Organic Chemistry Summary Reactions

FREE RADICAL MECHANISMS Free radical substitution What is free radical substitution? . . . Substitution reactions One atom in a molecule is replaced by another atom or group of atoms. Free radical substitution involves breaking a carbon-hydrogen bond in alkanes such as: methane

CH4

ethane

CH3CH3

propane

CH3CH2CH3

A new bond is then formed to something else. It also happens in alkyl groups like methyl, ethyl (and so on) wherever these appear in more complicated molecules. methyl

CH3

ethyl

CH3CH2

For example, ethanoic acid is CH3COOH and contains a methyl group. The carbon-hydrogen bonds in the methyl group behave just like those in methane, and can be broken and replaced by something else in the same way.

A simple example of substitution is the reaction between methane and chlorine in the presence of UV light (or sunlight). CH4 + Cl2 + HCl

CH3Cl

Notice that one of the hydrogen atoms in the methane has been replaced by a chlorine atom. That's substitution.

Free radical reactions Free radicals are atoms or groups of atoms which have a single unpaired electron. A free radical substitution reaction is one involving these radicals. Free radicals are formed if a bond splits evenly - each atom getting one of the two electrons. The name given to this is homolytic fission. Note: If a bond were to split unevenly (one atom getting both electrons, and the other none), ions would be formed. The atom that got both electrons would become negatively charged, while the other one would become positive. This is called heterolytic fission. Warning! "Fission" it just means "splitting". "Homo" and "hetero" are used in the sense of "same" (homo) or "different" (hetero). So, Homolytic Fission is splitting a bond to produce two particles which are the same in the sense that they both have a single unpaired electron (both are free radicals). Heterolytic Fission produces two particles which are different because one is a positive ion and the other a negative ion. To show that a species (either an atom or a group of atoms) is a free radical, the symbol is written with a dot attached to show the unpaired electron. For example: a chlorine radical

Cl

a methyl radical

CH3

Note: If you wanted to be fussy, the dot showing the electron really ought to be written next to the carbon atom in the methyl radical, because that's the atom with the unpaired electron - in other words as CH3. This isn't normally done unless the radical gets more complicated.

methane and chlorine /bromine. . . If a mixture of methane and chlorine is exposed to a flame, it explodes - producing carbon and hydrogen chloride. This isn't a very useful reaction! The reaction we are going to explore is a more gentle one between methane and chlorine in the presence of ultraviolet light - typically sunlight. This is a good example of a photochemical reaction - a reaction brought about by light. This reaction between methane and bromine happens in the presence of ultraviolet light - typically sunlight. Also a good example of a photochemical reaction.

Note: These reactions are sometimes described as examples of photocatalysis - reactions catalysed by light. It is better to use the term "photochemical" and keep the keep the word "catalysis" for reactions speeded up by actual substances rather than light.

CH4 + Cl2

CH3Cl + HCl

The organic product is chloromethane. (or bromomethane if reaction occurs with bromine) One of the hydrogen atoms in the methane has been replaced by a chlorine atom, so this is a substitution reaction. However, the reaction doesn't stop there, and all the hydrogens in the methane can in turn be replaced by chlorine atoms. [Multiple substitution]

The mechanism The mechanism involves a chain reaction. During a chain reaction, for every reactive species you start off with, a new one is generated at the end - and this keeps the process going. Species: a useful word which is used in chemistry to mean any sort of particle you want it to mean. It covers molecules, ions, atoms, or (in this case) free radicals. The over-all process is known as free radical substitution, or as a free radical chain reaction. Chain initiation: The chain is initiated (started) by UV light breaking a chlorine molecule into free radicals. Cl2

2Cl

Chain propagation reactions: These are the reactions which keep the chain going. CH4 + Cl + HCl

CH3

CH3 + Cl2 + Cl

CH3Cl

Chain termination reactions: These are reactions which remove free radicals from the system without replacing them by new ones. 2Cl

Cl

2

CH3 + Cl CH3Cl CH3 + CH3 CH3CH3 methylbenzene and chlorine . . . A Free Radical Substitution Reaction Methylbenzene has a methyl group attached to a benzene ring. The hexagon with the circle inside is the standard symbol for this ring. There is a carbon atom at each corner of the hexagon, and a hydrogen atom on each carbon apart from the one with the methyl group attached. The reaction we are going to explore happens between methylbenzene and chlorine in the presence of ultraviolet light - typically sunlight. This is a good example of a photochemical reaction.

The organic product is (chloromethyl)benzene. The brackets in the name emphasise that the chlorine is part of the attached methyl group, and isn't on the ring. One of the hydrogen atoms in the methyl group has been replaced by a chlorine atom, so this is a substitution reaction. However, the reaction doesn't stop there, and all three hydrogens in the methyl group can in turn be replaced by chlorine atoms. [Multiple substitution]

Important! There is another reaction which happens between methylbenzene and chlorine in the absence of light and in the presence of a number of possible catalysts. In that one, substitution happens in the benzene ring instead of in the methyl group. You will find this reaction discussed under electrophilic substitution reactions.

The mechanism The mechanism involves a chain reaction. During a chain reaction, for every reactive species you start off with, a new one is generated at the end - and this keeps the process going. The over-all process is known as free radical substitution, or as a free radical chain reaction. Chain initiation: The chain is initiated (started) by UV light breaking a chlorine molecule into free radicals. Cl2

2Cl

Chain propagation reactions: These are the reactions which keep the chain going.

Chain termination reactions: These are reactions which remove free radicals from the system without replacing them by new ones. If any two free radicals collide, they will join together without producing any new radicals. The simplest example of this is a collision between two chlorine radicals. 2Cl

Cl

2

Free radical addition polymerisation of ethene . . . A Free Radical Addition Reaction An addition reaction is one in which two or more molecules join together to give a single product. During the polymerisation of ethene, thousands of ethene molecules join together to make poly(ethene) commonly called polythene.

Conditions Temperature:

about 200°C

Pressure:

about 2000 atmospheres

Initiator:

a small amount of oxygen as an impurity

Note: The oxygen is sometimes described as a catalyst for the reaction. That's not strictly true. A catalyst can be recovered unchanged at the end of a reaction, but in this case the oxygen is used up. It gets incorporated into the polymer molecules - as you will see shortly.

The mechanism The over-all process is known as free radical addition. Chain initiation: The chain is initiated by free radicals, Ra , produced by reaction between some of the ethene and the oxygen initiator.

Chain propagation: Each time a free radical hits an ethene molecule a new longer free radical is formed.

Chain termination: Eventually two free radicals hit each other producing a final molecule. The process stops here because no new free radicals are formed.

Because chain termination is a random process, poly(ethene) will be made up of chains of all sorts of different lengths. alkenes with hydrogen bromide . . . The mechanism for the addition of hydrogen bromide to alkenes under free radical conditions. This is often known as the "peroxide effect". *Addition to symmetrical alkenes A symmetrical alkene is one like ethene where the groups at both ends of the carbon-carbon double bond are the same. The reaction happens at room temperature in the presence of organic peroxides or some oxygen from the air. Alkenes react very slowly with oxygen to produce traces of organic peroxides - so the two possible conditions are equivalent to each other.

The reaction is a simple addition of the hydrogen bromide. For example, with ethene: With a symmetrical alkene you get exactly the same product in the absence of the organic peroxides or oxygen - but the mechanism is different. The mechanism Hydrogen halides (hydrogen chloride, hydrogen bromide and the rest) usually react with alkenes using an electrophilic addition mechanism. However, in the presence of organic peroxides, hydrogen bromide adds by a different mechanism. With the organic peroxides present you get a free radical chain reaction. Chain initiation: The chain is initiated by free radicals produced by an oxygen-oxygen bond in the organic peroxide breaking. These free radicals extract a hydrogen atom from a hydrogen bromide molecule to produce bromine radicals. Chain propagation: A bromine radical joins to the ethene using one of the electrons in the pi bond. That creates a new radical with the single electron on the other carbon atom. That radical reacts with another HBr molecule to produce bromoethane and another bromine radical to continue the process. Chain termination: Eventually two free radicals hit each other and produce a molecule of some sort. The process stops here because no new free radicals are formed. *Addition to unsymmetrical alkenes An unsymmetrical alkene is one like propene where the groups at either end of the carbon-carbon double bond are different. The reaction happens under the same conditions as with a symmetrical alkene, but there is a complication because the hydrogen and the bromine can add in two different ways. Which way they add depends on whether there are organic peroxides (or oxygen) present or not.

Normally, when a molecule HX adds to a carbon-carbon double bond, the hydrogen becomes attached to the carbon with the more hydrogens on already. This is known as Markovnikov's Rule. Because the HBr adds on the "wrong way around" in the presence of organic peroxides, this is often known as the peroxide effect or anti-Markovnikov addition. In the absence of peroxides, hydrogen bromide adds to propene via an electrophilic addition mechanism. That gives the product predicted by Markovnikov's Rule. The free radical mechanism Chain initiation:This is exactly the same as in the ethene case above.

Chain propagation: When the bromine radical joins to the propene, it attaches so that a secondary radical is formed. This is more stable (and so easier to form) than the primary radical which would be formed if it attached to the other carbon atom.

That radical reacts with another HBr molecule to produce 1-bromopropane and another bromine radical to continue the process.

Chain termination Eventually two free radicals hit each other and produce a molecule of some sort. The process stops here because no new free radicals are formed. Why don't the other hydrogen halides behave in the same way? The reason that hydrogen bromide adds in an anti-Markovnikov fashion in the presence of organic peroxides is simply a question of reaction rates. The free radical mechanism is much faster than the alternative electrophilic addition mechanism. Both mechanisms happen, but most of the product is the one from the free radical mechanism because that is working faster. With the other hydrogen halides, the opposite is true. Hydrogen fluoride The hydrogen-fluorine bond is so strong that fluorine radicals aren't formed in the initiation step. Hydrogen chloride With hydrogen chloride, the second half of the propagation stage is very slow. If you do a bond enthalpy sum, you will find that the following reaction is endothermic.

This is due to the relatively high hydrogen-chlorine bond strength. Hydrogen iodide In this case, the first step of the propagation stage turns out to be endothermic and this slows the reaction down. Not enough energy is released when the weak carbon-iodine bond is formed.

In the case of hydrogen bromide, both steps of the propagation stage are exothermic.

ELECTROPHILIC ADDITION MECHANISMS MENU Addition to symmetrical alkenes Covers addition to symmetrical alkenes like ethene and cyclohexene. A symmetrical alkene has the same groups attached to both ends of the carbon-carbon double bond. What is electrophilic addition? . . . An explanation of the terms addition and electrophile, together with a general mechanism for these reactions. Background Electrophilic addition happens in many of the reactions of compounds containing carbon-carbon double bonds - the alkenes. The structure of ethene We are going to start by looking at ethene, because it is the simplest molecule containing a carbon-carbon double bond. What is true of C=C in ethene will be equally true of C=C in more complicated alkenes. Ethene, C2H4, is often modelled as shown on the right. The double bond between the carbon atoms is, of course, two pairs of shared electrons. What the diagram doesn't show is that the two pairs aren't the same as each other. One of the pairs of electrons is held on the line between the two carbon nuclei as you would expect, but the other is held in a molecular orbital above and below the plane of the molecule. A molecular orbital is a region of space within the molecule where there is a high probability of finding a particular pair of electrons. In this diagram, the line between the two carbon atoms represents a normal bond - the pair of shared electrons lies in a molecular orbital on the line between the two nuclei where you would expect them to be. This sort of bond is called a sigma bond. The other pair of electrons is found somewhere in the shaded part above and below the plane of the molecule. This bond is called a pi bond. The electrons in the pi bond are free to move around anywhere in this shaded region and can move freely from one half to the other.

Note: This diagram shows a side view of an ethene molecule. The dotted lines to two of the hydrogens show bonds going back into the screen or paper away from you. The wedge shapes show bonds coming out towards you.

The pi electrons are not as fully under the control of the carbon nuclei as the electrons in the sigma bond and, because they lie exposed above and below the rest of the molecule, they are relatively open to attack by other things. Electrophiles An electrophile is something which is attracted to electron-rich regions in other molecules or ions. Because it is attracted to a negative region, an electrophile must be something which carries either a full positive charge, or has a slight positive charge on it somewhere. Note: The ending ". . phile" means a liking for.

Ethene and the other alkenes are attacked by electrophiles. The electrophile is normally the slightly positive ( +) end of a molecule like hydrogen bromide, HBr.

Electrophiles are strongly attracted to the exposed electrons in the pi bond and reactions happen because of that initial attraction. You might wonder why fully positive ions like sodium, Na+, don't react with ethene. Although these ions may well be attracted to the pi bond, there is no possibility of the process going any further to form bonds between sodium and carbon, because sodium forms ionic bonds, whereas carbon normally forms covalent ones. Addition reactions In a sense, the pi bond is an unnecessary bond. The structure would hold together perfectly well with a single bond rather than a double bond. The pi bond often breaks and the electrons in it are used to join other atoms (or groups of atoms) onto the ethene molecule. In other words, ethene undergoes addition reactions. For example, using a general molecule X-Y . . .

Summary: electrophilic addition reactions An addition reaction is a reaction in which two molecules join together to make a bigger one. Nothing is lost in the process. All the atoms in the original molecules are found in the bigger one. An electrophilic addition reaction is an addition reaction which happens because what we think of as the "important" molecule is attacked by an electrophile. The "important" molecule has a region of high electron density which is attacked by something carrying some degree of positive charge. Note: When we talk about reactions of alkenes like ethene, we think of the ethene as being attacked by other molecules such as hydrogen bromide. Because ethene is the molecule we are focusing on, we quite arbitrarily think of it as the central molecule and hydrogen bromide as its attacker. There's no real justification for this, of course, apart from the fact that it helps to put things in some sort of logical pattern. In reality, the molecules just collide and may react if they have enough energy and if they are lined up correctly.

Understanding the electrophilic addition mechanism The mechanism for the reaction between ethene and a molecule X-Y It is very unlikely that any two different atoms joined together will have the same electronegativity. We are going to assume that Y is more electronegative than X, so that the pair of electrons is pulled slightly towards the Y end of the bond. That means that the X atom carries a slight positive charge. The slightly positive X atom is an electrophile and is attracted to the exposed pi bond in the ethene.

You are now much more likely to find the electrons in the half of the pi bond nearest the XY. As the process continues, the two electrons in the pi bond move even further towards the X until a covalent bond is made. The electrons in the X-Y bond are pushed entirely onto the Y to give a negative Y- ion.

Help! Why does the carbon atom have a positive charge? The pi bond was originally made using an electron from each carbon atom, but both of these electrons have now been used to make a bond to the X atom. This leaves the right-hand carbon atom an electron short - hence positively charged. Note also that we are only showing one of the pairs of electrons around the Y - ion. There will be other lone pairs as well, but we are only actually interested in the one we've drawn.

An ion in which the positive charge is carried on a carbon atom is called a carbocation or a carbonium ion (an older term).

In the final stage of the reaction the electrons in the lone pair on the Y- ion are strongly attracted towards the positive carbon atom. They move towards it and form a co-ordinate (dative covalent) bond between the Y and the carbon. Help! A co-ordinate (dative covalent) bond is simply a covalent bond in which both shared electrons originate from the same atom. The bond formed between the X and the other carbon atom was also a co-ordinate bond. Once a co-ordinate bond has been formed there is no difference whatsoever between it and any other covalent bond.

How to write this mechanism in an exam The movements of the various electron pairs are shown using curly arrows.

The reaction with hydrogen halides . . . Electrophilic addition reactions involving hydrogen bromide Alkenes react with hydrogen bromide in the cold. The double bond breaks and a hydrogen atom ends up attached to one of the carbons and a bromine atom to the other. In the case of ethene, bromoethane is formed.

With cyclohexene you get bromocyclohexane.

The structures of the cyclohexene and the bromocyclohexane are often simplified:

The mechanisms The reactions are examples of electrophilic addition. With ethene and HBr:

and with cyclohexene:

Electrophilic addition reactions involving the other hydrogen halides Hydrogen chloride and the other hydrogen halides add on in exactly the same way. For example, hydrogen chloride adds to ethene to make chloroethane:

The only difference is in how fast the reactions happen with the different hydrogen halides. The rate of reaction increases as you go from HF to HCl to HBr to HI. HF

slowest reaction

HCl HBr HI

fastest reaction

The reason for this is that as the halogen atoms get bigger, the strength of the hydrogen-halogen bond falls. As you have seen in the HBr case, in the first step of the mechanism the hydrogen-halogen bond gets broken. If the bond is weaker, it will break more readily and so the reaction is more likely to happen. The mechanisms The reactions are still examples of electrophilic addition. With ethene and HCl, for example:

This is exactly the same as the mechanism for the reaction between ethene and HBr, except that we've replaced Br by Cl. All the other mechanisms for symmetrical alkenes and the hydrogen halides would be done in the same way. The reaction with sulphuric acid . . . The mechanism for the reaction between ethene (and cyclohexene) and sulphuric acid.

The electrophilic addition reaction between ethene and sulphuric acid Alkenes react with concentrated sulphuric acid in the cold to produce alkyl hydrogensulphates. Ethene reacts to give ethyl hydrogensulphate.

The structure of the product molecule is sometimes written as CH3CH2HSO4, but the version in the equation is better because it shows how all the atoms are linked up. You may also find it written as CH3CH2OSO3H. All you need to do is to learn the structure of sulphuric acid, and after that the mechanism is exactly the same as the one with hydrogen bromide. As you will find out, the formula of the product follows from the mechanism in an inevitable way.

The mechanism for the reaction between ethene and sulphuric acid Sulphuric acid as an electrophile The hydrogen atoms are attached to very electronegative oxygen atoms which means that the hydrogens will have a slight positive charge while the oxygens will be slightly negative. In the mechanism, we just focus on one of the hydrogen to oxygen bonds, because the other one is too far from the carbon-carbon double bond to be involved in any way. The mechanism

Look carefully at the structure of the product so that you can see how it relates to the various formulae given earlier (CH3CH2OSO2OH etc). The electrophilic addition reaction between cyclohexene and sulphuric acid This time we are going straight for the mechanism without producing an initial equation. This is to show that you can work out the structure of obscure products provided you can write the mechanism. The mechanism for the reaction between cyclohexene and sulphuric acid

Having worked out the structure of the product, you could then write a simple equation for the reaction if you wanted to.

The reaction with bromine . . . The electrophilic addition of bromine to ethene Alkenes react in the cold with pure liquid bromine, or with a solution of bromine in an organic solvent like tetrachloromethane. The double bond breaks, and a bromine atom becomes attached to each carbon. The bromine loses its original red-brown colour to give a colourless liquid. In the case of the reaction with ethene, 1,2-dibromoethane is formed.

This decolourisation of bromine is often used as a test for a carbon-carbon double bond. If an aqueous solution of bromine is used ("bromine water"), you get a mixture of products. The presence of the water complicates the mechanism. The other halogens, apart from fluorine, behave similarly. (Fluorine reacts explosively with all hydrocarbons - including alkenes - to give carbon and hydrogen fluoride.) The mechanism for the reaction between ethene and bromine The reaction is an example of electrophilic addition. Bromine as an electrophile The bromine is a very "polarisable" molecule and the approaching pi bond in the ethene induces a dipole in the bromine molecule. If you draw this mechanism in an exam, write the words "induced dipole" next to the bromine molecule . The simplified version of the mechanism

The more accurate version of the mechanism In the first stage of the reaction, one of the bromine atoms becomes attached to both carbon atoms, with the positive charge being found on the bromine atom. A bromonium ion is formed.

The bromonium ion is then attacked from the back by a bromide ion formed in a nearby reaction.

The electrophilic addition of bromine to cyclohexene Cyclohexene reacts with bromine in the same way and under the same conditions as any other alkene. 1,2-dibromocyclohexane is formed.

The mechanism for the reaction between cyclohexene and bromine The reaction is an example of electrophilic addition. Bromine as an electrophile Again, the bromine is polarised by the approaching pi bond in the cyclohexene. Don't forget to write the words "induced dipole" next to the bromine molecule. The simplified version of the mechanism

The alternative version of the mechanism In the first stage of the reaction, one of the bromine atoms becomes attached to both carbon atoms, with the positive charge being found on the bromine atom. A bromonium ion is formed.

The bromonium ion is then attacked from the back by a bromide ion formed in a nearby reaction.

Addition to unsymmetrical alkenes Carbocations (carbonium ions) and their stability . . . All carbocations (previously known as carbonium ions) carry a positive charge on a carbon atom. The name tells you that - a cation is a positive ion, and the "carbo" bit refers to a carbon atom. However there are important differences in the structures of various types of carbocations. The different kinds of carbocations Primary carbocations In a primary (1°) carbocation, the carbon which carries the positive charge is only attached to one other alkyl group. Help! An alkyl group is a group such as methyl, CH3, or ethyl, CH3CH2. These are groups containing chains of carbon atoms which may be branched. Alkyl groups are given the general symbol R.

Some examples of primary carbocations include:

Notice that it doesn't matter how complicated the attached alkyl group is. All you are doing is counting the number of bonds from the positive carbon to other carbon atoms. In all the above cases there is only one such link. Using the symbol R for an alkyl group, a primary carbocation would be written box.

as in the

Secondary carbocations In a secondary (2°) carbocation, the carbon with the positive charge is attached to two other alkyl groups, which may be the same or different. Examples:

A secondary carbocation has the general formula shown in the box. R and R' represent alkyl groups which may be the same or different.

Tertiary carbocations In a tertiary (3°) carbocation, the positive carbon atom is attached to three alkyl groups, which may be any combination of same or different.

A tertiary carbocation has the general formula shown in the box. R, R' and R" are alkyl groups and may be the same or different.

The stability of the various carbocations The "electron pushing effect" of alkyl groups You are probably familiar with the idea that bromine is more electronegative than hydrogen, so that in a HBr bond the electrons are held closer to the bromine than the hydrogen. A bromine atom attached to a carbon atom would have precisely the same effect - the electrons being pulled towards the bromine end of the bond. The bromine has a negative inductive effect. Alkyl groups do precisely the opposite and, rather than draw electrons towards themselves, tend to "push" electrons away. This means that the alkyl group becomes slightly positive ( +) and the carbon they are attached to becomes slightly negative ( -). The alkyl group has a positive inductive effect.

The arrow shows the electrons being "pushed" away from the CH 3 group. The plus sign on the left-hand end of it shows that the CH 3 group is becoming positive. The symbols + and - simply reinforce that idea.

The importance of spreading charge around in making ions stable The general rule-of-thumb is that if a charge is very localised (all concentrated on one atom) the ion is much less stable than if the charge is spread out over several atoms.

You will see that the electron pushing effect of the CH3 group is placing more and more negative charge on the positive carbon as you go from primary to secondary to tertiary carbocations. The effect of this, of course, is to cut down that positive charge. At the same time, the region around the various CH3 groups is becoming somewhat positive. The net effect, then, is that the positive charge is being spread out over more and more atoms as you go from primary to secondary to tertiary ions. The more you can spread the charge around, the more stable the ion becomes.

Order of stability of carbocations primary < secondary < tertiary

The stability of the carbocations in terms of energetics When we talk about secondary carbocations being more stable than primary ones, what exactly do we mean? We are actually talking about energetic stability - secondary carbocations are lower down an energy "ladder" than primary ones. This means that it is going to take more energy to make a primary carbocation than a secondary one. If there is a choice between making a secondary ion or a primary one, it will be much easier to make the secondary one. Similarly, if there is a choice between making a tertiary ion or a secondary one, it will be easier to make the tertiary one. This has important implications in the reactions of unsymmetrical alkenes. If you are interested in these, follow the link below to the electrophilic addition reactions menu.

Why unsymmetric alkenes are a problem . . . The Problem The addition of H-X to an unsymmetrical alkene like propene An unsymmetrical alkene is one like propene or but-1-ene in which the groups or atoms attached to either end of the carbon-carbon double bond are different. For example, in propene there are a hydrogen and a methyl group at one end, but two hydrogen atoms at the other end of the double bond.

With these unsymmetrical alkenes, it is possible to get two different products during some addition reactions. During the addition of a molecule HX to propene, you could in principle get either this reaction:

or this one:

It depends on which way around you add the HX across the double bond. In fact, in most cases, it's mainly the second reaction which happens. The hydrogen atom becomes attached to the right-hand carbon atom as we've drawn it. Markovnikov's Rule When a compound HX is added to an unsymmetrical alkene, the hydrogen becomes attached to the carbon with the most hydrogens attached to it already.

Remember that the HX has to attach itself to the carbon atoms which were originally part of the double bond. So in this case, adding HX to CH3CH=CH2, the hydrogen is attached to the CH2 group, because the CH2 group has more hydrogens than the CH group. Notice that only the hydrogens directly attached count. The ones in the CH 3 group are totally irrelevant. The Mechanism HX as an electrophile In each of the cases we are interested in, X is more electronegative than hydrogen. That means that the bonding pair of electrons will be dragged towards the X end of the bond, and so the hydrogen becomes slightly positively charged. The slightly positive hydrogen is the electrophile, and is attracted to the pi bond in the propene. What happens next determines which way around the HX adds across the double bond. The two possible mechanisms In both possibilities, the pi bond breaks and the electron pair swings down to form a bond with the hydrogen atom. At the same time, the electrons in the H-X bond are repelled right down on to the X to give an X- ion. The difference lies in which way the electron pair in the pi bond swings. First possibility The electron pair moves to form a bond between the hydrogen and the left-hand carbon.

When the second stage of the mechanism happens, and the lone pair on X- forms a bond with the positive carbon atom, the product of this mechanism is not the one which Markovnikov's Rule predicts. Second possibility The electron pair moves to form a bond between the hydrogen and the right-hand carbon.

This time, the overall mechanism leads to the correct product. Why does one of these work better than the other? The second mechanism works much faster than the first, and so most of the product that you get is CH 3CHX-CH3. There will be small amounts of CH3-CH2-CH2X despite what Markovnikov says! The difference between the two mechanisms lies in the intermediates - the things formed at the half-way stage. In the mechanism that works best, you get a secondary carbocation formed as one of the intermediate ions.

In the slow mechanism which produces hardly any product, you get a primary carbocation formed instead.

It is much easier to form the secondary carbocation because it is more energetically stable. The activation energy for the reaction will be less, and so most of the reaction happens via that mechanism.

How to attack this sort of question in an exam Suppose you were asked for the mechanism for the addition of HX to some alkene you hadn't come across before. First, you need to decide whether the alkene is symmetrical or not. If it is symmetrical, there's no problem - it wouldn't matter which way around you added the HX. If it is unsymmetrical, you need to decide which way round the HX is going to add. For example, supposed you were asked for the mechanism for the addition of HX to but-1-ene, CH 3-CH2-CH=CH2. First, use Markovnikov's Rule to decide which carbon to attach the hydrogen to. In this case, the hydrogen would get attached to the CH2 end of the double bond, because that carbon has more hydrogens than the CH end. Now write the mechanism, taking care to draw the curly arrow showing the movement of the pi bond so that the hydrogen gets attached to that particular CH2 carbon.

If you are asked why the HX adds this way round, look at the carbocation formed as an intermediate and decide whether it is secondary or tertiary. Here it is a secondary ion. Then think about what sort of ion would be formed if the HX added the other way around. In this case that would be a primary ion.

Then say something like: "The secondary carbocation formed in this reaction is more energetically stable than the primary one which would be formed if the addition was the other way round, and so less activation energy is needed."

The reaction with hydrogen halides . . . An unsymmetrical alkene is one like propene in which the groups or atoms attached to either end of the carbon-carbon double bond are different. For example, in propene there are a hydrogen and a methyl group at one end, but two hydrogen atoms at the other end of the double bond. But-1-ene is another unsymmetrical alkene.

Electrophilic addition reactions involving hydrogen bromide As with all alkenes, unsymmetrical alkenes like propene react with hydrogen bromide in the cold. The double bond breaks and a hydrogen atom ends up attached to one of the carbons and a bromine atom to the other. In the case of propene, 2-bromopropane is formed.

This would normally be written in a more condensed form as

The product is 2-bromopropane. Note: There is another possible reaction between unsymmetrical alkenes and hydrogen bromide (but not the other hydrogen halides) unless the hydrogen bromide and alkene are absolutely pure. A different mechanism happens (a free radical chain reaction - not on UK A' level syllabuses) which leads to the hydrogen and bromine adding the opposite way round.

This is in line with Markovnikov's Rule which says:

When a compound HX is added to an unsymmetrical alkene, the hydrogen becomes attached to the carbon with the most hydrogens attached to it already. In this case, the hydrogen becomes attached to the CH2 group, because the CH2 group has more hydrogens than the CH group. Notice that only the hydrogens directly attached to the carbon atoms at either end of the double bond count. The ones in the CH3 group are totally irrelevant. The mechanism This is an example of electrophilic addition.

The addition is this way around because the intermediate carbocation (previously called a carbonium ion) formed is secondary. This is more stable (and so is easier to form) than the primary carbocation which would be produced if the hydrogen became attached to the centre carbon atom and the bromine to the end one. Electrophilic addition reactions involving the other hydrogen halides Hydrogen fluoride, hydrogen chloride and hydrogen iodide all add on in exactly the same way as hydrogen bromide. The only differences lie in the rates of reaction: HF

slowest reaction

HCl HBr HI

fastest reaction

This is because the hydrogen-halogen bond gets weaker as the halogen atom gets bigger. If the bond is weaker, it breaks more easily and so the reaction is faster. If the halogen is given the symbol X, the equation for the reaction with propene is:

Notice that the product is still in line with Markovnikov's Rule. The mechanism These are still examples of electrophilic addition. Again using X to stand for any halogen:

Again, the intermediate carbocation formed is secondary. This is more stable than the primary carbocation ion which would be formed if the hydrogen attached to the centre carbon atom and the X to the end one. If it is more stable it will be easier to make. The reaction with sulphuric acid . . . The electrophilic addition reaction between propene and sulphuric acid Alkenes react with concentrated sulphuric acid in the cold to produce alkyl hydrogensulphates. In the case of propene, the equation is:

This is in line with Markovnikov's Rule which says: When a compound HX is added to an unsymmetrical alkene, the hydrogen becomes attached to the carbon with the most hydrogens attached to it already.

In this case, the hydrogen becomes attached to the CH2 group, because the CH2 group has more hydrogens than the CH group. Notice that only the hydrogens directly attached to the carbon atoms at either end of the double bond count. The ones in the CH3 group are totally irrelevant. The mechanism This is an example of electrophilic addition.

The addition is this way around because it is easier to produce the secondary carbocation (carbonium ion) than the primary one which would be formed if the hydrogen became attached to the centre carbon atom and the rest of the sulphuric acid to the end one. The reaction with bromine . . . An unsymmetrical alkene is one like propene in which the groups or atoms attached to either end of the carbon-carbon double bond are different. For example, in propene there are a hydrogen and a methyl group at one end, but two hydrogen atoms at the other end of the double bond. But-1-ene is another unsymmetrical alkene.

The electrophilic addition of bromine to propene In common with all other alkenes, propene reacts in the cold with pure liquid bromine, or with a solution of bromine in an organic solvent like tetrachloromethane. The double bond breaks, and a bromine atom becomes attached to each carbon. The bromine loses its original red-brown colour to give a colourless liquid. In the case of the reaction with propene, 1,2-dibromopropane is formed.

The other halogens, apart from fluorine, behave similarly. (Fluorine reacts explosively with all hydrocarbons - including alkenes - to give carbon and hydrogen fluoride.) The mechanism for the reaction between propene and bromine The reaction is an example of electrophilic addition. The simplified version of the mechanism

The more accurate version of the mechanism

In the first stage of the reaction, one of the bromine atoms becomes attached to both carbon atoms, with the positive charge being found on the bromine atom. A bromonium ion is formed.

The bromonium ion is then attacked from the back by a bromide ion formed in a nearby reaction.

ELECTROPHILIC SUBSTITUTION MECHANISMS MENU What is electrophilic substitution? . . . Electrophilic substitution happens in many of the reactions of compounds containing benzene rings - the arenes. This is what you need to understand for the purposes of the electrophilic substitution mechanisms: 

Benzene, C6H6, is a planar molecule containing a ring of six carbon atoms each with a hydrogen atom attached.



There are delocalised electrons above and below the plane of the ring.



The presence of the delocalised electrons makes benzene particularly stable.



Benzene resists addition reactions because that would involve breaking the delocalisation and losing that stability. Benzene is represented by this symbol, where the circle represents the delocalised electrons, and each corner of the hexagon has a carbon atom with a hydrogen attached.

Electrophilic substitution reactions involving positive ions Benzene and electrophiles Because of the delocalised electrons exposed above and below the plane of the rest of the molecule, benzene is obviously going to be highly attractive to electrophiles - species which seek after electron rich areas in other molecules. The electrophile will either be a positive ion, or the slightly positive end of a polar molecule. The delocalised electrons above and below the plane of the benzene molecule are open to attack in the same way as those above and below the plane of an ethene molecule. However, the end result will be different.

If benzene underwent addition reactions in the same way as ethene, it would need to use some of the delocalised electrons to form bonds with the new atoms or groups. This would break the delocalisation and this costs energy. Instead, it can maintain the delocalisation if it replaces a hydrogen atom by something else - a substitution reaction. The hydrogen atoms aren't involved in any way with the delocalised electrons. In most of benzene's reactions, the electrophile is a positive ion, and these reactions all follow a general pattern. The general mechanism The first stage Suppose the electrophile is a positive ion X+. Two of the electrons in the delocalised system are attracted towards the X + and form a bond with it. This has the effect of breaking the delocalisation, although not completely.

The ion formed in this step isn't the final product. It immediately goes on to react with something else. It is just an intermediate.

There is still delocalisation in the intermediate formed, but it only covers part of the ion. When you write one of these mechanisms, draw the partial delocalisation to take in all the carbon atoms apart from the one that the X has become attached to.

The intermediate ion carries a positive charge because you are joining together a neutral molecule and a positive ion. This positive charge is spread over the delocalised part of the ring. Simply draw the "+" in the middle of the ring. The hydrogen at the top isn't new - it's the hydrogen that was already attached to that carbon. We need to show that it is there for the next stage. The second stage

Here we've introduced a new ion, Y-. Where did this come from? You have to remember that it is impossible to get a positive ion on its own in a chemical system - so Y- is simply the negative ion that was originally associated with X+. A lone pair of electrons on Y- forms a bond with the hydrogen atom at the top of the ring. That means that the pair of electrons joining the hydrogen onto the ring aren't needed any more. These then move down to plug the gap in the delocalised electrons, so restoring the delocalised ring of electrons which originally gave the benzene its special stability. Electrophilic substitution reactions not involving positive ions Halogenation and sulphonation In these reactions, the electrophiles are polar molecules rather than fully positive ions. These mechanisms are different from what's gone before (and from each other). The nitration of benzene . . . The mechanism for the formation of nitrobenzene from benzene. The electrophilic substitution reaction between benzene and nitric acid Benzene is treated with a mixture of concentrated nitric acid and concentrated sulphuric acid at a temperature not exceeding 50°C. As temperature increases there is a greater chance of getting more than one nitro group, -NO2, substituted onto the ring.

Nitrobenzene is formed.

or:

The concentrated sulphuric acid is acting as a catalyst. The formation of the electrophile The electrophile is the "nitronium ion" or the "nitryl cation", NO2+. This is formed by reaction between the nitric acid and the sulphuric acid.

The electrophilic substitution mechanism Stage one

Stage two

The Friedel-Crafts acylation of benzene . . . The mechanism for the substitution of an acyl group such as CH3CO into benzene. The electrophilic substitution reaction between benzene and ethanoyl chloride What is acylation? An acyl group is an alkyl group attached to a carbon-oxygen double bond. If "R" represents any alkyl group, then an acyl group has the formula RCO-. Acylation means substituting an acyl group into something - in this case, into a benzene ring. The most commonly used acyl group is CH3CO-. This is called the ethanoyl group. In the example which follows we are substituting a CH3CO- group into the ring, but you could equally well use any other alkyl group instead of the CH3. The facts The most reactive substance containing an acyl group is an acyl chloride (also known as an acid chloride). These have the general formula RCOCl. Benzene is treated with a mixture of ethanoyl chloride, CH3COCl, and aluminium chloride as the catalyst. A ketone called phenylethanone is formed. Note: Ketones: A family of compounds containing a carbon-oxygen double bond with a hydrocarbon group either side of it. In this case there is a methyl group on one side and a benzene ring on the other.

or better:

The aluminium chloride isn't written into these equations because it is acting as a catalyst. If you wanted to include it, you could write AlCl3 over the top of the arrow.

The formation of the electrophile The electrophile is CH3CO+. It is formed by reaction between the ethanoyl chloride and the aluminium chloride catalyst.

The electrophilic substitution mechanism Stage one

Stage two

The hydrogen is removed by the AlCl4- ion which was formed at the same time as the CH3CO+ electrophile. The aluminium chloride catalyst is re-generated in this second stage. The Friedel-Crafts alkylation of benzene . . . The mechanism for the substitution of an alkyl group such as CH3 into benzene. The electrophilic substitution reaction between benzene and chloromethane What is alkylation? Alkylation means substituting an alkyl group into something - in this case into a benzene ring. A hydrogen on the ring is replaced by a group like methyl or ethyl and so on. Benzene is treated with a chloroalkane (for example, chloromethane or chloroethane) in the presence of aluminium chloride as a catalyst. Any other alkyl group could be used in the same way. Substituting a methyl group gives methylbenzene - once known as toluene.

or better:

The aluminium chloride isn't written into these equations because it is acting as a catalyst. If you wanted to include it, you could write AlCl3 over the top of the arrow. Note: The methylbenzene formed is more reactive than the original benzene, and so the reaction doesn't stop there. You get further methyl groups substituted around the ring.

The formation of the electrophile The electrophile is CH3+. It is formed by reaction between the chloromethane and the aluminium chloride catalyst.

The electrophilic substitution mechanism Stage one

Stage two

The hydrogen is removed by the AlCl4- ion which was formed at the same time as the CH3+ electrophile. The aluminium chloride catalyst is re-generated in this second stage. An industrial alkylation of benzene . . . The mechanism for the substitution of an alkyl group such as CH3CH2 into benzene, by a reaction involving an alkene such as ethene. Note: The process described below is one of several ways of making alkylbenzenes like ethylbenzene from benzene and alkenes, using a variety of different catalysts and conditions. The electrophilic substitution reaction between benzene and ethene Industrially, alkyl groups can be substituted into a benzene ring using a variant on Friedel-Crafts alkylation. One possibility is that instead of using a chloroalkane with an aluminium chloride catalyst, they use an alkene and a mixture of aluminium chloride and hydrogen chloride as the catalyst. This is a cheaper method because it saves having to make the chloroalkane first. To put an ethyl group on the ring (to make ethylbenzene), benzene is treated with a mixture of ethene, HCl and aluminium chloride.

or better:

The aluminium chloride and HCl aren't written into these equations because they are acting as catalysts. If you wanted to include them, you could write AlCl3 and HCl over the top of the arrow. The formation of the electrophile The electrophile is CH3CH2+. It is formed by reaction between the ethene and the HCl - exactly as if you were beginning to add the HCl to the ethene.

The chloride ion is immediately picked up by the aluminium chloride to form an AlCl 4- ion. That prevents the chloride ion from reacting with the CH3CH2+ ion to form chloroethane. Note: It wouldn't matter if it did react, because chloroethane will react with benzene using a simple Friedel-Crafts alkylation reaction to give the product you want anyway.

The electrophilic substitution mechanism Stage one

Stage two

The hydrogen is removed by the AlCl4- ion which was formed at the same time as the CH3CH2+ electrophile. The aluminium chloride and hydrogen chloride catalysts are re-generated in this second stage. The electrophilic substitution reaction between benzene and propene The problem with more complicated alkenes like propene is that you have to be careful about the structure of the product. In each case, you can only really be sure of that structure if you work through the mechanism first. For example, the propyl group becomes attached to the ring via its middle carbon atom - and not its end one.

You still need a mixture of aluminium chloride and hydrogen chloride as catalysts. The formation of the electrophile When the propene reacts with the HCl, the hydrogen becomes attached to the end carbon atom. A secondary carbocation (carbonium ion) is formed because it is more stable than the primary one which would have been formed if the addition was the other way round.

Because the positive charge is on the centre carbon atom, that is the one which will become attached to the ring.

The electrophilic substitution mechanism Stage one

Stage two

Again, the hydrogen is removed by the AlCl4- ion. The aluminium chloride and hydrogen chloride catalysts are re-generated in this second stage. The halogenation of benzene . . . The mechanism for the substitution of atoms like chlorine and bromine into benzene rings. The electrophilic substitution reaction between benzene and chlorine or bromine Benzene reacts with chlorine or bromine in an electrophilic substitution reaction, but only in the presence of a catalyst. The catalyst is either aluminium chloride (or aluminium bromide if you are reacting benzene with bromine) or iron. Strictly speaking iron isn't a catalyst, because it gets permanently changed during the reaction. It reacts with some of the chlorine or bromine to form iron(III) chloride, FeCl 3, or iron(III) bromide, FeBr3.

These compounds act as the catalyst and behave exactly like aluminium chloride in these reactions. The reaction with chlorine The reaction between benzene and chlorine in the presence of either aluminium chloride or iron gives chlorobenzene.

or:

The reaction with bromine The reaction between benzene and bromine in the presence of either aluminium bromide or iron gives bromobenzene. Iron is usually used because it is cheaper and more readily available.

or:

The formation of the electrophile We are going to explore the reaction using chlorine and aluminium chloride. If you want one of the other combinations, all you have to do is to replace each Cl by Br, or each Al by Fe.

As a chlorine molecule approaches the benzene ring, the delocalised electrons in the ring repel electrons in the chlorine-chlorine bond.

It is the slightly positive end of the chlorine molecule which acts as the electrophile. The presence of the aluminium chloride helps this polarisation. The electrophilic substitution mechanism Stage one

Stage two

The hydrogen is removed by the AlCl4- ion which was formed in the first stage. The aluminium chloride catalyst is re-generated in this second stage. The sulphonation of benzene . . . The mechanism for reaction between benzene and concentrated sulphuric acid to produce benzenesulphonic acid The electrophilic substitution reaction between benzene and sulphuric acid There are two equivalent ways of sulphonating benzene: 

Heat benzene under reflux with concentrated sulphuric acid for several hours.



Warm benzene under reflux at 40°C with fuming sulphuric acid for 20 to 30 minutes.

Or:

The product is benzenesulphonic acid. The electrophile is actually sulphur trioxide, SO3, and you may find the equation for the sulphonation reaction written:

Note: Which version of this equation you use will depend on what question you are being asked. If the question refers to the reaction with sulphuric acid, then you must use that one. If the question refers to SO3 as the electrophile, then you could use this one.

The formation of the electrophile The sulphur trioxide electrophile arises in one of two ways depending on which sort of acid you are using. Concentrated sulphuric acid contains traces of SO3 due to slight dissociation of the acid.

Fuming sulphuric acid, H2S2O7, can be thought of as a solution of SO3 in sulphuric acid - and so is a much richer source of the SO3. Sulphur trioxide is an electrophile because it is a highly polar molecule with a fair amount of positive charge on the sulphur atom. It is this which is attracted to the ring electrons. The electrophilic substitution mechanism Stage one

Stage two

The second stage of the reaction involves a transfer of the hydrogen from the ring to the negative oxygen. Some substitution reactions of methylbenzene . . . Illustrates how to cope with the problem of substituting things into rings which already have something else attached. There are two problems you might come across: 

Whereabouts in the ring does the substitution happen? Does this make a difference to how you draw the mechanisms?



Can the group already attached to the ring get involved in any way?

Electrophilic substitution in methylbenzene The nitration of methylbenzene If you substitute a nitro group, -NO2, into the benzene ring in methylbenzene, you could possibly get any of the following products:

The carbon with the methyl group attached is thought of as the number 1 carbon, and the ring is then numbered around from 1 to 6. You number in a direction (in this case, clockwise) which produces the smaller number in the name - hence 2-nitromethylbenzene rather than 6-nitromethylbenzene.

In the case of methylbenzene, whatever you attach to the ring, you always get a mixture consisting mainly of the 2- and 4- isomers. The methyl group is said to be 2,4-directing, in the sense that it seems to "push" incoming groups into those positions. Isomers: Molecules which have the same molecular formula (i.e. contain exactly the same number and type of atoms), but with a different spatial arrangement of those atoms.

Some other groups which might already be on the ring (for example, the -NO 2 group in nitrobenzene) "push" incoming groups into the 3- position. We'll have a quick look at this later on this page. How to write the mechanism for the nitration of methylbenzene Reacting methylbenzene with a mixture of concentrated nitric and sulphuric acids gives both 2nitromethylbenzene and 4-nitromethylbenzene. The mechanism is exactly the same as the nitration of benzene. You just have to be careful about the way that you draw the structure of the intermediate ion. Making 2-nitromethylbenzene (the first step)

This just shows the first step of the electrophilic substitution reaction. Notice that the partial delocalisation in the intermediate ion covers all the carbon atoms in the ring except for the one that the -NO 2 group gets attached to. That is the only point of interest in this example - everything else is just the same as with the nitration of benzene. The hydrogen atom is then removed by an HSO4- ion - exactly as in the benzene case. Making 4-nitromethylbenzene (the first step)

Once again, the only point of interest is in the way the partial delocalisation in the intermediate ion is drawn - again, it covers all the carbon atoms in the ring apart from the one with the -NO 2 group attached. The electrophilic substitution reaction between methylbenzene and chlorine This is a good example of a case where what is already attached to the ring can also get involved in the reaction. It is possible to get two quite different substitution reactions between methylbenzene and chlorine depending on the conditions used. The chlorine can substitute into the ring or into the methyl group. Here we are only interested in substitution into the ring. This happens in the presence of aluminium chloride or iron, and in the absence of UV light. Substituting into the ring gives a mixture of 2-chloromethylbenzene and 4-chloromethylbenzene.

The mechanisms are exactly the same as the substitution of chlorine into benzene - although you would have to be careful about the way you draw the intermediate ion.

For example, the complete mechanism for substitution into the 4- position is: Stage one

Stage two

Electrophilic substitution in nitrobenzene Substitution into the 3- position (the first step) Methyl groups direct new groups into the 2- and 4- positions, but a nitro group, -NO 2, already on the ring directs incoming groups into the 3- position. For example, if the temperature is raised above 50°C, the nitation of benzene doesn't just produce nitrobenzene - it also produces some 1,3-dinitrobenzene. A second nitro group is substituted into the ring in the 3- position.

The mechanism is exactly the same as the nitration of benzene or of methylbenzene - you just have to be careful in drawing the intermediate ion. Draw the partial delocalisation to include all the carbons except for the one the new -NO2 group gets attached to.

In the second stage, the hydrogen atom is then removed by an HSO 4- ion - exactly as in the benzene case. This isn't shown because there's nothing new.

ELIMINATION MECHANISMS MENU Elimination reactions involving halogenoalkanes Elimination from 2-bromopropane . . . The formation of an alkene (propene) from 2-bromopropane. The elimination reaction involving 2-bromopropane and hydroxide ions 2-bromopropane is heated under reflux with a concentrated solution of sodium or potassium hydroxide in ethanol. Heating under reflux involves heating with a condenser placed vertically in the flask to avoid loss of volatile liquids. Propene is formed and, because this is a gas, it passes through the condenser and can be collected.

Everything else present (including anything formed in the alternative substitution reaction) will be trapped in the flask. The mechanism In elimination reactions, the hydroxide ion acts as a base - removing a hydrogen as a hydrogen ion from the carbon atom next door to the one holding the bromine. The resulting re-arrangement of the electrons expels the bromine as a bromide ion and produces propene.

Elimination from unsymmetrical halogenoalkanes . . . 2-bromobutane is an unsymmetric halogenoalkane in the sense that it has a CH3 group one side of the C-Br bond and a CH2CH3 group the other. You have to be careful with compounds like this because of the possibility of more than one elimination product depending on where the hydrogen is removed from. The basic facts and mechanisms for these reactions are exactly the same as with simple halogenoalkanes like 2-bromopropane. This page only deals with the extra problems created by the possibility of more than one elimination product. Background to the mechanism You will remember that elimination happens when a hydroxide ion (from, for example, sodium hydroxide) acts as a base and removes a hydrogen as a hydrogen ion from the halogenoalkane. For example, in the simple case of elimination from 2-bromopropane:

The hydroxide ion removes a hydrogen from one of the carbon atoms next door to the carbon-bromine bond, and the various electron shifts then lead to the formation of the alkene - in this case, propene. With an unsymmetric halogenoalkane like 2-bromobutane, there are several hydrogens which might possibly get removed. You need to think about each of these possibilities. Where does the hydrogen get removed from?

The hydrogen has to be removed from a carbon atom adjacent to the carbon-bromine bond. If an OH- ion hit one of the hydrogens on the right-hand CH3 group in the 2-bromobutane (as we've drawn it), there's nowhere for the reaction to go.

To make room for the electron pair to form a double bond between the carbons, you would have to expel a hydrogen from the CH2 group as a hydride ion, H-. That is energetically much too difficult, and so this reaction doesn't happen. That still leaves the possibility of removing a hydrogen either from the left-hand CH 3 or from the CH2 group. If it was removed from the CH3 group:

The product is but-1-ene, CH2=CHCH2CH3. If it was removed from the CH2 group:

This time the product is but-2-ene, CH3CH=CHCH3. In fact the situation is even more complicated than it looks, because but-2-ene exhibits geometric isomerism. You get a mixture of two isomers formed - cis-but-2-ene and trans-but-2-ene.

Cis-but-2-ene is also known as (Z)-but-2-ene; trans-but-2-ene is also known as (E)-but-2-ene. For an explanation of the two ways of naming these two compounds. Which isomer gets formed is just a matter of chance. Geometric isomerism: Isomerism is where you can draw more than one arrangement of the atoms for a given molecular formula. Geometric isomerism is a special case of this involving molecules which have restricted rotation around one of the bonds - in this case, a carbon-carbon double bond. The C=C bond could only rotate if enough energy is put in to break the pi bond. Effectively, except at high temperatures, the C=C bond is "locked". In the case of but-2-ene, the two CH3 groups will either both be locked on one side of the C=C (to give the cis or (Z) isomer), or on opposite sides (to give the trans or (E) one). Beware! It is easy to miss geometric isomers in an exam. Always draw alkenes with the correct 120° bond angles around the C=C bond as shown in the diagrams for the cis and trans isomers above. If you take a short cut and write but-2-ene as CH 3CH=CHCH3, you will almost certainly miss the fact that cis and trans forms are possible.

The overall result Elimination from 2-bromobutane leads to a mixture containing: 

but-1-ene



cis-but-2-ene (also known as (Z)-but-2-ene)



trans-but-2-ene (also known as (E)-but-2-ene)

Elimination v. substitution . . . The reaction between a halogenoalkane and hydroxide ions can lead to either an elimination reaction or nucleophilic substitution. The reactions Both reactions involve heating the halogenoalkane under reflux with sodium or potassium hydroxide solution. Nucleophilic substitution The hydroxide ions present are good nucleophiles, and one possibility is a replacement of the halogen atom by an -OH group to give an alcohol via a nucleophilic substitution reaction.

In the example, 2-bromopropane is converted into propan-2-ol. Elimination Halogenoalkanes also undergo elimination reactions in the presence of sodium or potassium hydroxide.

The 2-bromopropane has reacted to give an alkene - propene. Notice that a hydrogen atom has been removed from one of the end carbon atoms together with the bromine from the centre one. In all simple elimination reactions the things being removed are on adjacent carbon atoms, and a double bond is set up between those carbons. What decides whether you get substitution or elimination? The reagents you are using are the same for both substitution or elimination - the halogenoalkane and either sodium or potassium hydroxide solution. In all cases, you will get a mixture of both reactions happening - some substitution and some elimination. What you get most of depends on a number of factors. The type of halogenoalkane This is the most important factor. type of halogenoalkane

substitution or elimination?

primary

mainly substitution

secondary

both substitution and elimination

tertiary

mainly elimination

For example, whatever you do with tertiary halogenoalkanes, you will tend to get mainly the elimination reaction, whereas with primary ones you will tend to get mainly substitution. However, you can influence things to some extent by changing the conditions. The solvent The proportion of water to ethanol in the solvent matters. 

Water encourages substitution.



Ethanol encourages elimination.

The temperature Higher temperatures encourage elimination. Concentration of the sodium or potassium hydroxide solution Higher concentrations favour elimination. In summary For a given halogenoalkane, to favour elimination rather than substitution, use: 

heat



a concentrated solution of sodium or potassium hydroxide



pure ethanol as the solvent

The role of the hydroxide ions The role of the hydroxide ion in a substitution reaction In the substitution reaction between a halogenoalkane and OH- ions, the hydroxide ions are acting as nucleophiles. For example, one of the lone pairs on the oxygen can attack the slightly positive carbon. This leads on to the loss of the bromine as a bromide ion, and the -OH group becoming attached in its place.

The role of the hydroxide ion in an elimination reaction Hydroxide ions have a very strong tendency to combine with hydrogen ions to make water - in other words, the OH- ion is a very strong base. In an elimination reaction, the hydroxide ion hits one of the hydrogen atoms in the CH3 group and pulls it off. This leads to a cascade of electron pair movements resulting in the formation of a carbon-carbon double bond, and the loss of the bromine as Br -.

The dehydration of alcohols The dehydration of ethanol . . . Ethanol can be dehydrated to give ethene by heating it with an excess of concentrated sulphuric acid at about 170°C. Concentrated phosphoric(V) acid, H3PO4, can be used instead.

The acids aren't written into the equation because they serve as catalysts. If you like, you could write, for example, "conc H2SO4" over the top of the arrow. The mechanism - the full version We are going to discuss the mechanism using sulphuric acid. Afterwards, we'll describe how you can use a simplified version which will work for any acid, including phosphoric(V) acid. In the first stage, one of the lone pairs of electrons on the oxygen picks up a hydrogen ion from the sulphuric acid. The alcohol is said to be protonated.

The protonated ethanol loses a water molecule to give a carbocation (a carbonium ion).

Finally, a hydrogensulphate ion (from the sulphuric acid) pulls off a hydrogen ion from the carbocation.

The mechanism - a simplified version People normally quote a simplified version of this mechanism. Instead of showing the full structure of the sulphuric acid, you write it as if it were simply a hydrogen ion, H+. That leaves the full mechanism:

An advantage of this (apart from the fact that it doesn't require you to draw the structure of sulphuric acid) is that it can be used for any acid catalyst without changing it at all. For example, if you use this version, you wouldn't need to worry about the structure of phosphoric(V) acid. Note: Although most people probably write the mechanism in this form, it is actually quite misleading because it suggests the possibility of a free hydrogen ion in a chemical system. A free hydrogen ion is a raw proton and this is always attached to something else during a chemical reaction. The dehydration of more complicated alcohols . . . You have to be wary with more complicated alcohols in case there is the possibility of more than one alkene being formed. Butan-2-ol is a good example of this, with no less than three different alkenes being formed when it is dehydrated. Butan-2-ol is just an example to illustrate the problems

The basic facts and mechanisms for these reactions are exactly the same as with a simple alcohol like ethanol. This page only deals with the extra problems created by the possibility of more than one dehydration product. Background To make the diagrams less cluttered, we'll use the simplified version of the mechanism showing gain and loss of H+. Remember that the mechanism takes place in three stages: 

The alcohol is protonated by the acid catalyst.



The protonated alcohol loses a water molecule to give a carbocation (carbonium ion).



The carbocation formed loses a hydrogen ion and forms a double bond.

So, in the case of the dehydration of a simple alcohol like ethanol:

The dehydration of butan-2-ol The first two stages There is nothing new at all in these stages. In the first stage, the alcohol is protonated by picking up a hydrogen ion from the sulphuric acid.

In the second stage, the positive ion then sheds a water molecule and produces a carbocation.

The complication arises in the next step. When the carbocation loses a hydrogen ion, where is it going to come from? Where does the hydrogen get removed from? So that a double bond can form, it will have to come from one of the carbons next door to the one with the positive charge. If a hydrogen ion is lost from the CH3 group

But-1-ene is formed.

If a hydrogen ion is lost from the CH2 group

This time the product is but-2-ene, CH3CH=CHCH3. In fact the situation is even more complicated than it looks, because but-2-ene exhibits geometric isomerism. You get a mixture of two isomers formed - cis-but-2-ene and trans-but-2-ene.

Cis-but-2-ene is also known as (Z)-but-2-ene; trans-but-2-ene is also known as (E)-but-2-ene. Which isomer gets formed is just a matter of chance. The overall result Dehydration of butan-2-ol leads to a mixture containing: 

but-1-ene



cis-but-2-ene (also known as (Z)-but-2-ene)



trans-but-2-ene (also known as (E)-but-2-ene)

NUCLEOPHILIC ADDITION MECHANISMS MENU The addition of hydrogen cyanide to aldehydes and ketones . . . Aldehydes and ketones behave identically in their reaction with hydrogen cyanide, and so will be considered together - although equations and mechanisms will be given for both types of compounds for the sake of completeness. The reaction of aldehydes and ketones with hydrogen cyanide Hydrogen cyanide adds across the carbon-oxygen double bond in aldehydes and ketones to produce compounds known as hydroxynitriles. For example, with ethanal (an aldehyde) you get 2-hydroxypropanenitrile:

With propanone (a ketone) you get 2-hydroxy-2-methylpropanenitrile:

Note: When you are naming these compounds, don't forget that the longest carbon chain has to include the carbon in the -CN group. In both of the above examples, the longest carbon chain is 3 carbons - hence the "prop" in both names. The carbon with the nitrogen attached is always counted as the number 1 carbon in the chain. The reaction isn't normally done using hydrogen cyanide itself, because this is an extremely poisonous gas. Instead, the aldehyde or ketone is mixed with a solution of sodium or potassium cyanide in water to which a little sulphuric acid has been added. The pH of the solution is adjusted to about 4 - 5, because this gives the fastest reaction. The solution will contain hydrogen cyanide (from the reaction between the sodium or potassium cyanide and the sulphuric acid), but still contains some free cyanide ions. This is important for the mechanism.

Note: If the reaction is done using hydrogen cyanide itself, a little sodium hydroxide solution is added to produce some cyanide ions from the weakly acidic HCN. Again the pH of the solution is adjusted to around pH 5 - in other words, the sodium hydroxide is not added to excess. The rate of the reaction falls if the pH is any higher. Whichever set of reagents you use, the reaction contains the same mixture of hydrogen cyanide and cyanide ions. The mechanisms These are examples of nucleophilic addition. The carbon-oxygen double bond is highly polar, and the slightly positive carbon atom is attacked by the cyanide ion acting as a nucleophile. Nucleophile: A species (molecule or ion) which attacks a positive site in something else. Nucleophiles are either fully negative ions or contain a fairly negative region somewhere in a molecule. All nucleophiles have at least one active lone pair of electrons. When you write mechanisms for reactions involving nucleophiles, you must show that lone pair.

The mechanism for the addition of HCN to propanone

In the first stage, there is a nucleophilic attack by the cyanide ion on the slightly positive carbon atom.

The negative ion formed then picks up a hydrogen ion from somewhere - for example, from a hydrogen cyanide molecule.

The hydrogen ion could also come from the water or the H3O+ ions present in the slightly acidic solution. You don't need to remember all of these. One equation is perfectly adequate. Note: The product molecule here has been drawn differently from the one in the equation further up this page. It has been rotated through 90°. There is no reason why you can't do that if it makes the appearance of the mechanism easier to follow.

The mechanism for the addition of HCN to ethanal As before, the reaction starts with a nucleophilic attack by the cyanide ion on the slightly positive carbon atom.

It is completed by the addition of a hydrogen ion from, for example, a hydrogen cyanide molecule.

Note: Again, the product molecule looks different from the one in the equation further up this page. The central carbon atom still has the same four groups attached, but to make the mechanism easier to follow, they are simply arranged differently. That's not a problem - we're still talking about the same substance.

Optical isomerism in 2-hydroxypropanenitrile When 2-hydroxypropanenitrile is made in this last mechanism, it occurs as a racemic mixture - a 50/50 mixture of two optical isomers. It is possible that you might be exected to explain how this arises. Optical isomerism occurs in compounds which have four different groups attached to a single carbon atom. In this case, the product molecule contains a CH3, a CN, an H and an OH all attached to the central carbon atom. The reason for the formation of equal amounts of two isomers lies in the way the ethanal gets attacked. Ethanal is a planar molecule, and attack by a cyanide ion will either be from above the plane of the molecule, or from below. There is an equal chance of either happening.

Attack from one side will lead to one of the two isomers, and attack from the other side will lead to the other. All aldehydes will form a racemic mixture in this way. Unsymmetrical ketones will as well. (A ketone can be unsymmetrical in the sense that there is a different alkyl group either side of the carbonyl group.) What matters is that the product molecule must have four different groups attached to the carbon which was originally part of the carbon-oxygen double bond. The reduction of aldehydes and ketones . . . The reduction of aldehydes and ketones by sodium tetrahydridoborate Sodium tetrahydridoborate (previously known as sodium borohydride) has the formula NaBH4, and contains the BH4- ion. That ion acts as the reducing agent. There are several quite different ways of carrying out this reaction. Two possible variants (there are several others!) are: 

The reaction is carried out in solution in water to which some sodium hydroxide has been added to make it alkaline. The reaction produces an intermediate which is converted into the final product by addition of a dilute acid like sulphuric acid.



The reaction is carried out in solution in an alcohol like methanol, ethanol or propan-2-ol. This produces an intermediate which can be converted into the final product by boiling it with water.

In each case, reduction essentially involves the addition of a hydrogen atom to each end of the carbon-oxygen double bond to form an alcohol. Reduction of aldehydes and ketones lead to two different sorts of alcohol. The reduction of an aldehyde For example, with ethanal you get ethanol:

Notice that this is a simplified equation - perfectly acceptable to examiners. The H in square brackets means "hydrogen from a reducing agent". In general terms, reduction of an aldehyde leads to a primary alcohol. A primary alcohol is one which only has one alkyl group attached to the carbon with the -OH group on it. They all contain the grouping -CH2OH. Note: There is one exception to this. Methanol CH3OH is also a primary alcohol. Think of this as H-CH2OH.

The reduction of a ketone For example, with propanone you get propan-2-ol:

Reduction of a ketone leads to a secondary alcohol. A secondary alcohol is one which has two alkyl groups attached to the carbon with the -OH group on it. They all contain the grouping -CHOH. The simplified mechanisms The BH4- ion is essentially a source of hydride ions, H-. The simplification used is to write H- instead of BH4-.

Doing this not only makes the initial attack easier to write, but avoids you getting involved with some quite complicated boron compounds that are formed as intermediates. The reduction is an example of nucleophilic addition. The carbon-oxygen double bond is highly polar, and the slightly positive carbon atom is attacked by the hydride ion acting as a nucleophile. A hydride ion is a hydrogen atom with an extra electron - hence the lone pair. The mechanism for the reduction of ethanal In the first stage, there is a nucleophilic attack by the hydride ion on the slightly positive carbon atom. The lone pair of electrons on the hydride ion forms a bond with the carbon, and the electrons in one of the carbon-oxygen bonds are repelled entirely onto the oxygen, giving it a negative charge.

What happens now depends on whether you add an acid or water to complete the reaction. Adding an acid: When the acid is added, the negative ion formed picks up a hydrogen ion to give an alcohol.

Note: You may find that other sources write the hydrogen ion simply as H+. That's not good practice, because it suggests a free hydrogen ion. The hydrogen ion is actually attached to a water molecule as H3O+. Writing that makes the equation look more complicated. H+(aq) is a happy compromise.

Adding water: This time, the negative ion takes a hydrogen ion from a water molecule.

The mechanism for the reduction of propanone As before, the reaction starts with a nucleophilic attack by the hydride ion on the slightly positive carbon atom.

Again, what happens next depends on whether you add an acid or water to complete the reaction. Adding an acid: The negative ion reacts with a hydrogen ion from the acid added in the second stage of the reaction.

Adding water: This time, the negative ion takes a hydrogen ion from a water molecule.

NUCLEOPHILIC ADDITION / ELIMINATION MECHANISMS MENU What are acyl chlorides? Acyl chlorides (also known as acid chlorides) are one of a number of types of compounds known as "acid derivatives". This is ethanoic acid:

If you remove the -OH group and replace it by a -Cl, you have produced an acyl chloride.

This molecule is known as ethanoyl chloride and for the rest of this topic will be taken as typical of acyl chlorides in general. Acyl chlorides are extremely reactive. They are open to attack by nucleophiles - with the overall result being a replacement of the chlorine by something else. Nucleophiles A nucleophile is a species (an ion or a molecule) which is strongly attracted to a region of positive charge in something else. Nucleophiles are either fully negative ions, or else have a strongly - charge somewhere on a molecule. The nucleophiles that we shall be looking at all depend on lone pairs on either an oxygen atom or a nitrogen atom. Nucleophiles based on oxygen atoms We shall be talking about water and alcohols, taking ethanol as a typical alcohol.

Notice how similar these two molecules are around the oxygen atom. That's what turns out to be important. Nucleophiles based on nitrogen atoms We shall be considering ammonia and primary amines, taking ethylamine as a typical primary amine. A primary amine contains the -NH2 group attached to either an alkyl group (as it is here) or a benzene ring. As far as these reactions are concerned, the nature of any hydrocarbon attached to the nitrogen makes no difference. The nitrogen atom is the important bit.

Again, notice how similar these two molecules are around the nitrogen atom - and also how similar they are to the previous ones containing oxygen. Both types of molecule have an active lone pair of electrons

attached to one of the most electronegative elements. All of these molecules also have at least one hydrogen atom attached to the oxygen or nitrogen. Why are acyl chlorides attacked by nucleophiles? The carbon atom in the -COCl group has both an oxygen atom and a chlorine atom attached to it. Both of these are very electronegative. They both pull electrons towards themselves, leaving the carbon atom quite positively charged.

The overall reaction We are going to generalise this for the moment by writing the reacting molecule as "Nu-H". Nu is the bit of the molecule which contains the nucleophilic oxygen or nitrogen atom. The attached hydrogen turns out to be essential to the reaction. The general equation for the reaction is:

In each case, the net effect is that you replace the -Cl by -Nu, and hydrogen chloride is formed as well. Since the initial attack is by a nucleophile, and the overall result is substitution, it would seem reasonable to describe the reaction as nucleophilic substitution. However, the reaction happens in two distinct stages. The first involves an addition reaction, which is followed by an elimination reaction where HCl is produced. So the mechanism is also known as nucleophilic addition / elimination. The general mechanism The addition stage of the mechanism As the lone pair on the nucleophile approaches the fairly positive carbon in the acyl chloride, it moves to form a bond with it. In the process, the two electrons in one of the carbon-oxygen bonds are repelled entirely onto the oxygen, leaving that oxygen negatively charged.

Notice the positive charge that forms on the nucleophile. Just accept this for the moment. It's much easier to explain why that charge must be there if you have a real example in front of you. This is fully explained in the pages on the reactions involving water, ammonia and so on. The elimination stage of the mechanism This happens in two steps. In the first step, the carbon-oxygen double bond reforms. To make room for it, the electrons in the carbon-chlorine bond are repelled until they are entirely on the chlorine atom - forming a chloride ion.

Finally, the chloride ion plucks the hydrogen off the original nucleophile. It removes it as a hydrogen ion, leaving the pair of electrons behind on the oxygen or nitrogen atom in that nucleophile. That cancels the positive charge.

The reaction of acyl chlorides with water . . . Ethanoyl chloride is taken as a typical acyl chloride. Any other acyl chloride will behave in the same way. Simply replace the CH3 group in what follows by anything else you want. The reaction between ethanoyl chloride and water Ethanoyl chloride reacts instantly with cold water. There is a very exothermic reaction in which a steamy acidic gas is given off (hydrogen chloride) and ethanoic acid is formed.

The mechanism The first stage (the addition stage of the reaction) involves a nucleophilic attack on the fairly positive carbon atom by one of the lone pairs on the oxygen of a water molecule.

Note: Only one of the two lone pairs on the oxygen in water is shown. This is to avoid cluttering an already complicated diagram with things that aren't relevant.

The second stage (the elimination stage) happens in two steps. In the first, the carbon-oxygen double bond reforms and a chloride ion is pushed off.

That is followed by removal of a hydrogen ion by the chloride ion to give ethanoic acid and hydrogen chloride.

The reaction of acyl chlorides with alcohols . . . Ethanoyl chloride is taken as a typical acyl chloride. Any other acyl chloride will behave in the same way. Simply replace the CH3 group in what follows by anything else you want. Similarly, ethanol is taken as a typical alcohol. If you are interested in another alcohol, you can replace the CH 3CH2 group by any other alkyl group. The reaction between ethanoyl chloride and ethanol Ethanoyl chloride reacts instantly with cold ethanol. There is a very exothermic reaction in which a steamy acidic gas is given off (hydrogen chloride). Ethyl ethanoate (an ester) is formed.

The mechanism The first stage (the addition stage of the reaction) involves a nucleophilic attack on the fairly positive carbon atom by one of the lone pairs on the oxygen of an ethanol molecule.

Note: Only one of the two lone pairs on the oxygen in the ethanol is shown. This is to avoid cluttering an already complicated diagram with things that aren't relevant.

The second stage (the elimination stage) happens in two steps. In the first, the carbon-oxygen double bond reforms and a chloride ion is pushed off.

That is followed by removal of a hydrogen ion by the chloride ion to give ethyl ethanoate and hydrogen chloride.

The reaction of acyl chlorides with ammonia . . . Ethanoyl chloride is taken as a typical acyl chloride. Any other acyl chloride will behave in the same way. Simply replace the CH3 group in what follows by anything else you want. The reaction between ethanoyl chloride and ammonia Ethanoyl chloride reacts violently with a cold concentrated solution of ammonia. A white solid product is formed which is a mixture of ethanamide (an amide) and ammonium chloride.

Notice that, unlike the reactions between ethanoyl chloride and water or ethanol, hydrogen chloride isn't produced - at least, not in any quantity. Any hydrogen chloride formed would immediately react with excess ammonia to give ammonium chloride.

The mechanism The first stage (the addition stage of the reaction) involves a nucleophilic attack on the fairly positive carbon atom by the lone pair on the nitrogen atom in the ammonia.

The second stage (the elimination stage) happens in two steps. In the first, the carbon-oxygen double bond reforms and a chloride ion is pushed off.

That is followed by removal of a hydrogen ion from the nitrogen. This might happen in one of two ways: It might be removed by a chloride ion, producing HCl (which would immediately react with excess ammonia to give ammonium chloride as above) . . .

and

. . . or it might be removed directly by an ammonia molecule.

The ammonium ion, together with the chloride ion already there, makes up the ammonium chloride formed in the reaction. The reaction of acyl chlorides with primary amines . . .

Ethanoyl chloride is taken as a typical acyl chloride. Any other acyl chloride will behave in the same way. Simply replace the CH3 group in what follows by anything else you want. Similarly, ethylamine is taken as a typical amine. Any other amine will behave in the same way. Replacing the CH3CH2 group by any other hydrocarbon group won't affect the mechanism in any way. The reaction between ethanoyl chloride and ethylamine Ethanoyl chloride reacts violently with a cold concentrated solution of ethylamine. A white solid product is formed which is a mixture of N-ethylethanamide (an N-substituted amide) and ethylammonium chloride.

Notice that, unlike the reactions between ethanoyl chloride and water or ethanol, hydrogen chloride isn't produced - at least, not in any quantity. Any hydrogen chloride formed would immediately react with excess ethylamine to give ethylammonium chloride.

The mechanism The first stage (the addition stage of the reaction) involves a nucleophilic attack on the fairly positive carbon atom by the lone pair on the nitrogen atom in the ethylamine.

The second stage (the elimination stage) happens in two steps. In the first, the carbon-oxygen double bond reforms and a chloride ion is pushed off.

That is followed by removal of a hydrogen ion from the nitrogen. This might happen in one of two ways: It might be removed by a chloride ion, producing HCl (which would immediately react with excess ethylamine to give ethylammonium chloride as above) . . .

and

. . . or it might be removed directly by an ethylamine molecule.

The ethylammonium ion, together with the chloride ion already there, makes up the ethylammonium chloride formed in the reaction.