Alcohol

Ethanol is a primary alcohol. Heated with acidified potassium dichromate(VI), which gives nascent oxygen, it is first oxidised to ethanal and then to ethanoic acid: \ce{CH3CH2OH + [O] -> CH3CHO + H2O}, then \ce{CH3CHO + [O] -> CH3COOH}.

Hydrogenation and reduction add hydrogen, which is the reverse of oxidation. Hydration adds water to ethene and gives ethanol, the opposite direction. Fermentation of sugars makes ethanol, which is not the change asked for.

This is the same change as 1(i), with the options worded differently. Ethanol is oxidised, first to ethanal and then to ethanoic acid: \ce{CH3CH2OH -> CH3CHO -> CH3COOH}, using acidified \ce{K2Cr2O7} and heat.

Hydrogenation and reduction are the reverse of oxidation, and hydration gives an alcohol from an alkene, so none of options a, b or d gives an acid.

The Lucas test separates the three families by how quickly an oily layer of alkyl chloride appears. A tertiary alcohol gives the oily layer immediately, a secondary alcohol in 5 to 10 minutes, and a primary alcohol only on heating. So the tertiary alcohol reacts fastest.

Option a is wrong because primary alcohols are the slowest, and option b is the intermediate case. Option d is wrong because the three families react at different rates.

The iodoform reaction needs a \ce{CH3CH(OH)-} group or a \ce{CH3CO-} group. Ethanol (\ce{CH3CH2OH}), 2-propanol (\ce{CH3CH(OH)CH3}) and ethanal (\ce{CH3CHO}) all have one of these groups and give the yellow precipitate of \ce{CHI3}. The book lists ethanal among the compounds that give this test.

Methanol has only one carbon atom, so it has neither group and gives no precipitate.

Alcohols are aliphatic organic compounds that contain the hydroxyl group, \ce{-OH}, as their functional group. The general formula is \ce{R-OH}. An alcohol with one \ce{-OH} group is monohydric; one with two or more is polyhydric, for example \ce{1,2-ethanediol} (glycol).

Monohydric alcohols are classified by the carbon atom that carries the \ce{-OH} group:

1. Primary: that carbon is attached to one alkyl group, for example ethanol, \ce{CH3CH2OH}.
2. Secondary: that carbon is attached to two alkyl groups, for example 2-propanol, \ce{CH3CH(OH)CH3}.
3. Tertiary: that carbon is attached to three alkyl groups, for example 2-methylpropan-2-ol (tert-butyl alcohol), \ce{(CH3)3COH}.

Ethanol heated with concentrated \ce{H2SO4} at 140\,^\circ\mathrm{C} loses one water molecule from two ethanol molecules, which join through an oxygen atom to form diethyl ether. The acid is a catalyst: it supplies \ce{H+} at the start and takes it back at the end.

The \ce{-OH} group of an alcohol is less polar than the \ce{-OH} group of water. The alkyl group of an alcohol is electron releasing, so it pushes electron density towards the oxygen atom. This weakens the positive character of the hydrogen in the O-H bond, so the hydrogen is released less easily.

The alkoxide ion formed from an alcohol, \ce{RO-}, holds the negative charge less comfortably than the hydroxide ion. So the alcohol stays mostly un-ionised, and alcohols are weaker acids than water.

Two tests distinguish the three families: the Lucas test, which uses anhydrous \ce{ZnCl2} and concentrated \ce{HCl}, and mild oxidation with acidified \ce{K2Cr2O7}.

Lucas test: \ce{(CH3)3COH + HCl ->[{anhyd.} ZnCl2] (CH3)3CCl + H2O} (immediate); \ce{(CH3)2CHOH + HCl ->[{anhyd.} ZnCl2] (CH3)2CHCl + H2O} (5 to 10 minutes); \ce{CH3CH2OH + HCl ->[{anhyd.} ZnCl2][heat] CH3CH2Cl + H2O} (on heating).

Oxidation: \ce{CH3CH2OH + [O] -> CH3CHO + H2O} and then \ce{CH3CHO + [O] -> CH3COOH} for the primary alcohol; \ce{CH3CH(OH)CH3 + [O] -> CH3COCH3 + H2O} for the secondary alcohol. The tertiary alcohol has no hydrogen on the carbon that carries the \ce{-OH} group, so it cannot be oxidised.

Any two of the following three methods from the unit are enough:

1. Electrophilic addition of steam to ethene: \ce{CH2=CH2 + H2O ->[H3PO4][300\,^\circ C] CH3CH2OH}, at 60 to 70\,\mathrm{atm} pressure, with phosphoric acid as catalyst.
2. Reduction of ethanal: \ce{CH3CHO + 2[H] ->[NaBH4 or LiAlH4] CH3CH2OH}.
3. Substitution of bromoethane: \ce{CH3CH2Br + NaOH(aq) ->[heat] CH3CH2OH + NaBr}.

The book also gives hydrolysis of ethyl ethanoate with dilute acid or alkali and heat, and reduction of ethanoic acid with \ce{LiAlH4}, as further routes.

The \ce{-OH} group of ethanol is replaced by a chlorine atom. The products are chloroethane (ethyl chloride) and water, so this is a substitution reaction. Anhydrous \ce{ZnCl2} is the catalyst and the mixture is heated, because a primary alcohol reacts slowly.

Concentrated sulphuric acid removes water from ethanol. The product depends on the temperature:

1. At 180\,^\circ\mathrm{C}, one molecule loses water to give ethene: \ce{CH3CH2OH ->[{conc.} H2SO4][180\,^\circ C] CH2=CH2 + H2O}.
2. At 140\,^\circ\mathrm{C}, two molecules lose one water molecule between them to give diethyl ether: \ce{2CH3CH2OH ->[{conc.} H2SO4][140\,^\circ C] CH3CH2OCH2CH3 + H2O}.

When ethanol is heated with potassium bromide and concentrated sulphuric acid, the \ce{-OH} group is replaced by bromine and bromoethane is formed. The sulphuric acid reacts with KBr to give hydrogen bromide in the mixture, and the HBr does the substitution.

The by-product is potassium hydrogen sulphate, \ce{KHSO4}.

Oxidation of an alcohol removes hydrogen or adds oxygen. Ethanol is oxidised by a mild oxidising agent such as acidified potassium dichromate(VI), which gives nascent oxygen. The orange \ce{Cr2O7^2-} is reduced to green \ce{Cr^3+}, so the colour change shows that oxidation has taken place.

1. Ethanol to ethanal (a primary alcohol to an aldehyde): \ce{CH3CH2OH + [O] -> CH3CHO + H2O}.
2. Ethanal to ethanoic acid: \ce{CH3CHO + [O] -> CH3COOH}.

Overall: \ce{CH3CH2OH + 2[O] -> CH3COOH + H2O}.

Dehydration is the removal of a water molecule. In ethanol, the \ce{-OH} group and a hydrogen atom from the neighbouring carbon are lost as water, and a double bond forms between the two carbon atoms. Concentrated sulphuric acid, or heated \ce{Al2O3}, is used as the dehydrating agent.

The product, ethene, is an alkene, so dehydration converts an alcohol into an alkene.

Esterification is the reaction of an alcohol with a carboxylic acid, in the presence of concentrated sulphuric acid (or phosphoric acid) as catalyst, to give an ester and water. Ethanol and ethanoic acid give ethyl ethanoate. The oxygen of the ester comes from the alcohol.

The iodoform reaction is a test for the \ce{CH3CH(OH)-} group. Ethanol is warmed with excess iodine and excess sodium hydroxide, and a pale yellow precipitate of tri-iodomethane, \ce{CHI3}, forms. Sodium methanoate, \ce{HCOONa}, is the other product.

The product depends on the temperature. Concentrated sulphuric acid removes water from ethanol in two different ways, and the temperature decides which one happens.

The higher temperature favours the removal of water from within one molecule to form the alkene. At the lower temperature two molecules join through an oxygen atom to form the ether.

Oxidation of an alcohol to a carbonyl compound needs a hydrogen atom on the carbon that carries the \ce{-OH} group. This hydrogen is lost with the hydrogen of the \ce{-OH} group to form the \ce{C=O} group. A primary alcohol has two such hydrogens and gives an aldehyde; a secondary alcohol has one and gives a ketone.

A tertiary alcohol has three alkyl groups on that carbon and no hydrogen. A \ce{C=O} group cannot form without breaking carbon-carbon bonds, so mild oxidising agents leave it unchanged: the orange dichromate stays orange and the purple permanganate stays purple.

Use the iodoform test: warm each alcohol with excess iodine and excess sodium hydroxide. Ethanol has the \ce{CH3CH(OH)-} group and gives a pale yellow precipitate of \ce{CHI3}. Methanol, \ce{CH3OH}, has no such group and gives no precipitate.

Use the iodoform test. 2-propanol, \ce{CH3CH(OH)CH3}, has the \ce{CH3CH(OH)-} group and gives a yellow precipitate of \ce{CHI3}. 1-propanol, \ce{CH3CH2CH2OH}, does not, because its \ce{-OH} group is on a \ce{CH2} carbon and there is no \ce{CH3CH(OH)-} unit.

A second test is the Lucas test: 2-propanol forms an oily layer in 5 to 10 minutes, while 1-propanol forms one only on heating.

Reagent and condition: aluminium oxide, \ce{Al2O3}, as a heated catalyst. This is dehydration: the alcohol loses a molecule of water and forms an alkene. Ethanol gives ethene.

Reagent and condition: acidified potassium manganate(VII), \ce{KMnO4} with dilute \ce{H2SO4}, heated, with the carbonyl product collected by distillation.

A secondary alcohol is oxidised to a ketone, which is not oxidised further. The purple \ce{MnO4-} is reduced to colourless \ce{Mn^2+}.

Reagent and condition: sodium metal at room temperature. The hydroxyl compound reacts with sodium to form a sodium alkoxide and hydrogen gas, which is seen as bubbles.

For ethanol: \ce{2CH3CH2OH + 2Na -> 2CH3CH2ONa + H2}, giving sodium ethoxide. Water reacts with sodium in the same way, forming sodium hydroxide and hydrogen.

Reagent and condition: thionyl chloride, \ce{SOCl2}, with pyridine as the solvent. The \ce{-OH} group is replaced by chlorine, giving an alkyl chloride, and the gases sulphur dioxide and hydrogen chloride are given off.

For ethanol: \ce{CH3CH2OH + SOCl2 -> CH3CH2Cl + SO2 + HCl}, giving chloroethane.

Use the iodoform test. Warm the alcohol with excess alkaline iodine. A yellow precipitate of tri-iodomethane, \ce{CHI3}, confirms the \ce{CH3CH(OH)-} group. For example, butan-2-ol gives the test:

The R group is carried into the carboxylate \ce{RCOO-} (as its sodium salt), which stays in solution. If no yellow precipitate forms, the group is not present.

Add sodium metal. An alcohol contains the \ce{-OH} group, which reacts with sodium: hydrogen gas is released with bubbling (effervescence) and a sodium alkoxide is formed.

This test works for every alcohol, including tertiary alcohols, because it depends only on the O-H group. Mild oxidation with acidified \ce{K2Cr2O7}, with an orange to green colour change, is a further test, but it fails for tertiary alcohols.

Add sodium metal. An alcohol has an O-H group and gives off hydrogen gas, for example \ce{2CH3CH2OH + 2Na -> 2CH3CH2ONa + H2}.

An alkyl halide has no O-H group, so no hydrogen is evolved: bromoethane, \ce{CH3CH2Br}, gives no reaction with sodium.

Sodium gives hydrogen with both alcohols, so it cannot separate them. The iodoform test does: warm each with excess iodine and excess sodium hydroxide.

Ethanol has the \ce{CH3CH(OH)-} group and gives a yellow precipitate of \ce{CHI3}. Methanol has no such group and gives no precipitate.

Sodium gives hydrogen with both, so it cannot separate them. Use the iodoform test: 2-propanol gives a yellow precipitate of \ce{CHI3}, and 1-propanol does not.

The Lucas test also works: 2-propanol forms an oily layer in 5 to 10 minutes, while 1-propanol needs heating.

Heat the halogenoalkane with aqueous sodium hydroxide. The halogen atom is substituted by an \ce{-OH} group, giving an alcohol and a halide ion.

For example, bromoethane gives ethanol: \ce{CH3CH2Br + NaOH(aq) ->[heat] CH3CH2OH + NaBr}.

The unit gives these methods for any alcohol R-OH, with R the alkyl group:

1. With thionyl chloride in pyridine: \ce{ROH + SOCl2 ->[pyridine] RCl + SO2 + HCl}.
2. With hydrogen chloride and anhydrous zinc chloride: \ce{ROH + HCl ->[{anhyd.} ZnCl2] RCl + H2O}, heated for a primary alcohol.
3. With potassium bromide and concentrated sulphuric acid: \ce{ROH + KBr + H2SO4 -> RBr + KHSO4 + H2O}.
4. With phosphorus trichloride or pentachloride: \ce{3ROH + PCl3 -> 3RCl + H3PO3}, or \ce{5ROH + PCl5 -> 5RCl + H3PO4 + H2O}.

Oxidise a primary alcohol by heating it under reflux with acidified potassium dichromate(VI). The alcohol is first oxidised to the aldehyde, and the aldehyde is then oxidised to the carboxylic acid.

Refluxing keeps the aldehyde in the reaction mixture so that it is oxidised further. Distillation, by contrast, removes the aldehyde as it forms and stops at the aldehyde stage. A secondary alcohol gives a ketone, not an acid, so this conversion works only for primary alcohols.

Reduce the aldehyde with sodium borohydride, \ce{NaBH4}, or lithium aluminium hydride, \ce{LiAlH4}. The \ce{C=O} group gains two hydrogen atoms and a primary alcohol is formed.

Heat the alcohol with a carboxylic acid and concentrated sulphuric acid as catalyst. The ester and water are formed. The reaction is reversible, as the double arrow shows, so the mixture settles into an equilibrium.

This is esterification, with ethanol giving ethyl ethanoate.

Introduction

Alcohols are among the most useful intermediates in organic chemistry. They are made from several other families and can be converted into many others, so they link one family to another. The book notes their use in making perfumes, tinctures and many organic compounds.

Conversions of alcohols

How alcohols are themselves made

Alcohols are made from alkenes (steam addition, or cold dilute acidified \ce{KMnO4} to a diol), from halogenoalkanes (with aqueous \ce{NaOH}), from aldehydes and ketones (reduction), and from carboxylic acids and esters (reduction or hydrolysis). Alcohols are therefore both products and starting materials.

Evaluation

• Strength: one functional group gives access to alkenes, halides, carbonyl compounds, acids, esters and ethers, which makes alcohols a versatile starting point.
• Strength: the same reactant can give different products, such as ethene or diethyl ether from ethanol, so the choice of conditions controls the product.
• Limit: tertiary alcohols cannot be oxidised, so this route to carbonyl compounds and acids does not work for them.
• Limit: the conditions must be chosen with care. Dehydration at the wrong temperature gives the wrong product, and strong reagents such as \ce{LiAlH4} and \ce{SOCl2} need careful handling.

Conclusion

Alcohols are central to organic synthesis because one \ce{-OH} group leads to most other functional groups, provided the reagents and conditions are chosen correctly.

Step 1: unsaturation

\ce{C3H6O} has one degree of unsaturation, so A has one C=C or C=O bond, or a ring, in addition to the functional group. The saturated alcohol propan-1-ol, \ce{CH3CH2CH2OH}, is \ce{C3H8O} and does not match this formula.

Step 2: choosing A

A is oxidised to B, and B is oxidised again to C. A ketone cannot be oxidised further, so B cannot be a ketone. An aldehyde is oxidised once to a carboxylic acid, so B must be an aldehyde and A must be a primary alcohol.

Propanal, \ce{CH3CH2CHO}, has the formula \ce{C3H6O} but is itself an aldehyde: its oxidation gives propanoic acid, which cannot be oxidised again, so propanal cannot be A. The only primary alcohol with formula \ce{C3H6O} is prop-2-en-1-ol (allyl alcohol).

Step 3: the structures

Check: A, \ce{CH2=CHCH2OH}, has the formula \ce{C3H6O}, and each oxidation step removes two hydrogen atoms and adds one oxygen atom, as the equations show.