Allyl Groups
Allyl groups
A propylene group is often called an allyl group.
For instance, 3-bromopropene is commonly called allyl bromide.
The carbon position neighboring the double bond is called the allylic position.
Allyl carbocation, radical, and carbanion
This is an allyl carbocation.
The allyl carbocation has three carbon atoms that are each sp2 hybridized. They each contain one p-orbital for a total of three overlapping p-orbitals. Two p-orbitals are used to make the pi bond and one p-orbital is an empty p-orbital for the carbocation.
Two good resonance forms can be drawn for the allyl carbocation. What the allyl carbocation really looks like is a blend of these two resonance forms, the resonance hybrid. In the resonance forms, carbon atoms C1 and C3 have the positive charge on them. The hybrid shows carbon atoms C1 and C3 each have part of this positive charge or a partial positive charge, δ+. The resonance forms indicate a double bond between carbon atoms C1 and C2 in the first resonance form and a single bond between carbon atoms C1 and C2 in the second resonance form. Therefore, there is the electron density of 1.5 bonds between carbon atoms C1 and C2. The same can be said between carbon atoms C2 and C3.
The resonance hybrid can be explained by carefully studying the molecular orbital diagram of the allyl carbocation’s pi system. The allyl carbocation consists of three overlapping p atomic orbitals. These p-orbitals contain two electrons (two from the pi bond and zero from the carbocation). We indicate this on the atomic orbital, left side of the following diagram. Since there are three p atomic orbitals that we are overlapping, we must make three molecular orbitals. The lowest energy molecular orbital, 1, contains zero nodes, the middle one, called nonbonding, contains one node, and the highest energy molecular orbital, 2*, contains two nodes. Notice that the nodes are symmetrical.
There are always equal numbers of bonding and antibonding molecular orbitals. With three molecular orbitals for allyl carbocation, there is one bonding MO (𝜋1) and one antibonding MO (𝜋2*). The leftover MO is called nonbonding, and is always the middle energy. The nonbonding MO is the same energy as the original p atomic orbitals. With the one node for the nonbonding molecular orbital at the center carbon atom, we are left with two non-overlapping p-orbitals on carbon atoms C1 and C3. These are simply atomic p-orbitals. That is why it is the same energy and the p atomic orbitals and why we call this MO “nonbonding”. The bonding MO and antibonding MO are symmetrically distributed about the nonbonding MO. The amount of energy saved by making the 𝜋1 bonding MO is the same as the amount of energy lost in making a 𝜋2*antibonding MO.
We place the two pi system electrons into the molecular orbitals starting with the lowest energy MOs. We can place at most two electrons in any orbital. Therefore, we end up with two electrons in the 𝜋1 bonding molecular orbital. The 𝜋1 MO has two electrons that are shared over carbon atoms C1, C2, and C3. This is what the hybrid showed with 1.5 bonds between C1 and C2 and 1.5 bonds between C2 and C3. There are no electrons in the nonbonding MO. This nonbonding MO indicates that there are zero electrons in the p-orbitals on C1 and C3. This is shown in the partial positive charges on C1 and C3 of the resonance hybrid.
Allyl carbocation molecular orbital diagram
This is an allyl radical.
The allyl radical has three carbon atoms that are each sp2 hybridized. They each contain one p-orbital for a total of three overlapping p-orbitals. Two p-orbitals are used to make the pi bond and one p-orbital has one electron in it for the radical.
Two good resonance forms can be drawn for the allyl radical. What the allyl radical really looks like is a blend of these two resonance forms, the resonance hybrid. In the resonance forms, carbon atoms C1 and C3 have the free radical single electron on them. The hybrid shows carbon atoms C1 and C3 each have part of this radical or contain a partial radical, δ•. The resonance forms indicate a double bond between carbon atoms C1 and C2 in the first resonance form and a single bond between carbon atoms C1 and C2 in the second resonance form. Therefore, there is the electron density of 1.5 bonds between carbon atoms C1 and C2. The same can be said between carbon atoms C2 and C3.
The resonance hybrid can be explained by carefully studying the molecular orbital diagram of the allyl radical’s pi system. The allyl radical consists of three overlapping p atomic orbitals. These p-orbitals contain three electrons (two from the pi bond and one from the radical). We indicate this on the atomic orbital, left side of the following diagram. Since there are three p atomic orbitals that we are overlapping, we must make three molecular orbitals. The lowest energy molecular orbital, 𝜋1, contains zero nodes, the middle one, called nonbonding, contains one node, and the highest energy molecular orbital, 𝜋2*, contains two nodes. Notice that the nodes are symmetrical.
There are always equal numbers of bonding and antibonding molecular orbitals. With three molecular orbitals for allyl radical, there is one bonding MO (𝜋1) and one antibonding MO (𝜋2*). The leftover MO is called nonbonding, and is always the middle energy. The nonbonding MO is the same energy as the original p atomic orbitals. With the one node for the nonbonding molecular orbital at the center carbon atom, we are left with two non-overlapping p-orbitals on carbon atoms C1 and C3. These are simply atomic p-orbitals. That is why it is the same energy and the p atomic orbitals and why we call this MO “nonbonding”. The bonding MO and antibonding MO are symmetrically distributed about the nonbonding MO. The amount of energy saved by making the 𝜋1 bonding MO is the same as the amount of energy lost in making a 𝜋2*antibonding MO.
We place the three pi system electrons into the molecular orbitals starting with the lowest energy MOs. We can place at most two electrons in any orbital. Therefore, we end up with two electrons in the 𝜋1 bonding molecular orbital. We have one more electron that is placed into the nonbonding MO. The 𝜋1 MO has two electrons that are shared over carbon atoms C1, C2, and C3. This is what the hybrid showed with 1.5 bonds between C1 and C2 and 1.5 bonds between C2 and C3. There is one electron in the nonbonding MO. This nonbonding MO indicates that there is one electron shared by the p-orbitals on C1 and C3. This is shown in the partial radicals on C1 and C3 of the resonance hybrid. These p-orbitals are isolated and nonbonding. This is why this MO is called “nonbonding”.
Allyl radical molecular orbital diagram
This is an allyl carbanion.
The allyl carbanion has three carbon atoms that are each sp2 hybridized. They each contain one p-orbital for a total of three overlapping p-orbitals. Two p-orbitals are used to make the pi bond and one p-orbital has two electrons in it for the lone pair of electrons of the carbanion.
Two good resonance forms can be drawn for the allyl carbanion. What the allyl carbanion really looks like is a blend of these two resonance forms, the resonance hybrid. In the resonance forms, carbon atoms C1 and C3 have the lone pair of electrons (the negative charge) on them. The hybrid shows carbon atoms C1 and C3 each have part of this lone pair of electrons. There are also 1.5 bonds between carbon atoms C1 and C2 as well as between carbon atoms C2 and C3.
The MO diagram is very similar to the allyl free radical except there are a total of four pi electrons (two from the pi bond and two for the lone pair of electrons).
The MO diagram has two electrons in the nonbonding MO. This means the lone pair of electrons are shared over carbon atoms C1 and C3 as the resonance hybrid shows.
Allyl carbanion molecular orbital diagram
Stability of allylic cations
Allylic positions are important because they are especially reactive sites. For instance, when heat is applied to allyl bromide, or it is placed in an ionizing solvent, it can easily dissociate forming an allyl cation.
At first, this may seem surprising because a primary carbocation is formed, but the allyl cation is quite stable. In fact, its stability is similar to a secondary carbocation. We saw this in the previous section. The pi electrons of the allyl cation are stabilized, or lower in energy, because they end up in a bonding molecular orbital. Resonance forms can also be drawn for the allyl cation, spreading out the positive charge over two carbon atoms.
Another very stable position is next to a benzene ring with its double bonds.
Allylic Radicals
Like allyl carbocations, allyl radicals are also quite stable. They are resonance stabilized and the conjugated overlap allows for stabilized, lower energy pi molecular orbitals to be formed as we saw earlier.
When cyclohexene reacts with a bromine radical, the following product predominates.
Let’s analyze the hydrogen atoms that we could have replaced on cyclohexene. All of the hydrogen atoms on cyclohexene are secondary, yet their reactivity is very different. The hydrogen atoms on the double bond are called vinyl or vinylic hydrogen atoms. Hydrogen atoms on a carbon that neighbors the double bond are called allyl or allylic hydrogen atoms. This carbon is an allylic carbon atom. The other hydrogen atoms are regular alkyl hydrogen atoms. Allylic hydrogen atoms are the most reactive. Why is this since they are all secondary hydrogens?
Vinyl vs. allyl vs. alkyl protons
If we look at the radical that is formed when the allylic hydrogen atom is removed, we find that resonance forms can be drawn for it. This greatly stabilizes the free radical.
Free radical bromination at an allylic position
If a vinylic or alkyl hydrogen is removed, resonance forms cannot be drawn. Vinylic radicals are especially unstable. In a vinyl radical (much like in a vinyl cation), the electron deficient partially positive location is in an sp2 hybrid orbital and held closer to the positive carbon nucleus than it is in an sp3 hybrid orbital. This is a bad situation.
When we generate bromine radicals in the presence of a double bond, we need to be careful. If we did an initiation reaction with Br2 and ultraviolet light like we have already learned, we would run into problems. The problem is that alkenes react with Br2. We learned about this previously. Alkenes reacting with Br2 can result in the dibromination of the double bond.
How do we avoid this problem? We need to make sure we don’t have too much Br2 present in our reaction mixture. The best way to do this is to perform this reaction with NBS (N-bromosuccinimide) and not Br2. NBS generates Br2 in small amounts so the dibromination of the double bond does not happen.
Now that you know about allylic bromination, predict the product of the following reaction.
4. Fill in the blanks.
We call the carbon that neighbors a benzene ring a benzylic carbon atom. The hydrogen atoms on that carbon are benzylic hydrogen atoms.
We can now include these resonance-stabilized positions in a rank of the stability of free radicals.
vinyl ≈ methyl < 1° < 2° < 3° < allylic ≈ benzylic
Least stable Most stable
When a free radical reaction occurs in a compound with a pi-bond, you need to always consider the various resonance forms. Sometimes, if the molecule is symmetrical, the products we get are the same, so it does not matter. We saw this in the bromination of cyclohexene above and in the following.
So, if we draw the overall reaction, we only need to draw one product.
But, when the alkene is less symmetrical, it can lead to a mixture of products.
If we draw the overall reaction, we need to include both products, since they are different.
5.
(a) Identify the most reactive hydrogen atom(s) in the following molecules if they undergo free radical reactions.
b) For these molecules, draw the major monobrominated product(s) of each when they undergo a free radical bromination. Monobrominated means only one bromine atom is on the molecule.
SN2 of allyl bromide
The SN2, backside attack, reaction of a nucleophile with propyl bromide is a pretty good reaction since it is a primary halide. In the transition state, the attacked carbon atom is sp2 hybridized. The nucleophile attacks a p-orbital as the leaving group leaves from the p-orbital.
This reaction works quite well, but it does not work with Grignard reagents as the nucleophile.
Allyl bromide can also do an SN2 backside attack reaction. But, this SN2 works even better than propyl bromide. Why would this be a more reactive substrate if it is a primary bromide like propyl bromide? The reason is, in the transition state, the p-orbital formed as the nucleophile attacks and the leaving group leaves can overlap with the p-orbitals of the pi bond and form an extra stable, lower energy, transition state. This is because it is conjugated. In fact, this substrate is so reactive that allyl bromide will do an SN2 reaction with a Grignard reagent!
6. Draw the products of the following reactions.
a)
b)
c)
Answers
4.
5.
a)
b)
6.
a)
b)
c)
