How to Approach Multi-Step Synthesis Questions

Posted on November 1st, 2018

 

 

Professors often begin to expect students to perform synthesis (reactant to product) or retrosynthesis (product back to available reactants) on the exam.  This can be one of the most intimidating exercises because the products and reactant can look so different from each other it would seem at first glance its not possible to figure it out.  But here at StudyOrgo, we work to bring you clear explanations and clear-cut explanations to tackle many organic chemistry reactions and synthesis questions you will see in Orgo1 and Orgo2.

 

Let’s take a look at the retrosynthesis of 3,3-dimethylbutyne from 3,3 dmethylbutanol.  At first you may see the reagent has an alcohol group but maybe you wonder “how am I going to get to an alkyne, we haven’t even seen alkynes yet!”

The first step is to see which bonds have changed between the products and reactants and we can see the bond colored in red.  Since only one bond changed, this is relatively “easy” example.

Starting from an alkyne, remember that each double bond represents the presence of a pi bond that can be formed from an elimination reaction.  So, think “what can we eliminate that would have resulted in pi bond formation?”  Remembering E2 reaction mechanisms, any neighboring anti-periplanar proton and Br (leaving group) can be eliminate with a strong base to form a new pi bond.  In the case of the alkyne, we would need two leaving groups and two rounds of elimination to form an alkyne.  Thus, if we form two Br bonds to each of the carbons, we can then form the alkyne from 1,2-dibromo-3,3dimethylbutane.  In the forward reaction, we can do this with a strong base such as sodium ethoxide (NaOEt) in ethanol to form the alkyne.

Next, we have to form the haloalkane.  To do this, we would have to carry out an addition reaction.  Seeing that we need a Br on each carbon, it is not too difficult to realize that if we added Br2 across an alkene, then we can then from the haloalkane from 3,3 dmethylbutene.  In the forward reaction , we can do this by reacting the alkene with Br2 in chloroform to form the haloalkane.

Next, we have to form the alkene.  At this point we are looking for a way to come back to the starting material, which is 3,3 dmethylbutanol.  We could think of forming an alkene from 3,3 dmethylbutanol by performing another E2 elimination (or dehydration) to remove the alcohol and anti-periplanar proton which would form 3,3-dimethylbutene. In the forward reaction, we can do this by activating the alcohol with tosyl chlroide to make a very good leaving group and then reacting with a strong base such as potassium tert-butoxide (t-BuOK) to form the alkene.

 

And there you have it!  A three step synthesis from the alkyne product from a starting alcohol using reactions that are typically covered in the first few weeks of class.  Practice is the only way to learn these tricks and the more reactions you know, the easier it will be to find shortcuts to faster synthesis schemes and solving more complicated examples which will surely be on your final exam.  Sign up with StudyOrgo.com today to master over 180 organic chemistry reactions and learn the mechanisms fast!

Drawing Cycloalkanes

Posted on August 8th, 2018

 

One of the difficult things to learn in Orgo1 is 3D drawing on 2D paper.  This is as much a skill of art as science and can be very difficult for students to pick up, making the process harder to learn.  Here at StudyOrgo.com, we have come up with clear-cut explanations and descriptions of hard-to-learn concepts and organized them on our website so you can spend more time practicing and less time searching.

 

Cycloalkanes can be one of the hardest things to draw in this class.  First we will address the Haworth projection,

What is the chair conformation?

Because there are six sp3 hybridized carbons in a cyclohexane ring, the chair conformation becomes the most stable conformation. The ring can be twisted in several orientations that are all higher in energy and thus less stable, but because the ring can be twisted it can potentially become inverted, which would flip all of the substituents of the ring in the opposite order.  In the example below, if we twisted the red and green carbon in the direction of the arrows, the ring would become “flipped” have an equally stable conformation.  In solution, both are possible unless there is steric strain that prevents one conformation over the other.

 

 

 

 

 

 

How do I draw a chair conformation correctly?

First, draw the structure VERY BIG.  This will make it much more easy to draw in the substituents.  To draw this projection, 4 steps are needed:

Step 1: Draw a very flat and wide “V”.

Step 2: Draw two lines at an angle away from the “V”.

Step 3: Draw one line that comes up at a steeper angle than the “V” and then shorter and off of the center of the “V” (i.e. sorter than where the dotted line is).

Step 4: Draw the last line that closes the ring.

 

 

 

 

What is Axial and Equatorial?

Because two bonds to each carbon are required to from the cyclohexane ring, each carbon can have two “substituents” or R groups.  One will be parallel to the ring in the “equatorial position” the other will be orthogonal to the ring and will be in the “axial” position.  Let’s look at his example.  If we start with pink bonds in axial and blue bonds in equatorial, a ring flip will invert this and place the blue bonds in axial and pink bonds in equatorial.  Both exist in solution UNLESS there is steric hindrance present that makes one conformation less stableAny bulky group or electronegative group will prefer to be in the equatorial position. This is because the equatorial position places the substituent away from the ring where there is free space for the substituent.  On the other hand, bulky groups in the axial position will experience “clashing” with the other axial positions because space is limited.  The more substituents on the ring, the more important it is to consider this when drawing the “most stable” conformation.

 

 

 

What is Diaxial Strain?

Let’s look at chlorocyclohexane, there are two possible conformations.  In one case, the chlorine atom is in the axial position and is experiencing 1,3 diaxial strain with the other hydrogens in the axial position.  If we do a ring flip, we place the chlorine now in the equatorial position, and only hydrogens are in the axial position.  This would be the most stable conformation.

 

 

 

 

 

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How to Interpret Thermodynamics of Reactions

Posted on August 8th, 2018

In this post, we will discuss the concepts and theories of Thermodynamics.  While the topic sounds complicated, we at StudyOrgo have extensive experience instructing principles and reaction mechanisms frequently covered in Organic Chemistry. Sign up today for clear, detailed explanations of over 180 Orgo Chem reactions and reviews on conceptual topics!

The Equilibrium Constant

Remember from general chemistry that equilibrium constant (Keq) reflect the ratio of products to reactants in the following equation.

Keq = Products / Reactants = [C]*[D] / [A] * [B]

Thus, if one knows the concentration of product or reactant in solution and the Keq for the reaction, the other value can be also know.  One point of confusion is that these values should somehow be intuitively known.  They are calculated in the lab for each reaction, with each type of reactants.  Therefore, without the “empirical” data we wouldn’t know the Keq.  Fortunately, many experiments have been done in the world and Keq values are frequently found in tables and reference materials.

Spontaneity and Gibbs Free Energy

An important factor in choosing reactions in organic chemistry is knowing how “efficient” this reaction will be.  Since our goal to generate products, knowing the Keq for the reaction will allow us to predict how much product we can theoretically produce from a known starting concentration of reactants.  We can convert the Keq ratio into energy (kJ/mole) which is a more comparable measurement between reactions using a version of the Gibbs Free Energy equation.

dG = RT*ln[Keq]

Remember, that a release of energy (a negative dG) indicates a spontaneous and forward (towards product) reaction.  As a reference, let’s look at a table of corresponding dG and Keq values and how this relates to concentration of products.

Delta G (kj/mol) Keq value % products at equilibrium
+17 1×10^3 99.90%
+11 1×10^2 99%
+6 1×10^1 90%
0 1 50%
-6 1×10^-1 10%
-11 1×10^-2 1%
-17 1×10^-3 0.10%

When dG is negative (exothermic), the Keq is very high (products >> reactants) and we see that most values have a >90% concentration of products when the reaction reaches equilibrium. However, when dG is positive (endothermic), the Keq is very low (products << reactants) and we see most values have <10% products.

Therefore, when selecting reactions in your synthesis scheme you will likely want to choose reactions with the highest Keq and most exothermic to generate high concentration.

The reaction coordinate

Reaction coordinates are often used to visualize concepts of free energy between the reactants and products of reactions.  Below are examples of endothermic and exothermic reactions.

The Y-axis reflects the free energy of the reactants (A + B) and products (C + D).  The X-axis reflects the progress of the reaction, as the path between reactants and products will have several hills and valleys that reflect transition states and intermediates, respectively which we will cover in another article. The thermodynamics of a reaction can all be visualized from graphs like these.  You will see them a lot in the course!

Determining Gibbs Free Energy from reaction Coordinate

The difference between initial reactant (blue) and final product (green) free energies will give you the Gibbs Free Energy (dG) of the reaction.  Remember, negative = spontaneous and positive = not spontaneous.  The larger the value, the more complete the reaction goes to either side.  For example, a large positive value means low products (no reaction) and large negative value means high products (efficient reaction).

Here at StudyOrgo, we have developed easy to follow review study guides and exercise sets to help with reviewing all the concepts you will have to master to pass the course!  Check out www.studyorgo.com/summary.php for help with core topics in Orgo 1 that you will need to succeed for next semester!

How to Assign Stereoisomer Configuration

Posted on July 2nd, 2018

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

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Molecules, like your hands, that are not superimposable on their mirror images are called chiral objects, from the Greek word cheir  (meaning “hand”). All 3D objects can be categorized as either chiral or achiral. Chemical molecules are 3D objects and can also be classified in this way. To help you remember, chiral molecules are like hands and are NOT superimposable on its mirror image while achiral molecules are like tennis rackets: they are superimposable on their mirror images.

IUPAC recommended in 1996 that a tetrahedral carbon bearing four different groups be called a chirality center.  Many other names are commonly used include chiral center, stereocenter, stereogenic center, and asymmetric center.  They all mean the same thing, a carbon connected to 4 unique substituents that is not superimposable on its mirror image.

When a compound is chiral, it will have one opposite molecule; a non-superimposable mirror image, called its enantiomer (from the Greek word meaning “opposite”). The compound and its mirror image are said to be a pair of enantiomers. The word “enantiomer” is analogous to the word “twin”. When two children are identical twins, each one is said to be the (“evil…”) twin of the other. Similarly, when two compounds are a pair of enantiomers, each compound is said to be the enantiomer of the other.  A chiral compound will have exactly one enantiomers, all other molecules with the same molecular formula and constitutional arrangement are diastereomers, but are only seen when there are more than two chiral centers in a molecule.

In order to determine whether the stereocenter is in 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

  • This one is easy for most, but just look for any carbon with 4 substituents that are different.  Be careful to carefully count chain lengths and identify unique elements.

2. Assign priority

  • Step 1: Assign priority of bond based on atom atomic number of the element, highest (1) to lowest (4) weight.
  • Step 2: If two atoms are same, count the type of bonds connected to the carbon to find first point of difference
  • For 2-methyl-3-pantanol, oxygen is highest priority and hydrogen is lowest priority. However, 2 carbons are connected to the stereocenter, therefore count the number of bonds connected to each carbon center. In this case, the carbon with 2C and 1H has higher priority than the carbon with 1C and 2H.
  • For the haloalkane, oxygen is highest priority and hydrogen is lowest priority. However, 2 carbons are connected to the stereocenter, therefore count the number of bonds connected to each carbon center. In this case, the carbon with 1Br and 2C is higher priority than the carbon with 1F, 1Cl and 1C because Br is the heaviest and highest priority element.

 

3. Rotate the molecule so that Priority 4 atom is in the hashed wedge position.

  • In some cases, if you can’t flip the molecule in your head or on paper easily, assign the configuration to the stereocenter when the 4th position is NOT in the back of the paper position, and just reverse the assignment.  It works every time.

4. Determine the Priority sequence 1-2-3 rotates to the left (S) or the right (R).

 

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.

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.

Formation of Enols and Enolates

Posted on April 3rd, 2018

One question that comes up in organic chemistry often is “what is an enol or an enolate and how is it formed?”  These types of concepts are frequently covered quickly in class or not at all, but are very important for future reaction mechanisms.  We at Study Orgo have the combined experience of over 15 years of tutoring and teaching organic chemistry concepts to struggling students.  We have developed clear descriptions of reaction mechanisms and organic chemistry concepts to aid students in their studies.  Sign up today for access to over 180 reactions mechanisms and reviews!

The alpha carbon of a carbonyl, which is present in carboxylic acids, esters, ketones and aldehydes, are acidic which means the proton can be removed using a base.  In neutral or acidic conditions, this means the lone pairs on the C=O position can act as a weak nucleophile.

If the carbonyl oxygen can attack the alpha carbon C-H bond, it will abstract the hydrogen and perform a Keto-Enol tautomerization reaction that will lead to the resonance version of the carbonyl, which is the Enol (alkENE + alcohOL)

Enols – rearranging the pi bond and atoms of a carbonyl compound to an Enol

Catalyist: Acidic or Neutral Conditions to stabilize OH formation

Enols tautomers are generally unstable, preferring the “Keto” version 90-99% of the time versus the “Enol” version.  However, a catalytic amount of presence is sometimes enough to drive reactions forward if the mechanism requires the enol tautomer of the compound.

However, in some cases such as a beta diketone, shown below, the combined dipoles of two carbonyls makes the alpha carbon very acidic, meaning enol formation is very favorable.  In this case, it is 70-90% enol in solution.

 

Enolates – Deprotonating the alpha carbon and tautomerizing to the oxyanion

Catalyist: Strong Base to deprotonate the alpha carbon.

Like an Sn2 mechanism, a strong enough base will react with the acidic proton on the alpha carbon and deprotonate.  The electron density between the C-H bond will shift to make a new C=C bond, while the C=O electrons will be placed on the oxygen, creating and alkENE + alcohOL anion “ATE”) with a strong base to produce a stable carbanion.  The stability is due to the tautomerized structure that can be produced by placing the negative charge on the oxygen.

 

Enolates are generally forward reactions depending on the strength of the base.  How strong the base required depends on the pKa of the alpha C-H bond.  In the case of ketones, a strong base like LDA is required.  However, for beta dikeontes, a mild base like NaOH is enough to generate the enolate.

 

Formation of Enols and Enolates are an important source of carbon nucleophiles to make new C-C bonds in future reactions.

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