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nucleic acids and carbohydrates

. Now
we
know
that
our
molecular
structure
is
composed
of
three
implications
that
make
up 2
CH3CH2CH2
structures.

But
if
you
look
at
the
circled
implication
from
the
First
Signal,
there
 is
an
attachment
on
the
first
CH2
(‐CH2CH2).

This
means
that
something
attaches
to
this
 carbon.
 If
we
look
at
the
original
formula
given
to
us,
we
see
that
we
have
accounted
for
everything except
the
oxygen
[there
are
7
hydrogen
atoms
x
2
=
14
hydrogen
atoms
total;
there
are
3
 carbon
atoms
x
2
=
6
carbons
total

this
is
how
many
carbons
and
hydrogens
were
in
the
 given
molecular
formula].

Now
we
have
to
account
for
the
oxygen
atom.

Because
there
is
 an
attachment
to
the
CH2,
we
can
attach
the
oxygen
to
this
carbon.

Our
structure
will
 therefore
look
like
this: CH3CH2CH2‐O‐CH2CH2CH3 And
there’s
your
molecular
structure
for
the
molecule! Note:
Make
sure
to
check
your
work!

Check
the
formula,
integrals,
number
of
signals,
 splitting,
etc.
and
make
sure
they
are
all
correct.

Always
perform
an
atom
check
and
 compare
it
to
the
formula
given.

Additionally,
check
to
see
if
your
structure
agrees
with
the
 DBE
for
the
molecule.

 Mass
spectrum:
m/z
=
154
(M;
100%),
m/z
=
155
(11.23%),
and
m/z
=
156
(0.26%) IR
spectrum:
 1 H‐NMR:
2.1
ppm
(singlet;
integral
=
2),
1.9
ppm
(singlet;
integral
=
1),
and
1.1
ppm
(singlet,
 integral
=
6) Now
let’s
do
another
example
involving
mass
spectroscopy
and
infrared
spectroscopy. 1. We
first
need
analyze
what
the
information
the
mass
spectroscopy
is
telling
us.

The
first
 m/z
value
is
the
M
peak,
which
tells
us
how
many
nitrogen
atoms
are
present
in
the
 molecular
formula.

If
the
m/z
value
is
even,
then
there
are
zero
or
an
even
amount
of
 nitrogen
atoms.

If
the
m/z
value
is
odd,
then
there
are
an
odd
amount
of
nitrogen
atoms. In
this
example,
the
M
peak
has
an
intensity
of
154.

This
is
an
even
number,
so
there
 are
0
or
an
even
amount
of
nitrogens
in
the
structure.