Chapter 7 Notes

Chapter 7 Notes - Chapter
7
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Unformatted text preview: Chapter
7
 Alkenes
and
Alkynes
I:
 Properties
and
Synthesis
 Elimination
Reactions
of
Alkyl
Halides
 The
(E)‐(Z)
System
for
Designating
Alkene
 Diastereomers
 The
Cahn‐Ingold‐Prelog
convention
is
used
to
assign
the
groups
of
highest
priority
on
 each
carbon.

If
the
group
of
highest
priority
on
one
carbon
is
on
the
same
side
as
the
 group
of
highest
priority
on
the
other
carbon
the
double
bond
is
Z
(zusammen).

If
the
 highest
priority
groups
are
on
opposite
sides
the
alkene
is
E
(entgegen).
 In
Class
Problem:
 Using
the
(E)‐(Z)
and
the
(R)‐(S)
designations
give
the
IUPAC
names
for
each
of
 the
following:
 Relative
Stabilities
of
Alkenes
 Generally
cis
alkenes
are
less
stable
than
trans
alkenes
because
of
steric
hinderance
 or
steric
interactions.
 Heat
of
Hydrogenation
 The
relative
stabilities
of
alkenes
can
be
measured
using
the
exothermic
heats
of
 hydrogenation.

The
same
alkane
product
must
be
obtained
to
get
accurate
results.
 Heats
of
hydrogenation
of
three
butene
isomers:
 Overall
Relative
Stabilities
of
Alkenes
 The
greater
the
number
of
attached
alkyl
groups
(i.e.
the
more
highly
substituted
 the
carbon
atoms
of
the
double
bond),
the
greater
the
alkene’s
stability
 In
Class
Problem:
 Arrange
the
following
alkenes
in
order
of
decreasing
stability:
1‐pentene;
(E)‐2‐ pentene;
(Z)‐2‐pentene;
2‐methyl‐2‐butene.
 Cycloalkenes
 trans‐Cyclohexene‐
highly
 strained;
does
not
exist
at
 room
temperature
 Synthesis
of
Alkenes
via
Elimination
Reactions
 Two
most
important
means
for
synthesizing
alkenes.
 Dehydrohalogenation
 The
following
conditions
favor
reaction
by
an
E2
mechanism:
 (1)  A
secondary
or
tertiary
halide
should
be
used‐
steric
hindrance
will
inhibit
 substitution.
 (2)  When
a
primary
halides
are
employed,
a
bulky
base
should
be
used‐
steric
 hindrance
will
inhibit
substitution.
 (3)  Strong
and
non‐polarizable
bases
such
as
alkoxides
should
be
used‐
will
reduce
 SN1
and
E1
competition.
 (4)  Sodium
ethoxide
in
ethanol
and
potassium
tert‐butoxide
in
tert‐butyl
alcohol
are
 typically
used
to
promote
E2
reactions.
 (5)  Elevated
temperature
generally
favor
elimination
over
substitution.
 In
the
following
examples
of
E2
elimination
processes
only
one
alkene
product
 is
possible.
 What
is
the
major
reaction
pathway
when
two
or
more
possible
alkene
 products
may
be
formed?
 Zaitsev’s
Rule
 It
has
been
observed
that
formation
of
the
most
substituted
alkene
is
favored
when
a
 “small”
base
is
employed
in
an
E2
elimination
process.

For
instance,
some
alkyl
 halides
can
eliminate
to
give
two
different
alkene
products.
 Zaitzev’s
Rule:
when
two
different
alkene
products
are
possible
in
an
E2
elimination
 process,

the
most
highly
substituted
(most
stable)
alkene
will
be
the
major
product.

 This
is
true
only
if
a
“small”
base
such
as
ethoxide
is
used.
 The
transition
state
in
this
E2
reaction
has
double
bond
character.

The
stability
of
the
 transition
state
will
be
reflected
in
the
stability
of
the
product
which
is
being
formed.

 The
more
stable
the
product,
the
smaller
the
activation
energy
and
the
faster
the
 rate.

The
trisubstituted
alkene‐like
transition
state
will
be
most
stable
and
have
the
 lowest
ΔG‡.


 Kinetic
control
of
product
formation:

When
one
of
two
products
is
formed
because
 its
free
energy
of
activation
is
lower
and
therefore
the
rate
of
its
formation
is
higher
 the
reaction
is
said
to
be
kinetically
controlled.
 Formation
of
the
Least
Substituted
Alkene
 Using
a
Bulky
Base
 Bulky
bases
such
as
potassium
tert‐butoxide
have
difficulty
removing
sterically
 hindered
hydrogens
and
generally
only
react
with
more
accessible
hydrogens
(e.g.
 primary
hydrogens).
 When
an
elimination
yields
the
less
substituted
alkene,
we
say
that
it
follows
 the
Hofmann
rule.
 The
Stereochemistry
of
E2
Reactions:
 The
Orientation
of
Groups
in
the
Transition
State
 In
the
E2
transition
state
all
four
atoms
involved
must
be
in
the
anti
coplanar
 orientation.

This
stereochemistry
is
preferred
because
all
atoms
are
staggered
and
 there
is
maximum
orbital
overlap.
 In
a
cyclohexane
ring
the
eliminating
substituents
must
be
diaxial
to
be
anti
 coplanar.
 Neomenthyl
chloride
and
menthyl
chloride
give
different
elimination
products
because
 of
this
requirement.
 In
neomenthyl
chloride,
the
chloride
is
in
the
axial
position
in
the
most
stable
 conformation.

Two
axial
hydrogens
anti
to
chlorine
can
eliminate;
the
Zaitzev
product
 is
major.
 In
menthyl
chloride
the
molecule
must
first
change
to
a
less
stable
conformer
to
 produce
an
axial
chloride.

Elimination
is
slow
and
can
yield
only
the
least
 substituted
(Hoffman)
product
from
anti
elimination.
 In
Class
Problem:
 Write
structural
formulas
for
all
the
products
that
would
be
obtained
when
2‐ bromo‐3‐methylbutane
is
heated
 (a)  with
sodium
ethoxide
in
ethanol
 (b)  with
potassium

tert‐butoxide
in
tert‐butyl
alcohol
 Acid
Catalyzed
Dehydration
of
Alcohols
 Elimination
is
favored
over
substitution
at
higher
temperatures.

Typical
acids
used
in
 dehydration
are
sulfuric
acid
and
phosphoric
acid.

The
temperature
and
concentration
 of
acid
required
to
dehydrate

depends
on
the
structure
of
the
alcohol.

Primary
alcohols
 are
most
difficult
to
dehydrate,
tertiary
are
the
easiest.
 Primary
alcohol:
 Secondary
Alcohols:
 Tertiary
Alcohols:
 Rearrangements
of
the
carbon
skeleton
can
occur.
 Mechanism
for
Dehydration
of
Secondary
and
 Tertiary
Alcohols:
An
E1
Reaction
 Only
a
catalytic
amount
of
acid
is
required
since
it
is
regenerated
in
the
final
step
of
 the
reaction.
 Carbocation
Stability
and
the
Transition
State
 Stability
of
carbocations
is:
 The
second
step
of
the
E1
mechanism
in
which
the
carbocation
forms
is
rate
 determining.

The
transition
state
for
this
reaction
has
carbocation
character
 (Leffler‐Hammond
Postulate).

Tertiary
alcohols
react
the
fastest
because
they
have
 the
most
stable
tertiary
carbocation‐like
transition
state
in
the
second
step.
 The
relative
heights
of
ΔG‡
for
the
second
step
of
E1
dehydration
indicate
that
 primary
alcohols
have
a
prohibitively
large
energy
barrier.
 The
developing
positive
charge
is
most
effectively
delocalized
in
the
transition
state
 leading
to
a
tertiary
carbocation‐
stabilization
by
hyperconjugation.
 A
Mechanism
for
Dehydration
of
Primary
Alcohols:
 An
E2
Reaction
 Primary
alcohols
cannot
undergo
E1
dehydration
because
of
the
instability
of
the
 carbocation‐like
transition
state
in
the
2nd
step.

In
the
E2
dehydration
the
first
step
is
 again
protonation
of
the
hydroxyl
to
yield
the
good
leaving
group
water.

The
second
 step
is
a
concerted
E2
elimination.
 Carbocation
Stability
and
the
Occurrence
of
 Molecular
Rearrangements
 Rearrangements
During
Dehydration
of
Secondary
Alcohols:
Rearrangements
of
 carbocations
occur
if
a
more
stable
carbocation
can
be
obtained.
 Example
 The
first
two
steps
are
to
same
as
for
any
E1
dehydration.
 In
the
third
step
the
less
stable
2o
carbocation
rearranges
by
shift
of
a
methyl
group
with
 its
electrons
(a
methanide).

This
is
called
a
1,2
shift.
 The
removal
of
a
proton
to
form
the
alkene
occurs
to
give
the
Zaitzev
(most
substituted)
 product
as
the
major
one
 A
hydride
shift
(migration
of
a
hydrogen
with
its
electrons)
can
also
occur
to
yield
the
 most
stable
carbocation.
 Carbocation
rearrangements
can
lead
to
formation
of
different
ring
sizes.
 In
Class
Problem:
 Acid
catalyzed
dehydration
of
either
2‐methyl‐1‐butanol
or
3‐methyl‐1‐butanol
 gives
2‐methyl‐2‐butene
as
the
major
product.

Write
out
the
reactions
and
suggest
 reasonable
mechanisms.
 Synthesis
of
Alkynes
by
Elimination
Reactions
 Alkynes
can
be
obtained
by
two
consecutive
dehydrohalogenation
reactions
of
a
vicinal
 dihalide.
 Alkenes
can
be
converted
to
alkynes
by
bromination
and
two
consecutive
 dehydrohalogenation
reactions.
 Geminal
dihalides
can
also
undergo
consecutive
dehydrohalogenation
reactions
to
 yield
the
alkyne.
 The
Acidity
of
Terminal
Alkynes
 Acetylenic
hydrogens
have
a
pKa
of
about
25
and
are
much
more
acidic
than
most
other
 C‐H
bonds.
 The
relative
acidity
of
acetylenic
hydrogens
in
solution
is:
 Acetylenic
hydrogens
can
be
deprotonated
with
relatively
strong
bases
(sodium
amide
is
 typical).

The
products
are
called
alkynides.
 Replacement
of
the
Acetylenic
Hydrogen
Atom
of
 Terminal
Alkynes
 Sodium
alkynides
can
be
used
as
nucleophiles
in
SN2
reactions.

New
carbon‐carbon
 bonds
result
when
alkynides
are
reacted
with
electophilic
molecules
such
a
alkyl
halides.

 Only
primary
alkyl
halides
can
be
used
or
else
elimination
reactions
predominate.
 Hydrogenation
of
Alkenes
 Hydrogen
adds
to
alkenes
in
the
presence
of
metal
catalysts.
This
process
is
called
a
 reduction
or
hydrogenation.

An
unsaturated
compound
becomes
a
saturated
(with
 hydrogen)
compound.


 Heterogeneous
catalysts:
finely
divided
insoluble
platinum,
palladium
or
nickel
catalysts.
 Homogeneous
catalysts:
catalyst(typically
rhodium
or
ruthenium
based)
is
soluble
in
the
 reaction
medium.

An
example
is
the
Wilkinson’s
catalyst:
Rh[(C6H5)3P]3Cl.
 Hydrogenation:
The
Function
of
the
Catalyst
 The
catalyst
provides
a
new
reaction
pathway
with
lower
ΔG‡
values.
 In
heterogeneous
catalysis
the
hydrogen
and
alkene
adsorb
onto
the
catalyst
surface
and
 then
a
step‐wise
formation
of
C‐H
bonds
occurs.
 Both
hydrogens
add
to
the
same
face
of
the
alkene
(a
syn
addition).

Addition
to
opposite
 faces
of
the
double
bond
is
called
anti
addition.
 Hydrogenation
of
Alkynes
 Reaction
of
hydrogen
with
an
alkyne
using
regular
metal
catalysts
results
in
formation
 of
the
alkane.
 Syn
Addition

of
Hydrogen:
Synthesis
of
cis‐Alkenes
 The
P‐2
catalyst
nickel
boride
results
in
syn
addition
of
one
equivalent
of
hydrogen
to
a
 triple
bond.

An
internal
alkyne
will
yield
a
cis
double
bond.
 Lindlar ’s
catalyst
also
produces
cis‐alkenes
from
alkynes.
 Anti
Addition
of
Hydrogen:
Synthesis
of
trans‐Alkenes
 A
dissolving
metal
reaction
which
uses
lithium
or
sodium
metal
in
low
temperature
 (liquid)
ammonia
or
amine
solvent
produces
trans‐alkenes.

Net
anti
addition
occurs
by
 formal
addition
of
hydrogen
to
the
opposite
faces
of
the
double
bond.
 The
mechanism
is
a
free
radical
reaction
with
two
electron
transfer
reactions
from
the
 metal.

The
vinylic
anion
prefers
to
be
trans
and
this
determines
the
trans
stereochemistry
 of
the
product.
 In
Class
Problem:
 Starting
with
ethyne,
outline
the
synthesis
(Z)‐2‐hexene
and
(E)‐2‐hexene.

You
 may
use
any
needed
reagents.
 Structural
Information
from
Molecular
Formulas
and
 the
Index
of
Hydrogen
Deficiency
(IHD)
 Unsaturated
and
Cyclic
Compounds
 A
compound
with
the
general
molecular
formula
CnH2n
will
have
either
a
double
bond
or
a
 ring.
 A
compound
with
general
formula
CnH2n‐2
can
have
a
triple
bond,
two
double
bonds,
a
 double
bond
and
a
ring
or
two
rings.
 Index
of
Hydrogen
Deficiency:
the
number
of
pairs
of
hydrogen
atoms
that
must
be
 subtracted
from
the
molecular
formula
of
the
corresponding
alkane
to
give
the
molecular
 formula
of
the
compound
under
consideration.
 Example:
A
compound
with
molecular
formula
C6H12
 Hydrogenation
allows
one
to
distinguish
a
compound
with
a
double
bond
from
one
 with
a
ring.
 Compounds
Containing
Halogens,
Oxygen,
or
Nitrogen
 For
compounds
containing
halogen
atoms,
the
halogen
atoms
are
counted
as
if
they
 were
hydrogen
atoms.
 Example:
A
compound
with
formula
C4H6Cl2.

This
is
equivalent
to
a
compound
with
 molecular
formula
C4H8
which
has
IHD=1.
 For
compounds
containing
oxygen,
the
oxygen
is
ignored
and
IHD
is
calculated
based
on
 the
rest
of
the
formula.
 Example:
A
compound
with
formula
C4H8O
has
IHD
=
1
 For
compounds
containing
nitrogen,
one
hydrogen
is
subtracted
for
each
nitrogen
and
 the
nitrogen
is
ignored
in
the
calculation.
 Example:
A
compound
with
formula
C4H9N
is
treated
as
if
it
has
formula
C4H8
and
has
 IHD
=
1.
 ...
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