CARBON CHEMISTRY

REACTIONS IN ORGANIC CHEMISTRY

Introduction

Attention:
Before starting with the reactions in carbon chemistry, you better look back in module 04:
the Nomenclature of the carbon chemistry.
Module 4 is necessary to understand this module 11
After that, you can involve into the reactions of all those carbon containing molecules, carbonchemistry.
There are so many of them, that we must divide them as well as possible.

Content

1. Reactions in the carbon chemistry

1.1 Appearance and structure of organic substances

1.2 The C-atom and the organic reactions

1.3. reaction mechanism

2. Addition & eliminatione

2.1 What is addition?

2.2 Elimination

2.3 Fat/oil hardening; butter and margarine

2.4 Poly-addition

3. Substitution

4. Hydrolysis & Condensation

4.1 Condensation

4.2 Polycondensation

4.3 Saponification

5. Redox reactions in the carbon chemistry

5.1 Introduction

5.2 oxydation of the different organic groups

5.3 De oxydators Cu²⁺ and Ag⁺

6. Acid base reactions in carbonchemistry

6.1 Introduction

6.2 Phenol as an acid

6.3 Ampholyte and amino acid

1. Reactions in Carbon Chemistry

There are a number of characteristic differences between organic and inorganic reactions:

in general, organic reactions are slower.

They are slow. The reactivity of many oganic substances is less than the reactivity of many inorganic substances.

Many times they do not occur in watery environment.

Often they produce not only the desired end product, but also side products (polluting the main product)

connected to those last two remarks, we may say that organic reaction mostly have no high rendement, yield.

Question 1
Why do you think that organic reaction often do not occur in watery environment?

Often you must purify the final product, immediately after the organic reaction and the production of the final product. For that they can apply the fractional distillation.

Question 2
Imagine a product mix of 1-hexanol en 1-oktanol.
Explain why and how this mixture can be separated inte the components by fraction distillation.

The most important prime substances for the chemical industry of organic products are Coal, Crude Oil and Natural Gas.

Question 3
Explain, with your knowledge of geometry, why non of these substances contain the element Oxygen.

The most important components of crude oil are carbonhydrogens with relatively long chains (the quality of the cruede oil depends on that).
The various carbonhydrogens can be separated by means of distillation, but there is a problem:
Most molecules are too big and therefore a big part remains behind as tar and afphalt; and we get too few applicable products.
Look at the traffic: you need a lot of asphalt, but much more petrol and diesel (having much smaller molecules) to drive on that asphalt.
To improve the amount of smaller molecules in crude oil, there is a method of 'cracking':
Strong heating of the crude oil (please no oxygen present!) can break those big molecules in smaller ones.

We are talking about main processes (cracking and then distillation) of the oil raffinaries.
Specially the cracking process needs very good catalysts.

Question 4
What is meant with: "thermo catalysis"?

Goto answer 11-04

The products of the oil raffinary, those with the smaller molecules, in there turn, are raw materials for the chemical industry and often built up of the elements Hydrogen and Carbon: the Carbonhydrogens.
You might name them: carbonhydrides.
They are the raw materials vor many derivates of the chemical industry, applied for society, as: petrol, plastic, nylon, etc.

The alcanes have no big reactivity. Yes, they burn very good, undergo easily substitution reactions. Being saturated substances, they do not undergo addition reactions.
Addition is possible with alkenes and alkynes. Double and tiple bonds between carbon atoms are broken.
A very special carbonhydrogen is Benzene and its derivates.
Later in this course, we will talk about the very low reactivity of benzene.

1.1 Appearance and structure of organic compounds

[first have a look at module 4, in particular the story about the Carbon atom]

About the appearance of organic substances we may say that this is connected with:

the number of C-atoms per molecule
the number of unsaturated bondings
the character of substituents / special or functional groups

molecules with long chains are heavier and will tend to be solids
molecules with shorter chains tend to be liquid or gas.
the organic substances can contain more or less unsaturated bondings (double or triple) and the general rule is: the more saturated, the more solid the substance

in other words:
Unsaturated compounds will tend to be oily, where saturated substances are fattyer, more solid.
The molecules become less rigid when the number of double or triple bonds increases, and with that the relative number of H-atoms decreases.
Fat is more saturated than oil.
But at the same time you may not forget the possible polar character of a substance, which also has an important influence: the more polar, the less volatile.

Question 5
Ethane is a gas, but as soon as you substitute one H-atom with an OH group, (than you have ethanol = alcohol), then the substance becomes liquid.
Explain this phenomena.

More info about this topic you can find in module 4

att.: the character of a molecule determines also how such a molecule participates in any reaction.

1.2. The Carbon atom is special

Consider the next remarks:

N.B. about option 1:
In general the C-atom makes four sigma (σ)bonds with four other atoms, including other C-atoms.
Such a regular C-atom can be considered as an atom with four equal (sp³) bonds of the type σ.
This atom has four of these (single) bonds: overlap with other orbitals, i.e. with those of Hydrogen atoms.

N.B about substitution:
Substitution processes do not break this kind of bondings. They remain just sp³ with electrons of the type σ.

N.B. about double and triple bonds (options 2 and 3):
At Carbon atoms with sp² or sp hibridization, we have unsaturated bonds of the type σ and π. The π-bonds can suffer addition. During this addition process the number of electrons of the type π will diminish in favor of electrons of the type σ.

Question 6
Find in the chemistry books what they tell you about 'chemical bonding' between C-atoms.

1.3. reaction mechanism

A chemical reaction always has to do with: 'redistribution of valency electrons, at the moment that certain particles with enough energy have contact with one another.
In such a process, chemical bonds disappear and new are made. We know this from module 7.

The reaction is the result of a complex process, with various steps.
There always will be a first step (initiation), the next steps (continuation) and something like a termination, the final steps that stop the reaction.
All these steps toghether are the reaction mechanism.

In such a mechanism, there is always something like an 'attack' of certain particles at other particles. The attacking particles normally have sufficient energy for that (otherwise they would fail). They must be energy rich.

Besides that, catalysts have an important influence on the mechanism. They participate, yes, but at the end they remain unchanged.
In biochemistry we talk about biocatalysts or enzyms.

Catalysts can stimulate a process, or better: certain steps in a process.

In stead of starting a chemical process with a catalyst, it can also be done with light radiation, that makes a bonding unstable and thus a particle more reactive.
Active particles are made (with sufficient energy) that can continue the reaction and eventually also can stop the reaction.

In Carbon Chemistry we know two types of reaction mechanisms:

An attack is possible of a positive at a negative particle (electrophylic attack) or the opposite: an attack of a negative particle at a positive particle (nucleophylic attack).
The other, non polar, mechanism is that of 'radicals'. Uncharged particles can be very reactive because of unpaired electrons; charges do not play any role here.

2. Addition & Elimination

2.1 What is addition?

A carbon chain can have double or triple bonds between the C-atoms. In that case the substance is called: 'unsaturated'. This means, the double bonds can open and absorp C-atoms at both open ends.
Such a bonding opens (in one way or another) and the C-atoms at both sides have an open spot, an unpaired electron Another suitable atom can come and fill that open place.
Attention: we do not talk here about substitution where an atom is substituted, but about addition where a new chemical bond is made with such a C-atom.

Imagine: two people love each other and hold both hands. Then they lose one hand and connect to the hand of someone els.

Question 7
If you have models, try to demonstrate this type of reaction wiht those models, for example: ethene with chlorine.
And investigate also if the bond between the C-atoms remain their rotation flexibility.
If you have no models, then try nevertheless to respond the second question.

So addition is the reaction between an unsaturated substance that saturates itself with new connected atoms or atom groups.

We can approach this phenomena addition with the so called π- and σ-bonds.
These bonds exist in one molecule if the carbon chain has double or tripel bonds.
We can say: there is an overlap of hybride-orbitals of the sp² and/or sp type. In both bonding types σ (linear overlap) and π (parallel overlap). Only the π-bonds participate in the addition.

Example of a possible mechanism for addition: Bromium and ethene
Ethene has a double bond, so a place with a higher electron density (four electrons) and thus with a higher concentration of negative charge.
That negative charge is causes particularly by the π-electrons.
If a Bromine molecule approaches this bond, kind of polar induction is created in the Bromine molecule and something like an 'addition bridge' is formed.
Another Bromine atom kan now approache this bridge from the other side, as we see in de drawing.

Question 8
What can we observe during that reaction with Bromine water?

Goto answer 11-08

Alkenes can suffer addition with various substances, like:
Bromine and the other halogenes; Hydrogen; Water; and more.
Att: mostly a catalyst is needed for a good addition reaction.

Question 9
Describe all structures involved in the addition of propene with:

Water
Hydrogen
Acetylene gas

Question 10
Calciumcarbide (C₂H₂) is a white and solid substance with a special smell, very unstable, that spontaneously reacts with water.
The products of this reaction with water are: a gas with a sharp smell and a basic solution. If the gas is guided through Bromine water (a diluted solution of Bromine in water), the yellow colour slowly disappears.
Will there be a redox reaction in this process? Explain.
Goto answer 11-10

The ethene gas (acetylene gas) that is produced in this reaction, is a very energy rich substance. In the old days, it was applied in lamps/ torches. Today we use it in particular in welding.

Question 11
Ethyne is very open for addition; in fact it can suffer two times addition.
Explain this statement with structures and equations. If possible, demonstrate this reaction with models.

2.2 Elimination

Elimination is exactly the opposit of addition: a molecule splits into two new molecules, while some of those new molecules become insaturated, i.e get double of triple bonds.

Question 12
Propanol can be the prime material for propane in an elimination reaction.

Give this reaction in structures
what kind of substances can be helpful in this reaction?

Goto answer 11-12

2.3 Fat hardening; butter and margarine

Butter is a natural product of animal milk. Here we meet an animal (non vegetable) fat.
If we separate water and the proteins from the milk, we remain the animal fats: mostly a triester of glycerol and saturated fatty acids.
Butter is also such an ester of glycerol, but not exactly with a fatty acid, instead with butanoic acid. That acid, a liquid, can be set free and you will not like the smell. Keep butter in the freezer please.
A bit of this acid (butter acid) occurs in human sweat.

Milk is also the prime substance to make margarine. But in this production, you need also vegetable oil (i.e. palmoil).
A problem is that margarine is then a mix of oil(L) and milk(l), two liquids, one polar and the other non polar. In fact, this results in two problems:

Polar and non polar do not mix enough, and
Did you ever try to spread a liquid margarine on you slice of bread?

So what now?

In the old days, in the old margarine factories, they resolved the second problem by using less oil and more milk, trying to apply the so called fat hardening process.
Vegetarian oil contains a lot of unsaturated bondings and that's why they are liquid. So they simply reduced the number of double bonds by adding (addition of) Hydrogen. This made the oil more saturated, and consequently harder.
This was only possible with a catalys and under pressure. The Hydrogen connects at the opened positions in the C-chain.
Those days, they did not know the value of unsaturated fatty acids.
Nowadays we all know dat saturated fatty acids can harm you veins, support high blood pressure or even hart attacks. So they wanted to eliminate that hardenings process.
And how to resolve then the problem of margarine being a liquid?

Yes, it is possible to mix in a special way a polar and a non polar substance: stirr well and shake until alle components get the form of very little spherical droplets, all nicely mixed together. We call that an 'emulsion'.
Such en emulsion is more or less solid.
Unfortunately not stable. After some time it will return to the two liquids, one upon the other.
This can be resolved by using emulgators: substances that stabelize such an emulsion. This way the emulsion 'margarine' can be put into packets, in cups or whatsoever. It stays stable for many weeks or even months.

2.4 Poly-addition

First a summary about addition:
Addition is possible at insaturated bondings, i.e. at double or triple bonds in the carbon chain. A normal addition means that such a double bond opens, and atoms or atom groups can connect at the opened positions.
But: there exists the option that at the opened positions will connect exactly the same kind of unsaturated molecule.
In fact, they add to each other, they suffer addition of their own kind. Mind that here always a catalyst is needed.
This way, plastics can be formed, and that normally those plastics do no longer possess double bonds (these are consumed in the poly-addition).
The name of the monomere remains, but you just put the prefix: 'poly'.

All toghether a complex of reactions, according to a special reaction mechanism:

In one of another way, probably under influence of light (but it could be a catalyst), attack takes place at the double bond. That double bond is opened and connect itself on one side to the catalyst.
The other open side still remains open, and is ready to continue the attack on another molecule of the same type:

This part of the process is called: initiation:
Cl – Cl 2Cl· (reactive radical)
During the main reaction of the prorrogation / continuation, the chloring attacks the double bond:

Cl· + CH₂ = CH – CH₃ CH₂Cl – CH· – CH₃

This radical attacks the molecules that are present in good amounts: the other molecules of propene:
1. C₃H₆Cl· + C₃H₆ C₆H₁₂·
2. C₆H₁₂· + C₃H₆ C₉H₁₈·
3. etcetera
  This is a chain reaction causing enormous chains. Every time a bigger radical is made that every time again can catch a monomere. In this way macromolecules are formed.
The main reaction can be stopped if a radical (the unpaired electron) meets another radical, also with an unpaired electron
Those two electrons immediately form a normal covalent bond; the action is over and the polymerisation comes to an end.
There are different ways to close such a chain reaction:

Termination: C_nH_2n· + Cl· C_nH_2nCl

or:

C_nH_2n · + C_nH_2n· C_2nH_4n

N.B.
The example uses 'radicals': neutral particles with an unpaired electron.
Such a structure is rather unstable, and such an unpaired electron we indicate with a dot.

In general, the products of poly addition are: plastics in many variations, depending on the choice of the monomeres.
You can limit the reaction to one kind of monomeres, but also apply different monomeres in one process. Then the so called co-polymeres are produced.
An example is ‘TEFLON’ or TEFAL = poly-tetra fluor ethene, that is applied in the industry of cooking pans.
This is a badly absorbing polymere. In practice, in such a pan will food not easily burn.

Question 13
Give the structure of the 'monomere' of TEFLON (=ptfe).

Another example is PVC (PolyVinylChloride) = polychloroethaan. This is applied in huge amounts, for example in plumbing, water tubes, isolation materials, etc.
In another module we come back on this item.

Question 14
Give the formation reaction of PVC, starting with its monomere.

The technique of polymerisation

In factories, polymeres are mostly made in grain form. These grains then are treated in the more specialised industries to produce final products.
Often the grains are heated, and the substance becomes then softer to make it suitable for moulding.
Here we distinguish two different processes:

The (macro)molecules do not make in between bondings, no side chains appear. No threedimensional network is formed.
The product remains a certain flexibility. The material can be recycled: you kan grain it again, heat it up and again put into the moulds.
The molecules form a threedimensional network, making intermolecular bonds. For example if extra double bonds are/were available. These products will be hard, with no flexibility. Recycling is more difficult in this case.

Alkadienes and alkatrienes can also suffer polymerisation with the result: a substance with macromolecules that still remain unsaturated.
Such an unsaturated polymere still has properties of unsaturated substances, or: a certain flexibility (not liquid of course, the molecules are too big). You get something like rubber.
We can reduce the elasticity of such a substance by (partial) addition. This is applied, for example, in the vulcanisation process of car tires. Those tires would be too elastic and unfit to serve any car.
But at vulcanising (part of) the double bonds in the rubber are added, if possible with Oxygen atoms, but often and more effective with sulphur atoms. The material remains enough elasticity, but becomes harder.

Another special aspect at the formation of polymeres is the mixing of different monomeres, creating 'co-polymeres'. The molecules connect alternately.

Question 15
Give a structure of the co-polymere formed out of the monomeres ethene and propene.

Macromolecules of polymeres are so big, that the substance alway will be a solid. They might sometimes dissolve more or less in water, like proteins or polysaccharides. Then you will observe a certain cloudy mixture, not transparant, just because these molecules are so big.
The macromolecules are dissolved, but so big that you can 'see' them.

There is a special 'polymere chemistry' where the capacity exists to produce extremely specialised polymeres with very special applications.

The two most important types of polymerisation are the polyaddition and the polycondensation.

Benzene does not contain real double bonds. The 6 C-atoms of the ring are connected in another way (see module 4).
Addition of benzene is therefor not easy. The six bondings are very stable and do not allow being opened for an addition process.

3. Substitution

An H-atom at a carbon chain, or any atom group at such a chain, can be replaced by another atom or group of atoms. The substituent is the one that replaces the old one.
For substitution at an aliphatic chain, you heed light and heat.
For substitution at an aromatic chain of ring, you need a catalyst.
The alkanes can easily undergo any substitution process, where an H is substituted by another atom, for example, a Halogene.
The process is rather slow.
Substitution with, for example, Chlorine (Cl₂) is a slow process and needs light energy.

Question 16

Explain the word 'termination'; or: why do we use that word here?
A radical with an unpaired electron is very reactive. How come?

Question 17
If you have models, try to imitate the substitution of ethane with chlorine.

The nitril group

A – C≡N group is called "nitril", when the C is part of a main chain.
When a – C≡N group performs as a side chain (the C does not take part of the main chain), than the name 'cyanide' can be used, just like in the inorganic chemistry with cyanide ions: -CN^-
Nitril groups can be introduced / made in a substitution reaction of cyanide with a halogene alkane:

Question 18
Explain how we can consider this substitution as a method of chain enlarging.

N.B.
The Nitril group possesses a threefold bonding. With the consequence that not addition is possible, not just one or two, but even three times..
This is very special.

Question 19
A special rule is: more than one OH-group connected to the same C in a carbon chain is not stable; water will be produced.
Check that in the above reaction equation.

First we add three molecules of water to support the addition, but eventually one molecule of water will return. You can also see that finally an alkanoic acid and ammonia are formed, the two final products.

Substitution with the side product H₃PO₃

There is a methode to substitute an OH-group of a chain, by a halogene, using the reaction with 'phosphor trihalogenide'.

Substitute H-atoms of Benzene is easy going.
The stable structure of the ring does not lose its stable character in this process. The 6 C-atoms remain connected, without enay change.
Benzene suffers substitution with various groups: nitro, amino, alkyl, sulphon and others.

Question 20

In chemical structures, give the reaction of the substitution of Chlorine at toluene.
Mention also the reaction requirements.

Goto answer 11-20

4. Condensation & Hidrolysis

4.1 Condensation

Two molecules connect and form one new molecule, with water as a side product.

molecule 1 + molecule 2 new molecule + water

Example: the formation of an ester:

ethanol + ethanoic acid

ethyl ethanate + H₂O

in structural formulas

This is a reversible process, so the direct reaction is the condensation and the reverse reaction is called: hydrolysis.

The production of an ester, from an alcohol and a carboxylic acid can be accelerated with a catalyst. Therefor you can use, for example, an acid.
The H⁺-ions attack a negative position, and in that case we call the mechanism "electrophylic".
There are different mechanisms where negative particles attack positive positions of a molecule, and then we talk about 'nucleophylic'.

It happens that an alcohol reacts with a inorganis acid. For example, glycerol can react with nitric acid.
Normally glycerol reacts with fatty acids to produce fats and oils, but the reaction with nitric acid is comparable:

Question 21

The product has an official name: glycerol-trinitrate. But in practice you often hear the name: nitroglycerine

Give the reaction equation in molecular formulas
Expalin why that name of 'nitroglycerine' is not right according to the IUPAC rules.

Goto answer 11-21

Another example of condensation (and of the opposite, hydrolysis) is the production of ether and water out of the reactants alcolhol + (another) alcohol,
for example:

Ethanol + propanol ethoxy-propane + water.

This reaction needs a catalyst (can be an acid) as also higher temperatures.

4.2 Polycondensation

Many molecules connect to each other, where lots of water molecules are formed as a side product (or sometimes another small molecule)
A macro molecule is produced via a condensation mechanism.
example:
Glycol C₂H₄(OH)₂, with two OH-groups per molecule, can suffer a polycondensation process in two directions.

HO – CH₂ – CH₂ – OH

Each OH-group can react with the OH-group of another molecule. Every time water is formed.
The glycol molecule can extent to two sides.

The product is, in this case, a polyether, a solid, while the reactant glycol is a liquid.

Question 22

Try to imitate this polycondensation with models.
Explain why the polyether will be a solid.

Famous polycondensation products are:

Polyethers
Polyesters
Polypeptídes
Polysaccharides

Copolymerisation is the process where participate more than one kind of monomers.
In the industry many copolymeres are invented, made and applied.
For example, the material of videotapes are a copolymere product of the two monomeres: glycol (1,2-dihidroxyethane) and tereftalic acid (1.4-dicarboxylbenzene ).

Question 23
Give a part of the structure of that polymere above.

Question 24
Copolymerisation of the monomeres glycol and oxalic acid is possible, with a little bit of concentrated sulfuric acid as a catalyst.
This reaction starts with an attack of protons at oxalic acid, whereby a carbonium ion is formed (a C with a positive charge). This is the slow step of the reaction.

Try to complete the whole reaction mechanism, while you know that the two monomeres always connect alternately.
Explain the fact that the slowest step determines the total reaction rate.

Hydrolysis:

We already concluded that the hydrolysis process is exactly the opposit of condensation:

one molecule, using water, splits into two new molecules.

Question 25
Show what will be produced in the total hydrolysis of ethyl-acetate.

4.3 Saponification

Saponification is a process that applies hydrolysis of a fat in basic environment.
For that often animal fat is used. This fat suffers hydrolysis, and than the products glycerol and fatty acids are formed.

This way we can saponificate palm oil, and in basic environment the produced fatty acid is transformed into Sodium palmitate (a salt!). In fact, this is soap.
If as a base, we do not use NaOH, but KOH, the salt Potassium palmitate will be formed and this soap is softer.

5. Redox reactions in Carbon Chemistry

5.1 Introduction

Also in the carbon chemistry there are redox reactions. Many carbon compounds are capable to donate or capture electrons.
att.: talking here about oxydation, we do not actually talk about combustion with Oxygen.
Most compounds in carbon chemistry can suffer combustion and form carbon dioxyde and water, but this is not now the topic.
But yes, I give you first a short info about direct and indirect combustion; after that we go to the real redox reactions.

Complete combustion

A nice example of a complete combustion (although in steps) you can find in question 39.
Here is already the reaction:

Most Carbon bondings (with C and H and other elements) react (connect) strongly with Oxygen.
The most reactive element is here Hydrogen that immediately - in contact with Oxygen - creates water.
Then (but almost simultaneously) the Carbon reacts and forms CO₂. Possible present S or N or P or other elements can make more products like SO₂ and the Nitrogen oxydes N_xO_y. These (gaseous) products are responsible for air pollution.
Of course, the best petrol contains as less sulphur and nitrogen as possible.

Incomplete combustion

Takes place is there is insufficient Oxygen.
Of course H is the first to produce water, but maybe there is not enough Oxygen left over to burn the carbon atoms. There might be insufficient air, for example. Then CO (Carbon monoxyde) is formed, or even, the carbon remains unburned (soot).
So take care that you always provides the burning process with enough air.
Carbon monoxyde is very toxyc.

Oxydation number N_ox

In module 10 the redox reactions were treated, including the oxydation number. You know that during a redox reaction the oxydation numbers of some elements will change.
In redox reactions in the carbon chemistry, such an oxydation number not always can be easily defined. Where exactly occurs the transfer of electrons, that is the question, in particular regarding the carbon atoms.
The redox tables do not provide halfreactions of the carbon chemistry, so the changes of the oxydation number in these cases must be discovered by yourself.

As an example we take:

Normally an H in a compound has the oxydation number of +1 and the O has a N_ox=-2
These two data you know, and also that the whole molecule is neutral.
This way you can calculate that the oxydation number of the C in this case must be: -2.

Question 26
Calculate the oxydation number of C in the following compounds:

Ethane
ethanol
ethoxy-ethane
ethyl-ethan(o)ate
methane

Goto answer 11-26

Polariteit

Oxydising a non polar carbon chain is very difficult (not talking about combustion, with Oxygen). The division of valency electrons is in those chains very equilibrated and equal; there is no difference in electronegativity.
The presence of charges δ⁺ and δ^- can cause nucleophylic and electrophylic attacks with redoxreactions as a consequence.
As soon as there is an Oxygen atom connected to the chain, there is polarity that makes redox reactions (transfer of electrons) more easy to occur.
The Oxygen atom, or any other atom with a good electronegativity, causes the needed polarity in that substance.

But note: not only polarity is needed; there must also be a position where more Oxygen atoms (or other electronegative atoms) can be connected.
A general rule is:

A C-atom in a carbon chain can be oxydises when that C simultaneously:
has an O ánd an H where an extra O can be connected between the C and the H.

In other words: To oxydise a carbonchain, there must be already an O-atom at the C-atom to oxydise, and also, that same C-atom must have at least one H-atom.
Normally, during the oxydation, an O is connected between the C and the H:
H becomes then: OH / In fact an H is substituted by an OH.
Mind:
Ether molecules have an O-atom between two C-atoms. There is a certain symmetria, so less polarity; in this case oxydation becomes more difficult.

A bonding between two carbon atoms is - in normal oxydation reactions - difficult to break; carbon chains normally remain unchanged (execpt in the case of direct combustion with Oxygen, where carbondioxyde and water are made).

Question 27

Esplain if the substances below can be oxydised with acid dichromate:

butane
2-methyl,propanol-2
aceton
methoxy-ethane

Halfreactions

A number of compounds can be found in the redoxtable; although most of them cannot be found there.
You must be able to construct the half reactions without any table and found out how many electrons are transferred per oxydator or reductor. Normally you can find the halfreactions in those redox tables, but not in these cases.

Question 28
At the oxydation of 1-propanol with acid permanganate, this propanol changes into propanal.

Write the electronic formulas of those two organic compounds and compare the (total) number of valency electrons.
Write then the two half reactions in molecular formulas.

Have you seen that propanol has lost one pair of electrons? So it is a reductor.
To make the number of Oxygen atoms equal at both sides, you normally use water molecules here; to make the number of H-atoms equal, you use ions H⁺.
If these substances (water and H⁺) are needed as reactants (before the arrows), than water and/or acid should really be present as helping substances.

Oxydizing various functional groups

Alkanes

Alkanes are difficult to oxydise, exept of course the burning with Oxygen. Mind for example: natural gas, butagas, etcetera.

Question 29
Complete combustion of natural gas includes changes of the oxydation numbers.
Show these changes.

Alkanoles

Alkanoles are mostly oxydised with acid permanganate of with acid dichromate; i.e. with a rather strong oxydator.

Alkanales do not need such strong oxydators; weaker oxydators like Ag⁺ or Cu²⁺ are sufficiently strong to reach the aim; if at least some conditions are responded (see further). Alkanales can be oxydised more easy than alkanoles.

Oxydizing an alkanol, the first step gives a substance with two OH-groups at one C-atom (check that).
As said, such a compound with more OH-groups at one C-atom is not stable; of those two OH-groups, one molecule of water will be removed.
Than remains a double bonded O (alkanal, alkanone, alkanoic acid).

Question 30
Explain when - in this way - is formed an alkanal, een alkanol oo an alkanoic acid.

We also can burn directly alkanoles/alcohols, and of course we obtain the well known carbon dioxyde and water.
But now we are talking about the more subtile way with weaker oxydators and with special products:
the ion dichromate in acid environment can oxydise a primary alkanol in two steps (via the intermediate alkanal) up to the final product alkanoic acid.

Question 31
The following substances can be oxydised with dichromate? yes or no.
if yes, what is the product? And prove your answer.

2-propanol
Etanediol (glycol)

Question 32
statement: "2-methyl 2-propanol can be oxydised with permanganate-ions."
Is this statement true or false? Explain your answer.

Question 33
During the oxydation of 1-propanol with permanganate(aq), in the presence of sulfuric acid, the propanol will change into propanal.

Give the electronic formulas of the two organic substnces and compare the number of present valency electrons.
Also give the two half reactions in molecular formulas.

Att.:
During the oxydation of an alcohol (alkanol), the first step is the formation of a C-atom with two OH-groups (check that!).
Again: you know that such a situation is not stable. The two OH-groups at one and the same C-atom will create a water molecule; a C with a double bonded O remains.

Alkanales

For the oxydation of alkanales (rather easy) we often use acid permanganate or acid dichromate.
Yet these oxydators are stronger than needed in these cases.
An alkanal has an aldehyde group (-CHO) and the C=O bond is rather polar. that's why oxydators easily attack the δ⁺-part.
Weak oxydators, like silver- or copper(I) ions are stong enough to oxydise alkanales, whereby the product will be a carboxylic group, that means an acid.

So: it is easier to oxydise an alkanal than an alkanol.

Question 34

What will be the product when oxydising ethanal?
Idem for oxydising methanal (att.: here we have a very special situation!!)

Alkanones

In the case of alkanones, there is polarity between the C and the O of the C=O group, but that particular C-atom does not possess an extra H-atom where the oxydation shoult have to take place.
So it doesn't work. Only with very powerfull oxydators can be done something, and alone if the C - C bond is broken.

Question 35
2 Statements:

Alkanone contains a polar group
Alkanone cannot easily be oxydised

Are these statements true? Explain your answer without consulting the text.

Alkanoic acids

Organic acids normally cannot be oxydised easily, just like ethers and esthers.

Question 36
Explain why those acids cannot be oxydised and explain also why formic acid and oxalic acid are exeptions.

Question 37
With electronic formulas, show how oxalic acid (ethanoic diacid) can act as a reductor.

Question 38
2 Statements:

Methanoic acid can easily be oxydised
Ehanoic acid can easily be oxydised

Are these statements true? Explain.

Goto answer 11-38

Question 39
Study the reaction scheme below. It has five steps.

Of every step indicate if it is a redoxreaction or not.
If so, what could be a suitable oxydator?

N.B.
1. In these cases, oxydation comes with the introduction of an O-atom between a C and an H
2. mostly in organic substances, the presence of more OH-groups at one C is unstable
3. the final products are carbon dioxyde and water.
Try to imitate this reaction - if you have models.

Ethers

Yes, there is an O between two C-atoms, so there must be some polarity, but there is also great symmetry.
That's why we cannot expect big polarity, and therefore it will be more difficult to oxydise ethers.

Question 40

Explain the possibility of oxydation with acid dichromate of:

Butaae
2-methyl propanol-2
aceton
methoxymethane

It is difficult to oxydise organic acids. Not because there is no polarity, but because there is no H-atom connected to the polar C^δ+, where the O should be connected. The same for esthers.

Question 41
Methanoic acid and dicarbonic acid can easily be oxydised.

Explain why these two acids (also called formic acid and oxalic acid) are execption to the rule.
For the case of oxalic acid, give the electronic formulas of the reactant and of the product, and explain the difference.

5.3 The oxydators Cu²⁺ and Ag⁺

These two are rather weak oxydators. You can find them in the table, but mind you:
The Cu²⁺ has not the metalic Cu as product, but the ion Cu⁺.
The product of the oxydator Ag⁺ is the metalic Ag.
So: Both oxydators pick up one electron per ion and they do this only in basic environment.

Whenever we use the weak oxydator Ag⁺(aq) toghether with a certain supplement, we talk about:
AMMONIA SILVER SOLUTION (TOLLENS REAGENT = TR)
It is a mixture of Silvernitrate(aq) + ammonia(aq)

Whenever we use the weak oxydator Cu²⁺(aq) toghether with a certain supplement, we talk about:
FEHLINGS REAGENT (FR)

FEHLINGs Reagent

FR is a mixture of: copper(II)sulfate(aq) (sometimes indicated with Fehlings A) and a mixture of Sodium hydroxyde(aq) + Na-K-Tartrate(aq) (also indicated as Fehlings B)
The oxydator is the ion Cu²⁺ that dureing the reaction transfers into the ion Cu⁺.
Att.: these data are the basis for this reaction.
But the reaction conditions are a bit complicated:

This reaction only takes place in basic environment.
This reaction needs heat energy; heating is necessary.

We first need a solution of Copper(II)sulfate, CuSO₄(aq), known as Fehlings A

Because of the necessary basic environment we add some sodium hydroxyde, NaOH(aq).
But immediately you have a problem:
The Cu²⁺-ions react immediately with OH^- to form a precipitate of Copper(II)hydroxyde, Cu(OH)₂.

This has to be prevented.
And this is possible. For that we add another substance: Na-K-tartrate(aq). This substance captures the copper ions so they cannot longer react with the hydroxyde ions.

the mixture of the two solutions, NaOH(aq) and Na-K-tartrate(aq) we call Fehlings B.
We mix the Fehlings A + B only at the moment we need it, by mixing and thus creating the FR (which is not stable for a long time.
As soon as FR is added tot the oxydisable mixture (containing an organic reductor), heating is needed to have the necessary energy.
No need to have a strong reductor; a weak aldehyde group is enough, like in ethanal(aq).

In a test tube with a couple of nl ethanal, we add a couple of ml FR, and then we calmly heat up in a cool flame or in a waterbath. Slowly we see color change (green yellow red) and precipitation (the red substance is CuOH(s).

De Ammonia Silver Solution / the Tollens Reagent

In the case of the oxydator Ag⁺, the process is very similar to that with FR:

This reaction only takes place in basic environment.
This reaction needs heat energy; heating is necessary.

And also in this case, precipitation (now of AgOH) must be prevented. We do not use tartrate for that, but ammonia(aq).
Here the use of a very clean test tube is essential, because the product (Silver(s)) only precipitates on fat free glas walls.

In test tube 1 we add a couple of ml silvernitrate solution (AgNO₃(aq)).
To this solution we add - bit by bit - the ammonia(aq).
N.B.: ammonia as sucht is already basic, so there is no need to add NaOH.
For a moment you see a precipitate of AgOH, but soon this disappears when more ammonia is added.
Then the TR is ready for use as an oxydator.

A couple of ml of the TR is added to a couple of ml of the reductor solution. Again this reductor kan be something like ethanal or methanal (or glucose). There must be an aldehyde group in it.
You must stirr well and put the tube carefully in a water bath for some time. Now we can observe the formation a the silver mirror.

More about the function of Tartrate and the Ammonia

How do the ions Cu²⁺ or Ag⁺ succeed to maintain soluble in a basic solution?
How does tartrate protect the copper ions against the OH^--ions?
How do ammonia molecules protect the Silver ions against OH^--ions?
How do they fix that?

They surround those ions, the create a circle of protection.

Question 42
Try to explain in your own words some items about FR and TR:

the equilibria of half reactions and their positions
the influence of the basic environment
the influence of the temperature
how to prevent the precipitations
the observations you can do during the reactions.

Question 43
If you have at school of somewhere else the necessary substances and equipment)
Investigate if the following substances can be oxydised with FR or with TR:
(try to make the most beautyful silver mirror)

The substances to investigate are:

glucose

fructose

saccharose

a solution of saccharose, before and after being boiled for some time.

a solution of starch, before and after being boiled for some time.

If you have no lab conditions, try to do the experiments just by thinking.

These same reactants will be treated also in module 12, biochemistry.

6. Acid base reactions in Carbon Chemistry

6.1 Introduction

Much information about acid base reaction can be found in module 9. This paragraph deals with acid base reaction in the carbon chemistry.

acid + base

conjugated base + conjugated acid

Organic acids only donate H⁺ coming from the hydroxy groups (-OH), in most cases part of the carboxylic structure.
In general the organic acids are weak. One of the strongest organic acids is formic acid.
Tho carboxylic group owes ist acidity to the strong polarity within that group. It causes a strong repulsion between the C-atom and the H-atom that both have a certain positive charge (δ⁺).

Question 44
Write the structure of formic acid (methanoic acid) and invent any reason why this acid is somotimes used as a toilet cleaner.

Only Hydroxy groups can lose protons, but only if the to the same C of the hydroxy group also is connected another oxygen atom, or when this hydroxygroup is connected to a benzene ring.
Mind that H atoms, directly connected to a carbon chain never perform as donator of H⁺

Normal alcohol groups also do not donate H⁺-ionen, unless very agressive substances are involved, like Na or K.

Question 45
Write the names and structures of the substances you can obtain if the acids below dissociate one or more protons:

formic acid
stearic acid
phenol
1-propanol (in exceptional cases)

Question 46
Oxalic acid (two carboxyl groups directly connected to each other = H₂C₂O₄) is not only a stronger acid, but can also, easily, break in two parts where two molecules CO₂ are formed.
Explain this phenomena.

If the carbon chain of an organic acid is rather lon, we call it a fatty acid.
If in that chain also are included one or more double bonds, than this molecule is a (plurally) unsaturated fatty acid.

Question 47
Go find in tables / chemistry books / internet a fourfold unsaturated fatty acid and write the systematic name of it.

A NH₂-group, connected to a carbon chain is built up in such a way that the N-atom (more or less δ^-) still has a free electron pair, available for the capturing of particles without electrons, like the H⁺-ion.
If that happens, the amino group becomes positive: – NH₃⁺.

Question 48
Give the reaction of an amino group, in structures, capturing that H⁺ from acetic acid.

If an organic molecule (is a substance from the carbon chemistry) has an acid group as well as an amino group, than you have an amino acid.
Those two groups: carboxylic and amino, do not only react as acids or bases, but they also participate in condensation, when esters or peptides are formed.

Question 49
Find in the table with amino acids a simple amino acid and write the structures of this acid in strong acid and in strong basic environment.

Brought in acid environment (at low pH-values), an amino acid is positive and in basic environment (at high pH) the amino acid is negatively charged.
Somewhere between that high and low, there must be a pH value at which the amino acid molecules are neutral in average.
This special pH value is calles the the iso-electric point: het I.E.P.

Question 50
In the iso electric point the amino acid molecule can occur as a so called 'double ion'. Explain this.

Some complicated (aromatic) organic acids have - thanks to their structure - a certain color that changes if one or more protons are donated. These substances are fit to act as an acid base indicator. (see also module 9)

HIn

H⁺ + In^-
kleur 1 kleur 2

6.2 Phenol

There is a special type OH-group that can function as a weak acid. We don't talk here about ordinary aliphatic oH-groups, but about the aromatic OH-group connected to a benzene ring.
Such a benzen ring with an OH-group is called: phenol.

C₆H₅OH	+	OH^-		C₆H₅O^-	+	H₂O
acid		base		conjugated base	+	conjugated acid

In structures you see that phenol donates in the following way:

phenol phenolate
Special is that in phenolate all valency elctrons of Oxygen, toghether with the electrons op the type π in the ring, come to a kind of resonance state.
The negative ion becomes therefore more stable.
The consequence is that phenol stays with the tendency to donate a H⁺, much stronger than a normal aliphatic OH group.

Question 51
Explain why the strenghts of the substances has to do with a possible reversibility.

The hydroxyl group and hydroxy benzene

You already know that an OH group only can lose / donate H⁺ (react as an acid) if that C-atom at the same time is connected with a O (so talking about a carboxylic group).

The aliphatic hydroxyde group, that is an alcolhol group, normally is not an acid nor a base.
The alcohol group is not basic nor acid.

Only in the case of very agressive, reactive substance, like Sodium, the OH group can donate its H.

CH₃ – CH₂ – OH + Na·

CH₃ – CH₂ – O^- + Na⁺ + H·

As soon as 2 H·-radicals exist, a H₂-molecule can be formed, so a gas.
The other product is called Sodium ethanolate, a stong basic substance.

In structures:

Question 52

Please note very well what happens in the above reaction with the valency electrons.
Explain why CH₃ – CH₂ – O^- (+ Na⁺ ) has a strong basic character.
(where alcohol has no acid character at all)
The reaction is not considered as an equilibrium
Is this an acid base reaction or a redox reaction? Explain.

6.3 Organic ampholytes

There are organic molecules that contain in one and the same chain a carboxylic group (acid) and also an amino group (base). You know that this is an amino acid. Such an amino acid is amphoteric.
Thes molecules will be treated in module 12, the biochemistry.

Question 53

A very simple amino acid is Alanine (2-amino-pronanoic acid)

Give the structure and explain where and how the amphoteric reactions take place.
Explain the concept of 'double ion'.

Goto answer 11-53