Posts Tagged "organic chemistry"

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

Drawing Organic Molecules

Posted on August 24th, 2015

Drawing organic molecules is essential to getting a great grade in organic chemistry.  Often times, professors will deduct many points from students who understand the material just because their drawing are horrible!  Unfortunately, this class can be as much about art as the science but remember it is important not just to understand but to communicate that you understand!  Only lots of practice and these tips from StudyOrgo will guarantee you will get all the preparation you need to ace the next exam.

Sign up for a membership with StudyOrgo today to get help with all of your organic chemistry questions!

Two-dimensional line structures

The simplest drawings in organic chemistry are line drawings.  General chemistry often indicated each element bond in a molecule.  Organic chemists use complicated molecules and time is precious.  Therefore, a quick way to draw hydrocarbons was necessary. There are a few rules to help you draw the appropriate structures in organic chemistry.

  1. Draw carbons in a zigzag pattern, points are carbon atoms and lines are bonds. All other atoms bonded to carbon, unless otherwise stated, are hydrogens.
  2. Draw all bonds as far away as possible.
  3. Drawing single bond carbons in any direction is equivalent.
  4. Never, never, never draw more than 4 bonds to carbon!!!

drawing 1

Three-dimensional structures

Dash and wedge: One piece of information that is lost in the line drawing of molecules is the three-dimensional arrangement of the substituents around carbon atoms.  This becomes very important when dealing with stereocenters, which will have two possible enantiomers.  To describe this arrangement, chemists use the dash and wedge model.  Imagining the paper (or computer monitor in this case) is the mirror plane, the dash indicates a bond below the plane and the wedge indicates a bond above the plane.  Looking at the example below, we see that the stereochemistry of 1-chloroethanol is ambiguous when drawn as a line structure.  Drawn as a dash and wedge, it becomes very clear that the stereoconformation shown is R-1-chloroethanol.

drawing 2

Fischer projection: For acyclic molecules, especially those with many substituents, chemists will use the Fisher projection to make drawing the molecules more rapid.  The convention depicts bonds drawn from top to bottom are in the dash conformation while the bonds drawn from left to right are in the wedge conformation.  This easily, and quickly, indicates the stereochemistry without the cumbrous bold and dashed bonds, as shown to the right.

drawing 3

Haworth projection: For cyclic molecules, chemists utilize the Haworth projection. The projection depicts the ring on an angle, half above the plane of the paper and halve behind the plane of the paper.  This allows the substituents to be drawn in the plane of the paper.  Take for instance glucose, shown below.  The Haworth projection to the left is commonly drawn to depict whether the alcohol groups are above or below (alpha- and beta-, respectively) the plane of the ring.  A more comprehensive drawing places the ring in a chair conformation, which will indicate whether the substituents are in the equatorial or axial.  This allows one to determine the stereochemistry of each stereoisomer.

drawing 4

Newman projection: Recall that even with three-dimensional configuration of atoms, there is free rotation about sp3 hybridized bond.  While it is customary to draw the structure in the lowest energy arrangement, we sometimes consider other rotations about a C-C bond that could affect reaction mechanisms.  For this, we utilize the Newman projection.  Looking “down the barrel” of the sp3 bond, we place carbon #1 substituents in the front (in red) and carbon #2 substituents in the back (black).  The lowest energy conformation is to arrange the substituents of the two carbons in the ‘anti’ configuration, places Cl and Br opposite of each other.  We can rotate one of the atoms such that steric hindrance of the halogen orbitals causes strain.  It becomes higher in energy when at ‘gauche’, when the Cl and Br are not anti or in the highest energy ‘eclipsed’ configuration, when Cl and Br on top of each other.  Temporary rotation of the bonds can be important considerations when studying reaction mechanism in the future!

drawing 6

Acids in Organic Chemistry

Posted on July 20th, 2015

Key Concept #1: Identify the Bronsted Acids:  The first concept we have to keep in mind is there are two kinds of molecules, proton acceptors and proton donors.  Acids are considered proton donors. That means they have to give up a proton to another molecule.  Bases are therefore proton acceptors, meaning they have to accept a proton from another molecule, this requires at least a lone pair of electrons and usually a negative charge on the atom.

acids 1

Let’s look at the reaction above.  If we dissolve HCl gas into water, then the water becomes the Bronsted base because it accept the proton from HCl.  Since HCl is a strong acid, we know that this reaction far to the right.

Key Concept #2 – Determine [H3O+] in solution to measure acidity: As we dissolve Bronsted acids and bases into solution, the concentration of H3O+ in solution will change.  Remember though, that concentrations are in units of moles (M) and have a wide range of concentrations in solution (typically 10+1 to 10-15 M).  Since it is clumsy to use such large numbers, we transform this concentration into powers (the p in pH) of the concentration of H3O+ (the H in pH) in the equation below.

pH=-log[H3O+]

Now, let’s take a look the pH of some household items to relate what pH really means.  Milk has a pH of 6.7.  If we solve for [H3O+] in the equation above, we find this is 2×10-7M or 0.2mM.  This is very similar to water (pH=7) which is 1×10-7M or 0.1mM, so not very acidic.  Coca-Colaâ has a pH of 2.5, solving for [H3O+] we find the concentration of proton is 3.1×10-3M or 3,100.2mM or more appropriately denoted 3.1mM.  This tells us there is 15,000 times more acid in cola than milk or water.  So, it is important to keep in mind the change of pH by 1 unit equals 10-fold change in [H3O+].  There are even more dramatic changes in pH between household items that can help you to appreciate the relationship between pH and [H3O+] in the chart below.

acids 2

Key Concept #3 – Predicting the equilibrium of the reaction:  The numerical value associated with acidity is known as pKa, which is the equilibrium concentration of the acid and conjugate base. The higher the value, the more acidic the solution.  There are two key tips in predicting acidity; 1) equilibrium lies towards the weaker acid (a low pKa towards a higher pKa) and 2) equilibrium lies towards the most stable conjugate base.

Lets take a look at the following reaction.  If we know the pKa values of the acid and conjugate acid, we can easily see that acetic acid (pKa = 4.75) reacted with sodium hydroxide produces the conjugate acid water (pKa = 15.7).  Equilibrium will therefore shift to the right.

acids 3.jpg

Many times, however, you do not know the pKa value of the acid or the conjugate base.  Therefore we can predict equilibrium based on the stability of the conjugate base based on the follow four rules;

There are four factors to consider when comparing the stability of conjugate bases:

  1. Atom which has the charge—For elements in the same row of the periodic table, electronegativity is the dominant effect. For elements in the same column, size is the dominant effect.
  2. Resonance—a negative charge is stabilized by resonance.
  3. Induction—electron-withdrawing groups, such as halogens, stabilize a nearby negative charge via induction.
  4. Orbital—a negative charge in a sp-hybridized orbital will be closer to the nucleus and more stable than a negative charge in an sp3-hybridized orbital.

Below in the example, penanedione is reacted with sodium azide to make a enolate anion.  If we didn’t know the pKa of the acids, we could examine the base stability to determine equilibrium.  The azide anion as a negative charge on nitrogen and is somewhat stable. Although the negative charge on the conjugate base is carbon, which is less ideal than nitrogen, it has two neighboring carbonyls that contribute stability from resonance, inductive forces and the
sp-2 hybridization of the conjugate base.  For these reasons, we would predict the equilibrium to the right.  We in fact are correct since pKa of the acid, pentanedione, is 9 and the conjugate acid, ammonia, is 38.  Therefore, this reaction is 10^29 fold (i.e. pKa 38 – pKa 9) towards the right!

acids 4

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Naming Organic Molecules

Posted on June 18th, 2015

Naming molecules is a challenging exercise because there are many rules; the more complex the molecule, the more rules that need to be observed.  It is possible to name a single molecule in many different ways, but only one is correct!  These are easy points on quizzes and exams and can boost your grade if you know how to tackle them. We here at StudyOrgo have summarized the process down to 4 easy steps to name alkanes.  Follow along to determine the name of any molecule in your course.

Four discrete steps are required when assigning the name of an alkane:

  1. Identify the parent chain: Choose the longest chain. Looking at the first example below, several chain lengths are shown, however only the red trace contains the longest chain, therefore this is the correct parent chain. For two chains of equal length, the parent chain should be the chain with the greater number of substituents. In the second example below, two 7-carbon chains are possible, but only one gives 2 substituents, so it is therefore the correct parent chain!
    nomenclature 1
  2. Name the parent chain: In order to correctly name the parent chain, you will have to commit to memorizing the prefixes that denote how many carbons are in the parent chain. There is no easy way around it, so try covering the table up and recalling to help with memorization.
Number Of Carbons Parent Name AlkaneExample Number Of Carbons Parent Name AlkaneExample
1 meth methane 11 undec undecane
2 eth ethane 12 dodec dodecane
3 prop propane 13 tridec tridecane
4 but butane 14 tetradec tetradecane
5 pent pentane 15 pentadec pentadecane
6 hex hexane 20 eicos eicosane
7 hept heptane 30 triacont triacontane
8 oct octane 40 tetracont tetracontane
9 non nonane 50 pentacont pentacontane
10 dec decane 100 hect hectane

 

  1. Number the parent chain, identify and assign a location (carbon number) to each substituent: Just as with the parent chain prefixes, substituent prefixes are another terminology you will have t memorize! Give the first substituent the lower possible number. If there is a tie, choose the chain in which the second substituent has the lower number.  In the example of methyloctane, the methyl position could be at position 2 or 7. The lowest number prevails so 2-methyloctane is correct.
    nomenclature 2
Number Of Carbons In Substituent Terminology
1 Methyl
2 Ethyl
3 Propyl
4 Butyl
5 Pentyl
6 Hexyl
7 Heptyl
8 Octyl
9 Nonyl
10 Decyl
  1. Arrange the substituents alphabetically: Place the number corresponding to the position in front of each substituent. Note for the example below, there are four identical ethyl substituents, therefore we will label each position followed with the prefix tetra (di- for 2, tri- for 3, etc.) Note, the prefix is NOT included in alphabetization, as methyl comes after ethyl regardless of the “tetra” prefix in this example.
    nomenclature 3

And there you have it. With a little practice and help from StudyOrgo, you are well on your way to naming any organic molecule in your studies.  Sign up today for help with terminology, reaction mechanisms and more!

Studying for Summer Organic Chemistry

Posted on June 8th, 2015

One of the biggest challenges students can undertake is signing up for Organic chemistry in the summer.  How so?  Well, there are a few reasons.

First, the course time usually shrunk from 16 week to 8 weeks.  This means, longer classes and more frequent tests.  Secondly, the subject material is usually covered more quickly, but the content remains the same.  As such, students can feel overwhelmed and become exhausted studying the materials much more quickly than is typical.  But, we here at StudyOrgo have streamlined the process of studying organic chemistry mechanisms!

Our illustrative overviews with full mechanism descriptions and diagrams makes mastering any reaction in your class much easier.  And, with over 170 reactions, you’ll feel confident you’ll have everything you need to get the “A” in this course.  Here are a few tips for getting the right start to studying with help from StudyOrgo today!

  • Time management – Schedule your studying NOW! – Time management is the key to acing organic chemistry in the summer. 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 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.
  • 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 at will but cannot add them easily. 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 summer course.
  • Read ahead – Before each class, glance at the chapter to be covered that lecture beforehand. Don’t try to understand everything, just pay attention to the major words and phrases used and the ideas. This will allow you to pay more attention during class because you will already know what is being said, now you pay attention to the details. Most people are scrambling to write down notes and drawings in class, but not really paying attention.  Try it yourself, look at your classmates at the next class for a minute or two, they are usually feverishly writing!
  • Sign up with StudyOrgo – The Editors at StudyOrgo have painstakingly reviewed and prepared 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 StudyOrgo, you will have no trouble getting an A in Organic Chemistry this summer!