol
to
We need
carboxilic acid
alcohol
primary
a
and
Make
Step 2
Make a double bond to the oxygen
Step 3
If there
it is
is
a
hydrogen
a
as
He 504
reagent.
a
.
primary alcohol
Step 1
sure
NazCrO4
HzCrOy,
on
.
with
the carbon
the double bond
place an
,
Illustration
OH
O
HuCr Of
OH
OH
O
or
NazCrO4
H2SO4
OH
OH
HzCr04
G
O
OH
O
O
O
H2 204
or
↳ Mn04
oh
g
=
O
OH
O
(
(
aldehyde
H
a
alcohols
Primary
O
,
give you
OH
of
oth
there
.
H
O
I
I
oxidize
or
carboxilic
acid)
"onl
Alkoxide
If have
alcohol
an
and
Li orNa
CHyOH
Williamson
Cha
and
metal
a
lithium (Li)
such as
sodium (Nal
or
as
reagents
,
we
will
:
ether Synthesis
CH3CH2 I
CHz
:
H2SO4
add
we
almoxides
get
.
.
CHz CHzCHa
heat
H2SO4
OH
Mechanism
H
H
HcS04
2
you
:1
-
H
H8H
HS04
H
w
t
will give
elimination
HSO4
is
a
,
because
bad
nucleophile
Bad
nucleophile
Good
Protonation
leaving group
HC
HC)
substitution
SN1
OH
Cl
Mechanism
.
PirH :
H
Examples
&
ot
H
1
-
: Cl :
t
H
C
-
:
Br
OH
SN1
HBr
g
O
OH
I
HI
O
SN1
O
HC
OH
Cl
PBrz
SN2-primary
No carbonation
OH
:
SN2 :
inversion
inv
.
of
of
configuration
conf
.
Br
SOCI
SN2
OH
PBrz
C
mechanism
R
.
:
H
H
H
-
"Bri
·
Br i Br :
: :
R
"
Br
g
R
W
H
:
H
Bri
Examples
Br :
+
j
O
:
Bri
Br :
Tosyl chloride
H Br
t
P Brs
SN2
Br
OH
tosyl
C
OTS
O
SNI
Br
Br
OH
OH
C
S
g
g
Cl
Best
leaving
group
"O
H
Che
acids
and
Ether
Br:
CH30c :
CH 3
70
H
t
H
Chai
H
g
CH
,
C
C
1
zo
Chai
g
H
8
CH
C
,
Che
↑
Chy
H
H
Hiri
OH
HI
j'
H
C
+
CHo I
:
10
-
Less
HBr
...
O
30
Br
Chy
+
po
HF
O
yo
hindered
10
oH
+
CHy I
CH3 OH
Cha
+::
Che
i:
H
Che
- Chy)
CHz
- CH
:
Br :
F
H
+
CHOH
CHy
+
co
Br :
CHyOH
+
Br
C
CH3
-
·
Because
it is a
tertiary
carbon
Reduction
We get
alcohol from
an
ketones and aldehydes
.
We
use
NaBHy
LiAlty
,
,
and
H24d
as
reagents
.
Examples :
O
OH
O
1)
LiA)H4
NaBHy
H
OH
O
g
2) H20
OH
O
NaBHy
LiA)H4
OH
Ot
O
LiAlH4
OH
OH
O
O
O
O
LiAlH4
Octy
Acidic
O
basic conditions
and
conditions
Basic
O
-
Examples
OH
LiAlH4
octly
O
Of
Acidic conditions :
:
Negatively charged
Positively charged
Attacks less substituted
Attacks most substituted
side
side
.
:
OH
Basic
OH
CH30
g
Locty
OCHy
OH
g
CHzONa
mPCBA
OH
Basic
g
L OCHy
CH3
Epoxide
CHzSNa
g
CHy
Basic
OH
t
CH3 OH
Ht
used
mPCBA
(a peroxide) it
enantiomer
OH
acidic
we
t
enantiomer
O
Since
ScHz
OCHE
&
OH
CHzO
OH
metEpic is
#
* OCH2CHa
g
g
Of
O CH2CH3
O
is
anti
Grignard reaction
Griguard
reagents
Examples of
excellent
are
reagents
grighard
Mgt
:
Mg I
nucleophiles
as
well
strong
as
bases.
Aldehydes and ketones
with
grignard
:
Mg
Br
based
carbon
:
ir :
MgBr
MgCl
MgBr
of grignard reactions :
Examples
OH
O
MgBr
Basic
1-
g
-
Grignard's
OH
carbon s
OH
CHzMgBr
O
O
0
0
# Chy
Basic
O
.
: 0:
1) Mg
ChyCH2CH2 Br
CHyCHzCHuMgBr
2)
CHaCHiCHnig
5-
OH
OH
St
CH2CH 2 CHy
3) H20
O
O
CHz MgBr
H
g
g-
- H
H
2Cha
O
O
g
OH
1) CH3 MgBr
G
O
2) H20
O
OH
08
1) CHzCHz MgBr
:CHzCHa
2) H28
IR)cCuLi
Dialkyl
copper
is
a
Chair
nucleophile that reacts with alkylhalides and vynil halides.
weak
Cho
2
Chi
~
Tu culi
Examples
(Tuculito xd
:
(CH3)2CuLi
Br
=
KHyCHzlcCuLi
Cho :
OH
OH
Basic
St
Br
CHy
g
will give
you
alcohol.
1 2
,
double bonds.
two
Diene
addition
1 4
vs
.
Sp2
Sp2
Sp2
Sp2
Conjugated diene
1. 2
addition
I
is
Kinetic
addition
1. 4
addition
# is
is
is
it
and
a
are
Illustration
at
happens
type of
thermodynamic control
They
Cummulated
:
type of reaction that involves
a
control
1. 4
conjugated
It have different orbitals
It is all the way sp?
addition
Sp
Sp3
Not
:
1. 2
Sp2
Sp2
sp2
reaction
happens
it
and
that
placed in allylic position
in
both
temperatures
low
the
involves
addition
the
such
,
1 2 and
,
1, 4
,
functional
78 C
and
°
as
to the first and
groups
lower
This is faster.
.
of the functional
addition
high temperatures
at
the
of
such
addition
,
as
so
2507
.
they
are
to the first and fourth carbon.
groups
This
is
second carbon
.
slower
.
very stable
,
but
1, 4
addition
1. 2
addition
is
more
:
Br
-
HBr
t
H
H
Rearrangement of carbonation
H%
t
-
H
:
Examples
:
H Br
488
O
Hir
400
1.4
1. 4
Br
addition
addition
O
Br
Br
g
HBr
°
-
78 C
1, 2
addition
O
Br
1, 4
addition
stable
alder
Diels
diels
The
six
alder
reaction
is
the
reaction
between
a
conjugated diene
and
an
alkene
(dienophile
member rings
.
Illustration
4
same
"Y
carbons
Dienophile
Diene
Examples
:
Cy
Cu
t
Cy
t
CO2 Et
zu
CN
L
CO2 Et
COzEt
t
CO2 Et
CO2 St
COzEt
COzEt
COzEt
t
COzEt
CO2 Et
COzEt
CO2 Et
to form
unsaturated