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INTRODUCING ALCOHOLS

This page explains what alcohols are, and what the difference is between primary, secondary and tertiary alcohols. It looks in some detail at their simple physical properties such as solubility and boiling points. Details of the chemical reactions of alcohols are described on separate pages.

What are alcohols?

Examples

Alcohols are compounds in which one or more hydrogen atoms in an alkane have been replaced by an -OH group. For the purposes of UK A level, we will only look at compounds containing one -OH group.

For example:

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: If you aren't confident about naming organic compounds, then you really ought to follow this link before you go on.

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The different kinds of alcohols

Alcohols fall into different classes depending on how the -OH group is positioned on the chain of carbon atoms. There are some chemical differences between the various types.

Primary alcohols

In a primary (1°) alcohol, the carbon which carries the -OH group is only attached to one alkyl group.

Note: An alkyl group is a group such as methyl, CH₃, or ethyl, CH₃CH₂. These are groups containing chains of carbon atoms which may be branched. Alkyl groups are given the general symbol R.

Some examples of primary alcohols include:

Notice that it doesn't matter how complicated the attached alkyl group is. In each case there is only one linkage to an alkyl group from the CH₂ group holding the -OH group.

There is an exception to this. Methanol, CH₃OH, is counted as a primary alcohol even though there are no alkyl groups attached to the carbon with the -OH group on it.

Secondary alcohols

In a secondary (2°) alcohol, the carbon with the -OH group attached is joined directly to two alkyl groups, which may be the same or different.

Examples:

Tertiary alcohols

In a tertiary (3°) alcohol, the carbon atom holding the -OH group is attached directly to three alkyl groups, which may be any combination of same or different.

Examples:

Physical properties of alcohols

Boiling Points

The chart shows the boiling points of some simple primary alcohols with up to 4 carbon atoms.

They are:

They are compared with the equivalent alkane (methane to butane) with the same number of carbon atoms.

Notice that:

· The boiling point of an alcohol is always much higher than that of the alkane with the same number of carbon atoms.

· The boiling points of the alcohols increase as the number of carbon atoms increases.

The patterns in boiling point reflect the patterns in intermolecular attractions.

Note: If you aren't happy about intermolecular forces (including van der Waals dispersion forces and hydrogen bonds) then you really ought to follow this link before you go on. The next bit won't make much sense to you if you aren't familiar with the various sorts of intermolecular forces.

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Hydrogen bonding

Hydrogen bonding occurs between molecules where you have a hydrogen atom attached to one of the very electronegative elements - fluorine, oxygen or nitrogen.

In the case of alcohols, there are hydrogen bonds set up between the slightly positive hydrogen atoms and lone pairs on oxygens in other molecules.

The hydrogen atoms are slightly positive because the bonding electrons are pulled away from them towards the very electronegative oxygen atoms.

Note: If you want to be fussy, the diagram is slightly misleading in that it suggests that all of the lone pairs on the oxygen atoms are forming hydrogen bonds. In an alcohol that can't happen. Taking the alcohol as a whole, there are only half as many slightly positive hydrogen atoms as there are lone pairs. At any one time, half of the lone pairs in the total liquid alcohol won't have hydrogen bonds from them because there aren't enough slightly positive hydrogens to go around.

In the diagram, to show the 3-dimensional arrangement, the wedge-shaped lines show bonds coming out of the screen or paper towards you. The dotted bonds (other than the hydrogen bonds) show bonds going back into the screen or paper away from you.

In alkanes, the only intermolecular forces are van der Waals dispersion forces. Hydrogen bonds are much stronger than these and therefore it takes more energy to separate alcohol molecules than it does to separate alkane molecules.

That's the main reason that the boiling points are higher.

The effect of van der Waals forces . . .

. . . on the boiling points of the alcohols:

Hydrogen bonding isn't the only intermolecular force in alcohols. There are also van der Waals dispersion forces and dipole-dipole interactions.

The hydrogen bonding and the dipole-dipole interactions will be much the same for all the alcohols, but the dispersion forces will increase as the alcohols get bigger.

These attractions get stronger as the molecules get longer and have more electrons. That increases the sizes of the temporary dipoles that are set up.

This is why the boiling points increase as the number of carbon atoms in the chains increases. It takes more energy to overcome the dispersion forces, and so the boiling points rise.

. . . on the comparison between alkanes and alcohols:

Even if there wasn't any hydrogen bonding or dipole-dipole interactions, the boiling point of the alcohol would be higher than the corresponding alkane with the same number of carbon atoms.

Compare ethane and ethanol:

Ethanol is a longer molecule, and the oxygen brings with it an extra 8 electrons. Both of these will increase the size of the van der Waals dispersion forces and so the boiling point.

If you were doing a really fair comparison to show the effect of the hydrogen bonding on boiling point it would be better to compare ethanol with propane rather than ethane. The length would then be much the same, and the number of electrons is exactly the same.

Solubility of alcohols in water

The small alcohols are completely soluble in water. Whatever proportions you mix them in, you will get a single solution.

However, solubility falls as the length of the hydrocarbon chain in the alcohol increases. Once you get to four carbons and beyond, the fall in solubility is noticeable, and you may well end up with two layers in your test tube.

The solubility of the small alcohols in water

Consider ethanol as a typical small alcohol. In both pure water and pure ethanol the main intermolecular attractions are hydrogen bonds.

In order to mix the two, you would have to break the hydrogen bonds between the water molecules and the hydrogen bonds between the ethanol molecules. It needs energy to do both of these things.

However, when the molecules are mixed, new hydrogen bonds are made between water molecules and ethanol molecules.

The energy released when these new hydrogen bonds are made more or less compensates for that needed to break the original ones.

In addition, there is an increase in the disorder of the system - an increase in entropy. That is another factor in deciding whether things happen or not.

Note: If you haven't come across entropy before, don't worry about it. I mention it because the energy released when the new bonds are made isn't quite enough to compensate for breaking the old ones, meaning that the mixing process is endothermic. If it weren't for the increase in entropy, the solution wouldn't be formed.

To really understand this, you need to have studied entropy and free energy. If you should know about this, but aren't happy about the calculations involved, you might like to have a look at chapter 11 of my chemistry calculations book.

The lower solubility of bigger alcohols

Imagine what happens when you have got, say, 5 carbon atoms in each alcohol molecule.

The hydrocarbon chains are forcing their way between water molecules and so breaking hydrogen bonds between those water molecules.

The -OH end of the alcohol molecules can form new hydrogen bonds with water molecules, but the hydrocarbon "tail" doesn't form hydrogen bonds

That means that quite a lot of the original hydrogen bonds being broken aren't replaced by new ones.

All you get in place of those original hydrogen bonds are van der Waals dispersion forces between the water and the hydrocarbon "tails". These attractions are much weaker. That means that you don't get enough energy back to compensate for the hydrogen bonds being broken. Even allowing for the increase in disorder, the process becomes less feasible.

As the length of the alcohol increases, this situation just gets worse, and so the solubility falls.

THE MANUFACTURE OF ALCOHOLS

This page looks at the manufacture of alcohols by the direct hydration of alkenes, concentrating mainly on the hydration of ethene to make ethanol. It then compares that method with making ethanol by fermentation.

Manufacturing alcohols from alkenes

The manufacture of ethanol from ethene

Ethanol is manufactured by reacting ethene with steam. The catalyst used is solid silicon dioxide coated with phosphoric(V) acid. 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.

If you are interested in the reasons for the conditions used in this reaction, you will find them in the equilibrium section of this site by following this link.

If you are interested in the mechanism for the hydration of ethene, follow this link to the catalysis section.

Because these pages are (somewhat illogically!) in different parts of the site, use the BACK button (or HISTORY file or GO menu) on your browser to return to this page later.

The manufacture of other alcohols from alkenes

Some - but not all - other alcohols can be made by similar reactions. The catalyst used and the reaction conditions will vary from alcohol to alcohol. The only set of conditions you are going to need for UK A level purposes are those given above for manufacturing ethanol.

The reason that there is a problem with some alcohols is well illustrated with trying to make an alcohol from propene, CH₃CH=CH₂.

In principle, there are two different alcohols which might be formed:

You might expect to get either propan-1-ol or propan-2-ol depending on which way around the water adds to the double bond. In practice what you get is propan-2-ol.

If you add a molecule H-X across a carbon-carbon double bond, the hydrogen nearly always gets attached to the carbon with the most hydrogens on it already - in this case the CH₂ rather than the CH.

Note: The reason for this is dealt with in detail in the mechanism section of this site on a page about addition to unsymmetrical alkenes.

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The effect of this is that there are bound to be some alcohols which it is impossible to make by reacting alkenes with steam because the addition would be the wrong way around.

Making ethanol by fermentation

This method only applies to ethanol. You can't make any other alcohol this way.

The process

The starting material for the process varies widely, but will normally be some form of starchy plant material such as maize (US: corn), wheat, barley or potatoes.

Starch is a complex carbohydrate, and other carbohydrates can also be used - for example, in the lab sucrose (sugar) is normally used to produce ethanol. Industrially, this wouldn't make sense. It would be silly to refine sugar if all you were going to use it for was fermentation. There is no reason why you shouldn't start from the original sugar cane, though.

The first step is to break complex carbohydrates into simpler ones.

For example, if you were starting from starch in grains like wheat or barley, the grain is heated with hot water to extract the starch and then warmed with malt. Malt is germinated barley which contains enzymes which break the starch into a simpler carbohydrate called maltose, C₁₂H₂₂O₁₁.

Maltose has the same molecular formula as sucrose but contains two glucose units joined together, whereas sucrose contains one glucose and one fructose unit.

Yeast is then added and the mixture is kept warm (say 35°C) for perhaps several days until fermentation is complete. Air is kept out of the mixture to prevent oxidation of the ethanol produced to ethanoic acid (vinegar).

Enzymes in the yeast first convert carbohydrates like maltose or sucrose into even simpler ones like glucose and fructose, both C₆H₁₂O₆, and then convert these in turn into ethanol and carbon dioxide.

You can show these changes as simple chemical equations, but the biochemistry of the reactions is much, much more complicated than this suggests.

Yeast is killed by ethanol concentrations in excess of about 15%, and that limits the purity of the ethanol that can be produced. The ethanol is separated from the mixture by fractional distillation to give 96% pure ethanol.

For theoretical reasons, it is impossible to remove the last 4% of water by fractional distillation.

Note: If you are interested in this, you will have to read about it in a physical chemistry textbook under "azeotropic mixtures". Currently I haven't dealt with these on this site, and am unlikely to do so until at least 2005. I will add a link if or when I get around to it!

A comparison of fermentation with the direct hydration of ethene

THE DEHYDRATION OF ALCOHOLS

This page (a simple duplicate of a page in the section on alkenes!) looks at the dehydration of alcohols in the lab to make alkenes - for example, dehydrating ethanol to make ethene.

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.

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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.

REACTING ALCOHOLS WITH SODIUM

This page describes the reaction between alcohols and metallic sodium, and takes a very brief look at the properties of the alkoxide which is formed. We will look at the reaction between sodium and ethanol as being typical, but you could substitute any other alcohol you wanted to - the reaction would be the same.

The reaction between sodium and ethanol

Details of the reaction

If a small piece of sodium is dropped into some ethanol, it reacts steadily to give off bubbles of hydrogen gas and leaves a colourless solution of sodium ethoxide, CH3CH2ONa. Sodium ethoxide is known as an alkoxide.

If the solution is evaporated carefully to dryness, the sodium ethoxide is left as a white solid.

Although at first sight you might think this was something new and complicated, in fact it is exactly the same (apart from being a more gentle reaction) as the reaction between sodium and water - something you have probably known about for years.

Compare the two:

We normally, of course, write the sodium hydroxide formed as NaOH rather than HONa - but that's the only difference.

Sodium ethoxide is just like sodium hydroxide, except that the hydrogen has been replaced by an ethyl group. Sodium hydroxide contains OH- ions; sodium ethoxide contains CH3CH2O- ions.

Note: The reason that the ethoxide formula is written with the oxygen on the right unlike the hydroxide ion is simply a matter of clarity. If you write it the other way around, it doesn't immediately look as if it comes from ethanol. You will find the same thing happens when you write formulae for organic salts like sodium ethanoate, for example.

Using the reaction

There are two simple uses for this reaction:

To dispose of small amounts of sodium safely

If you spill some sodium on the bench, or have a small amount left over from a reaction, you can't just chuck it in the sink. It tends to react explosively with the water - and comes flying back out at you again!

It reacts much more gently with ethanol. Ethanol is therefore used to dissolve small quantities of waste sodium. The solution formed can be washed away without problems (provided you remember that sodium ethoxide is strongly alkaline - see below).

To test for the -OH group in alcohols

Because of the dangers involved in handling sodium, this is not the best test for an alcohol at this level. Because sodium reacts violently with acids to produce a salt and hydrogen, you would first have to be sure that the liquid you were testing was neutral.

You would also have to be confident that there was no trace of water present because sodium reacts with the -OH group in water even better than with the one in an alcohol.

With those provisos, if you add a tiny piece of sodium to a neutral liquid free of water and get bubbles of hydrogen produced, then the liquid is an alcohol.

Some simple reactions of alkoxide ions

This is going beyond the demands of UK A level, but you might come across the first example as a part of a bit of practical work. The second example is to reinforce the similarity between sodium ethoxide and sodium hydroxide.

Once again we will take the ethoxide ions in sodium ethoxide as typical. Essentially, ethoxide (and other alkoxide) ions behave like hydroxide ions.

Ethoxide ions are strongly basic

If you add water to sodium ethoxide, it dissolves to give a colourless solution with a high pH - typically pH 14. The solution is strongly alkaline.

The reason is that the ethoxide ions remove hydrogen ions from water molecules to produce hydroxide ions. It is these which produce the high pH.

Ethoxide ions are good nucleophiles

A nucleophile is something which carries a negative or partial negative charge which it uses to attack positive centres in other molecules or ions.

Hydroxide ions are good nucleophiles, and you may have come across the reaction between a halogenoalkane (also called a haloalkane or alkyl halide) and sodium hydroxide solution. The hydroxide ions replace the halogen atom.

In this case, an alcohol is formed.

Note: You will find the mechanism for this reaction in the mechanism section of this site.

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The ethoxide ion behaves in exactly the same way. If you knew the mechanism for the hydroxide ion reaction you could work out exactly what happens in the reaction between a halogenoalkane and ethoxide ion.

Compare this equation with the last one.

The only difference is that where there was a hydrogen atom at the right-hand end of the product molecule, you now have an alkyl group.

Two alkyl (or other hydrocarbon) groups bridged by an oxygen atom is called an ether. This particular one is 1-ethoxypropane or ethyl propyl ether. This reaction is a good way of making ethers in the lab.

Note: There are all sorts of ways of naming ethers. For UK A level purposes, the problem doesn't arise - you almost certainly won't have to name them.

If you have looked at the chemistry of halogenoalkanes, you may be aware that there is a competition between substitution and elimination when they react with hydroxide ions. Exactly the same competition occurs in their reactions with ethoxide ions. Just as the hydroxide ion can act as either a base or a nucleophile, exactly the same is true of alkoxide ions like the ethoxide ion.

You could read about the reactions of halogenoalkanes with hydroxide ions and work out for yourself what is going to happen in the possible elimination reaction if you used sodium ethoxide rather than sodium hydroxide. Because this is getting well beyond UK A level, I haven't given any detail for this anywhere on the site. The whole point about understanding chemistry (and especially mechanisms!) is that you can work things out for yourself when you need to!

REPLACING THE -OH GROUP IN ALCOHOLS BY A HALOGEN

This page looks at reactions in which the -OH group in an alcohol is replaced by a halogen such as chlorine or bromine. It includes a simple test for an -OH group using phosphorus(V) chloride.

Reactions involving hydrogen halides

The general reaction looks like this:

Reaction with hydrogen chloride

Tertiary alcohols react reasonably rapidly with concentrated hydrochloric acid, but for primary or secondary alcohols the reaction rates are too slow for the reaction to be of much importance.

A tertiary alcohol reacts if it is shaken with with concentrated hydrochloric acid at room temperature. A tertiary halogenoalkane (haloalkane or alkyl halide) is formed

Note: If you don't know what primary, secondary and tertiary alcohols are, you should read the introduction to alcohols before you go on. The terms primary, secondary and tertiary are used in exactly the same way with halogenoalkanes.

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Replacing -OH by bromine

Rather than using hydrobromic acid, you usually treat the alcohol with a mixture of sodium or potassium bromide and concentrated sulphuric acid. This produces hydrogen bromide which reacts with the alcohol. The mixture is warmed to distil off the bromoalkane.

Note: You will find practical details of this reaction on the page about preparation of halogenoalkanes. You don't need to read the beginning of that page because it is just a modification of this one!

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Replacing -OH by iodine

In this case the alcohol is reacted with a mixture of sodium or potassium iodide and concentrated phosphoric(V) acid, H₃PO₄, and the iodoalkane is distilled off. The mixture of the iodide and phosphoric(V) acid produces hydrogen iodide which reacts with the alcohol.

Phosphoric(V) acid is used instead of concentrated sulphuric acid because sulphuric acid oxidises iodide ions to iodine and produces hardly any hydrogen iodide.

A similar thing happens to some extent with bromide ions in the preparation of bromoalkanes, but not enough to get in the way of the main reaction. There is no reason why you couldn't use phosphoric(V) acid in the bromide case instead of sulphuric acid if you wanted to.

Note: If you are interested in the reactions between halide ions and concentrated sulphuric acid you could follow this link. In the present context, all you would need to do is read the beginning of that page.

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Reacting alcohols with phosphorus halides

Reaction with phosphorus(III) chloride, PCl₃

Alcohols react with liquid phosphorus(III) chloride (also called phosphorus trichloride) to make chloroalkanes.

Reaction with phosphorus(V) chloride, PCl₅

Solid phosphorus(V) chloride (phosphorus pentachloride) reacts violently with alcohols at room temperature, producing clouds of hydrogen chloride gas. It isn't a good choice as a way of making chloroalkanes, although it is used as a test for -OH groups in organic chemistry.

To show that a substance was an alcohol, you would first have to eliminate all the other things which also react with phosphorus(V) chloride. For example, carboxylic acids (containing the -COOH group) react with it (because of the -OH in -COOH), and so does water (H-OH).

If you have a neutral liquid not contaminated with water, and get clouds of hydrogen chloride when you add phosphorus(V) chloride, then you have an alcohol group present.

There are also side reactions involving the POCl₃ reacting with the alcohol.

Other reactions involving phosphorus halides

Instead of using phosphorus(III) bromide or iodide, the alcohol is usually heated under reflux with a mixture of red phosphorus and either bromine or iodine.

The phosphorus first reacts with the bromine or iodine to give the phosphorus(III) halide.

These then react with the alcohol to give the corresponding halogenoalkane which can be distilled off.

Reacting alcohols with sulphur dichloride oxide (thionyl chloride)

The reaction

Sulphur dichloride oxide (thionyl chloride) has the formula SOCl₂.

Traditionally, the formula is written as shown, despite the fact that the modern name writes the chlorine before the oxygen (alphabetical order).

The sulphur dichloride oxide reacts with alcohols at room temperature to produce a chloroalkane. Sulphur dioxide and hydrogen chloride are given off. Care would have to be taken because both of these are poisonous.

Why this reaction is useful

The big advantage that this reaction has over the use of either of the phosphorus chlorides is that the two other products of the reaction (sulphur dioxide and HCl) are both gases. That means that they separate themselves from the reaction mixture.

OXIDATION OF ALCOHOLS

This page looks at the oxidation of alcohols using acidified sodium or potassium dichromate(VI) solution. This reaction is used to make aldehydes, ketones and carboxylic acids, and as a way of distinguishing between primary, secondary and tertiary alcohols.

Important! It is pointless reading this page unless you are confident you know what primary, secondary and tertiary alcohols are. If you aren't sure, you must read the introduction to alcohols before you go on.

This page will also refer constantly to aldehydes and ketones. Follow this link if you haven't come across these compounds before.

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Oxidising the different types of alcohols

The oxidising agent used in these reactions is normally a solution of sodium or potassium dichromate(VI) acidified with dilute sulphuric acid. If oxidation occurs, the orange solution containing the dichromate(VI) ions is reduced to a green solution containing chromium(III) ions.

The electron-half-equation for this reaction is

Note: If you don't yet know about electron-half-equations just ignore this reference for now. If you should already know about them, but aren't very good at handling them, you might like to have a look at this link. It isn't particularly important for the purposes of the current page.

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Primary alcohols

Primary alcohols can be oxidised to either aldehydes or carboxylic acids depending on the reaction conditions. In the case of the formation of carboxylic acids, the alcohol is first oxidised to an aldehyde which is then oxidised further to the acid.

Partial oxidation to aldehydes

You get an aldehyde if you use an excess of the alcohol, and distil off the aldehyde as soon as it forms.

The excess of the alcohol means that there isn't enough oxidising agent present to carry out the second stage. Removing the aldehyde as soon as it is formed means that it doesn't hang around waiting to be oxidised anyway!

If you used ethanol as a typical primary alcohol, you would produce the aldehyde ethanal, CH₃CHO.

The full equation for this reaction is fairly complicated, and you need to understand about electron-half-equations in order to work it out.

In organic chemistry, simplified versions are often used which concentrate on what is happening to the organic substances. To do that, oxygen from an oxidising agent is represented as [O]. That would produce the much simpler equation:

It also helps in remembering what happens. You can draw simple structures to show the relationship between the primary alcohol and the aldehyde formed.

Important! This is not intended to suggest any sort of mechanism for the reaction - it is just a way of helping you to remember what happens.

If you are in the UK A level system (or its equivalent), it is highly likely that your examiners will accept equations involving [O]. To be sure, consult your syllabus, past papers and mark schemes. If you are studying a UK-based syllabus and haven't got any of these things, follow this link to find out how to get them.

Full oxidation to carboxylic acids

You need to use an excess of the oxidising agent and make sure that the aldehyde formed as the half-way product stays in the mixture.

The alcohol is heated under reflux with an excess of the oxidising agent. When the reaction is complete, the carboxylic acid is distilled off.

The full equation for the oxidation of ethanol to ethanoic acid is:

Note: This equation is worked out in detail on the page about electron-half-equations mentioned above, if you are interested.

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The more usual simplified version looks like this:

Alternatively, you could write separate equations for the two stages of the reaction - the formation of ethanal and then its subsequent oxidation.

This is what is happening in the second stage:

Secondary alcohols

Secondary alcohols are oxidised to ketones - and that's it. For example, if you heat the secondary alcohol propan-2-ol with sodium or potassium dichromate(VI) solution acidified with dilute sulphuric acid, you get propanone formed.

Playing around with the reaction conditions makes no difference whatsoever to the product.

Using the simple version of the equation and showing the relationship between the structures:

If you look back at the second stage of the primary alcohol reaction, you will see that an oxygen "slotted in" between the carbon and the hydrogen in the aldehyde group to produce the carboxylic acid. In this case, there is no such hydrogen - and the reaction has nowhere further to go.

Tertiary alcohols

Tertiary alcohols aren't oxidised by acidified sodium or potassium dichromate(VI) solution. There is no reaction whatsoever.

If you look at what is happening with primary and secondary alcohols, you will see that the oxidising agent is removing the hydrogen from the -OH group, and a hydrogen from the carbon atom attached to the -OH. Tertiary alcohols don't have a hydrogen atom attached to that carbon.

You need to be able to remove those two particular hydrogen atoms in order to set up the carbon-oxygen double bond.

Using these reactions as a test for the different types of alcohol

Doing the test

First you have to be sure that you have actually got an alcohol by testing for the -OH group. You would need to show that it was a neutral liquid, free of water and that it reacted with solid phosphorus(V) chloride to produce a burst of acidic steamy hydrogen chloride fumes.

Note: You will find the chemistry of the phosphorus(V) chloride test by following this link.

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You would then add a few drops of the alcohol to a test tube containing potassium dichromate(VI) solution acidified with dilute sulphuric acid. The tube would be warmed in a hot water bath.

Results for the various kinds of alcohol

Picking out the tertiary alcohol

In the case of a primary or secondary alcohol, the orange solution turns green. With a tertiary alcohol there is no colour change.

After heating:

Distinguishing between the primary and secondary alcohols

You need to produce enough of the aldehyde (from oxidation of a primary alcohol) or ketone (from a secondary alcohol) to be able to test them. There are various things which aldehydes do which ketones don't. These include the reactions with Tollens' reagent, Fehling's solution and Benedict's solution, and are covered on a separate page.

Note: You will find these tests for aldehydes by following this link.

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In my experience, these tests can be a bit of a bother to carry out and the results aren't always as clear-cut as the books say. A much simpler but fairly reliable test is to use Schiff's reagent. Schiff's reagent isn't specifically mentioned by any of the UK-based syllabuses, but I have always used it.

Schiff's reagent is a fuchsin dye decolourised by passing sulphur dioxide through it. In the presence of even small amounts of an aldehyde, it turns bright magenta.

It must, however, be used absolutely cold, because ketones react with it very slowly to give the same colour. If you heat it, obviously the change is faster - and potentially confusing.

While you are warming the reaction mixture in the hot water bath, you can pass any vapours produced through some Schiff's reagent.

· If the Schiff's reagent quickly becomes magenta, then you are producing an aldehyde from a primary alcohol.

· If there is no colour change in the Schiff's reagent, or only a trace of pink colour within a minute or so, then you aren't producing an aldehyde, and so haven't got a primary alcohol.

Because of the colour change to the acidified potassium dichromate(VI) solution, you must therefore have a secondary alcohol.

You should check the result as soon as the potassium dichromate(VI) solution turns green - if you leave it too long, the Schiff's reagent might start to change colour in the secondary alcohol case as well.

ESTERIFICATION

This page looks at esterification - mainly the reaction between alcohols and carboxylic acids to make esters. It also looks briefly at making esters from the reactions between acyl chlorides (acid chlorides) and alcohols, and between acid anhydrides and alcohols.

What are esters?

Esters are derived from carboxylic acids. A carboxylic acid contains the -COOH group, and in an ester the hydrogen in this group is replaced by a hydrocarbon group of some kind. We shall just be looking at cases where it is replaced by an alkyl group, but it could equally well be an aryl group (one based on a benzene ring).

A common ester - ethyl ethanoate

The most commonly discussed ester is ethyl ethanoate. In this case, the hydrogen in the -COOH group has been replaced by an ethyl group. The formula for ethyl ethanoate is:

Notice that the ester is named the opposite way around from the way the formula is written. The "ethanoate" bit comes from ethanoic acid. The "ethyl" bit comes from the ethyl group on the end.

Note: In my experience, students starting organic chemistry get more confused about writing names and formulae for esters than for almost anything else - particularly when it comes to less frequently met esters like the ones coming up next. Take time and care to make sure you understand!

A few more esters

In each case, be sure that you can see how the names and formulae relate to each other.

Notice that the acid is named by counting up the total number of carbon atoms in the chain - including the one in the -COOH group. So, for example, CH₃CH₂COOH is propanoic acid, and CH₃CH₂COO is the propanoate group.

Note: You can find more about naming acids and esters by following this link to a different part of this site.

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Making esters from carboxylic acids and alcohols

The chemistry of the reaction

Esters are produced when carboxylic acids are heated with alcohols in the presence of an acid catalyst. The catalyst is usually concentrated sulphuric acid. Dry hydrogen chloride gas is used in some cases, but these tend to involve aromatic esters (ones containing a benzene ring). If you are a UK A level student you won't have to worry about these.

The esterification reaction is both slow and reversible. The equation for the reaction between an acid RCOOH and an alcohol R'OH (where R and R' can be the same or different) is:

So, for example, if you were making ethyl ethanoate from ethanoic acid and ethanol, the equation would be:

Note: The mechanism for the esterification reaction is covered in the catalysis section of this site. It is not required for any UK A level (or equivalent) chemistry syllabus.

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Doing the reactions

On a test tube scale

Carboxylic acids and alcohols are often warmed together in the presence of a few drops of concentrated sulphuric acid in order to observe the smell of the esters formed.

You would normally use small quantities of everything heated in a test tube stood in a hot water bath for a couple of minutes.

Because the reactions are slow and reversible, you don't get a lot of ester produced in this time. The smell is often masked or distorted by the smell of the carboxylic acid. A simple way of detecting the smell of the ester is to pour the mixture into some water in a small beaker.

Apart from the very small ones, esters are fairly insoluble in water and tend to form a thin layer on the surface. Excess acid and alcohol both dissolve and are tucked safely away under the ester layer.

Small esters like ethyl ethanoate smell like typical organic solvents (ethyl ethanoate is a common solvent in, for example, glues).

As the esters get bigger, the smells tend towards artificial fruit flavouring - "pear drops", for example.

On a larger scale

If you want to make a reasonably large sample of an ester, the method used depends to some extent on the size of the ester. Small esters are formed faster than bigger ones.

To make a small ester like ethyl ethanoate, you can gently heat a mixture of ethanoic acid and ethanol in the presence of concentrated sulphuric acid, and distil off the ester as soon as it is formed.

This prevents the reverse reaction happening. It works well because the ester has the lowest boiling point of anything present. The ester is the only thing in the mixture which doesn't form hydrogen bonds, and so it has the weakest intermolecular forces.

Note: Follow this link if you aren't sure about hydrogen bonding.

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Larger esters tend to form more slowly. In these cases, it may be necessary to heat the reaction mixture under reflux for some time to produce an equilibrium mixture. The ester can be separated from the carboxylic acid, alcohol, water and sulphuric acid in the mixture by fractional distillation.

Note: Providing full details for organic preparations (including all the steps necessary in cleaning up the product) is beyond the scope of this site. If you need this sort of detail, you should be looking at an organic practical book.

Other ways of making esters

Esters can also be made from the reactions between alcohols and either acyl chlorides or acid anhydrides.

Making esters from alcohols and acyl chlorides (acid chlorides)

If you add an acyl chloride to an alcohol, you get a vigorous (even violent) reaction at room temperature producing an ester and clouds of steamy acidic fumes of hydrogen chloride.

For example, if you add the liquid ethanoyl chloride to ethanol, you get a burst of hydrogen chloride produced together with the liquid ester ethyl ethanoate.

Note: If you want to find out more about acyl chlorides, explore the acyl chlorides menu by following this link.

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Making esters from alcohols and acid anhydrides

The reactions of acid anhydrides are slower than the corresponding reactions with acyl chlorides, and you usually need to warm the mixture.

Taking ethanol reacting with ethanoic anhydride as a typical reaction involving an alcohol:

There is a slow reaction at room temperature (or faster on warming). There is no visible change in the colourless liquids, but a mixture of ethyl ethanoate and ethanoic acid is formed.

Note: If you want to find out more about acid anhydrides, explore the acid anhydrides menu by following this link.

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THE TRIIODOMETHANE (IODOFORM) REACTION WITH ALCOHOLS

This page looks at how the triiodomethane (iodoform) reaction can be used to identify the presence of a CH3CH(OH) group in alcohols.

Note: This reaction can also be used in testing for the CH₃CO group in aldehydes and ketones. You will find a link to this at the bottom of the page.

Doing the triiodomethane (iodoform) reaction

There are two apparently quite different mixtures of reagents that can be used to do this reaction. They are, in fact, chemically equivalent.

Note: It would be silly to learn both of these methods. Use whichever one your examiners want - find out by looking at past papers and mark schemes. If you haven't got these, go to the syllabuses page to find out how to get hold of them.

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Using iodine and sodium hydroxide solution

This is chemically the more obvious method.

Iodine solution is added to a small amount of an alcohol, followed by just enough sodium hydroxide solution to remove the colour of the iodine. If nothing happens in the cold, it may be necessary to warm the mixture very gently.

A positive result is the appearance of a very pale yellow precipitate of triiodomethane (previously known as iodoform) - CHI₃.

Apart from its colour, this can be recognised by its faintly "medical" smell. It is used as an antiseptic on the sort of sticky plasters you put on minor cuts, for example.

Using potassium iodide and sodium chlorate(I) solutions

Sodium chlorate(I) is also known as sodium hypochlorite.

Potassium iodide solution is added to a small amount of an alcohol, followed by sodium chlorate(I) solution. Again, if no precipitate is formed in the cold, it may be necessary to warm the mixture very gently.

The positive result is the same pale yellow precipitate as before.

Why the two reactions are equivalent: This reaction happens in three stages. In the first, the alcohol is oxidised to an aldehyde or ketone. In the first mixture, the iodine reacts with the sodium hydroxide solution to produce some sodium iodate(I). This is an oxidising agent.

In the second mixture, the sodium chlorate(I) already present is an oxidising agent.

After that the reaction happens in two further stages: first the aldehyde or ketone formed reacts with iodine, and the product of that reaction reacts with hydroxide ions. Iodine and sodium hydroxide is exactly what you are adding in the first method above.

In the second method, the sodium chlorate(I) solution is an oxidising agent, and oxidises the iodide ions in the potassium iodide to iodine. As well as any possible precipitate, you will also see the typical reddish-brown colour of iodine solution being formed during the reaction.

So although you didn't put any iodine into the mixture, it is made in situ. What about the hydroxide ions?

Sodium chlorate(I) solution is alkaline and contains enough hydroxide ions to carry out the second half of the reaction. Sodium chlorate(I) is alkaline because it reacts reversibly with water to form the weak acid chloric(I) acid together with hydroxide ions.

The chemistry of the triiodomethane (iodoform) reaction

What the triiodomethane (iodoform) reaction shows

A positive result - the pale yellow precipitate of triiodomethane (iodoform) - is given by an alcohol containing the grouping:

"R" can be a hydrogen atom or a hydrocarbon group (for example, an alkyl group).

If "R" is hydrogen, then you have the primary alcohol ethanol, CH₃CH₂OH.

· Ethanol is the only primary alcohol to give the triiodomethane (iodoform) reaction.

· If "R"is a hydrocarbon group, then you have a secondary alcohol. Lots of secondary alcohols give this reaction, but those that do all have a methyl group attached to the carbon with the -OH group.

· No tertiary alcohols can contain this group because no tertiary alcohols can have a hydrogen atom attached to the carbon with the -OH group. No tertiary alcohols give the triiodomethane (iodoform) reaction.

Summary of the reactions during the triiodomethane (iodoform) reaction

We will take the reagents as being iodine and sodium hydroxide solution.

This is being given as a flow scheme rather than full equations. You aren't likely to need the equation for the oxidation stage for UK A level purposes. The equations for the other two steps are given on a page about reactions of aldehydes and ketones. Follow the first link below if you are interested.