Cambridge AS & A Level9701

Alcohols (16)

Chemistry 9701 Chapter Notes

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Alcohols
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In this chapter

  1. 1Introduction to Alcohols
  2. 2Physical Properties and Hydrogen Bonding
  3. 3Oxidation of Alcohols
  4. 4Dehydration of Alcohols
  5. 5Other Key Reactions of Alcohols

1. Introduction to Alcohols

Alcohols are a homologous series of organic compounds containing the hydroxyl (–OH) functional group attached to a saturated carbon atom. Their general formula is CnH2n+1OH. The name of an alcohol is derived from the parent alkane by replacing the '-e' with '-anol'. For alcohols with three or more carbon atoms, a number is used to indicate the position of the –OH group, starting from the end that gives the lowest number. Alcohols are classified as primary (1°), secondary (2°), or tertiary (3°) based on the number of carbon atoms directly bonded to the carbon atom that carries the –OH group. A primary alcohol has one carbon atom attached, a secondary has two, and a tertiary has three.

General Formula: CnH2n+1OH

Key term

Functional Group: An atom or group of atoms within a molecule that is responsible for the characteristic chemical reactions of that molecule.

Examiner insight

Examiners expect you to be able to draw and interpret structural, displayed, and skeletal formulae for alcohols and correctly identify their class.

Common pitfall

Confusing the classification of an alcohol with the branching of the carbon chain. For example, 2-methylpropan-1-ol is a primary alcohol because the carbon attached to the -OH is only bonded to one other carbon, even though the overall molecule is branched.

Worked example 12 marks

An alcohol has the structural formula CH3CH(OH)CH2CH3. Give its systematic name and classify it as primary, secondary, or tertiary.

  1. 1

    Step 1: Identify the longest carbon chain containing the -OH group. Here, it is a chain of 4 carbon atoms, so the stem is 'butan-'.

  2. 2

    Step 2: Number the carbon chain from the end that gives the -OH group the lowest possible number. Numbering from the left gives the -OH group position 2. Numbering from the right gives it position 3. Therefore, we use 2.

  3. 3

    Step 3: Combine the parts to get the name: butan-2-ol.

  4. 4

    Step 4: To classify the alcohol, look at the carbon atom bonded to the -OH group (carbon-2). It is directly bonded to two other carbon atoms (carbon-1 and carbon-3).

  5. 5

    Step 5: Since it is bonded to two other carbon atoms, it is a secondary (2°) alcohol.

Worked example 22 marks

Draw the displayed formula for 2-methylpropan-2-ol and classify it.

  1. 1

    Step 1: The name 'propan-2-ol' indicates a 3-carbon chain with an -OH group on the second carbon.

  2. 2

    Step 2: The name '2-methyl' indicates a methyl (CH3) group also on the second carbon.

  3. 3

    Step 3: Draw the 3-carbon skeleton, place the -OH and CH3 groups on the central carbon, and then add hydrogen atoms to the other carbons to ensure each has 4 bonds. The structure is C(CH3)3OH.

  4. 4

    Step 4: The displayed formula shows every atom and every bond: A central carbon is bonded to three CH3 groups and one OH group.

  5. 5

    Step 5: The carbon atom attached to the -OH group is also attached to three other carbon atoms. Therefore, it is a tertiary (3°) alcohol.

Recap

  • Alcohols contain the hydroxyl (–OH) functional group.
  • The general formula for a saturated monohydric alcohol is CnH2n+1OH.
  • Alcohols are named with the suffix '-anol' and a number to show the position of the –OH group.
  • Primary (1°) alcohols have the structure RCH2OH.
  • Secondary (2°) alcohols have the structure RCH(OH)R'.
  • Tertiary (3°) alcohols have the structure RC(OH)R'R''.

Quick check

  1. What is the systematic name for the alcohol with the formula CH3CH2CH2OH?1 mark
  2. Is butan-2-ol a primary, secondary, or tertiary alcohol?1 mark

2. Physical Properties and Hydrogen Bonding

Alcohols have significantly higher boiling points than alkanes of similar relative molecular mass. For example, ethanol (Mr = 46) boils at 78 °C, while propane (Mr = 44) boils at -42 °C. This is due to the presence of strong intermolecular forces called hydrogen bonds. The oxygen atom in the –OH group is highly electronegative, creating a polar O-H bond. The slightly positive hydrogen atom (δ+) of one alcohol molecule is strongly attracted to the lone pair of electrons on the slightly negative oxygen atom (δ-) of a neighbouring molecule. A lot of energy is needed to overcome these hydrogen bonds, resulting in high boiling points. This same hydrogen bonding also explains why short-chain alcohols (like methanol, ethanol, propanol) are soluble in water. They can form hydrogen bonds with water molecules, allowing them to mix.

Key term

Hydrogen Bond: A strong type of intermolecular dipole-dipole attraction between a hydrogen atom covalently bonded to a highly electronegative atom (N, O, or F) and a lone pair of electrons on another nearby electronegative atom.

Fun fact

The 'legs' or 'tears' that form on the inside of a wine glass are due to the 'Marangoni effect'. Alcohol evaporates faster than water, causing the remaining liquid on the glass wall to have a higher surface tension, which pulls it up into droplets that then run down as tears.

Worked example 13 marks

Explain why the boiling point of propan-1-ol (97 °C) is much higher than that of butane (0 °C), given they have similar relative molecular masses (60 and 58 respectively).

  1. 1

    Step 1: Identify the intermolecular forces in each substance. Butane is a non-polar molecule and only has weak van der Waals forces between its molecules.

  2. 2

    Step 2: Identify the intermolecular forces in propan-1-ol. It has a polar O-H bond, so in addition to van der Waals forces, it has strong hydrogen bonds between its molecules.

  3. 3

    Step 3: Compare the strength of the forces. Hydrogen bonds are significantly stronger than van der Waals forces.

  4. 4

    Step 4: Relate the forces to the energy required for boiling. More energy is required to overcome the strong hydrogen bonds in propan-1-ol compared to the weak van der Waals forces in butane. Therefore, propan-1-ol has a much higher boiling point.

Recap

  • Alcohols have high boiling points due to strong intermolecular hydrogen bonds.
  • Hydrogen bonds form between the δ+ hydrogen of one –OH group and the δ- oxygen of another.
  • More energy is needed to overcome hydrogen bonds than van der Waals forces.
  • Short-chain alcohols are soluble in water because they can form hydrogen bonds with water molecules.
  • Solubility decreases as the hydrocarbon chain length increases, as the non-polar part of the molecule becomes more significant.

Quick check

  1. What is the strongest type of intermolecular force present in liquid ethanol?1 mark
  2. Why is hexan-1-ol less soluble in water than ethanol?1 mark

3. Oxidation of Alcohols

Alcohols can be oxidised using a suitable oxidising agent, most commonly acidified potassium dichromate(VI) (K2Cr2O7/H2SO4). The outcome of the reaction depends on the class of the alcohol and the reaction conditions. During the reaction, the orange dichromate(VI) ion (Cr2O7^2-) is reduced to the green chromium(III) ion (Cr^3+), providing a clear visual cue that oxidation has occurred.

  • Primary (1°) Alcohols: Can be oxidised twice. Mild oxidation, by warming the alcohol with the oxidising agent and immediately distilling off the product, yields an aldehyde. If the primary alcohol is heated under reflux with an excess of the oxidising agent, it is fully oxidised to a carboxylic acid.
  • Secondary (2°) Alcohols: Are oxidised by heating under reflux to form a ketone. Further oxidation is not possible under these conditions.
  • Tertiary (3°) Alcohols: Are resistant to oxidation by acidified potassium dichromate(VI). The orange solution remains orange when warmed.

Primary alcohol to aldehyde: RCH2OH + [O] → RCHO + H2O (Conditions: Distillation)

Primary alcohol to carboxylic acid: RCH2OH + 2[O] → RCOOH + H2O (Conditions: Reflux, excess oxidant)

Secondary alcohol to ketone: RCH(OH)R' + [O] → RCOR' + H2O (Conditions: Reflux)

Key term

Reflux: The process of heating a liquid in a vessel with an attached condenser to prevent the loss of volatile reactants or products, allowing the reaction to be carried out at a higher temperature for an extended period.

Examiner insight

Examiners frequently ask questions that require you to use oxidation reactions to distinguish between the different classes of alcohols. Be precise with the reagents, conditions, and expected observations.

Common pitfall

Forgetting the different conditions required to get an aldehyde versus a carboxylic acid from a primary alcohol. Distillation isolates the volatile aldehyde before it can be further oxidised.

Worked example 13 marks

Propan-1-ol is heated under reflux with excess acidified potassium dichromate(VI). Identify the organic product and write a balanced equation for the reaction using [O] to represent the oxidising agent.

  1. 1

    Step 1: Identify the alcohol. Propan-1-ol (CH3CH2CH2OH) is a primary alcohol.

  2. 2

    Step 2: Identify the conditions. 'Heated under reflux with excess oxidant' indicates strong oxidation.

  3. 3

    Step 3: Recall the product of strong oxidation of a primary alcohol. It is a carboxylic acid.

  4. 4

    Step 4: Name the specific product. The 3-carbon primary alcohol (propan-1-ol) will form the 3-carbon carboxylic acid, which is propanoic acid (CH3CH2COOH).

  5. 5

    Step 5: Write the balanced equation. Two atoms of oxygen from the oxidising agent are required. CH3CH2CH2OH + 2[O] → CH3CH2COOH + H2O.

Worked example 23 marks

A student has samples of three isomeric alcohols with the formula C4H10O: butan-1-ol, butan-2-ol, and 2-methylpropan-2-ol. Describe a simple chemical test to distinguish between them.

  1. 1

    Step 1: State the reagent and general procedure. Add a few drops of each alcohol to a separate test tube containing acidified potassium dichromate(VI) solution and warm gently in a water bath.

  2. 2

    Step 2: Describe the expected result for the tertiary alcohol. With 2-methylpropan-2-ol (a tertiary alcohol), there will be no reaction. The solution will remain orange.

  3. 3

    Step 3: Describe the expected result for the primary and secondary alcohols. With butan-1-ol (primary) and butan-2-ol (secondary), the orange solution will turn green, indicating oxidation has occurred.

  4. 4

    Step 4: Distinguish between the primary and secondary alcohols. Although both react, their oxidation products are different. A further test (e.g. with Tollens' reagent) on the products would be needed to distinguish the aldehyde (from butan-1-ol) from the ketone (from butan-2-ol), but for A-Level, simply identifying the tertiary alcohol is the key first step.

Recap

  • Oxidation of alcohols uses acidified potassium dichromate(VI), K2Cr2O7/H2SO4.
  • The colour change observed during oxidation is from orange (Cr2O7^2-) to green (Cr^3+).
  • Primary alcohols oxidise to aldehydes (distil) or carboxylic acids (reflux).
  • Secondary alcohols oxidise to ketones (reflux).
  • Tertiary alcohols are resistant to oxidation.

Quick check

  1. What is the organic product when propan-2-ol is heated with acidified K2Cr2O7?1 mark
  2. What observation would be made if 2-methylpropan-2-ol was warmed with acidified K2Cr2O7?1 mark

4. Dehydration of Alcohols

Dehydration is an elimination reaction where an alcohol loses a molecule of water to form an alkene. This reaction requires specific conditions: either heating the alcohol with a concentrated acid catalyst, such as concentrated sulfuric acid (H2SO4) or phosphoric acid (H3PO4), at around 170-180 °C, or passing the alcohol vapour over a hot catalyst, such as aluminium oxide (Al2O3) powder, at around 300 °C. The reaction involves the removal of the –OH group from one carbon atom and a hydrogen atom from an adjacent carbon atom.

General Equation: CnH2n+1OH → CnH2n + H2O

Example (Ethanol): CH3CH2OH → CH2=CH2 + H2O

Key term

Elimination Reaction: A type of organic reaction in which two substituents are removed from a molecule, typically from adjacent atoms, resulting in the formation of a double or triple bond.

Examiner insight

Be prepared to state both methods for dehydration (acid catalyst or hot Al2O3), as either could be specified or asked for in an exam.

Common pitfall

Confusing dehydration with oxidation. Dehydration removes H2O to form an alkene; oxidation reacts with an oxidising agent to form a carbonyl compound or carboxylic acid.

Worked example 13 marks

Draw the structure of the major organic product formed when butan-2-ol is dehydrated. State the necessary reagents and conditions.

  1. 1

    Step 1: Identify the reaction. Dehydration is the removal of H2O to form an alkene.

  2. 2

    Step 2: Identify the starting molecule: butan-2-ol, CH3CH(OH)CH2CH3.

  3. 3

    Step 3: The -OH group is removed from carbon-2. A hydrogen atom must be removed from an adjacent carbon (either carbon-1 or carbon-3).

  4. 4

    Step 4: Removing H from carbon-1 gives CH2=CHCH2CH3 (but-1-ene). Removing H from carbon-3 gives CH3CH=CHCH3 (but-2-ene). Both are possible products.

  5. 5

    Step 5: For A-Level, either product is usually acceptable unless you have learned Zaitsev's rule. But-2-ene is the major product. Let's show but-2-ene.

  6. 6

    Step 6: State the conditions: Heat with concentrated sulfuric acid (H2SO4) or concentrated phosphoric acid (H3PO4), OR pass vapour over hot aluminium oxide (Al2O3).

Recap

  • Dehydration of an alcohol is an elimination reaction that produces an alkene and water.
  • Conditions include heating with a concentrated acid catalyst (e.g., H2SO4 or H3PO4).
  • Alternatively, pass alcohol vapour over a hot catalyst like aluminium oxide (Al2O3).
  • The reaction removes the –OH group and a hydrogen from an adjacent carbon atom.

Quick check

  1. What is the name of the alkene produced by the dehydration of propan-1-ol?1 mark
  2. Name a suitable catalyst for the dehydration of ethanol.1 mark

5. Other Key Reactions of Alcohols

Besides oxidation and dehydration, alcohols undergo several other important reactions.

  1. Complete Combustion: Alcohols are flammable and burn in a plentiful supply of oxygen to produce carbon dioxide and water. This exothermic reaction makes them useful as fuels and in spirit burners. Example with ethanol: C2H5OH + 3O2 → 2CO2 + 3H2O.
  1. Reaction with Sodium: Alcohols react with reactive metals like sodium. The reaction is less vigorous than sodium's reaction with water. The O-H bond breaks, effervescence is observed as hydrogen gas is produced, and a salt called a sodium alkoxide is formed. This reaction demonstrates the weakly acidic nature of the alcohol's hydroxyl proton. Example with ethanol: 2C2H5OH + 2Na → 2C2H5ONa + H2. The product is sodium ethoxide.
  1. Substitution to form Halogenoalkanes: The hydroxyl group is a poor leaving group, but it can be substituted by a halide ion under acidic conditions. A common method is to heat the alcohol under reflux with a sodium halide (e.g., NaBr) and concentrated sulfuric acid. The acid first protonates the -OH group, turning it into -OH2+, which is a much better leaving group (it leaves as a neutral water molecule).

Complete Combustion: C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l)

Reaction with Sodium: 2ROH + 2Na → 2RONa + H2

Substitution with HBr: ROH + HBr → RBr + H2O

Key term

Sodium Alkoxide: The salt formed when an alcohol reacts with sodium, with the general formula RONa.

Examiner insight

The reaction with sodium is a classic test for alcohols. Be sure to use the term 'effervescence' to describe the production of hydrogen gas.

Worked example 12 marks

Write a balanced chemical equation for the complete combustion of propan-1-ol (C3H7OH).

  1. 1

    Step 1: Write the reactants and products. C3H7OH + O2 → CO2 + H2O.

  2. 2

    Step 2: Balance the carbon atoms. There are 3 carbons on the left, so we need 3 CO2 on the right. C3H7OH + O2 → 3CO2 + H2O.

  3. 3

    Step 3: Balance the hydrogen atoms. There are (7+1) = 8 hydrogens on the left, so we need 4 H2O on the right. C3H7OH + O2 → 3CO2 + 4H2O.

  4. 4

    Step 4: Balance the oxygen atoms. On the right, there are (3x2) + 4 = 10 oxygen atoms. On the left, one is in the alcohol, so we need 9 more from O2. This means 4.5 O2 molecules.

  5. 5

    Step 5: To get whole numbers, multiply the entire equation by 2. 2C3H7OH + 9O2 → 6CO2 + 8H2O.

Worked example 23 marks

Describe the expected observation when a small piece of sodium is added to a sample of dry ethanol, and name the two products formed.

  1. 1

    Step 1: Recall the reaction of alcohols with sodium. It is a redox reaction where sodium is oxidised and hydrogen is reduced.

  2. 2

    Step 2: Describe the observation. There will be steady effervescence or fizzing as a gas is produced. The sodium metal will dissolve as it reacts.

  3. 3

    Step 3: Identify the gas produced. The gas is hydrogen (H2).

  4. 4

    Step 4: Identify the other product. The sodium replaces the hydrogen of the -OH group to form a salt, sodium ethoxide (CH3CH2ONa).

  5. 5

    Step 5: State the names of the two products: Hydrogen and Sodium Ethoxide.

Recap

  • Alcohols undergo complete combustion to form CO2 and H2O.
  • The reaction of an alcohol with sodium produces hydrogen gas and a sodium alkoxide.
  • The reaction with sodium is a test for the presence of an -OH group.
  • The -OH group can be substituted by a halogen (e.g., Br) by reacting the alcohol with a hydrogen halide.

Quick check

  1. What are the two products of the complete combustion of methanol?1 mark
  2. What is the name of the salt formed when propan-1-ol reacts with sodium?1 mark

End-of-chapter exercise

Test yourself on the whole chapter. Work through these before moving on.

  1. An alcohol X has the molecular formula C4H10O. It is resistant to oxidation by acidified potassium dichromate(VI). Identify alcohol X by name and state its classification.2 marks
  2. Ethanol can be converted into ethene or ethanoic acid in two different reactions. For each conversion, state the reagent(s), conditions, and type of reaction.6 marks
  3. Draw the displayed formula for butan-2-ol. Predict the structure of the organic product formed when it is heated under reflux with acidified potassium dichromate(VI). Name this product.3 marks
  4. Explain, with reference to intermolecular forces, why the boiling point of ethanol (78 °C) is significantly higher than that of its isomer methoxymethane, CH3OCH3 (-24 °C).3 marks
  5. Write a balanced chemical equation for the reaction of sodium with propan-2-ol.2 marks
  6. Pentan-1-ol is a primary alcohol. State the two possible organic products from its oxidation. For the formation of each product, state the specific conditions required.4 marks
  7. An unknown compound Y is a liquid at room temperature and is soluble in water. When warmed with acidified potassium dichromate(VI) solution, the solution remains orange. When a small piece of sodium is added to Y, effervescence is observed. Deduce the class of compound that Y belongs to and explain your reasoning.3 marks
  8. Draw the skeletal formula for 3-methylbutan-2-ol and classify it as a primary, secondary or tertiary alcohol.2 marks
  9. Describe what you would observe when propan-1-ol is heated gently with acidified potassium dichromate(VI) in a test tube fitted for distillation.2 marks
  10. The dehydration of butan-1-ol can produce an alkene. Name the alkene and state one set of reagents and conditions to carry out this reaction.2 marks

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