Archive for the "Organic Chemistry General" Category

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.



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!



Review of Organic Chemistry 1 Terms

Posted on December 21st, 2015

Many of our students are unclear on a number of key terms that will be necessary to move forward in organic chemistry.  We have compiled a list of commonly misunderstood terms and explain them here.  With this review, our quiz mode review of all of the reactions you have learned and descriptions from, you will be sure to boost your final exam score and get a great grade in your class!

Isomers – There are two types of isomers in organic chemistry.

  • Constitutional isomers – two or more molecules with the same number of atoms but in a different geometrical arrangement (i.e. different connectivity).
  • Stereoisomers – molecules with the same geometrical arrangement (i.e. same connectivity) that are not superimposable on each other. For a carbon center (referred to as a stereocenter), this requires bonding to four different substituents!

chiral 1

  • Enantiomers – A pair of stereoisomers that are mirror images of each other.
  • Diastereomers – Any pairing of stereoisomers that are NOT mirror images of each other.
  • Meso compounds – A molecule with stereocenters that shows symmetry in reflection. Because of this symmetry, the molecule is considered achiral!

Screen Shot 2014-12-18 at 1.00.28 PM

Newman Projections – A way to visualize different rotational conformations of substituent comparing two carbon atoms that looks down their C-C bond, thus showing the alignment of the substituents.

  • Gauche conformation – when the angle between two substituents is 60°
  • Anti conformation – when the angle between two substituents is 180°
  • Eclipsed conformation – when two substituents overlap, or the angle between them is 0°

drawing 6

Stereoselectivity Reaction Terms – describes the reaction conditions that leads to UNEQUAL stereoisomer formation

  • Syn addition – addition of substituents on the same face of the place of symmetry across a double bond.
  • Anti addition – addition of substituents on the opposite face of the plane of symmetry across a double bond.
  • Inversion of configuration – substitution of a nucleophile with the opposite stereochemistry as the starting material (e.g., R ® S). This stereospecificty happens for ALL concerted mechanisms (SN2, E2, etc.)
  • Racemization (racemic mixture) – substitution of a nucleophile which produces both stereoisomers (e.g., R ® R + S). This stereospecificity happens for ALL carbocation (SN1, E1, etc.) intermediates.


Regioselectivity Terms – describes the reaction conditions that leads to UNEQUAL constitutional isomer formation.

  • Markovnikov selectivity – in alkene addition reactions, the placement of hydrogen will occur on the LEAST substituted carbon (a.k.a. carbon center greatest number of hydrogens).
  • Anti-Markovnikov selectivity – in alkene addition reactions, the placement of hydrogen will occur on the MOST substituted carbon (a.k.a. carbon center with fewest number of hydrogens).
  • Zaitsev product – in elimination reactions, the formation of the alkene with the MOST substituents is favored. (e.g., E2 elimination with a non-bulky base such as sodium ethoxide).
  • Hoffman product – in elimination reactions, the formation of the alkene with the LEAST substituents is favored. (e.g., E2 elimination with a bulky base, sodium tert-butoxide).

terms figure 2

Intermediate Terms – describes intermediates that are key to reaction progression.

  • Carbocation – formed any time a leaving group breaks a bond with carbon to generate a carbon center with 3 bonds and a positive charge.
  • Oxonium ion – formed any time an alcohol is protonated.
  • Mercurium ion – formed in oxymercuration/demercuration addition reactions
  • Bromonium ion – formed in halogenation addition reactions of alkenes
  • terms figure 3



Passing the First Test by the “Hard Professor”

Posted on September 29th, 2015

Many students have commented to us as that they have the “hard” organic chemistry professor. They are usually described as being picky graders, asking impossible problems and giving no feedback on what the student did wrong. But our experts at StudyOgo are here to let you know…there is no upper hand any professor has over their students.  The materials and principles of Orgo 1 have not changed in over 50 years!  The good news is that there is no question a professor can ask that isn’t straight out of your text book. So what makes them so hard? Here are a few types of professors and advice on how to meet or beat their course!


Problem #1: Bad presentation.
This is by far the most common problem of “hard” professors.  Hand-written notes, a ‘chalk-talk’ where they do more erasing than writing, or a PowerPoint with figures straight out of the text book but no explanations. Most of these teaching tools are not very useful for the confused Orgo student, because it causes more confusion than it clears up.  This leads to frustration and makes the student fall further behind.

  • Keep up to date with the material.  Divide your time over how many chapters of material you have and this will give yourself a deadline to complete the material.
  •  Read the book, as painful as it sounds, and each assigned chapter non-stop the first time through and the next day, go back to problematic sections for help.
  • Sign up with! Our team of experts as developed a custom presentation of difficult concepts in organic chemistry into an easy to understand format with a step-by-step breakdown and description of common reaction mechanisms in organic chemistry, complete with quiz-mode to check yourself once you think you have the hang of the reaction.  Check out free radical halogenation on our website!  Are you studying on the go? Check out the mobile app for mobile flashcards to pass the time on the train or bus!


Problem #2: Separating the A’s from the B’s.

You are likely in a class where Orgo Chem is a degree requirement.  Many professors will throw in “really hard” questions that terrify students and appears heartless.  Professors do this to assign A’s to the students who have kept up and followed along the whole time.  We believe you can be one of the few who aces these questions!

  • Check out the solution manual for your text book from the local library and try as many problems as you can on the material you find most difficult.  Remember; there are only so many ways a professor can ask you a question.  If you see a ton of practice problems, the probability of them asking a question you have already seen is extremely high. This means you will be ready for any question they ask.
  • At, we break down each mechanism in detail so when you practice your problem sets, you’ll have all the details.


Problem #3: High expectations.

Many times, professors will expect you to apply your knowledge to a problem you haven’t seen before.  After all, this is what scientists do every single day!  Since almost all professors are scientists, they often mix their research ideals with teaching, which can make it seem very hard.  But the experts at StudyOrgo know you can do it!

  • Learning organic chemistry is like building pyramid; the top will fall without a strong base. Go back to Chapter 1 and complete the assigned questions and DO NOT STOP until you can answer them all!  By the middle of the practice problems, you will start to feel like this isn’t so bad.
  • Then go on to try Chapter 2, Chapter 3, and so on and in no time you’ll be ready for the big test!
  • When confronted by these questions, think: “what is this question asking for that we covered already?”  When you come up with an answer, this can help you narrow down what concept to recall and help beat that “overwhelming” anxious feeling after reading the question.  After you relax, you’ll be ready because of all of you preparation!

Following these tips will allow you to pass any Orgo Chem class taught be even the most difficult professor. Although it might not seem like it in the moment, they want you to do well so get out there an impress them!

Free Radical Halogenation

Posted on September 22nd, 2015

Another common mechanism that is covered in the first weeks of organic chemistry is the free radical halogenation of alkanes.  This mechanism utilizes the homolytic cleavage (one electron per atom) property of halogens when exposed to heat or ionizing radiation (i.e. hv), which is a popular mechanism for future reactions in the course.  Radical halogens can extract the proton from a C-H bond to produce the corresponding acid and generate a radical carbon center.  In this article we will discuss all of the tips and tricks to getting an ‘A’ on your racical halogenation questions.  Sign up with StudyOrgo today for more in-depth mechanism coverage and answers to all of your organic chemistry questions!

Generating a radical halogen: there are THREE critical steps to free radical reactions.

1) Initiation: The Br2 single bond is broken by high energy ligh (hv) to form radicals placing one electron on each atom.

halogen 1

2) Propagation: (Hint: One radical reacts with a single bond to form another radical, thus propagating the radical species to drive the reaction forward.

  1. a) Radical Br abstracts one hydrogen from a C-H bond in propane to form radical propane and HBr.
    halogen 2
  2. b) Radical propane asbracts one Br from Br2 to form the bromoalkane and radical Br, thus restoring the reactants for another round as shown in step 2a.halogen 3

3) Termination: Any two radicals combine to form a single bond.  These species will be in low abundance. Hint: Radicals are destroyed by combining two radicals to form a single bond.  This eliminates the radical necessary for radical alkane formation (green boxes) as shown in step 2a and ends the reaction.

halogen 4

Regioselectivity: How to determine the major product

Radical bromination will always replace the C-H bond on the MOST substituted carbon center because the stability of the radical intermediate is higher with increasing substituents on the carbon center.

This selectivity is the same, but a weaker consideration, for radical chlorination which obeys Hammond’s Postulate, which says that stability of the radical center is outweighed by the extreme exothermicity of radical chlorination (compared to bromination), thus a mixture of chlorinated products is observed.

halogen 5

Stereoselectivity – How to determine the stereochemistry of carbon centers

Radial intermediates (step 2a product) produce a sp2-like hybridization orbital with the lone electron in the vacant 2p orbital, therfore attack of the radical electron on the C-H bond can take place from either side of the molecule.  The result will always produce a racemic mixture (or equal amount) of the two enantiomers.

halogen 6