Archive for December, 2014

Here at StudyOrgo.com, 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.

Students often mention to us they are confused about the rules of aromaticity and how best to study for difficult examples.  We at StudyOrgo.com 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.

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.

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.
      

• 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.
    

• 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.  

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

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 2^N 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.

This is just an example of the crystal-clear explanations you will receive as a member of StudyOrgo.com 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!

“I have an exam next week and I feel helpless right now. I feel like I do not understand anything at this point meanwhile I have had a good understanding for the past two exams. Will this site be beneficial to me since I have a short period of time to my exam?”

We receive comments like this frequently at StudyOrgo. To summarize the student’s dilemma, it sounds like the first two tests worth of material in organic chemistry were the basics: learning to name compounds, bond hybridization, sigma and pi bonding, maybe even some thermodynamics and bond energies. But now that you have covered these fundamental principles, it is time to put them into practice and learn about specific chemical reactions and their mechanisms. In this course thought, the quantity and variety of reactions can be just as overwhelming as their complexity. This is usually where it the course goes sideways for students, and a feeling of panic and despair replaces any confidence you felt about the material before.

When asked if this website will help a student in this predicament, our short answer is “YES!” What is likely this student’s trouble is either some important concepts critical to the new material were missed or not understood as well as previously thought. It happens all the time. Many times, students will tell us that the teacher “blew through” or “went really fast” over some material and the student missed it. After we re-learn the materials presented on the website, the student can be acing the practice problems!

What you need to move forward is to solidify the base of your pyramid of organic chemistry knowledge. You were off to a good start, but something went wrong somewhere, and we have the expertise to help. We at StudyOrgo have spent countless hours reviewing and preparing the material in the most crystal-clear and “get-to-the-point” manner as possible. We consult students and ask for their opinion on whether they get it based from the tutorials we present.

Once you are ready to move on past the basics, we take a step-by-step approach to going over every chemical reaction commonly taught in the course. We make a note to point out important points, Markonvnikov v. anti-Markovnikov selectivity, syn or anti addition, stepwise v. concerted mechanism, etc. By the time you finish the tutorial, you will be ready to practice you problem sets and you will be ready to ace the next test!