Master Organic Chemistry

Classic Exam Questions on Intramolecular Reactions

James Ashenhurst — Thu, 09 Apr 2026 14:43:54 +0000

Intramolecular Reactions

Classic exam questions.

Substitution Reactions
Williamson Ether Synthesis
Addition to Alkenes
Epoxide Opening
Wittig Reaction
Diels-Alder Reaction
Friedel-Crafts Alkylation and Acylation
Acetal Formation
Imine Formation
Ring-Closing Metathesis
Fischer Esterification
Aldol Reaction
Dieckmann Reaction (Intramolecular Claisen)
Malonic Ester Synthesis
Amines

Substitution Reactions

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Williamson Ether Reaction

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Alkene Additions

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Epoxide opening

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Wittig Reaction

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Diels Alder Reaction

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Friedel Crafts Alkylation and Acylation

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Acetal formation

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Imine formation

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Ring Closing Metathesis

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Fischer Esterification

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Aldol reaction

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Dieckmann

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Malonic Ester Synthesis

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Amines

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Two-Step Synthesis Problems – 20 Solved Examples

James Ashenhurst — Wed, 08 Apr 2026 17:44:02 +0000

Think about learning how to do synthesis two reactions at a time.

When you learn a new reaction, take a moment and just think of one extra reaction you could do with that product.

Example. You learn free-radical halogenation of alkanes to make alkyl halides. What’s one reaction of alkyl halides?

Elimination (with base) to give an alkene
Substitution (with nucleophile) to give (e.g.) a nitrile
later on, formation of a Grignard reagent.

Eventually, you will find yourself piecing together these two-step syntheses into longer chains of sequences.

Here are 20 examples of two-step synthesis questions that are great building blocks for bigger things (OK, some are 3 steps)

Chopping Off The Alkyl Group
Walking Over The Alkyl Halide
Making Molecules Alkynes of Ways
Trans to Cis? Not A One Step Process
Alkanes to Alcohols
Protect That Alcohol!
The Great Friedel-Crafts Workaround
Styrene
Polarity Reversal On A Ring
Fun With Carboxylic Acids… and Esters
Malonic Ester Madness
Cyclohexenone
Know What Amine?

Here are the relevant quizzes:

1. Chopping Off An Alkyl Group

Click to Flip

2. Walking Over The Alkyl Halide

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3. Alkynes of Ways To Make These

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4. Trans to Cis Is Not A Simple Operation

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5. Alkanes to Alcohols Via Alkenes

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6. Protect That Alcohol

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7. The Great Friedel Crafts Workaround

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8. Ethylbenzene to Styrene

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9. Polarity Reversal On An Aromatic Ring

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10. Carboxylic Acid Fun (and Esters)

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11. Malonic Ester Madness

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12. Cyclohexenone

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13. Know What Amine, Vern?

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Org 1, Midterm Number 1

James Ashenhurst — Sat, 07 Mar 2026 01:46:56 +0000

Org 1, Midterm 1: 10 Common Types of Exam Problems

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.

Click to Flip

2. Lewis structures

Expect to be able to draw a Lewis structure for simple molecules. Popular examples include diazomethane (CH₂N₂), nitromethane (CH₃NO₂), the carbonate ion (CO₃)2- and others. Here are a few examples:

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3. Formal Charges

Formal charge calculations are a staple. If you’re given the structure of a simple molecule, expect to be asked to quickly be able to calculate the formal charge of different atoms.

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Here’s the final boss:

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4. Geometry / Bond Angles

If you’re shown the chemical formula for a molecule, can you predict the geometry / bond angles at specific atoms? If you can determine hybridization, this should be straightforward.

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5. Bond types

Another wrinkle on hybridization questions is to be given the structure of a molecule and to identify the types of bonds. Not too tricky, but it starts with being able to assign hybridization. Here are some examples.

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6. Dipole moments

Not a huge component of midterms, but recognizing dipoles and dipole moments in molecules is a fundamental skill. Understand that atoms have different electronegativities and this leads to bond polarization. Furthermore, these small dipoles are vectors, and the vector sum affects the overall dipole moment of the molecule.

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7. Simple nomenclature

Identifying the longest chain in an alkane and properly assigning substituents and naming locants is important. So is being able to determine whether different carbons are primary, secondary or tertiary. This latter skill is absolutely crucial for later success in the course!

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8. Boiling / melting points

Here, just know the trends. Part 1 is recognizing the four intermolecular forces and how they affect boiling points. Part 2 is looking at other factors such as the molecular weight, geometry, and number of polar functional groups in each molecule to make a decision. Melting points may also make an appearance.

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9. Resonance

A fundamental subject. It’s important to be able to recognize resonance forms, first of all (don’t break single bonds! don’t move atoms!) and then to be able to rank them in order of their importance (full octets, put negative charges on least basic atoms, carbocations with adjacent alkyl groups are best…).

Even if you’re taking Org 2 these are worth revisiting, since resonance is a huge theme all throughout Org 2.

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10. Acid Base

Another fundamental subject, which is essentially asking, “what factors stabilize negative charge?” (there are 5 big ones). Understanding that acid-base equilibria favor formation of the most stable conjugate base.

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Click to Flip

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Click to Flip

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How Do We Know Methane (CH4) Is Tetrahedral?

James Ashenhurst — Fri, 25 Aug 2017 09:00:51 +0000

What Do The Valence Electrons Of Carbon Tell Us About The Bonding In CH₄?

(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

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

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.

Spectroscopy and Synthesis Quizzes

James Ashenhurst — Tue, 20 Aug 2024 18:40:14 +0000

Hey All!

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^-1and 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

Acetylides from Alkynes, And Substitution Reactions of Acetylides

James Ashenhurst — Wed, 01 May 2013 12:03:07 +0000

Alkylation of acetylides

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. 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:

Click to Flip

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).

Click to Flip

Draw the product of the reaction below:

Click to Flip

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

Click to Flip

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:

Click to Flip

A second, slightly more difficult synthesis question.

Click to Flip

More of the same thing!

Click to Flip

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!

Click to Flip

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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.

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
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.
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.
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)
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.

E and Z Notation For Alkenes (+ Cis/Trans)

James Ashenhurst — Thu, 03 Nov 2016 17:40:02 +0000

E and Z Notation For Alkenes

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

When do we use cis– and trans– Notation In Rings?
cis– and trans– Isomerism In Alkenes
Watch out for ambiguous names when geometrical isomerism is possible!
cis– and trans– isomerism in cyclic alkenes
When “cis“- and “trans‘” fails: E and Z Notation
E and Z Notation For Alkenes
Breaking Ties: The Method of Dots
Conclusion: E and Z Notation For Alkenes
Notes
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!

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

Alkene Stability

James Ashenhurst — Thu, 30 Apr 2020 15:46:10 +0000

Alkene Stability (And Instability)
What factors affect alkene stability? If you’ve studied elimination reactions, no doubt you’ve learned about Zaitsev’s Rule – about how elimination reactions generally favor the “more substituted” alkene.

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

Heat Of Hydrogenation As A Measure Of Alkene Stability
Stability of Alkenes Increases With Increasing Substitution
Heats Of Hydrogenation For Some Monosubstituted Alkenes
The Relative Stability of cis- and trans- Alkenes
Alkenes Stabilized By Conjugation: Resonance Energy
Alkene Stability: Summary
Notes
Bonus Topic #1: Why Is Alkyl Substitution Stabilizing?
Bonus Topic #2: trans-Cycloalkenes
Quiz Yourself!
(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 CH₂compared 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!

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

(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.

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.
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).
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
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
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.

Alkene Addition Reactions: “Regioselectivity” and “Stereoselectivity” (Syn/Anti)

James Ashenhurst — Tue, 22 Jan 2013 19:51:08 +0000

Alkene Addition Reactions: Regioselectivity and Stereoselectivity

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. 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.

(Advanced) References and Further Reading

[references]

Keto-Enol Tautomerism

James Ashenhurst — Tue, 21 Jun 2022 17:00:53 +0000

Keto-Enol Tautomerism

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

When Ketones Moonlight As Nucleophiles
Keto-Enol Tautomerism
Examples Of Keto-Enol Tautomerism
Properties of Enols
Keto-Enol Tautomerism: Mechanisms
Four Factors That Affect Keto-Enol Equilibria
Factor #1: Substitution
Factor #2: Conjugation/Resonance
Factor #3: Hydrogen Bonding
Factor #4: Aromaticity
Revisiting A Weird, “Unketone-like” Reaction
Summary And Conclusion
Notes
Quiz Yourself!
(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.

A nucleophile attacks carbon, forming C–Nu and breaking the C-O pi bond to give an alkoxide (O^–) This is nucleophilic addition.
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.

Not an actual aldehyde or ketone

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.

(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!

Master Organic Chemistry

Classic Exam Questions on Intramolecular Reactions

Substitution Reactions

Williamson Ether Reaction

Alkene Additions

Epoxide opening

Wittig Reaction

Diels Alder Reaction

Friedel Crafts Alkylation and Acylation

Acetal formation

Imine formation

Ring Closing Metathesis

Fischer Esterification

Aldol reaction

Dieckmann

Malonic Ester Synthesis

Amines

Two-Step Synthesis Problems – 20 Solved Examples

1. Chopping Off An Alkyl Group

2. Walking Over The Alkyl Halide

3. Alkynes of Ways To Make These

4. Trans to Cis Is Not A Simple Operation

5. Alkanes to Alcohols Via Alkenes

6. Protect That Alcohol

7. The Great Friedel Crafts Workaround

8. Ethylbenzene to Styrene

9. Polarity Reversal On An Aromatic Ring

10. Carboxylic Acid Fun (and Esters)

11. Malonic Ester Madness

12. Cyclohexenone

13. Know What Amine, Vern?

Org 1, Midterm Number 1

1. Hybridization (*crucial*)

2. Lewis structures

3. Formal Charges

4. Geometry / Bond Angles

5. Bond types

6. Dipole moments

7. Simple nomenclature

8. Boiling / melting points

9. Resonance

10. Acid Base

How Do We Know Methane (CH4) Is Tetrahedral?

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?

Notes

Spectroscopy and Synthesis Quizzes

Acetylides from Alkynes, And Substitution Reactions of Acetylides

1. Terminal Alkynes Are Acidic!

2. SN2 Reactions of Acetylides With Alkyl Halides: Finally, Some Carbon-Carbon Bond Formation!

3. Alkylation of Acetylides – Some Practice Questions

4. Practice Questions – Synthesis of Acetylenes

5. Reaction of Acetylides With Other Nucleophiles

6. Summary

Notes

Quiz Yourself!

(Advanced) References and Further Reading

E and Z Notation For Alkenes (+ Cis/Trans)

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

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

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

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

4. Cis– Trans- Isomerism For Cyclic Alkenes

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

6. The E and Z Notation For Alkenes

7. Breaking Ties: The Method of Dots

8. Conclusion: E and Z Notation For Alkenes

Notes

Quiz Yourself!

Alkene Stability

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. Summary: Stability of Alkenes

**1. Hybridization (crucial)**

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

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

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