2. The Covalent Bond in compounds that have purely non metals. For example in butane, which is C₄H₁₀, the gas compressed to a liquid in cigarette lighters, there are covalent bonds between the carbon atoms, and also between the carbon and hydrogen atoms.

In the above at 1. drawn non-metal portion of sodium benzoate, the covalent bonds occur between the carbon and carbon atoms in the hexagon and between the carbon and oxygen atoms outside the hexagon. Notice that in these structural formulas we rarely write C for carbon, but consider each cross section of lines a carbon atom. Thus, the hexagon consists of six carbon atoms.  In butane I showed how that works. 

Review the characteristics of each bond type and the forces that hold molecules together, such as hydrogen bonds, dipole-dipole forces, and London dispersion forces by clicking on this text.

Study your text book Bailey, Organic Chemistry Chapter 1.

Here is the first homework assignment that needs to be turned in on or before F 1/12/07 @ 12:15 by email or hard copy in PHY 77(RC). It is a good idea to print-out the assignment, you avoid skipping and missing problems that way.

From this point on I will write the date in front of the text so that you can see when lecture notes are added and where you need to be to keep up with the course and complete the assignments in time.

1/16/2007 II. Hybridization of carbon atoms. This subject will be discussed in depth in the Friday 1/19/07 review session. Study this in the week before that day.

1. Single bonded tetrahedral C atoms: sp³ hybridization.

Summarized: One electron from the 2s orbital combines with the three electrons in the 2p orbitals to form four equal orbitals that point to corners of a tetrahedron. So the 3 in sp³ has nothing to do with a power of three, it represents the three electrons in the p orbitals p_x, p_y and p_z that are each containing one electron. Therefore we write sp³.
 

Here is the second homework assignment that needs to be turned in on or before Friday 1/19/2007 @ 12:15 by email or hard copy in PHY 77(RC). These are problems on Lewis structures and some are on hybridizations of carbon. 

 

(1/17/2007) 2. C atoms with a trigonal planar shape and thus a 120⁰ bond angle, in fact all C atoms with one double bond exhibit sp² hybridization. One 2s and two 2p orbitals form three new orbitals that point to corners of a triangle, hence the trigonal planar shape. This occurs for example in tetrafluoroethylene, C₂F₄ a starting compound for the manufacturing of teflon, the polymer we use as the lineage in

fry pans and, by the way, C₂F₄ is one of the most slippery compounds we know. 

3. The linear C atoms, all C atoms with a triple bond as in acetylene, C₂H₂, the gas used for welding, sp hybridization. One 2s and one 2p orbital form two new orbitals pointing to "opposite sides" of a line, hence the linear structure of acetylene. The remaining 2p_y and 2p_z orbitals form pi bonds through sideways overlap. Also, C atoms with double bonds on each side and thus a 180⁰ bond angle show this type of hybridization.


The step from ground state to excited state in which one electron from the 2s orbital jumps into the 2p_z orbital costs a little bit of energy, which could be delivered by light (e.g.). Forming bonds produces a great deal of energy, which offsets the energy needed for the 2s-2p jump shown here:

View the steps for each of the three hybridizations of carbon.

Click here for a tutorial page on hybridizations of carbon.

Advise: Make sure by 1/19/2007 that you have read Bailey Chapter 1 completely.

(1/22/2007) Exercise: try to assign the type of hybridization in each of the carbon atoms in the following compounds.

Realize that bonds with heads-on overlap are called sigma bonds, σ bonds and sideways overlapping bonds are pi bonds, written as π bonds . π bonds occur in double and triple bonds only. We will take a look in an upcoming review session what the difference is shape wise. 

Good practice exercises can be found in Bailey: 1.30, 1.31, 1.32. I suggest you do those.

Summary of Hybridizations of Carbon

Starting 1/24/07 III. Organic Nomenclature

This material can also be studied and practiced from Bailey Chapter 2.

The homologic series of alkanes, C_nH_2n+2 are used as the basis for naming organic compounds. These compounds are produced during a fractional distillation of crude oil and petroleum. Gasoline is a mixture of alkanes that contain 5 to 10 C atoms. Here is the series from 1-10.

From here it continues with C₁₁H₂₄ undecane and C₁₂H₂₆ dodecane etc. using the familiar Greek counting numbers through nonadecane to C₂₀H₄₂: icosane.

As you probably know gasoline is rated with an octane number. That number is used to indicate the quality. The closer to 100 the better your gasoline. This number is based on the percentage isooctane versus heptane in gasoline. Heptane is a poor burning alkane in our engine that produces a lot of "knocking". The industry has given that compound an octane number 1. Isooctane, which is actually 2,2,4-trimethylpentane, which contains a total of eight C atoms, burns without knocking and has therefore an octane number 100. Looking at the correct systematical name for isooctane, 2,2,4-trimethylpentane, you notice that we have pentane as our stem name and three methyl groups as branches coming of pentane. Try to draw its structural formula. These branches are called alkyl groups. By taking away one hydrogen atom from the alkanes above you can determine the name and the formula of the branching alkyl groups. The suffix -ane of the alkane name changes into -yl, alkyl.

For example,

The propyl group can be attached to a main chain at its end C-atom and at its middle C-atom. Therefore we use the name propyl for -CH₂CH₂CH₃ and isopropyl for CH₃CHCH₃. As you notice, these two side chains are isomeric, both have the molecular formula C₃H₇. Used in line-bond formulas this will look like:

For the isopropyl group we use the letter i to determine the alphabetical order. In other words isopropyl comes before methyl.

Practice: Write the names of the following compounds. Key is: first find the longest continuous carbon chain, then write the lowest possible numbers for the branches, the side chains in it.

Answers: a. 2,3-dimethylbutane; b. 2,2-dimethyl-4-propylheptane; c. 2,3,3-trimethylpentane.

Answers: d. 4-isopropyloctane; e. 3-ethyl-4-isopropyl-2,2-dimethylheptane

Butyl groups. There are also four isomeric butyl groups that you need to know. All have the molecular formula C₄H₉, since they are derived from butane, C₄H₁₀ .

They are shown here:

Practice: Write the line-bond formulas of the following compounds.

a. 4-t-butyl-2,5-dimethylnonane

b. 1,3-di-sec.butylcyclohexane

The answers are shown here:

Rationale for structure a. above: Nonane is a chain with nine carbon atoms, C atoms number 2 and 5 carry a methyl group and C atom number 4 a t-butyl group.

for b: Cyclohexane is a ring with 6 C atoms and at positions 1 and 3 we have two secondary butyl groups. The secondary butyl group is, as the name says, attached with its second C atom to the ring. It is called a secondary butyl group, because the second C atom has two carbon neighbors.

__________

Friday 1/26/2007: no assignment due. In the review session we will individually work with a great nomenclature program on the lap top computers in PHY 82. It is essential that you get this down, so show up, or just stop by to checkoutthisawesomesoftware, yo

1/29/2007. Butyl groups. There are also four isomeric butyl groups that you need to know. All have the molecular formula C₄H₉, since they are derived from butane, C₄H₁₀ .

They are shown here:

Practice: Write the line-bond formulas of the following compounds.

a. 4-t-butyl-2,5-dimethylnonane

b. 1,3-di-sec.butylcyclohexane

The answers are shown here:

Rationale for structure a. above: Nonane is a chain with nine carbon atoms, C atoms number 2 and 5 carry a methyl group and C atom number 4 a t-butyl group.

for b: Cyclohexane is a ring with 6 C atoms and at positions 1 and 3 we have two secondary butyl groups. The secondary butyl group is, as the name says, attached with its second C atom to the ring. It is called a secondary butyl group, because the second C atom has two carbon neighbors.

__________

1/30/2007. Cycloalkanes, C_nH_2n. Cyclic alkanes are often used as solvents. They have ring structures, hence the term cyclic,  and therefore their general formula is C_nH_2n. The cycloalkanes that you are required to know are:

Now, let's use these structures to complete the nomenclature. You notice that we call the systematic organic nomenclature, IUPAC nomenclature. That acronym stands for International Union of Pure and Applied Chemistry, a group of people in charge with determining the names of new elements and new organic compounds.

The halogen substituents that we use in organic compounds are named with bromo, chloro, fluoro, iodo. Here are two examples of these alkyl halides. Ethyl chloride, or chloroethane is used by athletes to cool down the area after an injury. It evaporates quickly when applied to the skin using some local body heat. Ice works just as good...

More practice material.

        Answers: a. 1,1-dibromo-2-chloro-4-isobutylcyclohexane; b. 1,1-difluoro-3-isopropylcyclopentane; c. 3-ethyl-1-iodo-2,2,3-trimethylheptane; d. 2-cyclopropyl-3-fluoropentane.

   Here ends the material for Quiz 1, which is scheduled on Friday 2/2/2007 at 12:15pm in PHY 82. Sample Questions are posted, as well as the answers to them, so that you know what to expect.  Show up to check out the nomenclature software....., and first and foremost: email me with questions.

(2/5/2007)Now, let's use these structures to complete the nomenclature. You notice that we call the systematic organic nomenclature, IUPAC nomenclature. That acronym stands for International Union of Pure and Applied Chemistry, a group of people in charge with determining the names of new elements and new organic compounds.

Start reading Bailey Chapter 3. The halogen substituents that we use in organic compounds are named with bromo, chloro, fluoro, iodo. Here are two examples of these alkyl halides. Ethyl chloride, or chloroethane is used by athletes to cool down the area after an injury. It evaporates quickly when applied to the skin using some local body heat. Ice works just as good...

More practice material.

        Answers: a. 1,1-dibromo-2-chloro-4-isobutylcyclohexane; b. 1,1-difluoro-3-isopropylcyclopentane; c. 3-ethyl-1-iodo-2,2,3-trimethylheptane; d. 2-cyclopropyl-3-fluoropentane.

There is an excellent organic nomenclature computer program installed on the Reedley College (RC) server. You may practice nomenclature at RC in PHY 77 or PHY 82, and in the library at the North Centers (on Herndon Avenue) using the computers. There are also lap tops available at RC in PHY 77 and PHY 82. You work by yourself at your own pace. It is the way to go if you want to be proficient in nomenclature. Feel free to stop by.

(2/6/2007) IV. Isomerism.

Compounds that have the same molecular formula but different structural formulas are called isomers or isomeric compounds. For example, for the molecular formula C₄H₁₀ you can draw two different structural formulas: butane and 2-methylpropane. These molecules have different physical properties such as density, boiling point etc.

Under the two isomers for C₄H₁₀ above you see two different structures for compounds with the molecular formula C₅H₁₀. There are at least 8 more!

Answers for a and b respectively.

Here is the explanation on the proper use of cis and trans in the name of a compound. A special form of isomerism is more geometrical than constitutional. It is called cis-trans isomerism. It occurs when compounds have double bonds or rings. Here is an example:

You notice that the rigidity of the double bond and the ring, the fact that you cannot rotate the C-C bonds without breaking them, causes the molecule to possess cis-trans isomerism. Build these molecules with the Model Boxes in PHY 77 (RC) to verify this phenomenon. In alkenes it is the pi bond that hinders the free rotation. In cycloalkanes it is the ring structure that blocks free rotation.

Draw the structural formulas (line-bond is okay) for the following molecules. Properly use wedges and dotted lines in your formulas where needed.

a. cis-2-butene

b. trans-2-butene

c. cis-1,3-dihydroxycyclopentane (hydroxy group=OH)

d. trans-1-cis-3-tetrachlorobuta-1,3-diene

Answers will be given in the review session Friday. Please,  ask for it.

Primary, secondary, tertiary and quaternary C atoms and their use in nomenclature.

Primary C atoms have zero or one C neighbor. For example in methanol, CH₃OH and in ethane, C₂H₆ all the C atoms are primary C atoms. In cyclohexane all the C atoms have two carbon neighbors and therefore they are secondary C atoms. In 2-methylpropane the middle C atom is tertiary, because that carbon atom has three C neighbors. In the very symmetrical compound 2,2-dimethylpropane (draw it) you notice that the middle C atom has four C neighbors, therefore we call it a quaternary C atom. A simple concept! Here is an exercise to practice this.

Answer:

Primary, secondary, tertiary and quaternary C atoms.

This is the assignment that needs to be turned in on or before 2/9/2007 @ 12:30 pm by email or in PHY 77(RC) as hard copy.  

Some extra practice problems from Bailey are suggested here: Do 2:19 f, i, and k; 2.22 f; 2:25 n, and o; 

(2/8/2007)VI.  Alkenes.

Alkenes are compounds with a double bond, they are called unsaturated compounds, because they do not contain the maximum number of hydrogen atoms. Recall, C_nH_2n+2 is the general formula for alkanes, but alkenes have two hydrogen atoms less so their general formula is C_nH_2n. The nomenclature follows the same systematic rules as we have learned for alkanes. The lowest number is given to the double bond. Check out these examples:

Notice in the structure of limonene above that cycloalkenes also exist. The nomenclature for cycloalkenes follows the same rules as the one we learned for the cycloalkanes. 

Triple bonded compounds are called alkynes. They have 2 hydrogen atoms less than alkenes, therefore their general formula is C_nH_2n-2. Again, the nomenclature of alkynes is based on the same rules, now the ending of the name is -yne. Check out these two examples and notice that the nomenclature rules are very consistent.

A. Degrees of Un-saturation. A saturated compound has the maximum number of hydrogen atoms attached to carbon (or nitrogen for that matter). For alkanes we have used the general formula C_nH_2n+2. For cycloalkanes and alkenes this formula contains two hydrogen atoms less: C_nH_2n. The presence of a ring or a double bond causes that. We say: the compound has one degree of un-saturation. For example, 1-butene has one degree of un-saturation and cyclobutane also has one degree of un-saturation. Based on this theory how many degrees of un-saturation do the following compounds possess?

Answers: Compound a. 1,3-butadiene has two degrees of un-saturation, two double bonds, each one degree. You may also say: it lacks four hydrogen atoms to be completely saturated. 4 divided by 2 = 2⁰ of un-saturation.

b. cyclopentenylcyclooctane has three degrees of un-saturation: two rings and one double bond. 

c. benzene, a compound in gasoline and in polluted air, has four degrees of un-saturation: one ring and three double bonds.

 Using the molecular formula of benzene we can also calculate the degrees of un-saturation. Benzene is C₆H₆. Normal for a saturated six C-atom compound we have C₆H₁₄, C_nH_2n+2 for n = 6. Benzene has 14-6 = 8 H atoms less. Dividing this number(8) by two gives us the degrees of un-saturation: 4. Why dividing by two? Because for each ring or for each double bond we loose two H atoms. 

In situations that there are odd atoms in an organic compound use the following rules when the molecular formula is given. 

1. Replace each halogen atom, that is F, Cl, Br or I, with one hydrogen atom. For example, 1,2-dichloroethene, C₂H₂Cl₂ is equivalent to C₂H₄ and has therefore 1⁰ of un-saturation. 

2. Omit each oxygen atom. For example C₂H₅OH is equivalent to C₂H₆ with regards to its degrees of un-saturation: zero. 

3. For each N atom that is dropped from the formula subtract one hydrogen atom. Verify this in the formula of PABA, para-aminobenzoic acid, used in many sunscreens.

PABA is C₇H₇O₂N. After using the rules above, omitting the O atoms, and subtracting one H atom for each N atom dropped, the formula becomes C₇H₆. A saturated C-7 compound has (2x7) + 2 = 16 H atoms. Subtracting 16-6 = 10 H atoms, that is 10/2 = 5 degrees of un-saturation. Looking at the structural formula above we can verify the result: one ring , three C=C bonds and one C=O bond. Why do we subtract one H atom for each N atom dropped? Because nitrogen is trivalent and thus binds one extra hydrogen if inserted between carbon and another atom. (Oxygen is divalent and therefore can not bond extra H atoms, ergo conclusio we may drop oxygen atoms from the formula, when calculating un-saturation degrees).

Practice the degrees of un-saturation by doing the following problems.

1. a. Calculate the degrees of un-saturation in C₆H₆OFCl.

b. Draw a plausible (=acceptable, see MYTHBUSTERS on the Discovery Channel) structural formula for the compound in 1.a. with at least one ring.

2. Calculate the degrees of un-saturation in C₈H₉O₂N and suggest a plausible structural formula. The compound is aromatic.

Answers:

1.a. C₆H₆OFCl ~ C₆H₈. A saturated C-6 compound would have (2x6) + 2 = 14 H atoms, which is C₆H₁₄, recall C_nH_2n+2. Therefore the number of hydrogen atoms missing is 14-8 = 6, which means 6/2 = 3 degrees of un-saturation. One ring and two double bonds will do the trick.

b. A plausible structural formula is:

2. C₈H₉O₂N ~ C₈H₈. A saturated C-8 compound would have (2x8) + 2 = 18 H atoms, which is C₈H₁₈. Therefore the number of hydrogen atoms missing is 18-8 = 10, which means 10/2 = 5 degrees of un-saturation. We need an aromatic ring, which is 4 degrees of un-saturation. Adding one double bond gives us a possible answer.

Read Bailey Chapter 3 and do the problems 3.22 a, c, d, f, and h.; 3.25 b, c, and e.; 3.34 a, and d.; 3.41 all and 3.42 all.

(2/12/2007) Organic Reactions

Start reading in Bailey Chapter 4.

Organic reactions can be classified into three categories: addition, substitution and elimination.

1. In addition reactions we add atoms of a molecule to a compound with a double bond. For example, we can add two atoms of bromine to the carbon atoms of the double bond in propene. We have then formed 1,2-dibromopropane.

Or we can add hydrogenchloride gas to cyclopentene and produce 1-chlorocyclopentane. The two reactions are shown hereunder.

The dichloromethane, CH₂Cl₂ you notice above the arrow, is the solvent that does not take part in the reaction itself, but takes care of intimate contact between the reactants.

In Substitution Reactions we replace (substitute) an atom with another atom or group, for example in the reaction of 1-chlorocyclopentane with sodium hydroxide we can replace the chlorine atom with the hydroxide group.

In Elimination Reactions we delete a small molecule from the starting compound, for example we delete HCl from 2-chloropropane by using hot potassium hydroxide. The resulting compound is then propene.  

You may skip Bailey section 4.1C. Know the difference between elimination, substitution and addition reactions. Do problems 4.2 and 4.7. 

This is a great web site from a Canadian professor (with an interesting name:) Mombourquette

Check out his web site by clicking on his name, and read about the nomenclature from a different angle. With thanks to Jeffrey Velasquez.

Use the remainder of Chapter 4 as reading material that introduces you to Chapter 5. We will do a lot from that chapter.

(2/14/2007) Alkene Reactions. Bailey Chapter 5

Let's take a look at some important alkene addition reactions.

1. Hydrohalogenation, reactions of alkenes with HCl, HBr and HI.

The electrons in the pi bond of an alkene are highly reactive. When propene reacts with HBr the hydrogen atom attaches on one side of the double bond and the Br atom on the other side. Here is the result:

The Rule of Markovnikov: In reactions of alkenes with HX, in which X is an halogen, the H atom goes to the most hydrogen rich C atom of the double bond.

Do the following problems to practice this rule.

Friday 2/16 there is no Review Session. The college is not in session on Friday 2/16 and Monday 2/19 due to President's Day.

(2/20/2007) Alkene Reactions....continued

The reason behind the rule of Markovnikov can be explained with the reaction mechanism. The reaction mechanism is a representation of the steps in which the reaction proceeds. The above alkene addition goes in two steps which we can show using Lewis dot structures and curved arrows. View hereunder. [As a simplification we have left out the lone pairs around chlorine.]

The secondary carbo-cation that is formed in the first step above is more stable than a primary carbo-cation, that would be formed if the H atom would attach itself to the middle C atom of propene. Therefore the reaction follows this mechanism: via a 2⁰ carbo-cation. In many reactions you will notice that the highest carbo-cation is formed. In other words, the formation of a tertiary carbo-cation has preference over the formation of a secondary C⁺ ion. The stability of carbocations is:

This plays such an important role in organic reactions that carbo-cations formed in the first step practically always rearrange to a more stable structure. 

Here is another example. When 2-methyl-2-pentene is mixed with hydrogen bromide the following reaction takes place.

Notice that the rule of Markovnikov applies: the hydrogen atom of HBr went to the C atom of the double bond that carried one  H atom, the other C atom carried no H atoms. Try to write the reaction mechanism using Lewis dot structures for each compound and curved arrows. You will notice that a tertiary carbo-cation is formed. If the hydrogen would have been bonded by the other C atom a secondary carbo-cation would have been the result and this is less stable, therefore the reaction proceeded as shown.

The mechanism is shown here.

 Notice that the curved arrows start at the more negative site, at the pi bonds or the bromide ion and end at the more positive side the H atom or the carbo-cation. That is an important convention when drawing reaction mechanisms.

We will draw quite a few reaction mechanisms this semester to understand organic reactions better.

In the Review session of Friday 2/23 we will go over the Rule of Markovnikov and do more examples. There will be a worksheet on the subject. Keep on studying the following material before Friday!

(2/22/2007) Here is another important addition reaction in which solely trans products are formed. The addition of bromine to an alkene is shown here under.

2/22/2007. Rearrangement of carbo-cations. A carbocat ion formed during a reaction can rearrange itself to become more stable. We will study two types of rearrangements, a hydride, H^- shift and a methanide, a CH₃^- shift. A primary C+ ion can undergo a shift so that it becomes a secondary or tertiary C+ ion.

An advise is to draw the structure in full with all the C and H atoms. For example, in the following carbo-cation rearrangement:

Do the following exercises.

I. Draw the structural formulas formulas for the products of the following reactions.

a. 3-Methylcyclohexene and hydrogen chloride.

b. Methylenecyclohexane and hydrogen bromide.

c. Propene and hydrogen chloride.

d. Styrene and hydrogen bromide. Styrene = phenylethene or ethenylbenzene.

Answers to I. a, b, c, d above

II.  a. When 3-methyl-1-butene reacts with hydrogen iodide a H^- shift occurs. Show the step in the reaction mechanism in which the shift occurs.

b. When 3,3-dimethylcyclohexene reacts with hydrogen chloride a shift occurs. Show the step in the reaction mechanism in which the improvement of the C⁺ ion is visible.

3/1/2007 Here is a the Worksheet on the Markovnikov Rule we went over on 2/23. It contains a link to the answers for the last questions.

EXAM 1.

(3/5/07 - 3/8/2007) Read Bailey Chapter 5

With the following reactions we will complete this chapter.

Oxidations of alkenes.

Oxidations of alkenes with the well known oxidator KMnO₄, potassium permanganate. In basic environment and at lower temperatures a glycol (= a diol, with two adjacent OH groups) will be formed. For example, to make antifreeze, that is ethyleneglycol, we can treat ethylene, another name for the compound ethene, with potassium permanganate. See the first example, hereunder. Cyclopentene in the second example, forms obviously a cyclic diol, with the two OH groups in the cis position.

With potassium permanganate in acidic environment the alkene will be cleaved and acids will be formed when the carbons of the C=C have an H atom attached. Recall, carbon dioxide is an acid, although a very weak one. Notice, that we write the environment above the arrow and we leave the non-organic byproducts MnO₂ and H₂O out of the equation. These are conventions you will get used to as you go through ochem. 

I. Practice the oxidations with potassium permanganate by doing the following problems:

Answers are shown here under in the blue table.

Click here for More on alkene reactions you are required to know.

(3/9/07)Addition Reactions of Alkynes.

Alkynes follow the rule of Markovnikov when reacted with HCl or HBr. For example 1-butyne with excess HBr will produce 2,2-dibromobutane. The rule of Markovnikov is applied twice.

All other additions are straightforward, in other words with excess hydrogen gas 1-butyne would produce a saturated 4-C atom compound butane. Compressed to a liquid butane, C₄H₁₀ is used in cigarette lighters. And 1-butyne with excess chlorine gas forms 1,1,2,2-tetrachlorobutane.

(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:

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 example 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