Posts Tagged "organic chemistry"

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 StudyOrgo.com, 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

 

 

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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

 

 

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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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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.

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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

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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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