Archive for the "Tips" Category

Preparing for Organic Chemistry This Fall

Posted on August 5th, 2017

One of the questions we are repeatedly asked at StudyOrgo is “how do I to get ahead in organic chemistry this fall semester?” Many of you have heard that organic chemistry is a brutal class that does little but to depress your GPA. While it is true that this course is challenging, we here at StudyOrgo are devoted to helping you get the “A” you deserve!

Organic chemistry gets a bad name because it assumes that you are experts with regards to all of the general chemistry from freshman year, and you are now responsible to know it!  As an analogy, think of your chemistry courses as a pyramid to reaching your degree goals.  Organic chemistry is directly placed in the middle of the pyramid, it will be very important not only for the MCAT or DAT exams, but also for future advanced courses.  Organic chemistry is supported by General Chemistry, which is why you took it last year.  Fortunately, StudyOrgo is placed in the center of your pyramid base and we are here to help all of your organic chemistry questions.  Our simple and clear-cut explanations of reaction mechanisms and concepts will easily help you with anything you might struggle with this semester.  Here are a few tips on how to prepare today for the course this Fall.

chemistry pyramid

  • Open your text book –Read the title and abstract on the first page of each chapter and check out the number of pages. It will give you a very quick idea of what you will be learning about in each chapter and how much material you will be covering.
  • Look at a syllabus – Remember, your syllabus is an official contract between you and the professor. They must disclose what you are required to learn and how you will be graded. Professors can remove requirements but cannot easily add them. Use this to your advantage! Highlight the contents or reactions of the book that will be required and use this to focus your attention on while studying over the semester.
  • Schedule your studying! – Now that you know where the book is and a rough idea of what you are responsible for learning from the syllabus, take a calendar and divide the time you have to each test by the number of chapters. Schedule 2-3 hours a week to study and DON’T SKIP OR RESCHEDULE. Think of it as a doctor or dentist appointment – you just have to do it! Also, if you plan your studying ahead, you will be less likely to schedule something that gets in the way because you will already have penciled it in! Use your Smartphone calendar to send you alerts and reminders for your studying appointment.
  • Read ahead – If you have time this summer, read at least two chapters to get yourself ahead of the class. Don’t try to understand everything, just pay attention to the words used and the ideas. This will allow you to pay more attention and ask questions about the details in class instead of scrambling to write down notes and drawings.
  • Sign up with StudyOrgo – 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 consult students and ask for their opinion on whether they understand the material as presented. 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!

With a little time management and help from StudyOrgo, you will have no trouble getting an “A” in Organic Chemistry this year!

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How can I tell if a hydrogen is a wedge or a dash in a chair skeleton?

Posted on March 19th, 2017

 

“How can I tell if a hydrogen is a wedge or a dash in a chair skeleton?”

Here at StudyOrgo, we frequently get questions about topics in organic chemistry that are usually quickly covered, poorly described or expected that you know from previous courses.  These concepts are really important to understanding the more complex topics to come.  In this article, we will cover the concepts of stereochemistry descriptions using bold and wedged bonds.  This is just a preview of the detailed topics and materials available with your membership to StudyOrgo.com.  Sign up today!

The first thing we have to do is determine is how you want to orient you molecule.  Let’s take (1R, 2R) 1,2-dimethylcyclohexane for example.  If we orient the molecule to have the methyl groups on the right side, we see that we have two stereocenters available.  But the current drawing doesn’t indicate the stereochemistry yet.  That’s what the bold and hashed bonds will indicate.

Next, we have to visualize the cyclohexane ring in the chair conformation.  Remember, that the skeleton image shown above is more conveniently drawn, but loses the 3rd dimension information, so you have to put it back in the chair to determine which should be bolded and which should be wedge.

Next, we have to confirm that that the stereochemistry is correct.  To do this, you need to practice selecting most important substituents and rotating to assign stereochemistry.  Follow along with the examples below, using the blue and pink carbons shown.

At this point, you should be able to see how the hashed and bolded bonds are now appropriately drawn.  The pink stereocenter will be bolded, suggesting it is above the plane of the ring and the blue stereocenter will be hashed, suggesting it is below the plane.  Drawing the Newman Projection down the red bond shows that the methyl groups are “anti” to each other, making this a stable conformation.

 

 

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The SN1 Reaction

Posted on September 8th, 2015

Another reaction commonly covered in the first weeks of organic chemistry is the SN1 reaction. The SN1 reaction introduces you to repetitive concepts and rules you will encounter all semester, this time focusing on carbocation formation and reactivity. In this article, we will review the important topics of an SN1 reaction.  Sign up with StudyOrgo today to get detailed reaction mechanisms and explanations to prepare you to ace your next exam!

Alkyl halides as SN1 substrates

One of the most reactive molecules involving substitution reactions via SN1 are 2° and 3° alkyl halides.  However, there are a number of considerations to keep in mind to determine if this mechanism of substitution describes your reaction. First, let’s look at a simple SN1 reaction; a sec-butyl halide (a 2° methyl-ethyl carbon center).

sn1 figure 1

Carbocation formation and stability: Let’s break down the reaction name more simply.  The term SN1 reaction gives you 3 pieces of information, first the ‘S’ indicating ‘substitution’, the ‘N’ denoting the reaction involves a nucleophile and ‘1’ describing the process as unimolecular – meaning only the formation of the reactive substrate intermediate determines the rate of reaction.  This process is referred to as the rate determining step of the reaction, and can be thought of as the ‘bottleneck’ in the reaction. The leaving group will break the bond to carbon and take the electrons for the bond with it forming a carbocation intermediate.  Halogens are good leaving groups because of the inductive effects (or electron withdrawing potential) of the halogen atom and is the characteristic of good leaving groups. Carbocation formation is the first, and rate determining step, in the reaction.

sn1 figure 2

Product formation and racemization: Once the leaving group bond is broken, stability of the carbocation is the factor that determines if this mechanism occurs.  The more substituted the carbon center, i.e. 2° and 3°, the more stabilized the carbocation becomes as the positive charge becomes delocalized to the other carbons. Following formation of the carbocation, it will then react with the nucleophile.  Since the carbocation assumes a planar shape, attack by the nucleophile can occur from either side of the plane.  This leads to formation of a mixture of enantiomers, referred to as a racemic mixture.  This is in contrast to SN2 which will only produce the inverted stereoisomer of the reactant.

sn1 figure 3

Carbocation Rearrangement: As mentioned before, stability of the carbocation is the key step in determining rate and completion of SN1 reactions.  In some instances, the leaving group is bonded to a carbon center than neighbors a more substituted carbon center.  Let’s consider the reaction below, chloride leaves 2-chloro-3-methylpropane to form a 2° carbocation. The neighboring carbon center is 3°, and would make a more stable carbocation.  In this instance, the neighboring hydrogen will shift to the 2° carbocation to form a new 3° carbocation, which is much more stable in a process referred to as a 1,2-hydride shift. Attack of the methanol hydroxyl group on the carbocation followed by proton abstraction by chloride leads to formation of the 3-methoxy-3-methylpropane product.

sn1 figure 4

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The SN2 Reaction

Posted on August 31st, 2015

The start of first semester organic chemistry can be an information overload.  For the first few classes, you will review general chemistry concepts and then… the reactions start coming!  One of the first reactions that will be covered is the SN2 reaction, mainly because it is relatively straight forward and a good tutorial for how to describe reaction mechanisms.  In this article, we will review the important topics of an SN2 reaction.  Sign up with StudyOrgo today to get detailed reaction mechanisms and explanations to stay on top of your class!

Alkyl halides as SN2 substrates

One of the most reactive molecules involving substitution reactions are alkyl halides.  However, there are a number of considerations to keep in mind to determine if the SN2 mechanism describes your reaction. First, let’s look at a simple SN2 reaction; methyl chloride and NaOH to form methanol and NaCl.

sn2 figure 1

Let’s break down the reaction mechanism into the basic elements.  An SN2 reaction gives you 3 pieces of information, first the ‘S’ indicating ‘substitution’, the ‘N’ denoting the reaction involves a nucleophile and ‘2’ describing the process as bimolecular – meaning both the substrate and the nucleophile determine the rate of the reaction.  The hydroxide will attack the carbon center and form a new bond with carbon (which makes it the nucleophile) and the chlorine atom will leave the carbon center with the electrons from the C-Cl bond (which makes it the leaving group).

Inductive effects of leaving groups: Chloride is a good leaving group because of the inductive effects (or electron withdrawing potential) of the halogen atom.  This is the characteristic of good leaving groups.  The electronegativity of chlorine makes the carbon center slightly electrophilic, meaning it has a partial positive charge, which is strongly attracted to electron-rich nucleophiles.

sn2 figure 2

Strong bases as a nucleophile: In order to form a new bond with carbon, a good nucleophile has to be electron rich.  The strong basic properties of NaOH make the charge on oxygen negative, and thus a good nucleophile.  Likewise, the poor basic properties of Cl anion make it an excellent leaving group.  Below is a chart to help illustrate the contrasting properties of nucleophiles and leaving groups.

sn2 figure 3

Inversion of stereochemistry due to geometry of attack: Once the nucleophile attacks the carbon center, a partial formation of C-O bond and breaking of C-Cl bond occurs in a concerted (or instantaneous) fashion, depicted below.  Because the angle of attack for the nucleophile has to be opposite of the leaving group, the OH adds to the opposite side of the carbon center, causing an inversion of stereochemistry.  This is an important clue in determining if reactions occur using the SN2 mechanism.

sn2 figure 4

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Chirality and Assigning Stereochemistry to Molecules

Posted on August 11th, 2015

One of the most important skills to master in organic chemistry is the ability to assign stereochemistry.  We at StudyOrgo have devised clear cut explanations of these difficult concepts for students to maximize their time studying and learn difficult concepts quickly and easily. Sign up with StudyOrgo.com today for all of your organic chemistry studying needs!

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

Chirality is defined as any object in which the mirror images are not superimposable. A good example is your hands; they are mirror images but not superimposable. Translating this to organic molecules, a stereocenter is a carbon center with 4 unique substituents that are arranged such that the mirror image is not superimposable. Thus, they “look” like to different molecules although they have the same substituents. If we alter the arrangement of the substituents, we can always come up with 2 arrangements for each substituent, R or S configuration.  Thus, each stereocenter must have 2 stereoisomers.

chiral 1

In order to determine whether the sterecenter is the 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
  2. Assign priority based on atom atomic number, highest (1) to lowest (4) weight.
  3. If two atoms are same, move to next bond to find first point of difference
  4. Rotate the molecule so that Priority 4 atom is in the hashed wedge position.
  5. Determine the Priority sequence 1-2-3 rotates to the left (S) or the right (R).

chiral 2

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.

Screen Shot 2014-12-18 at 1.00.28 PM

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.

Let’s consider the biologically active form of testosterone, 5-DHT which is shown below.  We indicate that it has 7 stereocenters in the molecule.  Applying the 2n rule, we calculate 128 possible stereoisomer combinations.  That concludes that while testosterone has 1 enantiomer, it has 126 diastereomers and remember…only 5-DHT works on our bodies!

Screen Shot 2014-12-18 at 1.02.36 PM

 

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