Note: Apologies for the late article, but I had to spend several days writing this article in order to come up with a comprehensive list of tips to aid in the memorization of the organic reactions and mechanisms. I’ll be back to a regular posting schedule after this article.
Introduction
Learning and memorizing common organic chemistry reactions is not that hard, even though it may seem like so. There is no ‘best way to memorize organic chemistry reactions’; the best way to grasp reactions in OChem is to know the concepts. So in this article, I will teach you to:
- Recognise key reactive sites on each molecule (requires knowledge of functional groups).
- Recognise patterns in key reaction mechanisms, especially more complex ones (carbonyl addition, benzenes, radical reactions).
- Recognise common reagents and how they work (Grignard, Gilman etc.).
However, of course, I still demand that you know a few basic concepts of chemistry especially regarding electrons and how they work in mechanisms. Note that I will be doing an overview of these concepts before we begin. You must know:
- How electrons move to attack reactive sites.
- How to draw mechanisms for a reaction using arrow-pushing.
- How inductive and resonance effects are able to play a part in the movement of electrons.
So without further ado, let us start with the overview. So, how do electrons move to attack reactive sites (especially electron-deficient sites)? Remember that electrons usually move in pairs, with the exception of radical reactions.
Because of this, attack is also carried out in pairs, which are depicted by an arrow moving from the pair of electrons to a reactive atom (reactive site) for reaction (Fig. 1). The result is bond formation. An atom with an unshared pair of electrons may use this pair of electrons to attack at electron-deficient site.
Fig. 1: Drawing arrows to depict movement of electrons.
Sounds familiar? Yes, this is because we are referring to a nucleophile. If you are unaware of what a nucleophile is, I would suggest looking at the article linked above, which explains nucleophiles, and their counterparts, electrophiles, in great detail.
In what places would two electrons be present? In my opinion, two prominent examples exist, which are in unshared pairs of electrons or pi bonds. In both of these cases, two electrons are present that are available for donation (or attack).
While sigma bonds also have two electrons, the electrons are not available for donation, because the energy barrier is too high. Importantly, however, notice that many reactive organic compounds have either pi bonds or unshared pairs of electrons (or both, as can be seen for carbonyl compounds).
Finally, let us learn about how inductive and resonance effects work. Inductive effects simply refer to whether electron density is withdrawn from or donated to an atom. For example, in an alkyl halide, electron density is withdrawn from the carbon atom by electronegative chlorine, making it electron-deficient.
Since the carbon is electron-deficient, it is therefore prone to attack by a nucleophile (Fig. 2), which possesses electron density that can be donated. This makes it reactive. So this is how inductive effects play a part in dictating the movement of electrons.
Fig. 2: Attack by nucleophile on alkyl halide.
Resonance may also affect the movement of electrons. The transition between resonance forms (although this does not happen physically) involves the movement of electrons and are drawn out during mechanisms. For example, look at the resonance of benzene.
Tip 1: Recognize key reactive sites.
What do I mean by this? First, let us take a look at the following molecule (Fig. 3). For those who are interested, the molecule is known as 3-chloropropanoic acid and is used as a drug. Can we recognize the reactive sites on this molecule?
Fig. 3: Structure of 3-chloropropanoic acid.
Of course, when we look at organic chemistry, we are almost always looking at attacks by electrons on carbon atoms, since organic chemistry is about the study of carbon compounds. So the hint is that the reactive sites are usually carbon atoms.
And now, for the answer. This question is a relatively simple one, and if you didn’t identify the correct carbon atoms to be reactive sites, you should reread what I have stated above. There are two reactive sites on this molecule, as are circled in the diagram in Fig. 4.
Fig. 4: Reactive sites in 3-chloropropanoic acid.
As you may notice, both of the reactive sites are carbon atoms which are directly bonded to heteroatoms. Although this may not always be the case, it is likely to be so, due to the inductive effects discussed previously. However, because of inductive effects, it may also be possible for heteroatoms to affect non-adjacent atoms; we will look at this in more detail later.
Let us look at Fig. 4. We discuss the carbon bonded to the chlorine atom first. The chlorine atom withdraws electron density from the carbon atom via the inductive effect (since chlorine is more electronegative than carbon), causing the carbon to be electron-deficient.
So, why does this make the carbon a reactive site? Since the carbon is electron-deficient, it is thus susceptible to attack by a nucleophile, which possess the lone pair of electrons needed to form a bond with the carbon. The second factor is also that chlorine is a leaving group, but this will be discussed later.
The second reactive site on the molecule is the C=O oxygen (also known as the carbonyl oxygen). In this case, there are two factors which contribute to the reactivity of the carbon atom. These are the inductive effect and the presence of a pi bond.
Now, let us look at the factors you may need to identify in order to draw an accurate reaction mechanism (you will need to know where the reactant molecule attacks another reactant molecule, for example, and this can be done by identifying the reactive site). The factors, in no particular order, are listed below:
- Leaving groups
- Electronegative atoms
- Acids and bases
- Bond strength
- Point charges
- Resonance effects
Yes, there may be a lot of factors, but it will become second nature to you after you look at enough molecules. At that stage, you will already be able to identify reactive sites in <5 seconds. (Not guaranteed.) Just think about it this way: would you rather memorize 200+ reactions or memorize 6 key factors to look out for?
Without further ado, let us begin going through all the factors.
1st Factor: Leaving Groups
The concept of leaving groups is highly linked to the concept of electronegative atoms. Very often, the leaving group will be an electronegative atom as well. But before that, what exactly are leaving groups? Let us look at the diagram below of a substitution reaction.
The leaving group in this case would be the atom that departs from the carbon backbone at the end of the reaction. It is seen in substitution reactions. One property of a leaving group is that it has to possess a weak bond to the carbon atom. Why? So that it can easily ‘leave’!
For the leaving group to possess a weak bond, the cleavage of that bond should be energetically favorable. This means that the leaving group should be stable as a free atom. In general, when talking about organic chemistry, we consider atoms from Group 15, 16 or 17 (except for carbon).
These atoms, in order to obtain noble gas electron configuration, take on negative charges. Take, for example, P3- or O2-. Thus leaving groups will take on negative charges as free atoms. For them to be stable, they need to be electronegative, since this allows them to bear the negative charge better.
When we see a carbon atom bonded to an electronegative atom, this gives us an indicator that the electronegative atom is a leaving group. But before we can say that this atom is a leaving group, we still must look at the bond strength between the electronegative and carbon atoms.
The bond strength of a carbon-leaving group bond can vary not just based on inductive effects, but also from the presence of additional pi bonds, as in the case of a carbonyl functional group (Fig. 5). In the carbonyl, oxygen is not a leaving group because it is not that stable as a free atom and also because of the strong carbon-oxygen bond.
Fig. 5: Carbonyl functional group.
2nd Factor: Electronegative Atoms
Electronegative atoms can affect the reactivity of a carbon atom through two main ways (that I can think of now). The two ways are through the inductive effect and through being a leaving group, which are interconnected.
In the nucleophilic substitution reaction, a nucleophile attacks and forms a bond with a carbon (which is bonded to the leaving group). The leaving group then leaves. (It can also be the other way round.)
Remember how we said a nucleophile only attacks electron-deficient atoms? Well, this is true here as well. Because the heteroatom bonded to the carbon is electronegative, it withdraws electron density from the carbon atom, making the carbon atom electron-deficient. This is why it is attacked by the nucleophile.
3rd Factor: Acids And Bases
The factor of acids and bases is sometimes overlooked by students. When we talk about this factor, we do not usually look at carbon atoms as reactive sites; instead, we are looking at heteroatoms, most commonly oxygen or nitrogen.
We will talk about a simple example of a base which we might not expect to see. Look, again, at the example of the carbonyl in a carboxylic acid. On the surface, it appears that only the carbon atom is reactive, by the inductive effect. However, are you aware that even the oxygen atom can be reactive as well? Look at the example of the first and second steps of Fischer esterification.
In this reaction, the first step involves protonation of the oxygen atom! This is very interesting because the result is a positively-charged oxygen atom. This makes it a much better leaving group because after it leaves, it becomes the more stable OH- moiety, as opposed to the O2- moiety if the protonation had not taken place.
This unlocks the possibility of subsequent attack by the alcohol on the carbonyl carbon, setting the stage for the departure of one of the OH groups (since there are two after protonation from the carboxylic acid).
4th Factor: Bond Strength
Bond strength, in the end, is a culmination of the factors we discussed above. The strength of bond can vary based on many factors including the resonance and inductive effects we have discussed (resonance effects are discussed later).
Bond strength can also vary based on a wealth of other factors that are not significant enough to be discussed in a paragraph by themselves. But first, let us take note of how exactly bond strength affects the reactive site.
Naturally, reactions involve bond formation and bond breakage. For a carbon atom, for a bond to be formed to it, a bond needs to break away from it. This usually happens only when the bond is a weak one. As such, a carbon atom which possesses weak bonds can be considered a reactive site.
Some other factors can also affect the bond strength. These include the inductive effect, presence of only one bond as well as a large atomic radius. The inductive effect contributes to a bond’s polarity, making the bond more unstable.
The presence of a single sigma bond, of course, naturally makes the bond relatively weak as compared to double or triple bonds. Finally, the atomic radius matters because a bond formed between atoms of large atomic radius will have a large bond length, contributing to a low bond strength (Fig. 6).
Fig. 6: Weak bonds between large atoms.
5th Factor: Point Charges
When we looked at the inductive effect, we noted that the presence of a electron-deficient carbon atom made it a reactive site. Now let us look at something more ‘extreme’: what about actual (or formal) negative or positive charges on atoms?
These charges, no matter positive or negative, can greatly affect the reactivity of an atom. This is especially true for carbon. A positively-charged carbon atom, known as a carbocation, is formed during the first step of the SN1 reaction, and attack on this carbocation is a very fast step.
Negatively-charged carbon atoms tend to be slightly rarer than positively-charged carbon atoms. My memory only permits me to think of one case where negatively-charged carbon atoms may be produced: through an acid-base reaction.
For this to occur, the carbon atom must be acidic enough, as in the case of alkynes, where the carbon atom is weakly acidic. This allows it to undergo an acid-base reaction in the first step of acetylide formation, thus demonstrating the effect of acidity and basicity on reactivity.
In the second step of this reaction, an alkyl halide reacts with the acetylide ion formed to form a new alkyne (the negatively-charged carbon atom of the acetylide acts as a nucleophile in this case).
6th Factor: Resonance Effects
We have not yet discussed resonance effects, although we have looked at it in passing. Resonance is explained in detail in one of our previous articles, but we will attempt to give a brief introduction to resonance here as well.
Resonance suggests that a compound’s true structure is made up of several resonance forms. Although the real structure is static, it is more convenient to imagine that a structure rapidly transitions between resonance forms, unlocking new areas of reactivity, or perhaps making a compound more stable. Just look at the resonance of benzene (Fig. 7).
Fig. 7: Resonance forms of benzene.
Benzene, the simplest arene, is also exceedingly stable. This can be explained by the concept of aromaticity, which is linked, somewhat, to resonance. One reason why benzene is so stable is because electron density is evenly distributed around the ring (due to resonance), ensuring that there are no reactive sites.
Furthermore, remember when I said that the true structure is a combination of the two resonance forms? When we combine the two resonance forms of benzene, we obtain a hybrid where all carbon atoms possess 1.5 bonds with each other (1 sigma bond as well as a partial bond). This is another reason why it is so stable.
As we can see from here, the resonance makes benzene highly stable. There are other cases where resonance leads to point charges on some atoms, making them reactive (as in the transition from ketone to enolate, Fig. 8).
Fig. 8: Ketone to enolate.
Tip 2: Recognize patterns in reaction mechanisms.
Of course, some reaction mechanisms are slightly more difficult to decipher as compared to others. We compare two examples of reaction mechanisms: that of a simple nucleophilic substitution reaction and that of a Friedel-Crafts-type alkylation (Fig. 9). Which one would likely be easier to understand?
Fig. 9: Friedel-Crafts-type alkylations.
Naturally, it would be the former. The latter is a highly specific type of reaction which involves benzene, and as with many benzene mechanisms, it is difficult to remember. In this tip, we will look at 4 specific classes of mechanisms which encapsulate many reactions:
- Electrophilic aromatic substitution
- Nucleophilic aromatic substitution
- Free radical substitution
So, let us begin right away!
1st Mechanism: Electrophilic Aromatic Substitution
Electrophilic aromatic substitution reactions are not easy to understand at a glance mainly because the mechanism involves several steps. Not only that, because benzene possesses resonance forms, it is difficult for us to pinpoint the movement of electrons.
Nevertheless, all electrophilic aromatic substitution reactions involve the same kind of mechanism (Fig. 10). Benzene is a nucleophilic (since it is an electron-rich pi system), and thus it uses its pi electrons to attack the electrophile (arrow pointing from benzene to the electrophile, not the other way round!).
Fig. 10: Electrophilic aromatic substitution mechanism.
The mechanism begins with attack on the electrophilic atom by the pi electrons of one of the pi bonds on the benzene ring. After the pi bond breaks, a carbocation is formed. Finally, the intermediate is deprotonated, reforming the pi bond and this results in the final product.
Notice that in the electrophilic aromatic substitution mechanism, only 1 pi bond is ‘really’ affected in the reaction. The other 2 pi bonds seemingly do not participate. However, do note that resonance forms exist that stabilize the carbocation, although this is not indicated.
Also, I have a final note: usually, the reagents themselves are not electrophilic attack for benzene to attack. Instead, there must first be a reaction, usually resulting in a more electrophilic and positively-charged compound, that can then be attacked by benzene.
2nd Mechanism: Nucleophilic Aromatic Substitution
Next, we move on to nucleophilic aromatic substitution. In this type of substitution, benzene will be the electrophile instead of the usual nucleophile. In these cases, two mechanisms are possible.
In the first mechanism (Fig. 11), it is just as we would expect, with attack performed first by the nucleophile on one of the carbons forming the pi bond in benzene. The carbon here must be attached to a ‘leaving group’ - for example, a halide atom.
Fig. 11: Nucleophilic aromatic substitution mechanism.
The second step of the mechanism involves the departure of the leaving group, which occurs as the pi bond is reformed. The final product looks the same as if electrophilic aromatic substitution had occurred instead, so be sure to look at the reactants and determine which of the mechanisms is the correct one.
Just as a side note: Since the departure of the leaving group occurs in the second step only, how good the leaving group is does not matter here! So something that is stable as a free atom but with strong bonds to carbon can also act as a leaving group.
However, there is also another mechanism that may operate, known as the aryne or benzyne mechanism. In this mechanism, the leaving group departs in the first step instead of the second step. The mechanism is shown in Fig. 12.
Fig. 12: Nucleophilic aromatic substitution via arynes.
There is a key difference between the aryne and non-aryne mechanism: the aryne mechanism requires the leaving group to be excellent, and there must also be a base present. Through the mechanism, we can see why there needs to be a base. And since the leaving group departs in the rate-determining step, it has to be a good one.
The mechanism begins with an elimination, of the leaving group and a hydrogen atom. This forms an interesting intermediate known as an aryne (do note that the triple bond is a misleading one). It is then attacked by the nucleophile and then protonated to form the product.
3rd Mechanism: Free Radical Substitution
We have already looked at free radical substitutions comprehensively in a previous post. However, we will again outline the mechanism here for convenience sake. A typical free radical substitution reaction includes three types of steps: initiation, propagation and termination (Fig. 13).
Fig. 13: Mechanism of free radical substitution.
Fee radical substitution reactions involve free radicals, which are highly reactive particles possessing unpaired electrons.. A free radical can react with almost anything, and because of this, three types of steps are possible. The first step is the initiation step, which involves the homolytic cleavage of a bond into two separate parts, forming two different radicals.
In the propagation step, a radical attacks a stable molecule by abstracting a hydrogen (which also contains one electron). The radical becomes stable while the stable molecule becomes a radical. The new radical would be different from the old one in many cases, and thus it is called a propagation step.
In the final termination step, two radicals combine together, and this forms a stable molecule. No other radicals can be generated from here and thus it is known as a ‘termination step’. In Fig. 13 above, we observe that many electron movements are depicted with half-headed arrows, indicating the movement of one electron, while full-headed arrows depict the movement of two electrons.
Note: As for reagents, there is a huge list, so I will be discussing them only in the next article. (Anyway, this article is already getting way too long, in my opinion).
very well done! thank you so much
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