Preparing for Mechanism
Electronegativity, bond strengths, acids & bases, pKa values, pH, stereochemistry, regiochemistry
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If you’ve ever driven into a big city such as New York (or seen Manhattan from the Staten Island ferry like in this image) what you see is a collection of structures of different heights, shapes, and sizes. This is a 2-dimensional view of a very complex entity (part of NYC) that does not give any idea of what’s behind the front buildings and how everything in the city is connected.
This is how students often see mechanism; a bunch of unrelated structures and pathways that they have to memorize in order to pass the Organic sequence. This task is made a lot less onerous with a little organization of the ideas that are picked up in the Organic 1 course. We want to see the aerial “helicopter view” of the city/subject (below) and see how streets connect blocks and individual neighbourhoods in surprisingly organized and simple ways.
How To Get The Helicopter View
We need to know about topics such as Electronegativity and Bonding, which leads to an appreciation of Bond Strength and how to spot weak bonds that will be broken in a mechanism.
Understanding Acids and Bases and the pH of the reaction medium is key. This dictates what is allowed and present in a reaction and what is not.
Being comfortable with where charge prefers to be (negative in anions, positive in carbocations) will help to explain a lot about intermediate structures and product outcomes.
If we are solid on these basics, and know that only a limited number of things can happen as bonds form and break, the bigger mechanisms such as the Claisen condensation below become manageable.
1. Know Electronegativity Values
Periodic table trends are helpful, but actual Electronegativity values are even more so. What about the Mg to C bond? The trend says that C should be more electronegative but by how much? Knowing Mg is 1.2 and C is 2.5 (on the Pauling scale) says this is a very polar covalent bond in which the C is electron-rich and thus reactive. When we get into the weeds, e.g. B and Al in reducing reagents, knowing these numbers will explain the very different reactivities of compounds like NaBH4 and LiAlH4.
2. Know Some Bond Strengths
There is a general relationship between the strength of a bond and the relative atom size and electronegativity of the elements involved. Bonds between atoms of quite different sizes tend to be weak. An example would be C-O (350 kj/mol) and C-S (260 kj/mol) single bonds. Oxygen is directly above sulfur in the periodic table so O is smaller (and closer in size to C) and more electronegative (O = 3.5 vs S = 2.5). The C-S bond is weaker and easier to break. A C-N single bond (290 kj/mol) should be favoured over a C-Br bond (275 kj/mol), which will be manifested in SN2 reactions in which nitrogen nucleophiles can displace Br on alkyl bromide substrates.
In reactions involving double bonds, sigma bonds are generally stronger than pi bonds so addition to pi bonds to produce new single bonds is usually favoured. Many addition reactions are essentially irreversible at room temperature (e.g. hydroboration), partly because of the swapping of a weaker C=C pi bond (270 kj/mol) for (cumulatively) stronger C-H (413 kj/mol) and C-B (372 kj/mol) single bonds. Expect pi bonds to be reactive in many situations.
Bonds formed between atoms that each have at least one lone pair are always going to be weak. Bonds in peroxides (RO-OR) are easily broken (O-O single bond = 146 kj/mol), which correlates with electron repulsion between lone pairs on adjacent O atoms that are close in space. Likewise, the Br-Br bond (193 kj/mol) and Cl-Cl bond (239 kj/mol) will break easily, for example in the initiation step of a radical-based process.
See here for a general discussion of bond strengths.
3. Know About Acids & Bases
Acid-Base chemistry is where the discussion of mechanism begins in most Organic 1 courses. While this topic is covered in General Chemistry, its application in Organic seems to cause problems as the structures of the acids and bases involved are more complicated than the simple examples seen in the first year. The ideas, however, are the same.
Initially we are looking at protic acids in which the H will be bonded to a more electronegative element (e.g. OH in water and alcohols) or a bigger element (e.g. Br in HBr). HBr is a stronger acid than water because the H-Br bond is weaker and easier to break (363 kj/mol, very different sized atoms) than the O-H bond (467 kj/mol, atoms closer in size), but also because of the relative conjugate base stabilities. Br is much better at dealing with a negative charge (extra electrons) than O because it is bigger and the charge (lone pair) has more volume to occupy. The Br anion is thus much easier to form than hydroxide/alkoxide and this reaction favours the right-hand side completely.
Knowing what the products are in this situation we can make our first attempt at using curved arrows to show a mechanism that takes account of which bonds are formed and broken. The electron-rich hydroxide base is attracted to the electron-poor proton on H-Br (H EN = 2.1; Br = 2.8) so they interact to begin forming a bond. Since we know only one bond needs to form and one needs to break, we can be sure this won’t be a complicated mechanism. In fact, everything can happen at once in a concerted process (this is confirmed by kinetic data, which shows the reaction to be bimolecular). Two curved arrows only will be needed here to give the following mechanism.
So, what have we learned from the NaOH/HBr reaction? Hydroxide is reactive because the extra lone pair (i.e. the negative charge) is on a fairly small atom (O). In the presence of HBr, the hydroxide has a potential vehicle for becoming more stable through reaction. The electron-rich O is attracted to the electron-poor H and an O-H bond starts forming. As the introduction of more electrons breaks the “octet” rule at H, the H-Br bond must begin breaking, which makes sense as it’s weak. The H-Br bond pair of electrons becomes a lone pair on Br, which is able to handle that extra lone pair (i.e. the negative charge) because Br is reasonably electronegative but moreso because Br is large and can handle the four lone pairs. Overall, a weak H-Br bond is sacrificed for a stronger O-H bond and the negative charge ends up in a better place on Br.
This prompts an important question; where does charge (in this case negative charge) want to be? Generally, electronegative atoms prefer holding a negative charge and larger atoms are stable with that extra lone pair. Additionally, electron delocalization (resonance) greatly stabilizes negative charges and contributes to the favourability of certain species in reactions. In acid-base situations, if the charge in one of the bases is able to delocalize, that species will generally form and be favoured. In the example below, hydroxide reacts with the carboxylic acid to give products that are heavily favoured because the negative charge in the conjugate base is able to delocalize.
For a more in-depth look at Acid-Base ideas and Resonance see here.
4. Know pKa Values
Knowing actual pKa values (some examples below) will help in solving acid-base problems quickly. Again, trends are helpful but actual numbers will help us quantify equilibrium constants in individual steps to see if a reaction pathway (mechanism) stands a chance of actually happening. While pKa values are explained by a combination of factors (polarity, anion stability), it is convenient to have a set of numbers that include all of this background for use as a quick tool in problem solving. A more exhaustive list of pKa values will be needed later when we contemplate deprotonating weaker acids such as alkynes and amines.
For a more in-depth look at pKa values see here.
5. Understand pH
Outside of students just not studying and thus making a hash of mechanisms, more points are lost on mechanistic questions through a lack of understanding of pH than any other topic. This idea is introduced in high school and first year college chemistry classes, yet it becomes obvious at the second year level how little students actually understand about what species will, and will not, be present in a reaction flask at a certain pH level. This doesn’t need to be complicated, and we won’t be doing any calculations, but knowing what is present in base versus acid, and how that dictates mechanism, will clear up a lot of confusion.
For “neutral” read inert or unreactive first in that the substrate itself is unlikely to be attacked as there are no strong acids or bases present. The substrate will have time to “do its own thing” for example by breaking apart in carbocation-based reactions. Consider the SN1 reaction between a tertiary alkyl halide in an alcohol solvent. The pH of the mixture will be neutral when the chemicals are first introduced so there is no strong acid to be attacked or strong base to do the attacking. The alkyl halide is able to collapse to a carbocation, which then attracts the electron-rich O of the alcohol and an ether is formed. In the example below, the stable carboxylic acid does not react with water (or alcohols) at neutral pH as there is nothing that reactive present.
If we can identify the conditions in a reaction as being acidic (H2SO4, HBr, generic H3O+, etc.) we already know what is going to happen first; the organic substrate is going to attack the acid and the substrate will become activated. This is of major importance in acid-catalyzed reactions where neutral conditions are not sufficient for reactivity but adding an acid gets things going by activating the organic substrate. This shows up in dozens of mechanisms in Organic 1 and 2 and gives an element of predictability to how pathways unfold. Since we are dealing with a “positive” environment, any intermediates formed during these pathways will be positively charged (oxonium ions, carbocations, etc.).
See the reaction bottom left for an example where the substrate attacks the acid to yield a cationic oxonium ion intermediate, which may react further depending on what else is present. The environment in the bottom right example has changed to strongly basic so the “attacker” and “attackee” change. Now the negative (electron-rich) base can attack the (electron-poor) carboxylic acid proton as a way to become stable through deprotonation. Notice how the mechanism arrows have changed direction in going from acidic to basic conditions. In general, organic substrates attack acids while bases attack organic substrates. This is a reliable indicator of what will happen in other mechanisms in the different pH environments.
A note. In strong acid there cannot be any strong bases and in strong bases there cannot be any strong acids. For example, it is impossible to have hydroxide anion in sulfuric acid, and it is impossible to have H3O+ in hydroxide solution (they would be neutralized immediately). In neutral conditions there are no strong acids or strong bases. Carbocations will not form under basic conditions because the base is likely to attack the substrate quickly before a carbocation can form (e.g. in the E2 process).
6. Know Where You Are Going
This one may seem obvious but you aren’t going to be able to draw a mechanism successfully if you don’t know where you are going. This means knowing the expected products that form under certain conditions. In reality, chemists figure out mechanistic pathways after they know what a reaction produces using structural and stereochemical clues. Whenever you encounter a new reaction you should take a step back and look at the reactants, the conditions (acidic, basic, or neutral?), and the products. Assess which bonds are formed and broken, and then make sure that your mechanism includes a logical description of each of those events remembering that only so many things can happen in polar pathways.
For example, in the SN1 pathway for a tertiary alkyl bromide in alcohol, the C-Br bond has to break and a C-O bond has to form. This will be either concerted or stepwise (as are all reactions) and your study will tell you that the pathway is stepwise with formation of a carbocation. Loss of leaving group followed by nucleophilic attack gives the observed ether product. In the second semester, many students have that “aha!” moment in which mechanism clicks and they are able to predict products in new reactions because they recognize the limits of what can happen in different environments.
7. Any Stereochemical or Regiochemical Clues?
If the product of a reaction has a certain stereochemical or regiochemical identity, this is very helpful in working out how that product was formed. In the SN2 reaction the only product’s stereochemistry has inverted (stereospecific), which is only possible through a concerted attack of nucleophile at the back of the substrate (where the antibond space is located) and concomitant loss of leaving group. The only other option, stepwise progress through a carbocation, would produce a racemic mixture of products as the flat, prochiral intermediate could be attacked from two directions.
In the E2 reaction of an alkyl halide with KOt-Bu the less-substituted Hofmann regioisomer is the major product. This tells us that the reaction is irreversible since equilibration would give us the more stable Zaitsev outcome. The kinetically-controlled pathway then must be governed by steric considerations in transition states in which the very large base attacks the most accessible proton, which inevitably will be attached to the less crowded carbon. Working backwards from these clues helps us understand the factors governing mechanism and make its study much more rewarding than trying to memorize everything.
Pushing Curved Arrows
Organic mechanisms are simply a description of the bond-making and bond-breaking events that occur when starting materials are converted to products. The “curved arrows” used to describe these events were pioneered by Sir Robert Robinson in the UK and are now the convention used by chemists throughout the world. While undergraduate students often struggle with this idea it is as simple as driving on the correct side of the road; curved arrows always start at electron-rich areas and are pushed towards electron-poor areas.
If we think of electron density as a currency then the idea of “electron-rich” donating to “electron-poor” makes sense as long as we can identify sites of electron excess and electron deficiency. In Equation (i) the green curved arrow is used correctly to describe the lone pair from X (electron-rich) being used to form a covalent bond to Y (electron-poor). The green arrow in Equation (ii) is incorrect because Y does not have any extra electron density to share with X. In most of the reactions we study a curved arrow will start at a lone pair or pi bond.
Getting Started
Beginning with an acid-base reaction (Equation iii) we assess what bonds have been formed and broken and focus on the atoms involved. In this equation a new bond is formed between N and H and a bond is broken between H and O. The sodium ion does not change and is simply a spectator ion. The placement of the Na next to N on the left-hand side does tell us which atom will be the base.
N is negatively charged and therefore reactive (Equation iv). While the O in water also has two lone pairs, the O is neutral and much less reactive than a negative N. The N atom will try to become neutral by attacking water, in particular at one of the H atoms since they are electron-poor due to the electronegativity difference between O and H. This results in N becoming neutral and the charge being transferred to the more electronegative O.
To complete the mechanism for this process we must add curved arrows to describe all bonds being formed and broken during the reaction. The arrows must start at an electron-rich area and go towards an electron-poor area. Here the N is the aggressive electron-rich species so an arrow starts there and goes to H to form the N-H bond. To avoid breaking the octet rule, an arrow goes from the O-H bond to the O atom. Two events, two arrows.
If we assess what needed to be done in the acid-base example we can extrapolate that to organic mechanisms since the number of curved arrows will (typically) match the number of bonds that need to be formed and broken. The first carbon-based mechanism usually studied in the Organic sequence is the bimolecular substitution of a nucleophile on an electrophile with loss of leaving group, which actually boils down to one bond being formed and one being broken. Considering Equation (vi) it appears that the C-CN bond must form and the C-Br bond must break.
As with the acid-base reaction we identify the electron-rich species (cyanide anion) and the electron-poor species (C in the methyl group). One bond needs to form and one needs to break so, knowing that this reaction is bimolecular and concerted, the nucleophile is going to kick out the leaving group on carbon as described in Equation (vii). This simple preliminary analysis of what needs to happen in each reaction will take a lot of the mystery out of mechanism.
Essential Mechanisms
Consider the reaction between the secondary alkyl bromide shown and sodium cyanide in DMF as a polar, aprotic solvent. We know that this gives the inverted product of substitution in which the cyanide anion has replaced the bromide. We also expect the new C-C bond to be stronger than the C-Br bond that is lost. So, what clues lead to the 2 arrow mechanism shown below?
Firstly, with HCN having a pKa of about 9.2, the reaction is moderately basic because of the cyanide salt, which suggests that the basic cyanide will attack the organic substrate. Since there are no obviously acidic protons on the substrate, there will be no acid-base reaction, and the negative cyanide will be attracted to the electron-poor alpha carbon next to Br.
We only have two choices for how this mechanism plays out; either everything happens at once in a concerted pathway, or the Br breaks off first to give a carbocation, which then reacts with cyanide to give product. The second pathway would give enantiomeric products, i.e. a racemic mixture, because a flat carbocation would be formed. This does not happen so we focus on the concerted path.
The mechanism shown matches the paradigm described earlier. The basic reagent attacks the substrate, from the back where the antibond is, with the Br being displaced to maintain the octet at C. The negative charge has now been transferred from C to Br, which handles it better. This mechanism matches the experimental evidence, in which the stereochemistry is inverted, and the kinetic data that shows the reaction to be bimolecular.
Now we look at the reaction between a chiral tertiary alkyl bromide and methanol to give a racemic mixture of ethers. The pH of the reaction starts out as neutral, meaning there are no strong acids or bases present to react with the substrate. The alcohol is also not a strong nucleophile although the tertiary carbon is too crowded to be attacked anyway. Since we know that the Br is replaced, it has to break off at some point, so how does that happen to fit the observed racemic outcome?
In the SN2 pathway the Br was kicked out in a concerted process, here it must break off first, as part of a stepwise pathway, which agrees with the observed rate-determining step being unimolecular. That would generate a flat carbocation, which in this case would be prochiral and accessible from either face. Rather than the alcohol “attacking” the cation, we could think of the cation “recruiting” the weakly nucleophilic alcohol, which would be close by as it likely serves as the solvent in this solvolysis process.
Equal attachment from either side would lead to a racemic mixture of protonated ethers, which would then lose the extra proton to the solvent to furnish the ether products. The mechanism fits the observed data, and also the idea of assessing what can or will happen based on reaction pH. Here the neutral conditions meant the alkyl halide was not attacked and had time to collapse to the carbocation.
When we move to a hindered substrate under strongly basic conditions, we see a change in pathway to bimolecular elimination, i.e. the E2. Strongly basic conditions means the base attacks the substrate, however the SN2 is impossible as the alpa carbon is too crowded for approach. The base, here NaOCH3, is trying to become stable, by transferring its extra lone pair (i.e. the negative charge) to the Br leaving group, which can handle it better. Deprotonation at the beta carbon allows for a logical flow of electron density to form an alkene.
Smaller bases tend to give the more stable, more highly substituted Zaitsev alkene, while larger bases give the less substituted Hofmann outcome. Not only is this a useful method in synthesis, but it also tells us that the elimination is irreversible since establishment of an equilibrium, if the reaction was reversible, would give the thermodynamically favoured Zaitsev product. The outcome must therefore depend on transition state energies and be kinetically driven.
We know that basic conditions preclude carbocation formation, and that the reaction here in base has a bimolecular rate-determining step, so the reaction must be concerted. It is possible for all required bonds to be formed and broken at once, so a concerted pathway is the most logical option. Regioselectivity is important here but stereoselectivity is also observed with the trans isomer often being favoured, again due to preferred geometries in the possible transition states.
When the conditions change to acidic, alkyl halides are mostly inert, but alcohols will react to give alkenes, especially at higher temperatures. The Zaitsev regioisomer is typically favoured and the rate-determining step is known to be unimolecular. Elevated temperatures serve two purposes. Firstly, this boosts the entropic factor (TdeltaS) in the Gibbs equation, and secondly, this allows the thermodynamically unfavourable alkene(s) to be distilled off as they are formed. This method serves as a simple and reliable protocol for producing Zaitsev alkenes.
To come up with a mechanism, following the established rules means the substrate (alcohol) attacks the acid to generate a hydronium leaving group. Since we are not in basic medium, the leaving group is able to break off in the unimolecular rate-determining step to give the carbocation. This species may react with water as nucleophile to go back to the alcohol, or the water may serve as a weak base and deprotonate to give the alkene.
This whole process is stepwise and most likely an equilibrium. The alcohol will be favoured, since sigma bonds are generally stronger than pi bonds, requiring removal of the alkene from the mixture. This will then be replaced in the equilibrium and result in a high-yielding synthesis of the alkene.
In conclusion, knowing something about electronegativity tells us which bonds will be polarized. Knowing something about bond strengths helps us predict which bonds will break in a mechanism. Knowing relative atom size helps us appreciate where negative charges want to be. Understanding pH will guide us in deciding what is possible in a reaction, and also which direction the mechanism arrows should flow. Getting these basics worked out, and then applying them to new systems, is a reliable and rewarding way of studying and learning mechanism in the second year Organic sequence.