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Substitution vs Elimination

Introduction

We just spent the last 2 chapters talking about substitution vs elimination reactions, and, while it was briefly mentioned, we didn’t go into full detail on when each of these reactions typically occurs. In this chapter, we’ll do just that studying substitution vs elimination. Soon, you’ll be able to look at the nucleophile/base and the substrate for a reaction and identify whether those reagents will favor an SN2 reaction, an SN1 reaction, an E2 reaction, or an E1 reaction. We even provide an easy-to-read flow chart for reliably solving substitution versus elimination problems.

This chapter is often tricky for Organic Chemistry students as the way many textbooks present the topic is confusing, even appearing contradictory at times. The goal of this chapter is to simplify this complex material by breaking it into 3, very manageable parts.

Typically, an Organic Chemistry professor will give their students the substrate and the nucleophile/base and ask about which mechanism (SN1 mechanism, SN2 mechanism, E1 mechanism, or E2 mechanism) will be favored and the resulting product. Although we have learned about substitution vs elimination reactions separately up to this point, we will now put them together and learn how to distinguish between the two.

 

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E1 Reaction Mechanism

The E1 Reaction Mechanism

The E1 reaction mechanism is a 2-step reaction that also forms an alkene. The “1” in E1 refers to the rate law, which only depends on the concentration of the substrate.

In the first step of an E1 reaction mechanism, the leaving group leaves, forming a carbocation. In the second step, the base deprotonates a neighboring hydrogen to form a double bond between the carbocation and carbon that was just deprotonated.

A figure showing the 2-step E1 reaction mechanism and the E1 rate law

Notice that the base here is water, which has no negative charge and is a very weak base. This is consistent with most E1 reactions as E1 reaction mechanisms generally use a weak base. E1 reactions also tend to occur when the leaving group is on a tertiary carbon. An E1 reaction will never occur if the leaving group is on a primary carbon and will only rarely occur if the leaving group is on a secondary carbon.

 

Keep it Simple

In the E1 reaction mechanism, a carbocation intermediate is formed just like in an SN1 reaction. But in the SN1 mechanism section, we really focused on the carbocation rearrangement by hydride and methyl shifts. Why don’t we make such a big deal about that in the E1 section?

It’s because E1 reactions almost exclusively occur on tertiary carbons, so the molecule can’t ever rearrange to a more stable carbocation. SN1 reactions also prefer to occur on tertiary carbons, but they can also take place on secondary carbons, which opens up the door for rearrangements.

We also do not focus on the leaving group being anti-periplanar to a hydrogen. Because of the carbocation intermediate formed, this is not something to worry about; however, the hydrogen still must neighbor the leaving group.

 

Summary Points for E1 Reaction Mechanisms

  • 2-step mechanism that goes through a carbocation intermediate at the end of step 1.
  • Occurs almost exclusively when the leaving group is on a tertiary carbon and when the base involved is very weak.

 

 

 

E2 Reaction Mechanism

The E2 Reaction Mechanism

 

The E2 reaction mechanism is a 1-step mechanism that results in the formation of a carbon-carbon double bond.

 

Keep it Simple

Just like substitution reactions, the “2” in E2 reaction mechanism does not refer to the number of steps but rather to rate law of the reaction, which depends on both the concentration of the substrate and the base.

 

In the E2 reaction mechanism, the leaving group on the substrate leaves and a neighboring hydrogen is deprotonated. The electrons from the carbon-hydrogen bond fill in to form an alkene between the carbon that held the hydrogen and the carbon that was bonded to the leaving group.

Below is an example of an E2 reaction mechanism. The rate law is also shown below.

 

This figure illustrates the E2 reaction mechanism and the E2 rate law.

In an E2 reaction mechanism, the leaving group (the bromine here) leaves the molecule. At the exact same time, a base deprotonates a neighboring hydrogen. The neighboring hydrogen is a hydrogen that is connected to a carbon neighboring the carbon that is bonded to the leaving group (see figure below).

 

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Note that the base takes away the hydrogen and not any of the electrons of the carbon-hydrogen bond. The electrons in the carbon-hydrogen bond fill in to form the double bond between the carbon that had the hydrogen and the carbon that had the leaving group attached.

 

Keep it Simple
The base will NEVER deprotonate a hydrogen on the same carbon as the leaving group in an elimination reaction. This is a common mistake students make.

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Example
What is the product of the following E2 reaction? Draw the mechanism as well.

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Answer

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The first thing we must do is identify the leaving group. There are 2 halogens on the molecule, but remember from the Mechanism Preparation chapter, fluorine cannot be a leaving group because it cannot properly stabilize a negative charge; therefore, the leaving group is the chlorine.

After finding the leaving group, we then look at the neighboring carbons to see if there are hydrogens present. There are 3 hydrogens on the right neighboring carbon and none on the left neighboring carbon.

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All 3 hydrogens on the neighboring carbon on the right give the same product, so we can show the mechanism using any of the 3. We can now draw our E2 reaction mechanism and product:

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Anti-Periplanar

We just learned that the hydrogen that is deprotonated must be on a neighboring carbon; however, there still is 1 more stipulation about the deprotonated hydrogen: it must also be anti-periplanar to the leaving group. To understand what that means, let’s break down the term anti-periplanar.

Anti: Think back to Newman projections. Remember the anti conformation? This was the most stable position for 2 substituents as they were as far apart as possible. In E2 reactions, the hydrogen that will be deprotonated and the leaving group must be anti to one another.

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Periplanar: The neighboring hydrogen and the leaving group must also be in the same plane as one another. Using the same molecule in the Newman projection above redrawn into bond line drawing, we can see that both the leaving group and the hydrogen are in the plane of the page.

 

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The leaving group and the hydrogen can also be periplanar even if they are not in the plane of the page. Here’s an example below where the leaving group and the hydrogen are anti-periplanar, while being on dashes and wedges.

 

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To briefly summarize, the hydrogen that is deprotonated during an E2 reaction must be both on a neighboring carbon and anti-periplanar to the leaving group.

 

Example
Identify any hydrogen atom(s) that could be deprotonated if the molecule below underwent an E2 reaction.

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Answer

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In an E2 reaction mechanism, the deprotonated hydrogen must be (1) on a neighboring carbon and (2) anti-periplanar to the leaving group. Therefore, we must identify all the hydrogen atoms in this molecule that are neighboring and anti-periplanar.

The first step is to draw all the hydrogen atoms neighboring the leaving group as these are the hydrogen atoms we must further assess. The leaving group in this molecule is the chlorine.

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After explicitly drawing out each hydrogen, there is one hydrogen that is clearly anti-periplanar, shown in red below.

 

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But remember, there is free rotation around single bonds. So the molecule can rotate to make more neighboring hydrogens anti-periplanar. Try to do this with your model kit to see how each neighboring hydrogen can be rotated into the anti-periplanar position.

Because of free rotation around single bonds, all the neighboring hydrogens in this molecule can be rotated into an anti-periplanar position with the leaving group.

 

 

Determining the neighboring, anti-periplanar hydrogen(s) in a chair configuration can be very difficult. In fact, there might not even be a neighboring, anti-periplanar hydrogen. If there is no neighboring, anti-periplanar hydrogen, then the E2 reaction cannot occur. See below for an example of finding the anti-periplanar hydrogen atoms of a cyclohexane chair.

 

Example
Identify the neighboring, anti-periplanar hydrogens(s) in the molecule below.

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Answer

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First, let’s draw all neighboring hydrogens, remembering where the substituents are located in a chair conformation.

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Now we must see which, if any, of these hydrogens are anti-periplanar. As the bromine is pointed directly up, we need our hydrogen to be pointing straight down for it to be anti-periplanar. Using this approach, we can immediately see that there are 2 hydrogens pointing straight down, so these are the neighboring, anti-periplanar hydrogens.

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Because this molecule has anti-periplanar hydrogens, it would be able to undergo an E2 reaction in the presence of the right base. If this is difficult to see, I would highly recommend using a model kit because chair conformations can be difficult to fully understand spatially.

 

When dealing with E2 reactions using cyclohexane, you must draw the chair conformation to see if there are any neighboring anti-periplanar hydrogens. Even if you are explicitly given the 2-D cyclohexane, you should convert it into the 3-D chair prior to solving the E2 reaction.

 

Example
What is the E2 reaction product of the following? Draw the complete mechanism.

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Answer

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First, we must identify any neighboring, anti-periplanar hydrogens. To do this, we start by drawing out the implied hydrogens on the neighboring carbons:

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Now we assess each of these hydrogens to see if any are anti-periplanar. This reveals that none of these hydrogens are anti-periplanar (use a model kit if you are having trouble seeing this). Since an E2 reaction cannot occur without a neighboring, anti-periplanar hydrogen, does that mean no reaction will occur here? Normally yes, but since we are using a cyclohexane chair, we can perform a chair flip:

 

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After the chair flip, our chlorine is pointing straight up, so our anti-periplanar hydrogens must therefore be pointing directly down. After assessing the neighboring hydrogens, we can see we have 2 neighboring anti-periplanar hydrogens, both of which would give the exact same product:

 

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Keep it Simple
It can be difficult to tell whether groups are on the same side or on opposite sides of the double bond when doing an elimination reaction. For example, you might not be sure which of the products below the following elimination reaction will form.

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The two products are different molecules, so it’s important we know how to differentiate these two.

To solve difficult questions like these, I’d recommend rotating the leaving group and the neighboring hydrogen so they are anti-periplanar and in the plane of the page.

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From here, we perform the E2 reaction mechanism. In the final product, the groups that were on wedges will be on the same side of the double bond, while the groups that were on dashes will be on the other side of the double bond. Therefore, the final product looks like this:

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When does an E2 reaction occur?

E2 reactions occur when a strong base is present. Just like an SN2 reaction, E2 reactions prefer for the leaving group to be located on a primary carbon rather than a tertiary carbon; however, the preference is more flexible– E2 reactions can occur on a tertiary carbon if the base is strong enough. In the next chapter, we’ll talk more about assessing the reaction conditions to decide if a substitution reaction or an elimination reaction will occur, so don’t worry about this too much for now.

 

Summary Points for E2 Reactions

  • The rate of the E2 reaction depends on the concentration of both the substrate and the base.
  • Occurs with strong bases. There is a preference for the leaving group to be on a primary carbon, but, if the base is strong enough, an E2 reaction can still take place on a tertiary carbon. There will be much more on this next chapter.
  • There must be a neighboring anti-periplanar hydrogen to the leaving group in order for an E2 reaction to occur.

 

 

 

Sn1 vs Sn2

SN1 vs SN2

There is a hierarchy of 3 main factors that determines whether a reaction will undergo an SN1 vs SN2 reaction, 1 of which we have not yet discussed. Note that each of the factors is discussed below as though one is only trying to choose between an SN1 reaction and an SN2 reaction.

  1. The location of the leaving group: If the leaving group is on a tertiary carbon, the reaction will tend to undergo SN1 substitution. If the leaving group is on a primary carbon, the reaction will tend to undergo SN2 substitution. If the leaving group is on a secondary carbon, one must look at the remaining factors.
  2. The strength of the nucleophile: If the nucleophile is very strong due to a negative charge or high polarizability (e.g. I-), the reaction will tend to undergo SN2 reactions. If the nucleophile is weaker (e.g. H2O), then it will tend to undergo SN1 reactions.
  3. The solvent: This factor has not yet been covered. The solvent is the majority liquid in a solution. In the reaction below, CH3OH is the solvent.

 

A typical reaction used to demonstrate Sn1 vs Sn2 reactions

 

The type of solvent used plays a factor in whether the mechanism will be SN1 vs SN2. Polar protic solvents (solvents with an –OH) typically favor an SN1 mechanism, while polar aprotic solvents typically favor an SN2 mechanism. We’ll go over each in more detail below.

Polar protic solvents, solvents with an –OH, favor SN1 reactions. They are called polar protic because they are polar (due to the presence of an electronegative atom like oxygen on the molecule), and they are protic because they have an acidic proton that can be used to hydrogen bond with other molecules. They favor SN1 reactions because they are able to hydrogen bond with the carbocation that is formed after step 1 of the mechanism, which stabilizes the positive charge.

 

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Polar aprotic solvents do not contain any –OH groups, so it cannot undergo hydrogen bonding. Because of this, it is not a favorable solvent for a charged species, so polar aprotic solvents favor SN2 reactions.

Below is a list of common polar protic and aprotic solvents:

 

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Sn1 Reaction Mechanism

SN1 Reaction Mechanism

The SN1 reaction mechanism is a 2-step substitution reaction. Just as in SN2 reactions, the “1” in SN1 does not refer to the number of steps of the reaction, but rather to the rate law, which only depends on the concentration of the substrate. Therefore, the rate law of an SN1 reaction is completely independent of the concentration of the nucleophile.

 

Here is the SN1 reaction mechanism:

 

This image shows the 2 steps of the Sn1 reaction mechanism. It also illustrates the Sn1 rate law.

As you can see above, the 2 steps of the SN1 reaction mechanism are (1) the formation of a carbocation and (2) the nucleophilic attack on the substrate. This reaction does NOT follow the “bowling ball mechanism” of an SN2 mechanism, but rather, it is more of a cause and effect mechanism.

 

Cause: The leaving group on the substrate leaves, putting a positive charge on the substrate.
Effect: The electron-rich nucleophile is attracted to the newly-formed positive charge, resulting in nucleophilic attack at this location. A racemic mixture, or mixture of equal parts R and S enantiomers, is formed, so the final product lacks chirality.

 

Keep it Simple
Why does a racemic mixture always form in an SN1 reaction mechanism? This is because the mechanism goes through an sp2-hybridized intermediate, so it loses all chirality it initially might have had.
Let’s look at this concept more closely by looking at the SN1 reaction of the chiral molecule below.

 

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The first step will be the loss of bromine, the leaving group.

 

Note that a racemic mixture is formed during an Sn1 reaction mechanism.

When we form the carbocation, the hybridization changes from sp3-hybridized to sp2-hybridized, which has major consequences for stereochemistry. The newly formed sp2-hybridridized molecule is trigonal planar, meaning the bonds are flat in the plane of the page. In a trigonal planar carbocation, the nucleophile has a 50% chance of attacking the “front” of the molecule and a 50% chance of attacking the “back” of the molecule. Since the location of the nucleophile’s attack determines whether the molecule formed is R or S, a 50/50 mixture of R and S enantiomers will be formed.

In summary, the equal likelihood of the nucleophile attacking either side of the sp2-hybridized, trigonal planar molecule results in an achiral product.

 

Since carbocations are only stable when they are secondary or tertiary, SN1 reactions only occur on secondary or tertiary carbons. SN1 reactions cannot occur on a primary carbon because a primary carbocation isn’t stable enough to be formed.

 

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Because of the formation of a carbocation intermediate in the mechanism, SN1 reactions are prone to rearrangements, like hydride shifts and methyl shifts. Here’s an example of a hydride shift occurring after the formation of a carbocation in step 1 of the SN1 mechanism.

 

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When does an SN1 Reaction Occur?

SN1 reactions primarily occur on tertiary carbons, and sometimes secondary carbons, that have leaving groups attached to them. This is because the carbocation that is formed will be tertiary or sometimes secondary (and therefore fairly stable) when the leaving group leaves.

SN1 reactions do not need a particularly strong nucleophile because the positive charge formed in the first step will attract nucleophiles even as weak as water. Unlike the SN2 reaction in which the nucleophiles instigate the reaction, in an SN1 reaction, the nucleophile attacks only after a carbocation is formed. Therefore, the nucleophile in an SN1 reaction is typically weaker.

 

Summary Points for SN1 Reactions

  • Forms a carbocation in step 1, which is subject to hydride shifts and methyl shifts.
  • Occurs at tertiary and sometimes secondary carbons because carbocations can only stably exist on tertiary and sometimes secondary carbons.
  • Often uses weak nucleophiles. The presence of a positive charge on the substrate can attract even weak nucleophiles.
  • Forms a racemic mixture. Even if starting reactant is chiral, the end product will be achiral (assuming the stereogenic center is the site of reaction).

 

Keep it Simple
Notice that the keys points about an SN1 reaction are all dictated by the carbocation that is formed at the end of step 1. For example:
1. SN1 reactions only occur at tertiary and secondary carbons because those are the only places a carbocation can stably form.
2. SN1 reactions require only weak nucleophiles because the carbocation can attract even partially negatively charge species.
3. A racemic mixture is formed in an SN1 reaction because of the sp2-hybridized, trigonal planar intermediate formed by the carbocation.

 

 

Sn2 Reaction Mechanism

SN2 Reaction Mechanism

The SN2 reaction mechanism is a 1-step mechanism that substitutes one atom/group for another. In one step, the nucleophile attacks the substrate (or the electrophile), simultaneously causing the leaving group to leave the molecule. Here is the mechanism for an SN2 reaction:

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The rate law is dependent on both the concentration of the nucleophile and the electrophile/substrate.

 

Keep it Simple
Why is there a “2” in SN2 reactions if it’s a 1-step mechanism? It’s because this “2” doesn’t refer to the number of steps in the reaction at all. Instead, it refers to the fact that the rate law depends on the concentration of 2 reactants, both the nucleophile and the substrate.

Rate of SN2= k [Nucleophile][Substrate]

 

The nucleophile (the –OH) attacks the substrate on the carbon next to the leaving group (the bromine), causing the leaving group to leave. Think about it like 3 bowling balls:

 

This image shows a simplified version of the SN2 reaction mechanism, using a bowling ball analogy.

 

In order to displace the bromine, the –OH must attack from the exact opposite side of the bromine, which is known as backside attack. This leads to an important point: all SN2 reactions take place via backside attack. Think of the bowling ball analogy to understand why SN2 reactions must occur via backside attack.

Backside attack is important because it results in inversion of stereochemistry at the carbon of the attack. Therefore, all SN2 reactions result in inversion of stereochemistry at the carbon of the attack. This is particularly important if this carbon is a stereogenic center.

Let’s look at an example.

 

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Looking at the example above, the far-right carbon on the substrate, a stereogenic center, is the site of the nucleophilic attack (see why in the Keep it Simple below). To undergo the SN2 reaction mechanism, the -OH must attack the backside of the carbon indicated by the arrow below.

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The –OH assumes the position it attacks, while the deuterium and the hydrogen flip upward because of the attack of the –OH. This has inverted the stereochemistry. The mechanism for this reaction is shown below.

 

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Keep it Simple
A common question students ask is “How do I know which carbon the nucleophile attacks?” To answer this, let’s examine the molecule below:

 

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Because of induction, the electronegative chlorine has a partial negative charge, while the neighboring carbon has a partial positive charge.

 

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Because nucleophiles are electron-rich and often negatively charged, they will be attracted towards a positive charge, like the partial positive charge on the specified carbon atom. You can think of this like a magnet being attracted to another, oppositely polarized magnet.

It’s also important to note that a leaving group must be present at the carbon for a substitution reaction to occur.

In summary, it is the attraction of opposite charges that brings the nucleophile and the electrophile together. If a good leaving group is also present at this carbon, then a reaction can occur.

 

 

When does the SN2 reaction mechanism occur?

SN2 reactions occur with (1) strong nucleophiles at (2) very non-hindered carbons on the substrate (mostly primary and sometimes secondary carbons). We elaborate on these 2 factors below.

(1) The nucleophile itself must be strong as it is the instigator of the attack; it must actively go out and attack the substrate itself.

(2) If there are lots of other groups in the way (like with a tertiary carbon), the nucleophile won’t be able to make that “bowling ball” contact that is necessary for an SN2 reaction as the nucleophile will be blocked from its desired backside attack. Remember, an SN2 reaction always reacts via backside attack.

 

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As you can see from the green arrows above, the nucleophile must invasively attack the molecule as it must attack the carbon attached to the leaving group. Therefore, steric hindrance effects (like in a tertiary carbon) prevent the nucleophile from attacking where it wants.

 

Summary Points for SN2 Reactions

  • The rate law depends on the concentrations of both the substrate and the nucleophile.
  • Inversion of stereochemistry occurs due to the nature of the SN2 backside attack.
  • Occurs with strong nucleophiles because the nucleophile is the instigator of the reaction. The nucleophile must be strong enough to go out and attack a substrate.
  • Occurs at unhindered carbons, like primary or sometimes secondary carbons. This is because the nucleophile must invasively attack the substrate, which it would not be able to do if it was blocked from backside attack because of the steric hindrance of a tertiary carbon.

 

 

What Makes a Good Leaving Group?

What Makes a Good Leaving Group?

An atom or group of atoms often must leave so another group can take its place in organic chemistry reactions. The group that leaves is, not surprisingly, called a leaving group. This section will study the properties of leaving groups, and examine the question “what makes a good leaving group?” Let’s look at an example of a reaction, clearly indicating the leaving group:

In this reaction, the bromide is the leaving group

The bromine is able to leave because bromide (the negatively charged bromine atom) is stable enough to exist on its own when it leaves the molecule. This illustrates a critical point about leaving groups: the leaving group only leaves if it can exist on its own in a fairly stable state.

As a simple analogy for this, think of a daughter moving away from her parents. A 5-year-old daughter is not capable of living on her own, so she obviously cannot leave her parents. But a 20-year-old daughter can live on her own, so she is able to leave her parents to possibly start a job or go to college. Relating this back to chemistry, only the 20-year-old daughter has the stability to be able to leave on her own, so she can, in a chemical sense, be a leaving group.

So what makes a good leaving group? Or asked in another way: what are the properties of leaving groups, such as the bromine above, that make it able to exist on its own? The answer is simple: anything that will stabilize it! Remembering back to our list of properties that will stabilize a molecule in the acids/bases chapter:

 

 

Specifically, the large size of the bromide is able to dilute the negative charge, which stabilizes the atom as the charge is not overly concentrated in a small area. Since bromide is fairly stable because of this, it is able to be a leaving group.

Keep it Simple
As an analogy to understand how a large atom dilutes a negative charge, picture a cup of water and a gallon of water (representing the small and large atom respectively). If one drop of red food coloring (representing the single negative charge), is added to each amount of water, then the cup will be much more red than the gallon of water due to the concentration of the food coloring.

To apply this example to organic chemistry, imagine that we add a single negative charge to both a bromine and a fluorine atom, forming Br- and F-. The large size of the bromine is able to dilute the negative charge, which has a stabilizing effect, while the smaller fluorine atom’s negative charge will be more concentrated and therefore less stable.

What if we have a smaller atom, like fluorine? Fluorine is actually so small that it would not leave as the negative charge would be too concentrated and therefore unstable if it left as a leaving group. Cl-, Br-, and I- can exist on their own, so they can each be leaving groups.

What makes a good leaving group? Note that it all has to do with the stability of the leaving group.

Fun Fact
Teflon® is a very unreactive polymer with the chemical structure below. The “n” indicates that the portion within the parenthesis repeats over and over again, which is the definition of a polymer.

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Teflon® is often used as a coating on pans because it’s so unreactive (clearly, no one wants their pan reacting with their food!).

The lack of reactivity of Teflon® lies in the carbon-fluorine bonds. The fluorine will never leave because it is such a bad leaving group even at very high temperatures, so the pan remains unreactive and out of the food.

What is an Electrophile?

What is an Electrophile?

This section will be focused on 3 questions “what is an electrophile?” “what is hyperconjugation?” and “what is a methyl shift?” But first, let’s define electrophile.

 

Electrophiles lack a lone pair of electrons and are, to some degree, electron deficient. Because one of the atoms doesn’t have all the electrons it wants, electrophiles often, but not always, carry a positive charge. For example, a carbocation, or a positive charge located on a carbon atom, is a common finding in organic chemistry electrophiles.

What is an electrophile? This image shows 2 electrophiles. Note how each is electron deficient.

 

When dealing with a carbocation, there are two important concepts to know for mechanisms: hyperconjugation and hydride shifts.

 

What is Hyperconjugation?

We will now answer the question “what is hyperconjugation?” Hyperconjugation is the stabilization of a carbocation by neighboring carbon-hydrogen or carbon-carbon bonds. In general, we look at the number of carbon-hydrogen bonds on the carbon neighboring the carbocation to assess hyperconjugation. The more carbon-hydrogen bonds in this position, the more stable the carbocation.

It therefore follows that a tertiary carbocation is more stable than a secondary carbocation, which in turn is more stable than a primary carbocation.

What is hyperconjugation? As this image shows, it is the ability of neighboring carbon-hydrogen bonds to stabilize carbocations.

Hydride shift (and methyl shift)

As we just saw, carbocations prefer to have as many neighboring carbon-hydrogen bonds as possible. In fact, this is such a strong preference that hydrogen atoms, and sometimes even whole methyl groups, will shift, moving the carbocation to a more stable position. If a hydrogen moves to better stabilize the carbocation, it is called a hydride shift. If it’s a methyl group that moves, it is called a methyl shift. Take a look at the molecule below:

Picture39

The carbocation is on a secondary carbon, which is more stable than a primary carbocation but less stable than a tertiary carbocation. If a neighboring hydrogen moves via a hydride shift, the carbocation will get placed on a more stabilized tertiary carbon.

This figure shows an example of a hydride shift.

 

Keep it Simple
When drawing a hydride or methyl shift, notice that the arrows are slightly different than typical curved arrows. In a hydride or methyl shift, the tail of the arrow cuts through the bond that is moving, indicating the electrons of the bond are also moving. Some professors are strict about this while others are not, so be sure to check in with yours!

 

If there is no hydrogen available, then a methyl group can shift to try to move the carbocation to a more stable carbon.

This figure shows an example of a methyl shift.

The molecule prefers to make a hydride shift more than a methyl shift because of the large amount of energy that must be overcome for a methyl shift to occur. Therefore, methyl shifts only occur when no hydride shift is possible.

 

 

 

What is a Nucleophile?

What is a Nucleophile?

The basic question of this section will be “what is a nucleophile?” followed by a discussion of nucleophile strength. A nucleophile has at least one lone pair of electrons (often even more) that it uses to react with other atoms and molecules. Because nucleophiles tend to be electron-rich, it’s fairly common for them to have a negative charge.

 

What is a nucleophile? In this image, 2 examples are shown. Note that both have lone pairs of electrons on the nucleophilic atom.

 

Nucleophile Strength

We can assess the nucleophile stength by looking at (1) its polarizability and (2) its general stability. We will examine each of these two factors more closely.

 

1. Polarizability

Polarizability is the ability of an atom’s electron cloud to change shape to better attack another atom/molecule. This can therefore be viewed as the flexibility of the atom to distort its electron cloud. Critically, the more polarizable (or “flexible”) an atom or molecule is, the greater the nucleophile strength.

Polarizability is directly related to the size of the atom: the larger the atom, the more polarizable it is. This is because larger atoms have their outer valence electrons further away from the central nucleus, so these electrons experience more freedom to roam around and change shape. Polarizability therefore increases as you move down the periodic table.

Furthermore, polarizability is the most important factor in determining the strength of a nucleophile. Therefore, it is the first factor to assess in these types of questions.

As polarizability increases, nucleophile strength increases as well.

 

2. The general stability of the molecule

Stability is used as a tiebreaker when polarizability alone doesn’t give enough information on the nucleophilicity of an atom. To assess the general stability of the molecule, use the same method presented in the acid base chapter.

 

 

If a negative charge is on a highly electronegative atom or if the charge is spread out throughout the molecule, then the molecule is more stable. If a molecule is very stable, then it will be a weak nucleophile as it won’t want to react due to its high stability.

Furthermore, if the nucleophile has a negative charge, it is a stronger nucleophile than the same nucleophile in neutral form. This is because a neutral molecule is more stable than a charged molecule. As a result, the neutral molecule is less reactive (making it a weaker nucleophile), while the less stable charged molecule is more reactive (and therefore a better nucleophile).

Note that this is not as high priority in determining the strength of a nucleophile as polarizability and should only be used as a tiebreaker if polarizability alone doesn’t give enough information to determine the strength of the nucleophile.

Example
Which of the following is the strongest nucleophile?

Picture36

Answer
Molecule 1

There are two factors we must look at to assess the strength of a nucleophile: the polarizability and the stability.

First, let’s look at polarizability as this is the most important factor in determining the strength of a nucleophile. Molecule 1 and 2 are more polarizable than molecule 3 because sulfur is more polarizable than oxygen. Therefore, we can eliminate molecule 3.

We now look at the stability of each of the 2 molecules as the tie breaker to determine which is the stronger nucleophile. Since molecule 1 has a negative charge while molecule 2 is neutral, molecule 1 is less stable making it the stronger nucleophile.

 

In summary, this section described what is a nucleophile and how to determine nucleophile strength.

 

 

Drawing Fischer Projections

Drawing Fischer Projections

Fischer projections are yet another way to draw molecules. They are commonly used in Biochemistry as a simple way to depict sugars. However, drawing Fischer projections can be tricky at first. Here is what a Fischer projection looks like:

 

This figure is used for drawing Fischer projections. Specifically, glucose is shown in this example going from Fischer projection to bond line structure.

The horizontal bonds are coming out of the page as wedges, while the vertical bonds are going back into the page as dashes. Because all the horizontal bonds are wedges, the Fischer projection is drawn in the eclipsed configuration.

 

Many questions about drawing Fischer projections in Organic Chemistry require you to go from a Fischer projection to a bond line drawing. They can be tricky, but here is a method that vastly simplifies the process. This procedure works very well and requires almost no mental rotation, one of the hardest things about Orgo for many.

a. Start with a small turn of the molecule. It can be in either direction, but here I have shown it in only one orientation. This is the only step that requires some mental rotation:

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There is a carbon atom at each point where there is a dash and a wedge, so this molecule has 6 carbons (including the 2 explicitly shown).

Some people have trouble seeing the way the molecules flip. Likewise, some have trouble seeing how the atoms that were originally on the left side of the Fischer projection become wedges, while those on the right side become dashes. The following figure shows a right hand making precisely the same flip as our molecule. Note that the thumb is pointing to the left at the beginning, and after the flip, it points up.

 

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b. Rotate the molecule 90 degrees. Then, number each carbon.

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c. In step b, we saw that we had a 6-carbon chain (this can change of course depending on the question asked), so we draw a normal, 6-carbon chain in bond line form. Note the orientation after drawing the normal chain. Some are in their “up” conformation, while others are in their “down” conformation.

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If we apply this up/down system to the final structure in part “b,” notice that 1 and 6 are down, while 2, 3, 4, and 5 are up.

d.  Now we fill in the groups on each carbon. Notice that groups 1, 2, and 4 are in the same orientation for the drawings in b and c (1 is down in each, 2 is up, and 4 is up). This means that the way the atoms are arranged on this carbon stays the same (dashes stay dashes; wedges stay wedges). In 3, 5, and 6, the orientations are flipped (3 and 5 go from up to down, while 6 goes from down to up), so we must flip the dash and the wedge (wedges turn into dashes; dashes turn into wedges). So on carbon 3, the –OH is a wedge, but after flipping it downward, the –OH now becomes a dash and the –H becomes a wedge. No mental rotation is required here. Just remember that if the configuration changes, then the dashes and wedges change.

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From here, you can find whether each stereogenic center is R or S.