2.5 Introduction to Arrow Pushing in Reaction mechanisms

Please make sure you take a look at arrow pushing before reading this section, as the prior information should help with your understanding of what these curly arrows are portraying in a reaction mechanism.

When given the reactants and products

Most reaction mechanisms that one will come across are usually exactly like a simple acid/base reaction arrow pushing. Therefore if you get a strong handle on being able to push arrows in a simple acid/base reaction this will be a strong foundation moving forward in organic chemistry and other upper year courses.

To start let’s look at the simple acid/base reaction of acetic acid, CH₃CO₂H, and sodium hydroxide, NaOH. You probably have all seen before that these two species will undergo the following equation:

$$CH_3CO_2H+NaOH ⟶ CH_3CO_2Na+H_2O$$

To see better what is happening, let’s complete the Lewis structures for these two starting materials. First let’s start with acetic acid:

Now that we have the correct Lewis structures, let’s look at the overall reaction below and think about which covalent bonds are being created and broken.

So now that we know which covalent bonds are breaking and forming we can describe in words the changes that are occurring within the reaction and how our electrons are moving.

Once we can describe this process in words we can use curly arrows to describe this process. Remember a curly arrow starts with where the electrons are initially and shows where the electrons are moving to. Its like we explained things above in words and now we are explaining it with a visual with the arrows showing the electron movement.

Great now can you do the same for a similar reaction?

The key steps we took when pushing the arrows for the problem above was:

Determine which covalent bonds are being made and broken.
The positive spectator ions are not involved in the arrow pushing, they will just end up close to the new negative formal charge on the product side.
determine how to describe the electron movement in bond breaking and making in words.
Use your description you just created to draw arrows that visually show how the electrons are moving to make and break bonds. Remember arrows show the movement of electrons, so they start from either an atom with a formal negative charge (O in NaOH) or the most electronegative atom (N in NH₃)
Double check that the arrows used will give you the products on the right hand side of the equation.

When only given the reactants

When we are only given reactants we have to look at things more closely to determine where things will react. In doing this we have to look for the extremes. We are looking for the extremes in electron density. In most scenarios the region of highest electron density will “attack” the region of lowest electron density. When we say “attack” we are referring to where our arrow pushing will start.

Determining the region of highest electron density for the species that will be “attacking” in a reaction mechanism.

When talking about the region of highest electron density we will look for:

1^st: if present a negative formal charge will usually be the region of highest electron density. Keep in mind these special cases:
- If the negative formal charge is a spectator ion within the reaction and therefore is not involved in the arrow pushing (this is rare and most likely won’t be seen in first-year).
- This negative formally charged atom is resonance stabilized making the region of high electron density being located over a collection of atoms and one of these atoms that share the negative charge can attack instead of that specific one (example to be discussed later).
2^nd: If no negative formal charges are present in the reactants, then to determine the region of highest electron density we need to look for the most electronegative atom within the most polar bond.
3^rd: the product formed needs to be feasible/relatively stable and make sense based on your understanding of Lewis. Example: Although F is the most electronegative atom on the periodic table, neutral F will almost never be our starting point for a reaction, because in most scenarios it will mean we end up giving F a positive charge in our product.

For each of the following reactions a), b) and c) determine which atom will have the most electron density in each and therefore will be the species that will do the “attack” in that specific reaction mechanism.

Determining the region of lowest electron density for the species that will be “attacked” in a reaction mechanism.

When talking about the region of highest electron density we will look for:

1st: A species that has a positive formal charge will usually be the region of highest electron density. Keep in mind these special cases:
- If the positive formal charge is a spectator ion in the reaction, (this is extremely common, check to make sure that the species in not a group 1 or 2 metal cation, these are happy to stay the way they are, an example being Na⁺ in our first arrow pushing reaction above).
- If the positive formal charge is involved in resonance. In this case a separate atom that is involved in the resonance could be attacked instead in order to possibly create a more stable product (example to follow).
- If the positively charged species cannot make an additional bond. In this scenario usually an atom attached to this atom is attacked instead to remove a bond and give that atom a lone pair thus removing the formal charge. A common example of this is a positively charged N having a H removed in order to get back a lone pair:
2^nd: If no formal positive charges are present in the reactants, then we need to look for the most electropositive atom (atom with the least electronegativity) in the most polar bond.
3^rd: If there are more than one reactant in the reaction, then the region of lowest electron density (most electropositive region) should come from the other reactant (not the one where we located the most electron density) so the two species are reacting with each other.
4^th:Again, as stated above the product formed needs to be feasible. An example of something not feasible would be trying to bond with the NH₄⁺ above, it already has an octet and therefore cannot have an additional bond, the only way to neutralize the charge is to remove a bonded H.

For each of the following reactions a), b) and c) determine which atom will have the least electron density in each and therefore will be the species that will be “attacked” in that specific reaction mechanism.

Now that we have located the regions of high and low electron density for each reaction, use this understanding to answer the following questions:

You should notice that in the middle problem above we started with two species without any formal charges and ended up with two species that have formal charges. So in a sense we went from a stable to less stable scenario. Technically this reaction would be in equilibrium and favor the starting material, but for simplicity sake we did not show the equilibrium arrow for that reaction. A reason that the HCO₂H species would be relatively stable as HCO₂^– is due to the fact that their is resonance stabilization. The negative charge could exist on either one of the oxygens. To refresh yourself on resonance please click here.

In organic chemistry you will be applying these same concepts however your region of high electron density will be referred to as your nucleophile in the reaction (e.g. the base in an acid-base reaction) and your region of low electron density will be referred to as your electrophile (e.g. the acid in an acid-base reaction). To see an overview of this from an organic chemistry standpoint please click here.