Surviving Alkene Reactions
A logical approach
Surviving Reactions
• Seems like “surviving” is always the word that
best describes what has to happen in this
class… but I think with a little practice you’ll
find this isn’t all that difficult…
• Yes, its time for that memorization part so
many people dread. Organic chemistry has this
awful reputation and I think THIS is where it
all sprung from.
• Yes, you’ll need to remember some facts… but
even worse, you’ll need to APPLY these in order
to draw your product.
Where to begin…?
• There are several pieces that have to
come together when drawing the
product of a reaction.
• You have to maintain a specific carbon
skeleton for the molecule that contains
the alkene.
• You have to know what the reagents will
transform the alkene into.
• You have to know the details of the
pieces that need to be attached.
• Is the reaction “Markovnikov” or “NonMarkovnikov”?
• Is the reaction going to proceed “Syn”
or “Anti”?
• If regiochemistry is the concern, you’ll
need to determine which end of the
alkene is more substituted, versus less
substituted.
• If stereochemistry is the concern, you’ll
have to use wedges and dashes to
communicate if the pieces are attaching
on the same side or opposite sides.
• And, of course, when all is said and
done, you have to draw the actual
product…
• Hopefully this will help walk you through
the trials and tribulations of your first
reaction chapter…
• Let’s begin…
Understand your alkene
• Alkenes come in all shapes and sizes.
• When you want to do a reaction on an
alkene, you first have to identify with
the two carbons in the skeleton that
constitute the alkene.
• Depending on the reaction, you may
need to determine which end is the
more substituted end of the alkene.
• Let’s start here…
• Let’s look at an alkene so you can see
what needs to be seen.
• Identify how many carbons are
attached to each sp2 carbon of the
alkene.
• First, just identify the alkene carbons
(shown in blue).
C
C
• Then count carbons attached to each.
C
C
C
C
C
TWO
carbons
attached
C
C
ONE
carbon
attached
C
C
C
• The right-hand alkene carbon has two
carbons (in red) attached.
• The left-hand alkene carbon has only one
carbon (in green) attached.
TWO
carbons
attached
C
C
ONE
carbon
attached
C
C
C
• With this in mind, you should be able
to see that the right carbon is “more
substituted” with more alkyl groups.
• The left carbon is “less substituted”
with only one alkyl group attached.
• What about the next alkene?
end is more substituted?
Which
C
C
C
• The right side is more substituted
with two carbon groups. The left side
has only one carbon group.
• Try another…
• What about this alkene? Which end is
more substituted?
C
C
• The left side is more substituted with
two carbon groups. The right side
doesn’t have any carbon groups.
• Got the idea?
• Remember that any reaction that has
regiochemistry (i.e. that reacts
Markovnikov or Non-Markovnikov) will
need you to determine which end is
more (or less) substituted.
Reagents…
• For each reaction that is covered, there
are many with mechanistic information
that you need to remember (the “why is
it this or that?”).
• For each reaction, there are also some
quick and easy pieces to remember
(which will take some time to get
embedded in yours brains)…
Reagents…
• The Halogenation Reaction uses HX to
add “H” and “X” to the alkene in a
Markovnikov manner.
HCl
• HX can be HCl, HBr or H3PO4 and KI.
• “X” will attach to the more substituted
side.
Reagents…
• Answer:
HCl
H
• Need to remember:
H X Mark
Cl
Reagents…
• Addition of X2 will add “X” and “X” to
the alkene in an ANTI manner.
Cl2
• X2 can be Br2 or Cl2.
Reagents…
• Answer:
Cl
Cl2
Cl
• Need to remember:
X
X ANTI
Reagents…
• Addition of X2 and H2O will add “X”
and “OH” to the alkene in a
Markovnikov, ANTI manner.
Cl2
H2O
• X2 can be Br2 or Cl2.
Reagents…
• Answer:
Cl2
OH
Cl
H2O
• Need to remember:
X OH Mark ANTI
Reagents…
• Variation: Addition of X2 and ROH will
add “X” and “OR” to the alkene in a
Markovnikov, ANTI manner.
Br2
CH3OH
• X2 is Cl2 or Br2 and ROH can have any
alkyl group (R).
Reagents…
• Answer:
Br2
CH3O
Br
CH3OH
• Need to remember:
X
OR Mark
ANTI
Reagents…
• There are three reactions to make
alcohols from alkenes.
HO
H
Reagents…
• Addition of H2O and H+ will add “H”
and “OH” to the alkene in a
Markovnikov manner.
H2O
H2SO4
• H+ can come from any acid like H2SO4
or H3PO4.
Reagents…
• Answer:
H2O
HO
H2SO4
• Need to remember:
H OH
Mark
H
Reagents…
• Addition of 1. Hg(OAc)2, H2O and 2.
NaBH4 will add “H” and “OH” to the
alkene in a Markovnikov manner.
1. Hg(OAc)2, H2O
2. NaBH4
Reagents…
• Answer:
1. Hg(OAc)2, H2O
2. NaBH4
• Need to remember:
H
OH Mark
HO
H
Reagents…
• Addition of 1. BH3 and 2. NaOH, H2O2
will add “H” and “OH” to the alkene in
a NON-Markovnikov and SYN manner.
1. BH3
2. NaOH, H2O2
Reagents…
• Answer:
1. BH3
H
OH
2. NaOH, H2O2
• Need to remember:
H OH
Non-Mark
SYN
Reagents…
• Hydrogenation by the addition of H2,
Pd/C will add “H” and “H” to the alkene
in a SYN manner.
H2
Pd/C
Reagents…
• Answer:
H2
H
Pd/C
• Need to remember:
H
H SYN
H
Reagents…
• Dihydroxylation will add “OH” and
“OH” to the alkene in a SYN manner.
1. OsO4
2. NaHSO3, H2O
• Can use either the addition of KMnO4,
NaOH, H2O OR
• 1. OsO4 and 2. NaHSO3, H2O
Reagents…
• Answer:
OH
1. OsO4
2. NaHSO3, H2O
OH
• Need to remember:
OH
OH
SYN
Reagents…
• There are
reactions.
three
cyclopropanation
Reagents…
• Addition of CH2N2 will form a SYN
cyclopropane ring, with two hydrogens
on its apex.
CH2N2
Reagents…
• Answer:
CH2N2
H
H
• SYN addition for ring formation.
Reagents…
• Addition of KOH, CHCl3 will form a
SYN cyclopropane ring, with two
chlorines on its apex.
CHCl3
KOH
Reagents…
• Answer:
CHCl3
Cl
Cl
KOH
• SYN addition for ring formation
Reagents…
• Addition of CH2I2, Zn(Cu) will form a
SYN cyclopropane ring, with two
hydrogens on its apex. The hydrogens
can be replaced by one or two alkyl
groups.
CH2I2
Zn(Cu)
CH(CH3)I2
Zn(Cu)
Reagents…
• Answer:
CH2I2
H
H
Zn(Cu)
CH(CH3)I2
H
CH3
Zn(Cu)
• SYN addition for ring formation
Reagents…
• Addition of either of these last two
reagents will do an oxidative cleavage
reaction, splitting the molecule into
pieces. Every double bond must be split.
1. O3
O
2. Zn, H3O+
H
O
• Ozonolysis, 1. O3 and 2. Zn, H3O+, will
form aldehydes and ketones.
Reagents…
• In either of these last two reactions,
an oxidative cleavage reaction will
occur, splitting the molecule into
pieces. Every double bond must be
split.
O + O
Reagents…
• Addition of KMnO4, H2O or H3O+ will
form carboxylic acids, ketones or CO2
(H2CO3).
KMnO4
O
H2O or H3O+
KMnO4
H2O or H3O+
O
OH
HO
O
OH
O
Know your reagents…
• Make a single list of all of your reagents
and their important “must know” tidbits…
• Writing this list out will help you
remember all those little facts.
• Writing the list multiple times, over and
over again will REALLY ingrain it in your
memory.
• Start working on this early and the night
before the exam won’t be so painful!
Now, put it all together…
• Identify what the reagent is doing to
the alkene
• Then identify the carbons in your
alkene carbon skeleton
• Find the two carbons of the alkene in
your product skeleton
• Draw in the appropriate parts, with
correct
regiochemistry
and
stereochemistry.
• Let’s start with a simple example using
the reagent, HCl. HCl, like all HX
reactions, adds one H and one X (in
this case, Cl) to the ends of your
alkene in a Markovnikov fashion.
HCl
• Start by identifying the carbons in
your alkene carbon skeleton. Attach
all of the original carbons in a similar
zig-zag format (unless of course, its
an oxidative cleavage reaction).
• Identify the two carbons that were in
the original alkene. Your reaction will
occur between these two carbons.
• In the example here, the
substituted end is on the right:
ONE
carbon
attached
TWO
carbons
attached
more
• For the addition of HCl, the reaction
occurs Markovnikov-style with Cl
attaching to more substituted end.
HCl
H
Cl
• This would give you the final product
shown below:
• Or simply:
H
Cl
H
Cl
HCl
H
Cl
Let’s try a few more…
• The addition of X2 adds X and X in an
ANTI fashion.
• Try this problem:
Br2
• Find the two carbons of the alkene:
Br2
• Line up the pieces to be attached.
Notice that if the same piece is being
added to both ends, you don’t have to
worry about regiochemistry.
Br2
Br
Br
• Attach the pieces:
Br2
Br
Br
• And
finish
with
stereochemistry:
your
Br2
Br
Br
ANTI
Let’s do another…
• The addition of X2 and H2O adds X
and OH, Markovnikov style and in an
ANTI fashion.
• Try this problem:
Cl2, H2O
• Find the two carbons of the alkene:
Cl2, H2O
• Line up the pieces to be attached.
Markovnikov addition places the OH on
the more substituted end, X on the
less substituted end.
Cl2, H2O
Cl
OH
• Attach the pieces:
Cl2, H2O
OH
Cl
• And
finish
with
stereochemistry:
your
Cl2, H2O
OH
Cl
ANTI
And another…
• The addition of H2 with Pd/C adds H
and H in a SYN fashion.
• Try this problem:
H2, Pd/C
• Find the two carbons of the alkene:
H2, Pd/C
• Line up the pieces to be attached. No
regiochemistry since you are adding
the same thing to both ends.
H
H2, Pd/C
H
• Attach the pieces:
H
H2, Pd/C
H
• And
finish
with
stereochemistry:
your
H
H2, Pd/C
H
SYN
A different kind of reaction…
• The formation of cyclopropane rings
from alkenes occurs in a SYN fashion.
• Try this problem:
CH2N2
• Find the two carbons of the alkene:
CH2N2
• Form the cyclopropane (with H’s at
the apex) across the two carbons of
the alkene:
CH2N2
H
H
• And
finish
with
stereochemistry:
CH2N2
your
SYN
H
H
• Using CHCl3 and KOH forms a
cyclopropane with Cl’s on the apex.
• Using CH2I2 and Zn(Cu) forms a
cyclopropane with H’s on the apex.
Alternatively, you might have alkyl
groups, instead of H’s. For example,
you might use CH(CH3)I2 and form a
cyclopropane that has one H and one
CH3 group on the apex.
More
Flexibility!
Those last reactions…
• The oxidative cleavage reactions split
an alkene into two pieces.
• When using the ozonolysis reaction,
the alkene splits to form ketones
and/or aldehydes:
R
R
H
R
R
R
O
and
O
H
R
• Oxidatively cleave the following:
1. O3
2. Zn, H3O+
• It may help to draw in the H’s
attached to the alkene.
• Then STRETCH that alkene, and insert
the oxygen atoms that will form the
aldehydes or ketones. (This reaction
doesn’t happen this way but it may
help you draw the products.)
O
H
H
O
• Finish the cleavage of both the pi and
sigma bonds.
• The final products should be:
1. O3
O
2. Zn, H3O+
H
H
O
The final set of reactions…
• When using the KMnO4 reaction, with
either H2O or H3O+, the alkene splits
to form ketones and/or carboxylic
acids:
R
R
OH
R
R
R
O
and
O
OH
R
• Oxidatively cleave the following:
KMnO4
H
H2O or H3O+
• It may help to draw in the H’s
attached to the alkene.
• Then STRETCH that alkene, and insert
the oxygen atoms that will form the
carboxylic acids or ketones. Any
aldehydes that typically form for
ozonolysis are converted to carboxylic
acids with KMnO4.
O
H
OH
O
• The final products should be:
KMnO4
H2O or H3O+
H
O
OH
O
Other things to consider…
• As you get started, you will notice
that you spend a lot of your time
learning how to draw the products.
• You should also be able to look at a
reaction, given the starting material
and product, and determine what
reagent did the transformation shown.
Br
• Consider this reaction…
• What reagent was used?
What
changed when the starting molecule
turned into the product molecule?
Br
• View the alkene carbons – from both
the perspective of the starting alkene
and the product.
• Identify those carbons
• Then identify what changed on those
carbons.
Br
• Notice that a Br is now attached to
the left-hand side. An extra H is now
attached to the right-hand carbon.
• Ask yourself what reagent adds an H
and a Br and does so in such a way that
the Br is on the more substituted end?
• HBr, of course…
• Here’s another
reagent
is
transformation?
example
causing
–
Cl
OH
what
THIS
• Identify the carbons of the alkenes –
what pieces are adding to the alkene?
Cl
OH
Cl
OH
• The more substituted carbon (bottom)
now has an OH group, while the less
substituted carbon (top) has a Cl now
attached.
• The reagent that was used was Cl2 and
H2O.
Two final reactions to think
about…
• Don’t forget that you need to know
how to MAKE an alkene.
• An acid like sulfuric acid or phosphoric
acid is used to dehydrate an alcohol.
• A strong base like sodium hydroxide
or potassium tert-butoxide, KOtBu, is
used to dehydrohalogenate an alkyl
halide.
Synthesis…
• Synthesis problems require the
stringing together of a series of
reactions as one molecule is turned
into another molecule.
• Consider the following:
Br
Br
Br
Br
Br
Br
• Synthesis problems need to be worked
from both directions, forwards and
reverse.
• Identify what’s changing.
• You started with a single Br.
The
product has two Br, one on the original
position (note that I didn’t say the SAME
Br) and one on the carbon next door.
Br
Br
Br
• You do NOT know how to randomly add
Br’s to just any carbon.
• You actually only know ONE reaction that
alkyl halides do - dehydrohalogenate.
• You only know ONE reaction that adds
TWO halides (ANTI!) on two consecutive
carbons.
• Put the sequence together…
• First add base (like KOtBu or NaOH)
to form an alkene.
• Then add Br2 to add two Br’s in an
ANTI fashion.
Br
KOtBu
Br2
Br
Br
• Alkene reactions will go easier for you
if you start sooner not later for
studying all those bits and pieces you
need to know. This will also make
practicing problems easier, so you can
better learn to draw the parts and
new molecules.
• As always, if this has been helpful,
drop me a line at jdiscord@towson.edu
and let me know.