Notes/Chemistry/Paper 1/Hydroxy Compounds
CAIEAS Level9701§16

Hydroxy Compounds

Alcohols: six ways to make them; combustion, sodium and substitution; oxidation decided by the primary/secondary/tertiary count; the tri-iodomethane test; dehydration and esterification; and why alcohols are weaker acids than water.

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In the AS Halogen Compounds note, hydroxide ions replaced the halogen in a halogenoalkane and made an alcohol. This note studies alcohols themselves: compounds with an −OH-\text{OH} group on a saturated carbon. Alcohols link most of AS organic chemistry: alkenes, halogenoalkanes, carbonyl compounds, carboxylic acids and esters can all be turned into alcohols or made from them.

You start by naming and classifying alcohols and seeing how hydrogen bonding affects their properties. Then you learn six ways to make them, and their reactions: burning, sodium, swapping –OH for a halogen, oxidation, the tri-iodomethane test, dehydration and making esters. You finish with why alcohols are weaker acids than water. By the end you can give reagents, conditions, equations and observations for each.

Before you start you should be able to
  • Naming organic compounds up to C6, skeletal formulae and cis/trans isomerism (AS Introduction to Organic Chemistry: “Nomenclature (up to C6)”, “Types of formulae” and “Stereoisomerism: cis/trans and optical”)

  • Electrophilic addition to alkenes, Markovnikov's rule, and oxidation of alkenes by cold dilute acidified KMnO₄ (AS Hydrocarbons: “Electrophilic addition reactions and the bromine test”, “Oxidation of alkenes: diols and cleavage”, “Carbocation stability and Markovnikov addition”)

  • Primary/secondary/tertiary classification, substitution by NaOH(aq) and the five reagent sets that turn an alcohol into a halogenoalkane (AS Halogen Compounds)

  • Hydrogen bonding (AS Chemical Bonding: “Intermolecular forces: van der Waals' and hydrogen bonding”)

  • The positive inductive effect of alkyl groups (AS Hydrocarbons: “Carbocation stability and Markovnikov addition”)

  • Oxidation numbers and redox (AS Electrochemistry)

By the end of this page you can
  • Classify alcohols as primary, secondary or tertiary, including molecules containing more than one –OH group

  • Explain the boiling points and solubility of alcohols in terms of hydrogen bonding

  • Recall how alcohols are produced — electrophilic addition of steam with an H₃PO₄ catalyst; oxidation of alkenes by cold dilute acidified KMnO₄ to diols; heating halogenoalkanes with NaOH(aq); reduction of aldehydes/ketones with NaBH₄ or LiAlH₄ and of carboxylic acids with LiAlH₄; hydrolysis of esters with dilute acid or alkali and heat

  • Describe combustion of alcohols and construct balanced combustion equations

  • Describe the reaction of alcohols with sodium, write balanced equations for it, and use hydrogen volumes to count –OH groups

  • Describe substitution of the –OH group using HX(g), KCl plus concentrated H₂SO₄ or H₃PO₄, PCl₃ and heat, PCl₅, or SOCl₂

  • Describe oxidation with acidified K₂Cr₂O₇ or KMnO₄: distillation to stop at the aldehyde versus reflux to the acid; secondary → ketones; tertiary not oxidised; and the orange→green and purple→colourless colour changes used to distinguish them

  • State and use the tri-iodomethane reaction: alkaline I₂(aq) warmed gives a yellow precipitate of CHI₃ plus RCO₂⁻ for CH₃CH(OH)– compounds

  • Describe dehydration by heated catalysts such as Al₂O₃ or concentrated H₂SO₄/H₃PO₄ to alkenes, and esterification with carboxylic acids catalysed by concentrated H₂SO₄, exemplified by ethanol

  • Explain the acidity of alcohols relative to water in terms of the positive inductive effect of alkyl groups

01

Meet the alcohols

Syllabus requirement · §16.1.3(a)

“

classify alcohols as primary, secondary and tertiary alcohols, to include examples with more than one alcohol group

”

What an alcohol is

An alcohol is a compound in which a hydroxyl group, −OH-\text{OH}, is bonded to a saturated carbon atom. Alcohols with one –OH have the general formula CnH2n+1OH\text{C}_n\text{H}_{2n+1}\text{OH}: an alkane, CnH2n+2\text{C}_n\text{H}_{2n+2}, with one H replaced by OH.

To name an alcohol, find the longest chain that contains the C–OH carbon, number it so the –OH gets the lowest number, and use the ending -ol: CH3CH2OH\text{CH}_3\text{CH}_2\text{OH} is ethanol; CH3CH(OH)CH3\text{CH}_3\text{CH(OH)CH}_3 is propan-2-ol; (CH3)2CHCH2OH(\text{CH}_3)_2\text{CHCH}_2\text{OH} is 2-methylpropan-1-ol (the longest chain through the C–OH carbon is three carbons, numbered from the OH end). When there is more than one –OH, keep the e of the alkane and add di- or tri- with a number for each group: HOCH2CH2OH\text{HOCH}_2\text{CH}_2\text{OH} is ethane-1,2-diol (used as antifreeze), HOCH2CH(OH)CH2OH\text{HOCH}_2\text{CH(OH)CH}_2\text{OH} is propane-1,2,3-triol (glycerol).

The –OH group has two bonds that react, and knowing which one breaks sorts out the whole topic:

  • the O–H bond breaks when sodium reacts, when the alcohol is oxidised (together with a C–H bond on the same carbon), when it forms an ester, and when it acts as an acid;
  • the C–O bond breaks when –OH is swapped for a halogen and when water is removed to make an alkene.
THE HYDROXYL GROUP — two bonds, two kinds of reactionCH₃CH₂OHethanol C₂H₅OH — the circled –OH is the hydroxyl groupgeneral formula (monohydric): C𝑛H₂𝑛₊₁OHan alkane's C𝑛H₂𝑛₊₂ minus one H,plus OHO–H = the reactive handle (Na, [O], H⁺)C–O = the bond substitution swapsCLASSIFICATION — count the carbons bonded to the C carrying the –OHOHCCH₃HHprimary (1°)touches ONE carbonethanolOHCCH₃CH₃Hsecondary (2°)touches TWO carbonspropan-2-olOHCCH₃CH₃CH₃tertiary (3°)touches THREE carbons2-methylpropan-2-olpolyols count every –OH separately: ethane-1,2-diol HOCH₂CH₂OH is a diol whose two –OH groups are both primarycitric acid carries a tertiary –OH plus three –CO₂H — sodium reacts with all four

The hydroxyl group and its two reactive bonds, and how to classify an alcohol by counting the carbons bonded to the C–OH carbon.

Primary, secondary and tertiary

Find the carbon that carries the –OH. Count the carbon atoms bonded directly to it:

  • primary (1°): one other carbon (or none, in methanol) — ethanol CH3CH2OH\text{CH}_3\text{CH}_2\text{OH}, butan-1-ol CH3CH2CH2CH2OH\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{OH}
  • secondary (2°): two — propan-2-ol CH3CH(OH)CH3\text{CH}_3\text{CH(OH)CH}_3, butan-2-ol
  • tertiary (3°): three — 2-methylpropan-2-ol (CH3)3COH(\text{CH}_3)_3\text{COH}

Another way to see it: the C–OH carbon of a primary alcohol carries two H atoms, a secondary carries one H, and a tertiary carries none.

When a molecule has more than one –OH, classify each –OH separately. In ethane-1,2-diol both –OH groups are on end CH2\text{CH}_2 carbons, so both are primary. Glycerol, HOCH2CH(OH)CH2OH\text{HOCH}_2\text{CH(OH)CH}_2\text{OH}, has two primary –OH groups and one secondary. Citric acid has one tertiary –OH as well as three −CO2H-\text{CO}_2\text{H} groups.

This is the same count you used for halogenoalkanes. Here it decides what happens on oxidation: primary alcohols can go to an aldehyde and then a carboxylic acid, secondary alcohols to a ketone, and tertiary alcohols are not oxidised. The “Oxidation: products and colour changes” section explains why.

Classifying a mixed set

Classify each alcohol as primary, secondary or tertiary, giving systematic names:

(i) CH3CH2CH(CH3)CH2OH\text{CH}_3\text{CH}_2\text{CH(CH}_3)\text{CH}_2\text{OH} (ii) CH3CH2CH(OH)CH2CH3\text{CH}_3\text{CH}_2\text{CH(OH)CH}_2\text{CH}_3 (iii) (CH3)2C(OH)CH2CH3(\text{CH}_3)_2\text{C(OH)CH}_2\text{CH}_3 (iv) CH3CH(OH)CH(OH)CH3\text{CH}_3\text{CH(OH)CH(OH)CH}_3

Show full working
  1. 1

    (i) Classify. The C–OH carbon is the end CH2\text{CH}_2, bonded to one carbon → primary.

    Always find the C–OH carbon first and count only the carbons bonded to it. The rest of the chain doesn't matter for the class.

  2. 2

    (i) Name. The longest chain through the C–OH carbon is HOCH2–CH–CH2–CH3\text{HOCH}_2\text{–CH–CH}_2\text{–CH}_3 (four carbons), with a methyl branch. Numbering from the OH end puts OH on C1 and the methyl on C2 → 2-methylbutan-1-ol.

    The chain must contain the C–OH carbon. Students often count only the three carbons written in a row and miss that the branch point can lead into a longer chain.

  3. 3

    (ii) The C–OH carbon is bonded to an ethyl group on each side → two carbons → secondary. Five-carbon chain, OH on C3 from either end → pentan-3-ol.

    Two carbons attached means secondary, however big those groups are.

  4. 4

    (iii) The C–OH carbon carries two methyl groups and an ethyl group → three carbons → tertiary. Longest chain through it: four carbons, with OH and a methyl both on C2 → 2-methylbutan-2-ol.

    A tertiary C–OH carbon has no H on it. Check that as a second test: here it is bonded to C, C, C and O.

  5. 5

    (iv) Each of the two middle carbons carries an –OH and is bonded to two carbons → both –OH groups are secondary. Four-carbon chain with OH on C2 and C3 → butane-2,3-diol.

    With two –OH groups, classify each one on its own. The name keeps the 'e' of butane and adds 'diol' with both position numbers.

Answer

(i) primary, 2-methylbutan-1-ol (ii) secondary, pentan-3-ol (iii) tertiary, 2-methylbutan-2-ol (iv) diol, both –OH secondary, butane-2,3-diol

Hydrogen bonding: boiling points and solubility

The O–H bond is very polar, and the oxygen has two lone pairs. So the H of one alcohol molecule is attracted to a lone pair on the O of another: a hydrogen bond between molecules (AS Chemical Bonding). This explains two physical properties.

  • Boiling points are high. Ethanol (MrM_\text{r} = 46) boils at 78 °C, but propane (MrM_\text{r} = 44) boils at −42 °C. Both have similar van der Waals' forces, but only ethanol also has hydrogen bonds to break when it boils. Aldehydes and ketones cannot form hydrogen bonds with each other (they have no O–H), so they boil lower than the alcohols they come from. That is why an aldehyde can be distilled off as soon as it forms (see “Oxidation: products and colour changes”).
  • Small alcohols dissolve in water. Alcohol molecules form hydrogen bonds with water molecules, so methanol, ethanol and propanol mix with water in any proportion. As the hydrocarbon chain gets longer, more of the molecule cannot hydrogen bond, and solubility falls.

More –OH groups mean more hydrogen bonds: ethane-1,2-diol boils at 197 °C and glycerol at 290 °C.

Your turn

  1. 19701/14 M/J 2025 Q271 mark

    Which row shows a primary, a secondary and a tertiary alcohol?

    Options A–D as printed with the question.

    Options A–D as printed with the question.

    Stuck? Show hint

    For each drawn structure, ignore its column label and recount yourself — the wrong rows mislabel at least one structure.

    Show solution
    1. 1

      Row A: the primary entry is propan-1-ol ✓, and the secondary entry, HOCH2CH(OH)CH3\text{HOCH}_2\text{CH(OH)CH}_3, does have a secondary –OH on its middle carbon ✓. But the tertiary entry is glycerol, HOCH2CH(OH)CH2OH\text{HOCH}_2\text{CH(OH)CH}_2\text{OH}: its –OH groups are primary and secondary, none tertiary ✗.

      Check every entry in a row. One wrong entry is enough to reject the row.

    2. 2

      Row B: the primary entry is 2-methylpropan-1-ol, (CH3)2CHCH2OH(\text{CH}_3)_2\text{CHCH}_2\text{OH} ✓. But the secondary entry is (CH3)3COH(\text{CH}_3)_3\text{COH}: its C–OH carbon is bonded to three carbons, so it is tertiary ✗.

      Count the carbons on the C–OH carbon yourself; don't trust the column heading.

    3. 3

      Row C: the primary entry is propan-1-ol ✓. The secondary entry has a central carbon carrying H, CH₃ and two CH2OH\text{CH}_2\text{OH} groups. That central carbon has no –OH; both –OH groups are on end CH2\text{CH}_2 carbons, so both are primary ✗. The tertiary entry has the same problem.

      In a polyol, look at which carbon each –OH is actually on. A carbon with three branches is not tertiary unless the –OH is on that carbon.

    4. 4

      Row D: ethanol — C–OH carbon bonded to one carbon, primary ✓. Propan-2-ol — bonded to two, secondary ✓. 2-methylpropan-2-ol, (CH3)3COH(\text{CH}_3)_3\text{COH} — bonded to three, tertiary ✓. Answer D.

      Every wrong row mixes correct structures with one mislabelled one, so classify each structure on its own before choosing.

    Answer

    D — ethanol (primary), propan-2-ol (secondary), 2-methylpropan-2-ol (tertiary)

  2. 24 marks

    Propan-1-ol, CH3CH2CH2OH\text{CH}_3\text{CH}_2\text{CH}_2\text{OH} (MrM_\text{r} = 60), boils at 97 °C. Butane, C4H10\text{C}_4\text{H}_{10} (MrM_\text{r} = 58), boils at −1 °C.

    (a) Explain the difference in boiling points.
    (b) Propan-1-ol mixes completely with water, but hexan-1-ol is only slightly soluble. Explain why.

    Stuck? Show hint

    Similar Mr means similar van der Waals' forces. What extra force does only the alcohol have?

    Show solution
    1. 1

      (a) Name the extra force. Propan-1-ol molecules form hydrogen bonds between the H of one O–H group and a lone pair on the O of another molecule.

      Say what forms the hydrogen bond (O–H hydrogen to an O lone pair). 'It has hydrogen bonds' with no detail can lose the mark.

    2. 2

      (a) Compare with butane. Butane has only van der Waals' forces, similar in size to propan-1-ol's because the MrM_\text{r} values are nearly equal. Hydrogen bonds are stronger, so more energy is needed to separate propan-1-ol molecules → higher boiling point.

      Always link the stronger force to 'more energy needed to overcome it'. The forces broken on boiling are between molecules, not the covalent bonds inside them.

    3. 3

      (b) Propan-1-ol forms hydrogen bonds with water molecules, so it mixes with water.

      Solubility answers need 'with water': the alcohol must hydrogen bond to the water molecules.

    4. 4

      (b) Hexan-1-ol has the same single –OH but a much longer hydrocarbon chain, which cannot hydrogen bond with water. So much less of the molecule interacts with water, and solubility is low.

      One –OH group can only carry a short chain into water. Each extra CH₂ makes the molecule less like water.

    Answer

    (a) propan-1-ol has hydrogen bonds between molecules (as well as van der Waals' forces, similar to butane's); more energy needed to overcome them (b) propan-1-ol hydrogen bonds with water; hexan-1-ol's long hydrocarbon chain cannot, so it is much less soluble

The rest of this note

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Can you do all of these?

  • I can name alcohols, including diols and triols, and classify each –OH as primary, secondary or tertiary

  • I can explain the high boiling points and water solubility of small alcohols using hydrogen bonding

  • I can give reagents and conditions for all six ways to make an alcohol, and write reduction equations with [H]

  • I can write equations for combustion and for the reaction with sodium, and use gas volumes to find the number of –OH groups

  • I can give the five reagent sets that replace –OH by a halogen, the PCl₅ equation and its observation

  • I can predict the oxidation products of primary, secondary and tertiary alcohols, choose distillation or reflux, write [O] equations and state the colour changes

  • I can balance [O] equations for molecules with several groups: one H₂O for each alcohol group oxidised, none for an aldehyde → acid

  • I can use the tri-iodomethane test to find a CH₃CH(OH)– group, name both products, and combine it with the dichromate test to identify an alcohol

  • I can give conditions for dehydration and list every alkene formed, including cis/trans isomers

  • I can give conditions for esterification, write the equation and name the ester

  • I can explain why alcohols are weaker acids than water, why ethoxide is a stronger base than hydroxide, and which reagents react with alcohols (Na yes; NaOH and Na₂CO₃ no)