Final Exam notes Chem 8 Spring 2007 start here. 

(3/12/07) Aromatic Compounds, Chapter 6 Bailey. 

Benzene. Compounds with the formula C₆H₆ containing three alternating double bonds in a hexagonal ring system are called aromatic compounds. The stem name of these compounds is benzene. When benzene rings are mono-substituted, we use the same rules as in alkanes: look for the lowest locants and name the substituents in alphabetical order. When benzene rings are di-substituted an older nomenclature is still being used besides the familiar numbering. We use ortho for 1,2(=o), meta for 1,3(=m) and para for 1,4(=p) di-substitution. Only for di-substitution! Here are a few examples, in which we also incorporated the phenyl group when benzene is a part of the side chain.

Compounds that contain benzene rings are called arenes or aromatics. Many derivatives of these compounds are carcinogenic and were among the first ones identified as such. For example, in cigarette smoke we encounter many different poly-aromatic hydrocarbons (PAH's) such as benz[a]pyrene.

Biphenyl is interesting in the sense that we have it more than likely in our body as a polybrominated or polychlorinated compound.

PCB's, polychlorinated biphenyls, were produced between 1929 and 1972 (about 500,000 metric tons!) and used in transformers, coffee cups, baby bottles etc. It is highly fat soluble and tends to accumulate in the food chain, because fish feeding in PCB-contaminated waters and birds that feed on those fish have fairly large amounts in their fat tissue, and we are encouraged to eat birds and fish, because they contain healthy lean meat......... Production of toxic PCB's is banned since 1979 and currently the only real danger is in the mercury content of certain fish species. Shark and tuna are on top of the food chain, I would say: limit eating those; salmon, mackerel, trout, cat fish are lower placed in the food chain, therefore they contain less mercury and PCB's. Eating fish for its very low fat meat and extremely healthy oils everyday is not advised, twice a week is smart. 

(3/14/07) Reactions of Benzene. Bailey Chapter 6 continued.

The best way to study the reactions of benzene is to take the overview in Bailey in table 6.2 on page 169 and memorize the reactions.

Here is another example of an alkylation (Friedel-Crafts) including the reaction mechanism of this electrophilic substitution. The electrophile, the electron loving particle is the secondary propyl carbo-cation.

Overall reaction:

In the third step of the reaction mechanism you notice that the stable aromatic configuration of the ring is restored. This is typical for reactions of benzene and other aromatic compounds, such as toluene and phenol. Practice the reactions above doing the following problems.

Answers

Here is the homework assignment to be turned in Friday 3/16/2007, at 12:30 during the Review Session.

3-19-2007 Start doing the following problems from Bailey, and see me or email me this week if you have any questions:

Bailey: 5.29; 5.39 c; 5.40 a, b;

After studying these reactions including their mechanisms you should be able to grasp this for the other characteristic reactions of benzene.

To understand the relevancy of the group of compounds we are studying now read Bailey Connections 6.1, pages 158 and 159, an excellent expose on Cancer and Carcinogens around us.

3-20-2007 Benzene reactions become increasingly more interesting if a mono-substituted benzene ring is reacted again. For example, if phenol (recall, phenol=hydroxybenzene) is reacted with bromine and iron is used as catalyst. The bromine atom tends to go to the ortho and the para position and not to the meta position. In other words, the products are o-bromophenol and p-bromophenol. The amount of m-bromophenol formed is negligibly small. The best way to understand this is to go to this hand out: Reactions of monosubstituted benzene.

Do Bailey Problem 6.4 and 6.5. using o, m and p where appropriate. These are practice problems.

Here is a Worksheet on Benzene reactions that we will go over Friday 3/23/07, but it is also your homework for that day. It needs to be in at 12:30pm that day, if you can't make the Review Session. Do all the problems and then check the answers later this week when they are available on-line. Only question #1 will be available!

(3/29/2007) The next item of interest are the reactions of substituted benzene rings. The group attached to the benzene ring determines where the new incoming electrophile goes, refer to question 2 on the last work sheet. It says that alkyl groups will be activating and directing incoming groups to the ortho and para position. This is also done by OH groups, and NH₂ groups.

For example, when phenol reacts with isopropylchloride and AlCl₃ catalyst, o-isopropylphenol and p-isopropylphenol are formed, you will get a mixture of these two products, and not m-isopropylphenol. Actually, the amount of meta compound is so small that we neglect it.

This is the way we typically write this. Realize that this is not a complete and balanced equation. You would need a two in front of the isopropylchloride molecule and 2HCl as byproducts.

Here is a summary of the table 6.2 on page 169 from Bailey. One remark, we like to add SO₃ to sulfuric acid to get a better result of the sulfonation. It simply produces more SO₃⁺ electrophile when sulfur trioxide is present.

Activating groups such as OH (hydoxy), NH₂ (amino), OR (ether), R (alkyl) are ortho and para directing. Recall R= methyl, ethyl, propyl, butyl etc.

Deactivating groups such as NO₂ (nitro), SO₃H (sulfonic acid), COOH (carboxylic acid), COR (ketone), COH (aldehyde), COOR (ester), CF₃ (trifluoromethyl) are meta directing.   

Exception: The halogens, Br, Cl, and I are deactivating, but ortho and para directing. 

On your quiz you may use a table that I will provide you, so that these directions are easy to follow.

Here is an example of a meta directing reaction: when we mix nitrobenzene with acetyl chloride and AlCl₃ catalyst the product is meta-nitroacetophenone. 

This will help you determine the products of reactions of substituted benzenes, such as toluene, phenol, aniline. Try to find time to practice that in the coming days.

Good problems to practice are: Bailey 6.31, 6.42 and 6.43 from pages 184 and 185.

The last reactions of aromatic compounds you need to know are oxidation and reduction reactions, in short they are called redox reactions.

Read a great article on killer whales in the National Geographic of March 2005. Awesome pictures too! PCB's, perchlorobiphenyls from batteries, are abundant in killer whales, that eat mammals such as seals, dolphins and gray whales, and they are particularly damaging for their reproductive systems and their immune systems. The people from Kettleman City and Avenal are presently protesting against the intake of Biphenyl containing batteries in the largest landfill in the US along I-5 at Kettleman. In the meantime bring your used batteries to my lab, so that they will be properly recycled; don't throw them in the garbage, please. 

Here is the Worksheet II on Benzene reactions  Do all the problems and then check the answers later this week when they are available on-line.

The last reactions from this chapter are pertaining to oxidations and reductions of the side chain. For example, treating toluene with potassium permanganate produces benzoic acid.

This obviously is an oxidation. You have changed an ortho and para directing group into a meta directing moiety, the COOH group.

Here ends the material for Quiz 2 on 3/30/2007

(3/31/2007)A side chain in a benzene ring can also be reduced. For example, we treat ethyl phenyl ketone with zinc and hydrochloric acid and we produce propyl chloride as shown below. This is called a Clemmensen reduction.

Notice two things in the Clemmensen reduction: First of all, you don't loose any C atoms: in reactant and product there are three C atoms in the side chain. Secondly, you change a meta-directing group into an ortho and para directing one. That is very powerful for synthetic strategy!

(Starting 4/9/2007) X. Stereochemistry. Bailey Chapter 7.

Plane of Symmetry. Ideally, to study and understand this subject you need a model box. They are available at Reedley College in PHY 77, or in the bookstores at FCC or RC. You can also take gummy bears and tooth picks, no kidding, it works! As long as you make it visual for yourself.

If we were to build a tetrahedral compound that has four different groups attached to one carbon atom we could wind up with two different molecules. These molecules look completely the same drawn on paper, but in reality they are different. Realize, a paper is two dimensional and we exist and build this molecule in 3-D!

Here is an example. Build two molecules of bromochlorofluoromethane, a simple tetrahedral molecule and use three different colors for the halogens, use white for hydrogen and black for carbon. Chances are 50% that you have built two different molecules, that cannot cover each other. Amazingly, there are actually two different bromochlorofluoromethane molecules possible! That's cool! If you take a close look at these two molecules you will see that they are each other's mirror image. The carbon atom that has four different groups attached is called a stereo-center and the two isomers that you have built are called stereo-isomers.

Realize, that this molecule has no plane of symmetry. All objects (or body parts) that possess no plane of symmetry have a mirror image that is different. Your left hand has no plane of symmetry, therefore its mirror image is different. The mirror image of your left hand is a right hand and not another left hand! (Although many people have two left hands...but that is an entirely different story...) A wood screw or a shoe also does not possess a plane of symmetry, therefore their mirror images are different objects. Check this with some objects around you. A wine glass has a plane of symmetry and therefore its mirror image is the same object. A book has no plane of symmetry and its mirror image is entirely different with regards to the shape, but also the pictures and the letters! In short, molecules that have no plane of symmetry, that is exactly what we are studying here. This pertains to molecules with tetrahedral atoms (carbon, nitrogen etc) with four different groups attached, not to a planar carbon atom, such as the C=O group.

Unfortunately, the chemistry involved with this subject uses many Greek terms to explain this phenomenon. Stereochemistry is also called chirality, from the word cheir, which is Greek for hand (sic, get it?). The C atoms that have four different groups attached are named chiral C atoms. And mirror image isomers are called enantiomers.

A molecule with one chiral C atom has two enantiomers, two molecules that are each other's mirror image. There are many molecules in nature with more than one chiral C atom, glucose, cholesterol, prostaglandins to name a few. Each chiral C atom adds another two enantiomers to the total number of possible compounds. In other words, the answer to the question, how many different compounds exist when there are n chiral centers combined in one molecule, is 2ⁿ.

Before we continue let's take a look at four everyday life molecules with chiral centers that we use as medicines. You notice below how important it is that pharmaceutical companies make and sell the correct R or S form!! 

Six years ago, at the Royal Swedish Academy of Sciences the 2001 Chemistry Nobel Prize was awarded to three researchers, one from Japan and two from the U.S. They found ways to produce one form of mirror image drugs in very high yield and very high purity. The key to their winning research is that they developed the perfect catalyst for each synthesis. For example, the well known anti-inflammatory drug S-Naproxen, an important pain-killer, is currently being produced quite a bit cheaper than 5 years ago. Developing the optimal catalysts can be of great significance if we need to produce more and especially better antibiotics, for instance against anthrax and smallpox.

Here ends the homework assignment
 

(4/12/2007)Meso compounds. Take a close look at the molecule of 1,2-difluorocyclopentane shown here under. Build it if you have a model box available. You notice two interesting things. First of all it has two chiral centers. That means that you could build 2² = 4 different stereo-isomers. Amazingly, there are only three isomers possible, because the cis compound has a plane of symmetry and therefore its mirror image is the same.

Check this out for 2,4-pentanediol. Draw the stem of this molecule vertically and notice the plane of symmetry through C atom number 3. Next, try 1-fluoro-2,3-butanediol and notice that there is no internal plane of symmetry, a plane that slices the molecule into two equal halves. Draw all four stereo-isomers using wedges and dotted lines.

How do we distinguish between enantiomers, mirror image molecules physically? In other words, are there any physical constants that are different? Yes, there is one and only one. Enantiomers will bend polarized light in opposite directions. In other words, polarized light traveling through a solution with purely one isomer of trans-1,2-difluorocyclopentane in it rotates polarized light in opposite direction as its enantiomer. For example, + 7.2⁰ versus - 7.2⁰. Always the same value, but with opposite sign. That is the reason why chemists call these isomers sometimes optical isomers.

Polarimeters (Bailey section 7.3) are used in the grape and wine industry where sugar concentrations are measured with them. Recall, sugars are chiral molecules with many C^*'s. It speaks for itself that a meso-compound will produce no rotation (zero degrees: 0⁰) of polarized light, because the rotation caused by one half of the molecule would be canceled by the mirror imaged other halve.

Now do all other in-text problems from Bailey up till 7.12. Study the excellent examples that are written above each problem. You notice that at the end of each problem in bold they indicate what related problems you can find at the end of Chapter 7. For example, Problem 7.12 has a related problem: 7.25. Practice, practice is the name of the game.

4/16/2007 R-S nomenclature. The nomenclature for chiral compounds is based on the same mass priorities as we have seen in the E-Z nomenclature for alkenes. Working with the atomic number of the first neighbor to the chiral carbon we can easily assign priorities. Make sure you place the lightest group away from you. Let's practice that first with the following chiral molecule, 1-bromo-1-ethanol. The lightest group is hydrogen. The rotations shown are accomplished by (mentally) holding the CH₃ group in your hand and rotating the bottom three groups around the C-CH₃ axis.

You notice that by rotating the molecule we did not break any bonds and therefore we have not changed its configuration. Verify that we rotated the molecule every time 120⁰ and that all four structures above are exactly the same. Check this with a model box in class (or at home with tooth picks and gummy bears, this is not like MythBusters..... you want to try this at home.)

Now assign mass or atomic number priorities (that comes down to the same thing) to the four atoms around the chiral carbon.

Here are some practice problems.

Answers:

1.a. R-2-butanol b. (3S)-3-fluorobut-1-ene; c. (1R)-1-phenyl-1-ethanol

Here is another look at RS configuration explained from a different perspective, that might be helpful to you. Click on RS configuration.

 (4/18/2007) Racemic Mixture. A mixture of equal amounts of optical isomers is called a racemic mixture. For example, for the compound 1,2-difluorocyclopentane above: when equal amounts of the two enantiomers are mixed in one solution and we would measure the bending of the plane of polarized light, we would find 0⁰, because the effect of each enantiomer is canceled by the other.

This will become important when we look back at reactions between alkenes and bromine. For example, cyclopentene mixed with bromine produces trans-1,2-dibromocyclopentane. Referring to the mechanism, including the backside attack of the bromonium ion, it is easy to see that a racemic mixture of two enantiomers will be formed. It is shown below.

 
The enantiomer of trans-1,2-dibromocyclopentane will be formed as well, because the backside attack of the Br^- ion can also take place at the other C atom. The result is that both trans compounds are formed in equal amounts, which is called a racemic mixture. There is a rule that says, when the starting compounds of a reaction are non-chiral, the product will also be non chiral. Cyclopentene and bromine have obviously no chiral centers, so they are non-chiral. The product mixture is racemic, which means it contains two enantiomers that cancel each other's rotation in a polarimeter, so the chirality of the product mixture is also 0⁰. We are in compliance with the rule in green above.

(4/19/2007) Fischer Projections. Fischer projections are simplifications of a structural formula in which the chiral carbon center is shown as a cross hair and the four different groups attached to it are drawn North, South, East, and West. Let's take a look at R-lactic acid, the acid formed in our muscles on contraction of the sarcomers. You obtain the Fischer projection by pushing the groups onto the plain of the paper. The 3-D structure is now 2-D and easy to manipulate.

The Fischer projection on the right hand side shows the same four groups with the hydrogen atom away from you. Here are the rules for Fischer projection formulas.

1. You may rotate the formula 180⁰ or 360⁰ without changing the configuration: R remains R and S remains S.

2. A rotation of 90⁰ or 270⁰ changes the configuration from R into S or vice versa.

3. Swapping two groups changes R into S or vice versa. Apply the rules in the following exercises. Before you start doing these exercises you need to be aware of one more Greek term: diastereomers. This term is used when we compare, for example, cis-1,2-difluorocyclobutane with trans-1,2-difluorocyclobutane. The cis compound has an internal plane of symmetry and is therefore a meso compound. The trans compound is completely isomeric with the cis compound, but is obviously not its mirror image, it is not its enantiomer. In that case we use the beautiful Greek term diastereomers for these two compounds. Their relationship is not enantiomeric, but diastereomeric (.....just a seven-syllables-word).

Friday 4/20/2007 there is homework due. It is the work sheet shown below. Start it now. I will add one question on Friday and it is due at the end of the session at 12:50pm.

          3. Two questions on a meso-compound and a racemic mixture. Use examples of each to illustrate your             answers. Draw structural formulas of your examples.

          a. In what respect is a meso-compound similar to a racemic mixture?

          b. In what respect is a meso-compound different from a racemic mixture?

(4/23/2007)Substitution Reactions, Bailey Chapter 8.

Halogen atoms in alkyl halides can be substituted relatively easy when we use the proper reagents. For example, 1-chloropropane reacts good with sodium methoxide, CH₃ONa in acetone solvent. The reaction is shown below.

Since the concentrations of both species are determining the rate of this reaction we call this a second order substitution reaction, in short S_N2. The most important characteristics of an S_N2 reaction are:

1. The halogen is attached to a primary C atom, or an accessible secondary C atom.

2. The nucleophile is small and strong enough.

3. The configuration will change if the halogen is attached to a chiral C atom: R will become S and S will become R.

4. The solvent needs to be polar and preferably aprotic (=not containing OH or NH groups).

5. The leaving group needs to be polarizable (= having a good charge distribution). For example, I^-, Br^- and Cl^- are good leaving groups, since they are derived from strong acids. Triflate, mesylate and tosylate are even better leaving groups. Look up their formulas in Bailey of on-line.

Let's go through the first three and most important characteristics.

1. For example, the following alkyl halides can be substituted easily using an S_N2 reaction: methyl iodide, because the C atom has no C neighbors; ethyl bromide, because Br is attached to a primary C atom; 1-chloro-2-methylpropane, because Cl is attached to a primary C atom; 2-iodobutane, because the secondary C atom is very accessible, it has just small groups such as methyl and ethyl around it.

In short: all compounds that have an halide attached to a C atom with no C neighbors, one C neighbor, or two C neighbors with a lot of space around it, qualify.

2. Good nucleophiles are compounds that have lone pairs available such as HS^-, I^-, OCH₃^-, OH^-, CN^-, NH₃, acetylide ion etc. They should not be too bulky because other reactions will then compete with the substitution. These reactions are called eliminations and we will discuss them later. For example, t-butoxide is a very bulky nucleophile. Excellent nucleophiles are (add the lone pairs):

3. If R-2-iodobutane is reacted with potassium hydroxide the OH^- ion replaces the halogen with a backside attack so that S-2-butanol is the result. The configuration is changed from R to S, which is called inversion. This is a major characteristic of second order substitution reactions!! It is shown here including its complete reaction mechanism.

The backside attack of the OH^- ion causes the configuration to change. In the intermediate ion the OH and the I are in opposite positions attached to the chiral C atom. When I^- is leaving the other groups attached to the chiral carbon go through a flip over like an umbrella in the wind. Since the OH group is attached to the opposite side the configuration is now S, which is clearly visible in the mechanism above. Basically, this S_N2 reaction occurs in one quick step, which can be shown in a "one-hump- reaction-profile".

Practice the following S_N2 reactions and include inversion, if any. Write the equation using structural formulas and names under reactants and products. I made an answer page for you. You may access them by clicking the link under the yellow table.

Answers

S_N1 reactions. Substitutions of halogens at tertiary C atoms or at secondary C atoms that are sterically hindered follow another pathway. Let's look at the following example of 1-chloro-1-methylcyclopentane reacting with the ethylsulfide nucleophile.

The reaction above is called a first order nucleophilic substitution, because exclusively the concentration of the starting compound, 1-chloro-1-methylcyclopentane, determines the rate of the reaction. In short, these reactions are called S_N1 reactions. Another real important feature of these reactions is, that if the C atom to which the halogen is bonded is chiral, a mixture of equal amounts of R and S will be produced. This product mixture is called a racemic mixture.

Here we will write some problems from Bailey for you to practice this weekend.

(4/28/2007) Practice the following problems pertaining to S_N1 reactions. Use structural formulas.

Answers

(5/1/2007)  Read Bailey pages 235-239: F. Factors Influencing the Reaction Mechanism- S_N2 versus S_N1, and

G. S_N1 and S_N2: Summary.

Elimination reactions. Bailey 8.5

As always in chemistry (and in life...): there is competition. Substitutions are challenged by eliminations, which means that the halogen is released together with a hydrogen from the neighboring carbon and an alkene ensues. Recall,

Elimination reactions occur when the nucleophile is strongly basic, and the temperature is higher. Also, when the nucleophile is bulky (large) elimination predominates substitution.

For example, nucleophiles like the t-butoxy ion or the isopropoxy ion are so basic and bulky that they will look for a proton attached to a neighboring carbon atom, that is neighboring to the C-Cl, as is shown here.

In other words, besides the small amount of substitution product that would be formed here, the majority of the product is ethene (=ethylene), because the bulky isopropoxide ion is in fact a strong base that will attract an H⁺ from ethylene chloride. In one concerted step, similar to the S_N2 mechanism, the double bond is formed and the chlorine atom expelled as a chloride ion. This mechanism is called: E2, which stands for second order elimination.

(5/6/2007)To summarize: El and E2 will occur when the temperature is elevated to 80-100⁰C and the nucleophile is a strong base: alkenes are then the predominant products. E1 takes place just as S_N1, when sterically hindered C atoms are present.

This is where the lecture notes for Chem 8 end for this semester. The sample questions for the final will be posted this week. The final is on We 5/16, 2007 at noon (12pm) in PHY 77 at RC.