Acyclic
to
Cyclic
Carbohydrates:
A
Focus
on
Glucose

Having
trouble
seeing
carbohydrates
in
their
acyclic
and
cyclic
form?
Confused
at
the
connectivity
and
conformation?
Well,
you’ve
come
to
the
right
place.
We’ll
be
looking
at
the
acyclic
form
of
glucose
in
particular
and
how
the
acyclic
form
converts
to
the
cyclic.
To
go
about
this
topic,
the
best
idea
is
probably
to
first
build
a
model.
We
will
begin
by
building
the
structure
based
on
the
image
below.
When
looking
at
the
general
and
commonly
seen
acyclic
structure,
you
may
notice
that
both
the
hydrogen
atoms
and
the
alcohol
groups
project
towards
the
viewer.
This
may
confuse
some
people
who
are
used
to
the
presence
of
one
wedge
and
one
group
connected
with
dashed
lines.
It
also
may
appear
deceiving
because
the
image
presents
the
molecule
almost
as
if
it
were
planar.
Build
the
model
so
that
the
carbonyl
is
at
the
top
and
the
CH2OH
group
is
closest
to
the
bottom.
The
wedges
If
focusing
on
an
individual
carbon,
let’s
show
that
the
say
the
first
after
the
carbonyl,
you
will
hydrogen
see
that
the
carbons
connected
to
it
will
atoms
and
the
point
way
from
the
viewer
as
alcohol
groups
represented
with
dashed
lines
and
the
point
toward
other
groups
connected
will
point
the
viewer.
towards
the
viewer.
This
will
be
seen
in
the
built
model.

You
can
see
that
the
carbonyl
group
and
the
CH2OH
group
collide
into
one
another
After
building
the
model,
you
find
that
the
connected
carbon
atoms
form
a
ring
shaped
structure,
with
the
CH2OH
group
and
the
carbonyl
colliding
into
one
another.
Be
sure
to
rotate
the
atoms
so
that
the
hydrogen
atoms
and
the
alcohol
groups
of
the
carbon
stereocenters
point
toward
the
viewer.
If
you
were
to
look
at
each
individual
carbon
that
is
a
stereocenter
(the
carbons
excluding
those
in
the
carbonyl
and
in
the
CH2OH
group)
you
can
see
that
it
fits
the
image
above.
The
carbons
connected
to
the
particular
carbon
that
you
choose
to
focus
on
point
away
from
the
viewer
while
the
other
connected
point
toward
the
viewer.
Here
is
another
view
of
the
molecule
from
its
side.
From
this
angle,
which
is
a
different
angle
than
the
original
image
from
which
we
built
the
model
based
off
of,
we
can
see
a
clearer
image
of
the
colliding
carbonyl
and
CH2OH
group.
Already,
you
can
start
to
see
how
the
cyclic
form
will
take
place.
Switching
to
this
view,
we
will
now
convert
the
acyclic
form
to
the
cyclic
form.
Twist
the
molecule
as
needed
so
that
the
carbonyl
and
CH2OH
group
no
longer
collide
in
order
to
see
the
molecule
more
clearly.
Glucose
is
a
pyranose,
which
means
it
forms
a
six‐
membered
ring.
In
the
case
of
glucose,
the
carbon
of
the
carbonyl
will
bond
with
the
alcohol
of
the
carbon
just
above
the
CH2OH
group
in
order
to
make
a
six
membered
ring.
Remember
that
though
glucose
does
indeed
contain
six
carbons,
it
is
not
exactly
a
six
carbon
ring.
The
ring
will
consist
of
five
carbon
atoms
and
one
oxygen
atom.
The
sixth
carbon
will
be
an
equatorial
branch
connected
to
the
ring.
This
bonding
will
now
be
shown
through
images
of
the
model
for
clarity.
In
the
two
images
shown
above,
the
double
bond
from
the
oxygen
of
the
carbonyl
has
been
removed
and
replaced
with
a
yellow
“stick”
or
bond.
The
choice
of
a
different
colored
stick
is
only
so
that
it
will
be
easier
to
distinguish
and
find
the
anomeric
carbon
and
keep
track
of
its
location.
The
alcohol
group
of
the
carbon
nearest
to
the
CH2OH
group
has
also
been
rotated
so
that
the
oxygen
points
toward
the
anomeric
carbon.
You
may
notice
that
the
alcohol
has
lost
its
hydrogen
atom.
Do
not
worry
about
this
detail,
for
we
are
focusing
on
structure
rather
than
reactivity.
You
may
also
notice
that
a
stick,
representative
of
a
bond,
has
been
placed
on
the
ground
between
the
anomeric
carbon
and
the
oxygen
atom.
This
is
where
the
joining
will
take
place
to
convert
the
acyclic
molecule
to
the
cyclic
form.
The
oxygen
of
what
was
the
alcohol
group
of
the
carbon
above
the
CH2OH
group
has
now
been
joined
to
form
a
bond
with
the
anomeric
carbon.
This
is
the
top
view
of
the
cyclic
molecule.
By
tilting
the
molecule,
you
can
see
the
familiar
chair
conformation
that
is
often
drawn
in
Chem14C.
Notice
that
all
the
alcohol
groups
are
equatorial.
This
is
a
unique
characteristic
of
glucose.
The
oxygen
of
the
anomeric
carbon
should
be
an
alcohol
group,
but
that
change
is
not
shown
here
and
the
hydrogen
atom
has
been
omitted.
Works
Cited
http://www.chem.ucla.edu/harding/index.html