Archive for the "Organic Chemistry General" Category

Molecular Orbital Theory

Posted on June 5th, 2016

One of the most challenging concepts in conjugated system reactions is molecular orbital interactions, or MO theory.  The basics to this principle can be hard to grasp, but will be very informative in predicting the correct reaction conditions and outcome of the reaction if you understand them, which will give you a major advantage on future quizzes and exams.  We at StudyOrgo have devised a simple explanation of the basics to MO theory to help you with your study preparations.

Principles of MO Theory

The basic concept of MO theory is to describe the alternating patterns of orbitals that exist in pi bond systems.  We describe the orientation of each half of the dumbbell orbital centered around an atom as being bold or open. Remember, there are bonding and antibonding orbitals.  Essentially, orbitals with lower numbers of nodes (switches in the orientation of the dumbbell orbital which we show with the dashed line) behave more like bonding orbitals while orbitals with higher numbers of nodes behave like anti-bonding orbitals.  To determine what orbitals are occupied on your molecule, simply count how many pi bonds you have in your conjugated system.  Below is an example of butadiene.  It has 4 pi electrons, so we fill the lowest molecular orbitals first, two in each orbital.  Therefore, the highest occupied molecular orbital, or HOMO, in butadiene is #2.  As a result, the lowest unoccupied molecular obital, or LUMO, in butadiene is #3.  We can extend this to ethylene and see that with 2 pi electrons, molecular orbital #1 is the HOMO and molecular orbital #2 is the LUMO.

MO #1

Cycloadditions MO Theory

Remember that cycloaddition reactions describe the formation of new C-C sigma bonds through rearrangement of the pi electrons in a conjugated system.  Therefore, we need to use the molecular orbitals of the pi electrons to drive the reaction.  Let’s take the reaction of butadiene with ethylene, the most simplistic Diels Alder reaction (4+2 cycloaddition).  We start with the most conjugated molecule, butadiene and examine its HOMO.  Then we look at the least conjugated molecule, ethylene and examine its LUMO.  Why?  We need the electrons from a HOMO to flow from one molecule to another, so we have to use the LUMO orbital that is not occupied for this process.  We see that nodes of the HOMO and LUMO align; that is the open and closed halves of the dumbbells align.  This leads to a forward reaction and formation of the product, cyclohexene.

MO #2

However, in another example where we have a (2+2 cycloaddition), we see the HOMO and LUMO of ethylene do not align thus forbidding the reaction.  However, we have the ability to excite electrons from the HOMO to a higher orbital using light (hv) or heat, thus changing the HOMO.  We review this concept in detail in a previous article. When we do this, the HOMO of excited ethylene and LUMO of ground-state ethylene align and the reaction produces cyclobutane.

MO #3

We hope this description has helped clarify this concepts of MO theory. This is an example of the clear and simple expatiations we have prepared for over 175 reactions commonly seen in Orgo1 and Orgo2 courses.  Sign up today for your study preparation needs!

Tips on Sigmatropic and Electrocyclic Reactions

Posted on May 18th, 2016

One of the most common difficulties in Orgo 2 is the concept of sigmatropic and electrocyclic reactions.  Both of these reactions involve conjugated pi bond systems, which are multiple, adjacent carbon atoms with connected pi bond orbitals.  The Greek word sigmatropic literally means “sigma(bond) changing”, thus when presented with a conjugated pi bond molecule where bonds are broken and formed, the driving force of this reaction is carried out by moving the pi-bond system around the molecule.  Two types of sigmatropic reactions are common for this class.  First is the Cope rearrangement.  This is also referred to as a [3,3] sigmatropic rearrangement.  Many students struggle with with nomenclature, especially since the [x,x] nomenclature is most often presented on tests.  What does it exactly does [3,3] or [1,5] mean?

Lets look at a few examples.  Below is a typical Cope Rearrangement with [3,3] sigmatropic rearrangement.  We see that the reactant has 2 pi bond (or 4 pi electrons) in a pseudo-cyclic arrangement. Right now, each alkene has one substituent that is not an H.  if we move the pi-electrons around the ring, we can form a resonance structure (middle molecule) which is a complete ring.  The product is to break the sigma bond in red and form a sigma bond in green.  But why are there three arrows (meaning 6 electrons) involved in the mechanism?  This is because while in the reactant the red bond electrons are sigma, when the molecule rearranges, they will be used to form a new pi bond, while the reactant pi bonds will donate electrons to become the green sigma bond. Now one of the alkenes has two substituents, which is more favorable and drives the reaction to the right.

A question that comes up frequently is what the numbering of the rearrangement means?  To determine this, there a few easy steps to follow;

  1. Identify what bond will break and what bond will form.
  2. Draw an imaginary plane that intersects these two bonds.
  3. Count the atoms starting from the top of the plane from where the bond was broke to where it forms. This is the first number [3.x]
  4. Count the atoms starting from the bottom of the plane from where the bond was broke to where it forms.  This is the second number [x.3]
  5. Placing them together, we can see that a [3,3] sigmatropic rearrangement means a new bond is formed between atom 3 above and atom 3 below the imaginary plane.

The same is true for non-carbon atoms in the ring.  The second example looks at a [1,5] sigmatropic rearrangement of hydrogen (also called [1,5] hydride shift).

  1. Identify what bond will break and what bond will form.
  2. Draw an imaginary plane that intersects these two bonds.
  3. Count the atoms starting from the top of the plane from where the bond was broke to where it forms. In this case, there is only one atom, the hydrogen.  This is the first number [1.x]
  4. Count the atoms starting from the bottom of the plane from where the bond was broke to where it forms.  This is the second number [x.5]
  5. Placing them together, we can see that a [1,5] sigmatropic rearrangement means a new bond is formed between atom 1 above and atom 5 below the imaginary plane.

sigmatropic article 1

In cases where oxygen is present as an ether, the name of the reaction is changed to Claisen Rearrangement.  However, the mechanism is the same.  The aldehyde or ketone product is almost alway preferred to the ether reactant shown here.

sigmatropic article 2

 

Another question that is typically a major stumbling block of students is the rules for choosing reaction conditions for electrocyclic reactions.  This refers to the rotation of pi bonds  to form new sigma bonds or the rotation of sigma bonds to form new pi bonds.  The two atoms can rotate their orbitals in such a way that they rotate in the same directions (conrotory), which gives trans- orientation, or in opposite directions (disrotory), which gives cis- orientation.  The main point is that the substituents on the these atoms will be in different stereochemical arrangements.  When a chemist wants to choose the correct stereochemical outcome of the molecule, you have to choose the right conditions to get the substituents of these electrocyclic atoms in cis- or trans- to each other.

 

Below is a convenient table to help you identify the right condition for the reaction presented to you.

electrocyclic

 

 

How to approach synthesis problems

Posted on February 9th, 2016

Q: “I am in organic chemistry 2 and I am trying to learn how to predict the products.”

One of the most challenging, but ultimately the main goal, of organic chemistry is to take an abundant, cheap starting reactants and transform it into a biologically active or synthetically useful product.  In the two semesters of organic chemistry, you will learn reactions that will produce all of the functional groups that give molecules these useful properties. But as a chemist, it is very importantly to accurately predict how and in what yield these reactions will produce you desired products.

The stakes seem higher when you are given less than 2 hours to complete several multi-step synthesis problems that often ask to identify not only reagents but the mechanism of each step of the reaction.  We here at StudyOrgo have compiled advice and guidance for learning to tackle these problems.

  1. Relax!
  2. Understand that each step in a synthesis reactions can be broken down into two basic types of reactions;
    1. Change in the carbon-carbon skeleton: Take a look at the starting material and the final product. Count the carbons, is there a difference in number?  If so, then you will need a carbon-carbon bond breaking or addition reaction.  There are only a few examples of these reactions so that instantly narrows down reactions you have to focus on.
    2. Change in the identity or location of a functional group: This is the most common synthesis question. Remember, that the pathways from one functional group to another are limited.  Set up a reaction roadmap or use included with your StudyOrgo membership to help you memorize what reactions can be used to get you to your final functional group.
  3. There are often times MANY routes to get to a final product. ALL OF THEM ARE CORRECT, if in fact the route is possible.
    1. HINT: Professors will only ask you to use reactions they have covered in class. Make a list and indicate what the reaction does to help you memorize the products.
  4. Sound out the problem. Often time students have a viable synthesis approach, but get held up at one or two places and just leave the entire question blank. Remember, it is multistep, which means there are many places to get points!  Draw out what you know.  Worst-case scenario: you get partial credit.  Best case scenario: you realize what reaction is next and get full points!!
  5. Practice, practice, practice! Make sure to perform all of your practice problems assigned to you for each chapter.  It takes a lot of time and effort, but likely these examples will be used by your professor.  Also, the more examples you see the less surprises will be on the exam!
  6. Sign up with StudyOrgo for help! The Editors at StudyOrgo have spent numerous hours reviewing and preparing the material in the most crystal-clear and “get-to-the-point” manner as possible. We provide quick descriptions and in-depth mechanism explanations. Many of our reaction have multiple examples, so you can learn and then quiz yourself in our website! For the student on-the-go, we have also developed a mobile app (iOS and Android) provides all the functionality of the website! All of these benefits are included in your StudyOrgo membership!

 

 

Epoxide Reactions

Posted on February 3rd, 2016

Many students taking Orgo 1 have commented there are a few types of reactions the professors save to the end of the semester and cover quickly and “gloss” over or sometimes skip all together in the interest of time.  However, in Orgo 2, you will be responsible for all of the reactions necessary for multi-step synthesis (starting product known to get to unknown final product) and retrosynthesis (product known to get to unknown starting material) reactions.  We at StudyOrgo don’t want you to get stuck on trying to cram for exams by studying reactions that were poorly covered in your class.  In this article, we focus on epoxide formation and epoxide ring-opening reactions because of their usefulness in synthesis and industrial application.

Epoxide Formation

In order to form an epoxide, a electron-rich reagent is required, such as an alkene.  Formation of the epoxide occurs in the presence of a peroxide reactant, such as MCPBA or DET.  The choice of reagent depends on the stereo-specificity desired from the reaction.  MCPBA is used to add the epoxide symmetricaly across the double bond, so that substituents that are cis- or trans- remain in this configuration.  The Sharpless asymmetric addition allows the researcher to choose a stereoisomer of DET, referred to as (+) and (-).  This reagent allows for the attack of the peroxide intermediate on only one face of the alkene, allowing for the production of nearly pure enantiomeric excess product, generally >98%!

epoxides article 1

Ring-opening Reactions

Many of you have probably hear of epoxy-glue, which is a very strong binding agent.  The process usually involves mixing two reagents and you must quickly apply the mixture to the broken items before the expoxy hardens.  One tube will contain the resin, or epoxide, while the other contains the “hardening” agent, which is the nucleophile that will attack the epoxide.  In a ring-opening reaction, a molecule such as TETA, which contains 4 amino groups, will attack 4 equivalents of oxirane to produce a complex polymer, which is the basis for a strong glue.  Changing the size and complexity of the epoxide can allow for flexibility of strength, thermostability and rigidity!

epoxides article 2

We here at StudyOrgo have devoted countless hours to preparing complex reaction mechanisms in simple and easy-to-understand manner to help you maximize your studying.  Sign up with StudyOrgo for detailed explanations of epoxide reaction mechanisms and other essential Orgo 2 reactions today!

 

Review of Substitution and Elimination Reactions

Posted on January 5th, 2016

Happy New Year from StudyOrgo and congratulations on finishing the first semester of organic chemistry! Before you begin classes next semester, take a few days to review the main topics from Orgo1 to prepare you for the second semester of organic chemistry.  Not always advertised, this semester will be very different.  By now, your professors have covered all of the fundamental parts of reaction mechanisms and have covered a variety of mechanisms as examples. (i.e., arrow pushing, pKa, intermediates, transition states, etc.)  This semester, you will see hundreds of reactions that you will have to commit to memory and then recall them in synthesis and retrosynthesis reactions on the final, hence the harsh reputation of organic chemistry.

This winter break, we will review the key reaction mechanisms and hints of Orgo1. Sign up with StudyOrgo today to see all 175 reaction mechanisms, including detailed reaction mechanism explanations and diagrams!

The most commonly referenced reactions from Orgo1 are the substitution and elimination reactions. One of the hardest concepts is determining what reaction mechanism is predominant under certain conditions.  The BIGGEST factors in predicting reaction mechanism for these reactions is type of nucleophile and substrate substituent (1°, 2°, 3°).  It is important to remember that these reactions are usually ‘competing’ with each other, hence there will be major and minor products of different reaction mechanisms.  Your grade will hinge on predicting the MAJOR product.  In some cases, the substrate can only react in one way.

For instance, any haloalkane (1°, 2° and 3°) will react with a strong base, such as hydride (H-), to produce an elimination product. But in cases of a 2° substrate, there will be a mixture of substitution and elimination reactions, unless there is a methyl or hydride shift to produce a more stable tertiary carbocation (SN1 and E1 mechanisms).   In those cases, other factors help to predict stability of the transition state or intermediate such as solvent choices (polar protic vs. polar aprotic), strength of leaving group and steric effects.

To help you see the most likely result of reactions, we have compiled a useful table to help relate degree of substrate substitution to the choice of nucleophile.

Untitled

 

As you can see, the choice of nucleophile can dramatically change the reaction observed.  Key points are as follows;

  • strong bases will almost always give elimination products
  • weak nucleophiles and bases will almost always give substitution products
  • 1° will always undergo an SN2 or E2 mechanism
  • 3° will almost always undergo an SN1 or E1 mechanism

We hope that this learning aid will help you answer any questions you may have had about substitution and elimination reactions. We here at StudyOrgo have compiled hundreds of reactions with clear explanations to help you speed up your studying and get a great grade in organic chemistry.  Sign up today to get access to all of our reactions!