Archive for the "Organic Chemistry General" Category

The Diels Alder Reaction

Posted on February 11th, 2015

One reaction that always troubles students taking organic chemistry is the Diels-Alder reaction.  In this article, we will discuss the basics of the reaction and give you a preview of the clear-cut examples of organic chemistry reaction mechanisms available to StudyOrgo members!

The Reaction:

The Diels-Alder reaction is referred to as a pericyclic reaction, in that two reactants cyclize to become one ring.  The reaction utilizes a [4+2] cycloaddition electron mechanism; meaning that the first reactant has 4 pi-electrons and the second reactant has 2 pi-electrons.  These are named the diene (2 alkenes, or 4 pi-electrons) and the dieneophile (1 alkene, or 2 pi-electrons, that “seeks” the diene). The diene MUST be in the cis- conformation in order to cyclize, the trans- isoform would not form the ring… try it with your models or on paper!


The Mechanism:

Below is the reaction mechanism using arrow-pushing.  This mechanism occurs in a concerted, or one-step, process.  It is thought that the dienophile attacks the diene and rearranges the electron distribution to form the two new red C-C bonds and results in a new 2 pi-electron bond. Energetically, breaking 3 pi-bonds in the reactants and formation of the 2 sigma-bonds + 1 pi-bond in the product has a negative enthalpy value (deltaH), therefore the reaction is exothermic and spontaneous!



Most examples students encounter are more complex, where the dieneophile has substituents and the diene is already cyclized, thus forming a bridged product.  Depending on the orientation of the dienophile with respect to the diene, two products are possible.  Below is an example of such a reaction.


In this example, only one product is observed; the endo- product.  The selectivity for this reaction can be illustrated in the following diagram.  The addition of the reaction occurs via the “left-handed rule” that is, if you put your left hand thumb along the dieneophile and twist to the left (see red arrows), the rotation of the dienophile represents the stereochemistry of the substituents after the reaction is complete.  Two products are possible, the exo- and endo- products.  The reason endo- is preferred can be seen by looking at the Neumann Projection along the bolded bond.  In the exo- product, the Cl group is gauche to the Bridge group.  In the endo- product, the CL group is anti to the Bridge group.  Thus, steric hindrance is a major factor in determining the stereochemistry and since all reaction will deal with these steric factors, endo- will always be preferred!


Sign up today at StudyOrgo for more details and examples to improve your organic chemistry grade!

Deciphering 1H NMR Spectra

Posted on January 24th, 2015

Deciphering 1H-NMR Spectra

One of the most important concepts taught in organic chemistry is the method for determining the chemical structure of newly synthesized or unknown compounds. In this article, we will summarize the concept of proton NMR, the most common NMR information acquired by organic chemists.

While proton NMR is used every day in the real world by organic chemists, it is also tested every day in the real world by organic chemistry professors. That is why we are bringing you this exclusive simplified review on NMR.

Before you read on, if you like what you see, but are looking for explanations on different organic chemistry topics and reactions, check out how our signature online organic chemistry program works to help you earn higher grades.

Now if you are ready to read on about NMR- here ya go:

Placing an unknown sample in a strong magnetic field allow 1H nuclei (99.98% abundance)  to “resonate”, which is when their nuclear spins flip at a unique electromagnetic (EM) frequency (Hertz, Hz).  The instrument detects this and plots it on a graph in units of ppm.  This value is relative to an internal standard, tetramethylsilane (TMS) which is arbitrarily set at 0 ppm. The signals originating from your unknown sample are recorded as a net difference from the TMS reference.  The ranges for common functional groups are shown in the graph below.

Deciphering 1H NMR Spectra
Notice that protons (in red) neighboring or attached to electronegative elements (N,O,S,Cl,Br,I) have a large chemical shift (d).  This is colloquially referred to as “downfield” on the spectrograph. This is due to the deshielding effect, which is simply that the electron cloud around protons near electronegative or electron-withdrawing groups are smaller, so less EM radiation is required to resonate the nucleus.  Conversely, when the proton is connected to carbons, the chemical shift is <2 ppm.  This is because these proton nuclei are “shielded” from the magnetic field and higher energy is required for them to resonate and referred to as “upfield” on the spectrograph.  Once a NMR spectrograph is recorded, 4 pieces of information can be determined from the data as long as the chemical formula of the compound is known.

To illustrate the points, we will consider the following 1H-NMR spectrum of the C5H10O.

  1. Signal Count – Number of unique hydrogens

This is the easiest to interpret.  The number of peaks correspond to the number of unique, or chemically indistinguishable, hydrogen nuclei.  There are two peaks on the graph, therefore of the 10 hydrogens in the molecule, there are two types.

  1. Chemical Shift – Identity of neighbors

The two peaks on the spectrum are located at (d)2.42 and (d)1.07.  Remembering that the chemical formula includes 5 carbons and one oxygen, it is clear there is a good deal of symmetry to the molecule.  Furthermore, the only proton near an oxygen that exhibits a chemical shift ~ 2.5 with is neighboring a carbonyl group HCRC=O.  So far, we know our compound is symmetric and has a carbonyl group, therefore only 4 carbons can have carbons.

  1. Integration – Number of Equivalent Hydrogens

Integration is the calculation of the area beneath the peaks.  This information is determined by the computer software and would not be required for you to determine from visual inspection.  But if it is provided, it will instantly determine the number of “equivalent” (or chemically indistinguishable) hydrogens in your graph.  Looking at our spectra, we have integration of 33 + 48 = 81 cart units in total.  The integration fraction can be multiplied by the total number of hydrogens from the chemical formula to determine number of equivalent hydrogens.  For instance, at (d)2.42 [33/81 = 0.40 * 10 total hydrogens = 4 hydrogens at (d)2.42].  Therefore, we know that the protons near the carbonyl total 4.  Given the symmetry from the low peak count, we know these hydrogens must be two CH2 groups.  The same for the peak at (d)1.07 [48/81 = 0.60 * 10 total hydrogens = hydrogens at (d)1.07].  Since only 2 carbons are left, we must conclude these are terminal CH3 groups.

  1. Signal Splitting – Number of Non-equivalent Hydrogen Neighbors

Signal splitting occurs from a phenomenon of coupling, which occurs when NON-equivalent neighboring protons interfere with the resonance of a proton nuclei.  The degree of splitting occurs in an N+1 rule.  For instance, at d1.07 we have a triplet.  This indicates that there are 2 protons neighboring the CH3 group that are different.  This would have to be the protons from the CH2 group at (d)2.42.  Similarly, the peak at (d)2.42 as a quartet, indicating that 3 protons are neighboring the CH2 group that are different.  This would have to be the protons from the CH3 group.

And voila!  The chemical structure from the spectrum must be 3-propanone.  This is an example of the clear-cut explanations you will receive with your membership with StudyOrgo. With practice and help from StudyOrgo, you’ll be solving your organic reaction problems in no time!



Want to try it about before you purchase? No problem- check out our sample reaction flashcard page.

Ready to sign up? Sign up here.


Electrophilic Aromatic Substitution (EAS) Substitution Explained

Posted on December 29th, 2014

Here at, we have devoted a lot of effort to explain the mechanisms of organic chemistry reactions.  A popular second semester topic is electrophilic aromatic substitution (EAS).  Remember that the aromatic ring is made up of 6 pi-orbitals in a ring that is planar, which confers to it Huckel aromaticity.  This configuration is remarkably stable but under certain conditions, aromatic rings can undergo substitution reactions.



It becomes more complicated when the mechanism uses a benzene ring that already has substituents at one group.  For these situations, special rules are defined for 1) the orientation of the new group and 2) the rate of further addition on the ring.

First, let’s consider the rules of Directing Effects (i.e. where the electrophile will add on the ring).  Two possibilities exist; 1) ortho/para position or 2) meta position.  This will be determined by the identity of the substituent at position R (Figure 1).  Below in Figure 2 you will see a list of functional groups that will direct the electrophile to either of the possibilities.  In general: the more electron-rich the atom that is attached to the ring, the more ortho/para directing potential.



As for why this occurs, let’s consider resonance structures of the intermediates.  Take a benzene ring with a methoxy group attached (Figure 3, left).  If the electrophile attacks meta, there are 3 possible structures.  If the electrophile attacks para (ortho is exactly the same), there are 4 possible structures.  More important, the structure in green for ortho/para is most stable to complete the substitution mechanism.  This is because the oxygen group can donate electrons to the resonance.  In the meta position, oxygen cannot donate and it is therefore less favored. Now take a benzene ring with a nitro group attached (Figure 3, right).  Only 3 resonance structures are possible for either ortho/para or meta substitution.  But, the resonance structure in red for ortho/para is the least stable to complete the substitution mechanism because is harbors 2 cation centers next to each other, almost impossible to exist.  Therefore, meta is the most important contribution. This is the basis for determining the directing effects of EAS reactions!


Finally, a discussion on the reactivity of further substitution is necessary.  EAS reactions are classified as activating when describing the effect on the EAS reaction of occurring multiple times on the same ring.  For deactivating effects, the R group inhibits the possibility of further substitution.  These considerations mainly follow the rules of Inductive Effects.  To understand this concept, think of the conjugated aromatic ring system and consider what happens when either an electron-rich or electron-poor atom is attached to the ring (Figure 4, upper).


For a methoxy substituent, one lone pair of electrons on oxygen (electron rich) can add to the pi-orbital and contribute to resonance and stability of the conjugated system.  Because the source of electrons for this reaction is the benzene ring itself, this property of the methoxy group increases available electron density and the likeliness for a second substitution reaction to generate a di-substituted product.  However, for a nitro substituent (Figure 4, bottom), the nitrosium cation (electron poor) withdraws electrons from the conjugated system.  This reduces the availability of electron density for the mechanism to work and reduces the likeliness for a second substitution reaction.  The activating/deactivating effects of substituents are indicated by the arrow in Figure 1.

Remember, all of these considerations will affect the overall product identity and synthesis rate based on the starting reagent, but with a little logic and our helpful hints presented here, you are well on your way to acing the exam!

This explanation on electrophilic aromatic substitution pairs extremely well with out flashcards on Aromatic Compounds. Check out what’s covered in our unique program and learn how it works. When you are ready to get started sign up here.

Aromaticity Rules and Definition

Posted on December 23rd, 2014

Students often mention to us they are confused about the rules of aromaticity and how best to study for difficult examples.  We at have developed a clear and “get-to-the-point” presentation of the basics of aromaticity.  This is just a sample of the clear-cut explanations available on our website to our members.

Aromatic Compounds

Aromatic compounds are unusually STABLE and have important chemical and synthetic uses.  In fact, nucleic acids and amino acids that make up every cell rely heavily on the use aromatic compounds.  But, what makes a compound aromatic?  A short list of rules, discovered by Eric Huckel in the 1930’s, lists the properties of aromatic compounds.

The Huckel aromaticity rules are:

  1. Molecule is cyclic
  2. Have one p orbial per atom of the ring
  3. Be planar, in an sp2 hybridized orbital, over every atom of the ring
  4. Have a closed loop of 4n+2 pi-bond electrons, where n is equal to any integer (0,1,2,3,…)

But like most natural phenomenon, there exists a rule exactly the opposite.  Molecules that have an unusual INSTABILITY to them are anti-aromatic compounds.  They have similar rules to aromaticity, including:

  1. Molecule is cyclic
  2. Have one p orbial per atom of the ring
  3. Be planar, in an sp2 hybridized orbital, over every atom of the ring
  4. But, anti-aromatic compounds have a closed loop of 4n pi-bond electrons.

Below are the pi-orbital diagrams of benzene, the most identifiable aromatic compound.  Each of the three double bonds contributes 2 pi-electrons over 6 atoms, for a total of 4*1+2=6 electrons, in a ring, in a pi-orbital that is planar.  Therefore it is aromatic.  In contrast, hexatriene meets all of these criteria as well, but is not in a closed ring.  Hexatriene is therefore non-aromatic.  Finally, cyclobutadiene is the most identifiable anti-aromatic compound which is different only in that it has 4*1=4 pi-electrons, in a ring, in a pi-orbital that is planar.


Heterocyclic Aromatic Compounds

The diversity of compounds relies on using atoms other than carbon, however.  What about when atoms with lone pairs of electrons are involved?  A good rule of thumb is that if the atom is already participating in the pi-bond forming in the ring, then the lone pair of electrons are perpendicular to the ring and therefore are NOT participating to aromaticity.  A good example of this is pyrimidine, where both nitrogens are already contributing to the pi-bond ring and therefore, the lone pairs of electrons are not accessible.


However, there are many molecules where lone pair DO participate to aromaticity.  Below are several examples.  Take furan for example; oxygen has two lone pairs of electrons.  One of them is in a geometry parallel to the pi-bond system.  Therefore, these electrons DO participate in the pi-bond system and add 2 electrons the pi-bond count resulting in 4*1+2=6 electrons, therefore furan is aromatic.  Imidiazole is molecule that has two nitrogen atoms; one nitrogen participating in pi-bonding and not contributing lone pairs, while the other is not participating in pi-bonding but contributes electrons the pi-bond count.  The 4*1+2=6 electron count for imidazole renders it aromatic.


Aromatic Hydrocarbon Ions

Sometimes, carbocations and carbanions are produced in chemical reactions.  If these species are created in a cyclic conjugated system, it is possible that they can contribute to the formation of a Huckel aromatic compound, which gives the molecule added stability and special reactivity.
For example, cyclopentadiene is not aromatic because of the sp3 hybridized carbon at position 5 on the ring.  However, in the presence of strong base, cyclopentadiene can be deprotonated and cyclopentadienyl anion is generated.  The lone pair of electrons assumes a sp2 hybridized orbital, making the molecule planar, adding 2 more electrons to the ring to give 4n+2 pi-electrons and creating the 5th pi orbital necessary to complete Huckel’s Rule and results in an aromatic ion.

reating the 5th pi orbital necessary to complete Huckel’s Rule and results in an aromatic ion

Another example is formation of a carbocation, a common intermediate in substitution and elimination reactions.  Deprotonation of cycloheptatriene, a non-aromatic compound, at the sp3 hybridized position creates a sp2 hybridized orbital and, although this carbon’s pi-bond orbital is empty (a carbocation), it completes the 7th pi orbital necessary to complete the ring and maintains a 4n+2 electron count.  This carbocation, called tropylium ion, is now aromatic.


What is the Difference between Diastereomers and Enantiomers?

Posted on December 18th, 2014

This is a problem that plagues many students studying organic chemistry and it is one of the cornerstones of getting through the class.  Without a clear understanding of stereochemistry, determining the correct product for future reactions will be impossible, so let’s break it down into some simple concepts.

  • Concept 1 – in order to have stereoisomers, the molecule must be CHIRAL.
    • Remember in order to have chirality, molecules must have the following characteristics
      • Carbon center with 4 unique substituents, meaning they are chemically distinguishable from each other.
        Carbon center with 4 unique substituents
    • In this example, the molecule on the left has 3 red hydrogens.  These hydrogens are chemically indistinguishable from each other.  So the molecule is ACHIRAL.
    • The molecule on the right has 1 red hydrogen and 3 other unique substituents.  Therefore, the molecule is CHIRAL.
  • Concept 2 – Chrial molecules that have STEREOISOMERS.
    • Stereoisomers are molecules that have the same chemical formula, but differ in their arrangement at a chiral center.
      Stereoisomers are molecules that have the same chemical formula, but differ in their arrangement at a chiral center.
  • Concept 3 – Stereoisomers are identified by their “HANDEDNESS”, which refers to the arrangement of the substituents relative to their importance.
    • In general, elements of higher mass have higher priority.  Refer to our tutorial on chirality for more details.


  • Concept 4 – There are two types of STEREOISOMERS, enantiomers and diastereomers.
    • Enantiomers contain chiral centers that are non-superimposable & mirror images.  They only come in pairs!
    • Diastereomers contain chiral centers are non-superimposable but are NOT mirror images.  There can be many more than 2 depending on the number of stereocenters.

An easy way to remember enantiomers from diastereomers is to memorize the picture below.  In the case of 2 chiral centers, 4 stereoisomers are possible.  Only the exact opposites (diagonal arrows) are enantiomers and they therefore have a mirror image that is not superimposable.  The molecules with only one stereocenter that differs (parallel arrows) are diastereomers.  

only one stereocenter that differs (parallel arrows) are diastereomers

A biological example of this is saccharide (or sugar) chemistry and below is the enantiomers and diastereomers of threose.

A biological example of this is saccharide (or sugar) chemistry

While enantiomers can only come in pairs, many diastereomers can exist for a given molecule.  Let’s take, 5-DHT for example, the metabolically active form of testosterone.  This molecule has 7 stereocenters, using the 2N rule for determining the number of stereoisomers, which gives 128 possible combinations. But only one of them is the enantiomer of 5DHT!  The rest are diastereomers.

Screen Shot 2014-12-18 at 1.02.36 PM

This is just an example of the crystal-clear explanations you will receive as a member of about important, and often confusing and poorly explained, concepts in organic chemistry. Our site developers have listened to students’ concerns and have come up with clear visuals to our tutorials on organic chemistry topics.  Interested further?  Sign up today!