Archive for the "Reactions" Category

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 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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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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