Archive for the "Organic Chemistry General" Category

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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Naming Organic Molecules

Posted on June 18th, 2015

Naming molecules is a challenging exercise because there are many rules; the more complex the molecule, the more rules that need to be observed.  It is possible to name a single molecule in many different ways, but only one is correct!  These are easy points on quizzes and exams and can boost your grade if you know how to tackle them. We here at StudyOrgo have summarized the process down to 4 easy steps to name alkanes.  Follow along to determine the name of any molecule in your course.

Four discrete steps are required when assigning the name of an alkane:

  1. Identify the parent chain: Choose the longest chain. Looking at the first example below, several chain lengths are shown, however only the red trace contains the longest chain, therefore this is the correct parent chain. For two chains of equal length, the parent chain should be the chain with the greater number of substituents. In the second example below, two 7-carbon chains are possible, but only one gives 2 substituents, so it is therefore the correct parent chain!
    nomenclature 1
  2. Name the parent chain: In order to correctly name the parent chain, you will have to commit to memorizing the prefixes that denote how many carbons are in the parent chain. There is no easy way around it, so try covering the table up and recalling to help with memorization.
Number Of Carbons Parent Name AlkaneExample Number Of Carbons Parent Name AlkaneExample
1 meth methane 11 undec undecane
2 eth ethane 12 dodec dodecane
3 prop propane 13 tridec tridecane
4 but butane 14 tetradec tetradecane
5 pent pentane 15 pentadec pentadecane
6 hex hexane 20 eicos eicosane
7 hept heptane 30 triacont triacontane
8 oct octane 40 tetracont tetracontane
9 non nonane 50 pentacont pentacontane
10 dec decane 100 hect hectane

 

  1. Number the parent chain, identify and assign a location (carbon number) to each substituent: Just as with the parent chain prefixes, substituent prefixes are another terminology you will have t memorize! Give the first substituent the lower possible number. If there is a tie, choose the chain in which the second substituent has the lower number.  In the example of methyloctane, the methyl position could be at position 2 or 7. The lowest number prevails so 2-methyloctane is correct.
    nomenclature 2
Number Of Carbons In Substituent Terminology
1 Methyl
2 Ethyl
3 Propyl
4 Butyl
5 Pentyl
6 Hexyl
7 Heptyl
8 Octyl
9 Nonyl
10 Decyl
  1. Arrange the substituents alphabetically: Place the number corresponding to the position in front of each substituent. Note for the example below, there are four identical ethyl substituents, therefore we will label each position followed with the prefix tetra (di- for 2, tri- for 3, etc.) Note, the prefix is NOT included in alphabetization, as methyl comes after ethyl regardless of the “tetra” prefix in this example.
    nomenclature 3

And there you have it. With a little practice and help from StudyOrgo, you are well on your way to naming any organic molecule in your studies.  Sign up today for help with terminology, reaction mechanisms and more!

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Studying for Summer Organic Chemistry

Posted on June 8th, 2015

One of the biggest challenges students can undertake is signing up for Organic chemistry in the summer.  How so?  Well, there are a few reasons.

First, the course time usually shrunk from 16 week to 8 weeks.  This means, longer classes and more frequent tests.  Secondly, the subject material is usually covered more quickly, but the content remains the same.  As such, students can feel overwhelmed and become exhausted studying the materials much more quickly than is typical.  But, we here at StudyOrgo have streamlined the process of studying organic chemistry mechanisms!

Our illustrative overviews with full mechanism descriptions and diagrams makes mastering any reaction in your class much easier.  And, with over 170 reactions, you’ll feel confident you’ll have everything you need to get the “A” in this course.  Here are a few tips for getting the right start to studying with help from StudyOrgo today!

  • Time management – Schedule your studying NOW! – Time management is the key to acing organic chemistry in the summer. 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 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.
  • 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 at will but cannot add them easily. 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 summer course.
  • Read ahead – Before each class, glance at the chapter to be covered that lecture beforehand. Don’t try to understand everything, just pay attention to the major words and phrases used and the ideas. This will allow you to pay more attention during class because you will already know what is being said, now you pay attention to the details. Most people are scrambling to write down notes and drawings in class, but not really paying attention.  Try it yourself, look at your classmates at the next class for a minute or two, they are usually feverishly writing!
  • Sign up with StudyOrgo – The Editors at StudyOrgo have painstakingly reviewed and prepared 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 StudyOrgo, you will have no trouble getting an A in Organic Chemistry this summer!

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Organic Chemistry of Lipids

Posted on May 11th, 2015

Lipids are a fundamental part of biochemistry and draw many analogies to reactions with alkenes and alkanes.  In this article, we will review some basics of lipids and their organic chemistry implications.  Many of the reactions with lipids are simple hydrocarbon reaction mechanisms covered in Organic Chemistry 1.  With over 175 reaction mechanisms, StudyOrgo is devoted to making organic chemistry reaction mechanisms easy to learn and points out common places where errors are made.  Sign up with StudyOrgo.com today to get more in-depth coverage of common reactions for your next exam!

Lipids are any complex chain of hydrocarbons that may or may not contain additional functional groups. Some common classes of lipids are fatty acids and triglycerides, waxes, terpenes and steroids; examples of which are shown in Figure 1. 

Lipid Figure 1

Most lipids usually consist of a long hydrophilic tail coupled to a polar head group that undergoes reactions to elongate or shorted the hydrophobic tail.  We will cover some common organic chemistry reactions with each of these classes of lipids.

Complex Lipids – Fatty acids and Triglycerides

Complex lipids are capable of undergoing hydrolysis reactions.  The basic unit of a complex lipids are fatty acids, which are made up of a hydrophobic tail coupled to a carboxylic acid head group (Figure 2).  The length of the carbon chain denotes the lipid name.  In addition, the number of double bonds, or degree of unsaturation, also influences the name.  In biology, naturally occurring double bonds will always be found in the cis- configuration.

Lipid Figure 2

Hydrogenation of alkene groups to alkanes occurs at high temperatures, such as in deep fryers for cooking, and this side reaction leads to the isomerization of the double bond to the trans- conformation, hence trans-fatty acids.  These lipids are toxic in high quantities because they are unable to be metabolized by the cell, thus accumulating and undergoing oxidation reactions over time and theorized to promote inflammation and metabolic diseases.  Coupling of a fatty acids to the alcohol groups of glycerol forms the complex lipid, triglyceride (Figure 2).  This reaction occurs enzymatically in the cell but closely resembles the mechanism found in acid-catalyzed Fisher Esterification.

Lipid Figure 3

Simple Lipids – Sterols

Steroids are an integral part of cellular biology but are classified as simple lipids because they cannot be hydrolyzed (i.e. they do not have reactive carbonyl or carboxyl groups).  A lipid molecule having a tetracyclic shape in the arrangement similar to cholesterol is classified as a sterol.  Cholesterol serves as the building block of all steroids and substitution, addition and elimination of functional groups derives the variety of steroids found in the body. Cholesterol is synthesized from the 5 carbon lipid molecule isopentenyl phosphate.  Condensation of isopentenyl phosphate forms geranyl phosphate (10C) and farnesyl phosphate (15C) in a reaction mechanism that involves allylic carbocation and tertiary carbocation intermediates (Figure 3).  Condensation of two farnesyl phosphate molecules forms the 30 carbon intermediate squalene, which in several steps is converted to cholesterol.  Two important examples of steroids are the sex hormones found in humans, estradiol and testosterone, which regulate a wide range of biological functions.

Lipid Figure 4

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Carbohydrates in Organic Chemistry

Posted on April 24th, 2015

Many organic reactions you encounter have very practical uses.  In this section, we will describe the organic chemistry of carbohydrates.  Sign up today with StudyOrgo to see more illustrations and easy explanations to make your Organic Chemistry studying a success!

What’s in the Name?

Carbohydrates, or saccharides, are molecules of 3 carbons or more which contain at least one carbonyl group and one alcohol group.  Carbohydrates containing an aldehyde group are referred to as aldoses while carbohydrates with a ketone group are referred to as ketoses. Like all organic molecules, stereochemistry is an omnipresent consideration.  Recall that chiral, non-superimposable, mirror images will rotate plane-polarized light to the left or right, referred to as S and R respectively.  In carbohydrate chemistry, left- and right- rotation is referred to as L (levorotary) and D (dexorotary), but their meanings are exactly the same!  Remember that chirality is a key to life; as a matter of fact animal cells are only capable of utilizing the D-isoforms of saccharides while the L-isoforms are not metabolizable!

The simplest carbohydrate is glyceraldehyde (C=3) which has 1 stereocenter, therefore it has two enantiomers.  C=4 carbons have 2 stereocenters and two different arrangements of the OH groups, named Erythrose and Threose (diastereomers), but each has a mirror image (enantiomers), thus resulting in 4 stereoisomers. C=5 carbons has 3 stereocenters, thus giving 8 stereoisomers, the 4 diastereomers referred to as ribose, arabinose, xylulose and lyxose. Remember, for each OH arrangement, only 1 mirror image (the enantiomer, D or L) exists!

carbohydrate 1

Cyclization of Carbohydrates

Recall hemiacetal formation, which is the reaction of aldehydes or ketones with an alcohol to produce the hemiacetal functional group.  You will also recall that inter-molecular hemiacetal reactions are unfavorable.  However, intra-molecular hemiacetal formation, which is possible in carbohydrates, is very favorable (Figure 2). Thus, carbohydrates are typically drawn as a Hayworth projection. The aldose or ketose when cyclized is renamed to resemble the molecule pyran (5 carbon cyclized ether) or furan (4 carbon cyclized either).  Thus, for D-glucose the cyclized version is renamed D-glucopyranose to indicate a cyclic structure.  The ‘pyran’ is usually dropped in the vernacular of biochemistry but is an important distinction to make for the exam!

carbohydrate 2

 Anomers

The alcohol formed from the hemiacetal reaction is an important functional group in the organic chemistry of carbohydrates.  This carbon center, usually at position C1 or C2, is referred to as the anomeric position. It is also a stereocenter, and the resulting enantiomers are referred to as anomers.  When the OH is pointed above the plane of the ring, it is referred to as the beta- configuration while the OH pointed below the ring is the alpha- position.  All monosaccharides contain an anomeric carbon!

carbohydrate 3

Reducing Sugars

Equilibration back to the aldose or ketose makes the carbonyl susceptible to oxidation by reagents, such as silver nitrate, the main component of Tollen’s Reagent.  This reaction converts the carbonyl in saccharides to a carboxylic acid.  In the process, silver precipitates to form a mirror-like residue on the beaker. Thus, any saccharide that tests postive in this reaction must have an anomeric position that can equilibrate to the aldose or ketose configuration and is said to be a reducing sugar. Reducing sugars are useful carbohydrates in making polysaccharides, or polymers of carbohydrate units involving the formation of glycosidic bonds (Figure 4, sucrose red oxygen). However, when the anomeric position between two saccharides are linked together, the anomeric postion (also known as the reducing end) on both sugar units is unavailable.  Take for instance sucrose (table sugar); it is a disaccharide of glucose and fructose.  However since both reducing ends are used in the glycosidic bond, they will not react with Tollen’s Reagent, thus sucrose is said to be a non-reducing sugar.

carbohydrate 4

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