I’ve looked at a lot of first midterms for Org 1 at North American schools.  Here, I’ll share some of the common types of questions that come up, with examples. There won’t be much commentary here, mostly just a problem dump for now.

• Hybridization questions (crucial!)
• Lewis structures
• Formal charges
• Geometry / bond angles
• Bond types
• Dipole moment
• Simple nomenclature
• Boiling points / melting points
• Resonance questions
• Acid-base questions

1. Hybridization (*crucial*)

Expect to be able to determine the hybridization at pretty much any atom. One form this commonly takes is to be shown a large molecule and your job is to assign hybridization at various atoms. Remember the hybridization shortcut, which involves counting the number of attached (atoms + lone pairs).

• If it’s 4, the atom is sp³,
• If 3, it’s sp²,
• If 2, it’s sp. (If it’s 1, it’s probably hydrogen.)  Just be on the lookout for atoms with lone pairs adjacent to pi bonds.

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(Hint: since the dipole moment of CH₄ is zero, the answer is, “not enough”)

If the orbital  configuration  of carbon is 2s²2p² , then how can we use this information to figure out what the arrangement of the orbitals are in a simple organic molecule like methane (CH₄)?

It turns out that methane is tetrahedral, with 4 equal bond  angles of 109.5° and 4 equal bond lengths, and no dipole moment.

This brings up two questions. First, how do we know that CH₄ is tetrahedral? And secondly, how do  we reconcile this  electronic configuration (2s²2p² ) with the fact that we have four equal C–H bonds?

Table of Contents

1. The Electronic Configuration Of The Valence Electrons Of Carbon Is 2s²2p²
2. Can We Use This Information To Figure Out The Structure Of Methane (CH₄)? (Spoiler: No)
3. Maybe Methane (CH₄) Is Square Planar?
4. Disproving The Square Planar Structure Of CH₄ (1874) And Proposal Of A Tetrahedral Structure
5. Tetrahedral Carbons: Not A Popular Idea In 1874
6. So What Orbitals ARE Involved?
7. Notes

1. The Electronic Configuration Of The Valence Electrons Of Carbon Is 2s²2p²

In our review of atomic orbitals, we saw that the orbital configuration of the valence electrons of carbon is 2s²2p² as shown below:

Since the 2s orbital is lower in energy than 2p, it’s filled first. That means that there are two electrons in the 2s orbital, and a single electron in two of the three 2p orbitals. There’s also an empty 2p orbital.

[In addition, there are two electrons in the “inner shell” 1s orbital, which are not available for bonding].

2. Can We Use This Information To Figure Out The Structure Of Methane (CH₄)? (Spoiler: No)

So far so good. This is fine if we’re just talking about isolated carbon atoms.

But in order to be truly useful, we need to be able to relate the orbitals of carbon to the structure and bonding of actual organic compounds.

The simplest organic compound is methane, CH₄. So let’s bring four hydrogen atoms into the picture and try to apply what we’ve learned to come up with some hypotheses about the bonding in this molecule.

The 3 p-orbitals in carbon are all at 90 degrees to each other, along the x, y, and z axes.

Shouldn’t we expect that the structure of methane would have three C-H bonds for each of the p orbitals (at 90 degrees to each other) and then the fourth C-H bond attached to the 2s orbital? Since electron pairs repel, maybe we should put that C-H bond the maximum distance away from the other C-H bonds; this would give an H–C–H bond angle of 135°.

Following this logic would give a structure like this:

As it turns out, it can be shown that this proposal is wrong.

Why?

Dipole moment.

Recall that each C–H bond has a small dipole due to the difference in electronegativity between C (2.5) and H (2.2). We expect C to be partially negative and H to be partially positive. (See article: Dipoles and Dipole Moments)

• If the above structure accurately depicted the structure of methane, we’d expect methane to have 3 longer C–H bonds (to the 2p orbitals) and one shorter C–H bond (to the 2s orbital, which is closer to the nucleus)
• Furthermore, we’d expect 3 H–C–H bond angles of 90° and one H–C–H bond angle of 135°.
• When the vector sums of the C–H dipoles are added up in this structure, they would not all cancel out.
• We would therefore expect to observe a small, but measurable dipole moment for CH₄. [Note 1]

However, the measured dipole moment of CH₄ is zero. Therefore this cannot be the correct structure.

This tells us that all the bond lengths and bond angles in methane are identical.

3. Maybe Methane (CH₄) Is Square Planar?

Alright, you say. If all C-H bonds are of equal lengths and angles, why can’t CH₄ have the structure below, where all the bond angles are 90° and CH₄ is flat, in the plane of the page. (We call this structure “square planar”).

This was in fact the majority opinion for the arrangement of bonds around carbon until about 1880. Extremely brilliant chemists such as Berzelius went to their graves having no reason to doubt that methane was anything but flat.

However, we now know this to be wrong. Why?

4. Disproving The Square Planar Structure Of CH₄ (1874) And Proposal Of A Tetrahedral Structure

If methane is modified so that the central carbon is attached to four different groups, the molecule can exist as 2 different isomers that are non-superimposable mirror images (this is called “optical isomerism” and covered later in the course).

This is possible if the arrangement of 4 groups around the central carbon is tetrahedral, but not if the molecule is square planar. For example, the methane derivative bromochlorofluoromethane has four different groups around carbon and can be separated into two different isomers which rotate plane-polarized light in different directions. [As we’ll see later, these isomers are called “enantiomers”]

This  observation rules out the square planar structure. If carbon was square planar, the molecule would be flat, and be superimposable on its own mirror image, and only one isomer would be possible.

Jacobus Henricus van’t Hoff , a fellow at the veterinary college in Utrecht, was among the first to address the possibility of three-dimensional carbon. In his “La Chimie dans L’Espace” (1874) he noted that the arrangement of atoms in space has important practical consequences – a point that had been completely neglected to that point. van’t Hoff showed that a tetrahedral arrangement of four different groups around a carbon atom (which he called an “asymmetric carbon”) would give rise to two different isomers, and furthermore, this would explain why tartaric acid (with two asymmetric carbon atoms) existed in three forms (+, –, and meso).   [Source]

van’t Hoff’s work – which should be noted was purely theoretical –  was not well received in some circles.

5. Tetrahedral Carbons: Not A Popular Idea In 1874

The eminent German chemist Hermann Kolbe had this to say:

For his part, van’t Hoff flew to Stockholm on his Pegasus to receive the first Nobel Prize in Chemistry in 1901. [Note 2]

Incontrovertible proof for the tetrahedral arrangement of bonds around the carbon atom came in 1913 when Bragg determined the structure of diamond using X-ray crystallography and found it to be a tetrahedral network of carbon atoms with C-C-C bond angles of 109.5°.

6. So What Orbitals ARE Involved?

We now rationalize the tetrahedral arrangement of atoms around methane as being due to the repulsion of the bonding pairs of electrons with each other (a.k.a. VSEPR theory).

This doesn’t help us with understanding the orbitals involved in bonding, however.

If we accept that the arrangement of hydrogens around methane is tetrahedral, then how do we describe the bonding orbitals of methane, given what we know about the geometry of s and p orbitals?

After all, the 2p orbitals are all at 90 degrees to each other, but the bond angles in methane are 109.5°.

Furthermore, how do we account that each of the 4 bonds in methane are of identical lengths? What happened to the 2s orbital, for example?

Are the electrons in the C-H bonds considered to be in p orbitals or s orbitals? Or something else?

As it turns out,  the conventional treatment is to deal with the bonds around carbon as being in hybrid orbitals. 

More on that in the next post.

Next Post: Hybrid Orbitals

Notes

Note 1. Note that mono-deuterated methane (CH₃D, where D is deuterium, the heavy isotope of hydrogen) has a small dipole moment that has been measured. [ref]

Note 2: it should be noted that van’t Hoff’s Nobel Prize was for his contributions to physical chemistry, not organic stereochemistry.

I am indebted to this page covering van’t Hoff’s “The Arrangement of Atoms In Space” for historical perspective. Well worth reading in full.

From the same source: van’t Hoff’s tetrahedral models [from the Leiden history of science museum; source]

From the same author, some more historical perspective on the Kolbe/van’t Hoff spat:

Obviously Kolbe was silly to be so intemperate and spiteful. He was also short-sighted, and he guessed wrong. We can easily appreciate that in the court of history he got what was coming to him.

The challenge is properly to respect the indispensable contributions the attitude he was championing had made to the development of chemistry. It was by sticking close to careful experimental observations that chemistry had gotten where it was (and is). Kolbe was trying to keep science on a productive, intellectually justifiable path.

Read here for full.

If the site has had a consistent weak spot over the past few years, it’s been in the realms of spectroscopy and synthesis.

I’m happy to announce that this hole has at least been partially filled with a new selection of quizzes on these topics.

Spectroscopy Quizzes

(link)

The section on Spectroscopy starts with some exercises on Index of Hydrogen Deficiency (IHD), and then moves on to UV-Vis, IR, Mass Spectrometry and finally to ¹³C and ¹H NMR.

When teaching this topic, I’ve always been of the mind that students should start by calculating IHD first. With a bit of practice, it’s a simple calculation that sets the stage for the remaining topics. There is a lot of logic in structure determination and a quick IHD calculation helps you quickly rule out certain possibilities.

Next comes UV-Vis. The focus here is just on the key lesson that increasing conjugation results in a narrowing of the energy gap ΔE  between the highest-energy occupied molecular orbital (HOMO) and the lowest-energy unoccupied molecular orbital (LUMO). Since E = hν and λ = c / ν , this means the wavelength of light absorption increases as ΔE decreases. The goal of the quizzes is to learn to quantify conjugation lengths and correlate these with λ_max.

The section on infrared spectroscopy focuses on training students to recognize the main diagnostic bands in an IR spectrum – the OH stretch around 3200-3400 cm^-1 and the C=O stretch around 1700 cm^-1. Students who spend the first 15-30 seconds of IR analysis on looking at these two regions will be well-primed to then move on to looking at areas of secondary importance, such as the C-H stretch around 3000 cm^-1  and the acetylene C-H stretch around 3400 cm^-1. There are also exercises on recognizing nitriles, amides, amines, and carboxylic acids.

The exercises on mass spectroscopy focus on identifying the presence of halides (Cl and Br) by looking for their characteristic M+2 peaks, and roughly approximating the number of carbons in a sample by analyzing the strength of the M+1 peak.

¹³C NMR spectroscopy begins with quizzes on predicting the number of signals each molecule will generate. The second section then tries to prime students to look for the presence of sp² – hybridized carbon in the region above 100 ppm, and sp³ – hybridized carbon below 80 ppm. Questions on identifying molecules given the ¹³C spectrum are all multiple choice.

Finally, the section on ¹H NMR spectroscopy begins with identifying the number of signals a molecule will produce and then asks students to predict multiplicity and chemical shift. The final section provides some multiple choice questions on identifying which molecule is represented by each spectrum.

The “click to flip” quiz format is good for training students on key patterns and in applying logic, but it has its limitations. It does not lend itself well to classic structure determination problems that provide multiple pieces of information (IR, ¹³C, ¹H). Students who want more practice with these sorts of problems should probably check out these problems from CU Boulder.

Synthesis

The Practice Quizzes section now has over 100 new problems on Synthesis, divided roughly into Org 1 topics (alkenes, alkynes, nucleophilic substitution) and Org 2 topics (everything else up to amino acids).

These have a “roadmap” format where students must fill in the blanks to get to the desired product. To make things slightly easier, condensed formulae are provided.

Of course there can be multiple ways to solve each particular synthesis problem and I’ve tried to note situations where multiple alternatives exist.

I hope the instructor community and MOC Members will find these new resources useful.

Please let me know via this form if you spot any typos or items that need correction.

– James

• Terminal alkynes have unusually acidic C–H bonds (pK_a 25). Treatment with a strong base such as sodium amide (NaNH₂) gives an acetylide, the name for the conjugate base of a terminal alkyne.
• Acetylides are more stable than the conjugate bases of alkenes and alkanes due to the fact that the lone pair is held in an sp-hybridized orbital which has 50% s-character. Since s-orbitals are held closer to the positively charged nucleus than p-orbitals, the electrons in this orbital are more stable (i.e. have less potential energy)
• Acetylides are strong bases, but can also act as nucleophiles in nucleophilic substitution reactions (S_N2) with alkyl halides to form substituted acetylenes.
• These reactions work best for primary and methyl alkyl halides.
• Attempts to form C-C bonds via S_N2 reactions with secondary alkyl halides almost always results in elimination (E2) instead, due to the high basicity of the acetylide ion.
• The reaction of acetylides with alkyl halides one of the most important reactions you will learn in first semester organic chemistry because it provides a versatile way of forming C-C bonds and extending the carbon chain.
• This reaction is therefore a key entry point in planning the synthesis of various molecules, especially since the resulting alkynes can be hydrogenated to alkanes (and partially hydrogenated to alkenes, as we’ll soon see). [See article – Partial Hydrogenation of Alkynes to cis-Alkenes With Lindlar’s Catalyst]

Table of Contents

1. 1. Terminal Alkynes Are Acidic!
   2. Alkylations of Acetylides With Primary Alkyl Halides: Finally, Some Carbon-Carbon Bond Formation!
   3. Alkylation of Acetylides – Some Practice Questions
   4. Synthesis of Substituted Acetylenes – Practice Questions
   5. Other Reactions of Acetylides – Epoxide Opening and Addition to Aldehydes/Ketones
   6. Summary
   7. Notes
   8. Quiz Yourself!
   9. (Advanced) References and Further Reading

1. Terminal Alkynes Are Acidic!

Among hydrocarbons, terminal alkynes have a very special property.  Their C-H bonds are unusually acidic (pK_a 25).

The alkyne C-H bond is sp-hybridized. When C-H is deprotonated, the resulting carbanion is held in an orbital with 50% s-character. Since s-orbitals are closer to the nucleus than p-orbitals,  this means that the electrons experience greater stabilization from the positively charged nucleus than the conjugate bases of alkenes and alkanes.

Any factor which stabilizes a lone pair of electrons tends to reduce its basicity. (See article – Key Factors That Influence Acidity). In fact,  just thinking of “basicity” as a synonym for “lone-pair instability” can get you pretty far in organic chemistry! 

A common choice of base for deprotonating the C-H bond of acetylenes is sodium amide (NaNH₂), often used in its conjugate base, liquid ammonia (NH₃). NaNH₂ can also be used to deprotonate the great-granddaddy of all alkynes, acetylene itself.   [Note 1]

Note – don’t confuse NaNH₂/NH₃  [strong base!]  with sodium in ammonia,Na/NH₃  [reducing agent for triple bonds!]  [Note 2]

Acid-base reactions spontaneously proceed in the direction that gives weaker acids from stronger acids. (I lovingly call this the “Principle Of Acid-Base Mediocrity” – See article: How To Use a pKa Table).

Since we are proceeding from a stronger acid (terminal alkyne, pK_a 25) towards a weaker conjugate acid (NH₃, pK_a 38) the acid-base equilibrium here will be favorable.

On the other hand, the acid-base reaction between NaNH₂ and alkenes (pK_a 42) or alkanes (pK_a 50) is unfavorable since it would result in a stronger acid (NH₃, pK_a 38), as well as a stronger base. Remember – the stronger the acid, the weaker the conjugate base! 

2. S_N2 Reactions of Acetylides With Alkyl Halides: Finally, Some Carbon-Carbon Bond Formation!

OK. So we can make acetylides. Now what?

Well, acetylides are excellent nucleophiles.  They react with alkyl halides to give internal alkynes, in a reaction known as nucleophilic (aliphatic) substitution.

It is a substitution reaction because a new bond is formed (C-C) at the same carbon where a bond is broken (C-X, where X is a good leaving group). ( See article: What Makes a Good Leaving Group?).

More specifically, the substitution proceeds through an S_N2 mechanism (substitution, nucleophilic, bimolecular rate-determining step) since the C–C bond is being formed at the same time that the C–X bond breaks. The reaction occurs via donation of the nucleophile lone pair into the sigma* orbital of the C-X bond, often referred to as a “backside attack”. It results in inversion of configuration at the carbon, although inversion can only be observed with carbons bearing a chiral center. (See article: The S_N2 Mechanism)

The reaction works best for primary (and methyl) alkyl halides due to their lack of steric hindrance.

Secondary alkyl halides tend to give elimination (E2) instead of substitution, since there is more steric hindrance at a secondary carbon and acetylide is still a very strong base – even if it’s a weak base for a hydrocarbon!

All right. Perhaps you’ve already covered nucleophilic substitution reactions, and this reaction might not seem like such a big deal to you. Fair.

I would like to draw your attention, however, to the key bond that is formed in this reaction: C–C.

Up until this point, it’s unlikely you’ve covered any carbon-carbon bond forming reactions. If you’ve covered any at all, it might be the cyanide ion (e.g. NaCN) with alkyl halides. That isn’t so important for our purposes since we don’t cover reactions of cyano groups until later on in Org 2.

Since organic chemistry is ultimately the chemistry of carbon, having the ability to form a new C-C bond from a terminal alkyne via an S_N2 reaction is huge because it allows us to plan the synthesis of essentially any linear hydrocarbon from acetylene, provided we can partner it with primary alkyl halide.

(Those primary alkyl halides can themselves be made from various reactions with acetylene, a point we’ll get to later in this chapter!). 

The example below, for instance, shows the synthesis of 5-decyne:

This reaction is extremely versatile. Simply by changing the identity of the alkyl halide, we can  tack on pretty much any alkyl group we want –  so long as it’s primary – which gives us access to a huge variety of linear hydrocarbons!

3. Alkylation of Acetylides – Some Practice Questions

We’ll get to some synthesis applications a little further below. In the meantime, see if you can draw the product of this reaction:

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Here is another example of a reaction between an acetylide and an alkyl halide. Can you draw the product? (D is deuterium, the heavy isotope of hydrogen).

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Draw the product of the reaction below:

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In the reaction below, the acetylide is treated with an alkyl halide containing two leaving groups. Draw the product!

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4. Practice Questions – Synthesis of Acetylenes

As mentioned above in section two, the S_N2 reaction between acetylides and alkyl halides means that we can build up pretty much any linear alkyne from acetylene, provided that we have the necessary (linear) alkyl halides.

The questions below ask you to show how you would synthesize internal alkynes from acetylene and alkyl halides.

Here’s one synthesis problem:

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A second, slightly more difficult synthesis question.

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More of the same thing!

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5. Reaction of Acetylides With Other Nucleophiles

Acetylides don’t just react with alkyl halides! They are versatile nucleophiles with other electrophiles as well, although you might not see some of these reactions until later in your course, or perhaps in the second semester of a two-semester course.

Epoxides are 3-membered cyclic ethers with considerable ring strain (about 13 kcal/mol) (See article – Epoxides, The Outlier of the Ether Family).  Acetylides will react with epoxides at the least substituted position to form new C-C bonds (See Article: Epoxide Ring-Opening With Base)

Acetylides will also add to aldehydes and ketones through nucleophilic addition to the C-O pi bond. In this respect the reaction of acetylides is essentially identical to those of Grignard reagents.

6. Summary

• Acetylides react with primary and methyl alkyl halides to give new C-C bonds via nucleophilic substitution (S_N2 mechanism).
• They tend to give elimination with secondary alkyl halides.
• This is an extremely important reaction for first semester organic chemistry, as it allows for formation of longer carbon chains from acetylene.

In the next article in this series, we will show how the triple bond of alkynes can be partially hydrogenated to give alkenes. (See article: Partial Hydrogenation of Alkynes to Give Alkenes)

Notes

Note 1. Conditions for the deprotonation of acetylene are here. Note that acetylene is a gas, so it has to be bubbled through a solution containing NaNH₂ in ammonia. These days, it’s more common just to just purchase the conjugate base of acetylene (such as lithium acetylide, diethylamine complex) directly from a commercial supplier like Aldrich and weigh it out.

Note 2. By no means is NaNH₂ the only base used for deprotonating acetylenes, it just seems to be the textbook reagent of choice. Grignard and organolithium reagents are also often used to form acetylides.

Quiz Yourself!

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(Advanced) References and Further Reading

This is a pretty standard acid-base reaction, driven by the acidity of the sp-H atom. The utility lies in that this is still a robust method of C-C bond formation, and a useful way to introduce alkynyl groups if desired.

1. THE PREPARATION AND ALKYLATION OF METAL ACETYLIDES IN LIQUID AMMONIA*
   T. H. Vaughn, G. F. Hennion, R. R. Vogt, and J. A. Nieuwland
   The Journal of Organic Chemistry 1937 02 (1), 1-22
   DOI: 10.1021/jo01224a001
2. PREPARATION AND USE OF LITHIUM ACETYLIDE: 1-METHYL-2-ETHYNYL-endo-3,3-DIMETHYL-2-NORBORNANOL
   Mark Midland, Jim I. McLoughlin, and Ralph T. Werley Jr
   Org. Synth. 1990, 68, 14
   DOI: 10.15227/orgsyn.068.0014
   I was initially a little surprised that something like this was published so recently in Organic Syntheses, but reading the discussion gives some context. The selective formation of the monolithiated species from deprotonation of acetylene is tricky.
3. 1-PHENYL-1-PENTEN-4-YN-3-OL
   Lars Skattebøl, E. R. H. Jones, and Mark C. Whiting
   Org. Synth. 1959, 39, 56
   DOI: 10.15227/orgsyn.039.0056
   Alkynyl Grignards can also be formed by deprotonation of a terminal alkyne with a Grignard reagent, as this procedure demonstrates.
4. n-BUTYLACETYLENE
   Kenneth N. Campbell and Barbara K. Campbell
   Org. Synth. 1950 30, 15
   DOI: 10.15227/orgsyn.030.0015
   An extremely simple example of this reaction. The deprotonation is done with Na metal in liquid ammonia, and care has to be taken to avoid the conditions of dissolving metal reduction (the procedure states that the reaction should not turn blue)
5. Synthesis of Unsymmetrical Alkynes via the Alkylation of Sodium Acetylides. An Introduction to Synthetic Design for Organic Chemistry Students
   Jennifer N. Shepherd and Jason R. Stenzel
   Journal of Chemical Education 2006, 83 (3), 425
   DOI: 10.1021/ed083p425
   A nice paper that describes the adaptation of this reaction for undergraduate teaching labs.

• Unlike C–C single bonds, C–C double bonds can’t undergo rotation without breaking the pi bond
• One consequence of this is geometric isomerism – the existence of alkene stereoisomers that differ solely in how their substituents are arranged in space about the double bond
• In simple cases where there are two identical substituents on each carbon of the alkene, we can use cis– and trans– to designate the isomers where those substituents are on the same and opposite sides of the double bond, respectively.
• For geometric isomers that lack two identical substituents, we rank the two substituents on each end of the double bond according to the Cahn-Ingold-Prelog (CIP) rules.
• The Z isomer (“zusammen“, same) is the geometric isomer where the #1 ranked substituents are on the same side of the double bond. Mnemonic: “zee zame zide“
• E isomer (“entgegen“) is the geometric isomer where the #1 ranked substituents are on the opposite side of the double bond,

Table of Contents

1. When do we use cis– and trans– Notation In Rings?
2. cis– and trans– Isomerism In Alkenes
3. Watch out for ambiguous names when geometrical isomerism is possible!
4. cis– and trans– isomerism in cyclic alkenes
5. When “cis“- and “trans‘” fails: E and Z Notation
6. E and Z Notation For Alkenes
7. Breaking Ties: The Method of Dots
8. Conclusion: E and Z Notation For Alkenes
9. Notes
10. Quiz Yourself!

This post was co-authored with Matt Pierce of Organic Chemistry Solutions.  Ask Matt about scheduling an online tutoring session here.

Quick Review: cis– And trans- Isomerism (“Geometrical Isomerism”) In Rings

Earlier on our MOC series on cycloalkanes, we saw that a key feature of small rings is that they can’t be turned “inside out” without breaking bonds.(See post: Cycloalkanes – Dashes and Wedges)

One of the most important consequence of this is that it can lead to the existence of stereoisomers – molecules which share the same molecular formula and the same connectivity but have a different arrangement of atoms in space.

These two versions of  1,2 dichlorocyclopentane (below) are an example. They have the same connectivity – both are 1,2-dichlorocyclopentane –  but have different arrangements of their atoms in space. The chlorines are on the same side of the ring in the left-hand isomer (both “wedges”, coming out of the page)  and on the opposite sides  (one wedged, one dashed) on the right-hand isomer.

These two molecules cannot be interconverted through rotation  of the C-C bond without rupturing the ring (use a model kit and try, if you like). They are therefore isomers.

Molecules which have the same connectivity but different arrangement in space are known as stereoisomers. 

Specifically, the relationship between the two molecules above is that of diastereomers: stereoisomers which are not mirror images of each other. (See post: Types of Isomers)

These two molecules have different physical properties – different boiling points, melting points, reactivities, spectral characteristics and so on.

 [Just to note, the other subclass of stereoisomer is “enantiomers”. We apply this to two stereoisomers which are (non-superimposable) mirror images of each other. Also: keep in mind that the terms “diastereomer” and “enantiomer” denote comparative relationships, like the terms “brother” or “cousin”. ]

1. When Do We Use cis- And trans- Notation In Rings?

We use the terms cis- and trans–  to denote the relative configuration of two groups to each other in situations where there is restricted rotation.

[Side note: the “restricted rotation” is how cis- and trans- subtly differs from  syn and anti, which we use in cases where there is free rotation, such as the orientation of methyl groups in “eclipsed” and “staggered” butane. Bottom line: syn and anti forms can generally be interconverted through bond rotation: cis and trans forms cannot. ]

In nomenclature,  “cis” is used to distinguish the isomer where two identical groups (e.g. the two chlorines in 1,2-dichlorocyclopentane) are pointing in the same direction from the plane of the ring, and trans to distinguish the isomer where they point in opposite directions. [You might also hear organic chemists say, “the chlorines are cis to each other” or “the hydrogens are trans to one another”.]

A common name for these so-called “cis-trans” isomers is “geometric isomers”. Those scolds at IUPAC actually discourage the term “geometric isomers”, and for once, I agree:  the term is somewhat redundant and can cause confusion. In the rest of this post I’ll just use the term “cis-trans” isomers.

In order for cis- trans- isomerism to exist in rings, we need two conditions:

• two (and only two) carbons each bearing non-identical substituents above and below the ring
• the two carbons have at least one of those substituents in common

In 1,2-dichlorocyclopentane we saw that C-1 and C-2 each had non-identical substituents (H and Cl) above and below the ring, and they each had at least one substituent in common (in fact they have two substituents in common:  H and Cl ).

Here’s another example: cis- and trans– 1-ethyl-2-methylcyclobutane. Note that they each have two carbons which each bear non-identical substituents above and below the ring (H and CH₃; H and CH₂CH₃). They also have at least one substituent in common (H). So we can refer to cis-1-ethyl-2-methylcyclohexane as the isomer where the two hydrogens are pointing in the same direction, and trans where they point in opposite directions.

If you’ve covered chirality, you might also note an interesting fact: there are two ways to draw each of the cis- and trans– isomers, and they can’t be superimposed on each other. These are enantiomers, by the way. (See post: Enantiomers, Diastereomers or the Same)

So cis- and trans- doesn’t specify which enantiomer (it can be applied to either). It’s just describing the relative configuration of the two groups (H in this case). If we want to specify a particular enantiomer, we need to use the  Cahn-Ingold-Prelog (CIP) system of assigning R and S configurations, which provides us with the “absolute” configuration. In that case, cis– and trans- is redundant. (See post: Cahn-Ingold-Prelog System)

Because cis– and trans– is relative, it doesn’t work if the two carbons don’t share a common substituent. In that case you also have to use (R)/(S) .

We’re taking too long to go through rings here, so let’s just illustrate 2 examples where “cis” and trans” doesn’t work in rings and leave it there.

2. cis– and trans- Isomerism (Geometric Isomerism) In Alkenes

cis-trans isomerism  is also possible for alkenes.  As in small rings, rotation about pi bonds is also constrained: due to the “side-on” overlap of pi bonds, one can’t rotate a pi bond without breaking it. This stands in contrast to conventional sigma bonds (single bonds) in acyclic molecules, where free rotation is possible: witness 1,2-dichloroethane (below left).

Hence we can have molecules such as cis-1,2-dichloroethene [boiling point 60°C] and trans-1,2-dichloroethene [boiling point: 48°C] which can be separated from each other due to their differing physical properties.

We can also use the cis–trans nomenclature to distinguish isomers such as 2-methyl-3-hexene (above right). In the cis isomer, the two hydrogens are on the same side of the pi bond, and in the trans isomer, the two hydrogens are on the opposite side of the bond. [Note: this risks a “tsk-tsk” with accompanying finger-wag from IUPAC , but it nevertheless gets the right structure: see the Note 1 below for a digression as to why]

As with rings, the minimum requirement for cis-trans isomerism in alkenes is that each carbon is bonded to two different groups, and that the two carbons have at least one substituent in common. 

As with rings, cis-trans isomerism isn’t possible if one of the carbons of the double bond is attached to two identical groups, as with 1,1-dibromo-1-propene, below. Try it for yourself if you’re not convinced.

3. Watch Out For Ambiguous Names Where Cis/Trans Isomerism Is Possible

A quick digression: one consequence of our newfound appreciation of geometrical isomerism is that many simple-sounding molecule names  are actually ambiguous.

For instance, the descriptor “3-hexene” does not unambiguously describe a specific molecule.  [The same is true for 2-butene: try it! ]. To nail down the specific molecule,  we need to specify cis– or trans– 3-hexene.

Note that 1-hexene is still OK, since the 1-position of 1-hexene is attached to two identical groups (hydrogens) and thus no cis–trans isomers are possible.

4. Cis– Trans- Isomerism For Cyclic Alkenes

cis- and trans can also be applied to alkenes in rings. For example, on paper it’s possible to draw cis– and trans– cyclohexene, since the pi bond fulfills the requirements for cis- trans- isomerism. In reality, trans-cyclohexene is impossibly strained. Try kissing yourself on the tailbone. That will give you some idea of the strain involved in trying to accommodate a trans– double bond in  a six membered ring .  [Note 2]

For this reason, for ring sizes 7 and below, it’s safe to ignore writing “cis” : the configuration is assumed.

At ring sizes of 8 and above, we do need to put a cis– or trans- in the name, because the trans– isomer becomes feasible. (Imagine trying to kiss yourself on the tailbone if you had the neck of a giraffe: suddenly not impossible!)

5. A Solution For When “Cis” and “Trans” Fails: The E/Z System

We saw that cis and trans fails in rings when the two carbons lacked a common substituent. It also fails for alkenes under these circumstances.

Case in point: try to apply cis and trans to the alkene below:

See the problem?

In the absence of two identical groups, we have no reference point!

On the left, the chlorine is cis to Br and trans to F. But does that really justify calling the isomer “cis” ? How do we decide?

What we need is some way to determine priorities in these situations.

[note: some textbooks may still refer to this alkene as exhibiting “cis-trans isomerism” even though we must use E and Z]

6. The E and Z Notation For Alkenes

Thankfully, we can apply the ranking system developed by Cahn, Ingold, and Prelog for chiral centers (as touched on in this earlier post on (R)/(S) nomenclature) for this purpose.

The protocol is as follows:

• Each carbon in the pi bond is attached to two substituents. For each carbon, these two substituents are ranked (1 or 2) according to the atomic numbers of the atom directly attached to the carbon. (e.g. Cl > F )
• If both substituents ranked 1 are on the same side of the pi bond, the bond is given the descriptor Z (short for German Zusammen, which means “together”).
• If both substituents ranked 1 are on the opposite side of the pi bond, the bond is given the descriptor E (short for German Entgegen, which means “opposite”).

So Z resembles “cis” and E resembles “trans”  .  (Note:  they are not necessarily the same and do not always correlate: see Note 2 for an example of a cis alkene which is E . The E/Z system is comprehensive for all alkenes capable of geometric isomerism, including the cis/trans alkene examples above. We often use cis/trans for convenience, but E/Z is the “official”, IUPAC approved way to name alkene stereoisomers].

 One easy way to remember Z is to say “Zee Zame Zide” in a German accent. My way of doing it was pretending that the Z stands for “zis”. Whatever works for you.

Here’s a practical example:

As with chiral centers, ranking according to atomic number can result in ties if we restrict ourselves merely to the atoms directly attached to the pi bonds.

7. Breaking Ties: The Method of Dots

For instance, the alkene below presents us with a dilemma: one of the carbons of the alkene is attached to two carbon atoms. So how do we determine priorities in this case. How do we break ties?

In the case of ties, we must apply the method of dots.  Dots are handy placeholders which is why I like to use this method.

• Place a dot on each of the two atoms you are comparing.
• List the 3 atoms each atom is attached to, in order of atomic number.
• Compare the lists, much like you would compare a set of three playing cards. Just as a hand of (8, 8, 7) would beat (8, 7, 7), so would (C, C, H) beat (C, H, H).
• If the lists are identical, move the dots outward to the highest priority atom on the list.
• At the first point of difference, assign (E or Z).
• If there is no difference… then the groups are identical, and E / Z does not apply.

Here’s a practical example of the “method of dots”. 

Here’s a more complex example with multiple alkenes. In this case each pi bond is designated by a number with its own separate E or Z configuration.

OK, this was long. But hopefully useful.

Watch out for a future post in which we go into more detail on the “method of dots”.

8. Conclusion:  E and Z Notation For Alkenes

cis-trans-  is OK for describing simple alkene stereoisomers, but only works in certain cases. Furthermore,  it only gives relative configurations.  The E/Z system is comprehensive and describes the absolute configuration of the molecule.

See below for an example of an E alkene which is “cis” and a Z alkene which is “trans”. 

Just a reminder: this post was co-authored by Matt Pierce of Organic Chemistry Solutions.  Ask Matt about scheduling an online tutoring session here.

Notes

Note 1: It’s possible to have an alkene we’d describe as ‘cis’ be E and vice versa.

E/Z is the preferred, more comprehensive nomenclature since it describes absolute configuration, whereas cis- trans- merely describes relative configuration.

Note 2: trans-cyclopropene, trans-cyclobutene, and trans-cyclopentene have never been synthesized or observed. trans-cyclohexene is a laboratory curiosity, stable at a few degrees above absolute zero. trans-cycloheptene has an extremely short half-life at room temperature. trans-cyclooctene is a stable molecule [it also exhibits axial chirality, which is interesting! ].

Quiz Yourself!

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Become a  MOC member to see the clickable quiz with answers on the back.

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In this post we explore how increasing substitution at carbon increases the stability of alkenes, as well as the effects of conjugation and strain.

Table of Contents

1. Heat Of Hydrogenation As A Measure Of Alkene Stability
2. Stability of Alkenes Increases With Increasing Substitution
3. Heats Of Hydrogenation For Some Monosubstituted Alkenes
4. The Relative Stability of cis- and trans- Alkenes
5. Alkenes Stabilized By Conjugation: Resonance Energy
6. Alkene Stability: Summary
7. Notes
8. Bonus Topic #1: Why Is Alkyl Substitution Stabilizing?
9. Bonus Topic #2: trans-Cycloalkenes
10. Quiz Yourself!
11. (Advanced) References and Further Reading

1. Heat Of Hydrogenation As A Measure Of Alkene Stability

We might not spend as much discussing thermodynamics in here organic chemistry as you did in general chemistry, but that doesn’t mean the concepts have just gone away!

One area where we’ve previously seen the usefulness of thermodynamic data is the use of heat of combustion data to quantify ring strain. [See: Cycloalkanes – How To Calculate Ring Strain]. The heat of combustion for cyclopropane works out to about  166 kcal/mol per CH₂ compared to the heat of combustion for unstrained cyclohexane [157 kcal/mol per CH₂]. That “extra” heat of combustion seen in cyclopropane is attributed to the instability arising from the strain of bent C-C bonds far away from their ideal angle of 109.5°. That’s angle strain.

Another area of organic chemistry where thermodynamic studies are useful in the stability of alkenes.

Back in 1935, Prof. Kiasatakowsky  and co-workers at Harvard published a method for measuring the heat of hydrogenation of ethylene (aka “ethene”) as it was passed over a finely divided metal catalyst containing adsorbed hydrogen. [Note 1] Because hydrogenating a molecule is considerably more gentle than, say, BURNING it, the method tends to be more sensitive for determining subtle differences in enthalpies.

In a hydrogenation reaction, a C-C bond is broken, and two new C-H bonds are formed.

It was found that hydrogenation of ethylene released 32.5 kcal/mol (136 kJ/mol) of heat. [Note 2]

Once the heat of hydrogenation of ethene was obtained, the next logical step was to measure the heat of formation for a huge variety of other alkenes, and to see what patterns emerged from the data.

So what happens to the heat of hydrogenation when alkyl groups are added to the alkene?

2. Stability of Alkenes Increases With Increasing Substitution

Well, as you might imagine from someone who had invented a new technique, Kiastakowsky went to town on this, investigating the heat of hydrogenation of a huge variety of alkenes. [Note 3] In the following decades, even more data has been accumulated, which is easily obtainable (with references) from the NIST Chemistry Web Book.

For our purposes, there are six substitution patterns on an alkene (seven if you count ethene).

The most notable trend that was found is that the heat of hydrogenation decreases as C-H bonds are replaced with C-C bonds. 

So what does that mean? 

Since the same bonds are formed and broken in every hydrogenation reaction, the heat of hydrogenation is measuring the stability of each type of alkene.

This means that the lower the heat of hydrogenation, the greater the stability of the alkene.

The way to visualize “stability” here is to compare it to potential energy, much like a ball becomes more “unstable” with increasing height.

via GIPHY

So what we’re really saying here is that alkene stability increases with increasing substitution of hydrogen for carbon. 

[The image above uses heat of hydrogenation data for the series hex-1-ene, trans-hex-2-ene, cis hex-2-ene, 2-methylpent-1-ene, 2-methyl-pent-2-ene, and 2,3-dimethylbutene, which all share the molecular formula C₆H₁₂. ]

OK, you might ask. So, why does this happen?

The short answer is that substitution of alkyl groups on the alkene allows for donation of electron density between (full) C-C sigma orbitals and the (empty) C-C pi star orbital. It’s often not addressed in introductory courses, so we’ll push the explanation down to this footnote. [Bonus topic one]

3. Heats Of Hydrogenation For Some Monosubstituted Alkenes

Just for fun, let’s look at a series of mono-substituted alkenes. Nothing weird here, we’ll just go from propene up to hex-1-ene.

Note that the heat of hydrogenation is quite consistent for a series of linear, non-branched, monosubstituted alkenes.

4. The Relative Stability of cis- and trans- Alkenes

So what about disubstituted alkenes? There are three types (cis, trans, and 1,1-disubstituted) but let’s just concern ourselves with cis and trans here.

We all know by now that cis and trans alkenes should differ a little bit in stability because in a cis alkene the groups are held closer together (more strain!) and in a trans-alkene they are further apart. [For a good time, amaze your instructor and call it by its proper name:  1,2-strain]

Heat of hydrogenation data actually allows us to quantify the difference in stability between cis and trans alkenes.

For instance, compare cis– and trans– but-2-ene, or cis- and trans hex-2-ene. The difference in stability is about 1 kcal/mol, rounding up generously.

While a difference of 1 kcal/mol might not seem like a lot, it  isn’t *that* small – for an equilibrium at 25 °C, a difference of 1 kcal/mol will give you about an 80:20 ratio of products. [Note 4]

For a really good time you can pick something crazy like the cis– and trans- di t-butyl ethylene. [not the correct IUPAC name, but definitely more vivid than cis- and trans- 2,2,5,5-tetramethylhex-3-ene].

Here the trans is more stable than the cis by about 10 kcal/mol.

That’s a lot of strain.

5. Alkenes Stabilized By Conjugation: Resonance Energy

The stability of alkenes is also affected by conjugation. This is a really a topic for another chapter [specifically, see Conjugation and Resonance] where we talk about pi systems, but the bottom line is that the p-orbitals in adjacent pi-bonds can clump together forming larger “pi-systems”, which provides more “room” for electrons to roam, lowering their energy. [Note 5]

Heat of hydrogenation numbers allow us to quantify the effect of resonance stabilization. How so?

Take but-1-ene. As we saw above the heat of hydrogenation is about 30.1 kcal/mol.

Add a double bond, and you might expect the heat of hydrogenation to double as well. But it doesn’t! It’s actually a little bit less. [56.6 kcal/mol] . The difference  (that extra 3.6 kcal/mol of additional stabilization)  is called “resonance energy“.

The most dramatic example of resonance energy is found in the example of “cyclohexatriene” , which has an extra stabilization energy of 36 kcal/mol. That’s a sure sign that something highly unusual is going on with this molecule, which is better known as “benzene”. That “highly unusual” property is called aromaticity and it warrants its own chapter. [See: Introduction to Aromaticity]

6. Summary: Stability of Alkenes

Three key factors affect the stability of alkenes, and the influence of these factors can be measured through the enthalpy of hydrogenation.

• One important factor is the substitution pattern. As C-H bonds are replaced by C-C bonds, the stability of the alkene gradually increases in the order mono (least stable) < di < tri < tetrasubstituted (most stable).
• When hydrogenation liberates more energy than expected given the substitution pattern, that’s likely a sign of strain. This is exemplified in the difference in enthalpy of hydrogenation between cis- and trans- alkenes, where the trans- alkene is more stable by about 1 kcal/mol.
• When hydrogenation liberates less energy than expected given the substitution pattern, that’s a sign that some extra factor is stabilizing the molecule. Among commonly encountered factors, conjugation ranks high. The difference in energy between the “expected” heat of hydrogenation and the measured heat of hydrogenation is called the resonance energy. The conjugation of one pi bond with an additional pi bond is “worth” about 2-3 kcal/mol.

The increasing stability of alkenes with increasing substitution not only comes up in Zaitsev’s Rule, but also later in the course when you study Thermodynamic and Kinetic Control.

Notes

Note 1. It was a copper catalyst, after a lot of trial and error.  The advantage of measuring the heat of hydrogenation over the heat of combustion is that it is a more sensitive technique for measuring small energies.

Note 2. This number was first measured in 1935, remeasured in 1951, and so far as I am aware, has not been updated. See the entry in the NIST Chembook for ethylene.

Note 3. Standard heats of hydrogenation have been pulled from the NIST Chembook.

Note 4. Actually 82:18 at 298 K.   From delta G = -RT ln K, using delta G of 1000 cal, T = 298 K, R = 1.987 cal / mol•K .

Note 5. If you think of electrons as waves, a larger pi-system allows  for longer wavelengths,  and since energy is inversely proportional to wavelength, this means a lower overall energy of the electron.

And a big thank you to The Kraken for his steady hands in the stability GIF.

Note 6. What about alkynes (and allenes) ? Same trend. More substituted = more stable.

Great source for this is the NIST Chemical Webbook. Chapter 15 in this book (“Rearrangements involving Allenes”) provides a great overview.

Appendix 1: Why Does Increasing Substitution Increase Stability?

So why does increasing substitution at the alkene increase its stability? This is not an easy question to answer to an introductory audience in a few sentences, and given the time constraints of a typical course the answer you will generally get from an instructor will range from “it’s complicated” to “hyperconjugation” to “orbital mixing”. Very rarely you might get an MO diagram.

The unifying principle here is that full orbitals – even those from single bonds – can donate into empty (even antibonding) orbitals, and that this interaction is stabilizing.

In ethene (below left) all of the C-H bonds are in the plane of the alkene, and none can overlap with the pi bond.

When a methyl group is added, say, in propene, one of the C-H bonds can now align with the pi-system of the alkene. The pair of electrons from the C-H bond can then donate into the empty pi* orbital.

This can be visualized through “no-bond resonance”, below right, where a “resonance” form is shown with a broken C-H bond and a new C-C pi bond. [The quotation marks are to differentiate it from our traditional view of resonance where only pi-bonds are allowed to form and break]. 

This mixing results in a stabilization of the molecule. . Although CH₃ is in rapid rotation, at any given moment at least one of the C-H bonds will have the proper geometry to allow overlap with the pi system.

Predicted to slightly lengthen C-H and C-C pi and strengthen C-C sigma.

Appendix 2: trans-Cycloalkenes

99% of people reading this will never use this so it is going down in the footnotes.

In the vast majority of molecules you will encounter, the double bonds in rings are cis. Why? The most vivid answer is provided by trying to make them with a model kit.

That is not a happy double bond.

However at a ring size of 7, a trans double bond becomes more than transiently stable (albeit very short lived at 0°), and at a ring size of 8 there’s enough floppiness in the ring such that its boiling point can be measured [143°C !] . Larger ring sizes than 8 can easily accommodate a trans double bond.

The heat of hydrogenation can be used to quantify the stability of these rings (note that this is not the whole picture, since it doesn’t take entropy into account, and that can be quite significant).

At ring sizes of 11 and 12 the trans-isomer actually becomes more stable (when allowed to equilibrate with acid) but recall that anything involving equilibrium is ultimately a measure of delta G, and delta G also includes an entropy term (S). It turns out that the main factor in the increased stability of 11- and 12- membered trans-cycloalkenes is their greater entropy. See this reference.

Quiz Yourself!

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(Advanced) References and Further Reading

All heat of hydrogenation values cited here were obtained from the NIST Chemistry Web Book. Searching by CAS number never fails. Selected original references below.

1. Heats of Organic Reactions. I. The Apparatus and the Heat of Hydrogenation of Ethylene
   G. B. Kistiakowsky, H. Romeyn Jr., J. R. Ruhoff, Hilton A. Smith, and W. E. Vaughan
   Journal of the American Chemical Society 1935 57 (1), 65-75
   DOI: 10.1021/ja01304a019
   Prof. Kistiakowsky’s first (of many) papers on the heat of hydrogenation of organic molecules, where he describes the apparatus required to obtain accurate heat of hydrogenation data in painstaking detail. The results stand up.
2. Heats of Organic Reactions. IV. Hydrogenation of Some Dienes and of Benzene
   G. B. Kistiakowsky, John R. Ruhoff, Hilton A. Smith, and W. E. Vaughan
   Journal of the American Chemical Society 1936 58 (1), 146-153
   DOI: 10.1021/ja01292a043
   Contains the heat of hydrogenation for 1,3 butadiene, benzene, and other unsaturated molecules, including allene (71.0 kcal/mol).
3. Heats of Hydrogenation. IV. Hydrogenation of Some cis- and trans-Cycloölefins1
   Richard B. Turner and W. R. Meador
   Journal of the American Chemical Society 1957 79 (15), 4133-4136
   DOI: 10.1021/ja01572a042
4. Heats of hydrogenation. IX. Cyclic acetylenes and some miscellaneous olefins
   Richard B. Turner, A. D. Jarrett, P. Goebel, and Barbara J. Mallon
   Journal of the American Chemical Society 1973 95 (3), 790-792
   DOI: 10.1021/ja00784a025
5. RELATIVE STABILITIES OF cis- AND trans-CYCLONONENE, CYCLODECENE, CYCLOUNDECENE AND CYCLODODECENE
   Arthur C. Cope, Phylis T. Moore, and William R. Moore
   Journal of the American Chemical Society 1959 81 (12), 3153-3153
   DOI: 10.1021/ja01521a067
   A.C. Cope reported that when cis– and trans– cycloundecene (11-membered) and cyclododecene (12-membered) are allowed to equilibrate (by heating with catalytic TsOH)  the trans-double bond is favored at equilibrium (i.e. has lower Δ G)… even though trans-dodecene has a higher enthalpy (Δ H) than its cis-isomer. This is a helpful reminder that enthalpy (delta H) is just one part of the Gibbs equation (Δ G = Δ H – TΔ S), the trans-cycloalkenes have higher entropy (S) and this explains their greater stability.

• Almost all reactions of alkenes we will learn about can be classified as addition reactions 
• In an alkene addition reaction, a C-C pi bond breaks, and two new single bonds to carbon are formed
• Depending on the structure of the alkene and the bonds formed/broken in the reaction, a mixture of constitutional isomers may be formed. In alkene addition reactions, these are often called “regioisomers”.
• When formation of one regioisomer is favored this is called “regioselectivity“. The most common example of regioselectivity is “Markovnikov” selectivity observed in addition of H-X to alkenes, where the C-H bond forms on the “least substituted” carbon (fewest carbons directly attached) and the C-X bond forms on the “most substituted” carbon (most carbons directly attached). The opposite of “Markovnikov” regioselectivity is “anti-Markovnikov” regioselectivity, which is observed in two cases (hydroboration and free-radical addition of HBr).
• The stereoselectivity of alkene addition reactions is also very important!  The pi bond of an alkene is flat, and therefore has two faces where addition can occur. When the two new bonds to carbon are formed on the same face, this is called, “syn” addition. When the two new bonds to carbon are formed on opposite faces, this is called, “anti” addition.
• The stereochemistry of the addition reaction is highly dependent on the mechanism of the reaction.  Some reactions are selective for syn addition, some reactions are selective for anti addition, and some reactions provide a mixture of syn and anti addition (“unselective”).
• When learning a new alkene addition reaction, there are three key pieces of information you will need to successfully draw the products: 1) the bonds that form/break, 2) the regioselectivity, and 3) the stereoselectivity
• When drawing out the products of an alkene addition reaction, don’t forget to draw out the addition products from both the “top” and “bottom” faces. Depending on the structure of the alkene, and the nature of the reaction, this can lead to formation of constitutional isomers, enantiomers, diastereomers, or even identical products.
• Determining how the products of an alkene addition reaction are related is a very common type of exam problem!
  

Table of Contents

1. 1. Alkene Reactions Follow A Few Key Patterns
   2. Alkene Addition Reactions
   3. Regioisomers in Alkene Addition Reactions: Markovnikov vs. anti-Markovnikov Regioselectivity
   4. Syn and Anti Addition To Alkenes
   5. Stereoselectivity in Alkene Addition Reactions
   6. Draw Out The Product Of Addition To Each Face Of The Alkene, Then Determine Relationships
   7. Summary
   8. Notes
   9. Quiz Yourself!
   10. (Advanced) References and Further Reading

1. Alkene Reactions Follow A Few Key Patterns

In the articles that follow in this chapter, we are going to cover the main reactions of alkenes. In this article, I want to try to help you recognize the key pattern of bonds that form and break (“addition”) and some of the terms we use to describe the sub-patterns of “regioselectivity” (Markovnikov vs. anti-Markovnikov) and “stereoselectivity” (syn vs anti).

If you take a sneak peek at the Reaction Map of alkenes at the end of this chapter, you would be forgiven for thinking there are an intimidating number of alkene reactions to learn – over 20, depending on how you count.  [See article – Reactions of Alkenes]

However, it’s actually a lot easier than it seems, because these reactions follow predictable patterns. And there aren’t 20 patterns – there’s more like three major patterns (with two very minor patterns).

In each of these reactions the specific identity of the bonds that form on the two carbons of the alkene may change, but they will still obey the key patterns.

2. Alkene Addition Reactions

Almost all reactions of alkenes we will learn can be classified as “addition” reactions [Note 1 ]

In an addition reaction, the C-C pi bond is broken, and two new single bonds to carbon are formed.

The reactions below are specific examples of alkene addition reactions. At this point you don’t need to know how they work or anything else about them. We’ll cover them in detail in subsequent articles!

Just observe the pattern of bonds that form and break. We break a C-C pi bond (about 60 kcal/mol) and form two new single bonds to carbon.

If you’ve previously covered elimination reactions, the pattern above might look somewhat similar. In an elimination reaction, a new pi bond is formed, and two single bonds to carbon are broken.

In other words, addition is the exact opposite of elimination. hover for image or click this link.

Going forward, make sure you can see the “hidden” (or “implicit”) hydrogens because they will often not be drawn out. You’ll be expected to be able to “see” that a new C-H bond formed here even if it isn’t drawn out explicitly.

3. Regioisomers in Alkene Addition Reactions, and The Three Patterns Of Regioselectivity

Some addition reactions involve forming two identical single bonds to carbon, and others involve forming two different bonds to carbon.

If the two new single bonds to carbon are identical, then an addition reaction will form only one constitutional isomer.

In hydrogenation, for example, two new C-H bonds are formed and a C-C pi bond is broken. There is only one way of ordering the new C-H bonds, so only one constitutional isomer (regioisomer) can form. There can certainly be a mixture of stereoisomers (see below) which have the same connectivity, but have different arrangement of atoms in space. More on that in a bit. 

If the alkene addition reaction forms two different bonds to carbon, then there are two different ways of ordering them.

This will only result in formation of a single constitutional isomer if the alkene is completely symmetrical, such as cyclohexene.

hover to see it, or click this link.

If the alkene is not symmetrical – the vast majority of cases! –  a mixture of constitutional isomers are formed. We’ll discuss the possibility for stereoisomers below.

If this pattern were random, then we’d get a 1:1 mixture of products (constitutional isomers) – in other words, a reaction completely lacking in regioselectivity.

As it turns out, the reactions we are going to learn all have predictable patterns of regioselectivity and knowing the pattern will allow us to predict in advance which constitutional isomer will be formed as the major product.

The classic regioselectivity pattern is known as “Markovnikov” regioselectivity and was first observed back in 1870 by Victor Markovnikov in the addition of H–X to alkenes, where a new C-H and C-X bond is formed. [See Hydrohalogenation of Alkenes and Markovnikov’s Rule] (note that X here is a halogen, such as Cl, Br or I). 

These reactions tend to happen in a way such that the C–X bond forms to the alkene carbon bearing the most carbon substituents (the “most substituted” carbon) and the C-H bond forms to the carbon bearing the most hydrogens (“less substituted” carbon).

We’ll commonly use the term “Markovnikov-selective” to refer to this pattern of regioselectivity.

Markovnikov in 1870 had no idea why these reactions had the selectivity that they did.

We now know that the reaction proceeds through a carbocation intermediate. Since carbocations are unstable reaction intermediates with less than a full octet of valence electrons,  the reaction will favor the reaction pathway that passes through the most stable carbocation intermediate, which is generally the carbon which is bonded to the greater number of carbon substituents. [See article: Carbocation Stability] . This results in the observed product!

To see the mechanism, hover here or click this link.

Looking ahead a bit, we can apply the same rationale to reactions that pass through positively charged 3-membered ring intermediates such as bromonium ions. (See article – Halogenation of Alkenes). Attack of the nucleophile on these intermediates always occurs to the carbon best able to support positive charge, which can be thought of as another expression of “Markovnikov” selectivity.  hover to see or click this link.

If both sides of the alkene are attached to the same number of carbons, then the carbocation intermediates will be of roughly equal stability, and Markovnikov regioselectivity will not manifest itself.

A few reactions (hydroboration and free-radical addition) show the opposite pattern of regioselectivity, where the C-H bond ends up being bonded to the less substituted carbon and C-X to the more substituted carbon.

This is somewhat unimaginatively called “anti-Markovnikov” regioselectivity. For more details, see Hydroboration-Oxidation, and Free-Radical Addition of HBr To Alkenes. 

To sum up, every single one of the 20+ addition reactions you will learn can be grouped into one of these three patterns of regioselectivity:

• None, or not applicable (N/A)
• Markovnikov selective
• anti-Markovnikov selective

4. Syn and Anti Addition To Alkenes

The pi bond of an alkene is flat  (planar). The two sides of an alkene are called its faces.

When an alkene undergoes an addition reaction, the two new single bonds can either form on the same face of the alkene, or they can form on opposite faces.

When the two bonds form on the same face of the alkene, we call this syn addition. When they form to opposite faces of the alkene, we call this anti addition.

Back in the chapter on conformations, we used the terms syn and anti in Newman projections to refer to the orientation of two groups on adjacent carbons of a carbon-carbon single bond that can rotate freely. (See article: Newman Projection of Butane).  When they have a dihedral angle of 0° they are said to be syn. When they have a dihedral angle of 180° they are anti. [Note 2]

Note that syn addition and anti addition refers to the relative orientation of the groups immediately after addition has occurred.

Any subsequent bond rotation about the C-C bond will not change a syn addition product into an anti addition product (or vice versa).

It is tempting to refer to these orientations as “cis” and “trans”, and in some cases the cis and trans products are actually the products. [Note: some of the older chemical literature also uses “cis” and “trans” to refer to syn and anti addition.]

hover to see an example, or click this link.

However, cis and trans should really only be used in situations where the two groups are locked in place and cannot undergo bond rotation (i.e. geometric isomers).  Try to avoid using “cis” and “trans” addition for products that can undergo subsequent bond rotation.

5. The Three Categories of Stereoselectivity In Alkene Addition Reactions

The stereoselectivity of alkene addition reactions falls into three categories.

• Reactions that give exclusively syn addition products
• Reactions that give exclusively anti addition products
• Reactions that give a roughly equal mixture of syn and anti addition products.

Whether or not a given reaction gives exclusively syn, anti, or syn+anti products is a piece of information that had to be determined experimentally in each case. You will be told which reaction belongs in which category when we cover each one. 

Since you will be expected to draw the products of alkene addition reaction products, knowing which category each reaction falls into is extremely important!

Here’s an example of a reaction that is stereoselective for syn addition products. No products of anti addition are formed.

[See article: What’s A Racemic Mixture?]

On the other hand, halogenation of alkenes is stereoselective for anti addition products. No syn addition products are formed. [Note 3].

Addition of HX to alkenes is not stereoselective, and gives a roughly equal mixture of syn and anti addition products.

By combining the three patterns of regioselectivity with the three patterns of stereoselectivity, we can build a little 2 × 3 table with all the possibilities.

As it turns out, not every one of these possibilities is represented by reactions we will learn.

To skip ahead to the end of the chapter, there will really be three key alkene addition patterns to look out for:

• Reactions that go through carbocation intermediates, which tend to be Markovnikov-selective reactions that give a mix of syn + anti stereochemistry. (See article: Alkene Addition Pattern #1: The Carbocation Pathway).
• Reactions that go through a 3-membered ring intermediate, which tend to be Markovnikov selective (when possible) and give anti addition products. (See article: Alkene Addition Pattern #2: The 3-Membered Ring Pathway). 
• Reactions that go through a more or less concerted pathway, which mostly have no regioselectivity (except for hydroboration, which is anti-Markovnikov) and give syn addition products. (See article: Alkene Addition Pattern #3 : The “Concerted Pathway”).

Condensing the reactions of alkenes down into these 3 key buckets will simplify things a lot.

6. Determining Relationships Between Products In Alkene Addition Reactions

Once you have the pattern of bonds that form and break, the regioselectivity, and the stereoselectivity, there’s just one more piece of the puzzle left to address.

Remember that alkenes are flat, and in the vast majority of cases, addition can occur with equal likelihood to either face of the alkene.

Depending on the structure of the alkene, this can result in products that are stereoisomers (enantiomers or diastereomers).

You will need to practice drawing the product obtained for addition to each face, and then compare those two products to determine how they are related.

Remember “enantiomers, diastereomers, or the same?” questions from your chapter on stereochemistry?

Well, we’re going to do this again, in the context of alkene addition reactions. (See article: Enantiomers, Diastereomers or The Same – Two Methods For Solving Problems).

Trust me when I say that this is a very common type of exam question!

This is what makes organic chemistry “special” – concepts from previous chapters (like stereochemistry!) keep recurring and are integrated into the new material. 

Let me show you what I mean using addition of H-Br to alkenes as an example.

• In this reaction, a C-C pi bond breaks, and a C-H and C-Br bond forms.
• The reaction has Markovnikov regioselectivity (C-Br forms on most substituted carbon)
• The reaction gives a mixture of syn and anti addition products.

Now let’s explore the products that form when different alkenes undergo addition.

In the addition of H-Br to 1-methylcyclohexene, we can draw the products that are formed when addition happens to the top and bottom face.

In this case addition to the top face and the bottom face results in the same product (note the absence of any chiral center).

Now let’s try it on 1-butene.

In this case, applying the pattern to both faces gives 2-bromobutane. However, note the formation of a new chiral center. These two products are enantiomers of each other, and since they are formed in equal amounts, they constitute a racemic mixture. (See article: What’s A Racemic Mixture?).

Click to Flip

If the starting material has a pre-existing chiral center, look out for the possibility of forming diastereomers.

Click to Flip

Is there a simple trick for knowing immediately if a given reaction will give you enantiomers, diastereomers, constitutional isomers or even just the same molecule?

There is no simple trick. Just practice.

However, there’s good news. If you have already answered a lot of “enantiomers, diastereomers, or the same” type questions, then you’re already most of the way toward solving these types of problems.

The key thing here is just applying the pattern for each alkene addition reaction (bonds formed/broken, regioselectivity, stereoselectivity) to each of the two alkene faces, and then comparing the two products.

7. Summary

Lots of concepts covered here, but this lays the groundwork for everything we’ll discuss later in this chapter on alkene addition reactions.

• Every alkene addition reaction involves breaking a C-C pi bond and forming two new bonds to carbon
• There are three key classes of regioselectivity in alkene addition reactions: Markovnikov, anti-Markovnikov, and “none” (where the two bonds are identical).
• There are three key classes of stereoselectivity in alkene addition reactions: syn addition, anti addition, and syn + anti addition (“unselective”).
• With the pattern of bonds formed/broken, regioselectivity, and stereoselectivity in hand you should be able to draw the pattern of any alkene addition reaction.
• Just remember that alkenes are flat and addition can occur to either face of the alkene. Practice drawing out the products that are formed when addition happens to each of the two faces.
• Then compare these two products to determine whether they are enantiomers, diastereomers, or the same

Notes

Note 1. One reaction of alkenes that doesn’t fit into this pattern of addition reactions is oxidative cleavage (e.g. ozonolysis). In this reaction, both the C-C sigma bond and the C-C pi bond are broken, and two new bonds between each carbon and oxygen are formed. (See article: Ozonolysis of Alkenes).

Note 2. Technically, any two substituents with a dihedral angle between +30° and -30° are “syn” and any two substituents with a dihedral angle between +150 and -150° are “anti“.

Note 3. In cases where a very stable carbocation can be formed (such as on a benzylic carbon),  the 3-membered halonium ion can open to give a free carbocation, and anti stereoselectivity can be lost. See article: Halogenation of Alkenes. 

Note 4. Original definitions of syn and anti according to Klyne and Prelog.

Quiz Yourself!

Become a  MOC member to see the clickable quiz with answers on the back.

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Become a  MOC member to see the clickable quiz with answers on the back.

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(Advanced) References and Further Reading

[references]

Many ketones and aldehydes have an alter-ego “enol” form with completely different chemical properties than the familiar “keto” form.  In this article we’ll explore the structure and properties of this “enol” form, go through the mechanism for the keto-enol transformation, and describe some of the key factors that can affect the keto-enol equilibrium.

*mostly aldehydes and ketones, although it can occur in other species as well (e.g. acid halides)
NOTE: This is a 2022 update of an older post entitled, “Keto-Enol Tautomerism: Key Points”. 

Table of Contents

1. When Ketones Moonlight As Nucleophiles
2. Keto-Enol Tautomerism
3. Examples Of Keto-Enol Tautomerism
4. Properties of Enols
5. Keto-Enol Tautomerism: Mechanisms
6. Four Factors That Affect Keto-Enol Equilibria
7. Factor #1: Substitution
8. Factor #2: Conjugation/Resonance
9. Factor #3: Hydrogen Bonding
10. Factor #4: Aromaticity
11. Revisiting A Weird, “Unketone-like” Reaction
12. Summary And Conclusion
13. Notes
14. Quiz Yourself!
15. (Advanced) References and Further Reading

1. When Ketones Moonlight As Nucleophiles

Reactions of aldehydes and ketones: by this point we’ve pretty much seen ’em all. Right?

I mean, the carbonyl carbon of aldehydes and ketones is an electrophile.  So their reactions all follow the same pattern, more or less.

1. A nucleophile attacks carbon, forming C–Nu and breaking the C-O pi bond to give an alkoxide (O^–) This is nucleophilic addition.
2. Mild acid is added to protonate the alkoxide and form O–H.  [See here: Aldehydes and Ketones – 14 Reactions With The Same Mechanism ]

So that’s all there is to the reactions of aldehydes and ketones. Case closed?

Not quite. This is the article where we discover that most aldehydes and ketones live a double life.

By day, they are respectable electrophiles, undergoing addition with nucleophiles. But by night, maybe when the moon is full, they undergo a transformation into a completely different beast – one with a different structure, properties… and appetites.

Here’s an example of a reaction which doesn’t fit the normal “two-step” pattern.  Treating a ketone with acid and Br₂ gives…. a new C-Br bond next to the carbonyl carbon?

This seems decidedly un-ketone-like.  Recall that Br₂ is electrophilic (remember how it reacts with electron-rich alkenes and aromatic rings). 

So what the heck is our “electrophilic” ketone doing cavorting around with another electrophile? And forming bonds at….  a totally different place?

If Br₂ is the electrophile, then what was the nucleophile here? 

2. Keto-Enol Tautomerism

A clue to this aberrant chemical behavior is provided not by lycanthropy, but by the fact that many aldehydes and ketones are in equilibrium with a structural isomer known as the enol form (part alkene, part alcohol).

This behavior is known as keto-enol tautomerism.

The keto and enol forms are not resonance forms!   [Note 1 ]  They are structural isomers that can interconvert. This property is called tautomerism. Keto-enol tautomerism is the most commonly-encountered type of tautomerism, although there are others. [Note 2]

That means that the keto and enol forms, if separated, have very different physical and chemical properties, which will become important in a moment. [Note 3] 

We’ve actually encountered keto-enol tautomerism before, if only briefly. You may recall that alkynes treated with HgSO₄ and water give an enol intermediate, which then tautomerizes to a ketone. Similarly, alkynes undergo hydroboration-oxidation to give enols which can tautomerize to aldehydes. [See Hydroboration and Oxymercuration of Alkynes]

Your instructor may have glossed over the tautomerism part at the time, saying something like, “you’ll learn more about this in Org 2.” Well, the moment has arrived.

3. Some Examples of Keto-Enol Tautomerism

Here are a few more examples of keto-enol tautomerism.

You might recall that at the very beginning I said that keto-enol tautomerism happens in some aldehydes and ketones. Why “some” but not all? Because it can’t happen in cases where there is no hydrogen on the alpha carbon. [Note 4]

These aldehydes and ketones are known as “non-enolizable” aldehydes and ketones.

4. Properties of Enols

Might this enol form be responsible for those “weird” reactions of aldehydes and ketones we were talking about?

[NARRATOR: You think???  Why else would he be bringing this up?]

A closer look at the enol form may help us to understand some of its properties.

In the enol form,  a lone pair on oxygen is in conjugation with the C-C pi bond. [See post: Conjugation and resonance]

At first glance, we might think that having an electronegative oxygen attached to an alkene might make it more electron-poor, since it’s sucking electron density away through that C–O sigma bond.

However,  we can also draw a resonance form where a lone pair on oxygen attached to the pi-bond can donate a pair of electrons towards the ring, forming a new C–O pi bond.

In the process, the pair of electrons in the C–C pi bond moves over to become a lone pair on the adjacent carbon.

This is known as pi-donation and has the effect of making an attached pi bond more electron-rich and therefore more nucleophilic. [See post: Pi-Donation]

On balance, the pi-donation effect from oxygen tends to be much stronger than the inductive effect arising from electronegativity.

You may recall that OH is a strongly activating group for electrophilic aromatic substitution. [See: Activating and Deactivating Groups] You might even recall drawing resonance forms which show the electron density moving towards the ortho- and para- positions.

If you remember that, I’ve got good news. Enols behave in exactly the same way!

An attached -OH group has the effect of making the carbon adjacent to the C–O bond more electron rich and therefore more nucleophilic!

via GIPHY

This helps to untangle the mystery of why the new C–Br bond was formed on the “alpha carbon” in our example above.  In an enol, the alpha-carbon is strongly nucleophilic!

5. Keto-Enol Tautomerism: Mechanisms

This begs a question. If the keto and enol tautomers aren’t resonance forms, then there must be some process by which they interconvert. So how does this happen?

It might be instructive to show how to do it the wrong way first.

Click to Flip

Hold on. It looks so gosh-darn tempting to draw it all up as one concerted process. How can this possibly be wrong?

It’s wrong for the same reason that you can’t scratch your left elbow with your left hand. They are just too far apart to touch.

What’s needed is a “helper” molecule to transport the proton from one part of the molecule to the other. Water serves nicely. [Note 5]

In addition to water, interconversion of keto and enol tautomers is greatly assisted by the presence of acid or base. [Note 6] 

In the presence of acid, the carbonyl oxygen is protonated (Step 1, form O-H). Then, in the slow step, the alpha-carbon is deprotonated to give the enol (Step 2, break C-H, form C-C (pi), break C-O (pi). ). This gives us the enol.

For the acid-catalyzed conversion of enol to keto, hover here and an image will pop up or click on this link.

Keto-enol interconversion is also assisted by base. Here, deprotonation of the alpha carbon (Step 1, break C-H, form C-C (pi), break C-O pi) is the slow step, and protonation of the oxygen (Step 2, form O-H) is the fast step.

(The conjugate base of an enol is called an enolate. We’ll have a lot more to say about enolates shortly).

For the reverse process (base-catalyzed conversion of enol to keto) hover here or click on the link.

6. Four Factors That Affect Keto-Enol Equilibria

For most aldehydes and ketones, equilibrium strongly favors the keto form,  often by a factor of 10⁴ or more. This mostly has to do with the difference in bond strengths (C-O pi is a stronger bond than C–C pi , for more details see Note 7 ). The enol form is even less favored for carboxylic acids and esters. [Note 8]

That said, there are at least 4 key factors that can significantly influence the keto: enol ratio. It’s important to know them as they make good exam question material.

They are, in order from weakest to strongest influence:

• substitution –  enols are a type of alkene, and substituted alkenes are more stable
• conjugation – conjugation of the enol pi-bond with a neighboring pi system is stabilizing
• hydrogen bonding – intramolecular hydrogen bonding can stabilize the enol form
• aromaticity – if the enol is part of an aromatic ring, expect the enol form to dominate

Factor #1: Substitution

Remember Zaitsev’s rule? Eliminations tend to favor formation of the more substituted alkene, because they are more thermodynamically stable? [See: Stability of Alkenes]

The same trend applies to enols!

For example, which enol do you think will be favored at equilibrium?

Click to Flip

The more substituted an enol is, the more thermodynamically stable it tends to be. The difference often isn’t huge (1-2 kcal/mol) but recall that even a difference of 1 kcal/mol means an equilibrium ratio of about 80:20. [Calculated here]

(This is actually quite an important trend to be aware of, because you may soon be contrasting kinetic versus thermodynamic enolates.)

Compare the stability of the following aldehyde enols:

Click to Flip

Factor #2: Conjugation / Resonance

Click to Flip

π bonds are a little like Cheerios in milk: it’s more stable for them to connect together rather than hang out in isolation.  

Which leads us to ask:  which of these two ketones will have a greater preference for the enol form?

Click to Flip

The more favorable enol form will be that which allows conjugation with a neighboring pi system.

Factor #3: Hydrogen Bonding

This is an interesting one. Enols have an O-H bond, which is highly polarized; the hydrogen bears a partially positive charge and is capable of hydrogen bonding. If  a Lewis base (e.g. the oxygen of a carbonyl) is present nearby, an intramolecular hydrogen bond can result, which will stabilize the enol form.

Click to Flip

In some cases the proportion of enol form can be equal to or even greater than the amount of keto tautomer in solution.

(An NMR spectrum of 2,4-pentanedione, for example shows about a 1:1 mixture of the keto and enol tautomers) hover for image

[There is some dependence on solvent here. See Note 9]

Factor #4 : Aromaticity

The most powerful driving force to favor an enol tautomer is aromaticity.

Contrast these two ketones. Which would favor the enol tautomer more?

Click to Flip

Hopefully you can see that the aromaticity of phenol (with its resonance energy of >20 kcal/mol) greatly favors the enol form here. The keto tautomer can’t be detected in solution!

11. Revisiting An “Un-Ketone-Like” Reaction With Br₂

After all we’ve explored, I think we’re ready to go back at this “weird” reaction of a ketone one more time.

How might this reaction work?

We’re adding acid and forming a new bond to an electrophile at the alpha carbon. I think we can safely say that the first step is acid-catalyzed keto-enol tautomerism.

This gives us a nucleophilic enol, which can then react with Br₂ to form a new C-Br bond. Deprotonation then gives the neutral alpha-bromo ketone.

12. Summary and Conclusion

So what have we learned?

• Aldehydes and ketones that have a proton on the alpha carbon can participate in keto-enol tautomerism, where an equilibrium exists between two constitutional isomers – the keto and enol forms. Constitutional isomers that are in equilibrium are called “tautomers”.
• The enol tautomer is nucleophilic on the carbon adjacent to the C-OH bond (the “alpha-carbon”) and undergoes reactions with electrophiles
• Keto-enol tautomerism can be catalyzed with acid or base.
• Generally the keto tautomer is favored at equilibrium.
• Many structural factors that stabilize alkenes  such as substitution, conjugation, and participation in an aromatic ring may also help to stabilize the enol form. Also, the enol tautomer can be stabilized by a neighboring hydrogen bond acceptor (such as a ketone)

Notes

Some representative keto-enol ratios (collected from Carey & Sundberg and March’s Advanced Organic Chemistry)

Note 1 – Recall that resonance forms are not in equilibrium with each other – they are just ways of representing the distribution of pi-electrons in a molecule whose true structure is best thought of as a weighted hybrid of resonance forms. [See: Resonance Structures] 

In the transformation of a keto- to an enol- form, a C-H sigma bond and a C-O pi bond breaks, and an O-H sigma bond and C-C pi bond forms. This is another reason why these can’t be resonance forms – recall that we can’t break sigma bonds to interconvert resonance structures.

Note 2 – Keto-enol tautomerism is one of the most prominent types of tautomerism. Ring-chain tautomerism, which often happens in sugars, is another. [See: Ring Chain Tautomerism In Sugars]. A third is valence tautomerism, which occurs in certain molecules that lack a fixed structure.

Note 3 – The keto and enol forms have been separated for certain ketones. The first example was ethyl acetoacetate.

Note 4 – It is possible for certain alpha, beta unsaturated ketones that appear “non-enolizable” to be enolized at the gamma position.

Click to Flip

Note 5- It is possible in certain cases to prepare “pure” enols, uncontaminated with their keto tautomer. In these cases researchers are extremely careful to exclude water, which may act as a “proton shuttle” to interconvert the enol and keto forms.

Note 6 – See Ref 1 for more details on how acid accelerates the rate of keto-enol tautomerism.

Note 7 – A quick tabulation of the bonds that form and break using average values for their bond dissociation energies reveals at least a 10 kcal/mol difference between the keto and enol tautomers in most cases favoring the keto. See this quiz for an example.

Note 8. The proportion of enol tautomer in a solution of ethyl acetate was found to be less than one part in 10 million – several orders of magnitude below that found for ketones.

Note 9. There is some solvent dependence on the keto:enol ratio. In a non hydrogen-bonding solvent (like carbon tetrachloride) the enol:keto ratio for ethyl acetoacetate was measured to be around 98:2 . In a hydrogen-bonding solvent like water, however, there is more “competition” for accepting a hydrogen bond, and the enol: keto ratio drops to about 1 : 4. (See Ref 5)

Quiz Yourself!

Become a  MOC member to see the clickable quiz with answers on the back.  

Become a  MOC member to see the clickable quiz with answers on the back.

Become a  MOC member to see the clickable quiz with answers on the back.

Become a  MOC member to see the clickable quiz with answers on the back.

Become a  MOC member to see the clickable quiz with answers on the back.

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(Advanced) References and Further Reading

For a historical perspective on tautomerism, Structure and Mechanism in Organic Chemistry by C. K. Ingold is helpful.

For example, Ingold points to the coining of the term “tautomerism” by Laar (Berichte, 1885, 648 and Berichte, 1886, 19, 730) but mentions that Laar did not believe the keto and enol forms were separate species. [https://chemistry-europe.onlinelibrary.wiley.com/doi/abs/10.1002/cber.188601901165]

This was later conculsively established by separation of ethyl acetoacetonate into its ketonic and enolic components, which involved crystallizing the keto form at –80°C and showing it could be converted to the keto form (Berichte 1911, 44, 1138). Distillation of the pure (and more volatile) enol form was reported later (Berichte, 1921, 54, 579).

1. Mechanism of acid-catalyzed enolization of ketones
Gustav E. Lienhard and Tung-Chia Wang
Journal of the American Chemical Society 1969 91 (5), 1146-1153
DOI: 10.1021/ja01033a019

Study on the keto-enol equilibria of cyclohexanone  supports that the rate limiting step for acid-catalyzed enol → keto tautomerism is protonation of the beta-carbon (based on similarity of rates to those for hydrolysis of enol ethers). Rate-determining step for keto → enol tautomerism is deprotonation of carbon.

2. Tautomeric equilibria in acetoacetic acid
Karen D. Grande and Stuart M. Rosenfeld
The Journal of Organic Chemistry 1980 45 (9), 1626-1628
DOI: 10.1021/jo01297a017

From the abstract: “Tautomeric equilibria in acetoacetic acid have been examined by ¹H NMR and found to be strongly solvent dependent. Values for enol tautomer range from less than 2% in D₂O to 49% in CCl₄. Chemical shift data suggest that the enol tautomer is internally hydrogen bonded in the less polar solvents and that internal hydrogen bonding is unimportant for the keto tautomer.”

3. Generation of simple enols in aqueous solution from alkali metal enolates. Some chemistry of isobutyraldehyde enol
Y. Chiang, A. J. Kresge, and P. A. Walsh
Journal of the American Chemical Society 1986 108 (20), 6314-6320
DOI: 10.1021/ja00280a032

Gives an estimate for the equilibrium constant for keto-enol interconversion of isobutyraldehyde as K= 1.37 × 10^-4 , and estimates the pK_a of the enol form as 11.63 and the pK_a of the keto form as 15.49.

4. Kinetics and thermodynamics of keto-enol tautomerism of simple carbonyl compounds: an approach based on a kinetic study of halogenation at low halogen concentrations
Jacques Emile Dubois, Mohiedine El-Alaoui, and Jean Toullec
Journal of the American Chemical Society 1981 103 (18), 5393-5401
DOI: 10.1021/ja00408a020

Contains rate and equilibrium constants for the enolization of various ketones in aqueous solution (see Table 2, p. 5396), particularly cyclic alkanones and substituted acetophenones.

5. Solvent effects on keto-enol equilibria: tests of quantitative models
Sander G. Mills and Peter Beak
The Journal of Organic Chemistry 1985 50 (8), 1216-1224
DOI: 10.1021/jo00208a014

Interesting study on several different keto-enol tautomer pairs. Interesting note from the abstract: “In general, for the isomer pairs in which the enol cannot form an internal hydrogen bond, the equilibria appear to be controlled almost completely by the hydrogen-bonding basicity of the solvent” (emphasis mine).

6. Microwave spectroscopic study of malonaldehyde (3-hydroxy-2-propenal). 2. Structure, dipole moment, and tunneling
Steven L. Baughcum, Richard W. Duerst, Walter F. Rowe, Zuzana Smith, and E. Bright Wilson
Journal of the American Chemical Society 1981 103 (21), 6296-6303
DOI: 10.1021/ja00411a005

Malonaldehyde is the simplest beta-carbonyl aldehyde. This study uses microwave spectroscopy to ascertain the bond lengths in the internal hydrogen-bonded structure of the malonaldehyde enol tautomer.

7. Gas-phase acidities and heats of formation of 2,4- and 2,5-cyclohexadien-1-one, the keto tautomers of phenol
Christopher S. Shiner, Paul E. Vorndam, and Steven R. Kass
Journal of the American Chemical Society 1986 108 (19), 5699-5701
DOI: 10.1021/ja00279a006

The authors made the two keto tautomers of phenol in the gas phase (through a retro Diels-Alder reaction!) and studied their acidities and heats of formation. The heat of formation of the [2H]-tautomer of phenol is ca. 6 kcal/mol higher than that of phenol, whereas the heat of formation of the [4H]-tautomer is about 10 kcal/mol higher than phenol.

8. Superacidity and Superelectrophilicity of BF3−Carbonyl Complexes
Jianhua Ren, Christopher J. Cramer, and Robert R. Squires
Journal of the American Chemical Society 1999 121 (11), 2633-2634
DOI: 10.1021/ja9840899
The authors studied the acidity of acetaldehyde (pKa 17) complexed to boron trifluoride (BF3) and found that BF3 has an astounding impact on the acidity of acetaldehyde. Most studies were done in the gas phase, but a solution phase calculation estimates the acidity of acetaldehyde coordinated to BF3 as having a pKa of -7 ; a 24 order of magnitude increase in acidity!

Yes, there are a lot of reactions of carboxylic acid derivatives to learn!  In this article we’ll explore what is hands-down the most important pathway: nucleophilic acyl substitution.

There is good news, though:

• with negatively charged nucleophiles nucleophilic acyl substitution tends to follow a simple two-step mechanism (addition-elimination)
• furthermore, for all intents and purposes, NAS reactions behave a lot like acid-base reactions. If you recall that “stronger acid plus stronger base gives weaker acid plus weaker base”, (otherwise known as The Principle of Acid-Base Mediocrity)  you’ll be well on your way to being able to predict whether or not a given nucleophilic acyl substitution reaction will happen.

Table of Contents

1. Nucleophilic Acyl Substitution (NAS)
2. Three Nucleophilic Acyl Substitution Reactions (That Work Pretty Well)
3. The Addition-Elimination Mechanism for Nucleophilic Acyl Substitution With a Negatively Charged Nucleophile
4. Some Nucleophilic Acyl Substitutions That DON’T Work
5. NAS Reactions Are Favored When The Leaving Group Is A Weaker Base Than The Nucleophile
6. Some Quick Exercises
7. Carboxylic Acids… Are Acids!
8. Saponification
9. Intramolecular Nucleophilic Acyl Substitution
10. Grignard Reagents and LiAlH₄ Add Twice
11. Neutral Nucleophiles and Acid Catalysis
12. Summary
13. Notes
14. Quiz Yourself!
15. (Advanced) References and Further Reading

1. Nucleophilic Acyl Substitution

Nucleophilic acyl substitution is a reaction where a nucleophile forms a new bond with the carbonyl carbon of an acyl group with accompanying breakage of a bond between the carbonyl carbon and a leaving group.

This is classified as a substitution reaction because we are forming and breaking a bond on the same carbon. A carbon-nucleophile bond forms, and a carbon-leaving group bond breaks.

Although the mechanism is different, nucleophilic acyl substitution superficially resembles substitution reactions we’ve seen before such as nucleophilic aliphatic substitution and nucleophilic aromatic substitution. 

Hover to see other families of nucleophilic substitution reactions or click on the link.

In this article we will specifically cover examples of nucleophilic acyl substitution reactions with negatively charged nucleophiles. In a subsequent article we’ll cover cases where an acid catalyst is employed for nucleophilic acyl substitution, such as the Fischer esterification, acidic hydrolysis of esters, and many other examples.

(For neutral nucleophiles, as well as nucleophilic substitution under acidic conditions, see Addition-Elimination of Neutral Nucleophiles)

2. Three Nucleophilic Acyl Substitution Reactions (That Work Pretty Well)

Here are three classic examples of nucleophilic acyl substitution reactions that work well.

In the first, acid chlorides react with carboxylates (the conjugate base of carboxylic acids, acting as a nucleophile here) to give an anhydride. A new C-O bond is formed and a C-Cl bond is broken.

In the second example, we treat an anhydride with an alkoxide and form a new ester. A C-O bond is formed and a C-O bond is broken. In this case the carboxylate is our leaving group.

The third example involves the use of the bulky hydride source di-isobutyl aluminum hydride (DIBAL-H) to an ester at low temperature to form an aldehyde. A new C-H bond is formed and a C-O bond breaks.

At low temperature (–80° or so) this works well; higher temperatures result in over-reduction.  [Note 1]

3. The Addition-Elimination Mechanism for Nucleophilic Acyl Substitution With a Negatively Charged Nucleophile

So how might these reactions work? They’re not S_N2-type reactions, since S_N2 really only is effective on sp³ hybridized carbons. Nor have they been determined to be S_N1-type reactions where the leaving group leaves first and then the nucleophile attacks (although this describes the Friedel-Crafts acylation pretty well!)

Instead, experiments provide strong evidence for a two-step addition-elimination mechanism that proceeds through a tetrahedral intermediate. [Note 2 ]

The first step should be very familiar from the reactions of aldehydes and ketones. [See – The simple two-step pattern for 7 key reactions of aldehydes and ketones]

The carbon of a C=O group is electrophilic and reacts readily with the negatively charged nucleophile through an addition mechanism [Step 1, form C-Nu, break C-O (pi) ]

Now what? Through a reversal of the addition mechanism, which goes by the name elimination, the tetrahedral intermediate could revert back to starting materials, regenerating the starting acyl group.

Or, if we have a sufficiently good leaving group, elimination could result in re-formation of the C-O pi bond with loss of a leaving group.

[Step 2, form C-O (pi), break C-LG]

The result is that a substitution has occurred.

This two-step mechanism is known as addition-elimination and it is extremely common in the chemistry of carboxylic acid derivatives (i.e. acyl groups). In cases where a neutral nucleophile is used, or acid catalysts are present, some intervening proton transfer steps can lengthen the mechanism (e.g. PADPED) but the core process is still addition-elimination. (For more on carbonyl elimination – see Addition-Elimination)

4. Nucleophilic Acyl Substitutions That Don’t Work

So far, so good, right? We just add our nucleophile to the acyl group.  The nucleophile adds, the leaving group leaves, and we get our new substitution product. Simple!

Here’s two more examples of nucleophilic acyl substitution reactions.

You may have noticed that the reactions shown here are the exact opposite of the nucleophilic substitution reactions that “worked” in the section above, except we’ve just switched the identities of the nucleophile and the leaving group.

(Recall that a leaving group is just a nucleophile acting in reverse, and vice-versa). 

The trouble is,  although you can draw these up “on paper” as a nucleophilic acyl substitution reactions,  in the real world you could wait around for continental drift to make Pangaea II and they still wouldn’t happen.

So why is nucleophilic acyl substitution favored in one direction and not in the other?

5. NAS Reactions Are Favored When The Leaving Group Is A Weaker Base Than The Nucleophile

Let’s take a very brief trip back to Org 1, where we learned the Golden Rule of Acid-Base Reactions, otherwise known as the Principle of Acid-Base Mediocrity: the favored direction of an acid-base reaction is one where a stronger acid reacts with a stronger base to give a weaker acid and a weaker base. (See post: How to Use a pKa Table)

We intuitively know by this point that adding table salt to water will not result in a raging cloud of HCl gas and caustic sodium hydroxide, whereas adding concentrated HCl to pellets of NaOH is not something a sober person does without a lot of safety precautions.

So when trying to consider which direction is favored in a nucleophilic acyl substitution reaction, expect that it will favor formation of the weaker base.

Just like water flows downhill, NAS reactions will “flow” such as to give the more stable anion (weaker base).

(It might also help to think of  “weaker base” as another way of saying, “a better-stabilized electron pair”. )

It also helps us understand why nucleophilic acyl substitution reactions don’t work with aldehydes and ketones; it would result in a hydride ion (H-) or carbanion (R  -), both of which are very strong bases. [Note 3]

It might be helpful to put this into the form of a chart, if we chart pK_a (of the conjugate acid) versus nucleophile/leaving group.

Reactions that involve a “downhill” flow from nucleophile to leaving group are favored, whereas those that require an “uphill” conversion of a weaker base to a stronger base will be disfavored.

Another way of expressing the same idea is to make a chart like the following, where the weakest base (Cl-) is incapable of performing any nucleophilic acyl substitution reactions which might result in a stronger base:

For reasons that will soon become clear, HO(-) has been omitted from this chart.  [Note 4]

Hover on this link to see a simplified reaction energy diagram or click on this link.

6. Some Quick Exercises

To let this concept settle in, why not quiz yourself with some quick exercises?

First up, thioesters.

Click to Flip

[Note 5]

Second, an example of converting an ester into another ester, a reaction called, “transesterification”.

Click to Flip

You might ask – what if there isn’t a big pK_a difference between the nucleophile and leaving group? What if we are starting with the ethyl ester, for example, and want to convert it into the methyl ester?

Since the pK_a‘s are similar, one way to do it would be to “flood the zone” with nucleophile (i.e. use a large excess) to drive the equilibrium towards the desired product.

I’m not saying this is the best way (converting to an acid halide or anhydride followed by treatment with an alcohol would be superior) but it’s still workable.

7. Carboxylic Acids…. Are Acids!

Now that we have a handle on the key factor that affects nucleophilic acyl substitutions, let’s look at an example that should seem straightforward… but isn’t.

NaOH is the conjugate base of water (pK_a 14) and NaOCH₃ is the conjugate base of CH₃OH (pK_a 16).

So given everything we know, the reaction should slightly favor the product.. right?

Actually… no!

That’s because there’s an extra factor we need to consider first. Carboxylic acids are acids, and acid-base reactions are fast. (See: Acid-Base Reactions Are Fast)

So the first thing that happens is not addition, but deprotonation of the carboxylic acid to give a carboxylate.

If nucleophilic acyl substitution were to happen at this point, it would have to add to this carboxylate which would make a di-anion with two negative charges on the same molecule (if that sounds unstable –  it is!!).

Furthermore, the leaving group to make the ester would not be HO(-).

It would be O (^2-) !

With one notable exception [Note 6] , that won’t happen.

Note that carboxylic acids can be converted to esters, but that the reaction requires acid (e.g. the Fischer esterification). In this article we’re focusing on negatively charged nucleophiles.

8. Saponification of Esters With Base

If adding RO(–) to a carboxylic acid just results in deprotonation, what about the reverse: adding a hydroxide ion to an ester?

This is a well-known process known as saponification (or just, “basic hydrolysis”) and works extremely well.

The substitution portion of the reaction proceeds through the familiar addition–elimination process. This results in a carboxylic acid. However, since carboxylic acids are acids, and the reaction occurs under basic conditions, the resulting acid will rapidly be deprotonated by hydroxide to give the carboxylate.

Therefore to obtain the neutral carboxylic acid at the end of the reaction, one has to add acid during the workup step.

The name “saponification” comes from the classic use of this reaction in making soaps (carboxylates of long-chain fatty acids) by treating fats (which are esters of glycerol with long-chain acids)  with lye (a generic term for alkali metal hydroxides).

9. Intramolecular NAS

It’s always helpful to think about the intramolecular variant of any new reaction you learn. Nucleophilic acyl substitution is no exception!

If a five- or six-membered ring can be formed through nucleophilic acyl substitution,  it generally will. In this example, treatment with the strong base sodium hydride (NaH) results in an alkoxide that forms a new ring, displacing methoxide [CH₃O (–) ]

hover on this link to see the mechanism or click on this link.

Here’s a fun example.  What is happening here?

Click to Flip

(Hint: deprotonation isn’t the only way of making an alkoxide)

10.When The Nucleophile Adds Twice – Grignards and LiAlH₄

Grignard reagents and strong reducing agents like LiAlH₄ tend to add to carboxylic acid derivatives twice.

Grignard reagents (and organolithium reagents) give tertiary alcohols from esters, acid halides, and anhydrides.

Strong reducing agents (e.g. LiAlH₄) give primary alcohols from esters, acid halides, and anhydrides.

So how do these “double addition” reactions work?

Addition of the nucleophile R(-) or H(-) followed by elimination of the leaving group initially give a ketone (with Grignards) or an aldehyde (with LiAlH₄).

If the elimination process is fast relative to addition, then the aldehyde or ketone will be formed in the presence of the strong nucleophile before all the starting ester (or other carboxylic acid derivative) is consumed.

Since ketones (and aldehydes) are better electrophiles than esters (i.e. they react faster with nucleophiles), a situation results that I sometimes affectionately call a Cookie Monster Reaction:  the reaction forms a product that is more reactive towards the nucleophile than the starting material.

via GIPHY

 (because the Cookie Monster can’t stop after just one cookie).

This results in a second addition reaction occurring, which becomes an alcohol after addition of acid. (Even if one equivalent of Grignard is used, the product is still a tertiary alcohol and unreacted ester.)

So the final sequence ends up being addition–elimination–addition followed by a final workup (protonation).

Here is a specific example:

Click to Flip

It is possible to get addition to acid halides to stop at the ketone stage by using organocuprates (Gilman reagents) which are not nearly as ravenous as Grignard or organolithium reagents.

11. Neutral Nucleophiles and Acid Catalysis

This whole article has been confined to the reactions of negatively charged nucleophiles, such as Grignards, hydrides, amide bases, alkoxides, carboxylates, and halides.

We’ve seen that under these conditions,  the direction of the equilibrium is almost entirely dictated by the stability of the leaving group. This makes it extremely difficult to perform the substitution reactions of amides, for example, since R₂N(-) is such a strong base.

Is there another way?

Yes.  We should expect that adding acid would help with performing nucleophilic acyl substitutions, since the conjugate acid is always a better leaving group.

And that is indeed the case. See Addition-Elimination Reactions With Neutral Nucleophiles (Including Acid Catalysis)

The tradeoff here is that acidic conditions are not compatible with extremely basic nucleophiles such as Grignards and hydride reagents.

We’ll explore nucleophilic acyl substitution under neutral and acidic conditions in the next article.

12. Summary

Nucleophilic acyl substitution reactions with negatively charged nucleophiles proceeds through an addition-elimination mechanism.

• In the first step (addition)  the nucleophile attacks the carbonyl carbon, leading to a tetrahedral intermediate.
• In the second step (elimination) the carbon-oxygen pi bond is re-formed and a leaving group is displaced.
• Since we are forming and breaking a bond on the same carbon this is classified as a substitution reaction.
• The forward and backward reactions can be treated as an equilibrium. Like an acid-base reaction,  the equilibrium will favor the direction which produces the weakest base.
• Carboxylic acids do not generally undergo nucleophilic substitution under basic conditions since they are so easily deprotonated and the resulting leaving group (O2-) is highly basic.
• However, esters do undergo basic hydrolysis with hydroxide ions (“saponification”) to give carboxylic acids (after protonation).
• Intramolecular nucleophilic acyl substitution reactions can occur, especially when five- and six-membered rings can be formed.

Notes

[Note: This article significantly updates a previous article from 2011 entitled, “Simplifying the Reactions Of Carboxylic Acid Reactions, Part 1].

Note 1 – Technically the first step of DIBAL-H reduction is coordination of the carbonyl oxygen to the Lewis-acidic aluminum, followed by delivery of hydride. But for our purposes we can just look at it as addition of H(-) to the carbonyl carbon.

Note 2.  Let’s put the evidence for the mechanism in the form of two quizzes.

In the first experiment [Ref], an ester was treated with hydroxide containing isotopically labelled oxygen (¹⁸O versus “normal” ¹⁶O ). None of the recovered alcohol contained the isotopic label.

Click to Flip

The conclusion to draw here is that the carbonyl-oxygen bond is broken, not the alkyl-oxygen bond (e.g. in some kind of S_N2 process). Similar experiments have been done with esters attached to chiral alcohols and the hydrolyzed alcohols were found to maintain their optical purity.

In another classic experiment [Ref] an ester was treated with ¹⁸O labelled hydroxide and the reaction was stopped before completion.

The recovered starting ester showed some incorporation of ¹⁸O on the carbonyl oxygen. What conclusion can be drawn from this?

Click to Flip

Reversible addition of ¹⁸O labelled HO(-) leads to a tetrahedral intermediate. Assuming that the ¹⁸O and ¹⁶O labelled oxygen have essentially identical reactivity (they do, within a few percent), proton transfer followed by elimination of ¹⁶OH would lead to a starting ester enriched in ¹⁸O.  No exchange with the -OR group should occur.

This was a key experiment that established the existence of the tetrahedral intermediate.

Note 3. One notable exception to this is the final step of the Haloform reaction (see: The Haloform Reaction) where addition-elimination with NaOH as nucleophile results in expulsion of (-)CI₃ as a leaving group. This only works because the electron-withdrawing halogens help to stabilize the resulting carbanion.

Note 4. With enough heat, it is possible to perform the basic hydrolysis of amides, which has the appearance of being “uphill” since NH₂(-) is a stronger base than HO(-). However, as soon as the carboxylic acid is formed, deprotonation results in the carboxylate, which renders the reaction essentially irreversible since displacement by NH₂(-) would have to give O(^2-)

Note 5. Nature employs nucleophilic acyl substitution reactions extensively, and thioesters are important “acyl transfer” reagents in biochemistry.  The sulfur group of  Coenzyme A serves as the leaving group with a variety of nucleophilic acyl substitution reactions.

Of particular importance is the mevalonic acid pathway which leads to the biosynthesis of terpenes and steroids.

Note 6. Although Grignard reagents will merely deprotonate carboxylic acids to give carboxylates, organolithium reagents will add to carboxylates to give a stable tetrahedral intermediate that breaks down to a ketone after acidic workup.

Quiz Yourself!

Become a  MOC member to see the clickable quiz with answers on the back.

Become a  MOC member to see the clickable quiz with answers on the back.

Become a  MOC member to see the clickable quiz with answers on the back.

Become a  MOC member to see the clickable quiz with answers on the back.

Become a  MOC member to see the clickable quiz with answers on the back.

Become a  MOC member to see the clickable quiz with answers on the back.

Become a  MOC member to see the clickable quiz with answers on the back.

Become a  MOC member to see the clickable quiz with answers on the back.

(Advanced) References and Further Reading

March’s Advanced Organic Chemistry, Carey & Sundberg Part A (Chapter 3) and B (Chapter 7) are good advanced sources on this subject. Structure and Mechanism in Organic Chemistry by C. K. Ingold is valuable for historical context.

1. On the mechanism of hydrolysis. The alkaline saponifications of amyl acetate
M. Polanyi and A. L. Szabo.
Trans. Faraday Soc., 1934, 30, 508-512
DOI 10.1039/TF9343000508
Early study on the mechanism of acyl substitution employed isotopically labelled oxygen to identify acyl-O bond fissure as the dominant mechanism in ester hydrolysis.

2. Oxygen Exchange as Evidence for the Existence of an Intermediate in Ester Hydrolysis
Myron L. Bender
Journal of the American Chemical Society 1951 73 (4), 1626-1629
DOI: 10.1021/ja01148a063
Classic experiment for establishing the existence of a tetrahedral intermediate in nucleophilic acyl substitution. Bender performed the partial acidic hydrolysis of ethyl benzoate in ¹⁸O enriched water and recovered ¹⁸O-enriched ethyl benzoate. 

3. Chemical effects of steric strains—XIII: Kinetics of the reaction of sodium borohydride with carbonyl groups—a convenient tool for investigating the reactivities of aldehydes and ketones
H.C. Brown, O.H. Wheeler, K. Ichikawa
Tetrahedron, 1 (3), 1957, 214-220.
DOI: 10.1016/0040-4020(57)88041-7
Useful study on the rate constants for the addition of NaBH₄ to various aldehydes and ketones.

4. Mechanisms of Catalysis of Nucleophilic Reactions of Carboxylic Acid Derivatives.
Myron L. Bender
Chemical Reviews 1960 60 (1), 53-113
DOI: 10.1021/cr60203a005
Review on the mechanisms of nucleophilic acyl substitution.

• The reverse of nucleophilic addition to the C=O bond (giving a tetrahedral intermediate) is elimination of a leaving group from the tetrahedral intermediate to re-form the C=O bond.
• This is called, “1,2-elimination” or sometimes just “elimination” and is a key mechanism of the carbonyl functional group.
• Confusingly, eliminations of alkyl halides are also called 1,2-eliminations. See The E2 Mechanism or E1cB or the E1 Reaction for more on these.)
• Eliminations are generally favored when expulsion of a leaving group results in formation of a weaker base
• Acid catalysis is extremely helpful in promoting nucleophilic acyl substitution of carboxylic acids and amides, since elimination results in the much better leaving groups H₂O and NH₃  (as opposed to O^2- and (-)NH₂ under basic conditions)

Table of Contents

1. Elimination: The Second-Most Important Mechanism of the Carbonyl Group
2. Elimination in Carboxylic Acid Derivatives
3. What About Neutral Nucleophiles In Nucleophilic Acyl Substitution?
4. Acid Catalysis For Nucleophilic Acyl Substitution With Neutral Nucleophiles
5. Examples of Acid Catalysis
6. Summary
7. Notes
8. Quiz Yourself!
9. (Advanced) References and Further Reading

1. The Second-Most Important Mechanism Of The Carbonyl Group

Elimination (1,2-elimination) is an extremely important reaction mechanism of the carbonyl (C=O) group, which is present in such functional groups as aldehydes, ketones, carboxylic acids and carboxylic acid derivatives.

It’s the second most important reaction of carbonyls, after addition. In fact, it is the exact reverse of this nucleophilic addition reaction. (See post: Nucleophilic Addition)

For a reminder of the nucleophilic addition mechanism, hover here or click this link.

Elimination reactions tend to be favorable when the leaving group is a weaker base than the nucleophile. (See: What Makes A Good Leaving Group)

That’s why halide ions don’t successfully perform nucleophilic addition to aldehydes and ketones. The reaction is going uphill in terms of basicity. Since Cl(-) is a weaker base than O(-), elimination is much more favorable than addition. (See post: How to Use a pKa Table)

Elimination is unfavorable when it results in a stronger base being formed from a weaker base. This is why reduction of aldehydes and ketones with hydride reducing agents like NaBH₄ is irreversible (See: Sodium Borohydride).

The forward reaction for hydride reduction of aldehydes and ketones results in a stronger base (hydride, conjugate base of H₂, pK_a about 35) being converted to a weaker base (alkoxide, conjugate base of alcohol, pK_a about 16-18).

The opposite reaction (elimination) is about 20 pK_a units more disfavored from an acid-base perspective (See post: How to Use a pKa Table)

In general, a good rule of thumb is that if the nucleophile/base and leaving group are separated by more than 8 pK_a units, the reaction can be considered to be irreversible.  (See: A Handy Rule of Thumb For Acid-Base Reactions)

(However, there are some examples of reactions where H(-) can act as a leaving group in a concerted process – See Note 1)

The Principle of Acid-Base Mediocrity (“stronger acid plus stronger base gives weaker acid plus weaker base) keeps coming up again and again!

2. Elimination In Carboxylic Acid Derivatives

Besides aldehydes and ketones, addition and elimination also occur in the reactions of carboxylic acid derivatives such as acid halides, acid anhydrides, esters, and amides.

When nucleophilic addition occurs to a carboxylic acid derivative, it forms a tetrahedral intermediate with two potential leaving groups. [ but not the O- Note 2]

• The first potential leaving group is the original nucleophile; elimination of this from the tetrahedral intermediate would just give us back our starting material (after all, leaving groups are really just nucleophiles acting in reverse).
• The other potential leaving group is the X of the carboxylic acid derivative (e.g. (–)Cl for acid chlorides, (–)OCOR for anhydrides, (–)OR for esters, (–)NR₂ , (–NHR), –(NH₂) for amides)

When the nucleophile is a stronger base than the leaving group X(-) of the carboxylic acid derivative, then we will end up swapping out the X group of the carboxylic acid derivative. This is nucleophilic acyl substitution. (See post: Nucleophilic Acyl Substitution With Anionic Nucleophiles)

With negatively charged nucleophiles, nucleophilic acyl substitution is easy to perform on acid halides and acid anhydrides, more difficult to perform on esters,  and essentially impossible to achieve with amides, since that would require loss of the very basic leaving group NH₂(-), conjugate base of an amine (pK_a 35-38)

Click to Flip

Carboxylic acids generally don’t undergo addition-elimination under basic conditions since they will be deprotonated by strong base to give carboxylates, and the resulting leaving group would have to be the very strong base O(^2-)  (covered in more detail in this post: Transesterification)

3. What About Neutral Nucleophiles?

It’s not absolutely required to use basic nucleophiles for nucleophilic aromatic substitution, however.

Neutral nucleophiles are perfectly capable of performing some nucleophilic acyl substitutions.

One prominent case is that of acid halides and anhydrides, which are very sensitive to the presence of water.

Both of these functional groups can be hydrolyzed with water to give carboxylic acids.

After addition and proton transfer, the key step is elimination of the halide ion, resulting in formation of the carboxylic acid. This still follows the Principle of Acid-Base Mediocrity, since Cl(-) is a weaker base than H₂O (conjugate base of H₃O+, pK_a 0) . Likewise, water is at least comparable in base strength to carboxylic acids.

To see a mechanism for hydrolysis of acid anhydrides with water, hover here or click this link.

These reactions also work well with amines. The reactions between acid halides and amines are some of the best ways for making amides; this reaction is sometimes known as the Schotten-Baumann reaction (See Synthesis of Amides From Acid Halides).

Generally, at least two equivalents of amine are used, since one equivalent of HCl is generated. (If only one equivalent of amine were used, the reaction would not proceed to completion since the amine nucleophile would be protonated to give (non-nucleophilic) RNH₃(+). )

To see a mechanism for the formation of amides from acid halides,  hover here or click this link.

4. Acidic Conditions – Elimination Of The Conjugate Acid (A Better Leaving Group)

Addition of neutral nucleophiles to carboxylic acids has its limits, however. Water can displace weakly basic halides and carboxylates, but can’t displace the much stronger bases RO(-) or R₂N(-). (Amines, if heated might displace esters, but it requires a lot of heat – See Synthesis of Amides).

Aside from acid halides and anhydrides, most other carboxylic acids are inert under neutral conditions.

However, when an acid catalyst is added, it’s a different story altogether. It opens up a whole different set of reactions that don’t  happen otherwise.

We’ve previously seen an example of this in the synthesis of acetals (See post: Acetals and Hemiacetals) where elimination of one equivalent of ROH from a hemiacetal required an acid catalyst. (In the absence of acid, no elimination happens!).

For a reminder of the elimination mechanism in acetal synthesis, hover here or click this link.

This is also the case for carboxylic acid derivatives.

Recall that the whole reason amides and carboxylic acids are unreactive under basic and neutral conditions is because NH₂(-) and O(-) are such strong bases and therefore poor leaving groups.

But just imagine for a moment that we could run these reactions under acidic conditions.

Their leaving groups would no longer be NH₂(^–) and O(^2-) ; they could be their protonated cousins NH₃ and H₂O , which are much weaker bases and therefore much better leaving groups.

Suddenly, substitution reactions of carboxylic acids and esters becomes a plausible reaction!

Compare the leaving groups in these two reactions

Click to Flip

Or the leaving groups in these two reactions

Click to Flip

In each case, the protonated species is the better leaving group, because it is the weaker base.

The conjugate acid is a better leaving group, which is one important reason why acid catalysis helps with elimination reactions.

5. Examples Of Acid Catalysis In Nucleophilic Acyl Substitution

Here are two  examples of using acid as a catalyst in nucleophilic acyl substitution reactions, and how acid assists with loss of a leaving group.

A carboxylic acid can be converted into an ester under acidic conditions. This is the Fischer Esterification (See post: Fischer Esterification).

In the key step, a molecule of H₂O is eliminated from the tetrahedral intermediate, resulting in a protonated ester.

Similarly, amides can be converted to a carboxylic acid under acidic conditions. This is called acidic amide hydrolysis (See post: Amide Hydrolysis)

Likewise, the key step is elimination of the weak base NH₃ from the tetrahedral intermediate. In the absence of acid, no substitution occurs.

One limitation of acid catalysis for elimination is that we’re limited to using nucleophiles that aren’t irreversibly destroyed by strong acid.

This means that strongly basic nucleophiles like  NH₂(-), HO(-), RO(-), hydrides and Grignards are incompatible with acidic conditions, since acid-bases reactions are generally much faster than any reactions at carbon. (See post: Acid-Base Reactions are Fast),

6. Summary – Conclusion

• If nucleophilic addition is the most important reaction mechanism of carbonyls, then its opposite – elimination – ranks as the second-most important mechanism.
• It’s a key mechanism in nucleophilic acyl substitution reactions, in addition to its role in formation of acetals and imines that we saw previously in the chapter on aldehydes and ketones.
• The favorability of elimination is determined by the basicity of the leaving group. Good leaving groups are weak bases.
• Poor leaving groups can be transformed into good leaving groups through the addition of acid, since the conjugate acid of any species is a weaker base and therefore a better leaving group.
• Addition of acid is essential in the nucleophilic substitution reactions of carboxylic acids and amides such as the Fischer esterification and acidic amide hydrolysis.

Notes

Note 1. One example where hydride does leave in an elimination-type process is in the Canizarro reaction of aldehydes. The key step is transfer of a hydride from a (deprotonated) aldehyde hemiacetal to a second equivalent of the same aldehyde, resulting in an alkoxide and a carboxylic acid (which quickly undergo an acid-base reaction).

However, this reaction requires extremely basic conditions and heat (as well as requiring non-enolizable aldehydes)

Note 2.  But not O- , since this would have to leave as O^2-

Quiz Yourself!

Click to Flip

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(Advanced) References and Further Reading

1. S_N2 Mechanism for Alcoholysis, Aminolysis, and Hydrolysis of Acetyl Chloride
   T. William Bentley, Gareth Llewellyn, and J. Anthony McAlister
   The Journal of Organic Chemistry 1996 61 (22), 7927-7932
   DOI: 10.1021/jo9609844
   Although addition-elimination reactions are assumed in most cases, there are situations where nucleophilic acyl substitution results from direct attack at a carbonyl carbon with a nucleophile, especially in highly polar solvents where the leaving group is easily ionized.
2. Computational Studies of Nucleophilic Substitution at Carbonyl Carbon:  the S_N2 Mechanism versus the Tetrahedral Intermediate in Organic Synthesis
   Joseph M. Fox, Olga Dmitrenko, Lian-an Liao, and Robert D. Bach
   The Journal of Organic Chemistry 2004 69 (21), 7317-7328
   DOI: 10.1021/jo049494z
3. Estimates of hydride ion stability in condensed systems: energy of formation and solvation in aqueous and polar-organic solvents
   
   Craig A. Kelly  and  David R. Rosseinsky
   Phys. Chem. Chem. Phys., 2001 ,3,  2086-2090
   DOI: 10.1039/B010092G
   Source of the pK_a of H₂ as 35 (in DMSO)Formation of amides from acid halides (and anhydrides)
4. BENZOYL PIPERIDINE
   Marvel, C. S.; Lazier, W. A.
   Org. Synth. 1929, 9, 16
   DOI: 10.15227/orgsyn.009.0016
   This procedure from Organic Syntheses, a source of independently tested and reproducible synthetic organic laboratory procedures, is a classic Schotten-Baumann amide synthesis.The original Schotten-Baumann papers:
5. Ueber die Oxydation des Piperidins
   Schotten, C.
   Ber. 1884, 17 (2), 2544-2547
   DOI: 10.1002/cber.188401702178
6. Ueber eine einfache Methode der Darstellung von Benzoësäureäthern
   Baumann, E.
   Ber. 1886, 19 (2), 3218-3222
   DOI: 10.1002/cber.188601902348