Alkylide Anions: Making new C-C bonds with Alkynes

Posted on April 16th, 2019

 

Many students taking Orgo 2 have commented there are a few types of reactions the professors save to the end of the semester and cover quickly and “gloss” over or sometimes skip all together in the interest of time.  However, you will be responsible for all of the reactions necessary for multi-step synthesis (starting product known to get to unknown final product) and retrosynthesis (product known to get to unknown starting material) reactions.  We at StudyOrgo don’t want you to get stuck on trying to cram for exams by studying reactions that were poorly covered in your class.  In this article, we focus on how to make new C-C bonds using alkenes.

Alkynide Anion Formation

First, the terminal alkyne must be deprotonated with a strong base, such as sodium amide (NaNH2).  This proton can be removed because terminal alkyne hydrogens are more “acidic” than alkene or alkane hydrogens.  This forms the alkynide anion, which is a great nuleophile.

Next, the a alkynide anion must attack a primary or secondary carbon center to perform an Sn2 type reaction. This displaces the leaving group and forms the new C-C triple bond, which has now become an internal alkyne.

Once you have the the new alkyne, you can choose to reduce the alkyne to an alkene or alkane, or you can transform the alkyne into many other functional groups, including a ketone  or an alcohol by hydration.

Remember, alkynes are a great tool to build large carbon skeletons using simple haloalkanes.  Remember this reagent for the exam!  We here at StudyOrgo have devoted countless hours to preparing complex reaction mechanisms in simple and easy-to-understand manner to help you maximize your studying.  Sign up with StudyOrgo for detailed explanations of epoxide reaction mechanisms and other essential Orgo 2 reactions today!

 

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Huckel Aromaticity and Frost Circles

Posted on January 29th, 2019

 

As Orgo 2 lectures start up, there will be a number of concepts that will be covered quickly to speed up the class so you can learn lots of functional group transformations this semester. In this article, 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!

 

Aromatic compounds contain conjugated systems of pi bonds, meaning one pi bond after another, that are arranged in a ring.  This creates a current of electron density that circles the compound. If the pi electrons fill the bonding molecular orbitals, the compound becomes aromatic!  But if too many pi electrons are present, the non-bonding and antibonding orbitals become occupied, and the compound is now anti-aromatic.

 

Aromatic compounds are UNUSUALLY STABLE and have important chemical and synthetic uses.  How do we know if the compound is aromatic or not?

 

Here is a short list of rules lists the properties of aromatic compounds.
The Huckel aromaticity rules are:

  1. Molecule is cyclic
  2. Have one pi orbital per atom of the ring
  3. Planar, in an SP2 hybridized orbital, over every atom of the ring
  4. Have a closed loop of 4n+2 pi-bond electrons, where n is equal to any integer (0,1,2,3,…)

 

However, anti-aromatic compounds have an unusual INSTABILITY to them.
The Huckel anti-aromaticity rules are:

  1. Molecule is cyclic
  2. Have one p orbial per atom of the ring
  3. Be planar, in an sp2 hybridized orbital, over every atom of the ring
  4. But, anti-aromatic compounds have a closed loop of 4n pi-bond electrons.

 

Many times students are asked to describe the molecular orbitals of aromatic compound examples.  How do we know what they look like?  An easy way to describe the molecular orbitals of small conjugated rings is to draw a Frost Circle.  To do this, follow the steps below.

  1. Draw a circle
  2. Connect the carbon atoms to the circle such that one is always pointed at the bottom
  3. Draw a horizontal line at each carbon – this represents the pi bond molecular orbitals
  4. Draw a dotted line through the middle of the circle – this is the boundary between bonding and antibonding orbitals.
    1. Anything pi electrons occupying an orbital below the line are in a bonding orbital and are STABLE
    2. Any pi electrons on the line or above the line are in non-bonding & anti-bonding orbital and are UNSTABLE.

Below are the Frost Circles for many types of rings.

Before you begin, make sure the Huckel rules apply; do you have enough electrons? Is it SP2 hybridized?  Is it a ring? Is it planar?  Some aromatic rings are heterocyclic (elements other than carbon) and ionic (meaning one of the elements in the ring has a charge).  If you have the Huckel rules applied, then determine the number of pi electrons and fill in the Frost Circle.

Here is butane.

Three double bonds = 6 pi electrons.  Paired together, all of the pi electrons are in bonding orbitals so benzene is in fact aromatic.

 

Here is cyclooctatetraene.

Four double bonds = 8 pi elecrons, Pairing the first 6 electrons in the bonding orbitals, we place last 2 electrons in each of the non-bonding orbitals.  Because of these two electrons, this molecule is anti-aromatic!

 

Practice, practice, practice!  This is our suggestion to learn these tricks.  As you advance to aromatic reactions, it will be much easier to find shortcuts to faster synthesis schemes and solve 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!

 

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Study Ahead for Proton NMR

Posted on December 22nd, 2018

Reading the 1H-NMR Spectra

One of the most practical techniques covered for organic chemistry is the method for determining the chemical structure of unknown compounds. In this article, we will summarize the concept of proton NMR, the most informative NMR spectroscopy utilized by organic chemistry students.

During the second semester of organic chemistry, many professors will ask you to read 1H-NMR spectrum as a question or ask you to use the answer in other parts of the exam, which makes getting the correct structure a major obstacle in getting a great grade on the exam!

Why use proton NMR at all?  The answer is abundance. Virtually all organic chemistry molecules contain hydrogen, and since the NMR active isotope is 1H, which has a 99.98% abundance on Earth, it makes it a convenient and easily detected atom. An unknown sample in a strong magnetic field allow 1H nuclei to “resonate”, which means their nuclear spins flip at a unique electromagnetic (EM) frequency (Hertz, Hz).  The frequency at which the nuclear spin flips depends on the “chemical environment” the nucleus is in, meaning what other functional groups are in close proximity to the hydrogen atom.

The NMR instrument detects this and plots it on a graph in units of ppm.  This value is relative to an internal standard, tetramethylsilane (TMS) which is arbitrarily set at 0 ppm. The signals originating from your unknown sample are recorded as a net difference from the TMS reference.  The ranges for common functional groups are shown in the graph below.

Notice that protons (in red) neighboring or attached to electronegative elements (N,O,S,Cl,Br,I) have a large chemical shift (d).  This is referred to as “downfield” on the spectrograph. This is due to the deshielding effect, which is simply that the electron cloud around protons near electronegative or electron-withdrawing groups are smaller, so less EM radiation is required to resonate the nucleus.  Conversely, when the proton is surrounded by hydrocarbons, the chemical shift is <2 ppm.  This is because these proton nuclei are “shielded” from the magnetic field and higher energy is required for them to resonate and referred to as “upfield” on the spectrograph.  Once a NMR spectrograph is recorded, 4 pieces of information can be determined from the data as long as the chemical formula of the compound is known.

To illustrate the points, we will consider the following 1H-NMR spectrum of the C9H10O.

  1. Signal Count – Number of unique hydrogens

This is the easiest to interpret.  The number of peaks correspond to the number of unique, or chemically indistinguishable, hydrogen nuclei.  There are 4 peaks on the graph, therefore of the 10 hydrogens in the molecule, there are 4 types of protons.

  1. Chemical Shift – Identity of neighbors

This information refers to the chart shown above. Peaks C & D on the spectrum are located at (d)3.1 and (d)2.7, likely indicating C-H bonds near some electron-withdrawing groups. Further downfield on the graph, peak B is at 7.4, likely indicating an aromatic group and peak A at 9.9, likely indicating a proton of an aldehyde, which accounts for the one oxygen atom in the formula.  From this information, we now know there is an aromatic ring (6 carbons) and an aldehyde (1 carbon).  That leaves 2 carbons left, and since two types of C-H bonds from peaks C and D information, therefore we have accounted for all carbons and hydrogens, now we have to put them together.

  1. Integration – Number of Equivalent Hydrogens

Integration is the calculation of the area beneath the peaks.  This information is determined by the computer software and would not be required for you to determine from visual inspection.  But if it is provided, it will instantly determine the number of “equivalent” (or chemically indistinguishable) hydrogens in your graph.  Looking at our spectra, Peak A = 1 which would make sense for an aldehyde proton.  Peak B = 5, meaning an aromatic group with 5 hydrogens, that means only one substituent can be attached to the ring.  This is an important piece of information since we could think of many types of rings. Peak C = 2 and Peak D = 2, this is also helpful since we now know both carbons are internal, meaning they are not at the ends of the molecule (we would need 3 hydrogens to be a methyl at the end of the molecule).  With this information, you could likely predict the structure.

Signal Splitting – Number of Non-equivalent Hydrogen Neighbors

Signal splitting occurs from a phenomenon of coupling, which occurs when NON-equivalent neighboring protons interfere with the resonance of a proton nuclei.

Rule #1: starting from the hydrogen, count three nuclei away.  If you do not land on another proton, then there is no splitting contribution to the proton’s environment.

Rule #2: the degree of splitting occurs in an N+1 rule.  For instance, Peak A has a singlet, meaning 0+1. This indicates that there are NO protons neighboring the aldehyde. Peak B is a multiplet, and is indicative of an aromatic group given 5 hydrogens present.  Peak C and D are both triplets, meaning 2+1. This means each type of proton has two non-equivalent neighbors.  Thus, peak B and C are next to each other.

With all of this information, the only chemical structure from the spectrum that would make sense must be 3-phenylpropanal, shown above.  This is an example of the clear-cut explanations you will receive with your membership with StudyOrgo. With practice and help from StudyOrgo, you’ll be solving your organic reaction problems in no time!

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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!

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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 stable.  Any 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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