What is the difference between Sn2 and Sn1?

Posted on October 20th, 2017

 

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!

Substitution reactions involve the attack by an electron-rich element, referred to as the nucleophile, on an electron-poor atom, referred to as the electrophile.  As the reaction name suggests, we are substituting the nucleophile for another group on the electrophile atom, which is referred to as the leaving group.  The generic reaction looks like this.

In Substitution reactions, there are two mechanisms that will be observed.  An Sn2 and Sn1 reaction mechanism.

Sn2 reactions are bimolecular in rate of reaction and have a concerted mechanism.  The process involves simultaneous bond formation by the nucleophile and bond cleavage by the leaving group.  The transition state looks like this.  Because the reaction is concerted, Sn2 mechanisms will always lead to an inversion of stereochemistry!  For reactivity using an Sn2 mechanism, primary >> secondary >> tertiary carbon centers.

 

On the other hand, Sn1 reactions are unimolecular in rate of reaction and have a step-wise mechanism.  This process first involves bond cleavage by the LG to generate a carbocation intermediate.  The stability of carbocation formation will determine if Sn1 or Sn2 reactions occur.  In the second step, the electronegative nucleophile attacks the carbocation to form the product.  The steps look like this. Because the nucleophile can attach either side of the carbocation, which adopts an sp2-hybridized orbital with a trigonal planar geometry, an equal amount of inversion and retention is seen, referred to as a racemic mixture. For reactivity using an Sn1 mechanism, tertiary >> secondary >>> primary carbon centers.

The strength of nucleophiles used help to determine the reaction mechanism.  Strong bases will almost always proceed to Sn2 mechanism.  Weak nucleophiles will generally proceed to Sn1 mechanism when a stable carbocation is present.  Below is a list of nucleophile trends in order of nucleophile strength.

 

We hope that this learning aid will help you answer any questions you may have had about Sn2 and Sn1 reactions. We here at StudyOrgo have compiled hundreds of reactions with clear explanations to help you speed up your studying and get a great grade in organic chemistry.  Sign up today to get access to all of our reactions!

How Are Radical Ions Formed?

Posted on August 28th, 2017

Many students taking Orgo 1 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, in Orgo 2, you will be responsible for all of the reactions necessary for multi-step synthesis (starting product known to get to unknown final product) and retro-synthesis (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 will review the steps to radical ion formation used in a few reaction including free radical halogenation of alkanes.  To begin, the three steps to radical reactions are 1) Initiation, 2) Propagation and finally 3) Termination.  The formation of radicals always occurs in the Propagation step of the reaction.  Energy has to be supplied to a molecule to induce a reaction known as homolytic cleavage; the breaking of a bond where both atoms receive 1 electron.  Most reactions occur by heterolytic cleavage, which means 2 electrons that formed the bond being broke are given to one atom (negative) while the other atom loses them (positive).

There are two methods for initiating radicals, either heat (symbolized as delta) or light (symbolized as hv).   To show the homolytic cleavage during initiation, the convention is to draw a fishhook arrow (one sided barb to the arrow) to each atom receiving one of the electrons.  In Orgo 1, you mechanisms will be graded on the quality of your fish-hook arrow and what bond the electrons came from to what atoms they are going towards, so be very clear!!!

Below is an example of heterolytic vs homolytic cleavage and how to draw the arrows correctly!

There are two other steps to free radical halogenation that occur.

Once the radical is created, it will attack alkane bonds (C-H) of substrate molecules to create H-X and a new radical alkane.  This step is referred to as propagation, since the radical is transferred from one molecule to the other essentially. To complete halogenation of the alkane, the radical alkane will attack the abundant halogen (e.g. Cl2, Br2) to form a new C-X bond and generate another radical halogen, just like from the initiation step.

The final step will be termination, where one radical attacks another and now a new bond is formed and no radical product is made.

 

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.  Check out a membership to StudyOrgo.com and sign up today!

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!

In the reaction with ammonia, why is water the acid?

Posted on April 24th, 2017

In the reaction with ammonia, why is water the acid?

 

This is a great general chemistry reaction with important organic chemistry implications.  Another similar question is, why is water a base when reacted with an “acid”?  The answer to both question is: it’s all relative!  Here at StudyOrgo, we frequently get questions like this 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. 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.  Check out a membership to StudyOrgo.com and sign up today!

 

Remember that in terms of acids and bases, there are two definitions; the Bronsted and Lewis definition.  Bronsted acids are defined as proton donors, while Lewis acids are defined as electron acceptors.  Both are acids, but what we think of differently is whether protons or electrons are involved.  In an aqueous solution, general chemistry or biochemistry, we mainly think of the Bronsted definition because the dissociation of protons from acids changes the solution concentration of H+, which we interpret as a change in pH and the “acidity” of the solution.  We measure the dissociation of the protons from acids, or the acceptance of protons by bases, with a numerical value which is the pKa.  The greater the pKa, the weaker the acid and stronger the base.  The lower the pKa, the stronger the acid and weaker the base. An easy trick for thinking of pKa is, place the protonated version of the molecule on the left and think of it as a proton donor.  The pKa tells you how easy this donation will be, the lower the number the easier the proton donation.

Let’s look at the ammonia and water reaction; the pKa for water is defined as 14.  The pKa for ammonia is ~37.

Therefore, because the pKa of water is lower than ammonia, it is a stronger acid the ammonia and will donate protons to the ammonia base.  This reaction leads water to become the conjugate base OH- and ammonia to become the conjugate acid NH4+.  Interestingly, this is why ammonia is a caustic agent, it produces hydroxide that reacts with stains and microorganisms to effectively clean and sanitize household items.

 

This is important as an organic chemistry concept because the strength of acids and bases in terms of electrons, or the Lewis definition, is exactly how we think about mechanisms of bond breaking and bond forming; the flow of electrons.  The strength of Lewis acids and the conjugated acid can help identify which direction a reaction will proceed.  We can see that the reaction of water an ammonia is unfavorable, but enough of the reaction occurs in reality to significantly reduce the pH of water (pH~11 with ammonia).

Let’s look at an acetylene reacting with a base to generate an alkynide ion, a useful nucleophile for C-C bond formation.  If we try to react sodium hydroxide with alkyne, alkynide ion WILL NOT be formed.  This is because the conjugate acid product of the reaction is water, which has a pKa of 15, is STRONGER than the original acid acetylene (pKa = 25).  Reactions ALWAYS favor formation of the weaker acid, or in this case, the reactant side.

If we change our base to sodium amide, which is a much stronger base, alkynide ion WILL be formed.  This is because the conjugate acid product of the reaction is ammonia, which has a pKa of 38, and is a WEAKER conjugate acid than the original acid acetylene (pKa = 25).  In this case, the reaction favors the product side!

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.