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Nuclear Magnetic Resonance

Example of how to solve an 1H-NMR Spectroscopy problem Formula:
C6H14O H‐NMR:
3.3
ppm
(triplet;
integral
=
1) 
1.6
ppm
(sextet;
integral
=
1) 
0.90
ppm
(triplet;
integral
=
1.5) 1. The
first
step
is
to
calculate
the
DBE
(double
bond
equivalency)
value,
which
will
tell
us
how
 many
pi
bonds
are
present
or
if
there
is
a
potential
benzene
ring
present
in
the
structure. DBE
=
#
of
carbons
–
(# of
hydrogens
÷
2)
+
(#
of
nitrogens
÷
2)

+
1 







=
6
–
(14/2)
+
(0/2)
+
1
 







=
0
(this
means
there
are
no
pi
bonds
within
the
structure) 2. Knowing
that
the
integral
is
proportional
to
the
relative
number
of
equivalent
hydrogen
 atoms
present
within
the
structure,
we
can
use
the
integral
information
given
in
the
 problem
and
the
number
of
hydrogens,
also
given
in
the
problem
through
the
molecular
 formula,
to
find
the
factor
in
which
to
multiply
the
integral
by
so
as
to
determine
how
many
 hydrogen
atoms
are
present
within
each
signal. The
first
signal
had
an
integral
=
1 The
second
signal
had
an
integral
=
1 The
third
signal
had
an
integral
=
1.5  If
you
multiply
each
of
these
numbers
by
a
factor
of
four,
then
you
get: 4
hydrogen
atoms
for
the
first
signal 4
hydrogen
atoms
for
the
second
signal 6
hydrogen
atoms
for
the
third
signal 3. Now
you
can
determine
the
implications
for
each
integral
and
splitting
pattern. ‐ Before
you
find
each
implication,
you
need
to
find
the
number
of
neighbors
each
 signal
has.

This
is
revealed
through
the
splitting
pattern
given
in
the
problem. The
first
signal
is
a
triplet.

From
the
general
rule
stated
above,
the
signal
for
a
 proton
with
n
neighbors
is
split
into
n+1
lines.

Because
there
are
3
splits
in
the
 signal,
this
means
that
there
are
3‐1
neighbors,
which
equals
two
neighbors.

 The
second
signal
is
a
sextet.

This
means
that
there
are
6‐1
neighbors,
which
 equals
five
neighbors.