Chapter 9 Notes

Chapter 9 Notes - Chapter
9


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Unformatted text preview: Chapter
9
 Nuclear
Magnetic
Resonance
and
Mass
 Spectroscopy:
 Tools
for
Structure
Determination
 Introduction
 Spectroscopy:
the
study
of
the
interaction
of
energy
with
matter.

Energy
applied
to
 matter
can
be
absorbed,
emitted,
cause
a
chemical
change,
or
be
transmitted.

 Spectroscopy
can
be
used
to
elucidate
the
structure
of
a
molecule.
 Examples
of
Spectroscopy
 (1)
Infrared
(IR)
Spectroscopy
(Chapter
2).

Infrared
energy
causes
bonds
to
stretch
and
 bend.

IR
is
useful
for
identifying
functional
groups
in
a
molecule.
 (2)
Nuclear
Magnetic
Resonance
(NMR).

Energy
applied
in
the
presence
of
a
strong
 magnetic
field
causes
absorption
by
the
nuclei
of
some
elements
(most
importantly,
 hydrogen
and
carbon
nuclei).

NMR
is
used
to
identify
connectivity
of
atoms
in
a
 molecule.
 (3)
Mass

Spectrometry
(MS).

Molecules
are
converted
to
ions
by
one
of
several
methods
 (including
bombardment
by
a
beam
of
electrons).

The
ions
formed
may
remain
intact
(as
 molecular
ions,
M+),
or
they
may
fragment.

The
resulting
mixture
of
ions
is
sorted
by
 mass/charge
(m/z)
ratio,
and
detected.

Molecular
weight
and
chemical
formula
may
be
 derived
from
the
M+
and
M+1
ions.

Molecular
structure
may
be
deduced
from
the
 distribution
of
fragment
ions.

 The
Electromagnetic
Spectrum
 Electromagnetic
radiation
has
the
characteristics
of
both
waves
and
particles.

The
wave
 nature
of
electromagnetic
radiation
is
described
by
wavelength
(λ)
or
frequency
(ν). The
relationship
between
wavelength
(or
frequency)
and
energy
(E)
is
well
defined.
 Wavelength
and
frequency
are
inversely
proportional
(ν=
c/λ). The
higher
the
 frequency,
the
greater
the
energy
of
the
wave.

The
shorter
the
wavelength,
the
greater
 the
energy
of
the
wave.
 NMR
involves
absorption
of
energy
in
the
radiofrequency
range.
 Nuclear
Magnetic
Resonance
(NMR)
Spectroscopy
 The
nuclei
of
protons
(1H)
and
carbon‐13
(13C),
and
certain
other
elements
and
 isotopes,
behave
as
if
they
were
tiny
bar
magnets.

When
placed
in
a
magnetic
field
 and
irradiated
with
radio
frequency
energy,
these
nuclei
absorb
energy
at
frequencies
 based
on
their
chemical
environments.

NMR
spectrometers
are
used
to
measure
 these
absorptions.
 Chemical
Shift:

Peak
Position
in
an
NMR
Spectrum
 Nuclei
in
different
chemical
environments
in
a
molecule
will
absorb
at
slightly
different
 frequencies.

The
position
of
the
signals
in
the
spectrum
is
called
the
“chemical
shift.”

 There
are
two
reasons
for
differences
in
the
magnetic
environment
for
a
proton:
 (1)
The
magnetic
field
generated
by
electrons
circulating
around
the
nucleus
giving
the
 signal,
and
 (2)
Local
magnetic
fields
generated
by
electrons
elsewhere
in
the
molecule.
 NMR
Spectometers
 Continuous‐Wave
(CW)
NMR
Spectrometers
 This
is
the
oldest
type
of
NMR
spectrometer.

The
magnetic
field
is
varied
as
the
 electromagnetic
radiation
is
kept
at
a
constant
frequency.

Different
nuclei
absorb
the
 electromagnetic
energy
based
on
their
chemical
environment
and
produce
peaks
in
 different
regions
of
the
spectrum.
 Fourier
Transform
(FT)
NMR
Spectrometers
 The
sample
is
placed
in
a
constant
(and
usually
very
strong)
magnetic
field.

The
 sample
is
irradiated
with
a
short
pulse
of
radio
frequency
energy
that
excites
the
 nuclei
in
different
environments
all
at
once.

The
resulting
signal
contains
information
 about
all
of
the
absorbing
nuclei.

This
signal
is
converted
to
a
spectrum
by
a
Fourier
 transformation.

FT
NMR
allows
signal‐averaging,
which
leads
to
enhancement
of
real
 spectral
signals
versus
noise.

The
strong,
superconducting
magnets
used
in
FTNMR
 spectrometers
lead
to
greater
sensitivity
and
much
higher
resolution
than
continuous
 wave
instruments.
 Example:
1,4‐dimethylbenzene
 The
spectrum
is
measured
on
a
delta
(δ)
scale
in
units
of
parts
per
million
(ppm).

Lower
 frequency
is
to
the
left
in
the
spectrum;
these
absorptions
are
said
to
be
downfield.

 Higher
frequency
is
to
the
right
in
the
spectrum:
these
absorptions
are
said
to
be
upfield.

 The
small
signal
at
δ
0
corresponds
to
an
internal
standard
called
tetramethylsilane

(TMS)
 used
to
calibrate
the
chemical
shift
scale.

The
number
of
signals
in
the
spectrum
 corresponds
to
the
number
of
unique
sets
of
protons.

1,4‐dimethylbenzene
has
protons
 in
two
unique
environments
and
so
shows
two
signals.
 Integration
of
Peak
Areas.
The
Integral
Curve
 The
area
under
each
signal
corresponds
to
the
relative
number
of
hydrogen
atoms
in
 each
unique
environment
within
a
molecule.

The
height
of
each
step
in
the
integral
 curve
is
proportional
to
the
area
of
the
signal
underneath
the
step.
 Signal
Splitting
 The
signal
from
a
given
proton
will
be
split
by
the
effect
of
magnetic
fields
associated
 with
protons
on
adjacent
carbons.

Characteristic
peak
patterns
result
from
signal
 splitting
that
are
related
to
the
number
of
protons
on
adjacent
carbons.
 Example:
1,1,2‐trichloroethane
 Nuclear
Spin:
The
Origin
of
the
Signal
 The
nuclei
of
certain
elements
and
isotopes
have
spin
states
that
are
quantized.

1H
has
 a
spin
quantum
number
I =
1/2
and
has
allowed
spin
states
of
+1/2
or
‐1/2.

Other
 nuclei
with
I =
1/2
are
13C,
19F
and
31P
and
these
also
respond
to
an
external
magnetic
 field.

Nuclei
with
I
=
0
do
not
have
spin
(12C
and
16O)
and
do
not
respond
to
an
external
 magnetic
field.

The
nuclei
of
NMR‐active
nuclei
behave
like
tiny
bar
magnets.

In
the
 absence
of
an
external
magnetic
field
these
bar
magnets
are
randomly
orientated.

In
an
 external
magnetic
field
they
orient
either
with
(α
spin
state)
or
against
(β
spin
state)
the
 magnetic
field.
 Nuclei
aligned
with
the
magnetic
field
are
lower
in
energy
than
those
aligned
against
the
 field.

The
nuclei
aligned
with
the
magnetic
field
can
be
flipped
to
align
against
it
if
the
 right
amount
of
energy
is
added
(ΔE).

The
amount
of
energy
required
depends
on
the
 strength
of
the
external
magnetic
field.

The
stronger
the
external
magnetic
field,
the
 higher
the
radio
frequency
energy
required
to
flip
the
nuclear
spin.
 At
(a)
there
is
no
external
magnetic
field
and
therefore
no
energy
difference
between
the
 two
states.

At
(b)
the
external
magnetic
field
is
1.41
Tesla
and
energy
corresponding
to
a
 frequency
of
about
60MHz
is
needed
to
flip
between
the
spin
states.

At
(c)
the
external
 magnetic
field
is
7.04
Tesla
energy
corresponding
to
a
frequency
of
about
300MHz
is
 needed
to
flip
between
the
spin
states.
 Shielding
and
Deshielding
of
Protons
 Protons
in
an
external
magnetic
field
absorb
at
different
frequencies
depending
on
the
 electron
density
around
that
proton.

High
electron
density
around
a
nucleus
shields
the
 nucleus
from
the
external
magnetic
field.

Shielding
causes
absorption
of
energy
at
higher
 frequencies
(more
energy
is
required
for
this
nucleus
to
flip
between
spin
states)
‐
the
 signals
are
upfield
in
the
NMR
spectrum.

Lower
electron
density
around
a
nucleus
 deshields
the
nucleus
from
the
external
magnetic
field.

Deshielding
causes
absorption
of
 energy
at
lower
frequencies
(less
energy
is
required
for
this
nucleus
to
flip
between
spin
 states)
‐
the
signals
are
downfield
in
the
NMR
spectrum.
 Some
Generalizations
 (1)  Electronegative
atoms
draw
electron
density
away
from
nearby
protons
and
 therefore
deshield
them.
 (2)  (2)
Circulation
of
π
electrons
leads
to
a
local
induced
magnetic
field.

The
induced
 field
can
reinforce
or
diminish
the
external
field
sensed
by
a
proton
(depending
on
 the
location
of
the
proton),
causing
deshielding
or
shielding,
respectively.
 Alkene
and
aromatic
ring
hydrogens
are
deshielded
by
the
circulation
of
π
electrons.



 A
terminal
alkyne
hydrogen
is
shielded
by
the
circulation
of
π
electrons.



 Chemical
Shift
 Chemical
shifts
are
measured
in
relation
to
the
internal
reference
tetramethylsilane
(TMS).

 The
protons
of
TMS
are
highly
shielded
because
of
the
strong
electron
donating
capability
 of
silicon.

The
signal
for
TMS
is
well
away
from
most
other
proton
absorptions.
 The
δ
scale
for
chemical
shifts
is
independent
of
the
magnetic
field
strength
of
the
 instrument
(whereas
the
absolute
frequency
depends
on
field
strength)
 Thus,
the
chemical
shift
in
δ
units
for
protons
on
benzene
is
the
same
whether
a
60
MHz
or
 300
MHz
instrument
is
used.
 Chemical
Shift
Equivalent
and
Nonequivalent
Protons
 To
predict
the
number
of
signals
to
expect
in
an
NMR
spectrum
it
is
necessary
to
 determine
how
many
sets
of
protons
are
in
unique
environments.

Chemically
equivalent
 protons
are
in
the
same
environment
and
will
produce
only
one
signal.


 Homotopic
Hydrogens
 Hydrogens
are
chemically
equivalent
or
homotopic
if
replacing
each
one
in
turn
by
the
 same
group
would
lead
to
an
identical
compound.

The
six
methyl
hydrogens
are
 homotropic
and
the
two
vinyl
hydrogens
are
homotropic.
 Enantiotopic
and
Diastereotopic
Hydrogen
Atoms
 If
replacement
of
each
of
two
hydrogens
by
some
group
leads
to
enantiomers,
those
 hydrogens
are
said
to
be
enantiotopic.

In
the
absence
of
a
chiral
influence,
 enantiotopic
hydrogens
have
the
same
chemical
shift
and
appear
at
the

same
signal.
 If
replacement
of
each
of
two
hydrogens
by
some
group
leads
to
diastereomers,
the
 hydrogens
are
diastereotopic.

Diastereotopic
hydrogens
have
different
chemical
shifts
 and
will
give
different
signals.
 Signal
Splitting:
Spin‐Spin
Coupling
 The
signal
from
a
given
proton
will
be
split
by
the
effect
of
magnetic
fields
associated
 with
protons
on
adjacent
carbons.

Characteristic
peak
patterns
result
from
signal
 splitting
that
are
related
to
the
number
of
equivalent
protons
on
adjacent
carbons.

The
 effect
of
signal
splitting
is
greatest
between
atoms
separated
by
3
or
fewer
σ
bonds.
 Signal
splitting
is
not
observed
between
homotopic
or
enantiotopic
protons.
 Signal
splitting
occurs
only
when
two
sets
of
protons
have
different
chemical
shifts
(i.e.,
 are
not
chemical
shift
equivalent).

 The
magnetic
field
sensed
by
a
proton
 (Ha)
being
observed
is
affected
by
the
 magnetic
moment
of
an
adjacent
 proton
(Hb).

A
proton
(Hb)
can
be
 aligned
with
the
magnetic
field
or
 against
the
magnetic
field,
resulting
in
 two
energy
states
for
Hb.

The
 observed
proton
(Ha)
senses
the
two
 different
magnetic
moments
of
Hb
as
a
 slight
change
in
the
magnetic
field;
 one
magnetic
moment
reinforces
the
 external
field
and
one
substracts
from
 it.

The
signal
for
Ha
is
split
into
a
 doublet
with
a
1:1
ratio
of
peak
areas.

 The
magnitude
of
the
splitting
is
called
 the
coupling
constant
Jab
and
is
 measured
in
Hertz
(Hz).
 When
two
adjacent
protons
Hb
are
coupled
to
Ha,
there
are
four
possible
combinations
of
 the
magnetic
moments
for
the
two
Hbs.

Two
of
these
combinations
involve
pairings
of
 magnetic
moments
that
cancel
each
other,
causing
no
net
displacement
of
signal.

One
 combination
of
magnetic
moments
reinforces
and
another
subtracts
from
the
applied
 magnetic
field.

Ha
is
split
into
a
triplet
having
a
1:2:1
ratio
of
signal
areas
 When
three
adjacent
protons
are
coupled
to
Ha,
there
are
10
possible
combinations
 of
the
magnetic
moments
for
the
Hb’s.

Four
unique
orientations
exist
and
so
Ha
is
 split
into
a
quartet
with
intensities

1:4:4:1.
 The
general
rule
for
splitting
is
that
if
there
are
n
equivalent
protons
on
adjacent
atoms,
 these
will
split
a
signal
into
n
+
1
peaks.

Coupled
peaks
have
the
same
coupling
constants
 Jab.

Comparison
of
coupling
constants
can
help
with
the
analysis
of
complex
spectra.
 Several
factors
complicate
analysis
of
NMR
spectra:
 (1)
Peaks
may
overlap.
 (2)
Spin‐spin
coupling
can
be
long‐range
(i.e.,
more
than
3
bonds)
 (3)
Splitting
patterns
in
aromatic
groups
can
be
confusing
 For
instance,
a
monosubstituted
aromatic
ring
can
appear
as
an
apparent
singlet
or
a
 complex
pattern
of
peaks.
 Much
more
complex
splitting
can
occur
when
two
sets
of
adjacent
protons
split
a
 particular
set
of
protons.

In
the
system
below,
Hb
is
split
by
two
different
sets
of
 hydrogens
:
Ha
and
Hc.

Theortically
Hb
could
be
split
into
a
triplet
of
quartets
(12
peaks)
 but
this
complexity
is
rarely
seen.
 Example:
The
spectrum
of
1‐nitropropane
shows
splitting
of
Hb
into
only
6
peaks.
 Proton
NMR
Spectra
and
Rate
Processes
 An
NMR
spectrometer
is
like
a
camera
with
a
slow
shutter
speed.

The
NMR
 spectrometer
will
observe
rapid
processes
as
if
they
were
a
blur,
i.e.,
only
an
average
 of
the
changes
will
be
seen.

For
instance,
when
a
1H
NMR
spectrum
of
very
pure
 ethanol
is
taken,
the
hydroxyl
proton
is
split
into
a
triplet
by
the
two
adjacent
 hydrogens.

When
an
1H
NMR
of
“regular ”
ethanol
is
taken
the
hydroxyl
proton
is
a
 singlet.

Why?

Impure
ethanol
contains
acid
and
base
impurities
which
catalyze
the
 exchange
of
hydroxyl
protons.

This
rapid
exchange
is
so
fast
that
coupling
to
the
 adjacent
CH2
is
not
observed.

This
process
is
called
spin
decoupling.
 Spin
decoupling
is
 typical
in
the
1H
 NMR
spectra
of
 alcohols,
amines
and
 carboxylic
acids.

The
 proton
attached
to
 the
oxygen
or
 nitrogen
normally
 appears
as
a
singlet
 because
of
rapid
 exchange
processes.
 In
Class
Problem:
 A
compound,
C8H10O,
has
the
following
IR
and
1H
NMR
spectra.

What
is
the
 structure?
 Carbon‐13
NMR
Spectroscopy
 13C
(spin
=
½)
accounts
for
only
1.1%
of
naturally
occurring
carbon.

12C
has
no
magnetic
 spin
and
produces
no
NMR
signal.


 One
Peak
for
Each
Unique
Carbon
Atom
 Since
the
13C
isotope
of
carbon
is
present
in
only
1.1%
natural
abundance,
there
is
only
a
 1
in
10,000
chance
that
two
13C
atoms
will
occur
next
to
each
other
in
a
molecule.

The
 low
probability
of
adjacent
13C
atoms
leads
to
no
detectable
carbon‐carbon
splitting.

1H
 and
13C
do
split
each
other,
but
this
splitting
is
usually
eliminated
by
adjusting
the
NMR
 spectrophotometer
accordingly.

The
process
of
removing
the
coupling
of
1H
to
an
 attached
carbon
is
called
broadband
(BB)
proton
decoupling.

Most
13C
NMR,
therefore,
 consist
of
a
single
peak
for
each
unique
carbon.
 Off‐Resonance
Decoupled
Spectra
 Broad‐band
decoupling
removes
all
information
about
the
number
of
hydrogens
attached
 to
each
carbon.

Off‐resonance
decoupling
removes
some
of
the
coupling
of
carbons
to
 hydrogens
so
that
the
coupled
peaks
will
not
overlap.

Use
of
off‐resonance
decoupled
 spectra
has
been
replaced
by
use
of
DEPT
13C
NMR.
 DEPT
13C
NMR
 DEPT

(distortionless
enhanced
polarization
transfer)
spectra
are
created
by
 mathematically
combining
several
individual
spectra
taken
under
special
conditions.

The
 final
DEPT
spectra
explicitly
show
C,
CH,
CH2
,
and
CH3
carbons.

To
simplify
the
 presentation
of
DEPT
data,
the
broadband
decoupled
spectrum
is
annotated
with
the
 results
of
the
DEPT
experiments
using
the
labels
C,
CH,
CH2
and
CH3
above
the
appropriate
 peaks.
 Example:
1‐chloro‐2‐propanol
 (a)
The
broadband
decoupled
spectrum
and
(b)
a
set
of
DEPT
spectra
showing
the
 separate
CH,
CH2,
and
CH3
signals/
 13C
Chemical
Shifts
 Just
as
in
1H
NMR
spectroscopy,
chemical
shifts
in
13C
NMR
depend
on
the
electron
density
 around
the
carbon
nucleus.

Decreased
electron
density
causes
the
signal
to
move
 downfield
(desheilding).

Increased
electron
density
causes
the
signal
to
move
upfield
 (sheilding)
 Because
of
the
wide
range
of
chemical
shifts,
it
is
rare
to
have
two
13C
peaks
coincidentally
 overlap.

A
group
of
3
peaks
at
δ 77
comes
from
the
common
NMR
solvent
 deuteriochloroform
andcan
be
ignored
in
structure
analysis.
 Example:
13C
NMR
of
Pentane.
 Example:
13C
NMR
of
2‐Methylbutane.
 Example:
13C
NMR
or
2‐Pentanol.
 In
Class
Problem:
 Which
of
the
following
two
structures
is
consistent
with
the
13C
NMR
spectrum
 given
below?

 Introduction
to
Mass
Spectrometry
(MS)
 Mass
spectrometry
(MS)
involves
formation
of
ions
in
a
mass
spectrometer
followed
by
 separation
and
detection
of
these
ions
according
to
mass
and
charge.


A
mass
 spectrometer
produces
a
spectrum
of
masses
based
on
the
structure
of
a
molecule.


 A
mass
spectrum
is
a
plot
of
the
distribution
of
ion
masses
corresponding
to
the
 formula
weight
of
a
molecule
and/or
fragments
derived
from
it.

The
x‐axis
of
a
mass
 spectrum
represents
the
masses
of
ions
produced.

The
y‐axis
represents
the
relative
 abundance
of
each
ion
produced.

The
pattern
of
ions
obtained
and
their
abundance
is
 characteristic
of
the
structure
of
a
particular
molecule.
 The
Mass
Spectrometer
 One
common
type
is
the
Electron
Impact
Mass
Spectrometer
(EI
MS).

A
molecule
is
 bombarded
with
a
beam
of
high
energy
electrons.

An
electron
may
be
dislodged
from
 the
molecule
by
the
impact,
leaving
a
positively
charged
ion
with
an
unpaired
electron
 (a
radical
cation).

This
initial
ion
is
called
the
molecular
ion
(M+.)
because
it
has
the
 same
molecular
weight
as
the
analyte
.
 Examples:
 (1)  Ionization
of
an
alkane
with
only
sigma
bonds.
 (2)  Ionization
of
molecules
with
atoms
containing
lone
pair
and
pi
electrons.

 Ion
Sorting
 The
fragments
are
sorted
according
to
their
mass
to
charge
ratio,
(m/z).

Most
of
the
 fragments
detected
have
charge
+1;

the
net
effect
is
sorting
of
the
ions
by
mass
(m/z,
 where
z
=
+1).

The
charged
molecular
ion
(M+)
and
fragments
pass
through
an
analyzer
 that
sorts
the
ions
according
to
m/z.

One
method
of
sorting
involves
directing
the
ions
 through
a
curved
tube
that
passes
through
a
magnetic
field;
as
the
magnetic
field
is
 varied,
ions
of
different
m/z
values
successfully
traverse
the
tube
and
reach
the
detector.

 After
ion
sorting
the
results
are
plotted
as
a
spectrum
with
m/z
on
the
horizontal
axis
and
 relative
abundance
of
each
ion
on
the
vertical
axis.
 The
M+.
Ion
is
formed
by
loss
of
one
of
its
most
loosely
held
electrons.

I
nonbonding
 electron
pairs
or
pi
electrons
are
present,
an
electron
from
“one
of
these
locations”
is
 usually
lost
by
electron
impact
to
for
M+.

Loosely
held
nonbonding
electrons
on
nitrogen
 and
oxygen,
and
pi
electrons
in
double
bonds
are
common
“locations”
for
an
electron
to
 be
lost
(i.e.,
where
the
remaining
unshared
electron
in
M+.
“resides”).

In
molecules
with
 only
C‐C
and
C‐H
bonds,
the
“location”
of
the
lone
electron
cannot
be
predicted
and
the
 formula
is
written
to
reflect
this
using
brackets
(as
shown
on
the
previous
slide).
 Fragmentation
 Excess
vibrational
energy
is
imparted
to
the
molecular
ion
as
a
result
of
collision
with
 the
electron
beam
‐

this
causes
fragmentation.

Fragmentation
pathways
are
 predictable
and
can
be
used
to
determine
the
structure
of
a
molecule.

The
processes
 that
causes
fragmentation
are
unimolecular.

The
relative
ion
abundance
is
extremely
 important
in
predicting
structures
of
fragments.

The
fragmentation
pattern
is
highly
 characteristic
of
the
structure
of
the
molecule.
 The
Mass
Spectrum
 Data
from
a
mass
spectrometer
can
be
represented
as
a
graph
or
table.

The
most
 abundant
(intense)
peak
in
the
spectrum
is
called
the
base
peak
and
is
assigned
a
 normalized
intensity
of
100%.

The
masses
are
based
on
rounding
of
atom
masses
to
the
 nearest
whole
number
(in
low
resolution
mass
spectroscopy).

The
data
and
fragmentation
 patterns
for
propane
are
as
follows.
 Fragmentation
by
Cleavage
at
a
Single
Bond
 Cleavage
of
a
radical
cation
occurs
to
give
a
radical
and
a
cation
but
only
the
cation
is
 observable
by
MS.

In
general
the
fragmentation
proceeds
to
give
mainly
the
most
stable
 carbocation.

For
instance,
in
the
spectrum
of
propane
the
peak
at
29
is
the
base
peak
 (most
abundant)
100%
and
the
peak
at
15
is
5.6%.
 The
Mass
Spectrum
of
Ammonia
 The
small
peak
at
m/z
18
comes
from
 the
small
amount
of
15N1H3
because
of
 the
small
natural
abundance
of
15N
 compared
to
14N.

This
peak
is
called
 an
M+1
peak.
 The
exact
mass
of
certain
nuclides
is
shown
below.
 Determination
of
Molecular
Formulas
 and
Molecular
Weights
 The
Molecular
Ion
and
Isotopic
Peaks
 The
presence
of
heavier
isotopes
one
or
two
mass
units
above
the
common
isotope
yields
 small
peaks
at
M+.+1
and
M+.+2.
 The
intensity
of
the
M+.+1
and
M+.+2
peaks
relative
to
the
M
peak
can
be
used
to
confirm
a
 molecular
formula.

Example:
In
the
spectrum
of
methane
one
expects
an
M+.+1
peak
of
 1.17%
based
on
a
1.11%
natural
abundance
of
13C
and
a
0.016%
natural
abundance
of
2H.
 High‐Resolution
Mass
Spectrometry
 Low‐resolution
mass
spectrometers
measure
m/z
values
to
the
nearest
whole
number.


 High‐resolution
mass
spectrometers
measure
m/z
values
to
three
or
four
decimal
 places.

The
high
accuracy
of
the
molecular
weight
calculation
allows
accurate
 determination
of
the
molecular
formula
of
a
fragment.
 Example
 One
can
accurately
pick
the
molecular
formula
of
a
fragment
with
a
nominal
molecular
 weight
of
32
using
high‐resolution
MS.
 Example:
Spectrum
of
Hexane
 Example:
Spectrum
of
Neopentane
 Fragmentation
of
neopentane
shows
the
propensity
of
cleavage
to
occur
at
a
branch
 point
leading
to
a
relatively
stable
carbocation.

The
formation
of
the
3o
carbocation
is
 so
favored
that
almost
no
molecular
ion
is
detected.
 (1)
Carbocations
stabilized
by
resonance
are
also
formed
preferentially.

Alkenes
 fragment
to
give
resonance‐stabilized
allylic
carbocations.
 (2)
Carbon‐carbon
bonds
next
to
an
atom
with
an
unshared
electron
pair
break
readily
to
 yield
a
resonance
stabilized
carbocation,
where
Z
=
N,
O,
or
S
and
R
may
be
H.
 (3)
Carbon‐carbon
bonds
next
to
carbonyl
groups
fragment
readily
to
yield
resonance
 stabilized
acylium
ions.
 (4)
Alkyl
substituted
benzenes
often
lose
a
hydrogen
or
alkyl
group
to
yield
the
relatively
 stable
tropylium
ion.
 (5)
Other
substituted
benzenes
usually
lose
their
substitutents
to
yield
a
phenyl
cation.
 (6)
Fragmentation
can
occur
by
the
cleavage
of
two
bonds.

The
products
are
a
new
radical
 cation
and
a
neutral
molecule.

For
instance,
alcohols
usually
show
an
M+.‐18
peak
from
the
 loss
of
a
water
molecule.
 (7)
Cycloalkenes
can
undergo
a
retro‐Diels
Alder
reaction
(section
13.11)
to
yield
an
 alkadienyl
radical
cation.
 (8)
Carbonyl
compounds
can
undergo
a
McLafferty
Rearrangement
where
Y
may
be
R,
H,
 OH,
OR
etc.
 In
Class
Problem:
 A
student
has
two
flasks
each
containing
an
organic
liquid.

One
flask
contains
butyl
 propyl
ether
and
the
other
contains
butyl
isopropyl
ether.

Unfortunately
the
flasks
 are
not
labeled.

In
order
to
place
the
proper
label
on
each
flask
the
student
 obtained
the
mass
spectrum
of
each
of
the
liquids.

Identify
the
isomer
associated
 with
each
of
the
following
mass
spectra.


 In
Class
Problem:
 From
the
mass
spectrum,
IR
spectrum,
and
NMR
spectrum
determine
the
structure
of
 the
following
unknown
compound
of
formula
C5H10O
with
MW
=
86.
 (a)
Index
of
Hydrogen
Deficiency
(IHD):

For
C5H10O
omit
O
which
results
in
C5H10.

This
 is
consistent
with
1
deg
of
unsaturation:
carbonyl,
double
bond,
or
one
ring
could
be
 present.
 (b)
Mass
Spectrum:
 Analysis:
 (a)  MW
=
86
 (b)  Fragments
at
 
 (1)
71
(M‐15):
‐CH3,
‐CO
 
 (2)
58
(M‐28):
‐CH2CH2

 
 (3)
43
(M‐43):
‐C3H7,
‐C2H3O
 
 (4)
29
(M‐57):
‐C3H5O
 (b)
Infrared
Spectrum
 Analysis:
 (a)
Absorption
at
1716
indicates
a
carbonyl
 stretch:
probably
a
ketone.
 (b)
Absorptions
at
3000‐2850
indicate
C‐H
 alkane
stretches.
 (d)
NMR
Spectrum:
 Analysis:
 (a)  3H
singlet
at
2.13
ppm‐
no
adjacent
hydrogens:
 methyl
ketone,
benzylic,
acetylenic
 (b)  3H
triplet
at
0.93
ppm‐
two
adjacent
hydrogens:
 
 methyl
 (c)  2H
triplet
at
2.41
ppm‐
two
adjacent
hydrogens:
 
 methylene
adjacent
to
ketone
 (c)  2H
sextet
at
1.61
ppm‐
five
adjacent
hydrogens:
 
 
methylene
adjacent
to
a
methyl
and
another
 methylene 



 (d)
NMR
Spectrum
(continued):
 Proposed
Structure:
 Is
the
proposed
structure
consistent
with
the
mass
spectrum?

We
must
account
for
m/ z
at
43,
71,
29,
and
58.


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