Posts Tagged "organic chemistry"

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!

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.

Sign up with StudyOrgo today to get access over 180 chemical to reactions with detailed reaction mechanisms, explanations, tips and tricks to each mechanism.

How do you to tell when a hydrogen bond will occur?

Posted on March 7th, 2018

Hydrogen bonding is important for describing the driving force of reactions in organic chemistry and a very important concept for explaining the biochemistry of catalytic reactions that occur using protein as enzymes in biological systems.  In this post, we will discuss the rules and examples of hydrogen bond formation.  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!

Physical properties of molecules such as boiling and melting point, solubility and reactivity, are affected by the functional groups that make up the molecule. This can be explained by analyzing the type of intermolecular forces that are experienced between molecules.  Because these forces are not covalent, intermolecular forces are determined by the intensity of electrostatic forces which is what makes up each type of intermolecular force. As a review, the types of intermolecular forces are;

  • Van der Waals (London dispersion forces) – Weak, temporary dipole formed between hydrophobic C-H and C-C bonds.
  • Dipole-Dipole Interactions: – Strong, permanent dipole moments formed between atoms of functional groups containing bonds such as C=O, C=N, C-O, C-N, N-H and O-H bonds. The delta(-) side of one dipole is attracted to the delta(+) side of another molecule, forming a non-covalent electrostatic attraction.
  • Hydrogen Bonding: Sharing sharing of a hydrogen atom covalently attached to an electronegative element (typically O-H and N-H groups) between a lone pair of electrons on another electronegative element.

Two terms about hydrogen bonding that are key are;

  • The electronegative atom with the lone pair electrons is called the Hydrogen Bond Acceptor
  • The electronegative atom bonded to the hydrogen is called the Hydrogen Bond Donor
  • The Hydrogen Bond Donor must be aligned 180 degrees to the Hydrogen Bond Donor!

The more intermolecular forces the molecule has, the more energy will be required to disrupt these bonds when melting or boiling compounds, thus raising the observed temperatures from expected relative to their mass.  In addition, hydrogen bonds require polar bonds in the molecule and H-Bond Donor proton involved is protic (a donatable hydrogen atom). These are two terms that you will learn in the Sn1 mechanism.

Let’s look at ethanol as an example.  The hydrogen bonding occurs between the proton of one alcohol group and the oxygen lone pair electrons on another alcohol group.  This is a strong intermolecular force that holds the molecule in a complex 3D shape and makes it easier in reactions to attack the carbon connected to the O-H bond due to inductive effects, or the pulling of electrons away from the carbon.  Water is an extreme example, where all the atoms in the molecule participate in hydrogen bonding.  The oxygen lone pairs will accept a hydrogen from a neighboring molecule O-H.  Finally, acetic acid is another example.  Remember, that the H-Bond Acceptor can be any lone pairs, including those of C=O bonds.

 

These concepts are really important to understanding the more complex topics to come. With a membership to StudyOrgo, you will get even more tips and tricks on organic chemistry topics and detailed mechanisms with explanations.  Today’s blog is a preview of the detailed topics and materials available.

Preparing for Organic Chemistry This Fall

Posted on August 5th, 2017

One of the questions we are repeatedly asked at StudyOrgo is “how do I to get ahead in organic chemistry this fall semester?” Many of you have heard that organic chemistry is a brutal class that does little but to depress your GPA. While it is true that this course is challenging, we here at StudyOrgo are devoted to helping you get the “A” you deserve!

Organic chemistry gets a bad name because it assumes that you are experts with regards to all of the general chemistry from freshman year, and you are now responsible to know it!  As an analogy, think of your chemistry courses as a pyramid to reaching your degree goals.  Organic chemistry is directly placed in the middle of the pyramid, it will be very important not only for the MCAT or DAT exams, but also for future advanced courses.  Organic chemistry is supported by General Chemistry, which is why you took it last year.  Fortunately, StudyOrgo is placed in the center of your pyramid base and we are here to help all of your organic chemistry questions.  Our simple and clear-cut explanations of reaction mechanisms and concepts will easily help you with anything you might struggle with this semester.  Here are a few tips on how to prepare today for the course this Fall.

chemistry pyramid

  • 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 but cannot easily add them. 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 semester.
  • Schedule your studying! – Now that you know where the book is and a rough idea of what you are responsible for learning from the syllabus, 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 or dentist 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.
  • Read ahead – If you have time this summer, read at least two chapters to get yourself ahead of the class. Don’t try to understand everything, just pay attention to the words used and the ideas. This will allow you to pay more attention and ask questions about the details in class instead of scrambling to write down notes and drawings.
  • Sign up with StudyOrgo – The Editors at StudyOrgo have spent numerous hours reviewing and preparing 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 help from StudyOrgo, you will have no trouble getting an “A” in Organic Chemistry this year!

How can I tell if a hydrogen is a wedge or a dash in a chair skeleton?

Posted on March 19th, 2017

 

“How can I tell if a hydrogen is a wedge or a dash in a chair skeleton?”

Here at StudyOrgo, we frequently get questions about topics in organic chemistry that are usually quickly covered, poorly described or expected that you know from previous courses.  These concepts are really important to understanding the more complex topics to come.  In this article, we will cover the concepts of stereochemistry descriptions using bold and wedged bonds.  This is just a preview of the detailed topics and materials available with your membership to StudyOrgo.com.  Sign up today!

The first thing we have to do is determine is how you want to orient you molecule.  Let’s take (1R, 2R) 1,2-dimethylcyclohexane for example.  If we orient the molecule to have the methyl groups on the right side, we see that we have two stereocenters available.  But the current drawing doesn’t indicate the stereochemistry yet.  That’s what the bold and hashed bonds will indicate.

Next, we have to visualize the cyclohexane ring in the chair conformation.  Remember, that the skeleton image shown above is more conveniently drawn, but loses the 3rd dimension information, so you have to put it back in the chair to determine which should be bolded and which should be wedge.

Next, we have to confirm that that the stereochemistry is correct.  To do this, you need to practice selecting most important substituents and rotating to assign stereochemistry.  Follow along with the examples below, using the blue and pink carbons shown.

At this point, you should be able to see how the hashed and bolded bonds are now appropriately drawn.  The pink stereocenter will be bolded, suggesting it is above the plane of the ring and the blue stereocenter will be hashed, suggesting it is below the plane.  Drawing the Newman Projection down the red bond shows that the methyl groups are “anti” to each other, making this a stable conformation.