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Advanced Processing and Interpretation

In
this
case,
we
can
rule
out
C10H6N2 because
the
IR
spectrum
shows
that
there
is
a
carbonyl

in
Zone
4,
thus
revealing
an
oxygen
in
the
formula.

C10H6N2 lacks
an
oxygen.

You
can
also

rule
out
C10H202
because
it
does
not
correspond
with
the
integrals
given
in
the
1
H‐NMR.


The
integral
equals
9
(2+1+6),
and
H2 is
not
a
proportionate
ratio.

That
leaves
us
with

C10H180
as
our
formula.
C11
Possibility

154
–
(12
amu
per
C
x
11C)
=
22
amu
left
for
hydrogen,
nitrogen
and
oxygen
O N H Formula Does
it
work?
0 0 34 C10H34 No,
violates
H‐Rule
1 0 18 C10H180 Acceptable
2 0 2 C10H202 Acceptable
0 2 6 C10H6N2 Acceptable
O N H Formula Does
it
work?
C11H22
and
C11H8N
can
be
ruled
out
because
they
lack
an
oxygen.

C11H60
can
also
be
ruled

out
because
it
does
not
correspond
to
the
integral
given
to
us
on
the
1
H‐NMR.


Therefore,
our
final
formula
is
C10H18O.
The
last
step
is
to
calculate
the
DBE:
10
–
(18/2)
+
(0/2)
+
1
=
2
(possibly
two
rings
or
2
pi

bonds
or
a
ring
and
a
pi
bond)

  1. Now
we
have
to
analyze
the
IR
spectrum.

We
do
this
by
looking
at
each
zone
and
seeing
if

    there
are
peaks
that
correspond
to
different
functional
groups.
    Zone
1: Alcohol
O‐H:
absent;
no
peak
present

    Amine/Amide
N‐H:
absent;
no
peak,
no
nitrogen
in
formula
    Terminal
Alkyne
≡C‐H:
absent;
no
peak
in
zone
3
    Zone
2: Aryl/Vinyl
sp2
    
C‐H:
absent;
no
peak
>
3000
cm‐1
    Alkyl
sp3 C‐H:
present;
peaks
<
3000
cm‐1
    Aldehyde
C‐H:
absent,
no
peak
at
approx.
2700
cm‐1
    Carboxylic
Acid
C‐H:
absent;
no
broad
peak
    Zone
3: Alkyne
C≡C:
absent;
no
peak
    Nitrile
C≡N:
absent;
no
peak,
no
nitrogen
in
the
formula
    Zone
4: Carbonyl
C=O:
present,
peak
at
1716
cm‐1 
ketone
    Zone
5: Benzene
C=C:
absent;
no
peak,
not
enough
DBE
    0 0 22 C11H22 Acceptable
    1 0 6 C11H60 Acceptable
    0 1 8 C11H8N Acceptable
    Alkene
C=C:
absent;
no
peak
    From
the
IR
spectrum,
we
see
that
there
is
a
ketone
and
alkyl
sp3
    
C‐H
bonds
within
the

    structure.
  2. Now
we
have
to
analyze
the
1
    H‐NMR
given
to
us.

First,
put
all
given
information
into
a

    table:
    ppm Neighbors Integration #
H Implications
    2.1
ppm Singlet Integration
=
2
    1.9
ppm Singlet Integration
=
1
    1.1
ppm Singlet Integration
=
6
    Now
we
have
to
calculate
how
many
hydrogen
atoms
are
present
for
each
signal.

When

    you
add
up
each
integral
value,
you
get
a
total
of
9.

There
are
18
hydrogen
atoms
(as
seen

    in
the
molecular
formula),
so
it
is
a
1:2
ratio.

This
means
that
you
multiply
each
integral
by

    2
to
get
the
number
of
hydrogens:
    ppm Neighbors Integration #
H Implications
    2.1
ppm Singlet Integration
=
2 2×2
=
4
    1.9
ppm Singlet Integration
=
1 1×2
=
2
    1.1
ppm Singlet Integration
=
6 6×2
=
12
    Now
we
have
to
fill
in
the
implications:
    ppm Neighbors Integration #
H Implications
    2.1
ppm Singlet Integration
=
2 2×2
=
4 2x:
CH2








4x:
CH
    1.9
ppm Singlet Integration
=
1 1×2
=
2 CH2










2x:
CH
    1.1
ppm Singlet Integration
=
6 6×2
=
12 12x:
CH













4x:
CH3
    6x:
CH2