Nucleophilic Reaction

Nucleophilic Reaction

There is no denying the fact that when we study nucleophilic substitution and elimination reactions, we find that learners typically have very little trouble drawing each mechanism and predicting the products, so long as they are specifically told which reaction. But many learners find one aspect very challenging: predicting the winner of an S_N1/S_N2/E1/E2 competition. In our studies, The SN1/SN2/E1/E2-competition exam was the exam where many pre-health learners decided that their fate was sealed. In the years since then we have turned things around considerably. What did We learn? And what did We change?

This most important thing we learned was why learners were struggling with predicting which reaction dominates—learners were memorizing! We initially thought it was strange that learners were memorizing because we felt that the textbook we are using clearly explained the various factors that favor one reaction over another; and we heavily reinforced those ideas in class. But then it became apparent to me why learners still resorted to memorization. After treating each of those factors separately, the textbook we are using brought them together to summarize and make generalizations about the competition, as well as to provide exceptions. For example, if (1) the attacking species is both a strong nucleophile and a strong base, and (2) the substrate is primary, then SN2 will give the major product and E2 will give the minor product. An exception is when the attacking species is the tert-butoxide anion, in which case E2 will give the major product and SN2 will give the minor. As another example, if (1) the attacking species is a strong nucleophile but a weak base, and (2) the substrate is secondary, then both SN1 and SN2 can occur. In an aprotic solvent, however, we can expect SN2 to dominate.  But if the substrate is benzylic, the solvent is protic, and the leaving group is very good, then we can expect SN1 to dominate.

Not surprisingly, when a student sees that 15 pages of chemistry has been distilled down to a single page (or a single table) of summary/generalizations, it is incredibly attractive to try to memorize them. There are two major problems that arise. One is that, by doing so, learners circumvent the actual chemistry, so there ends up being no context for why a particular reaction dominates. This makes it easy for a student to forget a critical piece of information or become confused, especially given the large number of different combinations in which the factors can contribute. The second problem is that, in addition to memorizing the general rules, learners must also remember which scenarios require additional information, such as when to consider solvent, the bulkiness of the attacking species, or the benzylic nature of the substrate. In other words, learners feel that they must memorize exceptions in addition to the rules.

I’ve since had a huge effect on my learners’ success with the SN1/SN2/E1/E2 competition by providing a system for deciding the winner of the competition, which consists of the following steps:

(1)  Determine if the leaving group on the substrate is at least as good as F−. If so, then

(2)  Examine the type of carbon to which the leaving group is attached.

1. If the carbon is primary, then rule out SN1 and E1 unless the carbocation can be resonance-stabilized.
2. If the carbon is tertiary, then rule out SN2.

This system is also incorporated into my new textbook for two reasons. First, it gives learners something to hang their hat on. Yes, learners must still remember how to execute these steps, but the system for doing so does not change from one scenario to the next. The second great benefit is that when going through these steps, learners will frequently be reminded why. When deciding whether a leaving group is at least as good as F−, for example, they must apply arguments of charge stability. When the carbon is tertiary, which rules out SN2, learners will be reminded that it is due to steric hindrance. And it is this type of reinforcement that sets learners up for being in command of other reactions they will encounter in the course.

1. What do you mean by a leaving group “as good as F-“? What about OH- leaving group?
2. What do you mean by the “concentration of the attacking species”?
3. Also, doesn’t it matter if the attacking species is a good Nu or a good base? Or both? WEknow that something that is just a good base will only favor certain reactions. I’m still a little confused by which ones are good Nu’s, bases or both and if those are things we should just memorize…

1. A leaving group “as good as F-” essentially has to do with charge stability of the leaving group in the form in which it comes off. Cl-, for example, is a much better leaving group than F- because, being a larger atom, chlorine can better handle the charge. H2O is also a much better leaving group than F- because it is uncharged. HO- is a worse leaving group than F- because O is not as electronegative as F, meaning that the negative charge cannot be accommodated as well. There is also a nice correlation between how good a leaving group is and the pKa of its conjugate acid: The stronger the conjugate acid, the better the leaving group is. The pKa of HF, for example, is around 3 and the pKa of HCl is around -7. HCl is a much stronger acid than HF, so Cl- is a better leaving group than F-.

2. The attacking species is the species that can act either as a nucleophile or as a base. In short, the higher the concentration of the attacking species, the more that SN2 and E2 reactions are favored in the competition, whereas the lower the concentration of the attacking species, the more that SN1 and E1 reactions are favored. This can be understood from the rate laws of the respective reactions, but We like to add a more qualitative argument to the picture, which goes something like this: In an SN2 or E2 reaction, the attacking species is responsible for “forcing off” the leaving group (either directly or indirectly). Therefore, the greater the number of attacking species, the better that job can be accomplished. In an SN1 or E1 reaction, on the other hand, the attacking species must wait for the leaving group to have come off. Therefore, increasing the number of attacking species just means that there are more of them waiting.

3. We think this question relates very closely to the qualitative picture in #2 above. Strong nucleophiles favor SN2 over SN1 because they are good at forcing off the leaving group. Strong bases favor E2 over E1 for the same reason. An attacking species like HO-, for example, is a strong nucleophile and a strong base, so it tends to favor both SN2 and E2 over SN1 and E1. Cl-, on the other hand, is a strong nucleophile but a weak base, so it tends to favor SN2 over SN1 (strong nucleophile), but favors E1 over E2 (weak base). Now, identifying an attacking species as a strong base or strong nucleophile is important. There’s a bit more to it than this, but basically, strong nucleophiles tend to have a full negative charge, and strong bases are ones that are roughly as strong as HO- or stronger.

Energy Profiles

In the diagram below, we can clearly see that we need an input of energy to get the reaction going. Once the activation energy barrier has been passed, we can also see that you get even more energy released, and so the reaction is overall exothermic. If you had an endothermic reaction, a simple energy profile for a non-catalyzed reaction would look like this: Unfortunately, for many reactions, the real shapes of the energy profiles are slightly different from these, and the rest of this page explores some simple differences. What matters is whether the reaction goes via a single transition state or an intermediate. We will look at these two different cases in some detail. Energy profiles for reactions which go via a single transition state only This is much easier to talk about with a real example. The equation below shows an organic chemistry reaction in which a bromine atom is being replaced by an OH group in an organic compound. The starting compound is bromoethane, and the organic product is ethanol. During the reaction one of the lone pairs of electrons on the negatively charged oxygen in the -OH group is attracted to the carbon atom with the bromine attached. That's because the bromine is more electronegative than carbon, and so the electron pair in the C-Br bond is slightly closer to the bromine. The carbon atom becomes slightly positively charged and the bromine slightly negative. As the hydroxide ion approaches the slightly positive carbon, a new bond starts to be set up between the oxygen and the carbon. At the same time, the bond between the carbon and bromine starts to break as the electrons in the bond are repelled towards the bromine. At some point, the process is exactly half complete. The carbon atom now has the oxygen half-attached, the bromine half-attached, and the three other groups still there, of course. And then the process completes:
	Note:  These diagrams have been simplified in various ways to make the process clearer. For example, the true arrangement of the lone pairs of electrons around the oxygen in the first diagram has been simplified for clarity. The bromine also has 3 lone pairs as well as the bonding pair, but they play no part. And, of course, the other groups attached to the carbon have been left out in order to concentrate on what is important.
The second diagram where the bonds are half-made and half-broken is called the transition state, and it is at this point that the energy of the system is at its maximum. This is what is at the top of the activation energy barrier. But the transition state is entirely unstable. Any tiny change in either direction will send it either forward to make the products or back to the reactants again. Neither is there anything special about a transition state except that it has this maximum energy. You can't isolate it, even for a very short time. The situation is entirely different if the reaction goes through an intermediate. Again, we'll look at a specific example. Energy profiles for reactions which go via an intermediate For reasons which you may well meet in the organic chemistry part of your course, a different organic bromine-containing compound reacts with hydroxide ions in an entirely different way. In this case, the organic compound ionises slightly in a slow reaction to produce an intermediate positive organic ion. This then goes on to react very rapidly with hydroxide ions.
	Note:  If you haven't come across the use of curly arrows in organic chemistry yet, all you need to know for now is that they show the movement of a pair of electrons. In the first equation, for example, the bonding pair of electrons in the C-Br bond moves entirely on to the bromine to make a bromide ion. In the second equation, a lone pair on the hydroxide ion moves towards the positive carbon to form a covalent bond.
The big difference in this case is that the positively charged organic ion can actually be detected in the mixture. It is very unstable, and soon reacts with a hydroxide ion (or picks up its bromide ion again). But, for however short a time, it does have a real presence in the system. That shows itself in the energy profile. The stability (however temporary and slight) of the intermediate is shown by the fact that there are small activation barriers to its conversion either into the products or back into the reactants again. Notice that the barrier on the product side of the intermediate is lower than that on the reactant side. That means that there is a greater chance of it finding the extra bit of energy to convert into products. It would need a greater amount of energy to convert back to the reactants again. I've labelled these peaks "ts1" and "ts2" - they both represent transition states between the intermediate and either the reactants or the products. During either conversion, there will be some arrangement of the atoms which causes an energy maximum - that's all a transition state is.

Bonding

Bonding

• Compounds are formed when two or more atoms join together, forming Bonds. There are different types of bonds that occur between atoms which give rise to different properties.

Ionic Bonds

• Ionic Bonds form when electrons are transferred from one atom to another, forming charged Ions which are attracted to each other by Electrostatic Forces. Elements tend to loose or gain electrons, forming Ions, to get a 'full other shell'.

• Ionically bonded substances, such as Sodium Chloride, can from crystals known as Giant Ionic Lattices with bonds forming a network of connections between atoms.

• Giant Ionic Lattices have high melting and boiling points since atoms are held together by strong forces.

• Ionic substances can conduct electricity through the movement of charged Ions. However, they may only do so if the Ions are free to move around. Therefore Ionic substances conduct electricity when molten or dissolved, but not when in a solid state.

• Many Ionic Compounds dissolve in water. This is because the polar water molecules cluster around Ions, and so separate them from each other.

Covalent Bonds

• Covalent Bonds involve the sharing of electrons so that all atoms have 'full outer shells'.

• Sometimes in a Covalent Bond, both shared electrons come from the same atom. This is known as a Dative Covalent Bond. This often results in the formation of charged molecules.

Simple Molecular Structures

• Substances composed of relatively small covalently bonded structures are called Simple Molecular Structures.

• Simple Molecular Structures tend to have low melting and boiling points since the forces between molecules (intermolecular forces, which are van der Waals forces) are quite weak.

• They don't tend to conduct electricity, since there are no free charge carriers.

• They tend to be quite insoluble in water, but this depends on how polarised the molecule is. The more polar the molecules, the more water molecules will be attracted to them.

Giant Covalent Structures

• Sometimes, as is the case with Carbon, covalently bonded structures can form giant networks, known as Giant Covalent Structures. In these structures, each a network of bonds connects all the atoms to each other.

• These structures are usually very hard and have high melting and boiling points. This is because of the strong covalent bonds holding each atom in place.

• In general, Giant Covalent Structures cannot conduct electricity due to the fact that there are no free charge carriers. One notable exception to this is Graphite. This is a structure composed of 'sheets' of carbon atoms on top of each other. Electrons can move between the sheets and so carry electricity.

Metallic Bonding

• Metals form Giant Metallic Lattices. These are composed of positive metal ions surrounded by a 'sea' of Delocalized Electrons. The metal ions are attracted to the negative electrons.

• Metals tend to have high melting and boiling points because of the attraction between the metal ions and the electrons. The more Delocalised Electrons are present (because of a higher valency), the greater the melting and boiling points.

• Metals conduct electricity because the electrons are free to move and carry charge.

• Metals do not tend to dissolve, except in liquid metals, due to the strength of the attraction between the metal ions and the electrons.

• Metals are Malleable and Ductile. This is because there are no direct bonds between metal ions, so they can slide over each other.

The Shapes of Molecules

• Different molecules have different shapes. The shape of a molecule is dictated by the arrangement of Electron Pairs. This is because Electron Pairs repel each other. The molecule settles in a state where the Electron Pairs are furthest apart from each other.

• Different types of Electron Pairs have different repulsions. There occur Lone Pair Electrons (LP) and Bonding Pair Electrons (BP). The order of repulsion between different combinations, in order of decreasing repulsion, and so decreasing angle, is:
• LP - LP
• LP - BP
• BP - BP

• Common bond angles:
• 2 Electron Pairs on a single atom form 180° bond angles.
• 3 Electron Pairs on a single atom form 120° bond angles.
• 4 Electron Pairs on a single atom form 109.47(1220... = )° bond angles.
• 6 Electron Pairs on a single atom form 90° bond angles.
• 2 Lone Pairs and 2 Bonding Pairs on a single atom form a 104.5° bond angle.
• 1 Lone Pair and 3 Bonding Pairs on a single atom form a 107° bond angle.