Conjugation and Light

Absorbtion of UV light or light from the visible region of the electromagnetic spectrum can promote an electron into its excited state. Without going into MO theory here you can know that some of these transitions are more favored and some have small enough energy barriers that they can be promoted by longer wavelengths. Conjugated dienes have a sufficiently small barrier such that they can transition to excited state (absorb) in the UV and Visible regions.
Allowing our skin to absorb UV rays from the sun puts us at higher risk for skin cancer. The active ingredients in sunblock absorb in the UV region, acting as a shield for our skin. If the suns rays are absorbed by a compound on top of our skin, they will not be absorbed by our skin. A popular active ingtredient for sunblocks is PABA or para-amino benzoic acid. All of the double bonds of the aromatic ring are conjugated, in addition the carbonyl is in conjugation with the aromatic ring. Another active ingredient that has come on the market more recently is Parsol. Parsol is said to block more UVA rays. It takes 1ooX more UVA light to cause the same damage as UVB light, but over the course of your lifetime, it is probably prudent to protect yourself from both. Parsol contains two ketones that are in conjugation with two para substituted aromatic rings.

As you increase the amount of conjugated pi bonds the energy between ground state and excited state is reduced, the absorption is reduced to lower wavelengths in the visible region. When a compound absorbs in the visible region it takes on the color of the wavelengths it does not absorb. Beta-carotene has 11 conjugated double bonds and has a maximum absorbance at 454 nm, the blue region. White light that appears orange once blue has been removed. Beta-carotene gives carrots and many other vegetables their color. Indigo absorbs at the other region of the visible spectrum, around 675 nm. If it is absorbing red and yellow, it will appear blue under white light.

Functional Group Polarity

A molecules polarity depends primarily on

1. The extent to which it can hydrogen bond
2. The number of electronegative atoms
3. The polarizability of the bonds or atoms
4. The net dipole moment of the molecule.


Applying these rules you can make a list for determining the relative polarities of the functional groups when all else is similar.

Alkanes are the least polar. No hydrogen bonding, electronegative atoms, polarizability and the net dipole moment is small at best.
Alkenes are similar to alkanes in all respects except polarizability. The p orbitals of are large and polarizable, making this class of molecules more polar than alkenes.
Conjugated Polyenes and Aromatic Compounds are similar to alkenes but now the pi system (of p orbitals) is even larger, increasing polarity.
Ethers and alkyl halides are more polar than the above classes because of the introduction of a polar atom. Rules higher on the list take precedence over rules lower on the list.
Aldehydes, Ketones, and Esters have a large dipole moment due to the carbonyl. Along with the electronegative oxygens, this makes them more polar than an ether or alkyl halide.
Amines and Alcohols have the ability to hydrogen bond, the first and most important rule on the list. Although it is a close call, everything else similar, an alcohol would be more polar than an amine because oxygen is more electronegative (Rule 2).
Carboxylic Acids are the most polar functional group because they can hydrogen bond extensively, they have a dipole moment and 2 electronegative atoms.

Notice how large the difference is between esters and carboxylic acids-hydrogen bonding counts for a lot.



Obviously the rest of the molecule plays in to any decision. If you have a 30 carbon chain attached to an any group it will be significantly less polar, but this can give you a framework to make an educated decision. The rules aren't perfect, and sometimes lab experience can surprise you. Sometimes you have two compounds that you really couldn't make a good guess about because they seem so similar. In that case, you run the column and figure it out by NMR.

Distinguishing Aldehydes and Ketones using IR

The following two spectra are simple carbonyl compounds. Formaldehyde, the simplest aldehyde, and acetone, the simplest ketone. Students often come to me frustrated because they can not tell one carbonyl compound from the next, or the peak will be right between the two literature values. The aldehyde or ketone question is simple. In both you will see a very prominent C-O stretch around 1700cm-1 area.

Ketone

Aldehyde

But in the aldehyde you should also see see a peaks around 2820 and 2720cm-1. They often look like a doublet and are sometimes referred to as a Fermi doublet. These are the C-H stretches between the aldehydic proton and the carbonyl carbon. The presence of these peaks along with a carbonyl peak is a good indication that you have an aldehyde.

Guide to Separation Schemes

Step 1: Ask yourself which compounds are soluble in water at neutral pH. To determine this follow the general rules below.

a) For a given compound, if the ratio of carbon atoms to N-H, NH2, or OH groups is <> good bet that the compound is water soluble. Some examples:

b) If the compound is charged, then it is a good bet that the compound is water soluble. Some examples: c) Exceptions to Rule a, include compounds with very large dipole moments and <>

Step 2: Second, consider acid base chemistry that the compound can undergo. If you adjust the pH, can you form a salt, or reform a neutral compound from a salt. Knowing which compounds will and will not react when you are considering organic molecules is actually not terribly complicated. You just need to know approximate pKas for functional groups, and whether they will react with a strong or weak base, or an acid.

Carboxylic acids have a pKa around 5. This means that carboxylic acids can react with either a strong base like NaOH, or a weak base like NaHCO₃. This would form a carboxylic salt that would be soluble in water.

Alcohols generally have pKa’s around 17. Something with a pKa of 17 is not going to react with a strong base like NaOH. There is only one type of alcohol that can be deprotonated by NaOH-that is a phenolic alcohol. The pKa of a phenolic alcohol is significantly lower because of the aromatic ring, around 10. This is low enough to be deprotonated by NaOH, a strong base, but not by a weak base like NaHCO₃.

Amines have pKa’s around 35. They are bases, so they will not react with base. They can, however, become protonated by an acid to form the water soluble salt.

Neutral compounds remain soluble in organic solvents. Likewise if you do acid base chemistry that gets you back to the neutral compound it will no longer be soluble in water.

Step 3: Extraction is generally a preferred technique. If this cannot be used consider other methods.

a) Can a simple filtration be used? Is one a solid, and the other a liquid?

b) Is one of the compounds volatile? If the compound is low boiling it can be easily taken of by rotary evaporation. Some examples of volitle compounds are acetone, hexanes, ethyl acetate, methylene chloride, ethyl ether.

c) For remaining compounds, column chromatography would probably be a good choice. Column chromatography separates compounds based on polarity. An expert can separate almost anything on a column, even the most minor polarity differences.

Why Does Organic Chemistry Matter?

Why are we studying organic chemistry? Why are YOU studying organic chemistry? Do you know why it matters?
I am taking a learning class and this week we are studying neuroscience - how learning takes place in the brain. The most interesting thing to me is how many common tasks use both sides of the brain - people think that you are left-brained or right-brained and that is where your thinking takes place but your brain needs information from many sources to process information and perform a task.
Anyway, what does this have to do with organic chemistry? I have an autistic son who seems to have synapses that work sometimes and sometimes they do not. Impulses are controlled by the frontal cortex and he has some impulse control, but he takes medication that helps him with this, ritalin. How does a stimulant make you calmer? I am not a biochemist but I understand that ritalin helps to regulate his synapses - Briana might know more about this than I do. I want to know more about how this drug works in his brain and why.
I am grateful for organic chemists who study pharmaceuticals and synthesize them and make mine and my son's life better.

Reporting Spectral Data

In advanced laboratories, your spectral data is often the most important peice of data to prove that you have achieved your objectives. Ideally, you want as much data as possible to confirm that you have the compound that you claim you can. Often in intro labs we will just be using one type of spectroscopy in our lab period. Here is an example of a full spectral analysis of a compound I made. It includes 3 types of NMR as well as the IR data and MS data. There is a whole section about instrumentation that isn't included here.

(S)-2-allyl-2-fluoro-3,4-dihydronapthalen-2H-one

Colorless oil
BB1041

¹H NMR (400 MHz, CDCl₃) δ 8.09 (d, J = 8 Hz, 1H: aromatic CH), 7.55 (m, J = 8 Hz 1H: aromatic CH), 7.38 (t, J = 8 Hz, 1H: aromatic CH) 7.29 (m, 1H: aromatic CH), 5.92 (ddt, J = 8 Hz, 10 Hz, 17 Hz, 1H: CH=CH₂), 5.25 (d, J = 10 Hz, 1H: CH=CH(H)_cis, 5.21 (d, J = 17 Hz, 1H: CH=CH(H)_trans, 3.14 (m, 1H: diastereotopic benzylic CH₂), 3.04 (m, 1H: diastereotopic benzylic CH₂), 2.74 (m, 1H: diastereotopic allylic CH₂), 2.61 (m, 1H: diastereotopic allylic CH₂), 2.42, (m, 2H: diastereotopic cyclohexyl CH₂).

¹³C NMR (125 MHz, CDCl₃) δ 193.99 (d, J = 17.75 Hz: C=O), 142.71 (aromatic C), 134.08 (aromatic CH), 130.92 (aromatic C), 130.86 (=CH), 128.75 (aromatic CH), 128.31 (aromatic CH), 127.31 (aromatic CH), 119.89 (=CH2), 94.98 (d, J = 184.26 Hz: CF), 37.99 (d, J = 22.6 Hz, allylic CH₂), 31.91 (d, J = 22.6 Hz: cyclohexyl CH₂), 25.89 (d, J = 10.1 Hz, benzylic CH₂).

¹⁹F NMR (376MHz, CDCl₃) δ -158.79(m).

FTIR (CD₂Cl₂): ν_max 1701, 1604, 1238

HRMS calcd for C₁₃H₁₃OF [M + H] = 205.1029, found 205.1036

There are a few things that you would want to tweak for your lab reports, but also some lessons to learn. BB1041 is simply where I can find this in my notebook. The NMR data includes the strength of the instrument followed by the solvent. Then each peak is listed followed by the information about it. The IR data includes the solvent in paranthesis followed by a listing of the most significat peaks. For your lab reports, you should include a short analysis of those peaks. Yours should look something like this.

cyclohex-2-enone

Colorless oil

FTIR: (CD₂Cl₂): 3100(sp2 C-H strecthing), 2915(sp3 C-H stretching), 1680 (conj. ketone, C-O stretch), 1600 (C-C double bond stetching).

Solubility applied to Vitamins

Solubility is the extent which a compound dissolves in a solvent. Compounds tend to dissolve in compounds that have similar intermolecular forces. In my undergraduate class we simplify this into the rule of 5. Basically if there are five or less carbons to each polar (oxygen and nitrogen containing) functional roup then it is probably water soluble. If there are 6 or more carbons per polar functional group then the molecule is likely insoluble in water. It will most likely be soluble in hexane or another less polar solvent. There are exceptions for small molecules like acetone, if there are a small number (1 or 2) carbons with a polar functional group then it is probably at least slighltly soluble in water.

One reason why it is okay to load up on vitamin C when you are sick is because it is water soluble. If you intake too much vitamin C you will just end up excreting it, and it will not harm you. The reason for this is because vitamin C is water soluble. Looking at the structure you see a cyclic ester with 4 hydroxy groups. That is 5 carbons to 5 polar functional groups. It is soluble in water.

Another one that is soluble in water is Niacin. The carboxylic acid and the nitrogen are polar functional groups, there are only 6 carbons, so this too is water soluble.

Some examples of water insolube vitamins are vitamin K and vitamin A. These are not so easlily excreted. Ingesting a moderate excess of probably wont hurt you, but in large excess they can build up and cause problems. Story has it that some artic explorers died from eating polar bear liver which contain an unusual amount of vitamin A, which is just one more reason not to eat polar bear livers.

The structures of the non water soluble vitamins have a few polar functional groups, but they have long carbon chains, their carbon to polar functional group ratio greatly exceeds 5:1, therefore they are water insoluble.
We all know the latin root for vita, but apparantly the -amin part comes from amine. Polish chemist Cashimir Funk named them vitamines because he thought that all of these compounds would be amines. The name has since lost the e and Funk's hypthesis has been refuted, but the rest of the letters have remained.

Cis and Trans isomers

Beyond the basics of being able to recognize cis isomers and trans isomers, it is interesting to note the physical implications of the isomers. The best and most basic example of naturally occuring cis and trans isomers are the polymers of isoprene. Isoprene polymers occur naturally in rubber plants and gutta-percha plants. However they differ in the ways the isoprene units are arranged. Notice that the two double bonds of isoprene are connected by a single bond which allows for rotation. Polymers of isoprene can be connceted in either linkages or trans linkages. The highest priority chains will be the ones that link with the rest of the polymer.
The cis isomer looks like thisand the trans isomer repeats like this

The two isomers have vastly differing properties because of their geometries. The cis isomer cannot be packed tightly with another cis isomer, whereas chains of the trans isomers fit nicely together. The extent of cross-linking between two polymer chains is greatly affected by how compact the polymer chains can become. The cis isomer has fewer cross-links and the separate polymer chains can slip more-making a the stretchy, rubbery polymer we know from the rubber tree. The trans isomer is much harder. It is sometimes used in golf ball coverings.

Somewhere inbetween the two naturally occuring isomer polymers we have the rubber that we come in contact with most often. Industry has created a system of introducing sulfur cross-links between the cis polymer. Very soft rubber has a smaller percentage of sulfur cross-linkages, while harder rubber will contain more.

Beginning IR Problems

You have successfully isolated a compound you know to be one of three narcotic analgesics. How could you use IR to determine which compound you have isolated? What specific bands would you be looking to be present or absent?

The key to this problems is simply recognizing the fuctional groups. While the structures may be pretty similar overall there are key differences. For example heroine should display a carbonyl peak, while this would be true of oxycodone as well, in oxycodone you would also have an OH peak. Morphine should have no peaks in the carbonyl region.

Citation guide

What needs to be cited? When you summarize the findings of another author you need to cite it. If you use a figure (picture) from another source you need to cite it in the caption under or above the picture.

Journal

Author, A. B.; Author, C. D. J. Abbrev. 20XX, vol, xx-yy.
NOTE: No punctuation in journal abbreviations except periods. No conjunctions, articles, or prepositions in journal abbreviations. No comma or semicolon before or after journal titles.

Magazine with dates instead of volume numbers

Author, A. B.; Author, C. D. Magazine Abbrev. October 26, 1995, p. 20.

Book without editors:

Author, A. B.; Author, C. D. Book Title; Series Name and number; Publisher: City, STATE (2 letters), year; Vol. 1, pp xx-yy.

Book with editors, no authors named:

Book Title; Editor, A. B.; Editor, C. D., Eds.; Series Name and number; Publisher: City, STATE (2 letters), year; Vol. 2, pp xx-yy.

Book with editors and authors named:

Author, A. B. In Book Title; Editor, C. D.; Ed.; Series Name and number; Publisher: City, STATE (2 letters), year; Vol. 3, pp xx-yy.

Website

Website Home Page. http://websitename.com. (Accessed Month Day, Year).

for example something off blackboard without an author:

Blackboard CHEM 627 Course Materials Page . http://courseware.ku.edu. (Accessed June 23, 2008)

Example:

“Geldanamycin has been shown to reduce cell activity by binding to the ATP/ADP-binding domain of HSP 90, a chaperone protein¹.”

(at the end of text or as a footnote)

(1) Roe, M.S.; Chrisostomos, P; O’Brien, R.; Ladbury, J. E.; Piper, P. W.; Pearl, L. H. J. Med. Chem. 1999, 4, 260-266

Figures

Organic chemistry is a visual science. One scheme can summarize a great deal of information. Software is available to help chemists make their own schemes. ChemDraw is widely used among scientific professionals and is available through many academic institutions for free.

Just like tables, images should have names. If your image comes from a website, you need to cite it within the image caption. Images should not be excessively big, in ChemDraw once you highlight your molecule, you can go to object on the toolbar and choose Apply object Settings from ACS Document 1996. This seems to be a good choice for most purposes. If you have time, take a look at all the features offered by this program. Once you draw a molecule you can get its MW at the click of a button, you can also get the common name for the molecule, as well as spectra data predictions.

What types of images need to be in a lab report? If you are doing a reaction, that reaction needs to be included in the report. If we are isolating a molecule, you should include that molecule. If you have obtained data in the form of an image (for example: an IR spectra), that should also be included. In general, if a figure can be more informative, or serve as a graphical summary include it.

Example

Figure 10.3 Synthesis of Salicylic Acid

Figure 10.2 Menthyl Salicylate¹

(1) “Salicylic Acid,” The Columbia Encyclopedia, 6^th ed. New York, Columbia University Press, 2001.

Introduction

The "Introduction" of a laboratory report identifies the experiment to be undertaken, the objectives of the experiment, the importance of the experiment, and a general background (specific analysis goes in your discussion section) for understanding the experiment. Your measurable objective should be stated. This is never to familiarize yourself with a technique. This may be one of your personal goals, or the goal of your instructor, but it is not the scientific goal. The background often includes theoretical predictions for what the results should be.

Tips for writing a good introduction:

Move from general to specific: How does your experiment fit into a larger context? I agree that this can be a little difficult with some of the more technique driven labs, but it can be done with some creativity.

Engage the reader: Why should the reader care about what you did?

Review: Review the literature. Normally when you are doing original research, you review the studies that you based your research on. It may be more appropriate to review recent uses of whatever technique or reaction we studying in a lab report situation as it is not exactly original research.

Summarize the key concepts: Your discussion will provide full detail but you need to give your reader enough here so that they can what you were trying to do, even if they don’t understand the details of your discussion. Someone with a knowledge of general chemistry should be able to understand it. If there is a term that a person a level below you would not know, define it.

Learning from published research may be the best way to see all of this working together. Notice below how you are drawn in by a much bigger, more general, problem (stopping the virus that causes polio and herpes) than the one actually being investigated (alkylation of a ketone). Barry Trost is not the only person who can do this. If your lab involves formation of a Schiff base, for example, you could talk about the biological formation and importance of Schiff bases. If you are doing a simple technique, you can look current usage or adaptations of the technique.

Example:

Hamigeran B (1), a metabolite isolated from the poecilosclerid sponge Hamigera taragaenisis … exhibit 100% virus inhibition against both herpes and polio viruses with only slight cytotoxicity throughout the host cells¹. Its structural motif combined with the biological activity stimulates the development of a practical synthesis that can facilitate exploration of the chemical biology of such structural types represented by these scarce natural products. Nicolaou reported the first well known asymmetric sythesis… Our interest stemmed from the question of whether setting the stereochemistry of the quaternary center from which all remaining stereocenters may evolve might be achieved by Pd-catalyzed asymmetric allylic alkylation (AAA)⁵ of a ketone enolate …

Summarized from: Trost, B. M.; Pissot-Soldermann, C.; Chen, I.; Schroeder, G.M. “An Asymmetric Synthesis of Hamigeran B via a Pd Asymmetric Allylic Alkylation for Enantiodiscrimination.” J. Am. Chem. Soc. 2004, 126, 4480.

Practice Problems

The best way to learn organic chemistry is by challenging yourself with problems. It is not true that you need to spend late nights memorizing material straight from the book. Doing problems helps you to recognize patterns and find out where your understanding is lacking. Having graded countless exams throughout my graduate career, I have found that even the best memorizers tend to make small mistakes. The students who understand can write out the mechanisms, or draw attack arrows, and all their doodling seems to help.

I will be posting commentary for difficult problems as well as resources here. All posts will be open for comments, as a way to supplement for the inability to ask questions in the large lecture hall. Hopefully, through dialog, we can make organic chemistry a little more approachable.