Organic chemistry

 

The carbon

The most important element of organic chemistry is the carbon, C. A compound containing no carbon, cannot be an organic compound.

Without carbon life would be unthinkable. There would be no plants, animals, humans or microorganisms such as protozoa, bacteria and viruses.

 

In nature we find carbon not only in rocks such as limestone, CaCO3, and dolomite but also as elementary, graphite and diamond.

CaCO3, limestone mountains in Kurfirsten, Toggenburg, St. Gallen, Switzerland

EXERCISE: When lime, CaCO3 is mixed with hydrochloric acid, HCl, smoking carbon dioxide, CO2, escapes and calcium chloride, CaCl2, a commonly used drying agent in the preparative chemistry remains. (Filler of the drying tube). What is the stoichiometry?

Answer:

CaCO3 + 2HCl ===> H2O + CO2 + CaCl2

 

Graphite and diamond: these two fundamental modifications are initially known.

 

Graphite (picture above): This consists of honeycomb layers of hexagons. There are dehydrated benzene rings. The carbon is connected in graphite with 3 other carbon atoms. The fourth bonding electron is delocalized on the relevant layer. The layers are merely held together by weak forces, known as London or Van der Waals forces, so they can be moved mutually. Graphite passes along the surfaces of the electric current and is industrially used as a lubricant.

Diamond unit cell. (picture above) This modification is covalently bonded to 4 other carbon atoms. The basic cell is composed of 3 chair-shaped 6-membered rings. Similar to cyclohexane see picture on the top. Diamond is the hardest element (hardness 10) however its place in chemistry is of little access. Diamond, since neither free, nor non-bonding electrons are present, the electrical current does not conduct.

 

A third basic modification was discovered in 1985 by Curl, Kroto and Smalley.

The fullerene

These are spherical clusters of pentagons and hexagons.

The fullerene must have 12 isolated pentagons, this requires a geometric mathematical rule by Euler. The number of hexagons can be arbitrary.

The smallest possible cluster consists of 12 pentagons and 20 hexagons (C60)

It is said that doctoral candidates in Tucson, Arizona, should have detected fullerenes in soot.

The carbon atoms are 3-cohesive. The remaining non-bonding electrons are partially delocalized.

Unlike the diamond modification fullerenes are soluble and readily available to the chemistry.

 

Other carbons that are present elemental in nature and beeing mined are hard coal (75%-90% C) and brown coal (< 75% C). These two raw coals are refined to coke.

Biological materials beeing heated in the absence of air, carbon like soot and coke remain.

 

Nomenclature

In some universities, the nomenclature is treated rather subliminal.

In vocational and high schools you may even get an examination paper about substance nomenclature. Therefore, a subject that should not be neglected.

Here only the most essential rules are presented. However, many compounds have trivial names whose can hardly be denied. Who asking for salt in a restaurant, nobody will speak of sodium chloride or for the salad not ethanoic acid but vinegar is required. When highly concentrated liquor is asked, nobody calls for the ethanol.

Also, the welder works with acetylene ( “not” with ethyne) H-C = C-H

 

The rules are based on IUPAC (International Union of Pure and Applied Chemistry)

The most essential rules:

1. Determine the alkane group (Alk-ane, Alk -ene, Alk-yne, Alcohol, carboxylic acid, aldehyde, ketone, amine, amide etc, see table above).
2. Find the longest continuous carbon chain and determine the organic parent molecule.
3. substituents, such as organic residues (see below), (methyl, ethyl, propyl, butyl phenyl etc) or inorganic elements or compounds such as halides, metals etc are listed alphabetically.
4. places of substituents have the smallest possible numbers

 

Examples: CH3-CH2-CH2-CH2Cl

Rule 1: alkane group is an alkane

Rule 2: chain consists of 4 saturated carbon atoms, organic parent molecule with 4C = butane

Rule 3: a substituent is Cl, = chlorobutane

Rule 4: 1-chlorobutane (but not 4-chlorobutane)

 

CH3-CH2 = CHBr-CH2Cl

Rule 1: = Alken

Rule 2: = butene

Rule 3: = bromine, chlorine (alphabetic order!)

Rule 4: = 3-bromo-1-chloro-1-butene, (but not 4-chloro-2-bromo-4-butene)

 

Same substituents or multiple bonds existing more times in a compound, are described with Greek numbers:

CH2 = CH-CH = CH-CH3

1,3-pentadiene

HCCl3 = trichloromethane

 

Organic substituents ending in -yl

CH3-,  methyl

CH3CH2-,  ethyl

CH2 = CH2-,  ethenyl

(CH3) 2CH-,  i-propyl, isopropyl,

CH3 (CH2) 3-,  butyl

(CH3) 2CH-CH2-, i-butyl

C6H5-,  phenyl

 

Other substituents

-NH2,  amino

-ROH,  alkoxy

-RCOOH,  carboxy

CH3O-,  methoxy

CH3CH2O-,  ethoxy

S-,  thio

-SCH3,  methylthio

 

Example more complicate:

 

Essential amino acid: methionine

 

CH3-S-CH2-CH2-CH-NH2-COOH

 

Rule 1: Alkane group is carboxylic acid or alkanoic acid

Rule 2: parent compound is butyric acid

Rule3: substituents are ordered alphabetically: NH2 = amino, SCH3 = methylthio

Rule 4: start counting from COOH = 1, amino = 2, SCH3 = 4

This compound is chiral and like all the other amino acids left-handed, (= S, as sinistra), The second C of the chain is the chirality (CHNH2), therefore 2S. More about stereochemistry in the next chapter.

Systematic IUPAC name:

(2S) -Amino-4- (methylthio) butanoic acid

methionine

One of the 8 essential amino acid, which must be taken, because human body doesn’t produce itself.

Nomenclature of benzenes

 

Nomenclature will be handled in other lessons like Isomerie and stereochemistry.

 

isomerism

2 compounds have the same sum formula but a different structure, they are called Isomers.

If the atoms of two compounds are differently linked with the same sum formula, they are constitution isomers.

The cumulative formula is C3H6. Possible constitutional isomers are cyclopropane and propene, CH2 == CH–CH3.In complex chemistry, coordination isomers include, inter alia, ionization isomers.

Organic molecules that undergo a change in their spatial arrangement with respect to twisting are called conformational isomers. More on this in the next sub-chapter.

If 2 compounds have the same sum formula and also the same link but a different spatial arrangement, one speaks of stereoisomers.

Complex chemistry also distinguishes between hydratisomerism, coordination isomerism and bond isomerism, in addition to stereochemistry.

Here, a little more will be given in regard to organic chemistry:

 

stereoisomers

There are two Kinds:

• Diastereomers
• Enantiomers

In the case of diastereomers, the difference is in symmetry. As explained above, both compounds again have the same sum formula, but both are symmetrical. However, the spatial arrangement differs.

Important, both compounds have different chemical and physical properties. Such compounds are named with prefix cis- and trans- or Z and E (if unsaturated, see one of next chapters)).

 

For enantiomers, both substances with the same sum formula have no symmetry, differing as the left and right hand or unprofessionally spelled, are mirror image isometric. In contrast to diastereomers, enantiomers have the same physical properties, with one exception. If one of the enantiomers with linearly polarized light is irradiated, the plane of the light beam is turned to the right or to the left. The nomenclature proposed by Cahn, Ingold, and Prelog was developed. (R) -, = clockwise, (S) counterclockwise, (R = rectus, S = sinister). In the biochemistry of the sugars, one also finds the older abbreviations D / L, dextro / laevo = right / left, but do not have the same meaning.

 

The asymmetric compound is referred to as optically active or chiral. A compound which is called a symmetry center or a plane of symmetry (also called a plane of mirror)

is achiral, if both are missing it is chiral. Moreover, symmetrical substances can be covered.

Here are some examples of everyday life: Real estate:

This country house also has a gazebo on the left side and the same entrances also the left roof side is provided with a roof window. The back side is also symmetrical even if the pattern is different. This farmhouse therefore has a symmetry plane (mirrored) along the middle of the roof and is thus achiral.

 

These two houses have neither a symmetry plane nor a symmetry center. They are therefore chiral. Both houses are mirrored. One can call them enantiomers.

 

The following are some compounds as molecular models: hydrogen = white, carbon = black, chlorine = green and oxygen = red

Diastereomers: cis-1,2-dichloroethene and trans-1,2-dichloroethene

Are these enantiomers?

No! They are identical, this is 1,3-dichloropropane

 

But, are these enantiomers?

 

correct! They can not be covered. (R)- and (S)-trans-1,2-dichlorocyclopropane

 

2 symmetry planes: dichloromethane achiral

 

1,4-cyclohexanediol, has a symmetry center, therefore achiral

 

How to find out whether the chiral compound is rotating clockwise or counterclockwise?

First the chirality center must be determined. In organic chemistry this is usually a carbon atom. An organic compound may have several centers of chirality, such as the sugars. Only compounds with 1 chirality center are treated here.

Thereafter, the 4 substituents bound to the C are ordered according to certain priority rules, according to Cahn, Ingold and Prelog. Only the two most important ones are mentioned here. We do this using the lactic acid, its systematic name is: 2-hydroxypropanoic acid:

1.Priority is given to the atom with the higher order number. O = 16 comes before N = 14. For isotopes the atom with the higher mass has priority.

2. Consider the atom of the first sphere, which is bound directly to the C Atom or to the atom of the center of chirality. If there are several similar atoms in the first sphere, the atoms of the second sphere are arranged correspondingly. If a substituent contains double or triple bonds, it will be treated as if 2 or 3 atoms were present. C==O becomes O–C–C and C==C becomes C–C–C.

Lactic acid: Priority: 1. OH,  2. COOH,  3. CH3,  4. H

 

 

 

We have now determined the priorities and can now determine whether the compound is turning clockwise or conterclockwise. If a molecular model is present, it must be set up in such a way that the atom with the last priority, here the atom, H, is furthest away from the observer. If the first three substituents 1>2>3 are arranged in a clockwise direction, then the compound is rotated clockwise and is designated by the prefix (R)-. Prefix (S)- is given to the left-turning enantiomer.

If no molecular models are available or can not be created, the Fischer projections are used. Draw a cross with the vertical marking the main Carbon chain. At the top is the substituent, in which the numbering begins according to systematic nomenclature. This would be the COOH group. As described above, it si also possible there to find out whether the compound ist left- or rightturning.

 

Mesomerism

In high and grammar schools the term mesomerism also emerges. Some molecules in organic chemistry cannot be described correctly. It is assumed that the negative charge or also the electron pairs in multiple bonds are distributed in several atoms. Example: benzene and 1,3-pentadiene.

 

But also in inorganic chemistry.

In the chapter, molecules, see above: Here the structural formulas of carbonic acid, H2CO3 and nitric acid, HNO3, are shown.

 

2 times deprotonated, the carbonate anion, CO3 (-II) is formulated in 3 different limiting formulas. The molecule is planar and the charge, -2, cannot be detected accurately. The charge is delocalized.

Also for the compound HNO3, there are two possible limiting formulas, because one does not know, the negative charge, -I, is in the upper or lower oxygen.

 

In organic chemistry it is teeming with mesomeric structures. In universities and in higher textbooks of the organic

Chemistry, the word mesomerism hardly occurs. The compound is said to be resonance-stabilized.

Attention!

Sum formula C5H8

CH2 == CH – CH2 – CH == CH2 and CH2 == CH – CH == CH – CH3 are neither mesomers, nor enantiomers nor diastereomers, but merely constitution isomers!

 

Hydrocarbons

Hydrocarbons are understood as compounds of carbon, C  and hydrogen, H.

Basically it is to compare between to kinds of hydrocarbons:

• Saturated    
• Unsaturated

hydrocarbons.

 

Saturated hydrocarbons contain only single bonds. For example, methane, propane or dimethylpropane (neopentane) but also cyclohexane (see picture at the very beginning above) or bicyclo (4,4,0) decane (decalin) and bicyclo (2,2,1) heptane (norbornane) are saturated hydrocarbons. They are called alkanes.

Unsaturated hydrocarbons have double or triple bonds.

However, the main difference between single and double or triple bond is the possibility of movement. The C-C bond can be twisted.

The spatial arrangement with respect to the different positions by twisting and branching is called conformation. Molecules with different conformations are conformational isomers.

In the case of ethane, the methyl groups, -CH3, rotate around the C-C bond at room temperature, from the staggered to the ecliptic conformation, and back to the staggered one. The latter is energetically favored, because in the ecliptic conformation, the two opposite H atoms are mutually opposing each other.

Ethane  staggered conformation

Ethane ecliptic conformation

In the case of cyclohexane, the molecule continually changes from the chair- to the tub conformation passing the twist conformation each time and back. However, the chair shape is energetically favored.

Cyclohexane tub- or bath-conformation

Cyclohexane twist-conformation

In case of ethene, CH2 == CH2, the double bond does not allow any rotation. The molecule is therefore planar, or lies on a plane. Likewise, benzene, which has 3 double bonds planar and, in contrast to cyclohexane, has neither a chair nor a bath form. This has, however, other reasons, more about that below.

Unsaturated hydrocarbons with one or more double bonds are called alkenes:

Examples

Propene, 2-methyl-1,3-butadiene, (isoprene = a terpene), a-pinene (main component of turpentine), b-carotene (provitamin A) .Limones, cyclooctatetraene, benzene, azulene (the latter 2 compounds, benzene and azulene, are aromatics)

Hydrocarbons with several double bonds are also called polyenes. Whether one, several or no single bonds are present between 2 double bonds, a distinction is made between conjugated, isolated and cumulated double bonds. Butadiene and benzene are conjugated hydrocarbons and are resonance-stabilized (or mesomer cf. isomerism)

 

Unsaturated compounds with one or more triple bonds are called alkynes.

Ethyne, (acetylene) is also excluded in 3-fold binding, 2-butyne, butadiyne, benz-yne (is a very unstable)

Benz-yne, dehydrobenzene, is an aromatic but very unstable compound (see next sub-chapter)

The H atoms bound directly to the C with triple bonds are weakly acidic and can be deprotonated with strong bases such as NaNH2.

 

Aromatic hydrocarbons and Huckel’s rule

The word “aromatics” refers to some aromatic substances smelling in plants. In fact, these are benzene compounds such as, for example, caffeine and vanillin.

Aromatic hydrocarbons are rings or multiple rings with conjugated double bonds. The most important representative is the benzene.

Aromatic is a compound when:

1. The structure represents a ring.
2. has conjugated double bonds throughout the ring. Singel bonds may not be consecutive.
3. it is planar (representing a Level)

If one of these properties is missing, the compound is not an aromatic.

The electrons of the double bonds must be uniformly distributed (delocalized) in order for the multiple corner to lie on one plane. Ist charge density must be perpendicular above or below the edge of the ring so that the corners can not be moved.

In addition to benzene, there are also aromatic rings or multi-core benzenes, such as naphthalene, anthracene and phenanthrene.

Rings with more or less than 6 carbon atoms with conjugated double bonds are also aromatic, if the above mentioned rules are fulfilled.

Examples:

Azulene

1,3,5,6,7,9,11,13,15,17-cyclooctadecanones

(both compounds are mentioned on picture above)

Now, the above three rules are only applicable to a limited extent.

Cyclobutadiene and 1,3,5,7-cyclooctatetraene are not aromatics, why not?

The latter compound has already been briefly treated above for the conjugated double bonds: 1,3,5,7-cyclooctatetraene is not planar, it describes a tub.

In the case of cyclobutadiene, all of the 3 above mentioned rules are fulfilled, but ist is not an aromatic. Cyclobutadiene is also extremely unstable and decomposes at -200°C.

 

When is the compound aromatic?

Huckel’s rule

A simple mathematical formula reveals whether the compound is aromatic or not.

A cyclic compound with conjugated double bonds has 4n + 2 double bond eletrons. Benzene has 3 conjugated double bonds, 6 additional electrons are necessary.

The equation is

4n + 2 = 6

We solve for n and get

N = 4/4 = 1, a natural figure without decimals or fractions. Benzene is therefore an aromatic.

The same solution also supplies benz-in. (See picture above) The electron pair of the third bond of the tripel bond may not be counted, but is responsible for the ring being distorted. The molecular orbital theory will be treated in higher lessons.

How does this look like at cyclobutadiene? 4 additional electrons are required for the conjugated double bonds.

The equation is:

4n + 2 = 4

Solution after n

n = 2/4 = 1/2 or 0,5. Now we have a fraction or a decimal place, so cyclobutadiene is not an aromatic.

 

 

This also applies to 1,3,5,7-cyclooctatetraene with 8 double bond electrons.

4n + 2 = 8, n = 6/4 = 1,5, no aromatic as mentioned above.

 

Azulene has a total of 10 double bond electrons on both rings together.

4n + 2 = 10, n = 8/4 = 2. Yes, it is an aromatic compound.

In addition to aromatic hydrocarbons, heterocycles can also be aromatic. Pyridine is a simple example. Note that the electron pair on the Nitrogen is in the ring plane and does not contribute to be an aromatic.

In contrast, the electron pair in the pyrrole is perpendicular to the ring pland and is considered as an electron pair which contributes to the aromatic character. The Huckel’s equation is the same as for benzene.

Non aromatic rings meeting rule 2 can be made to an aromatic with help of Lewis bases and Lewis acids. But the molecule is now no longer of neutral charge.

 

 

Reactions and reaction mechanisms of organic chemistry

In this sub-chapter, basic principles on the reactions in organic chemistry are presented.

Important to know first is that most reactions are equilibrium reactions and often with mesomeric (resonance-stabilized) transition states.

Basics of the reaction mechanism:

It describes:

1. All reaction equations and transition states leading to the product.
2. The arrow always indicates where the electron pair is going.

 

Organic chemistry is divided into four reaction types:

(According to Peter Sykes, Reaction Mechanisms of Organic Chemistry)

1. Substitutions
2. Additions
3. Elimination
4. Rearrangements or shifts

Here there will be  treated briefly:

• The Markovnikov addition
• The nucleophilic substitution and elimination
• The electrophilic substitution on the benzene ring
• The oxidation and reduction
• The addition / elimination mechanism
• The polymerisation and radical substitution

It is recommended to keep an overview on the table of organic compounds and the one of the electronegativities. It is important to know that oxygen, O2 is more electronegative than nitrogen, N2 and N2 is more electronegative than carbon, O <N <C.

The following is a basic statement on the carbenium ions: If a hydrocarbon is removed with an atom and its electronpair (eg a chloride anion, Cl-), a carbocation occurs. One distinguishes between

1. CH3 +, methyl cation
2. RCH2 +, primary carbocation
3. R2CH +, secondary carbocation
4. R3C +, tertiary carbocation

Important: Stability in carbocations increases in this order from 1 to 4.

The reason for this is that C-C bonds are more easily polarized than C-H bonds.

For the sake of completeness, it should be mentioned that, conversely, the stability of carbanions decreases from 1 to 4. H3C-> RH2C-> RHC-> R3C-.

R is any atom (other than H) or alkyl.

 

Markovnikov addition

When propene is reacted with hydrogen bromide, 90% of 2-bromopropane is obtained. The mechanism proceeds as follows:

The proton, which here performs the function as a Lewis acid or as an electron pair acceptor, is preferably fixed at the end of the double bond. Consequently the more stable secondary carbocation occurs, especially because the proton is less electronegative than the bromide ion.

CH3-CH == CH2 + HBr  →  (CH3-CH+ -CH3) + Br-

The secondary carbenium ion (secondary carbocation) is preferably formed because it is more stable.

The more electronegative bromide ion therefore occupies position 2 and thus predominantly 2-bromopropane is obtained.

Anti-Markovnikov addition. It is also possible to prepare predominantly 1-bromopropane. However, this reaction proceeds according to a radical mechanism. It will be treated in higher courses.

 

Nucleophilic substitution

In the case of nucleophilic substitution in organic chemistry, one bound atom or one group is replaced by another at the carbon atom. Nucleophil means nucleus-loving (nucleus). The atom or the group still attached to the C is called the leaving group. The new atom or the new group to be bound to the carbon  is called nucleophile. Depending on the stability of the carbocation, there are 2 boundaries of reactions:

Direct examples:

NaOH + CH3Cl ===> H-O —— CH3 —— Cl ===> H-O-CH3

When methanol is produced from chloromethane, an activated complex is formed as a transition state. In this process, the C atom is attacked from behind or from the side which has been washed off from the chlorid ion, by a hydroxide ion, which acts as a Lewis base here.

white = H, black = C, green = Cl, red = O

When the complex is activated, the three H atoms are inverted. Imagine an umbrella in a strong wind. The binding of the chloride ion to be emitted is weakened after the 3 H atoms have been folded over. This is the second order nucleophilic substitution, since the rate-determining step is the activated complex. The reaction is bimolecular because it is an unstable carbocation. The abbreviation is Sn2.

The reaction proceeds differently when instead of 3 H, 3 methylene groups are bonded to the C atom. In the case of 2-chloro-2-methylpropane or t-butyl chloride, the hydroxide ion encounters a branching of 3 methyl groups and is thereby prevented until the chloride ion is emitted.

This results in a relatively stable tertiary carbocation, which is then rapidly attacked by the hydroxide ion from behind or from the front. The rate-determining step is the leaving group, here the chloride ion. The reaction proceeds unimolecular, which is called a Sn1 reaction.

 

Elimination

We remain with the t-butyl chloride, of which the chlorine atom has just been removed. Now the hydroxide ion renounces the attack on the tertiary carbocation and for this a proton of a methyl group is removed.

Result: 2-methylpropene has been get. It is the E1 mechanism, which is also unimolecular. Very good yields are obtained when the same reaction is carried out in aqueous ethanol* or with another weak base.

In the case of the bimolecular elimination, E2, strong bases in excess can immediately remove a proton of a methyl group from the 2-chloro-2-methylpropane even before the chloride ion has set off.

 

However, since the binding of the chloride ion is thereby weakened, the reaction proceeds bimolecularly, concertedly or simultaneously. The rate-determining step is, as with the Sn2 reaction, the activated complex.

About the leaving group:

Good leaving groups are halides, such as chlorine, bromine and iodine.

But not fluorine, F!

*For chemical synthesis in the laboratory, do not use firing alcohol. This alcohol is mixed with some pyridine, so that it can not be drunk.

 

The electrophilic substitution on the benzene ring (first substitution)

 

In the nucleophilic substitution, we have learned that the attacking compound must be a nucloephile, and that this is a Lewis base, an electron pair donor. The point of attack is the nucleus or an atomic nucleus. It is usually a carbon atom in organic chemistry because it is the least electronegative. More about that in the addition / elimination mechanism below.

The opposite happens, when an H atom is replaced by a substituent. This is an electrophilic substitution. The attacking atom or compound is an electrophile and acts as a Lewis acid, (electron pair acceptor). The electrophile catches an electron pair from the benzene ring.

 

Intermezzo

Let’s first explain briefly: What is a catalyst? A catalyst is an atom or a compound that accelerates the reaction without being consumed. The reaction occurs with less reaction heat.

There are 2 types of catalysts:

1. Homogeneous Catalyst
2. Heterogeneous Catalyst

The main difference is that the homogeneous catalyst participates in the reaction process and also temporarily bonds. These are used in the electrophilic substitution on aromatics. AlCl3, FeBr3, H2SO4 are important homogeneous catalysts.

In the heterogeneous catalyst, there are often elementary transition metals with traces of other elements that absorb reactants. The reactants merely touch the surface of the catalyzing metal, but do not bind with it. Iron, Fe, platinum, Pt, nickel, Ni, and platinum/rhodium, Pt/Rh are heterogeneous catalysts. The latter, Pt/Rh, is a constituent of the exhaust gas catalytic converter which converts nitric oxide, NO, and carbon monoxide, CO, into non-toxic gases such as carbon dioxide, CO2, and nitrogen, N2:

NO + CO  →  N2 + CO2

Example, the hydrogenation (H-addition) of alkenes with transition metals as catalyst, e.g. Platinum, Pt, palladium, Pd or nickel, Ni. When the surface is contacted with hydrogen, H2, the bond of H2 dissolves. Both H atoms are bound briefly to the metal surface before they are added to the alkene. The alkene also comes into contact with the metal surface with one of its double bonds.

If you perform organic chemical reaction tests with metals in the laboratory, do not use metal powders but metal filings because of the larger surface area.

 In biological chemistry, enzymes and coenzymes (vitamins) play the role of catalysts.

 

Benzene addition with homogeneous catalyst, what happens exactly?

We produce toluene from benzene. For this purpose, we use chloromethane, CH3Cl and aluminum trichloride, AlCl3 as a catalyst (Friedel-Crafts alkylation). The main reaction is:

 

AlCl3 acts as a Lewis acid (electron pair acceptor) and removes CH3Cl from a chlorine Atom.

CH3Cl + AlCl3 → CH3+  +  AlCl4-

This is the catalyst reaction.

CH3+ is very short-lived and now forms as a strong electrophilic group a complex with benzene, the π-complex (EPD/EPA-complex). The primary carbocation, CH3+, is attached to a particular ring carbon atom. This produces a mesomeric benzenium ion.

 

 

Here CH3 and H remain bound before the H-Atom is separated from a chloride anion of AlCl4+. See the final reaction:

 

The catalyst AlCl3 is regenerated and hydrochloric acid, HCl, is formed.

The  catalyst reactions are similar to the other  catalysts, FeBr3 and H2SO4. The electrophil, which must be a cation, is short-lived and very reactive.

 

Bromine addition, production of bromobenzene:

Br2 + FeBr3 → Br+ + FeBr4-

Bromine is heterolytically cleaved to Br+ and Br-. Br+,  acts as an electrophilic, and the latter, Br- , acts on FeBr3 analogously to AlCl3.

 

Preparation of nitrobenzene with H2SO4 as a catalyst:

HNO3 + 2H2SO4 → NO2+ + H3O+ + 2HSO4-

 

NO2+ is the electrophile.

Preparation of benzenesulfonic acid, an essential compound in preparative chemistry.

SO3 + H2SO4 → SO3H+ + HSO4-

Sulfur trioxide is protonated, SO3H+, and made electrophilic.

 

Second substitution on the benzene ring:

 

An H-atom has just been replaced by an electrophilic substituent on the benzene ring.

A further substituent is now to be introduced. The procedure is similar to that of the first substitution. The only question is: Which C-atom of the ring is the second substituent? This depends solely on the extent to which the electron density changes over the benzene ring by the first substituent. The electron density is increased by substituents with free electron pairs

The first atom bound to the carbon of the ring is decisive. Electronegativity also plays an important role.

In general, substituents which have one or more free electron pairs increase the electron density on the benzene ring.

Example: Second substitution at methoxy-benzene, reaction with HNO3

The halides, F, Cl, Br.I contribute somewhat less to the electron density since they are very electronegative.

The second substituent is located at position 2, ortho position or position 4, para position.

 

Substituents which do not have a free electron pair and are bound with a more electronegative atom, reduce the electron density by pulling it off.

In nitrobenzene, for example, with NO2 as a first substituent, the nitrogen does not have a free electron pair and is bound with 2 O atoms, which are more electronegative than nitrogen. N = 3, O = 3.4. The N atom has a positive charge.

The reaction proceeds more slowly.

The electron density is decreased  because the atom directly attached to the benzene ring does not have a free electron pair and the other bound atoms are more electronegative. The following first substitutes force the second substitute to the third or fifth position, so called meta-position. Example: Second substitution of benzoic acid with bromine.

Instead of excess charges, electron gaps are formed at the same points. These positively charged sites force the second electrophil to take a different position.

The second substituent is thus determined at position 3 or 5, meta position. By recording all mesomeric boundary structures, the meta-position can be derived.

 

The secondary substitution of toluene and generally alkylbenzene is a special case. The second substituent assumes ortho or para position, although the C atom has no free electron pair.

With our knowledge so far, this can be demonstrated by simulating all the mesomeric structures:

As with the Markovnikov addition described above, a tertiary carbo-cation is formed as a transition state, which is known to be relatively stable. The nitro group assumes ortho or para position.

However, the nitro group is more electron-selective than the methyl group. Therefore, 2 additional NO2 groups are easily substituted at the aromatic ring. Finally, trinitrotoluene is obtained. This powder serves as an explosive in hand grenades. It is easily produced in the laboratory.

 

Oxidation and reduction in organic chemistry

It is recommended to repeat chapter 7: Oxidation and reduction

Oxidation is a reaction with oxygen. However this is rather an alchemical definition. We take an example:

We burn methane:

CH4 + O2 → CO2 + H2O. This is an exothermic reaction, -800 kJ / mol

We now determine the oxidation numbers for the carbon atom. Work with the scheme of the electronegativities above!

H with 2,2 is less electronegative than C with 2,6, and O with 3,4 is more electronegative than C.

The oxidation numbers in organic chemistry are determined by assigning the bonding electrons to the more electronegative atom.

 

The oxidation number results from the difference between total bonding electons

plus any nonbonding electron pairs minus valence electrons.

Methane:

C has 8 bonding electrons. The four valence electrons are deducted from C. 8 – 4 = -4. Each hydrogen Atom has the oxidation number +1. Since methane is charge-neutral, the sum all of the oxidation numbers of the entire compound must be zero. -IV + (4*+I) = 0, or -4 + 4 = 0

What happens exactly when methane is going to be burnt?

At the end of the reaction, you get water, H2O and CO2. For the latter compound the oxygen atoms have a higher electronegativity (=3.4), which is why they are assigned to the bonding electrons. The oxidation numbers for each O-atom number are 4 binding electrons plus two nonbonding electron pairs (=4) minus 6 valence electrons. Total 8 – 6 = -2. For the C-atom, the oxidation number is +4. Control:

2 * -2 + 4 = 0. CO2 is charge neutral like methane.  Carbon has thus emitted a total of 8 electrons to the two oxygen atoms, thus being oxidized by 8 electrons.

But what happens exactly with this combustion?

 

Methane is first converted to methanol, then methanal (formaldehyde) and almost lastly methanoic acid (formic acid) has been got. Finally, CO2 is cleaved from the smallest carboxylic acid, formic acid. This is called decarboxylation. The two remaining H-Atoms are briefly converted to hydrogen H2 and a reaction with O2 as a last step is formed. Boiling gas reaction.

Oxygen is present in excess, otherwise it would be a catastrophe. On Mars, you could neither breathe nor drive. (with the exception of electric vehicules and solar vehicles or bringing with a bottle of Oxygen-gas). With a pressure of 6 millibar CO2 making a fire is out of the question.

 

Oxidation in preparative chemistry starts with alcohol. An aldehyde, a ketone or a carboxylic acid is synthesized from an alcohol. In the example above, we have already obtained methanol by the Sn2 reaction. Methanol is now going to be oxidized.

• Attention! Methanol is very toxic! Do not drink, it can cause blindness!

A very essential oxidizing agent is dichromate: K2Cr2O7, Na2Cr2O7 (in H2SO4 = Jones reagent) and CrO3 are commercially available. When Methanol or any other alcohol react with one of These oxidizing agents, formic acid or any other carboxylic acid is obtained directly. If desired to stop the oxidation in aldehyde or ketone, CrO3 must be complexed with pyridine.

With help of CrO3(pyridine)2 practically all alcohols can be oxidized to aldehydes or ketones in order to prevent reaction to caboxylic acid.

It is more difficult to obtain carboxylic acids from ketones because a C-C bond must be broken.

Another important oxidizing agent in organic chemistry is potassium permanganate, KMnO4, a wine-red solution.

 

Oxidizing agents in medicine:

As already mentioned in chapter 7, the blood sugar in the urine can be detected with silver nitrate, AgNO2 (Tollens Detection). Glucose is an Aldehyde, (an aldose) and can be easily converted into carboxylic acid. Silver is reduced and elementary silver separates out and adheres to the wall of the vessel. A silver mirror is formed.

Another reagent often used in medicine is the Fehling proof. This mixture consists of copper (II) hydroxyde, Cu(OH)2 and together with basic tartrate solution (alkaline salt of tartaric acid) forms a light blue complex compound. For sugar-positive values, copper (II) is reduced to copper (I), (it takes an electron). CuO2 is formed and the Color changes from light to yellow red.

 

 

Reductions in Organic Chemistry

 

In contrast to oxidations, carboxylic acids, aldehydes and ketones can be reduced to alcohols.

Important to know: The carbonyl group is polarized. This can also be explained by oxidation numbers. The total of 8 bonding and nonbonding electrons are assigned to the oxygen atom. This can also be seen in the resonance structure.

The C-Atom will only be attacked by nucleophiles. Thus. only electrophiles can be attached to the O-atom. This is usually a proton. Read more below.

There are two essential reducing agents:

• Sodium borohydride, (sodium tetrahydridoboranate), NaBH4
• Lithium aluminium hydride, (lithium tetrahydroalanate), LiAlH4

In these two frequently used reducing agents, a hydride ion is released.

Note: the hydride ion is not a proton! It is a hydrogen atom with an electron pair, it is an anion!

When both  compounds are dissolved in water, hydrogen, H2, an hydroxide are formed. The hydride ion, H–, of the reducing agent combines with a proton of the water.

NaBH4 + 4H-O-H → NaOH + B(OH)3 + 4H-H

LiAlH4 + 4 H-O-H → LiAl(OH)4 + 4H-H

LiAlH4 is more reactive than NaBH4. The synthesis with LiAlH4 should be carried out as anhydrous in diethyl ether,

CH3CH2-O-CH2CH3. NaBH4 can be dissolved in aqueous ethanol.

Care should be taken when working with these two reducing agents. Synthesis may only be carried out in a stateapprouved laboratory. The compound to be reduced in ether must first be cooled with dry ice (frozen CO2), ice, or liquid nitrogen so that the reaction does not precipitate too violently. The reducing agent must be attached dropwise to the reaction mixture in the chapel with a syringe. The chute sliding window of the chapel must be sunk so that only your hands reach the inside of the chapel. Hands are to be protected with rubber gloves.

What is the reaction mechanism?

Ethanol is reduced from acetaldehyde with NaBH4 in ethanol.

NaBH4 + CH3CHO → CH3CH2OH + Na+H3BOCH2CH3

 

Three further ethanal molecules can now be attacked. When all hydride anions are consumed, Natrium-tetraethoxyborate, (Na+B(OCH2CH3)4-) remains back.

With LiAlH4, a hydride ion also attacks the carbon atom of the carbonyl group. The oxygen atom of the carbonyl group is fixed at the position of the aluminum where a hydride ion has just released. This process is repeated until all 4 hydride ions on the aluminum are replaced by the carbonyl groups of the of the acetaldehyde molecules.

By reaction with water, (hydrolysis), 4 molecules of ethanol as well as lithium- and aluminum hydroxide are formed. (See figure above).

LiAlH4 is very reactive. Carboxylic acids can be directly reduced to alcohols.

 

Ethanol, CH3COH, can be prepared from acetic acid, CH3COOH. However, as experience shows, the reaction can come to a standstill. Lithiumethoxylate, CH3CO2-Li+ is often precipitated.

To prevent this, an ester is prepared by an addition/elimination mechanism. More about that you’ll learn in the next subchapter.

 

Wolff Kishner reduction (A name reaction)

Ketones are reduced to alkanes. The carbonyl group is eliminated.

Reduction of cyclohexanone to cyclohexane:

Hydrazine, N2H4, (85% dissolved in water) and sodium hydroxide solution, NaOH are required.

The cyclohexanone is disolved in this alcohol Ether, OHCH2CH2OCH2CH2OH, (boiling at 245°). The hydrazine solution is added and the whole mixture is heated with NaOH. The mixture is then treated with water and cyclohexane is obtained. (The carbonyl group ist eliminated.)

 

The addition / elimination mechanism

The proton, which is bound in the case of an aldehyde, for example, in the acetaldehyde at the carbon atom of the carbonyl group, can hardly be separated or replaced. One can also say hydrogen is a miserable leaving group. Only the oxidation makes the aldehyde to a carboxylic acid. Acetic acid is produced from acetaldehyde.

CH3CHO → CH3COOH

The reagents used are the oxidizing agents K2Cr2O7, Na2Cr2O7, CrO3, which has just been treated in the last sub-chapter.

Instead of the H-atom we now have a hydroxide group, a relatively good leaving group, which is easier to substitute.

This is done through the addition / elimination mechanism. In general, this can be represented as follows:

 

L is the leaving group, X is the nucleophile.

With the addition / elimination mechanism numerous syntheses in organic chemistry can be derived:

 

• The synthesis of acid chloride: CH3COOH → CH3COCl with thionyl chloride, SOCl2 or phosphorus pentachloride, PCl5.
• The synthesis of an amide, RCONH2 from acid chloride, CH3COCl.
• The ester synthesis e.g. Ethanol and acetic acid react with an acid as catalyst to ethyl acetate or nail varnish. See figure below for mechanism.
• Ester hydrolysis. (Ester saponification) from ethyl acetate, ethanol and acetic acid are again formed as a reaction. The catalyst used is a base.
• The Claisen condensation. Acetoacetate is formed from ethyl acetate.
• The aldol condensation or aldol addition. This is merely an addition reaction of 2 aldehydes. There is no elimination. However, the mechanism is similar. See drawing below

 

Synthesis of ethyl acetate acid catalyzed

Simplified mechanism.

The attack of ethanol without acid is less reactive. The proton is electrophilic and is preferably attached to the carbonyl oxygen. Thanks to this proton, the OH group, the leaving group, can be removed more easily, which takes an additional proton from the ethanol: one speaks of water as the leaving group. Finally, the catalysis proton of the original carbonyl group also decreases.

Ethyl acetate is a component of nail polish.

 

 

The acid-catalyzed ester hydrolysis or the back-reaction of the ester synthesis proceeds according to the same scheme. At the end, you get back acetic acid and ethanol.

These are typical equilibrium reactions. To shift the equilibrium further to the right it is advisable to use excess ethanol in the esterification and excess water in the case of ester hydrolysis. Only pure ethanol should be used (no fuel spray)

 

 

Ester hydrolysis, (saponification). From ethyl acetate base catalyzed e.g. with NaOH, you get  carboxylate (the deprotonated acetic acid) and ethanol. Here, the hydroxide anion as nucleophile attacks the carbonyl carbon.

Saponification is called ester hydrolysis, because soaps are produced during the hydrolysis of fats (= glycerol esters of higher carboxylic acids). Sodium salts of the fatty acids are hard soaps (core soaps), potassium salts are lubricating soaps.

 

 

 

The Aldol condensation

Synthesis of 3-hydroxybutanal

The starting compound is again acetaldehyde. 2 molecules react. An aldol is an aldehyde which is also an alcohol. Note again: the hydrogen atom at the C of the carbonyl group can only be replaced by oxidation. The aldol condensation is therefore only an addition reaction. This is mentioned here, because this mechanism also runs similarly and gives a small pre-taste to the next sub-chapter, polymerisation.

First, it should be noted that the methyl group, that a carbonyl carbon has as its neighbor, is relatively acidic. There, a proton can be removed with a base, with NaOH for example. This results in a mesomeric transition state and the acetaldehyde becomes a nucleophile. As in the addition / elimination mechanism, it can attack the carbon atom of the carbonyl group of the second acetaldehyde molecule. The mechanism is the same, except that the electron pair, which is pushed up to the oxygen atom, no longer descends because of this fixed H atom on the second acetaldehyde, which is why elimination is omitted. At this point, the removed methyl proton of the first acetaldehyde is returned. The hydroxide anion, OH–, the homogeneous catalyst as a base, is re-formed.

 

polymerization and radical substitution

Let’s come back to acetaldehyde. 2 molecules have just been allowed to react. This resulted in an aldol (see illustration above). Acetaldehyde is referred to as a monomer. If both molecules react with each other, a dimer is formed. Now the aldol can be reacted with a third, fourth, and other molecule, acetaldehyde, and then we have a polymer.

 

A1 + A2 + A3 + … Ax = A-A-A-A ……

 

The polymers include plastics such as plastic and rubber and natural materials such as starch, cellulose and proteins. The latter can consist of 20 different amino acids.

Production of macromolecules or polymers is carried out by the following mechanisms.

 

• Polyaddition

• Polycondensation

• Cationic and anionic polymerization

• Radical polymerization

 

The aldol condensation just mentioned is rather a polyaddition. A more complicated example is polyurethane, from 1,6-hexane diisocyanate and ethylene glycol.

In the polycondensation, one equivalent of water, H 2 O, is cleaved per linkage, and is achieved by the addition / elimination mechanism described above. Products can be:

• Nylon 66, a polyamide, from hexanedioic acid and 1,6-diaminohexane.

• Polyethylene terephthalate, PET, A polyester, of terephthalic acid and ethylene glycol

PET is part of the PET drink bottles. One can hardly be without it.

Caution: Petrol can only be kept in PET bottles for a short period of time at most 1 to 2 months. The O-Atoms form hydrogen bonds after five to six months with the hydrogen atoms of the antiknock petrol (branched heptanes and octanes), and some of the polymer also dissolves in solution. The gasoline get somewhat connected to PET. The gasoline loses its physical properties such as volatility, and can no longer be ignited. The result is a rather odorless and colorless liquid with an increased surface tension. (beautiful round drops are formed)

In the case of ionic polymerization, Lewis acids (cationic polymerization) and Lewis bases (anionic polymerization) are used. The polymerization of alkenes is particularly important. With a Lewis acid e.g. BF3 polypropylene are made from propylene. Condition is a double bond. Alkanes can not be polymerized. Lewis acid is indeed an electron pair acceptor as we know. Thus BF3 can fill its electron gap with an electron pair of the double bond of an alkene. The alkene mutates into a carbenium cation. We also know that the positive charge on the 2nd carbon atom of the propene must be present, because it is a secondary carbo-cation and thus more stable. At this point, another propene molecule can be applied. The carbon cation continues to propagate until the chain breaks.

Carbonyl compounds can be polymerized with the anionic polymerization. Polyethers are formed. An anion such as e.g. methoxide anion, CH3O- can be put in as starter substance.

 

Ethene can be synthesized more readily by radical polymerization because primary carbenium cations are more unstable.

What is a radical?

It is an atom or a compound with an unpaired nonbonding electron and is marked with a dot. A•

Chlorine, Cl2 (or bromine, Br2) is treated with light or UV to give 2Cl. split. We now have two chlorine radicals. The chlorine radical can replace a H-atom with alkanes. This is called a radical substitution:

The overall reaction

CH3 – CH3 + Cl2  →  CH3 – CHCl + HCl

 

The mechanism looks as follows:

Chain start:

From chlorine gas two radicals  arise:

Cl2 → 2Cl•

Propagation:

Further radicals are formed. An ethane radical is formed from ethane. This in turn reacts with chlorine. Chlorethane and a chlorine radical are formed

Cl• + CH3CH3 → CH3CH2• + HCl

CH3CH2• + Cl2 → CH3CH2Cl + Cl•

 Chain termination:

radicals are reconnected:

CH3CH2• + Cl• → CH3CH2Cl

Cl• + Cl• → Cl-Cl

In organic chemistry, alkanes can be converted directly into halogen-alkanes. In the Research laboratory, this is done with halogen-butanimide, (-succinimide). An important application is the anti-Markownikoff-addition. With N-bromo-butanimide, (N-bromosuccinimide, NBS) and hydrogen Bromide, HBr, Br2 is formed. Br2 is cleaved to bromine radicals, Br•, by light or UV.

 

Preparation of polyethylene:

The radical polymerization proceeds similarly:

 An ethenyl radical is formed from ethene as a monomer by means of a radical (radical initiator), which can add further ethene molecules. This occurs until the chain breaks, or until no ethene molecules are present.

For the sake of clarity, it is assumed that the polymer consists only of 2 units, a dimer.

For this purpose, the radicals used are, inter alia, peroxides, ROO., or azo compounds. Radicals should only be used sparingly.

 

Chain Start:

Ra2 → 2Ra•

Ra• + CH2 == CH2 →  RaCH2-CH•

Propagation:

Ra – CH2-CH• + CH2 == CH2 → Ra – CH2-CH2-CH2-CH•

Chain termination:

Ra-CH2-CH2-CH2 – CH• + Ra•  →  Ra-CH2-CH2-CH2-CH2-Ra

Ra• + Ra•  →  Ra2 *

* No radical should be present at the end of the polymerization. Since only a dimer is formed here, it is assumed that some radicals are still present.

 Radicals cannot be described as catalysts. Radicals are contained in the final product.

 Chain start and chain growth progresses relatively well because monomers can also absorb modified oxygen-radicals from the air.

 

How do we design a polymer?

Ethen CH2 == CH2

[-CH2-CH2-] x        polyethylene

 

Other polymers: 

• Chloroethene CH2 == CHCl,  [-CH2-CHCl-]x,  polyvinyl chloride (PVC)
• Fluoroethene CF2 == CF2,  [-CF2-CF2-]x,  polytetrafluoroethylene, (Teflon, pan coating)

 

Unsaturated polymers, manufacture of latex and rubber.

From dienes and alkynes, polymers are obtained which still contain double bonds.

In nature, latex is made of polyisoprene. Synthetic latex is produced by polymerization of butadiene, CH2 == CH-CH == CH2. The still remaining double bond of this polymer, [-CH2-CH == CH-CH2-] can be further crosslinked. By reacting with sulfur, S, technical term “vulcanization” of this latex, rubber is formed by crosslinking the chains.

 

Polybutylene, Latex (picture above)

Rubber (picture above)

 

Elemental sulfur is built up from S8 molecules. The structure is an 8-ring with 2 dimples.

In chemical reactions only the symbol S is written for the sake of simplicity.

Sulfur bridges also play an important role in biochemistry in the formation of proteins and polypeptides.

Sulfur is found in nature as the sulphides of transition metals: FeS, pyrite, PbS, galena, ZnS, sphalerite, and as alkaline earth sulphates: CaSO4, anhydrite, CaSO4*H2O, plaster, MgSO4, magnesium sulphate, and petroleum. With sulfur, benzene is obtained from cyclohexane, which is why quite a few benzene is found in petroleum.

Cyclohexan + S → Benzol + H2S

In the vicinity of active volcanoes, hydrogen sulfide, H2S escapes. (Smell of rotten eggs). There, elemental sulfur can also be released. Large stocks of elemental sulfur are found in the USA (Texas and Louisiana)

If sulfur is burned, SO2, sulfur dioxide, is formed, dissolved in water, you get the sulfurous acid, H2SO3. SO2 also produces sulfur trioxide, SO3 and is of great importance in organic preparative chemistry.

 

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