DRAWING ORGANIC MOLECULES

This page explains the various ways that organic molecules can be represented on paper or on screen - including molecular formulae, and various forms of structural formulae.

Molecular formulae

A molecular formula simply counts the numbers of each sort of atom present in the molecule, but tells you nothing about the way they are joined together.

For example, the molecular formula of butane is C4H10, and the molecular formula of ethanol is C2H6O.

Molecular formulae are very rarely used in organic chemistry, because they don't give any useful information about the bonding in the molecule. About the only place where you might come across them is in equations for the combustion of simple hydrocarbons, for example:

In cases like this, the bonding in the organic molecule isn't important.

Structural formulae

A structural formula shows how the various atoms are bonded. There are various ways of drawing this and you will need to be familiar with all of them.

Displayed formulae

A displayed formula shows all the bonds in the molecule as individual lines. You need to remember that each line represents a pair of shared electrons.

For example, this is a model of methane together with its displayed formula:

Notice that the way the methane is drawn bears no resemblance to the actual shape of the molecule. Methane isn't flat with 90° bond angles. This mismatch between what you draw and what the molecule actually looks like can lead to problems if you aren't careful.

For example, consider the simple molecule with the molecular formula CH2Cl2. You might think that there were two different ways of arranging these atoms if you drew a displayed formula.

The chlorines could be opposite each other or at right angles to each other. But these two structures are actually exactly the same. Look at how they appear as models.

One structure is in reality a simple rotation of the other one.

Note:  This is all much easier to understand if you have actually got some models to play with. If your school or college hasn't given you the opportunity to play around with molecular models in the early stages of your organic chemistry course, you might consider getting hold of a cheap set. The models made by molymod are both cheap and easy to use. An introductory organic set is more than adequate. Find them at www.molymod.com.

Alternatively , get hold of some coloured Plasticene and some used matches and make your own. It's cheaper, but distinctly messier!

Consider a slightly more complicated molecule, C2H5Cl. The displayed formula could be written as either of these:

But, again these are exactly the same. Look at the models.

The commonest way to draw structural formulae

For anything other than the most simple molecules, drawing a fully displayed

formula is a bit of a bother - especially all the carbon-hydrogen bonds. You can simplify the formula by writing, for example, CH3 or CH2 instead of showing all these bonds.

So for example, ethanoic acid would be shown in a fully displayed form and a simplified form as:

You could even condense it further to CH3COOH, and would probably do this if you had to write a simple chemical equation involving ethanoic acid. You do, however, lose something by condensing the acid group in this way, because you can't immediately see how the bonding works.

You still have to be careful in drawing structures in this way. Remember from above that these two structures both represent the same molecule:

The next three structures all represent butane.

All of these are just versions of four carbon atoms joined up in a line. The only difference is that there has been some rotation about some of the carbon-carbon bonds. You can see this in a couple of models.

Not one of the structural formulae accurately represents the shape of butane. The convention is that we draw it with all the carbon atoms in a straight line - as in the first of the structures above.

This is even more important when you start to have branched chains of carbon atoms. The following structures again all represent the same molecule - 2-methylbutane.

The two structures on the left are fairly obviously the same - all we've done is flip the molecule over. The other one isn't so obvious until you look at the structure in detail. There are four carbons joined up in a row, with a CH3 group attached to the next-to-end one. That's exactly the same as the other two structures. If you had a model, the only difference between these three diagrams is that you have rotated some of the bonds and turned the model around a bit.

To overcome this possible confusion, the convention is that you always look for the longest possible chain of carbon atoms, and then draw it horizontally. Anything else is simply hung off that chain.

It doesn't matter in the least whether you draw any side groups pointing up or down. All of the following represent exactly the same molecule.

If you made a model of one of them, you could turn it into any other one simply by rotating one or more of the carbon-carbon bonds.

How to draw structural formulae in 3-dimensions

There are occasions when it is important to be able to show the precise 3-D arrangement in parts of some molecules. To do this, the bonds are shown using conventional symbols:

For example, you might want to show the 3-D arrangement of the groups

around the carbon which has the -OH group in butan-2-ol.

Butan-2-ol has the structural formula:

Using conventional bond notation, you could draw it as, for example:

The only difference between these is a slight rotation of the bond between the centre two carbon atoms. This is shown in the two models below. Look carefully at them - particularly at what has happened to the lone hydrogen atom. In the left-hand model, it is tucked behind the carbon atom. In the right-hand model, it is in the same plane. The change is very slight.

It doesn't matter in the least which of the two arrangements you draw. You could easily invent other ones as well. Choose one of them and get into the habit of drawing 3-dimensional structures that way. My own habit (used elsewhere on this site) is to draw two bonds going back into the paper and one coming out - as in the left-hand diagram above.

Notice that no attempt was made to show the whole molecule in 3-dimensions in the structural formula diagrams. The CH2CH3 group was left in a simple form. Keep diagrams simple - trying to show too much detail makes the whole thing amazingly difficult to understand!

Skeletal formulae

In a skeletal formula, all the hydrogen atoms are removed from carbon chains, leaving just a carbon skeleton with functional groups attached to it.

For example, we've just been talking about butan-2-ol. The normal structural formula and the skeletal formula look like this:

In a skeletal diagram of this sort

there is a carbon atom at each junction between bonds in a chain and at the end of each bond (unless there is something else there already - like the -OH group in the example);

there are enough hydrogen atoms attached to each carbon to make the total number of bonds on that carbon up to 4.

Beware! Diagrams of this sort take practice to interpret correctly - and may well not be acceptable to your examiners (see below).

There are, however, some very common cases where they are frequently used. These cases involve rings of carbon atoms which are surprisingly awkward to draw tidily in a normal structural formula.

Cyclohexane, C6H12, is a ring of carbon atoms each with two hydrogens attached. This is what it looks like in both a structural formula and a skeletal formula.

And this is cyclohexene, which is similar but contains a double bond:

But the commonest of all is the benzene ring, C6H6, which has a special symbol of its own.

Note:  Explaining exactly what this structure means needs more space than is available here. It is explained in full in two pages on the structure of benzene elsewhere in this site. It would probably be better not to follow this link unless you are actively interested in benzene chemistry at the moment - it will lead you off into quite deep water!

Deciding which sort of formula to use

There's no easy, all-embracing answer to this problem. It depends more than anything else on experience - a feeling that a particular way of writing a formula is best for the situation you are dealing with.

Don't worry about this - as you do more and more organic chemistry, you will probably find it will come naturally. You'll get so used to writing formulae in reaction mechanisms, or for the structures for isomers, or in simple chemical equations, that you won't even think about it.

There are, however, a few guidelines that you should follow.

What does your syllabus say?

Different examiners will have different preferences. Check first with your syllabus. If you've down-loaded a copy of your syllabus from your examiners' web site, it is easy to check what they say they want. Use the "find" function on your Adobe Acrobat Reader to search the organic section(s) of the syllabus for the word "formula".

You should also check recent exam papers and (particulary) mark schemes to find out what sort of formula the examiners really prefer in given situations. You could also look at any support material published by your examiners.

Note:  If you are working to a UK-based syllabus and haven't got a copy of that syllabus and recent exam papers, follow this link to find out how to get them.

What if you still aren't sure?

Draw the most detailed formula that you can fit into the space available. If in doubt, draw a fully displayed formula. You would never lose marks for giving too much detail.

Apart from the most trivial cases (for example, burning hydrocarbons), never use a molecular formula. Always show the detail around the important part(s) of a molecule. For example, the important part of an ethene molecule is the carbon-carbon double bond - so write (at the very least) CH2=CH2 and not C2H4.

Where a particular way of drawing a structure is important, this will always be pointed out where it arises elsewhere on this site.

THE NAMES OF ORGANIC COMPOUNDS

This page explains how to write the formula for an organic compound given its name - and vice versa. It covers alkanes, cycloalkanes, alkenes, simple compounds containing halogens, alcohols, aldehydes and ketones. At the bottom of the page, you will find links to other types of compound.

Background

How this page is going to tackle the problem

There are two skills you have to develop in this area:

You need to be able to translate the name of an organic compound into its structural formula.

You need to be able to name a compound from its given formula.

The first of these is more important (and also easier!) than the second. In an exam, if you can't write a formula for a given compound, you aren't going to know what the examiner is talking about and could lose lots of marks. However, you might only be asked to write a name for a given formula once in a whole exam - in which case you only risk 1 mark.

So, we're going to look mainly at how you decode names and turn them into formulae. In the process you will also pick up tips about how to produce names yourself.

In the early stages of an organic chemistry course people frequently get confused and daunted by the names because they try to do too much at once. Don't try to read all these pages in one go. Just go as far as the compounds you are interested in at the moment and ignore the rest. Come back to them as they arise during the natural flow of your course.

Cracking the code

A modern organic name is simply a code. Each part of the name gives you some useful information about the compound.

For example, to understand the name 2-methylpropan-1-ol you need to take the name to pieces.

The prop in the middle tells you how many carbon atoms there are in the longest chain (in this case, 3). The an which follows the "prop" tells you that there aren't any carbon-carbon double bonds.

The other two parts of the name tell you about interesting things

which are happening on the first and second carbon atom in the chain. Any name you are likely to come across can be broken up in this same way.

Counting the carbon atoms

You will need to remember the codes for the number of carbon atoms in a chain up to 6 carbons. There is no easy way around this - you have got to learn them. If you don't do this properly, you won't be able to name anything!

code no of carbons

meth 1

eth 2

prop 3

but 4

pent 5

hex 6

Types of carbon-carbon bonds

Whether or not the compound contains a carbon-carbon double bond is shown by the two letters immediately after the code for the chain length.

code means

an only carbon-carbon single bonds

en contains a carbon-carbon double bond

For example, butane means four carbons in a chain with no double bond.

Propene means three carbons in a chain with a double bond between two of the carbons.

Alkyl groups

Compounds like methane, CH4, and ethane, CH3CH3, are members of a family of compounds called alkanes. If you remove a hydrogen atom from one of these you get an alkyl group.

For example:

A methyl group is CH3.

An ethyl group is CH3CH2.

These groups must, of course, always be attached to something else.

Types of compounds

The alkanes

Example 1:  Write the structural formula for 2-methylpentane.

Start decoding the name from the bit that counts the number of carbon atoms in the longest chain - pent counts 5 carbons.

Are there any carbon-carbon double bonds? No - an tells you there aren't any.

Now draw this carbon skeleton:

Put a methyl group on the number 2 carbon atom:

Does it matter which end you start counting from? No - if you counted from the other end, you would draw the next structure. That's exactly the same as the first one, except that it has been flipped over.

Finally, all you have to do is to put in the correct number of hydrogen atoms on each carbon so that each carbon is forming four bonds.

If you had to name this yourself:

Count the longest chain of carbons that you can find. Don't assume that you have necessarily drawn that chain horizontally. 5 carbons means pent.

Are there any carbon-carbon double bonds? No - therefore pentane.

There's a methyl group on the number 2 carbon - therefore 2-methylpentane. Why the number 2 as opposed to the number 4 carbon? In other words, why do we choose to number from this particluar end? The convention is that you number from the end which produces the lowest numbers in the name - hence 2- rather than 4-.

Example 2:  Write the structural formula for 2,3-dimethylbutane.

Start with the carbon backbone. There are 4 carbons in the longest chain (but) with no carbon-carbon double bonds (an).

This time there are two methyl groups (di) on the number 2 and number 3 carbon atoms.

Completing the formula by filling in the hydrogen atoms gives:

Note:  Does it matter whether you draw the two methyl groups one up and one down, or both up, or both down? Not in the least! If you aren't sure about drawing organic molecules, follow this link before you go on. Use the BACK button on your browser to return to this page.

Example 3:  Write the structural formula for 2,2-dimethylbutane.

This is exactly like the last example, except that both methyl groups are on the same carbon atom. Notice that the name shows this by using 2,2- as well as di. The structure is worked out as before:

Example 4:  Write the structural formula for 3-ethyl-2-

methylhexane.

hexan shows a 6 carbon chain with no carbon-carbon double bonds.

This time there are two different alkyl groups attached - an ethyl group on the number 3 carbon atom and a methyl group on number 2.

Filling in the hydrogen atoms gives:

Note:  Once again it doesn't matter whether the ethyl and methyl groups point up or down. You might also have chosen to start numbering from the right-hand end of the chain. These would all be perfectly valid structures. All you would have done is to rotate the whole molecule in space, or rotate it around particular bonds.

If you aren't sure about this, then you must read about drawing organic molecules before you go on.

If you had to name this yourself:

How do you know what order to write the different alkyl groups at the beginning of the name? The convention is that you write them in alphabetical order - hence ethyl comes before methyl which in turn comes before propyl.

The cycloalkanes

In a cycloalkane the carbon atoms are joined up in a ring - hence cyclo.

Example:  Write the structural formula for cyclohexane.

hexan shows 6 carbons with no carbon-carbon double bonds. cyclo shows that they are in a ring. Drawing the ring and putting in the correct number of hydrogens to satisfy the bonding requirements of the carbons gives:

The alkenes

Example 1:  Write the structural formula for propene.

prop counts 3 carbon atoms in the longest chain. en tells you that there is a carbon-carbon double bond. That means that the carbon skeleton looks like this:

Putting in the hydrogens gives you:

Example 2:  Write the structural formula for but-1-ene.

but counts 4 carbon atoms in the longest chain and en tells you that there is a carbon-carbon double bond. The number in the name tells you where the double bond starts.

No number was necessary in the propene example above because the double bond has to start on one of the end carbon atoms. In the case of butene, though, the double bond could either be at the end of the chain or in the middle - and so the name has to code for the its position.

The carbon skeleton is:

And the full structure is:

Incidentally, you might equally well have decided that the right-hand carbon was the number 1 carbon, and drawn the structure as:

Example 3:  Write the structural formula for 3-methylhex-2-ene.

The longest chain has got 6 carbon atoms (hex) with a double bond starting on the second one (-2-en).

But this time there is a methyl group attached to the chain on the

number 3 carbon atom, giving you the underlying structure:

Adding the hydrogens gives the final structure:

Be very careful to count the bonds around each carbon atom when you put the hydrogens in. It would be very easy this time to make the mistake of writing an H after the third carbon - but that would give that carbon a total of 5 bonds.

Compounds containing halogens

Example 1:  Write the structural formula for 1,1,1-trichloroethane.

This is a two carbon chain (eth) with no double bonds (an). There are three chlorine atoms all on the first carbon atom.

Example 2:  Write the structural formula for 2-bromo-2-methylpropane.

First sort out the carbon skeleton. It's a three carbon chain with no double bonds and a methyl group on the second carbon atom.

Draw the bromine atom which is also on the second carbon.

And finally put the hydrogen atoms in.

If you had to name this yourself:

Notice that the whole of the hydrocarbon part of the name is written

together - as methylpropane - before you start adding anything else on to the name.

Example 2:  Write the structural formula for 1-iodo-3-methylpent-2-ene.

This time the longest chain has 5 carbons (pent), but has a double bond starting on the number 2 carbon. There is also a methyl group on the number 3 carbon.

Now draw the iodine on the number 1 carbon.

Giving a final structure:

Note:  You could equally well draw this molecule the other way round, but normally where you have, say, 1-bromo-something, you tend to write the bromine (or other halogen) on the right-hand end of the structure.

Alcohols

All alcohols contain an -OH group. This is shown in a name by the ending ol.

Example 1:  Write the structural formula for methanol.

This is a one carbon chain with no carbon-carbon double bond (obviously!). The ol ending shows it's an alcohol and so contains an -OH group.

Example 2:  Write the structural formula for 2-methylpropan-1-ol.

The carbon skeleton is a 3 carbon chain with no carbon-carbon double bonds, but a methyl group on the number 2 carbon.

The -OH group is attached to the number 1 carbon.

The structure is therefore:

Example 3:  Write the structural formula for ethane-1,2-diol.

This is a two carbon chain with no double bond. The diol shows 2 -OH groups, one on each carbon atom.

Note:  There's no particular significance in the fact that this formula has the carbon chain drawn vertically. If you draw it horizontally, unless you stretch the carbon-carbon bond a lot, the -OH groups look very squashed together. Drawing it vertically makes it look tidier!

Aldehydes

All aldehydes contain the group:

If you are going to write this in a condensed form, you write it as -CHO - never as -COH, because that looks like an alcohol.

The names of aldehydes end in al.

Example 1:  Write the structural formula for propanal.

This is a 3 carbon chain with no carbon-carbon double bonds. The al ending shows the presence of the -CHO group. The carbon in that group counts as one of the chain.

Example 2:  Write the structural formula for 2-methylpentanal.

This time there are 5 carbons in the longest chain, including the one in the -CHO group. There aren't any carbon-carbon double bonds. A methyl group is attached to the number 2 carbon. Notice that in aldehydes, the carbon in the -CHO group is always counted as the number 1 carbon.

Ketones

Ketones contain a carbon-oxygen double bond just like aldehydes, but this time it's in the middle of a carbon chain. There isn't a hydrogen atom attached to the group as there is in aldehydes.

Ketones are shown by the ending one.

Example 1:  Write the structural formula for propanone.

This is a 3 carbon chain with no carbon-carbon double bond. The carbon-oxygen double bond has to be in the middle of the chain and so must be on the number 2 carbon.

Ketones are often written in this way to emphasise the carbon-oxygen double bond.

Example 2:  Write the structural formula for pentan-3-one.

This time the position of the carbon-oxygen double bond has to be stated because there is more than one possibility. It's on the third carbon of a 5 carbon chain with no carbon-carbon double bonds. If it was on the second carbon, it would be pentan-2-one.

This could equally well be written:

© Jim Clark 2000

THE NAMES OF MORE ORGANIC COMPOUNDS

This page continues looking at the names of organic compounds containing chains of carbon atoms. It assumes that you have already looked at the introductory page covering compounds from alkanes to ketones.

Note:  If you haven't already looked at that page, it would be a good idea to do so before you go on. The names on this second page aren't explained in quite as much detail as those on the introductory page - it assumes that you have already understood the main principles. If in doubt, follow this link first.

More types of organic compound

Carboxylic acids

Carboxylic acids contain the -COOH group, which is better written out in full as:

Carboxylic acids are shown by the ending oic acid. When you count the carbon chain, you have to remember to include the carbon in the -COOH group. That carbon is always thought of as number 1 in the chain.

Example 1:  Write the structural formula for 3-methylbutanoic acid.

This is a four carbon acid with no carbon-carbon double bonds. There is a methyl group on the third carbon (counting the -COOH carbon as number 1).

Example 2:  Write the structural formula for 2-hydroxypropanoic acid.

The hydroxy part of the name shows the presence of an -OH group. Normally, you would show that by the ending ol, but this time you can't because you've already got another ending. You are forced into this alternative way of describing it.

The old name for 2-hydroxypropanoic acid is lactic acid. That name sounds more friendly, but is utterly useless when it comes to writing a formula for it. In the old days, you would have had to learn the formula rather than just working it out should you need it.

Example 3:  Write the structural formula for 2-chlorobut-3-enoic acid.

This time, not only is there a chlorine attached to the chain, but the chain also contains a carbon-carbon double bond (en) starting on the number 3 carbon (counting the -COOH carbon as number 1).

Salts of carboxylic acids

Example:  Write the structural formula for sodium propanoate.

This is the sodium salt of propanoic acid - so start from that. Propanoic acid is a three carbon acid with no carbon-carbon double bonds.

When the carboxylic acids form salts, the hydrogen in the -COOH group is replaced by a metal. Sodium propanoate is therefore:

Notice that there is an ionic bond between the sodium and the propanoate group. Whatever you do, don't draw a line between the sodium and the oxygen. That would represent a covalent bond. It's wrong, and makes you look very incompetent in an exam!

In a shortened version, sodium propanoate would be written CH3CH2COONa or, if you wanted to emphasise the ionic nature, as CH3CH2COO- Na+.

Note:  The confusing thing about these salts (and even more so for the esters that are coming up next) is that they are named the wrong way round. In the formula, the sodium is at the end, but appears first in the name. Why?

Salts are always named with the metal first - think of sodium chloride or potassium iodide. So for consistency you would need to reverse the formula of sodium propanoate - NaOOCCH2CH3. But if you reverse the formula, you can't see immediately that it is related to propanoic acid. So you learn to live with the inconsistency.

Esters

Esters are one of a number of compounds known collectively as acid derivatives. In these the acid group is modified in some way. In an ester, the hydrogen in the -COOH group is replaced by an alkyl group (or possibly some more complex hydrocarbon group).

Example 1:  Write the structural formula for methyl propanoate.

An ester name has two parts - the part that comes from the acid (propanoate) and the part that shows the alkyl group (methyl).

Start by thinking about propanoic acid - a 3 carbon acid with no carbon-carbon double bonds.

The hydrogen in the -COOH group is replaced by an alkyl group - in this case, a methyl group.

Ester names are confusing because the name is written backwards from the way the structure is drawn. There's no way round this - you just have to get used to it!

In the shortened version, this formula would be written CH3CH2COOCH3.

Example 2:  Write the structural formula for ethyl ethanoate.

This is probably the most commonly used example of an ester. It is based on ethanoic acid ( hence, ethanoate) - a 2 carbon acid. The hydrogen in the -COOH group is replaced by an ethyl group.

Make sure that you draw the ethyl group the right way round. A fairly common mistake is to try to join the CH3 group to the oxygen. If you count the bonds if you do that, you will find that both the CH3 carbon and the CH2 carbon have the wrong number of bonds.

Acyl chlorides (acid chlorides)

An acyl chloride is another acid derivative. In this case, the -OH group of the acid is replaced by -Cl. All acyl chlorides contain the -COCl group:

Example:  Write the structural formula for ethanoyl chloride.

Acyl chlorides are shown by the ending oyl chloride. So ethanoyl chloride is based on a 2 carbon chain with no carbon-carbon double bonds and a -COCl group. The carbon in that group counts as part of the chain. In a longer chain, with side groups attached, the -COCl carbon is given the number 1 position.

Acid anhydrides

Another acid derivative! An acid anhydride is what you get if you dehydrate an acid - that is, remove water from it.

Example:  Write the structural formula for propanoic anhydride.

These are most easily worked out by writing it down on a scrap of paper in the following way:

Draw two molecules of acid arranged so that the -OH groups are next to each other. Tweak out a molecule of water - and then join up what's left. In this case, because you want propanoic anhydride, you draw two molecules of propanoic acid.

Amides

Yet another acid derivative! Amides contain the group -CONH2 where the -OH of an acid is replaced by -NH2.

Example:  Write the structural formula for propanamide.

This is based on a 3 carbon chain with no carbon-carbon double bonds. At the end of the chain is a -CONH2 group. The carbon in that group counts as part of the chain.

Nitriles

Nitriles contain a -CN group, and used to be called cyanides.

Example 1:  Write the structural formula for ethanenitrile.

The name shows a 2 carbon chain with no carbon-carbon double bond. nitrile shows a -CN group at the end of the chain. As with the previous examples involving acids and acid derivatives, don't forget that the carbon in the -CN group counts as part of the chain.

The old name for this would have been methyl cyanide. You might think that that's easier, but as soon as the chain gets more complicated, it doesn't work - as the next example shows.

Example 2:  Write the structural formula for 2-hydroxypropanenitrile.

Here we've got a 3 carbon chain, no carbon-carbon double bonds, and a -CN group on the end of the chain. The carbon in the -CN group counts as the number 1 carbon. On the number 2 carbon there is an -OH group (hydroxy). Notice that you can't use the ol ending because you've already got a nitrile ending.

Primary amines

A primary amine contains the group -NH2 attached to a hydrocarbon chain or ring. You can think of amines in general as being derived from ammonia, NH3. In a primary amine, one of the hydrogens has been replaced by a hydrocarbon group.

Example 1:  Write the structural formula for ethylamine.

In this case, an ethyl group is attached to the -NH2 group.

This name (ethylamine) is fine as long as you've only got a short chain where there isn't any ambiguity about where the -NH2 group

is found. But suppose you had a 3 carbon chain - in this case, the -NH2 group could be on an end carbon or on the middle carbon. How you get around that problem is illustrated in the next example.

Example 2:  Write the structural formula for 2-aminopropane.

The name shows a 3 carbon chain with an amino group attached to the second carbon.  amino shows the -NH2 group.

Ethylamine (example 1 above) could equally well have been called aminoethane.

Secondary and tertiary amines

You are only likely to come across simple examples of these. In a secondary amine, two of the hydrogen atoms in an ammonia molecule have been replaced by hydrocarbon groups. In a tertiary amine, all three hydrogens have been replaced.

Example 1:  Write the structural formula for dimethylamine.

In this case, two of the hydrogens in ammonia have been replaced by methyl groups.

Example 2:  Write the structural formula for trimethylamine.

Here, all three hydrogens in ammonia have been replaced by methyl groups.

Amino acids

An amino acid contains both an amino group, -NH2, and a carboxylic acid group, -COOH, in the same molecule. As with all acids the carbon chain is numbered so that the carbon in the -COOH group is counted as number 1.

Example:  Write the structural formula for 2-aminopropanoic acid.

This has a 3 carbon chain with no carbon-carbon double bonds. On the second carbon (counting the -COOH carbon as number 1) there is an amino group, -NH2.

© Jim Clark 2000

THE NAMES OF AROMATIC COMPOUNDS

This page looks at the names of some simple aromatic compounds. An aromatic compound is one which contains a benzene ring. It assumes that you are reasonably confident about naming compounds containing chains of carbon atoms (aliphatic compounds).

Note:  If you aren't sure about naming aliphatic compounds follow this link before you go on.

Naming aromatic compounds isn't quite so straightforward as naming chain compounds. Often, more than one name is acceptable and it's not uncommon to find the old names still in use as well.

Background

The benzene ring

All aromatic compounds are based on benzene, C6H6, which has a ring of six carbon atoms and has the symbol:

Each corner of the hexagon has a carbon atom with a hydrogen attached.

Note:  If you don't understand this structure, it is explained in full in two pages on the structure of benzene elsewhere in this site. Following this link could well take you some time!

The phenyl group

Remember that you get a methyl group, CH3, by removing a hydrogen from methane, CH4.

You get a phenyl group, C6H5, by removing a hydrogen from a benzene ring, C6H6. Like a methyl or an ethyl group, a phenyl group is always attached to something else.

Aromatic compounds with only one group attached to the benzene ring

Cases where the name is based on benzene

chlorobenzene

This is a simple example of a halogen attached to the benzene ring. The name is self-obvious.

The simplified formula for this is C6H5Cl. You could therefore (although you never do!) call it phenyl chloride. Whenever you draw a benzene ring with one other thing attached to it, you are in fact drawing a phenyl group. In order to attach something else, you have to remove one of the existing hydrogen atoms, and so automatically make a phenyl group.

nitrobenzene

The nitro group, NO2, is attached to a benzene ring.

The simplified formula for this is C6H5NO2.

methylbenzene

Another obvious name - the benzene ring has a methyl group attached. Other alkyl side-chains would be named similarly - for example, ethylbenzene. The old name for methylbenzene is toluene, and you may still meet that.

The simplified formula for this is C6H5CH3.

(chloromethyl)benzene

A variant on this which you may need to know about is where one of the hydrogens on the CH3 group is replaced by a chlorine atom. Notice the brackets around the (chloromethyl) in the name. This is so that you are sure that the chlorine is part of the methyl group and not somewhere else on the ring.

If more than one of the hydrogens had been replaced by chlorine, the names would be (dichloromethyl)benzene or

(trichloromethyl)benzene. Again, notice the importance of the brackets in showing that the chlorines are part of the side group and not directly attached to the ring.

benzoic acid (benzenecarboxylic acid)

Benzoic acid is the older name, but is still in common use - it's a lot easier to say and write than the modern alternative! Whatever you call it, it has a carboxylic acid group, -COOH, attached to the benzene ring.

Cases where the name is based on phenyl

Remember that the phenyl group is a benzene ring minus a hydrogen atom - C6H5. If you draw a benzene ring with one group attached, you have drawn a phenyl group.

phenylamine

Phenylamine is a primary amine and contains the -NH2 group attached to a benzene ring.

The old name for phenylamine is aniline, and you could also reasonably call it aminobenzene. Phenylamine is what it is most commonly for UK-based exam purposes.

Note:  In all cases where there is some possibility of alternative names, you need to know what your examiners are likely to call a particular compound. Refer to your syllabus and recent exam papers. If you are working to a UK-based syllabus for 16 - 8 year olds, and haven't got these, follow this link to find out how to get hold of them.

phenylethene

This is an ethene molecule with a phenyl group attached. Ethene is a two carbon chain with a carbon-carbon double bond. Phenylethene is therefore:

The old name for phenylethene is styrene - the monomer from which polystyrene is made.

phenylethanone

This is a slightly awkward name - take it to pieces. It consists of a two carbon chain with no carbon-carbon double bond. The one ending shows that it is a ketone, and so has a C=O group somewhere in the middle. Attached to the carbon chain is a phenyl group. Putting that together gives:

phenyl ethanoate

This is an ester based on ethanoic acid. The hydrogen atom in the -COOH group has been replaced by a phenyl group.

Note:  If you aren't happy about naming esters, follow this link before you go on.

phenol

Phenol has an -OH group attached to a benzene ring and so has a formula C6H5OH.

Aromatic compounds with more than one group attached to the benzene ring

Numbering the ring

Any group already attached to the ring is given the number 1 position. Where you draw it on the ring (at the top or in any other position) doesn't matter - that's just a question of rotating the molecule a bit. It's much easier, though, to get in the habit of drawing your main group at the top.

The other ring positions are then numbered from 2 to 6. You can number them either clockwise or anti-clockwise. As with chain compounds, you number the ring so that the name you end up with has the smallest possible numbers in it. Examples will make this clear.

Some simple examples

Substituting chlorine atoms on the ring

Look at these compounds:

All of these are based on methylbenzene and so the methyl group is given the number 1 position on the ring.

Why is it 2-chloromethylbenzene rather than 6-chloromethylbenzene? The ring is numbered clockwise in this case because that produces a 2- in the name rather than a 6-. 2 is smaller than 6.

Warning!  You will find all sorts of variations on this depending on the age of the book you look it up in, and where it was published. What I have described above isn't in strict accordance with the most modern interpretation of the IUPAC recommendations for naming organic compounds.

The names should actually be 1-chloro-2-methylbenzene, 1-chloro-3-methylbenzene, and so on. The substituted groups are named in alphabetical order, and the "1" position is assigned to the first of these - rather than to the more logical methyl group.

This produces some silly inconsistencies. For example, if you had the exactly equivalent compounds containing nitro groups in place of the chlorines, the names would change completely, to 1-methyl-2-nitrobenzene, 1-methyl-3-nitrobenzene, etc. In this case, the normal practice of naming the hydrocarbon first, and then attaching other things to it has been completely wrecked.

Do you need to worry about this? NO! It is extremely unlikely that

you would ever be asked to name these in an exam, and it is always easy to write a structure from one of these names - however illogical it may be! There is a simple rule for exam purposes. Unless you are specifically asked for the name of anything remotely complicated, don't give it. As long as you have got the structure right, that's all that matters.

2-hydroxybenzoic acid

This might also be called 2-hydroxybenzenecarboxylic acid. There is a -COOH group attached to the ring and, because the name is based on benzoic acid, that group is assigned the number 1 position. Next door to it in the 2 position is a hydroxy group, -OH.

benzene-1,4-dicarboxylic acid

The di shows that there are two carboxylic acid groups, -COOH, one of them in the 1 position and the other opposite it in the 4 position.

2,4,6-trichlorophenol

This is based on phenol - with an -OH group attached in the number 1 position on the ring. There are 3 chlorine atoms substituted onto the ring in the 2, 4 and 6 positions.

methyl 3-nitrobenzoate

This is a name you might come across as a part of a practical exercise in nitrating benzene rings. It's included partly for that reason, and partly because it is a relatively complicated name to finish with!

The structure of the name shows that it is an ester. You can tell that from the oate ending, and the methyl group floating separately from the rest of the name at the beginning.

The ester is based on the acid, 3-nitrobenzoic acid - so start with that.

There will be a benzene ring with a -COOH group in the number 1 position and a nitro group, NO2, in the 3 position. The -COOH group is modified to make an ester by replacing the hydrogen of the -COOH group by a methyl group.

Methyl 3-nitrobenzoate is therefore:

USING CURLY ARROWS IN REACTION MECHANISMS

This page explains the use of curly arrows to show the movement both of electron pairs and of single electrons during organic reaction mechanisms.

You can jump straight to the movement of single electrons further down this page if that is all you are interested in for the moment (for example, if you are currently working on free radical reactions).

Using curly arrows to show the movement of electron pairs

Curly arrows (and that's exactly what they are called!) are used in mechanisms to show the various electron pairs moving around. You mustn't use them for any other purpose.

The arrow tail is where the electron pair starts from. That's always fairly obvious, but you must show the electron pair either as a bond or, if it is a lone pair, as a pair of dots. Remember that a lone pair is a pair of electrons at the bonding level which isn't currently being used to join on to anything else.

The arrow head is where you want the electron pair to end up.

For example, in the reaction between ethene and hydrogen bromide, one of the two bonds between the two carbon atoms breaks. That bond is simply a pair of electrons.

Those electrons move to form a new bond with the hydrogen from the HBr. At the same time the pair of electrons in the hydrogen-bromine bond moves down on to the bromine atom.

There's no need to draw the pairs of electrons in the bonds as two dots. Drawing the bond as a line is enough, but you could put two dots in as well if you wanted to.

Notice that the arrow head points between the C and H because that's where the electron pair ends up. Notice also that the electron movement between the H and Br is shown as a curly arrow even though the electron pair moves straight down. You have to show

electron pair movements as curly arrows - not as straight ones.

The second stage of this reaction nicely illustrates how you use a curly arrow if a lone pair of electrons is involved.

The first stage leaves you with a positive charge on the right hand carbon atom and a negative bromide ion. You can think of the electrons shown on the bromide ion as being the ones which originally made up the hydrogen-bromine bond.

Note:  There are another three lone pairs around the outside of the bromide ion - making four in all. These aren't normally shown because they don't actually do anything new and interesting!

However, it is essential that you show the lone pair you are interested in as a pair of dots. If you don't, you risk losing marks in an exam.

The lone pair on the bromide ion moves to form a new bond between the bromine and the right hand carbon atom. That movement is again shown by a curly arrow. Notice again, that the curly arrow points between the carbon and the bromine because that's where the electron pair ends up.

That leaves you with the product of this reaction, bromoethane:

Note:  You can read a full description of this mechanism together with other similar reactions of ethene and the other alkenes by following this link.

Using curly arrows to show the movement of single electrons

The most common use of "curly arrows" is to show the movement of pairs of electrons. You can also use similar arrows to show the movement of single electrons - except that the heads of these arrows only have a single line rather than two lines.

shows the movement of an electron pair

shows the movement of a single electron

The first stage of the polymerisation of ethene, for example, could be shown as:

You should draw the dots showing the interesting electrons. The half arrows show where they go. This is very much a "belt-and-braces" job, and the arrows don't add much.

Whether you choose to use these half arrows to show the movement of a single electron should be governed by what your syllabus says. If your syllabus encourages the use of these arrows, then it makes sense to use them. If not - if the syllabus says that they "may" be used, or just ignores them altogether - then they are as well avoided.

There is some danger of confusing them with the arrows showing electron pair movements, which you will use all the time. If, by mistake, you use an ordinary full arrow to show the movement of a single electron you run the risk of losing marks.

STRUCTURAL ISOMERISM

This page explains what structural isomerism is, and looks at some of the various ways that structural isomers can arise.

What is structural isomerism?

What are isomers?

Isomers are molecules that have the same molecular formula, but have a different arrangement of the atoms in space. That excludes any different arrangements which are simply due to the molecule rotating as a whole, or rotating about particular bonds.

For example, both of the following are the same molecule. They are not isomers. Both are butane.

There are also endless other possible ways that this molecule could twist itself. There is completely free rotation around all the carbon-carbon single bonds.

Note:  Isomerism is much easier to understand if you have actually got some models to play with. If your school or college hasn't given you the opportunity to play around with molecular models in the early stages of your organic chemistry course, you might consider getting hold of a cheap set. The models made by molymod are both cheap and easy to use. An introductory organic set is more than adequate. Find them at www.molymod.com.

Alternatively , get hold of some coloured Plasticene and some used matches and make your own.

If you had a model of a molecule in front of you, you would have to take it to pieces and rebuild it if you wanted to make an isomer of that molecule. If you can make an apparently different molecule just by rotating single bonds, it's not different - it's still the same molecule.

Note:  It's really important that you understand this. If you aren't sure, then you must get hold of (or make) some models.

What are structural isomers?

In structural isomerism, the atoms are arranged in a completely different order. This is easier to see with specific examples.

What follows looks at some of the ways that structural isomers can arise. The names of the various forms of structural isomerism probably don't matter all that much, but you must be aware of the

different possibilities when you come to draw isomers.

Types of structural isomerism

Chain isomerism

These isomers arise because of the possibility of branching in carbon chains. For example, there are two isomers of butane, C4H10. In one of them, the carbon atoms lie in a "straight chain" whereas in the other the chain is branched.

Note:  Although the chain is drawn as straight, in reality it's anything but straight. If you aren't happy about the ways of drawing organic molecules, follow this link.

Use the BACK button on your browser to return to this page.

Be careful not to draw "false" isomers which are just twisted versions of the original molecule. For example, this structure is just the straight chain version of butane

rotated about the central carbon-carbon bond.

You could easily see this with a model. This is the example we've already used at the top of this page.

Pentane, C5H12, has three chain isomers. If you think you can find any others, they are simply twisted versions of the ones below. If in doubt make some models.

Position isomerism

In position isomerism, the basic carbon skeleton remains unchanged, but important groups are moved around on that skeleton.

For example, there are two structural isomers with the molecular formula C3H7Br. In one of them the bromine atom is on the end of the chain, whereas in the other it's attached in the middle.

If you made a model, there is no way that you could twist one molecule to turn it into the other one. You would have to break the bromine off the end and re-attach it in the middle. At the same time, you would have to move a hydrogen from the middle to the end.

Another similar example occurs in alcohols such as C4H9OH

These are the only two possibilities provided you keep to a four carbon chain, but there is no reason why you should do that. You can easily have a mixture of chain isomerism and position isomerism - you aren't restricted to one or the other.

So two other isomers of butanol are:

Note:  It's essential if you are asked to draw isomers in an exam not to restrict yourself to chain isomers or position isomers. You must be aware of all the possibilities.

You can also get position isomers on benzene rings. Consider the molecular formula C7H8Cl. There are four different isomers you could make depending on the position of the chlorine atom. In one case it is attached to the side-group carbon atom, and then there are three other possible positions it could have around the ring - next to the CH3 group, next-but-one to the CH3 group, or opposite the CH3 group.

Functional group isomerism

In this variety of structural isomerism, the isomers contain different functional groups - that is, they belong to different families of compounds (different homologous series).

For example, a molecular formula C3H6O could be either propanal (an aldehyde) or propanone (a ketone).

There are other possibilities as well for this same molecular formula - for example, you could have a carbon-carbon double bond (an alkene) and an -OH group (an alcohol) in the same molecule.

Another common example is illustrated by the molecular formula C3H6O2. Amongst the several structural isomers of this are propanoic acid (a carboxylic acid) and methyl ethanoate (an ester).

Note:  To repeat the warning given earlier: If you are asked to draw the structural isomers from a given molecular formula, don't forget to think about all the possibilities. Can you branch the carbon chain? Can you move a group around on that chain? Is it possible to make more than one type of compound?

Be careful though! If you are asked to draw the structures of

esters with the molecular formula C3H6O2, you aren't going to get a lot of credit for drawing propanoic acid, even if it is a valid isomer.

STEREOISOMERISM - GEOMETRIC ISOMERISM

Geometric isomerism (also known as cis-trans isomerism or E-Z isomerism) is a form of stereoisomerism. This page explains what stereoisomers are and how you recognise the possibility of geometric isomers in a molecule.

Further down the page, you will find a link to a second page which describes the E-Z notation for naming geometric isomers. You shouldn't move on to that page (even if the E-Z notation is what your syllabus is asking for) until you are really confident about how geometric isomers arise and how they are named on the cis-trans system.

The E-Z system is better for naming more complicated structures but is more difficult to understand than cis-trans. The cis-trans system of naming is still widely used - especially for the sort of simple molecules you will meet at this level. That means that irrespective of what your syllabus might say, you will have to be familiar with both systems. Get the easier one sorted out before you go on to the more sophisticated one!

What is stereoisomerism?

What are isomers?

Isomers are molecules that have the same molecular formula, but have a different arrangement of the atoms in space. That excludes any different arrangements which are simply due to the molecule

rotating as a whole, or rotating about particular bonds.

Where the atoms making up the various isomers are joined up in a different order, this is known as structural isomerism. Structural isomerism is not a form of stereoisomerism, and is dealt with on a separate page.

Note:  If you aren't sure about structural isomerism, it might be worth reading about it before you go on with this page.

What are stereoisomers?

In stereoisomerism, the atoms making up the isomers are joined up in the same order, but still manage to have a different spatial arrangement. Geometric isomerism is one form of stereoisomerism.

Geometric (cis / trans) isomerism

How geometric isomers arise

These isomers occur where you have restricted rotation somewhere in a molecule. At an introductory level in organic chemistry, examples usually just involve the carbon-carbon double bond - and that's what this page will concentrate on.

Think about what happens in molecules where there is unrestricted rotation about carbon bonds - in other words where the carbon-carbon bonds are all single. The next diagram shows two possible configurations of 1,2-dichloroethane.

These two models represent exactly the same molecule. You can get from one to the other just by twisting around the carbon-carbon single bond. These molecules are not isomers.

If you draw a structural formula instead of using models, you have to bear in mind the possibility of this free rotation about single bonds. You must accept that these two structures represent the same molecule:

But what happens if you have a carbon-carbon double bond - as in 1,2-dichloroethene?

These two molecules aren't the same. The carbon-carbon double bond won't rotate and so you would have to take the models to pieces in order to convert one structure into the other one. That is a simple test for isomers. If you have to take a model to pieces to convert it into another one, then you've got isomers. If you merely have to twist it a bit, then you haven't!

Note:  In the model, the reason that you can't rotate a carbon-carbon double bond is that there are two links joining the carbons together. In reality, the reason is that you would have to break the pi bond. Pi bonds are formed by the sideways overlap between p orbitals. If you tried to rotate the carbon-carbon bond, the p orbitals won't line up any more and so the pi bond is disrupted. This costs energy and only happens if the compound is heated strongly.

If you are interested in the bonding in carbon-carbon double bonds, follow this link. Be warned, though, that you might have to read several pages of background material and it could all take a long time. It isn't necessary for understanding the rest of this page.

Drawing structural formulae for the last pair of models gives two possible isomers.

In one, the two chlorine atoms are locked on opposite sides of the double bond. This is known as the trans isomer. (trans : from latin meaning "across" - as in transatlantic).

In the other, the two chlorine atoms are locked on the same side of the double bond. This is know as the cis isomer. (cis : from latin meaning "on this side")

The most likely example of geometric isomerism you will meet at an introductory level is but-2-ene. In one case, the CH3 groups are on opposite sides of the double bond, and in the other case they are on the same side.

The importance of drawing geometric isomers properly

It's very easy to miss geometric isomers in exams if you take short-cuts in drawing the structural formulae. For example, it is very tempting to draw but-2-ene as

CH3CH=CHCH3

If you write it like this, you will almost certainly miss the fact that there are geometric isomers. If there is even the slightest hint in a question that isomers might be involved, always draw compounds containing carbon-carbon double bonds showing the correct bond angles (120°) around the carbon atoms at the ends of the bond. In other words, use the format shown in the last diagrams above.

How to recognise the possibility of geometric isomerism

You obviously need to have restricted rotation somewhere in the molecule. Compounds containing a carbon-carbon double bond have this restricted rotation. (Other sorts of compounds may have restricted rotation as well, but we are concentrating on the case you are most likely to meet when you first come across geometric isomers.) If you have a carbon-carbon double bond, you need to think carefully about the possibility of geometric isomers.

What needs to be attached to the carbon-carbon double bond?

Note:  This is much easier to understand if you have actually got some models to play with. If your school or college hasn't given you the opportunity to play around with molecular models in the early stages of your organic chemistry course, you might consider getting hold of a cheap set. The models made by

molymod are both cheap and easy to use. An introductory organic set is more than adequate. Find them at www.molymod.com.

Alternatively , get hold of some coloured Plasticene and some used matches and make your own.

Think about this case:

Although we've swapped the right-hand groups around, these are still the same molecule. To get from one to the other, all you would have to do is to turn the whole model over.

You won't have geometric isomers if there are two groups the same on one end of the bond - in this case, the two pink groups on the left-hand end.

So . . . there must be two different groups on the left-hand carbon and two different groups on the right-hand one. The cases we've been exploring earlier are like this:

But you could make things even more different and still have geometric isomers:

Here, the blue and green groups are either on the same side of the bond or the opposite side.

Or you could go the whole hog and make everything different. You still get geometric isomers, but by now the words cis and trans are meaningless. This is where the more sophisticated E-Z notation

comes in.

Summary

To get geometric isomers you must have:

restricted rotation (often involving a carbon-carbon double bond for introductory purposes);

two different groups on the left-hand end of the bond and two different groups on the right-hand end. It doesn't matter whether the left-hand groups are the same as the right-hand ones or not.

Note:  The rest of this page looks at how geometric isomerism affects the melting and boiling points of compounds. If you are meeting geometric isomerism for the first time, you may not need this at the moment.

If you need to know about E-Z notation, you could follow this link at once to the next page. (But be sure that you understand what you have already read on this page first!)

Alternatively, read to the bottom of this page where you will find this link repeated.

The effect of geometric isomerism on physical properties

The table shows the melting point and boiling point of the cis and trans isomers of 1,2-dichloroethene.

isomer melting point (°C) boiling point (°C)

cis -80 60

trans -50 48

In each case, the higher melting or boiling point is shown in red.

You will notice that:

the trans isomer has the higher melting point;

the cis isomer has the higher boiling point.

This is common. You can see the same effect with the cis and trans isomers of but-2-ene:

isomer melting point (°C) boiling point (°C)

cis-but-2-ene -139 4

trans-but-2-ene -106 1

Why is the boiling point of the cis isomers higher?

There must be stronger intermolecular forces between the molecules of the cis isomers than between trans isomers.

Taking 1,2-dichloroethene as an example:

Both of the isomers have exactly the same atoms joined up in exactly the same order. That means that the van der Waals dispersion forces between the molecules will be identical in both cases.

The difference between the two is that the cis isomer is a polar molecule whereas the trans isomer is non-polar.

Note:  If you aren't sure about intermolecular forces (and also about bond polarity), it is essential that you follow this link before you go on. You need to know about van der Waals dispersion forces and dipole-dipole interactions, and to follow the link on that page to another about bond polarity if you need to.

Use the BACK button on your browser to return to this page.

Both molecules contain polar chlorine-carbon bonds, but in the cis isomer they are both on the same side of the molecule. That means that one side of the molecule will have a slight negative charge while the other is slightly positive. The molecule is therefore polar.

Because of this, there will be dipole-dipole interactions as well as dispersion forces - needing extra energy to break. That will raise

the boiling point.

A similar thing happens where there are CH3 groups attached to the carbon-carbon double bond, as in cis-but-2-ene.

Alkyl groups like methyl groups tend to "push" electrons away from themselves. You again get a polar molecule, although with a reversed polarity from the first example.

Note:  The term "electron pushing" is only to help remember what happens. The alkyl group doesn't literally "push" the electrons away - the other end of the bond attracts them more strongly. The arrows with the cross on (representing the more positive end of the bond) are a conventional way of showing this electron pushing effect.

By contrast, although there will still be polar bonds in the trans isomers, overall the molecules are non-polar.

The slight charge on the top of the molecule (as drawn) is exactly balanced by an equivalent charge on the bottom. The slight charge on the left of the molecule is exactly balanced by the same charge on the right.

This lack of overall polarity means that the only intermolecular attractions these molecules experience are van der Waals dispersion forces. Less energy is needed to separate them, and so their boiling points are lower.

Why is the melting point of the cis isomers lower?

You might have thought that the same argument would lead to a higher melting point for cis isomers as well, but there is another important factor operating.

In order for the intermolecular forces to work well, the molecules

must be able to pack together efficiently in the solid.

Trans isomers pack better than cis isomers. The "U" shape of the cis isomer doesn't pack as well as the straighter shape of the trans isomer.

The poorer packing in the cis isomers means that the intermolecular forces aren't as effective as they should be and so less energy is needed to melt the molecule - a lower melting point.

E-Z NOTATION FOR GEOMETRIC ISOMERISM

This page explains the E-Z system for naming geometric isomers.

Important!  If you have come straight here via a search engine (including the Google site search on the Main Menu of Chemguide), you should be aware that this page follows on from an introductory page about geometric isomerism. Unless you are already confident about how geometric isomers arise, and the cis-trans system for naming them, you should follow this link first. You will find links back to this current page at suitable points on that page.

The E-Z system

The problem with the cis-trans system for naming geometric isomers

Consider a simple case of geometric isomerism which we've already discussed on the previous page.

You can tell which is the cis and which the trans form just by looking at them. All you really have to remember is that trans means "across" (as in transatlantic or transcontinental) and that cis is the opposite. It is a simple and visual way of telling the two isomers apart. So why do we need another system?

There are problems as compounds get more complicated. For example, could you name these two isomers using cis and trans?

Because everything attached to the carbon-carbon double bond is different, there aren't any obvious things which you can think of as being "cis" or "trans" to each other. The E-Z system gets around this problem completely - but unfortunately makes things slightly more difficult for the simple examples you usually meet in introductory courses.

How the E-Z system works

We'll use the last two compounds as an example to explain how the system works.

You look at what is attached to each end of the double bond in turn, and give the two groups a "priority" according to a set of rules which we'll explore in a minute.

In the example above, at the left-hand end of the bond, it turns out that bromine has a higher priority than fluorine. And on the right-hand end, it turns out that chlorine has a higher priority than hydrogen.

If the two groups with the higher priorities are on the same side of the double bond, that is described as the (Z)- isomer. So you would write it as (Z)-name of compound. The symbol Z comes from a German word (zusammen) which means together.

Note:  I'm not getting bogged down in the names of these more complex compounds. As soon as I put the proper full

names in, the whole thing suddenly looks much more complicated than it really is, and you will start to focus on where the whole name comes from rather than on if it is a (Z)- or (E)- isomer.

If the two groups with the higher priorities are on opposite sides of the double bond, then this is the (E)- isomer. E comes from the German entgegen which means opposite.

So the two isomers are:

Summary

(E)- : the higher priority groups are on opposite sides of the double bond.

(Z)- : the higher priority groups are on the same side of the double bond.

Note:  I wish I could think of a really good way to remember this for non-German speakers - but I can't! If you have a way of remembering it which might help other students, please get in touch with me via the address on the about this site page.

Personally, I remember it as exactly the opposite of what you might expect from the shapes of the letters E and Z. In the case of the E, all the horizontal bits are on the same side of the letter; in the case of Z, the horizontal bits are on different sides. That's exactly the opposite of the arrangement of the groups in E and Z isomers. It's a fairly unsatisfying way to remember it, but it will have to do until someone gives me a better one!

Since writing this, I've had another suggestion which is Z = "zame zide" and E = "epposite". I think that would work well if English is your first language, but if it isn't, I'm not sure that you will understand what is going on! More suggestions would be welcome.

Rules for determining priorities

These are known as Cahn-Ingold-Prelog (CIP) rules after the people who developed the system.

The first rule for very simple cases

You look first at the atoms attached directly to the carbon atoms at

each end of the double bond - thinking about the two ends separately.

The atom which has the higher atomic number is given the higher priority.

Let's look at the example we've been talking about.

Just consider the first isomer - and look separately at the left-hand and then the right-hand carbon atom. Compare the atomic numbers of the attached atoms to work out the various priorities.

Notice that the atoms with the higher priorities are both on the same side of the double bond. That counts as the (Z)- isomer.

The second isomer obviously still has the same atoms at each end, but this time the higher priority atoms are on opposite sides of the double bond. That's the (E)- isomer.

What about the more familiar examples like 1,2-dichloroethene or but-2-ene? Here's 1,2-dichloroethene.

Think about the priority of the two groups on the first carbon of the left-hand isomer.

Chlorine has a higher atomic number than hydrogen, and so has the higher priority. That, of course, is equally true of all the other carbon atoms in these two isomers.

In the first isomer, the higher priority groups are on opposite sides of the bond. That must be the (E)- isomer. The other one, with the higher priority groups on the same side, is the (Z)- isomer.

And now but-2-ene . . .

This adds the slight complication that you haven't got a single atom attached to the double bond, but a group of atoms.

That isn't a problem. Concentrate on the atom directly attached to the double bond - in this case the carbon in the CH3 group. For this simple case, you can ignore the hydrogen atoms in the CH3 group entirely. However, with more complicated groups you may have to worry about atoms not directly attached to the double bond. We'll look at that problem in a moment.

Here is one of the isomers of but-2-ene:

The CH3 group has the higher priority because its carbon atom has an atomic number of 6 compared with an atomic number of 1 for the hydrogen also attached to the carbon-carbon double bond.

The isomer drawn above has the two higher priority groups on opposite sides of the double bond. The compound is (E)-but-2-ene.

A minor addition to the rule to allow for isotopes of, for example, hydrogen

Deuterium is an isotope of hydrogen having a relative atomic mass of 2. It still has only 1 proton, and so still has an atomic number of 1. However, it isn't the same as an atom of "ordinary" hydrogen, and so these two compounds are geometric isomers:

The hydrogen and deuterium have the same atomic number - so on that basis, they would have the same priority. In a case like that, the one with the higher relative atomic mass has the higher priority. So in these isomers, the deuterium and chlorine are the higher

priority groups on each end of the double bond.

That means that the left-hand isomer in the last diagram is the (E)- form, and the right-hand one the (Z)-.

Extending the rules to more complicated molecules

If you are reading this because you are doing a course for 16 - 18 year olds such as UK A level, you may well not need to know much about this section, but it really isn't very difficult!

Let's illustrate this by taking a fairly scary-looking molecule, and seeing how easy it is to find out whether it is a (Z)- or (E)- isomer by applying an extra rule.

Focus on the left-hand end of the molecule. What is attached directly to the carbon-carbon double bond?

In both of the attached groups, a carbon atom is attached directly to the bond. Those two atoms obviously have the same atomic number and therefore the same priority. So that doesn't help.

In this sort of case, you now look at what is attached directly to those two carbons (but without counting the carbon of the double bond) and compare the priorities of these next lot of atoms.

You can do this in your head in simple cases, but it is sometimes useful to write the attached atoms down, listing them with the highest priority atom first. It makes them easier to compare. Like this . . .

In the CH3 group:

The atoms attached to the carbon are H H H.

In the CH3CH2 group:

The atoms attached directly to the carbon of the CH2 group are C H H.

In the second list, the C is written first because it has the highest atomic number.

Now compare the two lists atom by atom. The first atom in each list is an H in the CH3 group and a C in the CH3CH2 group. The carbon

has the higher priority because it has the higher atomic number. So that gives the CH3CH2 group a higher priority than the CH3 group.

Now look at the other end of the double bond. The extra thing that this illustrates is that if you have a double bond, you count the attached atom twice. Here is the structure again.

So, again, the atoms attached directly to the carbon-carbon double bond are both carbons. We therefore need to look at what is attached to those carbons.

In the CH2OH group:

The atoms attached directly to the carbon are O H H.

In the CHO group:

The atoms attached directly to the carbon of the CH2 group are O O H.

In both lists, the oxygens are written first because they have a higher atomic number than hydrogen. In the CHO group list, the oxygen is written twice because of the C=O double bond.

So, what is the priority of the two groups? The first atom in both lists is an oxygen - that doesn't help. Look at the next atom in both lists. In the CH2OH group, that's a hydrogen; in the CHO list, it's an oxygen.

The oxygen has the higher priority - and that gives the CHO group a higher priority than the CH2OH group.

The isomer is therefore a (Z)- form, because the two higher priority groups (the CH3CH2 group and the CHO group) are both on the same side of the bond.

That's been a fairly long-winded explanation just to make clear how it works. With a bit of practice, it takes a few seconds to work out in any but the most complex cases.

One more example to make a couple of additional minor points . . .

Here's an even more complicated molecule!

Before you read on, have a go at working out the relative priorities of the two groups on the left-hand end of the double bond, and the two on the right-hand end. There's another bit of rule that I haven't specifically told you yet, but it isn't hard to guess what it might be when you start to look at the problem. If you can work this out, then you won't have any difficulty with any problem you are likely to come across at this level.

Look first at the left-hand groups.

In both the top and bottom groups, you have a CH2 group attached directly to the carbon-carbon double bond, and the carbon in that CH2 group is also attached to another carbon atom. In each case, the list will read C H H.

There is no difference between the priorities of those groups, so what are you going to do about it? The answer is to move out along the chain to the next group. And if necessary, continue to do this until you have found a difference.

Next along the chain at the top left of the molecule is another CH2 group attached to a further carbon atom. The list for this group is again C H H.

But the next group along the chain at the bottom left is a CH group attached to two more carbon atoms. Its list is therefore C C H.

Comparing these lists atom by atom, leads you to the fact that the bottom group has the higher priority.

Now look at the right-hand groups. Here is the molecule again:

The top right group has C H H attached to the first carbon in the chain.

The bottom right one has Cl H H.

The chlorine has a higher atomic number than carbon, and so the

bottom right group has the higher priority of these two groups.

The extra point I am trying to make with this bit of the example is that you must just focus on one bit of a chain at a time. We never get around to considering the bromine at the extreme top right of the molecule. We don't need to go out that far along the chain - you work out one link at a time until you find a difference. Anything beyond that is irrelevant.

For the record, this molecule is a (Z)- isomer because the higher priority groups at each end are on the same side of the double bond.

Can you easily translate cis- and trans- into (Z)- and (E)-?

You might think that for simple cases, cis- will just convert into (Z)- and trans- into (E)-.

Look for example at the 1,2-dichloroethene and but-2-ene cases.

But it doesn't always work! Think about this relatively uncomplicated molecule . . .

This is clearly a cis- isomer. It has two CH3 groups on the same side of the double bond. But work out the priorities on the right-hand end of the double bond.

The two directly attached atoms are carbon and bromine. Bromine has the higher atomic number and so has the higher priority on that

end. At the other end, the CH3 group has the higher priority.

That means that the two higher priority groups are on opposite sides of the double bond, and so this is an (E)- isomer - NOT a (Z)-.

Never assume that you can convert directly from one of these systems into the other. The only safe thing to do is to start from scratch in each case.

Does it matter that the two systems will sometimes give different results? No! The purpose of both systems is to enable you to decode a name and write a correct formula. Properly used, both systems will do this for you - although the cis-trans system will only work for very straightforward molecules.

© Jim Clark 2007

STEREOISOMERISM - OPTICAL ISOMERISM

Optical isomerism is a form of stereoisomerism. This page explains what stereoisomers are and how you recognise the possibility of optical isomers in a molecule.

What is stereoisomerism?

What are isomers?

Isomers are molecules that have the same molecular formula, but have a different arrangement of the atoms in space. That excludes any different arrangements which are simply due to the molecule rotating as a whole, or rotating about particular bonds.

Where the atoms making up the various isomers are joined up in a different order, this is known as structural isomerism. Structural isomerism is not a form of stereoisomerism, and is dealt with on a separate page.

Note:  If you aren't sure about structural isomerism, it might be worth reading about it before you go on with this page.

What are stereoisomers?

In stereoisomerism, the atoms making up the isomers are joined up in the same order, but still manage to have a different spatial arrangement. Optical isomerism is one form of stereoisomerism.

Optical isomerism

Why optical isomers?

Optical isomers are named like this because of their effect on plane polarised light.

Help!  If you don't understand about plane polarised light, follow this link before you go on with this page.

Simple substances which show optical isomerism exist as two isomers known as enantiomers.

A solution of one enantiomer rotates the plane of polarisation in a clockwise direction. This enantiomer is known as the (+) form.

For example, one of the optical isomers (enantiomers) of the amino acid alanine is known as (+)alanine.

A solution of the other enantiomer rotates the plane of polarisation in an anti-clockwise direction. This enantiomer is known as the (-) form. So the other enantiomer of alanine is known as or (-)alanine.

If the solutions are equally concentrated the amount of rotation caused by the two isomers is exactly the same - but in opposite directions.

When optically active substances are made in the lab, they often occur as a 50/50 mixture of the two enantiomers. This is known as a racemic mixture or racemate. It has no effect on plane polarised light.

Note:  One of the worrying things about optical isomerism is the number of obscure words that suddenly get thrown at you. Bear with it - things are soon going to get more visual!

There is an alternative way of describing the (+) and (-) forms which is potentially very confusing. This involves the use of the lowercase letters d- and l-, standing for dextrorotatory and laevorotatory respectively. Unfortunately, there is another different use of the capital letters D- and L- in this topic. This is totally confusing! Stick with (+) and (-).

How optical isomers arise

The examples of organic optical isomers required at A' level all contain a carbon atom joined to four different groups. These two models each have the same groups joined to the central carbon atom, but still manage to be different:

Obviously as they are drawn, the orange and blue groups aren't aligned the same way. Could you get them to align by rotating one of the molecules? The next diagram shows what happens if you rotate molecule B.

They still aren't the same - and there is no way that you can rotate them so that they look exactly the same. These are isomers of each other.

They are described as being non-superimposable in the sense that (if you imagine molecule B being turned into a ghostly version of itself) you couldn't slide one molecule exactly over the other one. Something would always be pointing in the wrong direction.

Note:  Unless your visual imagination is reasonably good, this is much easier to understand if you have actually got some models to play with. If your school or college hasn't given you the opportunity to play around with molecular models, you might consider getting hold of a cheap set. The models made by molymod are both cheap and easy to use. An introductory organic set is more than adequate. Find them at www.molymod.com.

Alternatively , get hold of some coloured Plasticene and some used matches and make your own.

What happens if two of the groups attached to the central carbon atom are the same? The next diagram shows this possibility.

The two models are aligned exactly as before, but the orange

group has been replaced by another pink one.

Rotating molecule B this time shows that it is exactly the same as molecule A. You only get optical isomers if all four groups attached to the central carbon are different.

Chiral and achiral molecules

The essential difference between the two examples we've looked at lies in the symmetry of the molecules.

If there are two groups the same attached to the central carbon atom, the molecule has a plane of symmetry. If you imagine slicing through the molecule, the left-hand side is an exact reflection of the right-hand side.

Where there are four groups attached, there is no symmetry anywhere in the molecule.

A molecule which has no plane of symmetry is described as chiral. The carbon atom with the four different groups attached which causes this lack of symmetry is described as a chiral centre or as an asymmetric carbon atom.

The molecule on the left above (with a plane of symmetry) is described as achiral.

Only chiral molecules have optical isomers.

The relationship between the enantiomers

One of the enantiomers is simply a non-superimposable mirror image of the other one.

In other words, if one isomer looked in a mirror, what it would see is the other one. The two isomers (the original one and its mirror image) have a different spatial arrangement, and so can't be superimposed on each other.

If an achiral molecule (one with a plane of symmetry) looked in a mirror, you would always find that by rotating the image in space, you could make the two look identical. It would be possible to superimpose the original molecule and its mirror image.

Some real examples of optical isomers

Butan-2-ol

The asymmetric carbon atom in a compound (the one with four different groups attached) is often shown by a star.

It's extremely important to draw the isomers correctly. Draw one of them using standard bond notation to show the 3-dimensional arrangement around the asymmetric carbon atom. Then draw the mirror to show the examiner that you know what you are doing, and then the mirror image.

Help!  If you don't understand this bond notation, follow this link to drawing organic molecules before you go on with this page.

Notice that you don't literally draw the mirror images of all the letters and numbers! It is, however, quite useful to reverse large groups - look, for example, at the ethyl group at the top of the

diagram.

It doesn't matter in the least in what order you draw the four groups around the central carbon. As long as your mirror image is drawn accurately, you will automatically have drawn the two isomers.

So which of these two isomers is (+)butan-2-ol and which is (-)butan-2-ol? There is no simple way of telling that. For A'level purposes, you can just ignore that problem - all you need to be able to do is to draw the two isomers correctly.

2-hydroxypropanoic acid (lactic acid)

Once again the chiral centre is shown by a star.

The two enantiomers are:

It is important this time to draw the COOH group backwards in the mirror image. If you don't there is a good chance of you joining it on to the central carbon wrongly.

If you draw it like this in an exam, you won't get the mark for that isomer even if you have drawn everything else perfectly.

2-aminopropanoic acid (alanine)

This is typical of naturally-occurring amino acids. Structurally, it is just like the last example, except that the -OH group is replaced by -NH2

The two enantiomers are:

Only one of these isomers occurs naturally: the (+) form. You can't tell just by looking at the structures which this is.

It has, however, been possible to work out which of these structures is which. Naturally occurring alanine is the right-hand structure, and the way the groups are arranged around the central carbon atom is known as an L- configuration. Notice the use of the capital L. The other configuration is known as D-.

So you may well find alanine described as L-(+)alanine.

That means that it has this particular structure and rotates the plane of polarisation clockwise.

Even if you know that a different compound has an arrangement of groups similar to alanine, you still can't say which way it will rotate the plane of polarisation.

The other amino acids, for example, have the same arrangement of groups as alanine does (all that changes is the CH3 group), but some are (+) forms and others are (-) forms.

It's quite common for natural systems to only work with one of the enantiomers of an optically active substance. It isn't too difficult to see why that might be. Because the molecules have different spatial arrangements of their various groups, only one of them is likely to fit properly into the active sites on the enzymes they work with.

In the lab, it is quite common to produce equal amounts of both forms of a compound when it is synthesised. This happens just by chance, and you tend to get racemic mixtures.

Note:  For a detailed discussion of this, you could have a look at the page on the addition of HCN to aldehydes

© Jim Clark 2000 (modified 2004)

INTRODUCING ALKANES AND CYCLOALKANES

This is an introductory page about alkanes such as methane, ethane, propane, butane and the rest. It deals with their formulae and isomerism, their physical properties, and an introduction to their chemical reactivity.

What are alkanes and cycloalkanes?

Alkanes

Formulae

Alkanes are the simplest family of hydrocarbons - compounds containing carbon and hydrogen only. They only contain carbon-hydrogen bonds and carbon-carbon single bonds. The first six are:

methane CH4

ethane C2H6

propane C3H8

butane C4H10

pentane C5H12

hexane C6H14

You can work out the formula of any of them using: CnH2n+2

Isomerism

All the alkanes with 4 or more carbon atoms in them show structural isomerism. This means that there are two or more different structural formulae that you can draw for each molecular formula.

For example, C4H10 could be either of these two different molecules:

These are called respectively butane and 2-methylpropane.

Note:  If you aren't confident about naming organic compounds, the various ways of drawing organic compounds, or structural isomerism, then you really ought to follow these links before you go on.

You should read the whole of the page about drawing organic molecules, but there is no need to read the other two beyond where they talk about alkanes.

Use the BACK button on your browser to return to this page.

Cycloalkanes

Cycloalkanes again only contain carbon-hydrogen bonds and carbon-carbon single bonds, but this time the carbon atoms are joined up in a ring. The smallest cycloalkane is cyclopropane.

If you count the carbons and hydrogens, you will see that they no longer fit the general formula CnH2n+2. By joining the carbon atoms in a ring, you have had to lose two hydrogen atoms.

You are unlikely to ever need it, but the general formula for a cycloalkane is CnH2n.

Don't imagine that these are all flat molecules. All the cycloalkanes from cyclopentane upwards exist as "puckered rings".

Cyclohexane, for example, has a ring structure which looks like this:

This is known as the "chair" form of cyclohexane - from its shape which vaguely resembles a chair.

Note:  This molecule is constantly changing, with the atom on the left which is currently pointing down flipping up, and the one on the right flipping down. During the process, another (slightly less stable) form of cyclohexane is formed known as the "boat" form. In this arrangement, both of these atoms are either pointing up or down at the same time.

Physical Properties

Boiling Points

The facts

The boiling points shown are all for the "straight chain" isomers where there are more than one.

Notice that the first four alkanes are gases at room temperature. Solids don't start to appear until about C17H36.

You can't be more precise than that because each isomer has a different melting and boiling point. By the time you get 17 carbons into an alkane, there are unbelievable numbers of isomers!

Cycloalkanes have boiling points which are about 10 - 20 K higher than the corresponding straight chain alkane.

Explanations

There isn't much electronegativity difference between carbon and hydrogen, so there is hardly any bond polarity. The molecules themselves also have very little polarity. A totally symmetrical molecule like methane is completely non-polar.

Note:  If you aren't sure about electronegativity and polarity, then you really ought to follow this link before you go on.

Use the BACK button on your browser to return to this page.

This means that the only attractions between one molecule and its neighbours will be Van der Waals dispersion forces. These will be very small for a molecule like methane, but will increase as the molecules get bigger. That's why the boiling points of the alkanes increase with molecular size.

Note:  If you aren't sure about Van der Waals forces, then you should follow this link before you go on.

Use the BACK button on your browser to return to this page.

Where you have isomers, the more branched the chain, the lower the boiling point tends to be. Van der Waals dispersion forces are smaller for shorter molecules, and only operate over very short distances between one molecule and its neighbours. It is more difficult for short fat molecules (with lots of branching) to lie as close together as long thin ones.

For example, the boiling points of the three isomers of C5H12 are:

boiling point (K)

pentane 309.2

2-methylbutane 301.0

2,2-dimethylpropane 282.6

The slightly higher boiling points for the cycloalkanes are presumably because the molecules can get closer together because the ring structure makes them tidier and less "wriggly"!

Solubility

The facts

What follows applies equally to alkanes and cycloalkanes.

Alkanes are virtually insoluble in water, but dissolve in organic solvents. The liquid alkanes are good solvents for many other covalent compounds.

Explanations

Solubility in water

When a molecular substance dissolves in water, you have to

break the intermolecular forces within the substance. In the case of the alkanes, these are Van der Waals dispersion forces.

break the intermolecular forces in the water so that the substance can fit between the water molecules. In water the main intermolecular attractions are hydrogen bonds.

Note:  If you aren't sure about hydrogen bonds, then you should follow this link before you go on.

Use the BACK button on your browser to return to this page.

Breaking either of these attractions costs energy, although the amount of energy to break the Van der Waals dispersion forces in something like methane is pretty negligible. That isn't true of the hydrogen bonds in water, though.

As something of a simplification, a substance will dissolve if there is enough energy released when new bonds are made between the substance and the water to make up for what is used in breaking the original attractions.

The only new attractions between the alkane and water molecules are Van der Waals. These don't release anything like enough energy to compensate for what you need to break the hydrogen bonds in water. The alkane doesn't dissolve.

Note:  The reason that this is a simplification is that you also have to consider entropy changes when things dissolve. If you don't yet know about entropy, don't worry about it!

Solubility in organic solvents

In most organic solvents, the main forces of attraction between the solvent molecules are Van der Waals - either dispersion forces or dipole-dipole attractions.

That means that when an alkane dissolves in an organic solvent, you are breaking Van der Waals forces and replacing them by new Van der Waals forces. The two processes more or less cancel each other out energetically - so there isn't any barrier to solubility.

Chemical Reactivity

Alkanes

Alkanes contain strong carbon-carbon single bonds and strong carbon-hydrogen bonds. The carbon-hydrogen bonds are only very slightly polar and so there aren't any bits of the molecules which carry any significant amount of positive or negative charge which other things might be attracted to.

The net effect is that alkanes have a fairly restricted set of reactions.

You can

burn them - destroying the whole molecule; react them with some of the halogens, breaking carbon-

hydrogen bonds;

crack them, breaking carbon-carbon bonds.

These reactions are all covered on separate pages if you go to the alkanes menu (see below).

Cycloalkanes

Cycloalkanes are very similar to the alkanes in reactivity, except for the very small ones - especially cyclopropane. Cyclopropane is much more reactive than you would expect.

The reason has to do with the bond angles in the ring. Normally, when carbon forms four single bonds, the bond angles are about 109.5°. In cyclopropane, they are 60°.

With the electron pairs this close together, there is a lot of repulsion between the bonding pairs joining the carbon atoms. That makes the bonds easier to break.

The effect of this is explored on the page about reactions of these compounds with halogens which you can access from the alkanes menu below.

© Jim Clark 2003

THE COMBUSTION OF ALKANES AND CYCLOALKANES

This page deals briefly with the combustion of alkanes and cycloalkanes. In fact, there is very little difference between the two.

Complete combustion

Complete combustion (given sufficient oxygen) of any hydrocarbon produces carbon dioxide and water.

Equations

It is quite important that you can write properly balanced equations for these reactions, because they often come up as a part of thermochemistry calculations. Don't try to learn the equations - there are far too many possibilities. Work them out as you need them.

Some are easier than others. For example, with alkanes, the ones with an even number of carbon atoms are marginally harder than those with an odd number!

For example, with propane (C3H8), you can balance the carbons and hydrogens as you write the equation down. Your first draft would be:

Counting the oxygens leads directly to the final version:

With butane (C4H10), you can again balance the carbons and hydrogens as you write the equation down.

Counting the oxygens leads to a slight problem - with 13 on the right-hand side. The simple trick is to allow yourself to have "six-and-a-half" O2 molecules on the left.

If that offends you, double everything:

Note:  You might well come across either version of these equations. The ones with the halves left in are often used in calculation work.

Forgive me if you find this last bit on equations unbearably trivial - not everybody does! Just be grateful that you have been well taught.

Trends

The hydrocarbons become harder to ignite as the molecules get bigger. This is because the bigger molecules don't vaporise so easily - the reaction is much better if the oxygen and the hydrocarbon are well mixed as gases. If the liquid isn't very volatile, only those molecules on the surface can react with the oxygen.

Bigger molecules have greater Van der Waals attractions which makes it more difficult for them to break away from their neighbours and turn to a gas.

Note:  If you aren't sure about Van der Waals forces, then you should follow this link before you go on.

Use the BACK button on your browser to return to this page.

Provided the combustion is complete, all the hydrocarbons will burn with a blue flame. However, combustion tends to be less complete as the number of carbon atoms in the molecules rises. That means that the bigger the hydrocarbon, the more likely you are to get a yellow, smoky flame.

Incomplete combustion

Incomplete combustion (where there isn't enough oxygen present) can lead to the formation of carbon or carbon monoxide.

As a simple way of thinking about it, the hydrogen in the hydrocarbon gets the first chance at the oxygen, and the carbon gets whatever is left over!

The presence of glowing carbon particles in a flame turns it yellow, and black carbon is often visible in the smoke. Carbon monoxide is produced as a colourless poisonous gas.

Why carbon monoxide is poisonous

Oxygen is carried around the blood by haemoglobin (US: hemoglobin). Unfortunately carbon monoxide binds to exactly the same site on the haemoglobin that oxygen does.

The difference is that carbon monoxide binds irreversibly - making that particular molecule of haemoglobin useless for carrying oxygen. If you breath in enough carbon monoxide you will die from a sort of internal suffocation.

.

THE HALOGENATION OF ALKANES AND CYCLOALKANES

This page describes the reactions between alkanes and cycloalkanes with the halogens fluorine, chlorine, bromine and iodine - mainly concentrating on chlorine and bromine.

Alkanes

The reaction between alkanes and fluorine

This reaction is explosive even in the cold and dark, and you tend to get carbon and hydrogen fluoride produced. It is of no particular interest. For example:

The reaction between alkanes and iodine

Iodine doesn't react with the alkanes - at least, under normal lab conditions.

Note:  I can't discover the truth about this! Some sources say that it doesn't react; others say that it reacts very slowly. I have found it impossible to find all the data I need to estimate whether there might be a temperature at which the reaction

becomes feasible. The information needed depends too much on what assumptions you make about the physical states of the reactants and products - for example, whether the iodine is present as a solid, a gas or a solution.

If you have any hard information, could you contact me via the address on the about this site page.

The reactions between alkanes and chlorine or bromine

There is no reaction in the dark.

In the presence of a flame, the reactions are rather like the fluorine one - producing a mixture of carbon and the hydrogen halide. The violence of the reaction drops considerably as you go from fluorine to chlorine to bromine.

The interesting reactions happen in the presence of ultra-violet light (sunlight will do). These are photochemical reactions, and happen at room temperature.

We'll look at the reactions with chlorine. The reactions with bromine are similar, but rather slower.

Methane and chlorine

Substitution reactions happen in which hydrogen atoms in the methane are replaced one at a time by chlorine atoms. You end up with a mixture of chloromethane, dichloromethane, trichloromethane and tetrachloromethane.

Note:  Follow this link if you aren't happy about naming organic compounds.

Use the BACK button on your browser to return to this page.

The original mixture of a colourless and a green gas would produce steamy fumes of hydrogen chloride and a mist of organic liquids. All of the organic products are liquid at room temperature with the

exception of the chloromethane which is a gas.

If you were using bromine, you could either mix methane with bromine vapour, or bubble the methane through liquid bromine - in either case, exposed to UV light. The original mixture of gases would, of course, be red-brown rather than green.

You wouldn't choose to use these reactions as a means of preparing these organic compounds in the lab because the mixture of products would be too tedious to separate.

The mechanisms for the reactions are explained on separate pages.

Note:  If you want the methane-chlorine mechanism, follow this link.

If you want the methane-bromine mechanism, follow this one.

Use the BACK button on your browser to return to this page.

Larger alkanes and chlorine

You would again get a mixture of substitution products, but it is worth just looking briefly at what happens if only one of the hydrogen atoms gets substituted (monosubstitution) - just to show that things aren't always as straightforward as they seem!

For example, with propane, you could get one of two isomers:

Note:  If you aren't sure about isomerism, you might like to follow this link.

Use the BACK button on your browser to return to this page.

If chance was the only factor, you would expect to get 3 times as much of the isomer with the chlorine on the end. There are 6 hydrogens that could get replaced on the end carbon atoms compared with only 2 in the middle.

In fact, you get about the same amount of each of the two isomers.

If you use bromine instead of chlorine, the great majority of the product is where the bromine is attached to the centre carbon atom.

The reasons for this are beyond UK A level chemistry.

Cycloalkanes

The reactions of the cycloalkanes are generally just the same as the alkanes, with the exception of the very small ones - particularly cyclopropane.

The extra reactivity of cyclopropane

In the presence of UV light, cyclopropane will undergo substitution reactions with chlorine or bromine just like a non-cyclic alkane. However, it also has the ability to react in the dark.

In the absence of UV light, cyclopropane can undergo addition reactions in which the ring is broken. For example, with bromine, cyclopropane gives 1,3-dibromopropane.

This can still happen in the presence of light - but you will get substitution reactions as well.

The ring is broken because cyclopropane suffers badly from ring strain. The bond angles in the ring are 60° rather than the normal value of about 109.5° when the carbon makes four single bonds.

The overlap between the atomic orbitals in forming the carbon-carbon bonds is less good than it is normally, and there is considerable repulsion between the bonding pairs. The system becomes more stable if the ring is broken.

© Jim Clark 2003

CRACKING ALKANES

This page describes what cracking is, and the differences between catalytic cracking and thermal cracking used in the petrochemical industry.

Cracking

What is cracking?

Cracking is the name given to breaking up large hydrocarbon molecules into smaller and more useful bits. This is achieved by using high pressures and temperatures without a catalyst, or lower temperatures and pressures in the presence of a catalyst.

The source of the large hydrocarbon molecules is often the naphtha fraction or the gas oil fraction from the fractional distillation of crude oil (petroleum). These fractions are obtained from the distillation process as liquids, but are re-vaporised before cracking.

There isn't any single unique reaction happening in the cracker. The hydrocarbon molecules are broken up in a fairly random way to produce mixtures of smaller hydrocarbons, some of which have carbon-carbon double bonds. One possible reaction involving the hydrocarbon C15H32 might be:

Or, showing more clearly what happens to the various atoms and bonds:

This is only one way in which this particular molecule might break up. The ethene and propene are important materials for making plastics or producing other organic chemicals. The octane is one of the molecules found in petrol (gasoline).

Catalytic cracking

Modern cracking uses zeolites as the catalyst. These are complex aluminosilicates, and are large lattices of aluminium, silicon and oxygen atoms carrying a negative charge. They are, of course, associated with positive ions such as sodium ions. You may have come across a zeolite if you know about ion exchange resins used in water softeners.

The alkane is brought into contact with the catalyst at a temperature of about 500°C and moderately low pressures.

The zeolites used in catalytic cracking are chosen to give high percentages of hydrocarbons with between 5 and 10 carbon atoms - particularly useful for petrol (gasoline). It also produces high proportions of branched alkanes and aromatic hydrocarbons like benzene.

For UK A level (and equivalent) purposes, you aren't expected to know how the catalyst works, but you may be expected to know that it involves an ionic intermediate.

Note:  You should check your syllabus to find out exactly what you need to know. If you are studying a UK-based syllabus and haven't got one, follow this link.

Use the BACK button on your browser to return quickly to this page.

The zeolite catalyst has sites which can remove a hydrogen from an alkane together with the two electrons which bound it to the carbon. That leaves the carbon atom with a positive charge. Ions like this are called carbonium ions (or carbocations). Reorganisation of these leads to the various products of the reaction.

Note:  If you are interested in other examples of catalysis in the petrochemical industry, you should follow this link. It will lead you to information on reforming and isomerisation (as well as a repeat of what you have just read about catalytic cracking).

Use the BACK button on your browser if you want to return quickly to this page.

Thermal cracking

In thermal cracking, high temperatures (typically in the range of 450°C to 750°C) and pressures (up to about 70 atmospheres) are used to break the large hydrocarbons into smaller ones. Thermal cracking gives mixtures of products containing high proportions of hydrocarbons with double bonds - alkenes.

Warning!  This is a gross oversimplification, and is written to satisfy the needs of one of the UK A level Exam Boards (AQA). In fact, there are several versions of thermal cracking designed to produce different mixtures of products. These use completely different sets of conditions.

If you need to know about thermal cracking in detail, a Google search on thermal cracking will throw up lots of useful leads. Be careful to go to industry (or similarly reliable) sources. You will find a Google search box at the bottom of the Main Menu (link below). Remember to search the whole web rather than Chemguide otherwise you will just end up back here again!

Thermal cracking doesn't go via ionic intermediates like catalytic cracking. Instead, carbon-carbon bonds are broken so that each carbon atom ends up with a single electron. In other words, free radicals are formed.

Reactions of the free radicals lead to the various products.

INTRODUCING ALKENES

This is an introductory page about alkenes such as ethene, propene and the rest. It deals with their formulae and isomerism, their physical properties, and an introduction to their chemical reactivity.

What are alkenes?

Formulae

Alkenes are a family of hydrocarbons (compounds containing carbon and hydrogen only) containing a carbon-carbon double bond. The first two are:

ethene C2H4

propene C3H6

You can work out the formula of any of them using: CnH2n

The table is limited to the first two, because after that there are isomers which affect the names.

Isomerism in the alkenes

Structural isomerism

All the alkenes with 4 or more carbon atoms in them show structural isomerism. This means that there are two or more

different structural formulae that you can draw for each molecular formula.

For example, with C4H8, it isn't too difficult to come up with these three structural isomers:

Note:  If you aren't confident about naming organic compounds, the various ways of drawing organic compounds, or structural isomerism, then you really ought to follow these links before you go on.

Use the BACK button on your browser to return to this page.

There is, however, another isomer. But-2-ene also exhibits geometric isomerism.

Geometric (cis-trans) isomerism

The carbon-carbon double bond doesn't allow any rotation about it. That means that it is possible to have the CH3 groups on either end of the molecule locked either on one side of the molecule or opposite each other.

These are called cis-but-2-ene (where the groups are on the same side) or trans-but-2-ene (where they are on opposite sides).

Note:  If you aren't confident about geometric isomerism, then it is essential that you follow this link before you go on.

Use the BACK button on your browser to return to this page.

Physical properties of the alkenes

Boiling Points

The boiling point of each alkene is very similar to that of the alkane with the same number of carbon atoms. Ethene, propene and the various butenes are gases at room temperature. All the rest that you are likely to come across are liquids.

In each case, the alkene has a boiling point which is a small number of degrees lower than the corresponding alkane. The only attractions involved are Van der Waals dispersion forces, and these depend on the shape of the molecule and the number of electrons it contains. Each alkene has 2 fewer electrons than the alkane with the same number of carbons.

Note:  If you aren't sure about Van der Waals forces, then you should follow this link before you go on.

You will find the boiling points of the alkanes explained in some detail on the introductory alkanes page. Everything said there applies equally to the alkenes.

You will find the way geometric isomerism affects melting and boiling points explained towards the bottom of the page you get to by following this link.

Use the BACK button on your browser to return to this page.

Solubility

Alkenes are virtually insoluble in water, but dissolve in organic solvents.

Note:  The reasons for this are exactly the same as for the alkanes. You will find a detailed explanation on the introductory alkanes page.

Use the BACK button on your browser to return to this page.

Chemical Reactivity

Bonding in the alkenes

We just need to look at ethene, because what is true of C=C in ethene will be equally true of C=C in more complicated alkenes.

Ethene is often modelled like this:

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.

Note:  Check your syllabus to see if you need to know how a pi bond is formed. If you are studying a UK-based syllabus and haven't got a copy of that syllabus, find out how to get one by following this link.

If you do need to know about the bonding in ethene in detail, follow this link as well.

Use the BACK button on your browser to return to this page.

The reactions of alkenes

Like any other hydrocarbons, alkenes burn in air or oxygen, but these reactions are unimportant. Alkenes are too valuable to waste in this way.

The important reactions all centre around the double bond. Typically, the pi bond breaks and the electrons from it are used to join the two carbon atoms to other things. Alkenes undergo addition reactions.

For example, using a general molecule X-Y . . .

The rather exposed electrons in the pi bond are particularly open to attack by things which carry some degree of positive charge. These are called electrophiles. If you explore the rest of the alkene menu, you will find lots of examples of this kind.

Note:  If you need to know about organic reaction mechanisms, it would be a good idea to read the page explaining the background to electrophilic addition before you start looking at individual cases from the alkenes menu.

In fact, if you are only really interested in mechanisms, then look at that page and then explore the rest of the electrophilic addition menu in the mechanisms section of this site.

© Jim Clark 2003

MAKING ALKENES IN THE LAB

This page looks at ways of preparing alkenes in the lab by the dehydration of alcohols.

Dehydration of alcohols using aluminium oxide as catalyst

The dehydration of ethanol to give ethene

This is a simple way of making gaseous alkenes like ethene. If ethanol vapour is passed over heated aluminium oxide powder, the ethanol is essentially cracked to give ethene and water vapour.

To make a few test tubes of ethene, you can use this apparatus:

It wouldn't be too difficult to imagine scaling this up by boiling some ethanol in a flask and passing the vapour over aluminium oxide heated in a long tube.

Dehydration of alcohols using an acid catalyst

The acid catalysts normally used are either concentrated sulphuric acid or concentrated phosphoric(V) acid, H3PO4.

Concentrated sulphuric acid produces messy results. Not only is it an acid, but it is also a strong oxidising agent. It oxidises some of the alcohol to carbon dioxide and at the same time is reduced itself to sulphur dioxide. Both of these gases have to be removed from the alkene.

It also reacts with the alcohol to produce a mass of carbon. There are other side reactions as well, but these aren't required by any current UK A level (or equivalent) syllabus.

The dehydration of ethanol to give ethene

Ethanol is heated with an excess of concentrated sulphuric acid at a temperature of 170°C. The gases produced are passed through sodium hydroxide solution to remove the carbon dioxide and sulphur dioxide produced from side reactions.

The ethene is collected over water.

WARNING!  This is potentially an extremely dangerous preparation because of the close proximity of the very hot concentrated sulphuric acid and the sodium hydroxide solution. I knew of one chemistry teacher who put several students into hospital by getting it wrong! That was many years ago before safety was taken quite so seriously as it is now.

The concentrated sulphuric acid is a catalyst. Write it over the arrow rather than in the equation.

Note:  You will find the mechanism for the dehydration of alcohols in the mechanism section of this site. You will also find a discussion of how to cope with questions about the dehydration of more complicated alcohols if you follow a link at the bottom of that page.

Use the BACK button (or the HISTORY file or GO menu) on your browser if you want to return to this page.

The dehydration of cyclohexanol to give cyclohexene

This is a preparation commonly used at this level to illustrate the formation and purification of a liquid product. The fact that the carbon atoms happen to be joined in a ring makes no difference whatever to the chemistry of the reaction.

Cyclohexanol is heated with concentrated phosphoric(V) acid and the liquid cyclohexene distils off and can be collected and purified.

Phosphoric(V) acid tends to be used in place of sulphuric acid because it is safer and produces a less messy reaction.

THE HYDROGENATION OF ALKENES

This page looks at the reaction of the carbon-carbon double bond in alkenes with hydrogen in the presence of a metal catalyst. This is called hydrogenation. It includes the manufacture of margarine from animal or vegetable fats and oils.

Hydrogenation in the lab

The hydrogenation of ethene

Ethene reacts with hydrogen in the presence of a finely divided nickel catalyst at a temperature of about 150°C. Ethane is produced.

This is a fairly pointless reaction because ethene is a far more useful compound than ethane! However, what is true of the reaction of the carbon-carbon double bond in ethene is equally true of it in much more complicated cases.

Note:  You will find the mechanism for this hydrogenation reaction described in some detail in the catalysis section of this site.

Use the BACK button on your browser to return to this page.

Margarine manufacture

Some margarine is made by hydrogenating carbon-carbon double bonds in animal or vegetable fats and oils. You can recognise the presence of this in foods because the ingredients list will include words showing that it contains "hydrogenated vegetable oils" or "hydrogenated fats".

The impression is sometimes given that all margarine is made by hydrogenation - that's simply not true.

Animal and vegetable fats and oils

These are similar molecules, differing in their melting points. If the compound is a solid at room temperature, you usually call it a fat. If it is a liquid, it is often described as an oil.

Their melting points are largely determined by the presence of carbon-carbon double bonds in the molecule. The higher the number of carbon-carbon double bonds, the lower the melting point.

If there aren't any carbon-carbon double bonds, the substance is said to be saturated. A typical saturated fat might have the structure:

Molecules of this sort are usually solid at room temperature.

If there is only one carbon-carbon double bond in each of the hydrocarbon chains, it is called a mono-unsaturated fat (or mono-unsaturated oil, because it is likely to be a liquid at room temperature.)

A typical mono-unsaturated oil might be:

If there are two or more carbon-carbon double bonds in each chain, then it is said to be polyunsaturated.

For example:

For simplicity, in all these diagrams, all three hydrocarbon chains in each molecule are the same. That doesn't have to be the case - you can have a mixture of types of chain in the same molecule.

Making margarine

Vegetable oils often contain high proportions of polyunsaturated and mono-unsaturated fats (oils), and as a result are liquids at room temperature. That makes them messy to spread on your bread or toast, and inconvenient for some baking purposes.

You can "harden" (raise the melting point of) the oil by hydrogenating it in the presence of a nickel catalyst. Conditions (like the precise temperature, or the length of time the hydrogen is passed through the oil) are carefully controlled so that some, but not necessarily all, of the carbon-carbon double bonds are hydrogenated.

This produces a "partially hydrogenated oil" or "partially hydrogenated fat".

You need to hydrogenate enough of the bonds to give the final texture you want. However, there are possible health benefits in eating mono-unsaturated or polyunsaturated fats or oils rather than

saturated ones - so you wouldn't want to remove all the carbon-carbon double bonds.

Note:  There is constant (and constantly changing) controversy about the health risks and benefits of eating the various types of fat. If you do a Google search on this topic, you will find all sorts of conflicting information. Look carefully at the reliability of the source of that information before you try to make up your mind. Just because it is on the internet doesn't mean that it is right!

You will find a Google search box on the Main Menu page (link below). Don't forget that you need to search the whole web and not just chemguide - otherwise you will just end up back here!

The flow diagram below shows the complete hydrogenation of a typical mono-unsaturated oil.

The downside of hydrogenation as a means of hardening fats and oils

There are some probable health risks from eating hydrogenated fats or oils. Consumers are becoming more aware of this, and manufacturers are increasingly finding alternative ways of converting oils into spreadable solids.

One of the problems arises from the hydrogenation process.

The double bonds in unsaturated fats and oils tend to have the groups around them arranged in the "cis" form.

Note:  If you don't know what this means, you will find more about it towards the bottom of the introductory page about esters. If you don't understand what cis and trans isomers are, you won't make any sense of what comes next!

Use the BACK button on your browser to return to this page.

The relatively high temperatures used in the hydrogenation process tend to flip some of the carbon-carbon double bonds into the "trans" form. If these particular bonds aren't hydrogenated during the process, they will still be present in the final margarine in molecules of trans fats.

The consumption of trans fats has been shown to increase cholesterol levels (particularly of the more harmful LDL form) - leading to an increased risk of heart disease.

Any process which tends to increase the amount of trans fat in the diet is best avoided. Read food labels, and avoid any food which contains (or is cooked in) hydrogenated oil or hydrogenated fat.

Note:  As I said earlier, ideas about this change all the time. It would be worth doing a Google search on trans fatty acids. The US Food and Drugs Administration (FDA) has a useful up-to-date fact sheet about them, and the Google search will find it for you amongst a lot of other information.

© Jim Clark 2003 (modified 2004)

THE HALOGENATION OF ALKENES

This page looks at the reaction of the carbon-carbon double bond in alkenes such as ethene with halogens such as chlorine, bromine and iodine. This is called halogenation.

Reactions where the chlorine or bromine are in solution (for example, "bromine water") are slightly more complicated and are treated separately at the end.

Simple reactions involving halogens

In each case, we will look at ethene as typical of all of the alkenes. There are no complications as far as the basic facts are concerned as the alkenes get bigger.

Ethene and fluorine

Ethene reacts explosively with fluorine to give carbon and hydrogen fluoride gas. This isn't a useful reaction, and you aren't likely to need it for exam purposes in the UK at this level (A level or equivalent).

Ethene and chlorine or bromine or iodine

In each case you get an addition reaction. For example, bromine adds to give 1,2-dibromoethane.

Note:  Follow this link if you aren't happy about naming organic compounds

Use the BACK button on your browser to return to this page.

The reaction with bromine happens at room temperature. If you have a gaseous alkene like ethene, you can bubble it through either pure liquid bromine or a solution of bromine in an organic solvent like tetrachloromethane. The reddish-brown bromine is decolourised as it reacts with the alkene.

A liquid alkene (like cyclohexene) can be shaken with liquid bromine or its solution in tetrachloromethane.

Chlorine reacts faster than bromine, but the chemistry is similar. Iodine reacts much, much more slowly, but again the chemistry is

similar. You are much more likely to meet the bromine case than either of these.

Note:  If you are interested in the mechanism for the addition of bromine to alkenes you will find it in the mechanism section of this site.

Use the BACK button on your browser to return to this page.

Alkenes and bromine water

Using bromine water as a test for alkenes

If you shake an alkene with bromine water (or bubble a gaseous alkene through bromine water), the solution becomes colourless. Alkenes decolourise bromine water.

The chemistry of the test

This is complicated by the fact that the major product isn't 1,2-dibromoethane. The water also gets involved in the reaction, and most of the product is 2-bromoethanol.

However, there will still be some 1,2-dibromoethane formed, so at this sort of level you can probably get away with quoting the simpler equation:

Note:  Check your syllabus, past papers and mark schemes to find out what you need to be able to do. If you are studying a UK-based syllabus (A level or equivalent) and haven't got any of these, follow this link to find out how to get hold of them. If you are working at a lower level, use the simpler equation.

If you are interested in the mechanism for this reaction, I'm afraid that you won't find it on this site because it isn't on any of the UK-based syllabuses. You can, however, deduce it fairly easily if you know the mechanism for the addition of pure bromine. Instead of the intermediate ion being attacked by a bromide ion, it is much more likely to be hit by a lone pair on an oxygen of a water molecule, followed by loss of a hydrogen ion from the product.

© Jim Clark 2003

ALKENES and HYDROGEN HALIDES

This page looks at the reaction of the carbon-carbon double bond in alkenes such as ethene with hydrogen halides such as hydrogen chloride and hydrogen bromide.

Symmetrical alkenes (like ethene or but-2-ene) are dealt with first. These are alkenes where identical groups are attached to each end of the carbon-carbon double bond. The extra problems associated with unsymmetrical ones like propene are covered in a separate section afterwards.

Addition to symmetrical alkenes

What happens?

All alkenes undergo addition reactions with the hydrogen halides. A hydrogen atom joins to one of the carbon atoms originally in the double bond, and a halogen atom to the other.

For example, with ethene and hydrogen chloride, you get chloroethane:

With but-2-ene you get 2-chlorobutane:

Note:  Follow this link if you aren't happy about naming organic compounds

Use the BACK button on your browser to return to this page.

What happens if you add the hydrogen to the carbon atom at the right-hand end of the double bond, and the chlorine to the left-hand end? You would still have the same product.

The chlorine would be on a carbon atom next to the end of the chain - you would simply have drawn the molecule flipped over in space.

That would be different of the alkene was unsymmetrical - that's why we have to look at them separately.

Conditions

The alkenes react with gaseous hydrogen halides at room temperature. If the alkene is also a gas, you can simply mix the gases. If the alkene is a liquid, you can bubble the hydrogen halide through the liquid.

Alkenes will also react with concentrated solutions of the gases in water. A solution of hydrogen chloride in water is, of course, hydrochloric acid. A solution of hydrogen bromide in water is hydrobromic acid - and so on.

There are, however, problems with this. The water will also get involved in the reaction and you end up with a mixture of products.

Warning!  The mechanism for this reaction is almost invariably given for the reaction involving the alkene and the simple molecules H-Cl or H-Br or whatever. In the presence of water, these molecules will

already have reacted with the water to produce hydroxonium ions, H3O+, and halide ions. The mechanism will therefore be different - involving an initial attack by a hydroxonium ion. Avoid this problem by using the pure gaseous hydrogen halide.

If you choose to follow this link to the mechanism, use the BACK button on your browser to return to this page.

Reaction rates

Variation of rates when you change the halogen

Reaction rates increase in the order HF - HCl - HBr - HI. Hydrogen fluoride reacts much more slowly than the other three, and is normally ignored in talking about these reactions.

When the hydrogen halides react with alkenes, the hydrogen-halogen bond has to be broken. The bond strength falls as you go from HF to HI, and the hydrogen-fluorine bond is particularly strong. Because it is difficult to break the bond between the hydrogen and the fluorine, the addition of HF is bound to be slow.

Variation of rates when you change the alkene

This applies to unsymmetrical alkenes as well as to symmetrical ones. For simplicity the examples given below are all symmetrical ones- but they don't have to be.

Reaction rates increase as the alkene gets more complicated - in the sense of the number of alkyl groups (such as methyl groups) attached to the carbon atoms at either end of the double bond.

For example:

There are two ways of looking at the reasons for this - both of which need you to know about the mechanism for the reactions.

Note:  If you should know about the mechanism, but are a bit uncertain about it, then you should spend some time exploring the electrophilic addition mechanisms menu before you go on, and then come back to this page later. You should look at the addition of hydrogen halides to unsymmetrical alkenes as well as symmetrical ones.

This will take you some time. Use the BACK button (or, more efficiently, the HISTORY file or GO menu) on your browser to return to this page.

If you don't need to know about the mechanisms, skip over the next bit!

Alkenes react because the electrons in the pi bond attract things with any degree of positive charge. Anything which increases the electron density around the double bond will help this.

Alkyl groups have a tendency to "push" electrons away from themselves towards the double bond. The more alkyl groups you have, the more negative the area around the double bonds becomes.

The more negatively charged that region becomes, the more it will attract molecules like hydrogen chloride.

Note:  If you aren't sure about pi bonds, you will find a simple mention of them in the introductory page on alkenes

You will find more about the electron pushing effect of alkyl groups on a page about carbocations in the mechanism section of this site. That is also important reading if you are to understand the next bit.

Use the BACK button on your browser to return to this page later.

The more important reason, though, lies in the stability of the intermediate ion formed during the reaction. The three examples given above produce these carbocations (carbonium ions) at the half-way stage of the reaction:

The stability of the intermediate ions governs the activation energy for the reaction. As you go towards the more complicated alkenes, the activation energy for the reaction falls. That means that the reactions become faster.

Didn't understand this?  You should have followed the link to the

page about carbocations mentioned above!

Use the BACK button on your browser to return to this page afterwards.

Addition to unsymmetrical alkenes

What happens?

In terms of reaction conditions and the factors affecting the rates of the reaction, there is no difference whatsoever between these alkenes and the symmetrical ones described above. The problem comes with the orientation of the addition - in other words, which way around the hydrogen and the halogen add across the double bond.

Orientation of addition

If HCl adds to an unsymmetrical alkene like propene, there are two possible ways it could add. However, in practice, there is only one major product.

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.

Warning!  Markovnikov's Rule is a useful guide for you to work out which way round to add something across a double bond, but it isn't the reason why things add that way. As a general principle, don't quote Markovnikov's Rule in an exam unless you are specifically asked for it.

You will find the proper reason for this in a page about the addition of hydrogen halides to unsymmetrical alkenes in the mechanism section of this site.

Use the BACK button on your browser if you want to return to this page.

A special problem with hydrogen bromide

Unlike the other hydrogen halides, hydrogen bromide can add to a carbon-carbon double bond either way around - depending on the conditions of the reaction.

If the hydrogen bromide and alkene are entirely pure

In this case, the hydrogen bromide adds on according to Markovnikov's Rule. For example, with propene you would get 2-bromopropane.

That is exactly the same as the way the other hydrogen halides add.

If the hydrogen bromide and alkene contain traces of organic peroxides

Oxygen from the air tends to react slowly with alkenes to produce some organic peroxides, and so you don't necessarily have to add them separately. This is therefore the reaction that you will tend to get unless you take care to exclude all air from the system.

In this case, the addition is the other way around, and you get 1-bromopropane:

This is sometimes described as an anti-Markovnikov addition or as the peroxide effect.

Organic peroxides are excellent sources of free radicals. In the presence of these, the hydrogen bromide reacts with alkenes using a different (faster) mechanism. For various reasons, this doesn't happen with the other hydrogen halides.

This reaction can also happen in this way in the presence of ultra-violet light of the right wavelength to break the hydrogen-bromine bond into hydrogen and bromine free radicals.

Note:  All this is explored in detail on the page about free radical addition of HBr to alkenes in the mechanism section of this site.

© Jim Clark 2003

ALKENES and SULPHURIC ACID

This page looks at the reaction of the carbon-carbon double bond in alkenes such as ethene with concentrated sulphuric acid. It includes the conversion of the product into an alcohol.

The addition of sulphuric acid to alkenes

The reaction with ethene

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.

Confused by all this? Don't be!

All you need to do is to learn the structure of sulphuric acid. A hydrogen from the sulphuric acid joins on to one of the carbon atoms, and the rest joins on to the other one. Make sure that you can see how the structure of the sulphuric acid relates to the various ways of writing the formula for the product.

Important!  Learn this structure for sulphuric acid. Sketch it over and over again until you can't possibly get it wrong.

Follow this link if you want the mechanism for this reaction.

Use the BACK button on your browser to return to this page.

The reaction with propene

This is typical of the reaction with unsymmetrical alkenes. An unsymmetrical alkene has different groups at either end of the carbon-carbon double bond.

If sulphuric acid adds to an unsymmetrical alkene like propene, there are two possible ways it could add. You could end up with one of two products depending on which carbon atom the hydrogen attaches itself to.

However, in practice, there is only one major product.

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.

Warning!  Markovnikov's Rule is a useful guide for you to work out which way round to add something across a double bond, but it isn't the reason why things add that way. As a general principle, don't quote Markovnikov's Rule in an exam unless you are specifically asked for it.

You can find more about this in the mechanism section of this site. You will find the mechanism for this reaction discussed in detail if you follow this link.

Use the BACK button (or HISTORY file or GO menu, if you have to explore several pages) on your browser if you want to return to this page.

Using these reactions to make alcohols

Making ethanol

Ethene is passed into concentrated sulphuric acid to make ethyl hydrogensulphate (as above). The product is diluted with water and then distilled.

The water reacts with the ethyl hydrogensulphate to produce ethanol which distils off.

Making propan-2-ol

More complicated alkyl hydrogensulphates react with water in exactly the same way. For example:

Notice that the position of the -OH group is determined by where the HSO4 group was attached. You get propan-2-ol rather than propan-1-ol because of the way the sulphuric acid originally added across the double bond in propene.

Using these reactions

These reactions were originally used as a way of manufacturing alcohols from alkenes in the petrochemical industry. These days, alcohols like ethanol or propan-2-ol tend to be manufactured by direct hydration of the alkene because it is cheaper and easier.

Note:  You can find out about the manufacture of alcohols by direct hydration by following this link.

ALKENES and POTASSIUM MANGANATE(VII)

This page looks at the reaction of the carbon-carbon double bond in alkenes such as ethene with potassium manganate(VII) solution (potassium permanganate solution).

Oxidation of alkenes

Experimental details

Alkenes react with potassium manganate(VII) solution in the cold. The colour change depends on whether the potassium manganate(VII) is used under acidic or alkaline conditions.

If the potassium manganate(VII) solution is acidified with dilute sulphuric acid, the purple solution becomes colourless.

If the potassium manganate(VII) solution is made slightly alkaline (often by adding sodium carbonate solution), the purple solution first becomes dark green and then produces a dark brown

precipitate.

Chemistry of the reaction

We'll look at the reaction with ethene. Other alkenes react in just the same way.

Manganate(VII) ions are a strong oxidising agent, and in the first instance oxidise ethene to ethane-1,2-diol (old name: ethylene glycol).

Looking at the equation purely from the point of view of the organic reaction:

Note:  This type of equation is quite commonly used in organic chemistry. Oxygen written in square brackets is taken to mean "oxygen from an oxidising agent". The reason for this is that a more normal equation tends to obscure the organic change in a mass of other detail - as you will find below!

The full equations are given below, although you probably won't need them.

The full equation depends on the conditions.

Under acidic conditions, the manganate(VII) ions are reduced to manganese(II) ions.

Note:  If you want to know how to write equations for redox reactions like this you could follow this link, and explore in the redox section of this site.

Use the BACK button (or HISTORY file or GO menu) on your browser to return to this page later.

Under alkaline conditions, the manganate(VII) ions are first reduced to green manganate(VI) ions . . .

. . . and then further to dark brown solid manganese(IV) oxide (manganese dioxide).

This last reaction is also the one you would get if the reaction was done under neutral conditions. You will notice that there are neither hydrogen ions nor hydroxide ions on the left-hand side of the equation.

Note:  You might possibly remember that further up the page it says that potassium manganate(VII) is often made slightly alkaline by adding sodium carbonate solution. Where are the hydroxide ions in this?

Carbonate ions react with water to some extent to produce hydrogencarbonate ions and hydroxide ions. It is the presence of these hydroxide ions that gives sodium carbonate solution its pH in the 10 - 11 region.

Complications

(. . . and you thought this was already complicated enough?)

The product, ethane-1,2-diol, is itself quite easily oxidised by manganate(VII) ions, and so the reaction won't stop at this point unless the potassium manganate(VII) solution is very dilute, very cold, and preferably not under acidic conditions.

That means that this reaction has little use as a way of preparing ethane-1,2-diol. Its only real use is in testing for carbon-carbon double bonds - and even then it isn't very good!

Note:  Ethane-1,2-diol is an alcohol, although unlike simple ones like ethanol it contains two -OH groups. The oxidation of alcohols is explored on another page if you are interested.

That page deals with the oxidation of alcohols by acidified potassium dichromate(VI) solution - a slightly less powerful oxidising agent than potassium manganate(VII). The essential chemistry will be the same, although manganate(VII) ions eventually oxidise ethane-1,2-diol all the way to carbon dioxide and water. You only need read the first part of that page about oxidation of primary alcohols.

Use the BACK button on your browser to return to this page if you choose to follow this link.

Using the reaction to test for carbon-carbon double bonds

If an organic compound reacts with dilute alkaline potassium manganate(VII) solution to give a green solution followed by a dark brown precipitate, then it may contain a carbon-carbon double bond. But equally it could be any one of a large number of other compounds all of which can be oxidised by manganate(VII) ions under alkaline conditions.

The situation with acidified potassium manganate(VII) solution is even worse because it has a tendency to break carbon-carbon bonds. It reacts destructively with a large number of organic compounds and is rarely used in organic chemistry.

You could use alkaline potassium manganate(VII) solution if, for example, all you had to do was to find out whether a hydrocarbon was an alkane or an alkene - in other words, if there was nothing else present which could be oxidised.

It isn't a useful test. Bromine water is far more clear cut.

Note:  You will find details of the use of bromine water in testing for carbon-carbon double bonds in the page about the reactions of alkenes with halogens.

THE DIRECT HYDRATION OF ALKENES

This page looks at the production of alcohols by the direct hydration of alkenes - adding water directly to the carbon-carbon double bond.

Manufacturing ethanol

Ethanol is manufactured by reacting ethene with steam. The reaction is reversible.

Only 5% of the ethene is converted into ethanol at each pass through the reactor. By removing the ethanol from the equilibrium mixture and recycling the ethene, it is possible to achieve an overall 95% conversion.

A flow scheme for the reaction looks like this:

Note:  This is a bit of a simplification! When the gases from the reactor are cooled, then excess steam will condense as well as the ethanol. The ethanol will have to be separated from the water by fractional

distillation.

All the sources I have looked at gloss over this, so I don't have any details. I assume it is a normal fractional distillation of an ethanol-water mixture.

You can find a full explanation of the reasons for the reaction conditions by following this link to the physical chemistry part of this site.

If you are interested in the mechanism for this reaction you will find it by following this link.

Use the BACK button (or HISTORY file or GO menu) on your browser if you want to return to this page later.

Manufacturing other alcohols

If you start from an unsymmetrical alkene like propene, you have to be careful to think about which way around the water adds across the carbon-carbon double bond.

Markovnikov's Rule says that when you add a molecule HX across a carbon-carbon double bond, the hydrogen joins to the carbon atom which already has the more hydrogen atoms attached to it.

Thinking of water as H-OH, the hydrogen will add to the carbon with the more hydrogens already attached. That means that in the propene case, you will get propan-2-ol rather than propan-1-ol.

The conditions used during manufacture vary from alcohol to alcohol. The only conditions you will need for UK A level purposes are those for making ethanol.

Note:  I haven't included the mechanism for the hydration of these more complicated alkenes anywhere on the site, but it isn't too difficult to work out for yourself if you know the mechanism for the hydration of ethene, and know about the stability of carbocations (carbonium ions).

The water adds to the propene in the way shown above because the secondary carbocation formed during the process is more stable than the primary one formed if the addition was the other way around.

These two pages are in two different parts of this site, so it would be best to come back to this page using the BACK button on your browser to get from one to the other if you are interested.

THE POLYMERISATION OF ALKENES

This page looks at the polymerisation of alkenes to produce polymers like poly(ethene) (usually known as polythene, and sometimes as polyethylene), poly(propene) (old name: polypropylene), PVC and PTFE. It also looks briefly at how the structure of the polymers affects their properties and uses.

Poly(ethene) (polythene or polyethylene)

Low density poly(ethene): LDPE

Manufacture

In common with everything else on this page, this is an example of addition polymerisation.

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.

The number of molecules joining up is very variable, but is in the region of 2000 to 20000.

Conditions

Temperature: about 200°CPressure: about 2000 atmospheres

Initiator: a small amount of oxygen as an impurity

Note:  If you want the mechanism for this reaction, you can get it by following this link to the mechanism section of the site.

Use the BACK button on your browser if you want to return to this page.

Properties and uses

Low density poly(ethene) has quite a lot of branching along the hydrocarbon chains, and this prevents the chains from lying tidily close to each other. Those regions of the poly(ethene) where the chains lie close to each other and are regularly packed are said to be crystalline. Where the chains are a random jumble, it is said to be amorphous. Low density poly(ethene) has a significant proportion of amorphous regions.

Note:  Various sources quote estimated values for the proportion of low density poly(ethene) which is crystalline. These vary from 50% to 75%. I have no idea what the correct value is!

One chain is held to its neighbours in the structure by van der

Waals dispersion forces. Those attractions will be greater if the chains are close to each other. The amorphous regions where the chains are inefficiently packed lower the effectiveness of the van der Waals attractions and so lower the melting point and strength of the polymer. They also lower the density of the polymer (hence: "low density poly(ethene)").

Note:  Follow this link if you don't understand van der Waals forces.

Use the BACK button on your browser to return to this page.

Low density poly(ethene) is used for familiar things like plastic carrier bags and other similar low strength and flexible sheet materials.

High density poly(ethene): HDPE

Manufacture

This is made under quite different conditions from low density poly(ethene).

Conditions

Temperature: about 60°CPressure: low - a few atmospheres

Catalyst: Ziegler-Natta catalysts or other metal compounds

Ziegler-Natta catalysts are mixtures of titanium compounds like titanium(III) chloride, TiCl3, or titanium(IV) chloride, TiCl4, and compounds of aluminium like aluminium triethyl, Al(C2H5)3. There are all sorts of other catalysts constantly being developed.

These catalysts work by totally different mechanisms from the high pressure process used to make low density poly(ethene). The chains grow in a much more controlled - much less random - way.

Properties and uses

High density poly(ethene) has very little branching along the hydrocarbon chains - the crystallinity is 95% or better. This better packing means that van der Waals attractions between the chains are greater and so the plastic is stronger and has a higher melting point. Its density is also higher because of the better packing and smaller amount of wasted space in the structure.

Note:  Despite the names, HDPE is only about 3% denser than

LDPE! HDPE has a density of about 0.95 g cm-3 compared with about 0.92 for LDPE.

High density poly(ethene) is used to make things like plastic milk bottles and similar containers, washing up bowls, plastic pipes and so on. Look for the letters HDPE near the recycling symbol.

Poly(propene) (polypropylene): PP

Poly(propene) is manufactured using Ziegler-Natta and other modern catalysts. There are three variants on the structure of poly(propene) which you may need to know about, but we'll start from the beginning with a general structure which fits all of them.

Note:  You should check your syllabus to find out exactly how much you need to know. It is pointless getting bogged down in this if you don't need to! If you are studying a UK-based syllabus and haven't got a copy of that syllabus, follow this link to find out how to get one.

Use the BACK button (or HISTORY file or GO menu) on your browser to return to this page later.

The general structure

If your syllabus simply mentions the structure of poly(propene) with no more detail, this is adequate.

The trick is to think about the shape of the propene in the right way:

Now line lots of them up in a row and join them together. Notice that the double bonds are all replaced by single bonds in the process.

In a simple equation form, this is normally written as:

The three variations on this structure

You have got to remember that the diagrams above are 2-dimensional. Real poly(propene) chains are 3-dimensional. There are three different sorts of poly(propene) depending in how the CH3

groups are arranged in space.

These are called isotactic, atactic and syndiotactic poly(propene). The commonly used version is isotactic poly(propene).

Isotactic poly(propene)

A bit of the isotactic poly(propene) chain looks like this:

Note:  Dotted lines show bonds going back into the screen or paper, and wedge shapes show them coming out towards you. If you aren't very happy about the various ways of drawing organic structures it might be worth following this link before you go on.

Use the BACK button on your browser to return to this page.

This very regular arrangement of the CH3 groups makes it possible for the chains to pack close together and so maximise the amount of van der Waals bonding between them. That means that isotactic poly(propene) is quite strong either as a solid object or when it is drawn into fibres.

This is the common form of poly(propene) which is used to make plastic crates and ropes amongst many other things. Look for the letters PP near the recycling symbol.

Atactic poly(propene)

In atactic poly(propene) the CH3 groups are orientated randomly along the chain.

This lack of regularity makes it impossible for the chains to lie closely together and so the van der Waals attractions between them are weaker. Atactic poly(propene) is much softer with a lower melting point.

It is formed as a waste product during the manufacture of isotactic poly(propene) and its uses are limited. It is used, for example, in road paint, in making roofing materials like "roofing felt", and in some sealants and adhesives.

Syndiotactic poly(propene)

Syndiotactic poly(propene) is a relatively new material and is another regularly arranged version of poly(propene). In this case, every alternate CH3 group is orientated in the same way.

This regularity means that the chains can pack closely, and van der Waals attractions will be fairly strong. However, the attractions aren't as strong as in isotactic poly(propene). This makes syndiotactic poly(propene) softer and gives it a lower melting point.

Because syndiotactic poly(propene) is relatively new, at the time of writing uses were still being developed. It has uses in packaging - for example, in plastic film for shrink wrapping food. There are also medical uses - for example, in medical tubing and for medical bags and pouches. There are a wide range of other potential uses - either on its own, or in mixtures with isotactic poly(propene).

Poly(chloroethene) (polyvinyl chloride): PVC

Poly(chloroethene) is commonly known by the initials of its old name, PVC.

Structure

Poly(chloroethene) is made by polymerising chloroethene, CH2=CHCl. Working out its structure is no different from working out the structure of poly(propene) (see above). As long as you draw the chloroethene molecule in the right way, the structure is pretty obvious.

The equation is usually written:

It doesn't matter which carbon you attach the chlorine to in the original molecule. Just be consistent on both sides of the equation.

The polymerisation process produces mainly atactic polymer molecules - with the chlorines orientated randomly along the chain. The structure is no different from atactic poly(propene) - just replace the CH3 groups by chlorine atoms.

Because of the way the chlorine atoms stick out from the chain at random, and because ot their large size, it is difficult for the chains to lie close together. Poly(chloroethene) is mainly amorphous with only small areas of crystallinity.

Properties and uses

You normally expect amorphous polymers to be more flexible than crystalline ones because the forces of attraction between the

chains tend to be weaker. However, pure poly(chloroethene) tends to be rather hard and rigid.

This is because of the presence of additional dipole-dipole interactions due to the polarity of the carbon-chlorine bonds. Chlorine is more electronegative than carbon, and so attracts the electrons in the bond towards itself. That makes the chlorine atoms slightly negative and the carbons slightly positive.

These permanent dipoles add to the attractions due to the temporary dipoles which produce the dispersion forces.

Note:  If you aren't happy about the various sorts of intermolecular forces, it is important to follow this link.

If you don't understand about electronegativity and polar bonds, then follow this one as well.

Use the BACK button (or HISTORY file or GO menu) on your browser to return to this page.

Plasticisers are added to the poly(chloroethene) to reduce the effectiveness of these attractions and make the plastic more flexible. The more plasticiser you add, the more flexible it becomes.

Poly(chloroethene) is used to make a wide range of things including guttering, plastic windows, electrical cable insulation, sheet materials for flooring and other uses, footwear, clothing, and so on and so on.

Poly(tetrafluoroethene): PTFE

You may have come across this under the brand names of Teflon or Fluon.

Structure

Structurally, PTFE is just like poly(ethene) except that each hydrogen in the structure is replaced by a fluorine atom.

The PTFE chains tend to pack well and PTFE is fairly crystalline. Because of the fluorine atoms, the chains also contain more electrons (for an equal length) than a corresponding poly(ethene) chain. Taken together (the good packing and the extra electrons) that means that the van der Waals dispersion forces will be stronger than in even high density poly(ethene).

Note:  There won't be any dipole-dipole attractions between the chains in addition to dispersion forces (unlike PVC). The fluorines are arranged regularly around the carbon backbone. Although each bond is very polar, overall they cancel each other out.

(In fact, there might even be some repulsions because the outsides of the chains all consist of slightly negative fluorine atoms, but these obviously aren't enough to have a significant effect on the strong dispersion forces. I haven't been able to find any mention of this either in textbooks or on the Web.)

You might like to read about CCl4 (a simple example of a non-polar molecule containing polar bonds) on the main page about electronegativity (different from the link above).

Use the BACK button on your browser to return to this page.

Properties and uses

PTFE has a relatively high melting point (due to the strength of the attractions between the chains) and is very resistant to chemical attack. The carbon chain is so wrapped up in fluorine atoms that nothing can get at it to react with it. This makes it useful in the chemical and food industries to coat vessels and make them resistant to almost everything which might otherwise corrode them.

Equally important is that PTFE has remarkable non-stick properties - which is the basis for its most familiar uses in non-stick kitchen and garden utensils. For the same reason, it can also be used in things like low-friction bearings

Note:  Despite an extensive Web search, I haven't found any convincing explanation for the non-stick properties of PTFE at the molecular level. Many sources talk about it in terms of surface tension or surface energy, which actually beg the question. Unless you can explain the origin of these in terms of attractions or repulsions at the molecular level, it seems to me that you aren't actually explaining anything just by using a posh-sounding term!

If you are interested, you will find a discussion of how far I've got in coming up with an answer to this on a page about the non-stick properties of PTFE in the section of questions that I can't answer to my satisfaction!

© Jim Clark 2003

EPOXYETHANE ( ETHYLENE OXIDE )

This page looks at the manufacture of epoxyethane from ethene, and then at some of the products that are made from epoxyethane.

The manufacture of epoxyethane

Conditions

Temperature: about 250 - 300°C

Pressure: about 15 atmospheres

Catalyst: silver

Problems and hazards during manufacture

The main problem comes in controlling the temperature. The reaction is exothermic and so the temperature will tend to rise unless it is carefully controlled.

At higher temperatures the ethene burns in the oxygen to produce carbon dioxide and water which means that the temperature would increase even more - and the whole thing get completely out of hand!

Two hazards during manufacture come from the nature of epoxyethane. It is

poisonous and carcinogenic (cancer producing); highly inflammable or explosive in contact with air.

The reactivity of epoxyethane

Ring strain

The reason that epoxyethane is so reactive is that bonding pairs in the ring of atoms in the molecule are forced very close together. The bond angles are about 60° rather than about 109.5° when carbon atoms normally form single bonds.

The overlap between the atomic orbitals in forming the carbon-carbon and carbon-oxygen bonds is less good than it is normally, and there is considerable repulsion between the bonding pairs. The system becomes more stable if the ring is broken.

When epoxyethane reacts a carbon-oxygen bond is always broken and the ring opens up.

Uses of epoxyethane

Manufacture of ethane-1,2-diol (ethylene glycol)

Acid catalysed hydrolysis of epoxyethane

Epoxyethane reacts with water in the presence of an acid catalyst (very dilute sulphuric acid) at a temperature of about 60°C. Ethane-1,2-diol is produced.

A large excess of water is used to try to prevent the product from reacting with the original epoxyethane. Ethane-1,2-diol is an alcohol (because it contains simple -OH groups), and alcohols react with epoxyethane (see below).

Even in the presence of a large excess of water, this reaction happens as well:

The product is still an alcohol, and similar reactions can also lead to quite long chains.

Uses of ethane-1,2-diol

Ethane-1,2-diol is used as an antifreeze in car engines. It is added to the cooling water to prevent it from freezing under very cold conditions.

Ethane-1,2-diol is also used in the manufacture of polyesters such as poly(ethylene terephthalate). You may have come across this as a fibre used to make clothes (perhaps under the brand name Terylene), or as a clear material used to make plastic drinks bottles (PET).

The reaction of epoxyethane with alcohols

This is a reaction which students at this level often find difficulty remembering. It is actually probably easier to work out than remember. Think of it as an extension of the reaction with water.

Alcohols have the formula R-OH, where R is an alkyl group. Water can be thought of as H-OH.

The reaction of epoxyethane with water can be colour-coded like this:

Now do the same thing with the alcohol:

Product molecules of this type are used as solvents.

Notice that the product is still an alcohol. It has an -OH group at the right-hand end of the molecule. If the epoxyethane is in excess, the reaction can continue. (In fact, it continues to some extent even if the epoxyethane isn't in excess.)

The product from this reaction is again an alcohol, and can go on to react with even more epoxyethane! What you get eventually is a chain with a structure:

Compounds of this type are used as plasticisers (added, for example, to PVC to make it more flexible) or as non-ionic surfactants (detergents). To make the surfactant, you would start with a fairly long chain alcohol to produce a molecule such as:

RADICAL CHAIN MECHANISM

FOR REACTION OF METHANE WITH Br2

Step 1 (Initiation) Heat or uv light cause the weak halogen bond to undergo homolytic cleavage to generate two bromine radicals and starting the chain process.

Step 2 (Propagation) (a) A bromine radical abstracts a hydrogen to form HBr and a methyl radical, then (b) The methyl radical abstracts a bromine atom from another molecule of Br2 to form the methyl bromide product and another bromine radical,  which can then itself undergo reaction 2(a) creating a cycle that can repeat.

Step 3 (Termination) Various reactions between the possible pairs of radicals allow for the formation of ethane, Br2 or the product, methyl bromide. These reactions remove radicals and do not perpetuate the cycle.

Isomerism in Organic Chemistry

Isomerism is the phenomenon whereby certain compounds, with the same molecular formula, exist in different forms owing to their different organisations of atoms. The concept of isomerism illustrates the fundamental importance of molecular structure and shape in organic chemistry.

Structural Isomerism

Structural Isomers have different structural formulae because their atoms are linked together in different ways.

It arises owing to:

arrangement of Carbon skeleton

e.g. The formula C4H10 represents two possible structural formulae, butane and methylpropane:

position of Functional group

e.g. propan-1-ol and propan-2-ol

different Functional groups

e.g. the molecular formula C2H60 represents both ethanol and methoxymethane.

Cyclic alkanes are isomeric with alkenes, e.g. cyclopropane and propene

Isomer Map

Stereoisomerism

Stereoisomers have the same structure and bond order but their atoms and groups of atoms are arranged differently in space. They have different spatial arrangements and their molecules are not superimposable. There are two types:

Geometric Isomerism

Involves a double bond, usually C=C, that does not allow free rotation about the double bond (unlike a C-C single bond). They are not superimposable.

e.g. cis-but-2-ene and trans-but-2-ene

Stereoisomers may possess quite different physical properties, such as melting point, density and solubility in water (look up those of maleic acid and fumaric acid). Ring structures and other steric factors also result in geometric isomerism.

Isomer Map

Optical isomerism

Involves an atom, usually carbon, bonded to four different atoms or groups of atoms. They exist in pairs, in which one isomer is the mirror image of the other.

e.g. butan-2-ol

These isomers are referred to as enantiomers. The central carbon atom to which four different atoms or groups are attached, is called an asymmetrical carbon atom.

Enantiomers have identical physical constants, such as melting points and boiling points, but are said to be optically active since they can be distinguished from each other by their ability to rotate the plane of polarised light in opposite directions. A mixture of enantiomers in equal proportions is optically inactive, and is called a racemic mixture. Use an organic chemistry textbook to find out more about optical isom

Case study 1a.3 The chain isomerisation of alkanes

Chain Isomerization is used in the petrochemical industry to produce more branched alkanes with a higher octane number, from linear alkanes, for fuels more suitable for petrol engines. The proportions of the 'isomers', as well as the hydrocarbon chain length, in crude oil does not match specific market demands. Straight chain alkanes are heated with a suitable catalyst to break up the chains and more branched alkanes, as well as lower alkanes are formed on recombination of the fragments (see examples below).

For a given carbon number of an alkane, the more branched the alkane, the higher the octane number. The higher the octane number of a fuel/molecule, the less the tendency it has to cause auto-ignition resulting in 'knocking' or 'pinking' in the car engine.

(1) is linear heptane,

a chain isomer of C7H16, and octane number = 0.

(2) is 2,2,4-trimethylpentane,

a highly branched chain isomer of C8H18, octane number = 100 (used to be called 'iso-octane').

The known tendency of a mixture of (1) and (2) to auto-ignite are compared with other fuels/molecules to give them their individual 'octane rating'.

Using skeletal formula, one possible isomerisation reaction of pentane C5H12 is shown below. They are reversible reactions, so changing reaction conditions, can change the position of the equilibrium.

(3) pentane, octane number 62 (4) 2-methylbutane, octane number 93

Case study 1b.1 Positional isomers of molecular formula C3H7Br

In the uv light catalysed reaction of bromine and propane gases, the free radical substitution reaction can produce two initial mono-substitution products. [full mechanism]

CH3CH2CH3 + Br2 {CH3CH2CH2Br or CH3CHBrCH3} + HBr

(1) 1-bromopropane, , bpt 71oC, primary halogenoalkane,

(2) 2-bromopropane,  , bpt 59oC, secondary halogenoalkane, 

They are both low boiling colourless liquids, but the more compact molecule (2) has a lower boiling point. Chemically they are very similar e.g. both undergoing all the nucleophilic substitution reactions with ammonia, cyanide ion, and hydroxide ion etc. In the case of the latter, (1) would give the primary alcohol, propan-1-ol and (2) would give the secondary alcohol, propan-2-ol. [lots of named halogenoalkane structures]

For higher bromoalkanes e.g. 1-bromobutane CH3CH2CH2CH2Br and 2-bromobutane CH3CH2CHBrCH3, another chemical difference will show up refluxing with ethanolic potassium hydroxide by which, following an elimination reaction,

1-bromobutane can only form but-1-ene CH3CH2CH=CH2, but 2-bromobutane can form two isomeric elimination products, namely

but-1-ene CH3CH2CH=CH2, and but-2-ene CH3CH=CHCH3, i.e. you can eliminate either side of the C-Br bond.

Case study 1b.2 Two linear positional isomers of molecular formula C5H10

(1) , pent-1-ene, bpt 30oC, 

(2) , pent-2-ene, cis bpt 37oC, trans bpt 36oC,

They are very similar physically e.g. relatively non-polar volatile colourless liquids, with similar low boiling points. Chemically similar e.g. all the usual electrophilic addition reactions of any alkene, though may, or may not be, some 'isomeric consequences' as regards both their formation or addition reaction products and some examples are outlined below. Unlike pent-1-ene, pent-2-ene can also exist as geometric isomers.

Both can be formed in cracking pentane or higher alkanes in which various isomers of C5H10 would be formed. In the laboratory they can be made by elimination reactions e.g. (a) the 'dehydration' of isomeric pentanols  with conc. sulphuric acid or (b) or by hydrolysing bromoalkanes  by refluxing with ethanolic potassium hydroxide. The formation of more than one isomer of the pentenes depends on the position of the -OH in alcohols or the -Br in bromoalkanes e.g.

(a)(i) CH3CH2CH2CH2CH2OH ==> CH3CH2CH2CH=CH2 + H2O

Pentan-1-ol (above) can only give 1 isomer, pent-1-ene, but pentan-2-ol (below)

(a)(ii) CH3CH2CH2CHOHCH3 ==> {CH3CH2CH2CH=CH2 or CH3CH2CH=CHCH3} + H2O

can give 2 isomers, pent-1-ene and pent-2-ene, because elimination of a -H (as well as the -OH) can take place either side of the >CH-OH group from an adjacent C-H, which is not possible with pentan-1-ol with the -OH group on the end carbon.

(b)(i) CH3CH2CH2CH2CH2Br + KOH ==> CH3CH2CH2CH=CH2 + H2O + KBr

1-bromopentane can only give 1 isomer, pent-1-ene (above), but 2-bromopentane (below)

(b)(ii) CH3CH2CH2CHBrCH3 + KOH ==> {CH3CH2CH2CH=CH2 or CH3CH2CH=CHCH3} + H2O + KBr

can give two isomers, pent-1-ene and pent-2-ene, because elimination of a -H (as well as the -Br) can take place either side of the >CH-Br group from an adjacent C-H, which is not possible with 1-romopentane with the -Br group on the end carbon.

You can also derive many other isomers from the molecular formula C5H10 e.g. methylbutenes (chain/positional isomers wrt pentenes), methylcyclobutane and dimethylcyclopropanes (both chain/functional group isomers wrt pentenes).

Case study 1b.5 The alcohols based on C4H10O or C4H9OH

The molecular formula C4H10O can lead to a multitude of isomers including different 'types' or 'classes' of alcohols based on the formula C4H9OH, resulting in some differences in physical and chemical properties which are summarised below.

(1) butan-1-ol, bpt 118oC, is a primary alcohol and oxidised to an aldehyde (butanal) using aqueous sulphuric acid/potassium dichromate(VI). Its fully linear structure gives it the highest boiling point.

(2) butan-2-ol, bpt 100oC, is a secondary alcohol and oxidised to a ketone (butanone) using aqueous sulphuric acid/potassium dichromate(VI).

(3) 2-methylpropan-1-ol, bpt 108oC, is primary alcohol and oxidised to an aldehyde (2-methylpropanal) using aqueous sulphuric acid/potassium dichromate(VI)

(4) 2-methylpropan-2-ol, bpt 83oC, is a tertiary alcohol and not readily oxidised because the strong C-C chain would have to be broken. It gives the lowest boiling point because it has the most compact structure (for explanation see case study 1a.1)

(1) and (2) are positional isomers of each other (same chain structure), as is (3) wrt (4). The pairs (8)/(9) and (10)/(11) are chain isomers of each pair. From the molecular formula C4H10O you can also derive three ethers, (5) ethoxyethane, (6) 1-methoxypropane and (7) 2-methoxypropane.

(5) , (6) ,  (7)

Case study 1b.6 Halogenoalkane isomers of C4H9Cl

From this molecular formula, both chain and positional isomers can be derived, as well as illustrating the three classes of halogenoalkanes and a few physical and chemical differences are summarised below. The alcohols formed by hydrolysis of C4H9Cl are considered in case study 1b.5 above. Skeletal formulae are used in the examples below.

(1) 1-chlorobutane, bpt 79oC, , a primary halogenoalkane, in a HCl elimination reaction only but-1-ene is formed. The most 'linear' structure gives the highest boiling point. On hydrolysis with aqueous sodium hydroxide, the primary alcohol butan-1-ol is formed.

(2) 2-chlorobutane, bpt 67oC, , a secondary halogenoalkane, in HCl elimination but-1-ene and but-2-ene are formed. On hydrolysis the secondary alcohol butan-2-ol is formed.

(3) 1-chloro-2-methylpropane, bpt 68oC, , a primary halogenoalkane, in HCl elimination 2-methylpropene is formed. On hydrolysis the primary alcohol 2-methylprop-2-ol is formed.

(4) 2-chloro-2-methylpropane, bpt 51oC, , a tertiary halogenoalkane, in HCl elimination 2-methylpropene is formed. The most 'compact' structure gives the lowest boiling point.  On hydrolysis the tertiary alcohol 2-methylprop-2-ol is formed.

They are all volatile colourless liquids and all undergo the usual nucleophilic substitution reactions of any halogenoalkanes. There can chemical differences in  e.g. (4) is likely to react via the 2 step 'unimolecular' SN1 carbocation mechanism (carbocation stability is tert > sec > prim), and (1) is more likely to go by the SN2 'bimolecular' one step mechanism. Also, with ethanolic/aqueous sodium hydroxide, con-current elimination is much more likely with tertiary halogenoalkanes than primary ones. See also bromobutanes for an example of differences in elimination products, mechanisms of haloalkane reactions and [lots of halogenoalkane structures]

Case study 1c.1 Functional group isomers of C2H6O

(1) ethanol, an alcohol, bpt 79oC,

(2) methoxymethane, an ether, bpt -25oC,

The highly polarised Oδ--Hδ+ bond arises from the difference in oxygen/hydrogen electronegativity (O>>H). This results in alcohol molecules being much more polar than ethers and the formation of 'hydrogen bonding' between alcohol molecules (however misnamed!). Hydrogen bonding is the strongest intermolecular force and the resulting increased inter-molecular forces raises the boiling point of alcohols quite considerably compared to the isomeric ether. The Cδ+-Oδ- is polar, but the two polarities of the C-O-C linkage tend to cancel each out. The lower alcohols tend to be more soluble in the highly polar solvent water (water-alcohol H bonding) than the less polar ether molecules are.

Their structural differences leads to quite different chemical reactions and products, apart from combustion! Alcohols have a diverse chemistry via the C-OH group which ethers lack giving them quite a limited chemistry. However the lack of reactivity/chemistry of ethers makes ethers very useful as solvents for other reactants!

(a) Alcohols like (1) react with carboxylic acids to form esters via the -OH group, ethers cannot.

CH3CH2OH + CH3COOH ==> CH3COOCH2CH3 + H2O

Ethanol forms the ester ethyl ethanoate when heated with ethanoic acid and a little conc. sulphuric acid.

(b) Alcohols can be dehydrated to form alkenes, ethers cannot.

CH3CH2OH ==> CH2=CH2 + H2O

Ethanol forms ethene when heated with conc. sulphuric acid.

(c) Alcohols rapidly react with sodium metal

2CH3CH2OH + 2Na ==> CH3CH2O-Na+ + H2 

Ethanol Case study 1c.3 Functional group isomers of C3H6O

Some physical similarities e.g. low boiling colourless polar liquids or gases, (1) and (2) also show chemical similarities, as do (3) and (4), but there are significant chemical differences between all four shown below.

(1) , propanal (an aldehyde), bpt 49oC, adds HCN to give hydroxynitrile, gives yellow-orange ppt with 24DNPH, produces the primary alcohol, propan-1-ol, on reduction, readily oxidised to propanoic acid, gives silver mirror with ammoniacal silver nitrate and red-brown ppt with Fehlings/Benedicts reagent. I2 reaction?

(2) , propanone (a ketone), bpt 56oC,  adds HCN to give hydroxynitrile, gives yellow-orange ppt with 24DNPH, produces secondary alcohol, propan-2-ol, on reduction, NOT readily oxidised, NO silver mirror with ammoniacal silver nitrate and NO red-brown ppt with Fehlings/Benedicts reagent. I2 reaction?

(3) , prop-2-ene-1-ol (a bi-functional molecule alkene/alcohol or enol), bpt 97oC, higher bpt due to hydrogen bonding via -OH (not possible with 1 and 2 above), gives electrophilic addition reaction of Br2, H2O, HI etc. like any other alkene, reacts with sodium to give H2 and forms esters with carboxylic acids or acid chlorides just like alcohols do, NO reaction with ammoniacal silver nitrate, Fehlings/Benedicts reagent or 24DNPH.

(4)  , cyclopropanol (an alicyclic secondary alcohol), bpt ?, very unstable, difficult to study, and readily isomerises to (1) propanal (see case study 1c.4 below. Theoretically has the chemistry of a secondary alcohol e.g. oxidised to the ketone cyclopropanone, reacts with sodium to give H2, forms esters with carboxylic acids or acid chlorides.

(5) , methoxyethene (a bi-functional ether-alkene), bpt 5oC, gives electrophilic addition reaction of Br2, H2O, HI etc. like any other alkene, but no aldehyde, ketone or alcohol chemistry.

(6) , 1,2-epoxypropane (a cyclic-ether), bpt 35oC, no alkene, aldehyde, ketone or alcohol chemistry.

(7) , 1,3-epoxypropane (a cyclic-ether), bpt 49oC,  no alkene, aldehyde, ketone or alcohol chemistry.

Again, for a 'small' molecular formula, C3H6O2 packs quite an isomeric punch! but don't worry too much, (1) to (2) are ones whose detailed structure, naming, physical properties and chemical reactions you should be very familiar with. (3) to (5) you should cope with in a functional group concept Q and (6) to (7) I wouldn't worry too much about!

forms the salt sodium ethoxide and hydrogen.

INTRODUCING ARENES (AROMATIC HYDROCARBONS)

This page looks at the structures and physical properties of the simplest arenes (benzene and methylbenzene), together with a very brief introduction to their reactivity. Much of this is covered in detail elsewhere on the site - in sections on bonding and mechanisms, for example. You will find links to all of these pages.

What are arenes?

Arenes are aromatic hydrocarbons. The term "aromatic" originally referred to their pleasant smells, but now implies a particular sort of delocalised bonding (see below).

The arenes are based on benzene rings. The simplest of them is benzene itself, C6H6. The next simplest is methylbenzene (old name: toluene) which has one of the hydrogen atoms attached to the ring replaced by a methyl group - C6H5CH3.

The structure of benzene

The structure of benzene is covered in detail in two pages in the organic bonding section of this site. It is important to understand these thoroughly to make sense of benzene and methylbenzene chemistry. Unless you have read these pages recently, you should spend some time on them now before you go any further on this page.

Note:  You will find two pages dealing with the bonding in benzene. One deals with the Kelulé structure, and the other with the modern delocalised structure.

You should read both of these pages starting with the one about the Kelulé structure. You will find a link to the second page at the bottom of the first one.

This is likely to take you some time because you will probably have to follow up other links as well in order to fully understand the second page. Don't try to short-cut this.

Use the BACK button (or more likely, the HISTORY file or GO menu) on your browser to return to this page later.

What you need to understand about benzene is:

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

atom attached.

The six carbon atoms form a perfectly regular hexagon. All the carbon-carbon bonds have exactly the same lengths - somewhere between single and double bonds.

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

This diagram shows one of the molecular orbitals containing two of the delocalised electrons, which may be found anywhere within the two "doughnuts". The other molecular orbitals are almost never drawn.

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.

The structure of methylbenzene (toluene)

Methylbenzene just has a methyl group attached to the benzene ring - replacing one of the hydrogen atoms.

Attached groups are often drawn at the top of the ring, but you may occasionally find them drawn in other places with the ring rotated.

Note:  If you have to count up the hydrogens in a diagram of this kind, don't forget that there isn't a hydrogen atom at any corner of the hexagon where there is something else attached. The molecular formula of methylbenzene, for example, is C7H8. Check that you get that answer from this diagram!

Physical properties

Boiling points

In benzene, the only attractions between neighbouring molecules are van der Waals dispersion forces. There is no permanent dipole on the molecule.

Benzene boils at 80°C - rather higher than other hydrocarbons of similar molecular size (pentane and hexane, for example). This is presumably due to the ease with which temporary dipoles can be set up involving the delocalised electrons.

Note:  If you aren't happy about van der Waals dispersion forces then you should follow this link before you go on.

Use the BACK button on your browser to return to this page.

Methylbenzene boils at 111°C. It is a bigger molecule and so the van der Waals dispersion forces will be bigger.

Methylbenzene also has a small permanent dipole, so there will be dipole-dipole attractions as well as dispersion forces. The dipole is due to the CH3 group's tendency to "push" electrons away from itself. This also affects the reactivity of methylbenzene (see below).

Melting points

You might have expected that methylbenzene's melting point would be higher than benzene's as well, but it isn't - it is much lower! Benzene melts at 5.5°C; methylbenzene at -95°C.

Molecules must pack efficiently in the solid if they are to make best use of their intermolecular forces. Benzene is a tidy, symmetrical molecule and packs very efficiently. The methyl group sticking out in methylbenzene tends to disrupt the closeness of the packing. If the molecules aren't as closely packed, the intermolecular forces don't work as well and so the melting point falls.

Solubility in water

The arenes are insoluble in water.

Benzene is quite large compared with a water molecule. In order for benzene to dissolve it would have to break lots of existing hydrogen bonds between water molecules. You also have to break the quite strong van der Waals dispersion forces between benzene molecules. Both of these cost energy.

The only new forces between the benzene and the water would be van der Waals dispersion forces. These aren't as strong as hydrogen bonds (or the original dispersion forces in the benzene), and so you wouldn't get much energy released when they form.

It simply isn't energetically profitable for benzene to dissolve in water. It would, of course, be even worse for larger arene molecules.

Reactivity

Benzene

It has already been pointed out above that benzene is resistant to addition reactions. Adding something new to the ring would need you to use some of the delocalised electrons to form bonds with whatever you are adding. That results in a major loss of stability as the delocalisation is broken.

Instead, benzene mainly undergoes substitution reactions - replacing one or more of the hydrogen atoms by something new. That leaves the delocalised electrons as they were.

Note:  The delocalisation is, in fact, broken during a substitution reaction, but reforms at the end. If you are interested in the mechanisms for benzene's substitution reactions you could follow this link.

Should you want to come back to this page later, it would probably be best to use the HISTORY file or GO menu on your browser.

Methylbenzene

You have to consider the reactivity of something like methylbenzene in two distinct bits:

For example, if you explore other pages in this section, you will find that alkyl groups attached to a benzene ring are oxidised by alkaline potassium manganate(VII) solution. This doesn't happen in the absence of the benzene ring.

The tendency of the CH3 group to "push" electrons away from itself also has an effect on the ring, making methylbenzene react more quickly than benzene itself. You will find this explored in other pages in this section as well.

MANUFACTURING ARENES

This page looks at the manufacture of arenes such as benzene and methylbenzene (toluene) by the catalytic reforming of fractions from petroleum (crude oil).

Catalytic reforming

What is reforming?

Reforming takes straight chain hydrocarbons in the C6 to C8 range from the gasoline or naphtha fractions and rearranges them into compounds containing benzene rings. Hydrogen is produced as a by-product of the reactions.

For example, hexane, C6H14, loses hydrogen and turns into benzene. As long as you draw the hexane bent into a circle, it is easy to see what is happening.

Similarly, methylbenzene (toluene) is made from heptane:

The process

The feedstock

The feedstock is a mixture of the naphtha or gasoline fractions and hydrogen. The hydrogen is there to help prevent the formation of carbon by decomposition of the hydrocarbons at the high temperatures used. The carbon would otherwise contaminate the catalyst.

The catalyst

A typical catalyst is a mixture of platinum and aluminium oxide. With a platinum catalyst, the process is sometimes described as "platforming".

Temperature and pressure

The temperature is about 500°C, and the pressure varies either side of 20 atmospheres.

Converting some of the methylbenzene into benzene

Methylbenzene is much less commercially valuable than benzene. The methyl group can be removed from the ring by a process known as "dealkylation".

The methylbenzene is mixed with hydrogen at a temperature of between 550 and 650°C, and a pressure of between 30 and 50 atmospheres, with a mixture of silicon dioxide and aluminium oxide as catalyst.

NITRATION OF BENZENE AND METHYLBENZENE

This page looks at the facts about the nitration of benzene and methylbenzene. The mechanisms for these reactions are covered elsewhere on the site, and you will find links to these.

The nitration of benzene

Nitration happens when one (or more) of the hydrogen atoms on the benzene ring is replaced by a nitro group, NO2.

Benzene is treated with a mixture of concentrated nitric acid and concentrated sulphuric acid at a temperature not exceeding 50°C. The mixture is held at this temperature for about half an hour. Yellow oily nitrobenzene is formed.

You could write this in a more condensed form as:

The concentrated sulphuric acid is acting as a catalyst and so isn't written into the equations.

At higher temperatures there is a greater chance of getting more than one nitro group substituted onto the ring. You will get a certain amount of 1,3-dinitrobenzene formed even at 50°C. Some of the nitrobenzene formed reacts with the nitrating mixture of concentrated acids.

Notice that the new nitro group goes into the 3 position on the ring. Nitro groups "direct" new groups into the 3 and 5 positions. The reasons for this "directing effect" are beyond UK A level.

Note:  The numbering on the ring goes clockwise around the ring starting with number 1 at the top.

It is also possible to get a third nitro group attached to the ring (in the 5 position). However, nitro groups make the ring much less reactive than the original benzene ring. Two nitro groups on the ring make its reactions so slow that virtually no trinitrobenzene is produced under these conditions.

Note:  Follow this link if you want the mechanism for the nitration of benzene.

You will find the mechanism for the nitration of nitrobenzene (the reaction producing 1,3-dinitrobenzene) at the bottom of the page you will get to by following this second link.

If you choose to follow either of these links, use the BACK button (or the HISTORY file or GO menu) on your browser to return to this page later.

The nitration of methylbenzene (toluene)

Methylbenzene reacts rather faster than benzene - in nitration, the reaction is about 25 times faster. That means that you would use a lower temperature to prevent more than one nitro group being substituted - in this case, 30°C rather than 50°C. Apart from that, the reaction is just the same - using the same nitrating mixture of

concentrated sulphuric and nitric acids.

You get a mixture of mainly two isomers formed: 2-nitromethylbenzene and 4-nitromethylbenzene. Only about 5% of the product is 3-nitromethylbenzene. Methyl groups are said to be 2,4-directing.

For 2-nitromethylbenzene:

. . . and for 4-nitromethylbenzene:

Note:  You will find the mechanism for the nitration of methylbenzene by following this link.

Use the BACK button (or HISTORY file or GO menu) on your browser to return to this page later.

Just as with benzene, you will get a certain amount of dinitro compound formed under the conditions of the reaction, but virtually no trinitro product because the reactivity of the ring decreases for every nitro group added. From an experimental point of view this is just as well. Trinitromethylbenzene used to be called trinitrotoluene or TNT!

The reason for the 2,4-directing effect of the methyl group is beyond UK A level, but the increased reaction rate can be explained up to a point.

The reactivity of a benzene ring is governed by the electron density around the ring. Methyl groups tend to "push" electrons towards the ring - increasing the density, and so making the ring more attractive to attacking reagents.

This is actually a simplification. In order to understand the rate effect properly you have to think about the stability of the intermediate ions formed during the reactions, because this affects the activation energy of the reactions. This is also the basis for the directing effect and is again beyond UK A level.

Warning!  I have deliberately chosen not to use the strict

IUPAC names for the compounds in this section, because I see them as illogical and inconsistent with the older names for these compounds. The names for the compounds with one nitro group substituted into methylbenzene should be methyl-2-nitrobenzene and methyl-4-nitrobenzene. The reason for these names is that you should strictly name the attached groups in alphabetical order.

It seems to me to be illogical on three counts. Firstly, it breaks up the name of the hydrocarbon, so that it is no longer immediately obvious that you are talking about a derivative of methylbenzene.

Secondly, it doesn't relate to the older names for these compounds. Methyl-4-nitrobenzene, for example, used to be called para-nitrotoluene. The "para" refers to the 4 position. You could equally have called it 4-nitrotoluene. My name of 4-nitromethylbenzene is a direct translation of this. I think that's entirely logical!

Thirdly, you can end up with completely different names for compounds which are structurally similar. For example, if you think about substituting chlorines into these positions instead of nitro groups you will have to completely change the names for purely alphabetical reasons. That's silly!

I may well be in a minority of one on this, but I am sticking to my guns on it. In fact, IUPAC is much more flexible about these things than they are sometimes given credit for. I doubt if it would worry them, although it might upset your teachers or lecturers. Obviously, if you are working in a system which still calls methylbenzene "toluene", none of this bothers you!

HALOGENATION OF BENZENE AND METHYLBENZENE

This page looks at the reactions of benzene and methylbenzene (toluene) with chlorine and bromine under various conditions. The mechanisms for several of these reactions are covered elsewhere on the site and you will find links to these other pages.

The halogenation of benzene

Substitution reactions

Benzene reacts with chlorine or bromine in the presence of a catalyst, replacing one of the hydrogen atoms on the ring by a chlorine or bromine atom.

The reactions happen at room temperature. 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, FeCl3, or iron(III) bromide, FeBr3.

These compounds act as the catalyst and behave exactly like aluminium chloride, AlCl3, or aluminium bromide, AlBr3, 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, written more compactly:

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:

Note:  Follow this link if you want the mechanism for the halogenation of benzene.

Use the BACK button on your browser to return to this page later.

Addition reactions

In the presence of ultraviolet light (but without a catalyst present),

hot benzene will also undergo an addition reaction with chlorine or bromine. The ring delocalisation is permanently broken and a chlorine or bromine atom adds on to each carbon atom.

For example, if you bubble chlorine gas through hot benzene exposed to UV light for an hour, you get 1,2,3,4,5,6-hexachlorocyclohexane.

Bromine would behave similarly.

The chlorines and hydrogens can stick up and down at random above and below the ring and this leads to a number of geometric isomers. Although there aren't any carbon-carbon double bonds, the bonds are still "locked" and unable to rotate.

One of these isomers was once commonly used as an insecticide known variously as BHC, HCH and Gammexane. This is one of the "chlorinated hydrocarbons" which caused so much environmental harm.

Note:  Geometric isomerism is explored on another page, although only with regard to carbon-carbon double bonds.

It isn't too difficult to relate this to the ring case. For example, the chlorine atoms in the 1 and 2 positions on the ring could both be pointing up above the ring (or down below it) or one could be pointing up and the other one down. That corresponds exactly to the cis and trans positions that you are probably familiar with in carbon-carbon double bonds.

Use the BACK button on your browser to return to this page if you choose to follow this link.

The halogenation of methylbenzene

Substitution reactions

It is possible to get two quite different substitution reactions between methylbenzene and chlorine or bromine depending on the conditions used. The chlorine or bromine can substitute into the ring or into the methyl group.

Substitution into the ring

Substitution in the ring happens in the presence of aluminium chloride (or aluminium bromide if you are using bromine) or iron, and in the absence of UV light. The reactions happen at room temperature.

This is exactly the same as the reaction with benzene, except that you have to worry about where the halogen atom attaches to the ring relative to the position of the methyl group.

Methyl groups are 2,4-directing, which means that incoming groups will tend to go into the 2 or 4 positions on the ring - assuming the methyl group is in the 1 position. In other words, the new group will attach to the ring next door to the methyl group or opposite it. The reason for the directing effect is beyond UK A level.

With chlorine, substitution into the ring gives a mixture of 2-chloromethylbenzene and 4-chloromethylbenzene.

With bromine, you would get the equivalent bromine compounds.

Note:  You will find the mechanism for this reaction half way down the page you will get to by following this link.

Use the BACK button on your browser to return to this page later.

Substitution into the methyl group

If chlorine or bromine react with boiling methylbenzene in the absence of a catalyst but in the presence of UV light, substitution happens in the methyl group rather than the ring.

For example, with chlorine (bromine would be similar):

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. However, the reaction doesn't stop there, and all three hydrogens in the methyl group can in turn be replaced by chlorine atoms.

That means that you could also get (dichloromethyl)benzene and (trichloromethyl)benzene as the other hydrogen atoms in the methyl group are replaced one at a time.

If you use enough chlorine you will eventually get (trichloromethyl)benzene, but any other proportions will always lead to a mixture of products.

Note:  You will find the mechanisms for these reactions spread over 3 pages (depending on how much detail you want) which you can get to by following this link

Use the BACK button (or HISTORY file or GO menu) on your browser to return to this page.

Addition reactions

I haven't been able to track down anything similar to the reaction between benzene and chlorine in which six chlorine atoms add around the ring.

That perhaps isn't surprising. Chlorine adds to benzene in the presence of ultraviolet light. With methylbenzene under those conditions, you get substitution in the methyl group. That is energetically easier because it doesn't involve breaking the delocalised electron system.

Whether you would get addition to the ring if you used a large excess of chlorine and did the reaction for a long time, I don't know. Once all the hydrogens in the methyl group had been substituted, perhaps you might then get addition to the ring as well.

None of this is needed for any of the UK A level syllabuses.

Note:  If you have any solid information on this (either that the reaction definitely does or definitely doesn't happen), I would be grateful if you could contact me via the address on the about this site page.

FRIEDEL-CRAFTS REACTIONS OF BENZENE AND METHYLBENZENE

This page gives details of the Friedel-Crafts reactions of benzene and methylbenzene (toluene). The mechanisms for some of these reactions are covered elsewhere on the site and you will find links to those as you go along.

Friedel-Crafts acylation

Friedel-Crafts acylation of benzene

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, and in this case the reaction is sometimes called "ethanoylation". In the example which follows we are substituting a CH3CO- group into the ring, but you could equally well use any other acyl group.

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. The mixture is heated to about 60°C for about 30 minutes.

A ketone called phenylethanone (old name: acetophenone) is formed.

Note:  Don't worry too much about the name "phenylethanone" - all that matters is that you can draw the structure.

You may find slight variations on the conditions for this

reaction. Various recipes I have found vary the temperature and time - for example by having a slightly lower temperature for a longer time.

or, if you want a more compact form:

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 (see below).

Note:  You will find the mechanism for this reaction in the mechanisms section of this site.

Use the BACK button on your browser to return to this page later.

Friedel-Crafts acylation of methylbenzene (toluene)

The reaction is just the same with methylbenzene except that you have to worry about where the acyl group attaches to the ring relative to the methyl group.

Normally, the methyl group in methylbenzene directs new groups into the 2- and 4- positions (assuming the methyl group is in the 1- position). In acylation, though, virtually all the substitution happens in the 4- position.

Note:  The reason for the 2,4-directing effect of the methyl group is beyond UK A level. The reason that you get virtually none of the 2- isomer in this instance is because of the size of the incoming acyl group. Everything gets too cluttered (and therefore less stable) if you try to put the acyl group next door to the methyl group.

Friedel-Crafts alkylation

Friedel-Crafts alkylation of benzene

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.

The facts

Benzene reacts at room temperature with a chloroalkane (for example, chloromethane or chloroethane) in the presence of aluminium chloride as a catalyst. On this page, we will look at substituting a methyl group, but any other alkyl group could be used in the same way.

Substituting a methyl group gives methylbenzene.

or:

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. You can improve your chances of just getting monosubstitution by using a large excess of benzene.

You won't have to worry about this for UK A level purposes.

You will find the mechanism for this reaction in the mechanisms section of this site.

Use the BACK button on your browser to return to this page later.

Friedel-Crafts alkylation of methylbenzene (toluene)

Again, the reaction is just the same with methylbenzene except that you have to worry about where the alkyl group attaches to the ring relative to the methyl group.

Unfortunately this time there is a problem! Where the incoming alkyl group ends up depends to a large extent on the temperature of the reaction.

At 0°C, substituting methyl groups into methylbenzene, you get a mixture of the 2-.3- and 4- isomers in the proportion 54% / 17% / 29%. That's a higher proportion of the 3- isomer than you might expect.

At 25°C, the proportions change to 3% / 69% / 28%. In other words the proportion of the 3- isomer has increased even more. Raise the temperature some more and the trend continues.

The reason for this is again beyond UK A level.

Note:  The problem in this case lies in the fact that the methyl groups attaching to the ring can fall off again and reattach somewhere else in the presence of the aluminium chloride. You can get equilibria set up between the various isomers.

The reason for the 2,4- directing effect of the methyl group in methylbenzene lies in the fact that the 2- and 4- isomers form faster than the 3- isomer. However, in this case, the 3- isomer is the most thermodynamically stable of the three. If you raise the temperature, or allow more time, the equilibria set up favour the most stable product.

You do NOT have to worry about this for A level purposes.

Friedel-Crafts alkylation industrially

The manufacture of ethylbenzene

Ethylbenzene is an important industrial chemical used to make styrene (phenylethene), which in turn is used to make polystyrene - poly(phenylethene).

It is manufactured from benzene and ethene. There are several ways of doing this, some of which use a variation on Friedel-Crafts alkylation. What follows is the method required by the UK A level Exam Board, AQA.

The reaction is done in the liquid state. Ethene is passed through a liquid mixture of benzene, aluminium chloride and a catalyst promoter which might be chloroethane or hydrogen chloride. We are going to assume it is HCl (because that's what AQA want!).

Promoters are used to make catalysts work better.

There are two variants on the process. One (the Union Carbide / Badger process) uses a temperature no higher than 130°C and a pressure just high enough to keep everything liquid.

The other (the Monsanto process) uses a slightly higher temperature of 160°C which needs less catalyst. (Presumably - although I haven't been able to confirm this - the pressure would

also need to be higher to keep everything liquid at the higher temperature.)

or:

Again, 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.

Note:  You will find the mechanism for this reaction in the mechanisms section of this site. This will show exactly what the HCl and aluminium chloride are doing in the reaction.

Use the BACK button on your browser to return to this page later.

ASSORTED REACTIONS OF BENZENE AND METHYLBENZENE

This page gives details of some reactions of benzene and methylbenzene (toluene) not covered elsewhere in this section. It deals with the combustion, hydrogenation and sulphonation of benzene and methylbenzene (toluene), and with the oxidation of side chains attached to benzene rings.

Remember that benzene, methylbenzene and similar hydrocarbons based on benzene rings are collectively known as arenes.

Combustion

Like any other hydrocarbons, benzene and methylbenzene burn in a plentiful supply of oxygen to give carbon dioxide and water. For example:

For benzene:

. . . and methylbenzene:

However, for these hydrocarbons, combustion is hardly ever complete, especially if they are burnt in air. The high proportion of carbon in the molecules means that you need a very high proportion of oxygen to hydrocarbon to get complete combustion. Look at the equations.

As a general rule, the hydrogen in a hydrocarbon tends to get what oxygen is available first, leaving the carbon to form carbon itself, or carbon monoxide, if there isn't enough oxygen to go round.

The arenes tend to burn in air with extremely smoky flames - full of carbon particles.

You almost invariably get incomplete combustion, and the arenes can be recognised by the smokiness of their flames.

Hydrogenation

Hydrogenation is an addition reaction in which hydrogen atoms are added all the way around the benzene ring. A cycloalkane is formed. For example:

With benzene:

. . . and methylbenzene:

These reactions destroy the electron delocalisation in the original benzene ring, because those electrons are being used to form bonds with the new hydrogen atoms.

Although the reactions are exothermic overall because of the strengths of all the new carbon-hydrogen bonds being made, there is a high activation barrier to the reaction.

The reactions are done using the same finely divided nickel catalyst that is used in hydrogenating alkenes and at similar temperatures (around 150°C), but the pressures used tend to be

higher.

Note:  Pressure values quoted can be anywhere from about 20 atmospheres to 200 atmospheres. I have no way of knowing which is right!

This hydrogenation reaction is important in estimating the delocalisation energy for benzene. You can find more about this by following this link to a page about the Kekulé structure for benzene.

Use the BACK button on your browser to return to this page later.

Sulphonation

Sulphonation involves replacing one of the hydrogens on a benzene ring by the sulphonic acid group, -SO3H.

The sulphonation of benzene

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. Fuming sulphuric acid, H2S2O7, can usefully be thought of as a solution of sulphur trioxide in concentrated sulphuric acid.

Or:

The product is benzenesulphonic acid.

Note:  You will find the mechanism for this reaction by following this link.

Use the BACK button on your browser to return to this page.

The sulphonation of methylbenzene

Methylbenzene is more reactive than benzene because of the

tendency of the methyl group to "push" electrons towards the ring. Exactly how this increases the rate of reaction is beyond UK A level - it is rather more complicated than just an increase in the electron density of the ring.

The effect of this greater reactivity is that methylbenzene will react with fuming sulphuric acid at 0°C, and with concentrated sulphuric acid if they are heated under reflux for about 5 minutes.

As well as the effect on the rate of reaction, with methylbenzene you also have to think about where the sulphonic acid group ends up on the ring relative to the methyl group.

Methyl groups have a tendency to "direct" new groups into the 2- and 4- positions on the ring (assuming the methyl group is in the 1- position). Methyl groups are said to be 2,4-directing. The origin of this directing effect is also beyond UK A level.

So you get a mixture which mainly consists of two isomers. Only about 5 - 10% of the 3- isomer is formed.

The main reactions are:

and:

In the case of sulphonation, the exact proportion of the isomers formed depends on the temperature of the reaction. As the temperature increases, you get increasing proportions of the 4- isomer and less of the 2- isomer.

This is because sulphonation is reversible. The sulphonic acid group can fall off the ring again, and reattach somewhere else. This tends to favour the formation of the most thermodynamically stable isomer. This interchange happens more at higher temperatures.

The 4- isomer is more stable because there is no cluttering in the molecule as there would be if the methyl group and sulphonic acid group were next door to each other.

Side chain oxidation in alkylbenzenes

An alkylbenzene is simply a benzene ring with an alkyl group attached to it. Methylbenzene is the simplest alkylbenzene.

Alkyl groups are usually fairly resistant to oxidation. However, when they are attached to a benzene ring, they are easily oxidised by an alkaline solution of potassium manganate(VII) (potassium permanganate).

Methylbenzene is heated under reflux with a solution of potassium manganate(VII) made alkaline with sodium carbonate. The purple colour of the potassium manganate(VII) is eventually replaced by a dark brown precipitate of manganese(IV) oxide.

The mixture is finally acidified with dilute sulphuric acid.

Overall, the methylbenzene is oxidised to benzoic acid.

Interestingly, any alkyl group is oxidised back to a -COOH group on the ring under these conditions. So, for example, propylbenzene is also oxidised to benzoic acid.

Note:  These are flow schemes - deliberately not full equations. To be honest, I don't really know whether you could write an accurate single equation for anything more complicated than a methyl group attached to the ring. In other cases, you will certainly get some carbon dioxide produced, but possibly some other organic molecules as well depending on the conditions.

The separation of the benzoic acid is beyond the scope of this site. The sulphuric acid converts benzoate ions (formed under the alkaline conditions) into benzoic acid. The problem lies in separating solid benzoic acid from solid manganese(IV) oxide.

This reaction has some similarities to the oxidation of alkenes by potassium manganate(VII), which you might also like to have a look at if it is on your syllabus.

Use the BACK button on your browser to return to this page later if you choose to follow this link.

Part IV AROMATIC HYDROCARBONS (Arenes)

The five reactions described are electrophilic substitution reactions involving the generation of a powerful electrophile (electron pair acceptor) which subsequently attacks the electron rich Π ot dnet senerA .dnob elbuod eht fo metsys nortcele )ip( eht swolla snoitutitsbus esuaceb ,noitidda naht rehtar ,noitutitsbus ogrednu .tcatni niamer ot gnir enezneb elbats yrev

The electrophilic substitution of an arene - nitration

for benzene : C6H6 + HNO3 ==> C6H5NO2 + H2O    [see mechanism 19 below] for methyl benzene: C6H5CH3 + HNO3 ==> O2NC6H4CH3 + H2O The nitrating mixture consists of concentrated nitric acid (source of the nitro group -

NO2) and concentrated sulphuric acid which acts as a catalyst and as a strong acid.

mechanism 19 - electrophilic substitution in the nitration of the benzene ring

[mechanism 19 above] Benzene is converted into nitrobenzene, when R = H. When R = CH3, methylbenzene will form a mixture of the three possible substitution

products methyl-2/3/4-nitrobenzene, o and methyl-3-nitrobenzene is the minority product, the mechanism above

would show the formation of one of the major products methyl-2-nitrobenzene.

Step (1) The sulphuric acid protonates the nitric acid (strong acid, but weaker than H2SO4)

Step (2) The protonated nitric acid loses a water molecule via a sulphuric acid molecule, to generate the electrophile, the nitronium ion, NO2

+. This is a much more powerful electrophile, i.e. electron pair acceptor, than the original nitric acid, and is needed to attack the very stable aromatic ring of benzene.

o Steps (1) and (2) can be written as: 2H2SO4 + HNO3 ==> NO2+ + H3O+ +

2HSO4- 

Step (3) An electron pair from the delocalised ip electrons of the benzene ring forms a C-N bond with the electron pair accepting nitronium ion forming a highly unstable carbocation. It is very unstable because the stable electron arrangement of the benzene ring is partially broken to give a 'saturated' C (top right of ring).

Step (4) The hydrogensulphate ion (HSO4-, formed in step (1), abstracts a proton

from the highly unstable intermediate carbocation to give the nitro-aromatic product and reform the sulphuric acid catalyst as well as the stable benzene ring.

o Note that the hydrogensulphate ion, HSO4- has been shown in the style of

-:O-SO2-OH, to emphasize the importance a lone pair on the oxygen abstracting the proton from the aromatic carbocation.

GENERAL COMMENT to compare aromatic electrophilic substitution with the electrophilic addition with alkenes:

o Like alkenes, arenes are susceptible to electrophilic attack because of the high electron density of the ip electrons involved in the carbon-carbon bonding and both show little reactivity towards nucleophilic reagents.

o However two points should be considered because of the particular stability of the aromatic (benzene) ring.

1. This makes aromatic compounds less reactive than alkenes, which readily undergo addition rather than substitution.

2. Aromatics do not usually undergo addition because this will remove the stability conferred on the molecule by the benzene ring. By under going substitution rather than addition, the stable aromatic ring is preserved.

FURTHER COMMENTS o The overall nitration reaction is the substitution of -H by -NO2 

 

 

The electrophilic substitution of an arene - chlorination (example of aromatic halogenation)

C6H6 + Cl2 ==> C6H5Cl + HCl    [see mechanism 21 below] Chlorine is bubbled into a mixture of the arene and anhydrous aluminium chloride

catalyst. Other catalysts like anhydrous iron(III) chloride can be used, and they are collectively known as halogen carriers.

mechanism 21 - electrophilic substitution by halogen in a benzene ring

[mechanism 21 above] When R = H, benzene forms chlorobenzene. o Step (1) The non-polar and uncharged chlorine molecule is not a strong

enough an electrophile to disrupt the ip electron system of the benzene ring. The aluminium chloride reacts with a chlorine molecule to form a positive chlorine ion Cl+ which is a much stronger electron pair accepting electrophile and a tetrachloroaluminate(III) ion (either this or an AlCl3-Cl2 complex - details not needed for A level).

o Step (2) An electron pair from the delocalised ip electrons of the benzene ring forms a C-Cl bond with the electron pair accepting positive chlorine ion forming a highly unstable carbocation. It is very unstable because the stable electron arrangement of the benzene ring is partially broken to give a 'saturated' C (top right of ring).

o Step (3) The tetrachloroaluminate(III) ion, formed in step (1), abstracts a proton from the highly unstable intermediate carbocation to give the chloro-aromatic product, hydrogen chloride gas and reform the aluminium chloride catalyst.

Also consider C6H5CH3 + Cl2 ==> ClC6H4CH3 + HCl when R = CH3, methylbenzene forms a mixture of chloro-2/3/4-methylbenzene.

o chloro-3-methylbenzene is the minority product and the mechanism above would show the formation of chloro-2-methylbenzene.

FURTHER COMMENTS o The overall halogenation reaction is the substitution of -H by -Cl  o Bromination can be carried in the same way by mixing bromine, the

aromatic hydrocarbon (arene) with a halogen carrier catalyst such as anhydrous AlBr3 or FeBr3.

o Why do aromatic compounds tend to react by electrophilic substitution BUT alkenes tend to react by electrophilic addition?

They both interact with electrophiles because they both have 'electron rich' electron pair donating bonding systems i.e. the >C=C< double bond in alkenes and the delocalised ∏ electrons of the benzene ring, but the benzene ring has a particularly high stability which is preserved on substitution. For the same reason alkenes are generally more reactive than arenes.

o If methyl benzene is reacted with chlorine in the presence of uv light, substitution takes place in the alkyl side chain. In other words it behaves like an alkane and undergoes a free radical substitution reaction. The initial product is chloromethylbenzene, C6H5CH2Cl, and further substitution products can be formed C6H5CHCl2 and C6H5CCl3. This illustrates the significance of changing reaction conditions which function via a different mechanism to give a different product.

initiation: Cl-Cl ==> Cl. + Cl. 

chain propagations: Cl. + C6H5CH3 ==> HCl + C6H5CH2

. then C6H5CH2

. + Cl-Cl ==> C6H5CH2Cl + Cl.  chain terminations:

2C6H5CH2. ==> C6H5CH2CH2C6H5 or 2Cl. ==> Cl2 or C6H5CH2

.

+ Cl. ==> C6H5CH2Cl

 

 

The electrophilic substitution of an arene - alkylation (Friedel-Crafts reaction)

C6H6 + R3C-Cl ==> C6H5-CR3 + HCl (R = H, alkyl, aryl)   [see mechanism 23 below] The arene is refluxed with a chloroalkane and anhydrous aluminium chloride catalyst.

mechanism 23 - electrophilic substitution by an alkyl group in the benzene ring

[mechanism 23 above] If R' = H, benzene would form methylbenzene if chloromethane was used.

o Step (1) The weakly polar and uncharged halogenoalkane molecule is not a strong enough an electrophile to disrupt the ip electron system of the benzene ring. The aluminium chloride reacts with the halogenoalkane molecule to form a carbocation which is a much stronger electron pair accepting electrophile than the original acid chloride (either this or an AlCl3-R3Cl complex - details not needed for A level).

o Step (2) An electron pair from the delocalised ∏ electrons of the benzene ring forms a C-C bond with the electron pair accepting carbocation forming a second highly unstable carbocation. It is very unstable because the stable electron arrangement of the benzene ring is partially broken to give a 'saturated' C (top right of ring).

o Step (3) is a proton transfer, as the tetrachloroaluminate(III) ion [formed in step (1)], abstracts a proton from the highly unstable intermediate carbocation to give the alkyl-aromatic product, hydrogen chloride gas and reform the aluminium chloride catalyst.

If R' = CH3 methylbenzene: C6H5CH3 + R3C-Cl ==> R3C-C6H4CH3 + HCl A mixture of polysubstituted alkyl aromatic compounds are formed.

o e.g. using chloromethane, 1,2- or 1,3- or 1,4-dimethylbenzene will be formed, so if R=H, the mechanism above would show the formation of 1,2-

dimethylbenzene.

FURTHER COMMENTS o The overall alkylation reaction is the substitution of -H by -CR3  o Bromoalkanes can also be used for alkylation, but more expensive. Similar

catalysts can be used e.g. anhydrous AlBr3 or FeBr3.

 

 

The electrophilic substitution of an arene - acylation (Friedel-Crafts reaction)

for R = H, benzene: C6H6 + R'COCl ==> C6H5COR' + HCl   [see mechanism 25 below]

Benzene is refluxed with an acid chloride and anhydrous aluminium chloride catalyst and a ketone is formed.

mechanism 25 - electrophilic substitution by an acyl group in the benzene ring

[mechanism 25 above] If ethanoyl chloride, CH3COCl, was used (R=CH3-), benzene forms phenylethanone, C6H5-CO-CH3.

o Step (1) Although the acid chloride molecule is polar, it is still not a strong enough electrophile to disrupt the ip electron system of the benzene ring. The aluminium chloride reacts with an acid chloride molecule to form a carbocation (acylonium ion, RCO+) which is a much stronger electron pair accepting electrophile than the original acid chloride (either this or an AlCl3-RCOCl complex - details not needed for A level).

o Step (2) An electron pair from the delocalised ip electrons of the benzene ring forms a C-C bond with the electron pair accepting carbocation forming a second highly unstable carbocation. It is very unstable because the stable electron arrangement of the benzene ring is partially broken to give a 'saturated' C (top right of ring.

o Step (3) is a proton transfer, as the tetrachloroaluminate(III) ion [formed in step (1)], abstracts a proton from the second highly unstable intermediate carbocation to give the ketone product, hydrogen chloride gas and reforming the aluminium chloride catalyst.

for R = CH3, benzene: C6H5CH3 + R'COCl ==> R'COC6H4CH3 + HCl o and again there is the potential to form three position isomers by substituting

in the 2, 3 or 4 position on the ring. FURTHER COMMENTS

o The overall acylation reaction is the substitution of -H by RCO

 

 

The electrophilic substitution of an arene - sulphonation

for R = H, benzene: C6H6 + SO3 ==> C6H5SO2OH  [see mechanism 44 below] or C6H6 + H2SO4 ==> C6H5SO2OH + H2O Benzene is heated with concentrated sulphuric acid or even better, 'fuming' sulphuric

acid, which has a higher sulphur trioxide content and more efficient at introducing the sulphonic acid group into the benzene ring.

mechanism 25 - electrophilic substitution by an acyl group in the benzene ring

[mechanism 44 above]   o Step (1) Sulphur trioxide is formed (or already present). It is a powerful

electrophile, i.e. electron pair acceptor because of the effect of the three very electronegative oxygen atoms bonded to the central sulphur atom.

o Step (2) An electron pair from the delocalised ip electrons of the benzene ring forms a C-S bond with the electron pair accepting sulphur trioxide forming a second highly unstable carbocation. It is very unstable because the stable electron arrangement of the benzene ring is partially broken to give a 'saturated' C (top right of ring).

o Step (3) A hydrogensulphate ion removes a proton and the complete benzene ring is reformed giving the anion of the aromatic sulphonic acid.

o Step (4) is a proton transfer to give the sulphonic acid. for R = CH3, benzene: C6H5CH3 + H2SO4 ==> CH3C6H4SO2OH + H2O

o and again there is the potential to form three position isomers by substituting in the 2, 3 or 4 position on the ring, the mechanism diagram shows the formation of methyl-2-benzenesulphonic acid.

FURTHER COMMENTS o The overall sulfonation reaction is the substitution of -H by -SO2OH

 

 

The orientation of products in aromatic electrophilic substitution reactions

Certain groups, already present, can increase the electron density of the benzene ring and make the aromatic compound more reactive towards electrophiles such as those described above. However the effect seems to enhance the reactivity at the 2 and 4 substitution positions more than the 3 substitution position.

o Groups that increase reactivity are e.g. -CH3, -Cl, -OH, -NH2, -NHCOCH3, and favour substitution at the 2 and 4 positions (typically 90-100% combined).

o They all, by some means, have a small, but significant, electron donating (+I inductive effect) on the ring of ip electrons.

o For example, methyl benzene is significantly more reactive than benzene and when nitrated, over 90% of the products are either methyl-2-nitrobenzene or methyl-4-nitrobenzene.

Certain groups, already present, can decrease the electron density of the benzene ring and make the aromatic compound less reactive towards electrophiles such as described above. However the effect seems to decrease the reactivity at the 2 and 4 substitution positions more than the 3 substitution position.

o Groups that decrease reactivity, by some means, are e.g. -NO2, COOH, -CHO, -SO2OH, and favour substitution at the 3 position (typically 70-90%) and their effect does fit in with them all being strongly electronegative groupings giving a -I inductive effect.

o For example, nitrobenzene is much less reactive than benzene and on nitration, 93% of the product is 1,3-dinitrobenzene.