cyclopropanation reaction is a reaction in which an alkene reacts with a molecule called a carbene
to form a product that has a cyclopropane or a triangle where the pi bond used to be,
here we show it with a CH2 and here we have it in skeletal structure just showing a triangle.
This reaction is stereospecific just like with epoxidation where if you start with a trans
alkene you'll get a trans product and if you start with a cis alkene you get a cis product,
as you see in this three dimensional drawing of the same product, not forgetting that we
have hydrogen atoms here also trans to each other. Wait, what's a carbene?
Carbenes are something you've likely not seen before now, a carbene is a very interesting,
unhappy, unstable molecule as follows. We have a carbon atom with two bonds,
in this case I'll show hydrogens, one lone pair and that's it. If you count it up,
you'll notice that we have a total of six electrons, not eight in carbon’s octet,
leaving us with an empty P orbital but despite the incomplete octet, this is still a neutral molecule
which we can prove using the formal charge shortcut that I have linked below,
carbon should have four directly attached is four for a formal charge of zero.
But wait, if you use the hybridization shortcut that I teach, link below, you'll see that having
just three groups around the carbon gives us SP2 hybridization with an empty P orbital. All of
these things come together to give us a molecule that is both a nucleophile and an electrophile,
a nucleophile that wants to attack using its lone electron pair and an electrophile that
invites an attack into its empty P orbital. Sounds insane? Actually yes, that's why it’s so reactive
and that's why we need special circumstances when working on this reaction, you will likely
learn two common cyclopropanation reactions. the first uses a haloform with a carbene intermediate,
for this reaction we start with an alkene, here we're showing trans-2-butene and react it with a
haloform such as chloroform, CHCl3 in the presence of a strong base like tert-butoxide to give us the
carbene intermediate for a final product that has our cyclopropane but with two halogens attached
The second is the Simmons-Smith reaction using a carbenoid rather than a carbene intermediate,
this reaction begins with an alkene in this case trans-2-butene,
reacting with CH2I2 and a Zinc Copper couple to give us the cyclopropane product.
The intermediate as we'll see when we look at the mechanism is not quite a carbene because
carbon will have a complete octet thanks to the help of the zinc copper. So how does this work?
Let's jump into the reactions. Because the carbene is so unstable we don't react the
alkene with a carbene, instead we create the carbene right as we're doing the reaction,
in this case we start with chloroform and bring in a strong base like tert-butoxide.
Tert-butoxide uses one of its lone electron pairs to grab a hydrogen atom off of the chloroform
and collapse those electrons back onto carbon, this gives us a very unhappy carbon and a
tert-butanol on the side even though it has a complete octet, it has a negative charge.
The negative charge is stabilized by the highly electronegative halogens, in this case chlorine
via the inductive effect, so much so that one of the chlorines ultimately gets so excited to the
point where it completely takes away the bonding electrons and breaks away from the molecule,
this gives us a Cl- just floating around somewhere in solution, now that it has
that extra lone pair that it stole from carbon as well as a neutral molecule that has a carbon atom,
the two lone electrons from the initial hydrogen and the two remaining halogens.
This is called a dihalocarbene in that we have two halogens, in this case two chlorines sitting on
the carbene, the carbon with only six electrons in its valence made by having two bonds and one
lone pair, this reaction using a strong base to grab a hydrogen off a carbon and ultimately kick
out one of the halogens should remind you of another elimination reaction that you've seen
except for one major difference This is called an alpha elimination
reaction because both hydrogen and the halogen come off of the same carbon atom,
in contrast to the beta elimination that you've seen for the E1 and E2 reactions
where hydrogen and halogen come off of neighboring one off the alpha and one off the beta carbons.
Now that we have a carbene, let's take a look at the alkene. For the sake of this reaction,
lets take a look at cis-2-butene reacting with dichlorocarbene. Remember that the carbene has
an empty P orbital which likes to get attacked and has the lone electron pair that likes to do
the attack and so the reaction happens as follows, the alkene has a nucleophilic pi bond meaning the
pi electrons want to attack and attack they will right into the empty P orbital of the carbene.
At the very same time, the carbene will use its lone electrons which are also nucleophilic
to attack the other carbon of that pi bond. This is not a step one, step two reaction,
this is a concerted mechanism where both nucleophilic electron pairs
the pi bond and the carbene are going to attack each other at the very same time,
what do we get? The carbon skeleton retains its stereochemistry, that means the two methyl
groups are still cis to each other. We have a bond between this carbon and the carbene carbon
where the pi bond attacked and another bond that we showed in purple electrons where the carbene
attacked our molecule right back, and of course let's not forget our two halogens
sitting on top of the cyclopropane. If you're asked to show the stereochemistry,
consider drawing these two methyl groups either both on dashes or both on wedges to show that
they are still cis to each other because we started with a cis alkene. The Simmons-Smith
reaction is slightly different because we're not reacting with a carbene, we're using a carbenoid,
a carbenoid is very similar to a carbene but it’s lacking some of those properties namely an empty P
orbital because of an incomplete octet but let's back it up, for this reaction, we're going to
start with CH2I2 as the source of carbon and react it with the Zinc Copper couple, the zinc is
going to insert itself between the carbon and the iodine to give us a molecule that looks like this.
This right here would be our carbene except instead of a lone pair,
it has the halogen attached, instead of an empty P orbital, we have that zinc iodine also attached.
Why is this so important compared to a carbene, this carbenoid is actually very stable.
The mechanism for the Simmons-Smith reaction is very similar to what we've already seen
in that we have a concerted attack with one extra step. When the pi bond reaches out to attack the
carbon, Iodine gets knocked loose in the process and in order for carbon to attack the pi bond,
it has to let go of the zinc iodine, the way to think of this is imagine you're sitting at a ball
game, you're relaxing, enjoying the game, eating ice cream. Suddenly, the ball comes right at you
without thinking, what is your reflex reaction, you lift your hand to your face,
you want to protect yourself. What happens to your ice cream well, let's just say you're
not going to finish it are you. When carbon gets attacked, it has to let go of its ice cream.
Let's take a look at what happened to every piece in this reaction.
Our skeleton remains the same, that means the cis alkene still looks cis but now we have a red bond
to the blue carbon where the pi bond attacked and a purple bond where the carbenoid attacked back,
we still have two hydrogen atoms because those bonds are hardest to break
so they haven't gone anywhere and somewhere in solution we have that I-Zn-I, as a side product
from everything that we let go. If you're asked to show the stereochemistry,
given that we started with a cis alkene, you can draw both of the methyl groups on dashes or wedges
to show that they're still cis to each other. For even more on alkene reactions, including
my entire video series alkene practice quiz and cheat sheet, visit my website link below
or go to leah4sci.com/alkene-reactions, the link again, leah4sci.com/alkene-reactions