= CH – CH3
Cl – CH·
This radical attacks the molecules that are present in good amounts: the other molecules of propene:
The main reaction can be stopped if a radical (the unpaired electron) meets another radical, also with an unpaired electron
- C3H6Cl· + C3H6
- C6H12· + C3H6
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.
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:
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.
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.
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.
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.
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.
- Explain the word 'termination'; or: why do we use that word here?
- A radical with an unpaired electron is very reactive. How come?
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:
Explain how we can consider this substitution as a method of chain enlarging.
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.
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.
Goto answer 11-20
- In chemical structures, give the reaction of the substitution of Chlorine at toluene.
- Mention also the reaction requirements.
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:
The product has an official name: glycerol-trinitrate.
But in practice you often hear the name: nitroglycerine
Goto answer 11-21
- Give the reaction equation in molecular formulas
- Expalin why that name of 'nitroglycerine' is not right according to the IUPAC rules.
Another example of condensation (and of the opposite, hydrolysis) is the production of ether and water out of the reactants alcolhol + (another) alcohol,
Ethanol + propanol ethoxy-propane + water.
This reaction needs a catalyst (can be an acid) as also higher temperatures.
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.
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.
- Try to imitate this polycondensation with models.
- Explain why the polyether will be a solid.
Famous polycondensation products are:
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 ).
Give a part of the structure of that polymere above.
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.
We already concluded that the hydrolysis process is exactly the opposit of condensation:
one molecule, using water, splits into two new molecules.
Show what will be produced in the total hydrolysis of ethyl-acetate.
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.
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.
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 NxOy.
These (gaseous) products are responsible for air pollution.
Of course, the best petrol contains as less sulphur and nitrogen as possible.
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 Nox
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 Nox=-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.
Calculate the oxydation number of C in the following compounds:
Goto answer 11-26
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:
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.
has an O ánd an H where an extra O can be connected between the C and the H.
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.
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).
Esplain if the substances below can be oxydised with acid dichromate:
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.
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 are difficult to oxydise, exept of course the burning with Oxygen.
Mind for example: natural gas, butagas, etcetera.
Complete combustion of natural gas includes changes of the oxydation numbers.
Show these changes.
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).
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.
The following substances can be oxydised with dichromate? yes or no.
if yes, what is the product? And prove your answer.
- Etanediol (glycol)
statement: "2-methyl 2-propanol can be oxydised with permanganate-ions."
Is this statement true or false? Explain your answer.
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.
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.
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.
- What will be the product when oxydising ethanal?
- Idem for oxydising methanal (att.: here we have a very special situation!!)
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.
Are these statements true? Explain your answer without consulting the text.
- Alkanone contains a polar group
- Alkanone cannot easily be oxydised
Organic acids normally cannot be oxydised easily, just like ethers and esthers.
Explain why those acids cannot be oxydised and explain also why formic acid and oxalic acid are exeptions.
With electronic formulas, show how oxalic acid (ethanoic diacid) can act as a reductor.
Are these statements true? Explain.
- Methanoic acid can easily be oxydised
- Ehanoic acid can easily be oxydised
Goto answer 11-38
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?
- In these cases, oxydation comes with the introduction of an O-atom between a C and an H
- mostly in organic substances, the presence of more OH-groups at one C is unstable
- the final products are carbon dioxyde and water.
- Try to imitate this reaction - if you have models.
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.
Explain the possibility of oxydation with acid dichromate of:
- 2-methyl propanol-2
It is difficult to oxydise organic acids. Not because there is no polarity, but because there is no H-atom connected to the polar Cδ+, where the O should be connected. The same for esthers.
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.
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)
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:
We first need a solution of Copper(II)sulfate, CuSO4(aq), known as Fehlings A
- This reaction only takes place in basic environment.
- This reaction needs heat energy; heating is necessary.
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:
And also in this case, precipitation (now of AgOH) must be prevented. We do not use tartrate for that, but ammonia(aq).
- This reaction only takes place in basic environment.
- This reaction needs heat energy; heating is necessary.
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.
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.
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:
If you have no lab conditions, try to do the experiments just by thinking.
- a solution of saccharose, before and after being boiled for some time.
- a solution of starch, before and after being boiled for some time.
These same reactants will be treated also in module 12, biochemistry.
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+).
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.
Write the names and structures of the substances you can obtain if the acids below dissociate one or more protons:
- formic acid
- stearic acid
- 1-propanol (in exceptional cases)
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.
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+.
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.
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.
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
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.
In structures you see that phenol donates in the following way:
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.
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.
- 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.
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.
A very simple amino acid is Alanine (2-amino-pronanoic acid)
Goto answer 11-53
- Give the structure and explain where and how the amphoteric reactions take place.
- Explain the concept of 'double ion'.