How to Interpret Thermodynamics of Reactions

Posted on August 8th, 2018

In this post, we will discuss the concepts and theories of Thermodynamics.  While the topic sounds complicated, we at StudyOrgo have extensive experience instructing principles and reaction mechanisms frequently covered in Organic Chemistry. Sign up today for clear, detailed explanations of over 180 Orgo Chem reactions and reviews on conceptual topics!

The Equilibrium Constant

Remember from general chemistry that equilibrium constant (Keq) reflect the ratio of products to reactants in the following equation.

Keq = Products / Reactants = [C]*[D] / [A] * [B]

Thus, if one knows the concentration of product or reactant in solution and the Keq for the reaction, the other value can be also know.  One point of confusion is that these values should somehow be intuitively known.  They are calculated in the lab for each reaction, with each type of reactants.  Therefore, without the “empirical” data we wouldn’t know the Keq.  Fortunately, many experiments have been done in the world and Keq values are frequently found in tables and reference materials.

Spontaneity and Gibbs Free Energy

An important factor in choosing reactions in organic chemistry is knowing how “efficient” this reaction will be.  Since our goal to generate products, knowing the Keq for the reaction will allow us to predict how much product we can theoretically produce from a known starting concentration of reactants.  We can convert the Keq ratio into energy (kJ/mole) which is a more comparable measurement between reactions using a version of the Gibbs Free Energy equation.

dG = RT*ln[Keq]

Remember, that a release of energy (a negative dG) indicates a spontaneous and forward (towards product) reaction.  As a reference, let’s look at a table of corresponding dG and Keq values and how this relates to concentration of products.

Delta G (kj/mol) Keq value % products at equilibrium
+17 1×10^3 99.90%
+11 1×10^2 99%
+6 1×10^1 90%
0 1 50%
-6 1×10^-1 10%
-11 1×10^-2 1%
-17 1×10^-3 0.10%

When dG is negative (exothermic), the Keq is very high (products >> reactants) and we see that most values have a >90% concentration of products when the reaction reaches equilibrium. However, when dG is positive (endothermic), the Keq is very low (products << reactants) and we see most values have <10% products.

Therefore, when selecting reactions in your synthesis scheme you will likely want to choose reactions with the highest Keq and most exothermic to generate high concentration.

The reaction coordinate

Reaction coordinates are often used to visualize concepts of free energy between the reactants and products of reactions.  Below are examples of endothermic and exothermic reactions.

The Y-axis reflects the free energy of the reactants (A + B) and products (C + D).  The X-axis reflects the progress of the reaction, as the path between reactants and products will have several hills and valleys that reflect transition states and intermediates, respectively which we will cover in another article. The thermodynamics of a reaction can all be visualized from graphs like these.  You will see them a lot in the course!

Determining Gibbs Free Energy from reaction Coordinate

The difference between initial reactant (blue) and final product (green) free energies will give you the Gibbs Free Energy (dG) of the reaction.  Remember, negative = spontaneous and positive = not spontaneous.  The larger the value, the more complete the reaction goes to either side.  For example, a large positive value means low products (no reaction) and large negative value means high products (efficient reaction).

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How to Assign Stereoisomer Configuration

Posted on July 2nd, 2018

Chirality is an important aspect of life.  This is so because many of metabolites used in living cells, in particular amino acids that form enzymes, are also chiral. Chirality contributes asymmetry to molecules, allowing them the ability to recognize “handedness” and further add to the complexity and specificity of reactions. Organic chemists must pay constant attention to the chirality of molecules both before and after reactions, less the compounds lose their biological or chemical activity.

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Molecules, like your hands, that are not superimposable on their mirror images are called chiral objects, from the Greek word cheir  (meaning “hand”). All 3D objects can be categorized as either chiral or achiral. Chemical molecules are 3D objects and can also be classified in this way. To help you remember, chiral molecules are like hands and are NOT superimposable on its mirror image while achiral molecules are like tennis rackets: they are superimposable on their mirror images.

IUPAC recommended in 1996 that a tetrahedral carbon bearing four different groups be called a chirality center.  Many other names are commonly used include chiral center, stereocenter, stereogenic center, and asymmetric center.  They all mean the same thing, a carbon connected to 4 unique substituents that is not superimposable on its mirror image.

When a compound is chiral, it will have one opposite molecule; a non-superimposable mirror image, called its enantiomer (from the Greek word meaning “opposite”). The compound and its mirror image are said to be a pair of enantiomers. The word “enantiomer” is analogous to the word “twin”. When two children are identical twins, each one is said to be the (“evil…”) twin of the other. Similarly, when two compounds are a pair of enantiomers, each compound is said to be the enantiomer of the other.  A chiral compound will have exactly one enantiomers, all other molecules with the same molecular formula and constitutional arrangement are diastereomers, but are only seen when there are more than two chiral centers in a molecule.

In order to determine whether the stereocenter is in the R or S configuration, there are a series of steps to follow.

1. Identify the stereocenter as 4 unique substituents attached to the chiral center

  • This one is easy for most, but just look for any carbon with 4 substituents that are different.  Be careful to carefully count chain lengths and identify unique elements.

2. Assign priority

  • Step 1: Assign priority of bond based on atom atomic number of the element, highest (1) to lowest (4) weight.
  • Step 2: If two atoms are same, count the type of bonds connected to the carbon to find first point of difference
  • For 2-methyl-3-pantanol, oxygen is highest priority and hydrogen is lowest priority. However, 2 carbons are connected to the stereocenter, therefore count the number of bonds connected to each carbon center. In this case, the carbon with 2C and 1H has higher priority than the carbon with 1C and 2H.
  • For the haloalkane, oxygen is highest priority and hydrogen is lowest priority. However, 2 carbons are connected to the stereocenter, therefore count the number of bonds connected to each carbon center. In this case, the carbon with 1Br and 2C is higher priority than the carbon with 1F, 1Cl and 1C because Br is the heaviest and highest priority element.


3. Rotate the molecule so that Priority 4 atom is in the hashed wedge position.

  • In some cases, if you can’t flip the molecule in your head or on paper easily, assign the configuration to the stereocenter when the 4th position is NOT in the back of the paper position, and just reverse the assignment.  It works every time.

4. Determine the Priority sequence 1-2-3 rotates to the left (S) or the right (R).


Lastly, an important concept to keep in mind is that as molecules become more complex, they also can acquire more stereocenters.  Keeping in mind that each stereocenter can produce 2 stereoisomers, we describe possible stereoisomerism using the 2n rule. Let’s examine a molecule with 2 stereocenters, following the 2n rule that gives us 22=4 stereocenters.  The possible combinations are listed below.

We now introduce the last concept to stereochemistry which is the difference between enantiomers and diastereomers.  Enantiomers are molecules with exactly opposite stereoisomers.  For example, the enantiomer of the molecule with stereochemistry R,R would be S,S.  The relationship between molecule R,R and R,S is what is described as diastereomers, which differ in some but not all stereocenters.

Formation of Enols and Enolates

Posted on April 3rd, 2018

One question that comes up in organic chemistry often is “what is an enol or an enolate and how is it formed?”  These types of concepts are frequently covered quickly in class or not at all, but are very important for future reaction mechanisms.  We at Study Orgo have the combined experience of over 15 years of tutoring and teaching organic chemistry concepts to struggling students.  We have developed clear descriptions of reaction mechanisms and organic chemistry concepts to aid students in their studies.  Sign up today for access to over 180 reactions mechanisms and reviews!

The alpha carbon of a carbonyl, which is present in carboxylic acids, esters, ketones and aldehydes, are acidic which means the proton can be removed using a base.  In neutral or acidic conditions, this means the lone pairs on the C=O position can act as a weak nucleophile.

If the carbonyl oxygen can attack the alpha carbon C-H bond, it will abstract the hydrogen and perform a Keto-Enol tautomerization reaction that will lead to the resonance version of the carbonyl, which is the Enol (alkENE + alcohOL)

Enols – rearranging the pi bond and atoms of a carbonyl compound to an Enol

Catalyist: Acidic or Neutral Conditions to stabilize OH formation

Enols tautomers are generally unstable, preferring the “Keto” version 90-99% of the time versus the “Enol” version.  However, a catalytic amount of presence is sometimes enough to drive reactions forward if the mechanism requires the enol tautomer of the compound.

However, in some cases such as a beta diketone, shown below, the combined dipoles of two carbonyls makes the alpha carbon very acidic, meaning enol formation is very favorable.  In this case, it is 70-90% enol in solution.


Enolates – Deprotonating the alpha carbon and tautomerizing to the oxyanion

Catalyist: Strong Base to deprotonate the alpha carbon.

Like an Sn2 mechanism, a strong enough base will react with the acidic proton on the alpha carbon and deprotonate.  The electron density between the C-H bond will shift to make a new C=C bond, while the C=O electrons will be placed on the oxygen, creating and alkENE + alcohOL anion “ATE”) with a strong base to produce a stable carbanion.  The stability is due to the tautomerized structure that can be produced by placing the negative charge on the oxygen.


Enolates are generally forward reactions depending on the strength of the base.  How strong the base required depends on the pKa of the alpha C-H bond.  In the case of ketones, a strong base like LDA is required.  However, for beta dikeontes, a mild base like NaOH is enough to generate the enolate.


Formation of Enols and Enolates are an important source of carbon nucleophiles to make new C-C bonds in future reactions.

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How do you to tell when a hydrogen bond will occur?

Posted on March 7th, 2018

Hydrogen bonding is important for describing the driving force of reactions in organic chemistry and a very important concept for explaining the biochemistry of catalytic reactions that occur using protein as enzymes in biological systems.  In this post, we will discuss the rules and examples of hydrogen bond formation.  We at StudyOrgo have extensive experience instructing principles and reaction mechanisms frequently covered in Organic Chemistry. Sign up today for clear, detailed explanations of over 180 Orgo Chem reactions and reviews on conceptual topics!

Physical properties of molecules such as boiling and melting point, solubility and reactivity, are affected by the functional groups that make up the molecule. This can be explained by analyzing the type of intermolecular forces that are experienced between molecules.  Because these forces are not covalent, intermolecular forces are determined by the intensity of electrostatic forces which is what makes up each type of intermolecular force. As a review, the types of intermolecular forces are;

  • Van der Waals (London dispersion forces) – Weak, temporary dipole formed between hydrophobic C-H and C-C bonds.
  • Dipole-Dipole Interactions: – Strong, permanent dipole moments formed between atoms of functional groups containing bonds such as C=O, C=N, C-O, C-N, N-H and O-H bonds. The delta(-) side of one dipole is attracted to the delta(+) side of another molecule, forming a non-covalent electrostatic attraction.
  • Hydrogen Bonding: Sharing sharing of a hydrogen atom covalently attached to an electronegative element (typically O-H and N-H groups) between a lone pair of electrons on another electronegative element.

Two terms about hydrogen bonding that are key are;

  • The electronegative atom with the lone pair electrons is called the Hydrogen Bond Acceptor
  • The electronegative atom bonded to the hydrogen is called the Hydrogen Bond Donor
  • The Hydrogen Bond Donor must be aligned 180 degrees to the Hydrogen Bond Donor!

The more intermolecular forces the molecule has, the more energy will be required to disrupt these bonds when melting or boiling compounds, thus raising the observed temperatures from expected relative to their mass.  In addition, hydrogen bonds require polar bonds in the molecule and H-Bond Donor proton involved is protic (a donatable hydrogen atom). These are two terms that you will learn in the Sn1 mechanism.

Let’s look at ethanol as an example.  The hydrogen bonding occurs between the proton of one alcohol group and the oxygen lone pair electrons on another alcohol group.  This is a strong intermolecular force that holds the molecule in a complex 3D shape and makes it easier in reactions to attack the carbon connected to the O-H bond due to inductive effects, or the pulling of electrons away from the carbon.  Water is an extreme example, where all the atoms in the molecule participate in hydrogen bonding.  The oxygen lone pairs will accept a hydrogen from a neighboring molecule O-H.  Finally, acetic acid is another example.  Remember, that the H-Bond Acceptor can be any lone pairs, including those of C=O bonds.


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IR Spectroscopy Review

Posted on January 14th, 2018

Studying ahead for Organic Chemistry this Spring semester is a good way for getting the best grade this semester and keeping up with the rigorous course work in Orgo 2.  Most students find the pace of this class very challenging compared to other courses.  This is because while there is a lot of information to learn, it also builds on previous concepts from Orgo 1, a course most students want to forget!  In your time before classes begin, consider reading ahead or brushing up on some concepts that were covered late last semester to give you a boost right away.

But don’t worry!  StudyOrgo has you covered.  Our Editors have spent years tutoring and teaching Organic Chemistry to students and we have seen all of the pitfalls common to the first few weeks of the semester.  Our online platform allows members to learn organic chemistry concepts and mechanisms quickly and the material presented in an easy-to-follow format. Follow along with us and sign up with StudyOrgo today to help prepare you for all of your Organic Chemistry questions.

One of the concepts you will need to have mastered quickly in Orgo 2 is the usefulness of Infrared (IR) Spectrometry.  In this article, we will break down the key concepts and give you all the info you need to master this technique quickly.

IR Spectroscopy Principles

A usefulness of using light for analysis is it is relatively non-destructive and cheap to produce.  We can see that there are many regions of electromagnetic radiation that we can use for molecular predictions.  For instance, long wavelengths like microwaves are used in NMR for determining atomic structure.  Longer wavelengths like IR and UV/Vis region are used for predicting functional groups and can help you identify unknown compounds.

Infrared radiation causes vibrating the bonds between atoms.  This is a similar principle to how heating up molecules works, and you may have heard of infrared imaging systems, which can measure the relative temperature of molecules.  This technology works on a similar principle to IR spectroscopy, but we use it in the lab to determine functional groups.

Higher energy regions of the IR spectrum (larger wavenumbers) will cause stretching of bonds where lower energy regions (smaller wavenumbers) will cause bending and twisting.  Thus, depending on the type of bond and the atoms involved in the bond, we can predict what they are based on which wavenumber region they absorb.

There are 3 pieces of information you can get from the IR spectrum of samples.

Signal Region

In the graph above, we can see the IR spectrum for isopropanol.  We see that the range of wavenumbers (inversely proportional to wavelength), there are multiple regions that are causing unqiue-looking signals.  The functional groups are alkane and alcohol.  Almost all your moleucles will have C-C and C-H bonds, so many o fhtese signals are not useful.  But an alcohol is unqiue, with a large peak at 3400.

Signal IntensitySome signals will be weak and some will be strong, as we see in the figure above.  This has to do with how efficiently the region is being absorbed by the molecule.  If it is a strong signal it should be consistent and easy to detect.  However some bonds are not as efficient.  With these signals, sometimes you miss them because they are too weak.  So if you think you may have a functional group but the peak is not there, remember it maybe “invisible” if it is a weak signal. 

Signal ShapeMost types of bonds will absorb in a very narrow region of the IR, giving the typical narrow signal shape. Only a few types of bonds will case large regions of the spectrum to absorb, causes a broad signal.  These always are functional groups that can undergo hydrogen bonding, such as O-H, N-H, C=O, etc.  Typcially OH and COOH functional groups are very broad, while carbonyls are amines are more broad than ususal.

Common signals used to predict functional groups


Analytical Chemistry – Infrared (IR) Spectroscopy

We have placed a typical carboxylic acid spectrum on top of this trend chart to help you learn the various signals that you will encounter.  Signals in the “fingerprint” region are useful for matching you unknown to a known standard, too usually too “noisey” to predict anything.  You can see that the broad OH peak at 3400 cm-1 and the C=O peak at 1710 cm-1 falls very neatly in the trend.

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