Keilin (1925) was studying the mosquitoes responsible of the malaria. With a spectromicroscope he looked at the legs of one mosquito and saw that they change of colour. At rest the muscles are oxidised but when they are excited the muscles consume the oxygen and are thus less oxidised. The colour is given by the cytochrome, a molecule that changes of colour in function of the oxidation.

This molecule belongs to the mitochondrion. The mitochondrion is a kind of bacteria hosted by eukaryotes.

They have their own membrane, composed of an inner membrane and of a porous outer membrane. The matrix is inside the inner membrane and the intermembrane space is also called cristae where the membrane is protruding. Similarly to the haemoglobin, the cytochrome is a protein that possesses a haem the iron of which can change of oxidation state. This property is used in the respiratory chain. Cytochrome belongs to the inner membrane of the mitochondrion along which it can move and it separate the glycolysis from the citric acid cycle. The cycle of the citric acid takes place in the matrix of the mitochondrion while glycolysis takes place outside of the mitochondrion.

Respiratory chain

The respiratory chain can be summarised by the following figure and table:

 In the inner membrane we find 4 complexes – NADH-ubiquinone reductase, succinate-ubiquinone reductase, ubiquinol-cytochrome c reductase and cytochrome oxidase – that make redox reactions and transfer electrons from one side of the membrane to the other side (sometimes protons instead of electrons). One reaction at one side of the membrane (matrix or intermembrane space) is always coupled with one reaction at the other side of the membrane. One reaction of reduction is coupled with a reduction of oxidation and vice-versa. The whole phenomenon involves transfers of electrons inside the membrane and transfers of protons from one side of the membrane to the other side. Note that some complexes belong to the membrane but only face one of its sides.

Q is the ubiquinone, a cofactor that belongs to the membrane. It interacts with the cytochrome c in a loop of oxidoreduction. One can see that the oxygen is involved by the complex IV. To resume the loop, we can write two reactions of oxidoreduction, one with the NAD and one with the FAD.

However O₂ cannot directly interact with FADH₂ and NADH+H⁺ and this loop is thus necessary. For redox, the free energy of Gibbs is

To determine the difference of potential ∆E°, we look at the couples involved in the reaction:

For FAD we have 43.3kcal/mol. There is a transfer of protons involved in the process but there must be an equilibrium between the concentrations of charge inside the cell and outside the cell. The protons can move out of the matrix of the mitochondrion through some protein complexes that generate ATP from the flow of protons (bottom of the figure). The F₁-ATPase protein is composed of two parts, F₀ in the membrane and F₁ in the matrix of the mitochondrion. F₀ pumps the protons from the intermembrane space towards the matrix. F₁ receives the energy from the transport and transforms ADP into ATP. Those species are charged negatively and require a transport through the membrane (top of the figure). One ADP can go in the matrix only if one ATP is moving out of the matrix. The inorganic phosphates P_i are also transported through the membrane.

The energy to form the ATP comes from the gradient of pH. It generates a proton motive force.

Where the indexes mat and ext respectively matrix and its exterior (the intermembrane space). The gradient of pH generates a difference of electric potential ΔV.

Combining the two effects, we have

Dividing this expression by nF, we obtain the expression of a variation of free energy ∆p associated to the transfer of one mole of protons (in volt).

The coefficient 2.303RT/nF of the gradient of pH has a value of 59mV at 25°C.

We can measure the pH and the difference of potential: ∆V=0.14V and ∆pH=-1.4.

It corresponds to a variation of free energy equal to

To form one mole of ATP, we need 7.3kcal. In the respiratory chain, there are 4 complexes that oxidise the NADH and the FAD.  They respectively form the equivalent of 11 and 8 protons by loop.

We can write a global equation of the “combustion” of a pyruvate during the citric acid cycle and the respiratory chain:To summarise, the oxidation of one NADH generates ~3 ATP and the oxidation of FAD generates ~2ATP.

The 15 ATP come from

• 4 NADH (one before the citric acid cycle and 3 during the cycle) à12ATP,
• from 1 FADH₂ à 2ATP
• and from 1 GTP»1 ATP.

To form one pyruvate, glycolysis forms 2 ATP and uses 2 NAD that are reduced into NADH+H⁺.

So considering 3 ATP by NADH, it corresponds to the production of 8 ATP. Added to the 15 ATP of each pyruvate (x2), we obtain 38ATP by glucose. In comparison to the combustion of the glucose (∆G°’=-686kcal/mol), it represents about 40% of the potential energy (38×7.3kcal/mol=277kcal/mol).

The NAD involved in the glycolysis and that produces a NADH is not at the same place (cytosol) that the NADH required in the cellular respiration (matrix of the mitochondrion). The NADH cannot enter freely into the mitochondrion because it is charged. There is a system of shuttle that may differ for different cells and that will modify the NADH to allow its passage through the membrane, then modify it again so it can be used there. In the brain and in muscles, it is a shuttle of glycophosphate that we will next explain. The shuttle is basically composed of two reactions: one in the cytosol and one in the matrix of the mitochondrion:

The shuttle can be represented like this:

If you remember, dihydroxyacetone-3P was part of the glycolysis (red rectangle in the following figure): it is one of the two trioses phosphate produced from fructose -1,6-diphosphate. The fact that this triose is involved in the shuttle does not really decrease the yield of the glycolysis significantly: the cell needs a given quantity of dihydroxyacetone for the shuttle but the molecules are regenerated in the matrix of mitochondria and can then return to the cytosol. So they just have to be produced once. The rest of the production is turned into glyceraldehyde 3P for the glycolysis.

The FADH₂ is oxidised during the respiration and the dihydroxyacetone can move back to the cytosol. The shuttle has a small cost: it oxidises one NADH into NAD in the cytosol while the reaction in the matrix involves FAD/FADH₂. In term of energy, it represents one ATP used because all the processes don’t take place at the same place.

The malate-aspartate shuttle

The same cofactors (NADH+H⁺/NAD⁺) are used so the yield does not change with this shuttle (38 ATP). The mechanism is based on the reducing power of the oxaloacetate.

Reduced into malate (1), it is transported in the matrix of the mitochondrion (2) through the malate-α-ketoglutarate transporter if one α-ketoglutarate is available to make the displacement in the opposite direction. The reverse reaction, the oxidation of the malate into the oxaloacetate is made in the matrix (3). Yet, the oxaloacetate cannot goes back out of the membrane. To do so, it is transformed into an amino acid, the aspartate, by a reaction of transamination (4):

It is done here by the aspartate transaminase with the use of the glutamate as amino acid. As a result, the oxaloacetate is transformed into aspartate that can move through the membrane through a transporter requiring the passage in the opposite direction of a glutamate (5). In the cytosol, it is turned back into the oxaloacetate with the exact same reaction (6) than at the other side of the membrane.

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The principle of the following exercises is pretty simple (but the exercises are not): We give you a series of reactions that are following each other but we will only give you some of the reactants and/or of the products. From them you have to reconstruct the whole process and find out which molecule and which structure hide behind which letter.

1. In this first exercise you have the structure of the initial reactant (A) and the composition of the first product B. You have to determine the structure of B, the product obtained by the first reaction. Then from B you have to determine the structure of C, then the compound D and its structure, etc.

Correction

A→B: Two sets of protons are more acid than the others: the protons in α of carbonyl. The protons from CH₃ are less acid than the CH₂ because CH₂ is between two carbonyls. It is thus there that the base attacks. The negative carbon attacks next the formaldehyde. A rearrangement occurs after the neutralisation to obtain B and H₂O.

B→C: The first step is identical than for the previous reaction and the two molecules merge together. The attack is done on the sp² carbon and we obtain a structure that is stabilised by a H bond forming a cycle between the OH and the C=O.

C→D: This time the proton (denoted *) is not attacked by the base because removal of one methylic proton allow the formation of a cycle of 6 carbons. In the second step, the acid catalyses the loss of a water molecule to obtain a double liaison conjugated with the carbonyl.

D→E: The double liaison is reduced to obtain a cyclohexane.

E→F: The esters are turned into carboxylic acids with a basic catalysis. CO₂ is lost if we increase the temperature.

F→G: Zn reduces selectively a ketone into an alkyl chain.

G→H: The diazonium carbene (check nom) is able to take the proton from the acid. The carboxylate attacks next the carbene to form a methyl ester and to free N₂. CH₃OH leads to the same result but the process is different: There is a nucleophilic substitution on the carbonyl to replace -OH by -OCH₃.

2. This exercise focuses on reactions on aromatics and on carboxylic acids and their derivatives.

Correction

A→B: The electrophile character of Br has to be enhanced by an acid of Lewis to do the reaction.

B→C: A nitro group is added to the cycle. There is already one substituent on the cycle so we have to determine if the nitro group is added in ortho, meta or para. A halogen orientates the reaction on the ortho/para positions. The para position should be favoured because of the steric hindrance on the ortho position.

C→D: the reduction is limited to the nitro group that turns into an amine. To reduce completely the aromatic ring we have to heat up the solution to 300°C.

D→E: NaNO₂ does not add a nitro group on the ring. It leads to the removal of the amine and the formation of an arenium. This very reactive species reacts with water to replace the amine by a hydroxyl group –OH.

E→F: The base takes the proton of the p-bromophenol. There is then a nucleophile attack by the anion on CH₃I to obtain the p-bromometoxybenzene.

A→G:  It is the chloromethylation reaction. During this reaction, the formaldehyde and the chlorydric acid form a chloromethanol stabilised by ZnCl₂. The acid protonates the alcohol and the ring can attack it to reject water and bind CH₂Cl.

G→H: A simple SN₂ by CN^– followed by its transformation into a carboxylic acid. This transformation is done by successive attacks of water molecules on the carbon bond to the nitrogen.

H→I: SOCl₂ is a molecule that allows us to obtain an acyl chloride from an acid. It cannot be done with HCl or Cl₂ because Cl^– is a better leaving group than OH^–.  The reaction is followed by the formation of a primary amide.

I→J: The amide is reduce into an amine by LiAlH₄. LiAlH₄ can generates H^– that attacks the carbonyl.

 E+G→K: A base takes the proton from the bromophenol to obtain a stronger nucleophile. A SN₂ takes place between the two species to merge them into one molecule.

G→M: The second step of the reaction leads to the formation of the carboxylic acid as it was the case in the reaction G->H. The missing element on M is the nitro group in meta. This position is favoured because of the mesomeric captor effect of the COOH through the CH₂. The effect is however smaller than for a mesomeric captor directly in contact with the aromatic ring.

3.  In this exercise the last product of a long series of reaction is given. It is the direct product of a reaction of ozonolysis. You have thus to go backwards in the reactions, starting from the end to find the reactants of each reaction. The formulas of most of the molecules are given. G₁ and G₂ are isomers.

Correction

I→…: One of the products, the oxalic acid, is a carboxylic acid and one reactant is the water. We can thus guess that the reaction is a reaction of substitution on a derivative of carboxylic acid. The other product of the reaction is one methanol molecule. One carboxylic acid was thus an ester before the reaction. Only one methanol is generated by the reaction so only one of the two acids of the oxalic acid was an ester.

H→I: The ozonolysis cuts a molecule at double liaisons and leads to the formation of carboxylic acids in presence of an oxidant. There are 4 acid groups in the products of this reaction and 2 esters. The 4 acid groups indicate that a bigger molecule was cut down at two places. The double liaisons were thus conjugated with the esters. It is thus an example of reaction that involves a part of the conjugated system and not all of it. We don’t know if the double liaisons are cis or trans.

G→H: G₁ and G₂ are two isomers. There are two other information that we may consider to find the isomers. CH₃I, Ag₂O and delta are the reactants of the Hoffmann reaction. This reaction breaks one C-N bond and forms a double liaison on this carbon. It is thus one of the pi liaisons that the nitrogen was bound. The second information is that the nitrogen is no more on the product, meaning that it had only one liaison with the chain. The second product of the reaction, N(CH₃)₃, confirms that. The two isomers are thus different from the carbon on which N(CH₃)₂ was bound. It can be the carbon in α or in β of carbonyl.

F→G: It is the same reaction than G->H but the nitrogen is still on the molecule after the reaction. It means that it was bound somewhere else on the molecule. The location is where the pi liaison stands. The molecule had thus a cycle of 6 atoms prior to the reaction. Contrarily to the species H, the cycle F is not symmetric. It is why we can obtain two isomers G₁ and G₂.

E→F: CH₂N₂ and MeOH/HCl are two techniques to replace a carboxylic acid by a methylic ester. The species E has thus two carboxylic acids.

D→E: KMnO₄ acts like the ozone. The two carboxylic acids were thus forming one bridge of the cycle. This bridge also forms a cycle of 6 atoms at the side of the nitrogen (and a cycle of 8 carbons with the other side).

C→D: If we check the compositions of the reactant C and of the product D, we see that there is a difference of H₂O. The role of H₂SO₄ was thus to remove this molecule of water from C with the formation of a double liaison. The hydroxyl group could be at two places (in α or β of the bridged carbons). At this point, we cannot say which position is correct but the reaction A->B is only possible with the hydroxyl group in β of the bridged carbon. The species is thus symmetric and achiral as is the species A.

A→C: A classic reaction of reduction. A gets H₂ in the process and we can assume that the hydroxyl group was a ketone before the reduction.

A→B: The base is there to remove a proton in α of carbonyl. Those protons are acid in reason of the tautomerism enol-ketone. The carbanion attacks one benzaldehyde on its carbonyl and water is lost after this attack.  The double liaison is in α of carbonyl and forms a long resonance chain with the phenyl. This reaction can be repeated on the other side of the carbonyl to obtain the product B.

4. There are a few specific reactions in this exercise (mainly BàE). You have the formulas of all the compounds but only the structure of the compound K to start with. HNO₃, ΔT is a reactant that breaks the C-C liaison of a ketone to obtain two carboxylic acids. The mechanism is unknown.

Correction

H→K: As explained in the wording, HNO₃ breaks a ketone into two acids.  As the two acids are present in the product, H is a cycle with a ketone on it. The cycle has 6 carbons.

H→I: This reaction leads to the formation of an oxime, i.e. a base of Schiff where R=OH. The mechanism involves a nucleophilic substitution by the nitrogen on the carbonyl to reject one water molecule.

I→J: This reaction is the transposition of Beckmann. Under acid conditions, the OH of the oxime is protonated and water is freed. The cationic nitrogen is attacked next by a carbon in alpha of the oxime, placing the nitrogen into the chain. The water comes back to attack the carbocation and to form an amide.

G→H: A simple reaction to change an ester into a carboxylic acid. The group is then removed from the molecule by an elevation of temperature.

F→G: A proton in α on carbonyl is taken by the strong base. The formed anion attacks next the ester to form a cycle of 6 carbons wearing one ketone and one methylic ester.

E→F: The carboxylic acids are replaced by methylic esters.

D→E: Ag₂O is able to oxidize an aldehyde into a carboxylic acid.

C→D: Pb(AcO)₄ is a compound that reacts with cis-glycols. Pb exchanges one equivalent of acetic acid to bind with one oxygen. The process is slow but it will also bind with the second OH to form a cycle of 5 atoms. This cycle breaks to generate 2 ketones that are now separated. The same mechanism is obtained with HIO₄. As there is only one product, the he reactant is thus a cyclic cis-glycol.

B→C: The osmium tetroxide is a reactant that affects specifically C=C and that generates an osmic ester, a bit like the complex formed by Pb(AcO)₄ with the cis-glycol. Na₂SO₃/H₂O removes the osmium from the molecule to obtain the cis-diol. KMnO₄ can do the same but we have to be in basic and cold conditions or we will obtain a diacid. The reactant is thus a cycle of 7 carbon with a double liaison between two of the carbons.

A→B: This reaction is simply the removal of a molecule of water from the cycle that gives a double liaison. We can deduce this from the difference of composition between the reactant and the product: C₇H₁₄O-C₇H₁₂=H₂O.

The post Chapter 2 : Organic chemistry – Exercises appeared first on BORZUYA UNIVERSITY.

The electrons of highest energy can be excited and reach orbitals of higher energies if they receive the exact amount of energy ΔE separating the two energy levels, in the form of heat of or photons. We say that the electron passes from its ground state to an excited state.

When an excited electron goes back to its ground state, it generates a photon with the same energy ΔE=hν. The frequency of the photon is characteristic of the atom or of the molecule from which it is emitted. The frequency ν, as a unit of measure, is the amount of times that a process occurs over a period of one second. In the case of photons, it is the amount of waves that passes by a point over one second (units: hertz). The wavelength λ=c/ν is the length of one wave, or the distance between two summits, in meter. The wavenumber n is the reciprocal of the wavelength (1/λ, units: cm^-1).  The difference between the frequency and the wavenumber is that wavenumber has nothing to do with the velocity of the wave. This makes it useful for situations where the velocity or frequency are not fixed.

For instance, when we heat up a metal, it changes of colour because the excited electrons emit photons within the range of the visible. The spectre of the visible is composed of all the colours that humans can see. Each colour corresponds to a photon of a given wavelength. The violet is the visible colour with the smallest wavelength (390nm) (or the largest wavenumber) and the red is the visible colour that has the largest wavelength (780nm) (or the smallest wavenumber). Ironically the colour that we see on a given material is not the colour that the material absorbs but all the other colours that haven’t been absorbed and that are thus reflected by the material towards our eye.

Radiations are not limited to the visible spectre. There are photons with wavelength above 780nm and below 390nm. Radiations with a wavenumber smaller than the visible spectre are called infrared (the red being the colour with the smallest wavenumber). If we continue to decrease the wavenumber (or increase the wavelength) we reach the microwaves and then the radio waves. On the other side of the visible spectre we find the ultraviolet radiations (the violet having the largest wavenumber of the visible spectre), then the X-rays and finally the γ-rays. As the energy of a photon is E=hν, the photons gain energy when we go from the radio waves towards the gamma rays. X-rays have enough energy to extract a deep electron (not an electron of valence) from an atom. No need to say the damages that gammas rays from radioactive elements can do.

The electronic transitions we discussed previously correspond to radiations in the UV/visible spectre. Those transitions are not the only transitions that can take place. Between each electronic level we find vibration levels and between the vibration levels we find rotational levels. The transitions between vibration levels are in the range of the IR while rotational transitions are in the range of the microwaves. These transitions lead to an emission of heat and that’s exactly how the microwaves oven works: the food is bombarded by radiations in the range of microwaves and the molecules of the food rotate, generating the heating of the food.

Spectroscopy

The absorption of radiations of a solution depends thus on its composition. We can whether measure how a solution absorbs a given radiation or how this solution emits photons when its molecules are excited.

Absorption spectroscopy

The absorbance A of a solution is its capacity to block/absorb radiations. The absorbance A(λ) depends on the wavelength of the radiation.

If we consider an incoming ray with an intensity I₀ that is heading towards a cell of length b containing a solution with a concentration c, the intensity I of the ray behind the cell is decreased by dI proportionally to the length of the cell (a longer cell absorbs more) and to the concentration of the solution (a more concentrated solution absorbs more).

E is the coefficient of absorbance of the solution that does not depends on the concentration nor on the size of the cell and dc is the concentration of solution through which the ray passed through, i.e. the difference of concentration between inside and outside the cell.  We can rewrite this expression as

and we can integer dI and dc between the emission of the ray and its detection, i.e. from I=I₀ to I=I and from c=0 to c=c. It gives

The absorbance is thus given by the law of Beer-Lambert

We can also define the transmittance that is T=I/I₀.

The absorbance is thus directly proportional to the concentration of the solution. Deviations from the linearity can be observed for concentrated solutions (c≠activity) or if some molecules of the sample are involved in an equilibrium.

The walls of the cell are made to let a maximum of the ray pass. They are usually in quartz (transparent between 200 and 380nm). A variation of the intensity of the ray can however be observed but can be determined using a white solution, i.e. a solution that only contain the solvent, with c=0 or a reference. The detector can be setup to define T=1 (or A=0) with the white solution. As the absorbance is directly proportional to the concentration of the solution we can determine the concentration of our sample if we measured the absorbance of our white solution and of an etalon solution the concentration of which is known.

If several species are in the solution, the absorbance is the sum of the individual absorbances.

To determine the concentrations, we need to repeat the experiment at several wavelengths. Don’t forget that epsilon depends on the wavelength.

We can thus determine c₁ and c₂ if the ε_i are known.

Experimental setup

The instrument is composed of 5 major parts:

• A stable source of radiant energy,
• A transparent device destined to contain the sample,
• A system to select a range of the spectre,
• A detector that can transform the luminous signal into an electric signal,
• A device to treat the signal.

The source of the ray has to be a monochromatic lamp, sending only photons with a given wavelength. It can be for instance a hollow cathode lamp or a laser source.  However, there is always a small range of wavelengths that is emitted. The impact can be limited if the wavelength is so that the ε coefficients don’t vary too much on this range of wavelengths.

The cell is usually in quartz and its size is usually 1cm long for spectroscopy of solutions. For gases the thickness is between 1 and 100mm but the ray can pass several times in the cell with the use of mirrors, giving optical paths up to 120m. It is also possible to allow a flow of gas.

The selection of the wavelength is done by a combination of prisms/mirrors and thin gaps. The prisms separate the wavelengths of the incident light and the gaps only allow a small range of wavelengths to pass toward the detector.

The detector can also introduce some error in the measure. The relative intensity of those errors is large when A→0 or when it A is large. Again, a good choice of wavelength limits the experimental errors.

Molecular spectroscopy

Another way to use the absorption properties of solutions is to sweep a range of wavelengths and to detect the absorbance. It requires a source of light able to change of wavelength. If there were only the electronic levels, we would obtain discrete rays corresponding to the exact energy of the transition between the ground state of the electron and one excited state. However, the rotational and vibrational sublevels transform the rays into bands. The bands are thinned if we perform the spectroscopy of gases. In the case of UV-visible absorption spectroscopy, this technique is mostly used quantitatively. For instance we can follow the evolution of a reaction through the spectre of absorbance. If the products and the reactants are absorbing at different wavelengths, the intensity of the peaks characteristic of the products will increase over time while the peaks of the reactants decrease in size.

Chromophore: it is an organic covalent unsaturated group that absorbs in the UV (ex: C=C, C=O, NO₂) with electrons that can be delocalised. The photons excite electrons that jump between energy levels that are extended π orbitals, created by a series of alternating single and double bonds, often in aromatic systems.

Auxochrome: A saturated group with free electrons that modifies the intensity and the wavelength of the absorption when bound to a chromophore.

Bathochromic shift: displacement of the absorption towards larger wavelengths (displacement towards the red) due to a solvent effect or a substitution.

Hypsochromic shift: displacement of the absorption towards smaller wavelengths (displacement towards the blue) due to a solvent effect or a substitution.

Hyperchromicity: increase of the intensity of the absorption.

Hypochromicity: decrease of the intensity of the absorption.

The post Chapter 12 : Spectroscopic methods appeared first on BORZUYA UNIVERSITY.

Instead of addition reactions, we observe reactions of substitution. The mechanism involves two steps.

• the first step is the electrophile attack of one pi liaison on an electrophile. As this step implies the loss of aromaticity of the substrate, this step is slow and is the determining step of the reaction. The product of this step is called an intermediate of Wheland or an arenium ion.
• the second step involves the removal of a proton from the cycle to regenerate the aromaticity. This step is fast.

Most of the electrophiles have to be activated to allow the first step of the reaction. It is done by a Lewis acid.

Halogenation

FeBr₃ and AlCl₃ can be used as Lewis acids to bind a halogen (Br and Cl respectively) on a benzene ring.

They enhance the electrophile character of the halogen that can now be attacked by the aromatic substrate.

FeBr₄^– acts like a base to take a proton and close the ring.

In the following table, we can find the energies of liaison involved in the process for the halogens. On the left, we have the energies for the reactants and in the middle column we have the energies for the products. The enthalpy of reaction is shown on the right column.

If we take a look at the variation of energy involved by a halogenation, we see that the enthalpy of reaction is positive for iodine. The reaction is thus not done. The reaction is highly exothermic for F₂. In fact, this reaction is explosive. For Cl and Br, we need to use catalysts (the Lewis acids).

Sulfonation and nitration

SO₃ and NO₂⁺ are electrophile enough to bind with an aromatic ring.

Nitration

The nitrate can be reduced selectively to obtain an amine.

Sulfonation

The sulphur atom is electrophile because of the electrocaptor character of the oxygen’s. However, the reaction is reversible in presence of water to form sulphuric acid. This process is exothermic and we should keep an eye on it.

We produce detergents from the benzenesulfonic acid.

And from that point we can produce sulfonamides that are usually good antibiotics.

The first of this kind was the Prontosil (4-[(2,4-Diaminophenyl)azo]benzenesulfonamide) in 1932, developed by Domagk at Bayer, who won a Nobel prize in medicine for it in 1939. The research program was designed to find dyes that might act as antibacterial drugs in the body. The discovery and development of this first sulfonamide drug opened a new era in medicine. Other examples of antibacterial agents follow:

Alkylation of Friedel-Crafts

This reaction, as its name mentions, allows to bind a chain on an aromatic ring. The chain that is to be added has to have an electrophile carbon. Again, we need a Lewis acid to do this reaction. The first step is the activation of the halogenoalkane by the Lewis acid.

Next, the cycle attacks the alkane, ejecting the halogenated Lewis acid. The product of this step is called an intermediate of Wheland or an arenium ion. This step is the determining step of the reaction, involving the opening of the ring due to the electrophile attack. A proton is taken by the halogenated Lewis acid to give the aromaticity back with the recovering of the Lewis acid and the liberation of an acid. This step is fast with regard to the previous step. As a global result, we added a new carbon chain on the phenyl.

The halogen on the added chain is not mandatory. An alcohol can do the job as well, or other precursors of carbocations.

The alkylation can be intramolecular if it results in a new cycle.

Limitations of the method

The alkyl group that has been added to the phenyl is an inductive donor group. It means that it gives some of its charge to the phenyl, increasing its ability to attack electrophilic carbons. As a result, the alkylation process does not stop after the addition of one chain if there is still places on the phenyl to greet new ones (we will see a bit later on which spot which group can be added).

A second limitation is the rearrangement of the carbocation (as usual).

Acylation of Friedel-Crafts

The difference between the alkylation and the acylation is that for the second one the electrophile is an acylium cation, leading to the addition of -C=O-R on the phenyl.

The acylium ion is obtained with the help on a Lewis acid

The rest of the process is identical.

However, there is no polyacylation because the carbonyl is a mesomeric and inductive captor, taking electrons from the cycle. It is possible to reduce the ketone with the reaction of Clemmensen or of Wolff-Kishner.

Regioselectivity

A benzene that already wears a group can be attacked on 3 non-equivalent positions: the ortho, meta and para positions. Some groups orient the reaction towards the ortho/para positions and other groups towards the meta position.

Mesomer effect

The mesomeric effect is the possibility for a heteroatom to share one lone pair with nearby atoms. For instance, O and N are mesomeric donors. They can stabilise carbocations by the sharing of one of their lone pairs that counterbalances the positive charge. They also enhance the nucleophile character of double liaisons. We note the mesomeric donors +M.

Groups can also be mesomeric acceptors and are noted –M. For instance the NO₂ group is a mesomeric acceptor because it has a resonance form that can keep the additional charge.

Note that the oxygen can be mesomeric donor or acceptor depending on the group it is in and of the structure of the molecule.

Inductive effect

When two atoms are of different electronegativity, there is a displacement of electrons from the least electronegative atom to the most electronegative one. Groups or atoms can thus take or give some electrons from or to the nearby atoms. Atoms or groups that take electrons from the rest of the molecule are inductive acceptors and are noted –I. They destabilise carbocations and positive charge and can enhance the electrophile character of the atom they are bond with. For instance, the carbon of a carbonyl give a part of its charge to the oxygen and wears a partial positive charge δ⁺. A nucleophile is thus prone to attack this carbon.

An atom that is less electronegative than the carbon will give a part of its charge to the chain. They are called inductive donors and noted +I. The simplest inductive donor is H that is less electronegative than the carbon. However, H is taken by convention as the neutral intensity of the inductive effect. –CH₃ is an inductive donor because the inductive effect of the 3 hydrogen’s is transmitted through the carbonic chain (over 2-3 carbons). As a result, the alkyl groups are inductive donors. The effect is stronger for -C(CH₃)₃ than for –CH₃ because the amount of hydrogen atoms that can share their electrons.

The hybridation of the carbon is something to be taken into account to determine the inductive effect. The electrons in the s orbitals are more bound to the atom than the ones from the p orbitals. As a result, the electronegativity of sp² carbons is slightly higher than the electronegativity of sp³ carbons.

Regioselectivity in function of the substituents of aromatic rings

The mesomeric and inductive effects are cumulative and a nitrogen in a NH₂ group is simultaneously a mesomeric donor and inductive acceptor (+I-M). The mesomeric effect always prevails in intensity.

• inductive donors

The groups that are donors of electron by hyperconjugaison (or inductive donors) are activating, i.e. favour the addition of new groups on the carbon. The substitution is oriented towards the ortho and para positions because the carbocation which has been made during the first step of the substitution can be stabilised by the inductive donor.

It is not possible to place the positive charge on the spot of the group if the substitution was to be made on the meta position. Usually, the para position is favoured to the ortho position because of the steric hindrance. However some groups such as NO₂ seem to show more interest on the ortho position.

• inductive captors

Inductive electrocaptors are deactivating and orientate the reaction in meta. The reason is that in the other two positions, the carbocation can be at the feet of the CF₃ group that wants to take more electrons, destabilizing furthermore the carbocation.

• mesomeric donors

Mesomeric electrodonors are activating and orientate the reaction in ortho/para because there is an additional resonance form.

Note that NH₂, as well as the oxygen from the ether, is a mesomeric donor and an inductive captor. The mesomeric effect is always more important than the inductive one.

• mesomeric captors

Groups that are electrocaptor by resonance are deactivating and orient the reaction in meta.

• halogens

Halogens are deactivating but orient the reaction in ortho and para.

The steric hindrance here is very important and explain the difference of population between the ortho and para reactions.

If there were two groups on the phenyl, the effects are additive and the substitution is made on the most activated/ less deactivated positions.

Both methyl groups of the xylene (dimehtylbenzene) shown above are activating and orientate in ortho/para. As both groups are activating, this reactant is more reactive than the toluene. The positions 2, 4 et 6 are thus favoured. The positions 4 and 6 are slightly more probable because there is less steric hindrance. In the case of the following xylene, the slots are almost equivalent.

In the case of deactivating groups, the same reflexion is made. The COOH’s orientate in meta and the position 5 is the least deactivated position.

Nucleophilic substitutions

The nucleophilic substitution on aromatic rings is slower than the electrophilic substitution and that substitutions on sp³ carbons.

The reason is that the cycle is already full of electrons. Moreover, a sp² carbon is more electronegative than a sp³ carbon. It is thus difficult to add a nucleophile (that loves positive charges) on a phenyl and the presence of a captor group on the phenyl is required to stabilise the carbanion. The leaving group has to be a good one.

Several mechanism are possible to add a nucleophile on an aromatic substrate. One of them is the Addition-Elimination.

The activating group (NO₂ here), is necessary to stabilise the negative charge of the carbanion formed during the first step of the reaction which is the slowest one. The intermediate complex is called intermediate complex of Meisenheimer.

The halogen (as leaving group) reactivity sequence is opposite to the one for SN₂. On an aliphatic chain, the cleavage of the C-X liaison is made during the determining step of the SN₂. On an aromatic, the cleavage doesn’t take place during the determining step. Moreover, the halogens take electrons from the cycle and the small halogens are thus more reactive than the big ones. As a result, we have the following reactivity sequences:

Aliphatic: F<<Cl<Br<I

Aromatic: F>>Cl>Br>I

A heterogeneous atom in the cycle (O or N) can play the role of the captor group.

An application of this mechanism is the determination of the terminal amino acid of peptides. The peptide reacts with an aromatic compound through its terminal amine group.

Once the two substrates are bound together, we hydrolyse the peptide bonds (i.e. the amides). All the amino acids are now separated but only one of them is bound to the aromatic group.

SN₁ mechanism

This mechanism is mostly used to produce salts of arenediazonium. This species is obtained from aniline C₆H₇N with NaNO₂ in an acidic environment.

The mechanism is the following:

Arenediazonium salts are stables at low temperatures and lose N₂ at higher temperatures, freeing a phenyl cation that can react with a nucleophile.

Other halogens than iodine don’t give good results because of secondary reactions. To add Cl, Br or F on an aromatic cycle, we use the Sandmeyer reaction, using copper salts like CuCl, CuBr or CuCN. The mechanism is a bit more complex and involves radicals.

Mechanism involving a benzyne

Normally, halogenoarenes cannot make usual SN₂ or SN₁ reactions. However, in very harsh conditions of temperature and of pH, it is possible to force such reactions.

The mechanism involves a triple liaison in the cycle, the species being called a benzyne, The existence of this intermediate has been shown by an isotopic labelling. The C bound to the halogen is an isotope and we observe a racemic melange as product:

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A carboxylic acid is more acid than the corresponding alcohol because the conjugated base, the carboxylate, is stabilised by resonance (pKa≈5 vs pKa≈17).

The oxygen of the carbonyl is more basic than the other one because the positive charge is more stabilised.

Synthesis of carboxylic acids

There are several ways to form carboxylic acids:

• by oxidation:

Carboxylic acids can be made by the oxidation of aldehydes or of alcohols. We can use CrO₃ or KMnO₄ as oxidants for instance.

• by Grignard

A reactant of Grignard can react with CO₂ and then with water to form a carboxylic acid

• from nitriles

In acid environments, nitriles react with water to generate new acids and free a molecule of ammoniac.

This method can be used to extend a carbonic chain. Once a carboxylic acid is obtained, we reduce it to obtain an alcohol. The –OH group is removed by nucleophilic substitution and the whole loop restarts. Each loop adds one carbon to the chain.

The derivatives of carboxylic acids are

Reaction of Addition-Elimination

Carboxylic acids and their derivatives are attacked by nucleophiles on the carbon, as it is done for ketones and aldehydes. However, this addition is followed by an elimination.

It is thus the substitution of one group by the nucleophile at the condition that the group L that was previously on the carbonyl is a better leaving group than the nucleophile.

Cl^–>RCOO^–>RO^–>OH^–>NR₂^–>C^–

As for ketones and aldehydes, the reaction can be catalysed by an acid or a base.

Note that the addition-elimination reaction for carboxylic acids is in competition with acid-base reactions. OH^– is not a very good leaving group and if the nucleophile is a strong base, the carboxylate is formed. In the case of a nucleophile which is not a good base, there is a competition between the AE and the AB reactions.

Formation of the acyl chloride

From what we wrote just above, we cannot replace RCOOH by RCOCl because Cl^– is a better leaving group than OH^–. However, there is a trick to make this reaction possible: we use thionyl chloride (SOCl₂). The oxygen bound to the hydrogen in COOH is a nucleophile and can attack the sulphur atom, expelling a chlorine that catches the proton on its way. The newly formed chlorydric acid can now react with the thioester, eliminating a molecule of SO₂ and of HCl.

The same kind of mechanism is used to produce acid bromine from carboxylic acids and PBr₃ (phosphorous tribromide). Yet, those reactions don’t work with HCOOH because the products HCOCl and HCOBr are unstable. They decompose in carbon monoxide and the corresponding acid.

Formation of esters

Carboxylic acids react with alcohols to produce esters.

Strong acids catalyse this reaction.

RO^– is a slightly better leaving group than OH^– so we want to displace the equilibrium towards the right by the use of an excess of alcohol or carboxylic acid and by the removal of water from the system. If we want to do the hydrolysis of an ester, i.e. the reverse reaction, we use an excess of water. It was shown that the hydrolysis of esters works this way (attack of the water on the carbonyl and not an attack of the water on R’) by the use of isotopes for the esteric oxygen.

Another way to obtain esters is the use of diazomethane, reaction that we have seen at the beginning of the elimination chapter.

Intramolecular esterifications are more favourable than intermolecular ones for entropic reasons and are especially favourable if a cycle of 5 or 6 atoms is formed.

This kind of cycle is called a lactone (δ-lactone if a cycle of 6 atoms, γ-lactone if a cycle of 5, etc.). Lactones are sensible to the hydrolysis.

Formation of amides

Under normal conditions, an amine reacts with a carboxylic acid to form a salt (ammonium carboxylate) through an acid-base reaction. Once the amine has taken the proton, it cannot attack the carbonyl anymore.

If we heat up the system, the reaction goes back and another process, thermodynamically more favourable occurs, leading to the formation of an amide. In an amide, the nitrogen is not basic.

Intramolecular reactions are possible and lead to the formation of lactams.

Reactions of carboxylic acids and their derivatives

We have shown several reactions of the carboxylic acids. It is often interesting to consider to transform the carboxylic acid into an acyl chloride prior to the desired reaction. Acyl chloride are usually more reactive and this additional step in the process can increase the global yield of the reaction.

Reactions of acyl chlorides

From the acyl chloride, we can obtain many other functional groups.

Reactions of esters

As ester are less reactive than the acyl chlorides, less reactions are available. We can hydrolyse the ester, exchange the esteric chain with another one or generate an amide.

An additional reaction is to use a Grignard reactant on an ester to obtain a tertiary alcohol. The ketone is more reactive than the ester so the reaction continues.

Reactions of amides

Peptide bonds can be hydrolysed.

Reactions of nitriles

The carbon atom from a nitrile is at the same oxidation state than the one of the carboxylic acid. A nitrile in presence of water and acid leads to the formation of a carboxylic acid and the rejection of ammoniac.

Nitriles can also react with reactants of Grignard but the reaction stops at the formation of ketones.

Nitriles can also be reduced into primary amines by LiAlH₄.

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Its pKa is around 20 (for comparison, the hydrogen of a –C-H has a pKa of 40, a =C-H a pKa of 37 and –O-H a pKa of 18). The result is the enolate ion. It is stabilised by resonance with the carbonyl.

Tautomerization

The enolate ion can react by its oxygen or the carbon in α.

Enols are usually unstable and tautomerize into the corresponding carbonyl. It is the enol-ketone tautomerism. The tautomerization is done by acidic or basic catalyse.

The difference of energy is about ΔG=15kcal in favour or the ketone form. The enol-ketone equilibrium can be displaced if one form is stabilised by the structure of the molecule.

The tautomerization allows to modify the conformation of the molecule. A cis molecule tautomerizes to become trans and decrease the steric hindrance.

We can also exchange hydrogen’s for its isotopes because of the tautomerism.

The chemical properties of isotopes are almost identical (because given by the electrons), but some physical properties can be different between isotopes of the same element. The speed of reaction and the ebullition temperature are two examples of properties that change depending of the isotope. It is thus possible to separate the species wearing isotopes by distillation or by centrifugation.

Halogenation

The hydrogen can also be exchanged with halogens through acidic or basic catalyse. In this case, the results are different depending on which catalyst we use.

– Acid catalysis:

Once one halogen is on the molecule, it is harder to form the enol because it is more difficult for C=O to catch a proton due to the inductive captor effect of the halogen.

– Basic catalysis:

During the basic catalyse, we don’t have to protonate the carbonyl and the halogenation can take place as many times as there are α protons.

Alkylation

We have seen in the previous section that NaH alone doesn’t reduce carbonyls. However it has an effect on the carbon in α of carbonyls. It is a good way to introduce an alkyl group at this position.

A strong base can also take the proton α of carbonyl.

One problem is that the reaction can continue if there are several hydrogen’s in α of carbonyl and polyalkylation is frequent, leading to a racemic mix of products.

Moreover, the regioselectivity is low, giving an additional possibility of product.

To avoid polyalkylation, we can replace the oxygen by pyrrolidine (C₄H₉N) to obtain the corresponding enamine that has the advantage of a low polyalkylation.

This protection is also effective for aldehydes that are subject to the aldol condensation reaction in basic environments.

Aldol condensation

If we add a few NaOH in a solution of acetaldehyde, dimers are formed. The first step is the deprotonation in α of carbonyl by the base to obtain the enolate ion that attacks next another aldehyde, leading to the formation of a β-hydroxyaldehyde. The β-hydroxyaldehyde is stabilised by its internal hydrogen bond.

If we heat up the solution, the molecule loses a water molecule to form an enone (double liaison conjugated with a carbonyl).

Condensation of Claisen

The analogue reaction for esters is called the condensation of Claisen. NaOH is not used in the present case because it would cleave the ester. We can use sodium ethoxide instead for instance to obtain the enolate ion.

As for the aldol condensation, the enolate ion attacks another ester to obtain β-ketoesters.

The condensation of Claisen can be made with two different esters (mixt condensation of Claisen) or between two esters in the same molecule (intramolecular condensation of Claisen).

The intramolecular reaction is favoured in comparison to the intermolecular one because it does not decrease the entropy.

Properties of α,β unsaturated aldehydes and ketones

Such species are stabilised by resonance.

They are way more stable than β,γ-unsaturated aldehydes/ketones that rearrange themselves into the α,β unsaturated compound in presence of a base.

Obviously, the double liaison and the carbonyl groups can go through reactions independently, such as addition reactions for example.

But additions can also be made in several cases on the conjugated system. Those are 1,4 additions.

The nucleophilic attack is possible on the carbon of the double liaison because it is depleted by the conjugation with the carbonyl.

The Michael’s addition is the 1,4 addition of an enolate ion on an enone:

A Michael’s addition can be followed by an intramolecular aldol condensation, the reaction is known as the annulation of Robinson.

The post Chapter 9 : carbanions in α of carbonyl appeared first on BORZUYA UNIVERSITY.

The two fragments A and B that are removed can be removed from the same carbon, in which case we talk about a 1,1 elimination, from two adjacent carbons (1,2 elimination or beta elimination) or two carbons separated by one carbon (1,3 elimination). The 1,2 elimination is the most common one.

Elimination 1,1

The two fragments are removed from one single carbon. It leads to the formation of a higly reactive carbene.

This reaction is possible because the chlorine atoms take the electrons from the carbon. Because of that, the C-H bond is destabilised and a strong base can remove the proton.

The carbon of the carbene is a singlet and is less stable than a triplet. It will thus react quickly with double liaisons to obtain a cyclopropane chain. A triplet would not be able to attack the π liaison.

The reaction is stereoselective (on the same side of the plane).

The smallest carbene is the methylene that can be produced from diazomethane (check nom anglais). The diazomethane is toxic and explosive so the reaction has to be controlled.

It can also be used to produce methyl esters.

Elimination 1,2

This kind of elimination leads to the formation of new π liaisons. It is the opposite of a reaction of addition. We will only discuss the main reaction of β elimination but there can be other mechanisms involved to obtain such a reaction.

Similarly to the substitutions, where there is a mechanism with a kinetic of order 1 (SN₁) and one of order 2 (SN₂), there is an elimination of order 1 (E₁) and of order 2 (E₂).

E₁ mechanism

The first step of the reaction is the departure of one anion and the formation of a carbocation. This step is slow and is determining the speed of the reaction.

The second step is the capture of a proton on the adjacent carbon by a base. It leads to the formation of the π liaison to have a neutral species.

We can know if an elimination is of order 1 from

• the order of the kinetic of reaction (order 1).
• the isotopic effect: if the hydrogen is replaced by a deuterium, we don’t see any modification of the speed of reaction. It is so because the hydrogen doesn’t take part in the determining step. If it was involved, the speed would be divided by 5 to 8.

• the stability of the carbocation: the stability of the carbocation (degree of substitution or hyperconjugaison) affects the speed of reaction.
• rearrangements: explained by the existence of a carbocation.

E₂ mechanism

The kinetics is of order 2 and involves a reaction in one step, where the base takes the proton and the departure of the leaving group are simultaneous.

This reaction is only made if the proton and the leaving group are in anti positions.

The argument in favour of the E₂ mechanism are

• a kinetic of order 2
• a huge isotopic effect
• no rearranging
• huge influence of the energy of the C-X bond (X^– is the leaving group)

Competition between the mechanisms

As for the nucleophilic substitutions, there is a competition between the two mechanisms. It is not always 100% of E₁ but it can be a mix of both, eventually leading to a racemic melange of products. The strength of the base and the steric hindrance influence which elimination mechanism is used: a strong base favours the E₂ mechanism while a weak base and steric hindrance favours the E₁ mechanism.

Now, the molecules with a good leaving group can often also be the target of a substitution reaction in presence of nucleophiles. There is thus a competition between E₁, E₂, SN₁, SN₂ and eventually other reactions. The three main parameters to consider to determine which reactions should occur are the strength of the base, its nucleophilicity and the steric hindrance.

Nucleophiles that are bad nucleophiles have good yields with SN₂ with primary and secondary substrates. If the steric hindrance is larger, the main product becomes the SN₁ product. If the nucleophile is a strong base, the main product is the SN₂ but if the steric hindrance increases, the elimination becomes more important.

Nucleophiles that are strong bases and that are hindered favour the elimination.

To resume:

Regiochemistry of the elimination

When the carbon wearing the leaving group is not at one extremity of a chain, there can be several possible products. We can predict which hydrogen will be removed.

• If the generated double liaison can be conjugated with other π liaisons, those products will be favoured.

• In the case of bridged substrates, the product never involves a π liaison on a bridgehead carbon.

•  For E₁ eliminations: the elimination is determined by the stability of the formed olefins. The most substituted C=C is thus favoured. It is the rule of Zaitsev.

– There can be exceptions because of steric hindrance

• for E₂ eliminations, one H in anti is required
  • the rule of Zaitsev is followed if the leaving group is not charged (most substituted C=C)
    • 
  • it is the opposite it the leaving group is charged: it is the rule of Hofmann
    • in such an elimination, it is the most acidic proton that is taken by the base, i.e. the most substituted carbon because it is stabilised by the mesomeric donor effect of alkyl chains.
      • it can be a good way to determine the position of a nitrogen in a molecule, if it is a primary, secondary, etc.

Configuration of the product

In the case of one E₂ elimination, the reaction is stereospecific and there is thus only one possible configuration.

In the case of one E1 elimination, the product is more stable if the voluminous substituents are in trans.

Dehydratation of alcohols

This elimination does not occur in basic or neutral environments: OH^– is a bad leaving group and we would form –O^– in presence of a strong base.

In an acidic environment, the elimination can be done by E₁ or by E₂. The E₁ is favoured in the case of substituted products and the E₂ is the main mechanism for primary carbons.

Reactivity: tertiary>secondary> primary

E₁            E₁                E₂

Fragmentation of 1,3 diols

It is one case of 1,3 elimination during which the molecule is cleaved

The post Chapter 8 : Reactions of elimination appeared first on BORZUYA UNIVERSITY.

 As two carbons are involved in the reaction, the addition of the new group can give several products, with more or less (stereo)selectivity. Moreover, the transformation into sp³ carbons also introduces a regioselectivity that often has to be considered.

Consequently, we will discuss in this section the methods and the rules that have to be followed to obtain the desired product during the addition of an electrophile group on double or triple liaisons.

Catalytic hydrogenation

It is, in the principle, the easiest addition on a C=C liaison because there is no new group on the product. The sp² carbons are reduced into sp³ carbons. Yet, this reaction is not spontaneous despite the fact that it is an exothermic reaction. The activation energy to break the H-H liaison is huge and a catalyst is required to perform the reaction.

As catalyst, we can use platinum (Pt), Raney’s Nickel (Ni) and Palladium on C (Pd/C).

The have the ability to dissociate the hydrogen atoms and to fix them on their surface, making them available for the addition on the alkene. As detailed in the kinetics section, the catalyst decreases the energy of activation of the reaction and the process can be depicted in 4 steps:

– approach and binding of the reactants

– displacement on the surface

– reaction(s)

– departure of the products

At the first step, the reactants take place on the surface of the heterogeneous catalyst. The amount of reactant that can take place on the catalyst depends on its surface. It is thus advised to use small particles of catalyst to increase the ratio surface/volume (or surface/mass).  The alkene places itself parallel to the surface, in the direction of the π liaison.

There is no reason that the reactants immediately bind on nearby spots. The second step is thus that the reactants are moving on the surface. They will eventually be in contact with each other so that the reaction can take place.

The third step (and eventually the next ones) is the reaction between the reactants. One activated hydrogen attacks one sp² carbon while the π liaison is now connecting the second carbon with the platinum. The next step of the reaction is that the liaison between the carbon and the platinum attacks the remaining hydrogen.

As last step, the alkane leaves the surface of the catalyst, letting space for reactants to bind.

The hydrogenation is stereoselective, i.e. the place of the groups on the carbons are not equivalent: at the level of the C=C liaison, the molecule is planar.  If the hydrogen atoms are fixed on the same side (but not on the same carbon) of the molecule. It is a syn addition. As the added groups are hydrogen’s, the product is thus cis (see nucleophilic substitutions about cis and trans products).

If the groups were added on the opposite sides of the molecules (still not on the same carbon), the addition would be anti and the product would be trans. This product is not observed because the reactants are bound to the catalyst and cannot move freely during the reaction. Both hydrogen’s are thus on the same side of the molecule before the reaction and after the reaction.

This is of importance only if, after the reaction, the newly sp³ carbon cannot rotate. It will be the case when the liaison is involved in a cycle or if there are voluminous groups. Moreover, the fact that we obtain only the cis product does not mean that only one product is possible. The reason is that there is one plane at each side of the π liaison (for a total of 2). If the two planes are equivalent, we will obtain a 50-50 racemic mix of two cis products.

If the planes are not equivalent, one product is favoured: the hydrogen’s are fixed on the least occupied side because the alkene places itself on the catalyst so that the voluminous groups are far from the catalyst.

Soluble catalysts can also be used for this reaction. It is the case for the Wilkinson’s catalyst chloro(tris(triphenylphosphine)rhodium(I).

It also lead to the cis products.

In terms of reactivity, the substituted carbons react slower than carbons with less groups because of the steric hindrance: the hydrogen’s are not very mobile on the surface of the catalyst and the π liaison has to be nearby to obtain the hydrogenation.

The alkynes also react faster than alkenes, allowing us to hydrogenate all the alkynes into alkenes before the formation of alkanes. We can thus limit the hydrogenation to the alkenes if we use a bad catalyst and put one equivalent of H₂ by alkyne. We will obtain the alkane with 2 equivalents of H₂ by alkyne.

Again, it will lead to the cis alkene. To obtain the trans alkene, we use sodium in ammoniac. The sodium generates a radical on one carbon and a negative charge on the other one.

The trans product is favoured for steric reasons.

Aromatic products can difficultly be hydrogenated with H₂/cat (at 150°C) but it can also be reduced differently by the reduction of Birch, involving Na, NH₃ and an alcohol.

Other electrophile additions

Addition of HX

The other reactants HX containing a hydrogen atom can more easily be broken (or are already ionic) and do not require a catalyst. In the case of a halogen acid, because of their nucleophilic character, the π liaisons which are weaker than σ liaisons tend to attack electrophile molecules. In this case, the electrophile target is the proton. It is called an electrophile attack. The nucleophile X^– attacks the positive carbon (or carbocation) afterwards.

This reaction does not involve stereoselectivity because there are two distinct steps but involves regioselectivity: it is the rule of Markovnikov that says that the halogen is fixed on the most substituted carbon, or inversely that the proton is fixed on the least substituted carbon. Personally, I prefer to remember that the most stable carbocation is formed. Remember that carbocations can rearrange the molecule to form the most stable carbocation.

Addition of water

Water is added to alkenes in presence of sulphuric acid.

The sulphuric acid is chosen because the sulphate is less nucleophilic than the water because the charge is stabilised by resonance. HCl has to be avoided because Cl^– is more nucleophilic than H₂O. Moreover, all the steps are reversible. It means that the newly generated alcohol can be dehydrated in the case of an excess of acid.

The presence of a carbocation explains some rearrangements of alkenes that can be observed in acidic solvents.

Addition of X₂

Halogen’s are added to C=C to obtain trans products.

It is strange that something as charged in electrons as halogens are used as electrophiles for this reaction. The reason is that the liaison between the halogens is polarisable. A halogen will thus feel the electrons of the π liaison and transfer part of its electrons to the other one. The partially positive halogen can thus be attacked by the π liaison and forms an intermediate halonium that will be attacked by the negative halogen on the other side of the C-C liaison.

This reaction is stereoselective (always trans) and stereospecific (the result depends on the arrangement of the reactant). In presence of other nucleophiles, such as water, the reaction begins normally but there is a competition between X^– and the nucleophile.

This reaction is slower on alkynes because the intermediate halonium is less stable. The halogenation is thus hard to control: the alkenes will react faster than the alkynes.

The oxymercuration reaction

This reaction is a useful way to form alcohols or ethers from alkenes. It involves mercurial acetate and water/alcohol in THF as solvent. The nucleophilic opening of the water/alcohol is done in anti and on the most substituted carbon (so there is more space for the mercurial acetate).

The oxymercuration follows the Markovnikov rule but there is no carbocation involved in the process. As all the additions, it is regioselective.

Water leads to the formation of an alcohol and we can use an alcohol instead of the water to obtain an ether.

Hydroboration reaction

BH₃ (borane) is a Lewis acid that is stable under the form of a dimer and can replace its 3 hydrogen by 3 carbon chains in THF (tetrahydrofuran (CH₂)₄O). Once the chains are bound, it is possible to remove the boron by an oxidation with H₂O, NaOH and H₂O₂ to obtain as many alcohols.

This reaction does not follow the Markovnikov rule (anti-Markovnikov) and is stereoselective. The mechanism of reaction involves the empty p orbital of the boron that interacts with the π liaison to form a complex. This complex allows a transition state with 4 centres wherein the liaisons move.

The Markovnikov rule applies when the hydrogen is positively charged. In the present case, we add H^– and BH₂⁺.

The removal of the boron is made with retention of configuration and is stereospecific. The boron is less electronegative than the carbon and is thus giving its electrons. NaOH takes a proton from the H₂0₂ to form HOO^– that attacks the boron, transferring its negative charge. There is a rearrangement occurring that puts an oxygen between the chain and the boron.

The water and OH^– take action next to remove the boron.

This addition can also be done on alkynes to form aldehydes through tautomerization.

Reaction of alkenes with peroxycarboxylic acids

The peroxycarboxylic acids are generally unstable. The MCPBA and the MMPP are however stable peroxyacids vastly used in laboratory.

One of the oxygen of the peroxycarboxylic acids is electrophile. Those acids can thus react with alkenes to form epoxides and the carboxylic acid.

The addition is thus a concerted reaction, is stereospecific and is syn. The reaction is made on the less hindered plane of the alkene. The reactivity increases with the amount of substituents that share their electrons with the π liaison (alkyl chains for example).

Ozonolyse

The ozone can break double liaisons. As a first step, it opens the double liaison to give the intermediate called molozonide.

It breaks to form ozonide that can next be oxidized or reduced.

The following table gives the products of the ozonolyse as a function of the substituents of the double liaison.

The ozonolyse is thus a good way to obtain ketones and carboxylic acids from alkenes. The method is also useful to determine to position of a double liaison in a molecule when we cannot easily determine it with usual techniques.

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Polymers are derivatives from petroleum and are made of small blocks called monomers. Monomers are small molecules, usually gaseous, wearing a functional group that can give some properties to the polymer. The monomers composing a polymer are added one by one to the polymer that grows in length. We will see the reaction a bit later. Due to their huge size, the polymers are solid. To write the formula of a polymer, we write the smallest constituting block between brackets, that block being present n times in the polymer. Note that for one reactor, the lengths of the produced polymers are not all identical. We will obtain a distribution of size and n is thus not a fixed value.

Here are some of the usual polymers and the monomers they are made of.

The properties of a polymeric material are given by the interactions between the chains of polymer. Chains can be linear of branched. It is possible to make polymers with several different monomers. There are several ways to obtain such polymers. The monomers in the chain can be put

As written above, the properties of the polymers depend on the interactions between the chains. If there is no interaction at all, the chains freely move along each other, without resistance and if we stretch the material by two opposite sides, it will expend and break. The same happens if chains have physical attractions between each other. Those can easily be broken by a heating or by force. This kind of material is called thermoplastic.

Liaisons can be made between the chains. If there are only a few of them, when we stretch the material it will expend up to a point where the liaisons block any further movement. They prevent the material to break as easily as a thermoplastic one. Such a material is called an elastomer.

If there are more than a few liaisons, the material becomes more rigid and the rigidity increases when we heat it up. It is called a heat-hardening material.

Stress-strain curve

The elasticity/rigidity of polymers are determined by a stress-strain curve. A machine stretches the material while the applied force and the stretching (or strain) of the material are measured. The material should have the geometry of a cylinder to avoid problems related to symmetry. Instead of force, we measure the pressure (or stress) R=F/S where S is the surface of the base of the cylinder.

In a first time, the deformation is elastic and the variation of the length of the sample ΔL is directly proportional to the applied force F. The slope of this curve gives the module of Young E that is a characteristic of the elasticity (in the given direction) of the material. If the force was to be removed, the material would get back to its initial form. It is thus a reversible process.

Next, the curve softens and the deformation is no more an elastic deformation but a plastic deformation: the material is deformed by the applied force and cannot get back to the initial state. The variation of slope can be slight or massive. This region is called the strain hardening region and the point where the slope changes is called the elasticity limit R_e. The curve should rise up to a maximum R_m called the ultimate tensile strength. Yet all the nodes don’t always break at the same time and we can see the apparition of plateaux on the curve. At R_m, the liaisons binding the polymers are all broken. The experiment continues until the material is broken in two distinct pieces.

The ductility refers to the way a material breaks. Heat-hardening materials breaks at once, after a short elastic region (with a large Young’s module E): because of all the liaisons, the material is not flexible but when the stress is large enough, the material breaks directly in two pieces. In the case of ductile materials, the sample becomes thinner and longer before the fracture. As a result, the curve continues further after R_m.

The parameters that influence the properties of polymers are

• The temperature: by the interactions between the chains. At some point, called the temperature of glass transition T_G, the properties of the polymer change drastically. The temperature of glass transition T_G is an interval of temperature characteristic, for a given material, of the transition between a solid and breakable state and a flexible and elastic state. T_G depends on all the following parameters and on the pressure.
• Crystallinity: it refers to the order in the solid. The interactions and the density in the polymer are not identical everywhere in the material. The crystalline areas are areas where the chains are well packed and have strong interactions. The chains in amorphous areas are not well packed. One single chain can be simultaneously in an amorphous area and in a crystalline area.

• Length of the chains: it increases the amount of interactions
• Monomer(s): the group that the monomer wears influences the interactions between the chains
• Reticulation: it is the amount of chemical liaisons that we make in the material
• Speed of deformation: an elastic material can break if we stretch it fast enough
• Additives: we can add some molecules in the material to bind the chain together, or to avoid it to happen.

Synthesis

There are several ways to make polymers. We can subdivide them in two types: the polyaddition and the polycondensation. The difference is that there is that the Polyaddition only generate the polymer while the polycondensation also generate small molecules.

Polyaddition

It can be subdivided in two big categories: radical polymerisation (uncharged polymerisation) and ionic polymerisation (anionic, cationic or coordinative polymerisation). They are all similar in the process and we will only show the radical polymerisation here.

The synthesis is composed of 3 main steps: the initiation, the propagation and the termination step.

Initiation step

An initiator is mixed with the monomers. In the case of the radical polymerisation, the initiator is a radical. This radical can be generated by thermal decomposition or by photolysis.

For instance, benzoyl peroxide (BPO) can be homolytically cleaved by an elevation of temperature or by a laser of a given frequency.

In fact, the produced radical rearranges to free CO₂.

The radicals react with one monomer, transferring the radical to the other side of the molecule.

From that point, we have a beginning of chain that wears a radical and that is able to attack monomers. It is the propagation step:

As there is still a radical at the end of this reaction, it goes on and on, increasing the length of the chain by one monomer each time the reaction takes place. A chain stops to grow when the radical is neutralised. It is the termination step. During this step, two growing chains react together. It can happen two ways:

Termination: the two chains merge together

Disproportionation: the ends of the chains wearing the radical have a redox reaction together. As a reminder, the usual disproportionation reaction is a specific type of redox reaction in which a species is simultaneously reduced and oxidised to form two different products.

In the case of polymers:

Polycondentation

The process is quite different from the polyaddition. In this case, the monomers wear a functional group at both extremities. The functional groups can react with the ones of the other monomers to condensate together. For instance, a carboxylic acid condensates with ah alcohol to form an ester and water.

If the alcohol was replaced by an amine group, we would obtain an amide that is also the usual bound between peptides and thus called a peptide bond.

The production of polymers is an exothermic process: we create new bonds. However it decreases the disorder as there are less molecules in the system. As a result, the production of polymers has to be performed in a given range of temperatures. The temperature can be approximated from the type and the concentration of monomers. Indeed, except at the very beginning of the reaction, we have that the concentration of two polymers of similar sizes are approximately equal.

The post Chapter 6: Polymers appeared first on BORZUYA UNIVERSITY.

Kinetics is a field of the chemistry that studies the speed of reactions. The speed of a reaction can depend on the conditions of the reaction. For instance, when we put H₂(g) and O₂(g) we don’t produce water spontaneously.

This reaction has a very negative ΔG⁰. However, if we produce a spark, the reaction will go very fast.

We have seen that the temperature has an influence on the equilibrium of reactions. At SCTP, the formation of ammonia has a negative ΔG⁰ but it becomes positive for temperatures larger than 434K.

Without temperature variation one reaction can be speeded up if we use a catalyst. We will discuss catalysts later in this chapter.

It can thus be important to know the speed of reactions in function of several parameters in order to optimise the production of one particular molecule, the destruction of some waste or any other process.

Speed of reaction

The speed of a reaction is the variation of concentration of the involved species in function of the time. It is always positive. As there can be several different stoichiometric coefficients, we have to agree on the definition of the speed of a reaction because the concentrations don’t vary the same way. To obtain this speed, we divide the variation of concentration over time of the corresponding species by the stoichiometric coefficient. In the reaction

we consider only one speed of reaction which is

The concentrations of reactants decrease over time and the speed is of the opposite sign than the variation of their concentration.

During the reaction, the speed depends on the concentrations of all the involved species.

The coefficients α, β and γ are the order of the reaction with regard to the corresponding species. α is the order of reaction with regard to NO₂, β to NO and γ to O₂. The order of the reaction with regard to one species is often their stoichiometric coefficient but not systematically. The sum α+β+γ is the global order of the reaction.

At the equilibrium the speed of the reaction in one direction is equal to the speed of the reaction in the other direction.

It is not true to say that v=0. The two reactions have the same speed.

At the beginning of the reaction, we can neglect the concentrations of the products in the expression of the speed with β=γ=0. The speed at the beginning of the reaction is thus simply

Integral method

We can determine the order of the reaction with regard to one species by keeping constant the concentrations of the other reactants. Basically, we put a large excess of reactants except for the species that we want to study. This way, only one concentration varies significantly. During the reaction, we detect the variation of the concentration of the target species over time.

We focus on the period of time just after the beginning of the reaction. Before it, there is no variation to be detected (the reactant is probably not even in the solution) and later the products can play a role in the speed of reaction.

If the reaction is of order 1 with regard to the target species, the speed is

And thus we can find a straight line when we plot ln[A] in function of the time, the slope of which is –ak

If the order=2, the ln[A] won’t give a straight line but 1/[A] does. In this case, the slope is ak.

The order can also be equal to zero, meaning that the concentration of the reactant does not matter (except for very small concentrations). That means that the reaction is limited by something else. We often observe order zero reactions in presence of catalyst.

A catalyst is a species that affects the reaction but is not consumed during the reaction. The most common catalysts are solid but some catalysts are liquid. On solid catalysts, the reactants can bind on the catalyst. Because of this new liaison, the liaisons in the reactant are weaker and the reactivity is improved. The reaction is thus limited by the available surface on the catalyst. The solution is mixed during the reaction so that there is always reactants in the vicinity of the catalyst. Otherwise the reactants would be further and further of the catalyst as the time goes on. The evolution of the concentration would be more complex in this case. We will discuss furthermore about catalysts later.

Method of the initial speeds

Instead of using excesses of reactants, we can do several experiments and vary one concentration each time. The reaction

is obviously not a simple reaction. The speed depends simultaneously on the three reactants BrO₃^–, Br^– and H⁺. We first do the reaction with identical concentrations [BrO₃^–] = [Br^–] = [ H⁺] = 0.1M. We analyse the evolution of the concentration of one species as a function of time and we only consider the variation of concentration directly after the beginning of the reaction to determine a speed of the reaction.

In a second time, we do the same reaction but modify the concentration of one reactant. The speed of reaction will change consequently and we can determine the order for each species. Ultimately when we know every order we can determine the value of k.

Model of collisions

The reactions are decomposed in elementary events. In this case the order is the stoichiometric coefficient (usually 1 or 2). Elementary events are reactions that are done in one step: everything is done during the collision between the two molecules.

For instance, the reaction between the carbon monoxide CO and NO₂ is an elementary event: the oxygen atom is transferred from NO₂ to the CO during a collision between the two species.  When the impact occurs and the molecules were in the good orientation, the liaison N-O weakens and simultaneously a liaison is formed between  C and O

The collisions between molecules don’t always lead to a new product and there are some conditions for the reaction to happen. The intermediate of reaction OCONO can be obtained only if the collision comes from a given angle: the intermediate is linear. In addition to this spatial limitation, to transfer the atom of oxygen, the liaison between O and N needs to be stretched, requiring some energy. Finally there is a variation of the number of moles of gas during this reaction: NO is a liquid and therefore there is a decrease of entropy that has to be overpassed.

There is thus a minimum of energy that is required to obtain the reaction. This energy is called the energy of activation. One can represent the evolution of a reaction this way:

The reactants, taken separately have a given amount of energy. The products of the reaction have a lower energy. The difference of energy between the products and the reactants is the enthalpy of reaction. To obtain the products, the energy of activation is similar to a barrier that has to be climbed. At the summit of this barrier is the excited state, called this way because it is high in energy. After the excited state, the energy decreases down to the products energy that should be lower than the energy of the reactants.

To do the reaction in the other direction, forming CO and NO₂, the energy of activation is much larger and the products are energetically less favourable. The difference between the two energies of activation is also equal to the enthalpy of reaction.

To surpass the energy of activation, the molecules must have enough kinetic energy. If it is the case, they do successful collisions. Remember that the kinetic energy of molecules depends on the temperature. If the temperature gets higher, a larger population of the molecules will have a kinetic energy larger than the energy of activation.

In the expression of the speed of reaction, we find these dependencies in the constant of speed k

The energy parameter is in the exponential and the spatial parameter and the frequency of the collisions are found in the constant A.

Note that slow molecules can also react if they have a large potential energy, for example a lot of energy of vibration. Instead of the kinetic energy, the potential energy is used.

As a result, if two NOBr are correctly aligned, they can form Br2 even if their kinetic energy is low because of the vibration of the liaisons Br-N.

Eventually the N-Br liaisons can be stretched simultaneously, approaching the two Br from each other and facilitating the formation of Br₂.

The speed of the reaction to form hydroiodic acid from hydrogen and iodine may let us think that it is an elementary reaction.

Indeed, the exponents of the concentrations are the stoichiometric coefficients of the reactants. However, this reaction is not an elementary reaction. To have been one, the molecules would need to be aligned in one specific way and the liaisons have to break and to form at the same time. It means that 4 events have to happen at the same time in addition to the specific angle of collision.

Yet the reaction is done with a speed that let us think about an elementary event. The complete reaction is the succession of elementary events occurring one after each other.

The first step is the homolytic dissociation of I₂. It is done with the help of a catalyst metal.

It was observed that the composition of the recipient can affect the speed of the reaction. The dissociation is thus done at the surface of the recipient. After the dissociation, I can now react with H₂ to form H₂I that can react with the second I from the first reaction to produce 2 HI.

While the last reaction is complete, the first and second reactions are reactions of equilibrium and can thus go in both directions.

The speed of the global reaction is determined by the slowest elementary event. This slow step is called the determining step and is here the last reaction:

How do we find that this speed is equal to the one obtained experimentally?

To obtain such a thing, we use two hypotheses:

• the equilibrium reactions are almost at the equilibrium: the third reaction is slower than the two first reactions and we assume that the difference of speed is such that the first reactions (almost) have the time to reach the equilibrium. As a result, we can say that for those two reactions, the speed in one direction equals the speed in the opposite direction

The ratio between the constants of reaction k₁/k_-1 is the global constant K₁.

We can do the same for the second equation:

We can now replace the concentrations of v₃ with the expressions that we found just now

The constant that was found experimentally is thus a combination of the three constants of reaction.

• the concentration of H₂I is constant:

H₂I is the product of the second reaction and the reactant of the third reaction. By this assumption, we assume that as soon as H₂I is produced, it will be consumed by the next step of the reaction or turns back into H₂ and I by the reverse of the second reaction.

We can insert this in the expression of v₃.

Given that v₁=v_-1 (from the first hypothesis), [I]²=K₁[I₂] and

We have here the same expression than the one we obtained with the first hypothesis in the numerator. We can thus make the assumption that the concentration of H₂I is constant if k₃[I]/k_-2<<1, i.e. that the concentration of I is small and that k₃<<k_-2. It is thus only correct at the beginning of the reaction.

We might think that the hydrogenation of Br₂ follows the same kinetics but we are far from the true. Indeed,

This reaction is also a series of elementary steps

Catalyst

A catalyst is a species that is involved in a reaction but is not consumed or produced during the reaction. It helps a reaction to take place by a diminution of the energy of activation.

A catalyst can be in the same phase than the reactants, in this case we talk of a homogeneous catalyst, or in a different phase, as for the reaction between I₂ and H₂ where the catalyst is the surface of the recipient, a solid. In this second case, we talk of a heterogeneous catalyst. There are at least 4 stages for the action of the catalyst. We will see them through the catalytic hydrogenation of the ethylene.

• adsorption and activation of the reactants: the species that are in the vicinity of the heterogeneous catalyst may bind to it. In term of entropy, the binding is quite neutral: the species comes from a liquid to bind on a surface where a lot of spot are possible. There is thus a lot of possible configurations on the surface.

• Migration of the reactants on the surface: the reactants must be near to react together

• Reactions

• Desorption of the products

In the case of a homogeneous catalyst, the catalyst is just a species that intervene during the reaction but is not consumed by it. In high levels of the atmosphere, NO acts as a catalyst in the destruction of the ozone.

Case of the ozone

The formula of the ozone is O₃. This molecule is sensible to UV beams that breaks it into O₂ and a radical O^.. A radical is a species or a molecule that has has unpaired valence electrons and that is very reactive. They are usually the result of a homolytic dissociation, i.e. the electrons of the liaison are equally shared between the two atoms. It is so reactive that it can break nearby molecules to form a liaison. We use radicals to produce polymers (that will be seen in the corresponding chapter) and also in antibacterial products. The good point of using ozone as an antibacterial product is that it rejects oxygen O₂ as waste, so it is clean for the environment.

The destruction of ozone is done in two steps: the first step is the dissociation of a radical O^. under the beaming of UV photons.

The radical can react with the O₂ (reverse reaction) to get back to ozone or react with another ozone molecule to form more O₂.

In total, we obtain 3 O₂ from 2 O₃ molecules.

The radical O^. is not a catalyst of the reaction even if we don’t find it in the global reaction because it is produced and consumed during the reaction. It is not a species that is already present in the system from the beginning and that remains in the system once the reaction is complete. The speed that we find experimentally is

It is not the speed that we could expect from a one step reaction. We can do the kinetic analysis of the reaction. The determining step of this reaction is the second reaction.

The concentration of the radical is difficult to determine, but we can still make the same assumptions we made with the reaction between H₂ and I₂: the concentration of the intermediate product doesn’t change and the first reaction is at the equilibrium:

We can insert this in the expression of the speed

Freon’s

Freon’s are small species full of halogens, CCl₂F₂ for instance that have a very high heat capacity and are nonpolar and inert. There have been used as refrigerant in the fridges for a long time and we thought that, as those species were inert in the labs we had no specific care to apply to them. However Freon’s are deteriorated by UV rays. In dumps, the tubes containing the Freon eventually break and the gas goes up high in the atmosphere where it is not inert anymore. The deterioration of Freon’s due to UV’s leads to the formation of radicals that will interact with the ozone.

If we sum the two last equations, we get the determining step of the deterioration of the ozone alone.

The Cl^. is thus a catalyst of the destruction of the ozone and the hole in the ozone layer started to grow. The problem is that the Freon gas takes a very long time (years) to reach the higher atmosphere and at the moment we observed their effect, a lot of Freon’s were already climbing. Now the use of Freon’s is forbidden and the hole in the ozone layer is slowly decreasing in size.

Another product having an effect on the ozone is some nitrogen products of the exhaust pipes of cars. During the combustion of the fuel, nitrogen monoxide and dioxide can be formed.

Nitrogen dioxide is deteriorated by light.

Low in the atmosphere, and at the ground level, NO acts as a catalyst and generates peaks of ozone. Higher, NO catalyses the destruction of the ozone.

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