This video tutorial explains how to draw chemical reaction mechanisms by illustrating electron movement using curved arrows, based on the transformation from starting materials to products. It emphasizes understanding the fundamental rules of electron flow to predict and represent reaction pathways.
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Welcome to Chem Help ASAP. In this video we'll do some problems
that require us to draw chemical mechanisms
based on the structure of a starting material and product.
Go to the description of this video to find links to printable copies of these
problems as well as printable solutions. This
video also contains links to other videos that might help
you with this material. Let's review some ideas before we do the
problems. First, when we draw mechanisms, we are showing electron movement,
not atom movement. Second, we show electron movement with arrows.
These can be curved arrows -- sometimes are called
curly arrows. These arrows, since they are moving electrons, always start at a
source of electrons -- either a bond or
a lone pair. Finally all movement must be captured with an arrow. Any
bonding changes must have an arrow associated with it.
Alright! So those are the ground rules. Let's look at the problems.
Here are our first two problems. in this problem, if you notice, here
are starting materials on the left and then over to the right
what we have is this new OH bond, and our chlorine
is now chloride. It's lost its bond. So we need to account for
the formation of this bond and the breaking of our HCl bond, and we're
going to do that by moving electrons. So we need to start this reaction with
electrons. We're going to use lone pair electrons on oxygen. You don't
need to draw both of them in, but it does... oxygen does have two electrons.
We're going to start our arrow where those electrons are.
Those electrons are going to go to make a new bond to hydrogen.
Well hydrogen can only make one bond, so if you're going to make a new bond to
hydrogen, you better break an old bond. And that broken bond, those electrons,
will go to chlorine. Chlorine doesn't mind picking up extra
electron density because it's very electronegative. So now we have an
arrow that corresponds to the formation of the O-H bond, and we have an arrow a
second arrow that corresponds to the breaking of the H-Cl bond.
And what we see on the right is the outcome of that reaction. Now this oxygen
has three bonds. When oxygen has three bonds, it typically
has a positive formal charge, and now this chlorine has a negative
charge. We can draw in one or multiple lone pairs, and
there's a chlorine with a full full octet in its
valence shell. Now this is a reversible reaction.
This is an acid-base reaction, and it can go backwards. So for the
backwards reaction, we show the chlorine getting that
hydrogen back, using its lone pair. And then, when we
break this O-H bond to go back to starting materials, we start the
arrow in the middle of that O-H bond. That's where the electrons are,
and we get the lone pair back on oxygen. So let's look at the next question. In
the next question, we're going to make a new N-H bond in one
of our products, and we will have broken an O-H bond. So again, we're going to need
a lone pair, and we're going to use a lone pair on
nitrogen. And we start our arrow where the electrons are and point
our arrow to where the electrons finish. They're going to finish on that hydrogen.
So we're going to make an N-H bond. Now that hydrogen can only make one bond.
We are going to break this bond to hydrogen
and those electrons end up on that oxygen. And so we can show this in our
product. Now we have four N-H bonds on our nitrogen, and this oxygen
would have an extra lone pair. We can draw them all in
because it started with two, and now it's going to finish with three lone pairs
and have a formal charge. This, again, is an acid-base reaction. It's
reversible. So, to go backwards, we could have our
oxygen lone pair reach over and grab that N-H.
And that would break that N-H bond, and that's going to become a lone pair on
the nitrogen in the starting material. So those are our first two problems.
Okay! Two more problems. This top problem is kind of a
special thing. Notice it has this weird headed arrow. This is called a resonance
arrow. Resonance is a process where electron density gets
delocalized through a molecule. You may or may not have encountered the idea of
resonance yet, but resonance is not a chemical reaction.
But, since it is movement of electron density, we can
represent it with our our mechanism arrows. So formally this is
not a chemical reaction, but we are showing the movement of
electrons. So in this process, as you can see on the left, we
have a C=O double bond which becomes a C-O single bond. So we're going to break
that C-O pi bond, that second bond. Furthermore
we're going to form a new bond between carbon and nitrogen.
So we start where the electrons are. This is... We have a nitrogen lone pair.
Just because it's not drawn, don't forget about it. And we're going to use those
electrons to make that new C-N pi bond. So we go
from a lone pair to a bonding pair. This carbon already has four bonds, so if
we're going to break... if we're going to form a new bond to that
carbon, we better break an old one. And the bond that will break is the C-O
bond, and those electrons will go up on oxygen. So this oxygen started
with two lone pairs, and now in the product
it's going to have three lone pairs and it has a negative charge. The nitrogen
lost the lone pair to form a bond, and so it therefore has a positive formal
charge. Resonance... I wouldn't say this is
reversible but instantaneously these electrons,
or simultaneously, they're going back and forth so we can actually reverse the
process. Take a lone pair on oxygen, make a new C-O
pi bond, and then break the C-N pi bond and get
the lone pair back. Now this lower reaction is actually a
chemical reaction. This is something called an SN2 reaction. You
will see plenty of these if you keep on studying organic chemistry,
and as you can see, we have broken this C-O bond and we formed in the
product a C-Br bond. So, to show this mechanism, we need
electrons that can show the formation of the bond. Start where the electron
density is. We'll start with this anion -- plenty of electron density on that
bromine. And that arrow will go point to that
carbon. Now that carbon... I haven't shown all the
bonds, but it already has four bonds. If we're going to make another bond to
carbon, we're going to break an old one. And that bond will be to the C-O bond.
It's then, when we need to break, and we'll put those electrons right on
that oxygen. As a result this oxygen will have a lone
pair (actually has two), and we get bromine, and this bromine is
now down to three lone pairs because it used one of its lone pairs
to make a bond. So that's an example of an SN2 reaction. You'll see plenty of
those, but that, that's the electron movement. Just by
seeing the starting materials and products, we can know what the electron
movement must be even if we don't know the name of the reaction.
Here are two more reactions. So in this top reaction, what we're doing is we are
going from a structure that has a C-O bond on the left,
and it goes to having no C-O bond on the right. So we're going to break that bond,
and here is the bond of interest. And we are simply going to break that bond and
we're going to put the electrons on the oxygen.
Often when you see atoms that have a positive charge, they end up accepting
electrons, and sure enough that's exactly what this oxygen does.
It takes those electrons -- that bonding pair from the bond -- and it's going to
claim it as a lone pair. So this oxygen actually has to get a
full octet. It has a lone pair, so in the product we get water, and that
oxygen has two lone pairs. Now the carbon has lost its bond. That
carbon only has three bonds and therefore it has a positive charge. It's
a carbocation. This is a reversible process, so the
oxygen can then turn around and use this lone
pair and reform that C-O bond. That's
a one arrow mechanism. Now in this
lower reaction, let's put the lone pairs on this oxygen. We're going to make an O-H
bond from protonation of an oxygen with HBr.
So we start where the electrons are. This oxygen behaves as a base.
That lone pair goes after that hydrogen. That hydrogen already has a bond. We're
going to break an old bond to that hydrogen,
and who's going to accept the electrons from this H-Br bond?
It's going to be the electronegative bromine. So
when we look at our products we can have, our oxygen now only has one lone
pair. Bromine has four lone pairs -- started with
three. Let's put those in. This is technically a
reversible process, and you could imagine the bromine with
one of its lone pairs turning around and grabbing that
hydrogen right back from oxygen, breaking that O-H
bond and reforming that second oxygen lone pair.
Okay, two more reactions. In this reaction we're going to make this - the C-N [C-C!!]
bond in the product on the right. So this carbon, which has a negative charge
has a lone pair... Let's give nitrogen it's lone pair too.
This is actually cyanide... would be the full lewis structure for
cyanide. There's a triple bond in there. The nitrogen doesn't play a role in this
mechanism but the carbon does. The carbon has electron density, and it
is going to make a new bond to that carbon. Well that carbon has a
problem. It's already got four bonds, so if we're going to make a new bond to
that carbon, we need to break an old bond and it's going to be the
double bond between oxygen and carbon. And those electrons from that
double bond are going to land right up on oxygen. So
oxygen started with two lone pairs, and it's going to finish with three.
And the carbon lone pair became a new C-C bond to the central carbon of the
of the molecule. Technically this is a reversible
reaction -- probably favors the right. But this oxygen can use a lone pair to
reform that C=C double bond and push the carbon back out with its
lone pair of electrons. So that is a reversible process. Now in
this bottom case, this is kind of an interesting one. We
have a double bond, and let's go ahead and draw, draw in our full
hydrogens. So that carbon technically has two hydrogens, and it's easy to miss but
based on the structure, that carbon now has three hydrogens.
Where did it get that third hydrogen? Well it got it from,
from this hydronium ion in the middle of the screen.
And so the electron density for that to happen comes from the pi bond, the C=C pi
bond. So that pi bond will break to give you
two electrons, and with those two electrons, we're going to make a new bond
to this hydrogen. Now hydrogen only makes one bond, so as soon as we make a new
bond, we better break the old bond. And those electrons in that O-H bond go
on to the oxygen. The oxygen is positively charged so it's
asking for electrons. When we give it
to the oxygen through breaking that bond, now what happens is,
since we have we have a positive charge here in our left side, we better have a
positive charge somewhere on the right. And we do. It's this carbon
right here. This is it. Lost a bond and got nothing in return.
Therefore that carbon has a positive formal charge.
This is technically an acid-base reaction. All we did was move a proton,
and it this will be a reversible process like acid-base reactions are.
So we can take one of our lone pairs on oxygen.
We can make a new O-H bond, and now we'll take
the electrons in this C-H bond, and we'll make a new bond --
a C=C pi bond. And that actually gets the positive charge back on oxygen as our
starting materials show. Okay! Here are our last problems. This top
process is actually another example of resonance.
Note that we have this special double-headed arrow, and what
it looks like all we do is we flip around a double bond. And
that's kind of what we do. We're going to change the position of this double bond,
This bonding pair in the double bond is going to swing over to
make a new carbon-carbon pi bond. And as a result
this carbon right here is going to lose a bond, so the charge,
the formal charge, goes from being on the left hand side of the molecule
over to the right hand side. And so this is a single arrow. We're
moving a single bond and as a result, the formal charge
also moves. This is not a reversible process
but *instantaneously* these are two simultaneous states, and so we can
think of conversion back to the state on the left
by moving the bond back. And that'll get the positive charge back on the left.
Now in the bottom structure, this is a little bit more involved reaction.
This is called a halogenation, a dibromination,
at least the start of it.
Again, you probably haven't encountered this mechanism, but just by looking at
the bonding changes, we can infer how the electrons would
move in order to get us to that, to that end product. This thing right
here on the right is called a bromonium ion --
one of the staples of organic 1 classes. So
if you notice, we need to make two bonds to our bromine. That means we need two
arrows -- at least two arrows -- going to and from
this bromine on the left
to the the molecule that's on the far left.
So we're going to start this off by having the pi
bond attack the bromine. Now bromine typically only makes one bond, and so we
can have this C-Br bond break -- I'm sorry, not C-Br --
the Br-Br bond break to make our bromide. So we can account
for this. That bromide started with three lone
pairs, and it's gonna have four in the end.
Fill those in. Now, so we've accounted for one of our bonds (we have two bonds)...
we've accounted for one of them but not the other one. To make this other
bond, we're going to use a lone pair on bromine,
and we're going to have it attack back to our molecule. And so we have two
arrows between our starting carbon containing molecule
and our bromine. And now we have two bonds
between that carbon-containing molecule and bromine, and that gives us our ring.
Bromine has two lone pairs. It formally has a positive charge,
and that's how we make a bromonium ion. Again you've probably never seen this
reaction, but we can draw... we can postulate what the mechanism
would be or speculate on how arrows would allow us to get to this
product from this reaction. So this is the
the value of these mechanistic arrows. It helps us understand
patterns in chemical reactions so we can start to make predictions on how these
reactions work. I hope you found this video helpful.
Please subscribe, like, or leave a comment or question.
In particular I'd like to hear if there are any different types of problems you
would like to see in future videos. Thank you for watching and remember that
links to printable copies of these problems
and solutions are in the video description. Take care.
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