Common dilemma in organic chemistry as you move through the latter parts of Org 1 and then into Org 2:

When more than one reaction is possible, how do you know which one will happen? In a steel cage match between an acid base reaction and other types of reactions, which wins?

Here’s a good rule of thumb to keep in mind: acid-base reactions are fast, relative to other reactions. 

Three examples – these are common trick questions, by the way:

 Example 1. Acid-base versus nucleophilic substitution reactions (SN2 reactions).

Here, note that our nucleophile (the conjugate base of an alkyne, pKa 25) can remove the proton of an alcohol (pKa ~15) or perform an SN2 reaction on the primary alkyl halide. With a difference of 10 pKa units between the alkyne and the alcohol, the acid-base reaction between the deprotonated alkyne (“acetylide”, stronger base) to produce a deprotonated alcohol (“alkoxide”, weaker base) is extremely favorable. And since acid-base reactions are fast, relative to other reactions, the preferred first reaction here is deprotonation of the alcohol to give the conjugate base (“alkoxide”)

Bonus question: what would be the final product of this reaction, after the deprotonation? Answer below.

Example 2. Acid-base reaction versus addition to a carboxylic acid

Grignard reagents are very good nucleophiles – reacting with carbonyl compounds such as ketones, aldehydes, and esters. But as the conjugate bases of alkanes (pKa ~ 50) they are also extremely strong bases. When combined with a carboxylic acid (pKa ~4 or 5) the result is not an addition to the carbonyl, but an acid base reaction (45 pKa units makes for a pretty favorable reaction!). It’s always helpful to remember that carboxylic acids… are acids.

Example 3 – Another Grignard

The same applies for reactions of Grignard reagents with molecules that have hydroxyl groups in addition to  aldehydes or ketones. If merely one equivalent is added, the first thing to happen will be deprotonation of the alcohol, which is faster than addition to the ketone carbonyl carbon. It’s only after addition of a second equivalent of Grignard that addition to the ketone will occur.

So what’ s going on?

What’s going on here is an application of a handy principle in chemistry called the Principle of Least Motion. 

Simply stated, it’s this.

Acid-base reactions on “heteroatoms” (that means atoms other than carbon,  such as O, N, and S) generally require very little reorganization of the nuclei in the structure. Therefore these reactions are fast, relative to reactions where the nuclei have to move or be reorganized.

Think about removing a proton from an O-H. After loss of hydrogen, the oxygen gains a new lone pair. But its hybridization doesn’t change – it started as sp3, and it’s still sp3. So the nuclei (other than the H, of course) don’t significantly change positions in these reactions. No extra motion, in other words.

However when bonds are formed or broken at carbon – such as in the SN2 reaction or in additions to carbonyl carbons – a lot of atomic furniture has to get rearranged. For instance, the SN2 proceeds through a backside attack, which means that the geometry of the molecule changes from tetrahedral to trigonal bipyramidal (that’s the  5-coordinate transition state) and then back to tetrahedral. In addition to carbonyl compounds, we’re changing the hybridization of carbon from sp2 to sp3. That requires a shift from trigonal planar to tetrahedral geometry.

 Bottom line: acid base reactions on oxygen, sulfur, or nitrogen are fast. So long as the acid base equilibrium is reasonable [How to use a pKa table, a handy rule of thumb for acid-base reactions] do them first.

P.S. It’s interesting that Grignard reagents (the conjugate bases of alkanes, pKa ~50) don’t usually deprotonate aldehydes (pKa ~18) or ketones (pKa ~20). That’s another application of this principle. Removing a proton from an aldehyde or ketone requires breaking a C-H bond, and the resulting base (called an “enolate”) will undergo a change in hybridization from sp3 to sp2. Hence, it’s slow.

P.P.S. After the first acid-base reaction, the deprotonated alcohol can then do an SN2 reaction on the primary alkyl bromide.

Last time I talked about pKa and how it’s the closest thing we have to a universal measurement of  the strengths of all kinds of different acids and bases.  I also referred to a post on how to use a pKa table (key lesson: stronger acid plus stronger base gives weaker acid and weaker base).

How far can we push this? Let’s look at three illustrative examples.

On one extreme, we have one mole of a really strong acid – let’s say hydrochloric aid (HCl), pKa –8. And to it, we add (slowly!) a solution of water containing one mole of sodium hydroxide (the conjugate base of water, pKa 15.7).

HCl and NaOH react to give water and NaCl . How favorable is this reaction? We can make a rough estimate. The pKa of HCl is -8. Sodium hydroxide is the conjugate base of H2O (pKa 15.7). That’s a difference of about 24 pKa units – and since each pKa unit represents one order of magnitude, this reaction is favorable with an equilibrium constant of about 10 to the power of 24.

For all intents and purposes, a reaction with an equilibrium constant this huge is irreversible.

That is to say that HCl and NaOH are completely consumed when they react together, giving only H2O and NaCl.

 What about the other extreme: the reaction of methanol (pKa of 15.2) with sodium hydroxide (the conjugate base of water, pKa 15.7)?

Neither side of the acid-base reaction is strongly favored. Here we’re dealing with a very small difference in pKa – only 0.5 pKa units. So the equilibrium constant here would only be about 10 to the power of 0.5 —> 3.16  toward giving the weaker acid (water) and the weaker base (CH3O(–).

At equilibrium we’d expect to have a mixture of about 75% CH3O(-) [the weaker base] and 25% HO(-)  [the stronger base].

In other words, both species are present in solution.

 So how far can we stretch this? In between these two extremes, at what point does  a reaction become irreversible for practical purposes?

There’s no hard and fast rule on this. But for practical purposes, a good rule of thumb is about 10 pKa units. 

That is to say, if the difference in pKas between an acid and a base is about 10 pKa units or less, it is useful to consider their acid-base reaction to be in equilibrium.

Think about what that means – a ratio of one molecule in 10 billion can make the difference in a reaction!

One in 10 billion might not sound like a lot. But when you consider that a mole contains 10 to the power of 23 molecules, and each of them are colliding millions of times per second, the odds aren’t really as bad as they look. These reactions can happen. If your only chance of buying a private jet rested on you winning a Powerball lottery – but you were able to fill out hundreds of thousands of entries per second, every second, you’d be at the G5 dealer by next Tuesday.

Here’s an example you’ll see in Org 2. The Claisen condensation begins with the deprotonation of an ester (pKa ~24) by an alkoxide ion (conjugate base of an alcohol, pKa ~15). That’s disfavored by about 10 to the power of 9, since we’re going from a weaker base (alkoxide) to a stronger base (deprotonated ester, a.k.a ester enolate).

Even though there’s only a small amount of the deprotonated ester present at equilibrium, this can be enough to get the reaction to go! You can take my word for it – that this rule of thumb applies  - and leave it there. Or if you’d prefer to go through an actual application of this concept, I’ll finish up with that.

Slightly advanced – an application of how this works. 

One application of this concept can be found in the Claisen condensation of esters. The Claisen condensation involves the addition of a deprotonated ester (an “enolate”) to another equivalent of an ester, through an addition-elimination reaction (more detail here and here).

The first step is deprotonation of the ester by an alkoxide ion [in this case CH3O(-) ] as  mentioned above. This enolate can then attack a second equivalent of ester, which then eliminates an equivalent of alkoxide ion. This reaction is also potentially reversible. However, the new product – a “beta-keto ester” – has considerably more acidic protons than the starting ester, and an acid-base reaction between it (pKa 12) and alkoxide (pKa 15) is quite favorable.

The equilibrium eventually favors the final product, because the conjugate base of the di-ester (pka 12) is considerably weaker than methanol (pKa ~15).

In other words, even though the first step is extremely disfavored, this is made up for by the fact that there is a very good “driving force” for the subsequent reaction.

Next – let’s answer the question “so what?”. Why do acid-base reactions matter? That’s next.

So last time we went through all the different trends that affect acidity. The bottom line is that any factor which stabilizes the conjugate base will result in increased acidity. 

Now it’s fine to know trends – it’s extremely important, in fact – but what do you do when you want to compare the acidity of two molecules that aren’t connected by a trend? For instance, resonance increases the acidity of alcohols, but by how much? More than you’d increase acidity by changing OH to SH? What if you added an electron withdrawing group?

It’s impossible to sort out all the opposing variables this way. Trends are only qualitative guides to this topic.

In order to answer this question we’ll need some hard data, and there’s only one way to get it in organic chemistry – through measurement.

Let’s look at hydroiodic acid,  H–I. In solution (let’s use water)  H–I will protonate water to give H3O(+) and I(-). The reverse reaction also operates. [I (–) reacts with H3O(+) to give H-I back]. So this reaction is an equilibrium. 

Here’s the useful part. We can measure the equilibrium constant of this reaction, and that will tell us about the relative acidity of H-I. The specifics of how we can measure this are another story, but for our purposes, let me assure you that these types of equilibrium constants can be measured accurately.

For H-I, it turns out that the equilibrium constant for this reaction is about 10,000,000,000. That means for every molecule of HI, there are 10 billion molecules of I(-). In other words, H-I donates its proton very readily to give its conjugate base, which is a prime example of a strong acid.

So what? Well, we can do this not just for H-I, but for ANY species with a hydrogen. This equilibrium is the “ruler” by which we can measure any acid – by its equilibrium constant.*

We call this equilibrium constant Ka – the “acidity constant”.

Let’s look at 3 examples. Note that these are somewhat oversimplified in that the solvent has been left out.*

Now comparing acidity by numbers with lots of exponents after them is not the most convenient way to do things. So instead, we’ve taken to using a logarithmic scale. These are common – the Richter scale is logarithmic, for instance – an 8.0 magnitude quake is 10 times more powerful than a 7.0 magnitude quake.

For acidity, the number we use is called “pKa”. It’s obtained by taking the logarithm of the acidity constant Ka, and we arbitrarily decide to multiply it by negative one [the vast majority of Ka values are less than 1, and we're generally more comfortable dealing with positive numbers].

So for H-I, the pKa is -10 (representing a very strong acid), that for methanol is 15 (a weak acid), and for methane, it’s 50 (an extremely weak acid).

That scale comprises 60 orders of magnitude. That’s a huge number!!!!!

To give you an idea of the scale, of pKa, this is the range of the smallest value for length (the Planck length, 10 to the -35 m), to the width of the known universe (~10 to the 26 meters). This website has a phenomenal animation of this.

Anyhow, these measurements have been done for thousands of different molecules now. The result is a big table that allows us to compare the acidity of all kinds of different functional groups. Here’s an example of a pKa table from a previous post.

The pKa table is your friend. In one document, it gives you information on the scope and magnitude of a wide range of chemical behavior – the strongest of the strong acids, and the weakest of the weak acids. And since the stronger the acid, the weaker the conjugate base, it also provides information about basicity.

Next Post: How to Use A pKa Table

* N.B. The choice of solvent for these measurements is important due to the “leveling effect”. For instance  Water is commonly used as a solvent for molecules with a pKa value less than zero and no greater than about 15, since there can be no stronger acid in water than H3O(+) and no stronger base than HO(-). To get some of the more extreme values other solvents must be used, such as dimethyl sulfoxide (DMSO). If you end up in a more advanced organic chemistry course, look at some of these values with a harder eye through that lens. Strongly recommended – Evans’ pKa table gives values in H2O and DMSO. And there is the heroic efforts of Hans Reich at Wisconsin in compiling pKa values.

This time of year one of the most common problems students come to me with is how to do synthesis problems.

Like this:

 There’s a lot of different ways to go about learning how to do synthesis – and I’ll have a lot more to say about that in future posts – but today’s post addresses one of the simplest and most effective ways of learning how to do it.

It’s all about maps.

Maps are easy to understand – they make distances and relationships concrete.

Take a map of Texas. Imagine you’re in Laredo, and you want to go to Dallas. How can you get there? You have choices. A map helps you see the possibilities.

At one level of organic synthesis you can think of functional groups as being like cities on a map, and reactions that link them are like roads.

Let’s look at a reaction map for ketones (not comprehensive)

If you look at all these reactions – forward and backward – you can link functional groups to each other through these types of reactions.

You can build reaction maps for other functional groups. Here’s one for secondary alcohols.

The more reactions you learn, you’ll see that some types of functional groups (like ketones) are very “busy” – like central hubs,  there’s lots of reactions that link to them (and link from them). On the other hand, other functional groups are a little like Laredo: dusty, one-horse towns in the middle of nowhere. (Of all the functional groups you learn about, ethers fall into this category the best).

Let’s go back to our problem.  If you identify the functional groups involved, it can help you to identify what types of reactions are possible for getting you from your starting point to the destination.

We’re asked to go from an aldehyde to a tertiary alcohol.

Go back to the reaction map. Look for the tertiary alcohol as a product. Then work backwards. How do you get there? One way is from a ketone, in a Grignard reaction. Then think backwards from there – how to get there? One way is to oxidize a secondary alcohol. And if you trace back secondary alcohols, you can get there from the aldehyde and a Grignard reaction.

Once you know what reactions to use, it’s much easier to design your synthesis. Here, the problem is identifying what alkyl groups to use in each of the two Grignard reactions.

Here’s one key way in which roadmaps and reaction maps are different, however. In real life, you want to pick the shortest route from, say, Laredo to Dallas, especially if your car’s A/C is on the fritz in the Texas heat. And yes, in the lab, chemists will choose the shortest, most efficient route from one starting material to the final product. Thankfully, for our purposes – which is ultimately just an intellectual exercise just done “on paper” – the key lesson is just to get there. If your synthesis is longer than it needs to be – even if you end up driving through Moose Jaw, Saskatchewan on your way to Dallas -  don’t worry about it too much at this point. Efficiency is a goal for later.

P.S. Here is an an excellent reaction map by Adam of the Chemistry Blog

P.S.S. “I got busted in Laredo…. for reasons that I’d rather not disclose“

Today’s Reaction Friday is all about keto-enol tautomersm, the chemistry equivalent of Jekyll and Hyde.

Included:

• Tautomers are constitutional isomers (not resonance forms)
• The same factors which stabilize alkenes stabilize the enol form
• How to draw the mechanism for tautomerism (and how NOT to draw it!)

A lot of students I talk to have questions about solvents, so I’ve decided to put together a reference post on them.

Solvents can cause considerable confusion in reactions, because they’re listed along with the reagents of a reaction but often don’t actually participate in the reaction itself. And to be honest, a lot of instructors (myself included) are less than consistent about when to include solvents and when not to. So the whole exercise can come across as somewhat arbitrary: when do you know when to include the solvent?

Let’s back up. What’s a solvent, anyway?

A solvent is a liquid that serves as the medium for a reaction. It can serve two major purposes:

1. (Non-participatory) to dissolve the reactants. Remember “like dissolves like” ? Polar solvents are best for dissolving polar reactants (such as ions); nonpolar solvents are best for dissolving nonpolar reactants (such as hydrocarbons).
2. Participatory:  as a source of acid (proton), base (removing protons), or as a nucleophile (donating a lone pair of electrons). The only class of solvents for which this is something you generally need to worry about are polar protic solvents (see below).

OK. So what does “polar” and “non-polar” mean?

• Polar solvents have large dipole moments (aka “partial charges”); they contain bonds between atoms with very different electronegativities, such as oxygen and hydrogen.
• Non polar solvents contain bonds between atoms with similar electronegativities, such as carbon and hydrogen (think hydrocarbons, such as gasoline).  Bonds between atoms with similar electronegativities will lack partial charges; it’s this absence of charge which makes these molecules “non-polar”.

There are two common ways of measuring this polarity. One is through measuring a constant called “dielectric constant” or permitivity. The greater the dielectric constant, the greater the polarity (water = high, gasoline = low).  A second comes from directly measuring the dipole moment.

There’s a final distinction to be made and this causes confusion. Some solvents are called “protic” and some are called “aprotic”.  What makes a solvent a “protic” solvent, anyway?

• Protic solvents have O-H or N-H bonds. Why is this important? Because protic solvents can participate in hydrogen bonding, which is a powerful intermolecular force. Additionally, these O-H or N-H bonds can serve as a source of protons (H+).
• Aprotic solvents may have hydrogens on them somewhere, but they lack O-H or N-H bonds, and therefore cannot hydrogen bond with themselves.

Nonpolar solvents: 

These solvents have low dielectric constants (<5) and are not good solvents for charged species such as anions. However diethyl ether (Et2O) is a common solvent for Grignard reactions; its lone pairs are Lewis basic and can help to solvate the Mg cation.

“Borderline” Polar aprotic solvents

These solvents have moderately higher dielectric constants than the nonpolar solvents (between 5 and 20). Since they have intermediate polarity they are good “general purpose” solvents for a wide range of reactions. They are “aprotic” because they lack O-H or N-H bonds. For our purposes they don’t participate in reactions: they serve only as the medium.

Polar aprotic solvents

These solvents all have large dielectric constants (>20) and large dipole moments, but they do not participate in hydrogen bonding (no O-H or N-H bonds). Their high polarity allows them to dissolve charged species such as various anions used as nucleophiles (e.g. CN(-), HO(-), etc.). The lack of hydrogen bonding in the solvent means that these nucleophiles are relatively “free” in solution, making them more reactive. For our purposes these solvents do not participate in the reaction.

Polar protic solvents

Polar protic solvents tend to have high dielectric constants and high dipole moments. Furthermore, since they possess O-H or N-H bonds, they can also participate in hydrogen bonding. These solvents can also serve as acids (sources of protons) and weak nucleophiles (forming bonds with strong electrophiles).

They are most commonly used as the solvent for their conjugate bases. (e.g. H2O is used as the solvent for HO(-); EtOH is used as the solvent for EtO(-). )

These types of solvents are by far the most likely to participate in reactions. There are many examples (too many to list) where a polar protic solvent such as water, methanol, or ethanol can serve as the nucleophile in a reaction, often when a strong electrophile (such as an acid) is present. So if you see this type of solvent, be on the lookout.

I’m sure I missed something or something wasn’t clear. If you’d like something expanded on, please leave a comment!

Source for data: Wikipedia

Let’s review what’s been talked about so far in this series on acid-base reactions:

An acid base reaction involves the donation of a proton (H+) from an acid to a base.

• The species which loses H+ is the acid
• The species which gains H+ is the base
• The conjugate base is what becomes of the acid after it loses H+
• The conjugate acid is what becomes of the base after it gains H+

All else being equal, charged species are more unstable than neutral species. Since an acid is becoming more negative upon loss of a proton, the stability of the new lone pair on the conjugate base is a key factor in determining how favorable the reaction will be.

In other words:

• any factor which stabilizes the conjugate base will increase the acidity.
• any factor which destabilizes the conjugate base will decrease the acidity. 

For our purposes, a roughly equivalent word for “stability”  is “basicity”. The more “unstable” the pair of electrons on a species is, the more basic it is. Stabilizing a lone pair lowers the basicity; destabilizing the lone pair increases basicity.

Last time we went through the effect of electronegativity on acidity and basicity. But there are at least six more factors that can act to stabilize negative charge in organic chemistry. So let’s go through each of these in turn, and explicitly  tie together the relationship between the stability of negative charge and acidity/basicity.

1. The less charge the better.

All else being equal, the lower the charge density, the more stable a species is. So stability increases as we go from O(2-) to HO(-) to H2O. This is demonstrated in the acidity/basicity trends of water and related species, shown here. H3O(+) is the most acidic (most stable conjugate base); HO(-) is the least acidic (least stable conjugate base).

2. Polarizability

High charge densities tend to be less stable than low charge densities. The more “spread out” a charge can be, the more stable it will be. So as we go down the periodic table from F(-) to I(-), notice that the magnitude of the negative charge doesn’t change, but the volume that it occupies does. Iodide ion, being considerably larger than fluoride ion (206 pm vs. 119 pm) is more stable, which means that H-I is a stronger acid than H-F.

3. Electron withdrawing groups

The last post described how stability of a negative charge increases with increasing electronegativity of the atom. Adding electron withdrawing groups to an atom can have the a similar effect to that of increasing electronegativity. For example, in the molecule below, the stabilization of the negative charge increases with every hydrogen that is substituted for chorine.

4. Resonance

Resonance is another means by which negative charge can be dispersed in a molecule. If the conjugate base has a charge which can interact with adjacent double bonds or p orbitals, its stability will increase. This leads to increased acidity of the conjugate acid.

5. Orbitals

Increasing the “s-character” of an orbital has a similar effect as that of increasing electronegativity and adding electron withdrawing groups; it brings the negative charge closer to the positively charged nucleus, which , according to Coulomb’s Law, is a favorable interaction. So as we go from sp3 (alkane) to sp2 (alkene) to sp (alkyne) hybridization, the stability of the negative charge increases. So alkynes are remarkably acidic compared to alkanes.

6. Aromaticity

A special case is that of aromaticity – usually dealt with in the first few weeks of second-semester organic chemistry – which is a special type of stability exhibited by some conjugated molecules. Without going into details, I’ll just say that if the conjugate base is aromatic, this will lead to a tremendous increase in acidity relative to similar (but non-aromatic) molecules.

I know this has been yet another long post but hopefully it at least ties together the  issue of charge stability and acidity/basicity. Whenever the issue of stability comes up, so should the issue of acidity and basicity. These items are woven together.

So far I’ve just talked about trends. But what do you do when you need to compare the acidity of two species that aren’t related by a trend, like, propanol and H2S?  How do you proceed? In this case we are going to resort to the results of experimental measurements, and in the next post I’ll start to discuss how we do that.

Next Post: Walkthrough of Acid-Base Reactions (4) – pKa

When you have two hydrogens attached to a single carbon, they can have three different types of relationships. We call them “homotopic”, “enantiotopic”, and “diastereotopic”.

First of all, when is this important?

1.  Certain reactions directly replace hydrogens with other atoms. For example, free radical chlorination replaces C-H bonds with C-Cl bonds. So understanding these principles help in understanding what potential types of products you could obtain from these reactions.
2. In Nuclear Magnetic Resonance (NMR) these relationships determine whether or not you these hydrogens are in the same “chemical environment”. In other words, whether or not they have the same or different signals.

 Homotopic

Take a molecule like ethane. Let’s label (with color) two different hydrogens, blue and red. Next, let’s replace each of these hydrogens in turn with a different atom. In this example it could be chlorine (Cl) but really this can be done with any atom or group (except hydrogen of course).

Replace the red H and the blue H in turn with Cl and compare the molecules that are formed. Ask: how are these molecules related?

In this case they are both chloroethane. Since the two molecules are the same, the two hydrogens are said to be “homotopic”. Replacement of either gives rise to the same product.

Enantiotopic

Let’s look at butane next; specifically, the second carbon of butane.  Replacement of the red  H with Cl leads to (R)-2-chlorobutane, while replacement of the blue H with Cl leads to (S)-2-chlorobutane. These hydrogens are therefore not homotopic. Since enantiomers are obtained here, these two protons are therefore enantiotopic.

Note that the CH3 protons of butane are homotopic; it’s only the C-2 (and C-3) hydrogens of butane that are enantiotopic.

Diastereotopic

It’s also possible to have diastereotopic protons. Look at the alkene below. Replacement of the red H with Cl leads to the E-alkene, while replacement of the blue H with Cl leads to the Z-alkene. What’s the relationship between these two compounds? They’re diastereomers – stereoisomers, but not mirror images. So the two protons are said to be diastereotopic. 

There’s another potential situation which can lead to diastereotopic protons. Look at the molecule below – R-2-butanol. Replacement of the red H leads to the (R, R) product. Replacement of the blue H leads to the (R, S) product. Therefore, these two products are diastereomers, and the two protons are diastereotopic.

 When does it matter?

Two situations:

1. In free radical chorination – say, of butane – on the second carbon (C-2), substitution of C-H with Cl will lead to a mixture of stereocenters. It’s important to recognize when this can happen.
2.  (Most common) – In NMR spectroscopy:

• homotopic protons have the exact same chemical shift
• enantiotopic protons have the same chemical shift in the vast majority of situations. However, if they are placed in a chiral environment (e.g. a chiral solvent) they will have different chemical shifts.
• diastereotopic protons have different chemical shifts in all situations

P.S. I’m writing about this because a reader suggested this topic through my feedback form. Have a suggestion for a post? Send feedback here (anonymous if you like)

Last time I started writing about acid-base reactions. We looked at this list of stabilities of anions going across the topmost row of the periodic table.

 Fluoride ion is the most stable in this series because it’s the most electronegative; carbon is the least stable because it’s the least electronegative.

Because of this, we were able to say that H-F was the most acidic, because it had the most stable conjugate base.

And H-CH3 (methane)was the least acidic, because it had the least stable conjugate base.

Let’s look at the flip side of this reaction. Instead of starting with HF, H2O, H3N, and CH4 and asking how likely they are to donate a proton to a common base (water in our example) , imagine we start with the anions [ F-, HO-, H2N- and H3C- ] and have them take a proton away from  a common acid (such as water).

Which reactions would be most favorable? Which would be least favorable?

The same principle applies. The less stable the anion, the more likely the reaction will be to proceed to completion. 

So in this case, the reaction of F- with H2O would be the least favored, because F- is the most stable. And the reaction of H3C- with H2O would be the most favored, becuse H3C- is the least stable.

[A clarification: these are equilibrium reactions. So what I mean by favored here is the extent to which the equilibrium would favor the products on the right]

Notice the role that each of these anions plays in these reactions: it is accepting a proton from water, so in other words it is acting as a base.

Therefore, our whole discussion of the  “stability” of anions,  for lack of a better term, goes by another name you’re familiar with: basicity. 

In other words:

• the more stable a lone pair of electrons is, the less basic it will be.
• the less stable a lone pair of electrons is, the more basic it will be.

Let’s tie these two posts together with a common thread:

•  For any group of acids, H-X (where X can literally be anything), the strongest acid will have the most stable conjugate base. Since stability is inversely correlated with basicity, another way of putting it is:
• The stronger the acid, the weaker the conjugate base.
• Today’s post is about how the opposite is also true: The weaker the acid, the stronger the conjugate base.

Next time, we’ll apply this framework to other stability trends we’ve discussed previously.

P.S. One last note: a common misconception students have is that “weak acids are strong bases”. Not true! Methane (CH4) is a weak acid, but it can’t act as a base – it doesn’t have a lone pair.

The proper way to say it is that “weak acids have strong conjugate bases”. So the conjugate base of CH4, CH3(-) is an extremely strong base.

Next Post: Walkthrough of Acid-Base Reactions (3) – Acidity Trends

The Fischer esterification is one of the most important reactions of carboxylic acids. Treatment of carboxylic acids with an alcohol in the presence of acid catalyst leads to the formation of esters, along with the elimination of a molecule of water. It therefore falls into the category of  ”condensation reactions”.
The reaction is an equilibrium. Driving the reaction forward is usually accomplished by using the alcohol as solvent. In practice, a method of sequestering the water is through the use of a Dean Stark trap.
Although the alcohol is generally the solvent in the reaction, it’s possible to perform “intramolecular” reactions where the carboxylic acid and alcohol are present on the same molecule.

Acid helps in two ways: 1) to make the carbonyl carbon more electrophilic, and 2) protonation of OH gives OH2, which is a superior leaving group (i.e. weaker base) in the elimination step.

The reaction goes through the five-step PAPED mechanism described here.