Hydroxy compounds

Classifying alcohols as primary, secondary and tertiary

An alcohol is a compound in which an -OH group is attached to a saturated carbon atom, that is, a carbon atom joined to other atoms by single bonds only. The -OH group is called the hydroxy group. Alcohols are classified by how many carbon atoms are attached to the carbon atom that carries the -OH group, not by how many carbon atoms the molecule contains in total. Primary (1 degree): that carbon is attached to one other carbon atom, or to none at all, as in methanol. Secondary (2 degree): it is attached to two other carbon atoms. Tertiary (3 degree): it is attached to three other carbon atoms.

Propan-1-ol, CH3CH2CH2OH, is primary because the CH2OH carbon is joined to one carbon. Propan-2-ol, CH3CH(OH)CH3, is secondary because the carbon holding the -OH group is joined to two methyl groups. 2-methylpropan-2-ol, (CH3)3COH, is tertiary because that carbon carries three methyl groups and no hydrogen atom. When you are shown a displayed or skeletal formula, find the carbon with the -OH on it and count the carbon atoms bonded directly to it; that single number fixes the class.

Many alcohols contain more than one hydroxy group, and the class is then worked out one group at a time. Ethane-1,2-diol, HOCH2CH2OH, contains two primary alcohol groups and is the main ingredient of antifreeze. Propane-1,2,3-triol (glycerol), HOCH2CH(OH)CH2OH, contains two primary groups and one secondary group, which is why oxidising agents attack it in more than one place. Butane-2,3-diol, CH3CH(OH)CH(OH)CH3, contains two secondary groups. A molecule can therefore have a mixture of classes, and each group oxidises in the way its own class predicts.

The classification matters because it controls the chemistry. The first step in the oxidation of an alcohol is the removal of the hydrogen atom that sits on the carbon carrying the -OH group, so a primary or secondary alcohol has a hydrogen to lose and a tertiary alcohol does not. That single difference explains the whole pattern of oxidation products, and it is the basis of the test that picks a tertiary alcohol out of a group of unknowns.

Making alcohols from alkenes: steam and cold manganate(VII)

The industrial route to ethanol is the acid-catalysed hydration of ethene: ethene is heated with steam, using phosphoric acid, H3PO4, as the catalyst, at a high temperature (about 300 degrees C) and a high pressure (about 60 atm). The catalyst is held on a solid support, so it stays in the reactor, does not contaminate the product and can be used again. The reaction is an electrophilic addition, the same mechanism you met with hydrogen halides: the pi bond of the alkene attacks a proton, giving a carbocation, then a lone pair on the oxygen of a water molecule attacks the positively charged carbon, and the protonated alcohol finally loses a proton to give the alcohol. Nothing is lost from the alkene, so every atom of the C=C ends up in the product.

Which alcohol you obtain depends on the alkene. Ethene is symmetrical, so hydrating it gives only ethanol. An unsymmetrical alkene such as propene can give two alcohols, and the major product comes from the more stable carbocation, so propene gives mainly propan-2-ol with some propan-1-ol. The same reaction carried out in the laboratory by bubbling the alkene into concentrated sulphuric acid and then adding water works for the same reason, but the industrial version is faster and cleaner.

A second route from an alkene uses an oxidising agent instead of water. Shaking or bubbling the alkene with cold, dilute, acidified potassium manganate(VII), KMnO4(aq) with dilute sulphuric acid, oxidises the double bond and adds a hydroxy group to each carbon, giving a diol. Ethene gives ethane-1,2-diol and propene gives propane-1,2-diol. The purple colour of the manganate(VII) ion disappears as it reacts, so the same test also shows that a compound contains a C=C.

The words cold and dilute are the conditions that matter. Under these conditions the carbon skeleton stays intact and both new -OH groups are added. Warm, concentrated or strongly acidic manganate(VII) attacks the double bond far more aggressively and breaks the molecule at the C=C, so you no longer isolate a diol.

Making alcohols from halogenoalkanes, carbonyl compounds, acids and esters

From a halogenoalkane. Warming a halogenoalkane with NaOH(aq) hydrolyses the carbon-halogen bond by nucleophilic substitution: the lone pair on the oxygen of a hydroxide ion attacks the delta-positive carbon, and the halide ion leaves. Bromoethane gives ethanol: CH3CH2Br + NaOH(aq) -> CH3CH2OH + NaBr. The aqueous solvent matters, because hot ethanolic sodium hydroxide takes the other route and eliminates HBr to give an alkene instead of an alcohol.

From an aldehyde or a ketone. Both contain the polar C=O group, in which the carbon is delta-positive and the oxygen is delta-negative. A hydride ion, H-, from sodium tetrahydridoborate(III), NaBH4, in water or ethanol, or from lithium tetrahydridoaluminate(III), LiAlH4, in dry ether, attacks that delta-positive carbon, and the alkoxide ion that forms is protonated by water or dilute acid to give the alcohol. An aldehyde gives a primary alcohol (ethanal gives ethanol) and a ketone gives a secondary alcohol (propanone gives propan-2-ol). Only the C=O group is reduced, so a C=C elsewhere in the molecule survives.

From a carboxylic acid. A carboxylic acid must be reduced all the way from -COOH to -CH2OH, which is a harder job. LiAlH4 in dry ether, followed by the careful addition of dilute acid, does it and gives a primary alcohol; ethanoic acid gives ethanol. NaBH4 is not a strong enough reducing agent for a carboxylic acid. Dry ether is used because LiAlH4 reacts violently with water, so any water present would destroy the reagent before it reached the acid.

From an ester. Esters contain the -COO- group, which is hydrolysed by hot water to give back the carboxylic acid and the alcohol: CH3COOCH2CH3 + H2O <-> CH3COOH + CH3CH2OH. This is a reversible reaction, and with dilute acid as the catalyst the mixture settles at an equilibrium in which much of the ester is unchanged. If dilute alkali is used instead, the carboxylic acid that is formed is neutralised to its salt, so it cannot re-form the ester and the hydrolysis goes to completion: CH3COOCH2CH3 + NaOH -> CH3COONa + CH3CH2OH. Alkaline hydrolysis is therefore the route to use when you want the maximum amount of alcohol back.

• Alkene + steam, H3PO4 catalyst: electrophilic addition, gives an alcohol.
• Alkene + cold dilute acidified KMnO4: gives a diol.
• Halogenoalkane + NaOH(aq), heat: nucleophilic substitution, gives an alcohol.
• Aldehyde or ketone + NaBH4 or LiAlH4: gives a primary or a secondary alcohol.
• Carboxylic acid + LiAlH4 in dry ether: gives a primary alcohol.
• Ester + dilute acid (reversible) or dilute alkali (complete), heat: gives the alcohol back.

Oxidation: distillation against reflux

The oxidising agents for alcohols are acidified potassium dichromate(VI), K2Cr2O7(aq) acidified with dilute sulphuric acid, and acidified potassium manganate(VII), KMnO4(aq) also acidified with dilute sulphuric acid. Both are mixtures of an oxidising agent and an acid, and in each case the mixture is heated with the alcohol. The alcohol is oxidised and the oxidising agent is reduced, so it must be present in excess if you want every molecule of the alcohol to react; the shorthand + [O] is used for the oxygen that is added.

A primary alcohol is oxidised in two stages. The first gives an aldehyde, and the second gives a carboxylic acid: ethanol + [O] -> ethanal + H2O, then ethanal + [O] -> ethanoic acid. The aldehyde is easy to oxidise again, so which product you obtain depends entirely on the apparatus. In distillation the flask is closed by a condenser set at a downward angle, and the aldehyde, which boils far below the alcohol, boils off and is collected in a cooled receiver as it is formed, so it escapes further oxidation. In refluxing the condenser is clamped vertically, so any vapour condenses and runs straight back into the flask; the aldehyde stays in the flask with the excess oxidising agent and is oxidised on to the carboxylic acid. Propan-1-ol therefore gives propanal by distillation and propanoic acid by reflux.

A secondary alcohol is oxidised to a ketone and stops there: propan-2-ol + [O] -> propanone + H2O. A ketone is not oxidised further by acidified dichromate(VI) at all, because the carbon of the C=O group carries no hydrogen atom and a carbon-carbon bond would have to be broken instead, which these mild oxidising agents cannot do.

A tertiary alcohol is not oxidised by acidified dichromate(VI) or acidified manganate(VII). Oxidation starts by removing the hydrogen atom that is attached to the carbon carrying the -OH group, and in a tertiary alcohol that carbon is joined to three carbon atoms instead, so it has no hydrogen atom to lose. Nothing happens, and the solution stays orange, which is exactly the observation that identifies a tertiary alcohol.

The tri-iodomethane (iodoform) reaction

The reagent is alkaline aqueous iodine: iodine, I2(aq), together with sodium hydroxide solution, warmed gently with the alcohol. If the alcohol contains the group CH3CH(OH)-, that is a carbon carrying an -OH group which is also joined to a CH3 group, a pale yellow precipitate of tri-iodomethane, CHI3, appears. The ion that is formed at the same time is the carboxylate ion RCO2-, where R is whatever other group was attached to the CH3CH(OH)- carbon. The syllabus gives the group as CH3CH(OH)-R, so the test is really a test for that unit, not for alcohols in general.

Ethanol, CH3CH(OH)H, reacts, because the CH3CH(OH)- group is there with R = H; the ion formed with the precipitate is the methanoate ion, HCO2-. Propan-2-ol, CH3CH(OH)CH3, reacts, and here R = CH3, so the ion is the ethanoate ion, CH3CO2-. Butan-2-ol, CH3CH(OH)CH2CH3, reacts with R = CH2CH3, so the ion is the propanoate ion, CH3CH2CO2-. Methanol, propan-1-ol, butan-1-ol and pentan-3-ol all lack the group and give no precipitate.

What happens in the flask explains the name. Iodine in alkaline solution is an oxidising agent, so it first oxidises the alcohol to the corresponding carbonyl compound: ethanol is oxidised to ethanal and propan-2-ol to propanone. The CH3CO- group of that product then reacts with more iodine and alkali, splitting off the CH3 as tri-iodomethane and leaving the carboxylate ion behind. That is why the same yellow precipitate also appears with ethanal and with methyl ketones such as propanone and butanone, and why the reaction is not a test for an alcohol as such.

The value of the test in the exam is that it separates pairs of alcohols that the dichromate test cannot. Both methanol and ethanol give the orange-to-green change with acidified dichromate(VI), because both are primary alcohols, and both propan-1-ol and propan-2-ol are oxidised as well; only the iodine test tells the members of each pair apart.

Combustion, reaction with sodium and conversion to halogenoalkanes

Combustion. Alcohols burn in a good supply of air to give carbon dioxide and water, releasing a great deal of energy because the C-C, C-H and O-H bonds of the alcohol and the O=O bonds of the oxygen are replaced by the far stronger C=O bonds of carbon dioxide and O-H bonds of water: C2H5OH + 3O2 -> 2CO2 + 3H2O. Ethanol burns with a clean blue flame and is used as a fuel in its own right and blended with petrol, and the longer the carbon chain the more energy a mole of the alcohol releases. In a poor supply of oxygen the combustion is incomplete, and carbon monoxide and soot are formed instead, which is why the supply of air must be controlled in a burner.

Reaction with sodium. The O-H bond is polar, so the hydrogen atom carries a partial positive charge, and sodium, an excellent electron donor, gives an electron to that hydrogen; hydrogen atoms then pair up and escape as hydrogen gas: 2CH3CH2OH + 2Na -> 2CH3CH2ONa + H2. The sodium sinks in the ethanol (ethanol is less dense than water but sodium is denser than both), there is a steady effervescence of hydrogen, and as the sodium disappears the white solid sodium ethoxide, CH3CH2ONa, forms; the solid is left behind if the solution is evaporated. The reaction is real evidence that the -OH group is acidic, because it is the O-H hydrogen, and not a C-H hydrogen, that is released.

Conversion to a halogenoalkane. Every one of these reagents replaces the -OH group by a halogen. Passing the hydrogen halide gas in, for example HBr(g), gives bromoethane and water: CH3CH2OH + HBr -> CH3CH2Br + H2O. The mechanism is nucleophilic substitution: a lone pair on the oxygen of the alcohol picks up the proton from HBr, converting the -OH into -OH2+, which is a good leaving group; the bromide ion then attacks the delta-positive carbon and a water molecule leaves. HBr can also be made in the flask by adding sodium bromide or potassium bromide to the alcohol with concentrated sulphuric acid or concentrated phosphoric acid; the syllabus also allows KCl with concentrated H2SO4 or concentrated H3PO4 to make HCl for the same purpose.

Phosphorus and sulphur reagents. With PCl3 and heat, three alcohol molecules react with one of phosphorus trichloride: 3C2H5OH + PCl3 -> 3C2H5Cl + H3PO3. With PCl5 the reaction is fast at room temperature and gives two other products: C2H5OH + PCl5 -> C2H5Cl + POCl3 + HCl; the white solid dissolves and misty fumes of hydrogen chloride are given off. With SOCl2 (thionyl chloride) the products are C2H5OH + SOCl2 -> C2H5Cl + SO2 + HCl. SOCl2 is the reagent of choice for a pure sample because both of its by-products are gases, so they bubble out of the mixture and leave a pure liquid chloroalkane, whereas POCl3 is a liquid that must be separated by distillation.

• Combustion: alcohol + O2 -> CO2 + H2O; a poor air supply gives carbon monoxide and soot.
• Sodium: 2CH3CH2OH + 2Na -> 2CH3CH2ONa + H2; steady effervescence, less vigorous than with water.
• HBr(g): CH3CH2OH + HBr -> CH3CH2Br + H2O, by nucleophilic substitution.
• KCl or KBr with concentrated H2SO4 or H3PO4: makes the hydrogen halide in the flask.
• PCl3 and heat; PCl5 at room temperature; SOCl2: all replace -OH by Cl.
• SOCl2 is preferred because SO2 and HCl are gases and leave a pure product.

Dehydration to an alkene and esterification

Dehydration is the removal of the elements of water from a molecule. Passing ethanol vapour over heated aluminium oxide, Al2O3, or heating the alcohol with excess concentrated sulphuric acid or concentrated phosphoric acid, gives ethene: C2H5OH -> C2H4 + H2O. Al2O3 is a solid catalyst, so it stays in the tube and leaves the gaseous alkene clean. The reaction is the reverse of the hydration of an alkene, which is why the same sort of catalyst works for both directions. An unsymmetrical alcohol can lose water in more than one way and so give a mixture of alkenes, with the more substituted alkene usually the major product, so butan-2-ol gives both but-1-ene and but-2-ene.

Esterification is the other reaction of an alcohol with a concentrated acid, but here the acid is a carboxylic acid and the sulphuric acid is only a catalyst. An alcohol and a carboxylic acid heated together with concentrated sulphuric acid react to form an ester and water: CH3COOH + C2H5OH <-> CH3COOC2H5 + H2O, giving ethyl ethanoate. Because the reaction is reversible and slow, the mixture is heated under reflux for some time, the concentrated sulphuric acid speeds it up, and the water it helps to remove keeps the equilibrium pushed towards the ester. Using an excess of one of the reactants has the same effect.

The two reactions are easy to confuse in an exam, because both use concentrated sulphuric acid and heat. Ask what the alcohol is being heated with: if a carboxylic acid is present you are making an ester, and if the alcohol is being heated on its own over a hot solid catalyst or with excess concentrated acid you are dehydrating it to an alkene. The ester line is also the one that shows the -OH group of the alcohol and the -OH group of the carboxylic acid combining and losing water.

Esters made this way are real substances with real uses: ethyl ethanoate is a sweet-smelling liquid used as a solvent in glues and nail varnish remover and, in tiny amounts, as a flavour. The smell of many fruits and of some wines and spirits is a mixture of esters made in exactly this way.

The acidity of alcohols compared with water

Water and an alcohol both have an O-H bond that can release a proton, so both are acids, but both are extremely weak: H2O <-> H+ + OH-, and CH3CH2OH <-> H+ + CH3CH2O-, where CH3CH2O- is the ethoxide ion. Neither liberates carbon dioxide from a carbonate, neither turns litmus red in the way a carboxylic acid does, and their pKa values are so close that comparing them needs a careful argument rather than a number. In that comparison, water is the slightly stronger acid and ethanol the slightly weaker one.

The reason is the alkyl group. An alkyl group such as CH3 or C2H5 is electron-releasing, so it pushes electron density towards the oxygen atom it is attached to. In the ethoxide ion that extra electron density intensifies the negative charge that is already on the oxygen, and a more concentrated negative charge is a less stable arrangement than the one in the hydroxide ion, where no alkyl group is present to push electrons in. Because the ethoxide ion is the less stable of the two anions, it pulls a proton back from the water far more readily, so the equilibrium for ethanol lies further to the left than the equilibrium for water and ethanol releases fewer protons.

Two pieces of laboratory evidence support this. First, sodium reacts with ethanol with a steady effervescence of hydrogen, but noticeably less violently than it does with water, which shows that ethanol supplies hydrogen ions less readily. Second, the trend continues along the series: a secondary alcohol, with two electron-releasing alkyl groups, is a slightly weaker acid than a primary alcohol, and a tertiary alcohol is weaker again, because each extra alkyl group pushes more electron density onto the oxygen of the alkoxide ion.

The argument also works in reverse, which is a good check on your understanding. If an electron-withdrawing group is placed near the -OH, it pulls electron density away from the oxygen, spreads out the negative charge on the alkoxide ion, stabilises it, and makes the alcohol a stronger acid. 2-chloroethanol, ClCH2CH2OH, is a stronger acid than ethanol, and is in fact slightly stronger than water. The same reasoning explains, in the next part of the course, why phenol, in which the -OH group sits directly on a benzene ring, is a much stronger acid than an alcohol: the negative charge on the phenoxide ion is delocalised into the ring and spread over several carbon atoms, and phenol is even able to dissolve in sodium hydroxide solution to give sodium phenoxide, while an alcohol does not react with sodium hydroxide at all.