Cambridge AS & A Level9701

Alcohols (32)

Chemistry 9701 Chapter Notes

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In this chapter

  1. 1Structure, Naming, and Classification
  2. 2Physical Properties: Boiling Points and Solubility
  3. 3Oxidation of Alcohols
  4. 4The Tri-iodomethane (Iodoform) Test
  5. 5Dehydration of Alcohols
  6. 6Other Key Reactions

1. Structure, Naming, and Classification

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. They are named by replacing the '-e' of the parent alkane with '-ol'. For chains of three or more carbons, a number is used to indicate the position of the -OH group, starting from the end that gives the lowest number. Alcohols are classified based on the carbon atom bonded to the -OH group. A primary (1°) alcohol has one alkyl group attached to this carbon. A secondary (2°) alcohol has two alkyl groups. A tertiary (3°) alcohol has three alkyl groups.

General Formula: CnH2n+1OH

Primary (1°): RCH2OH

Secondary (2°): R2CHOH

Tertiary (3°): R3COH

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 name all structural isomers of a given alcohol molecular formula and classify each one correctly.

Common pitfall

When classifying alcohols, students mistakenly count the number of carbons in the whole molecule, instead of counting only the carbons directly bonded to the carbon atom that carries the -OH group.

Worked example 13 marks

An alcohol has the molecular formula C4H10O. Draw the structural formula for butan-2-ol and classify it as primary, secondary, or tertiary.

  1. 1

    Step 1: Identify the parent chain. 'Butan-' means a 4-carbon chain.

  2. 2

    Step 2: Identify the functional group and its position. '-2-ol' means an -OH group is on the second carbon atom.

  3. 3

    Step 3: Draw the carbon skeleton and add the -OH group. C-C-C-C. Add OH to the second carbon: CH3-CH(OH)-CH2-CH3.

  4. 4

    Step 4: Add hydrogen atoms to satisfy the valency of each carbon. The structural formula is CH3CH(OH)CH2CH3.

  5. 5

    Step 5: Classify the alcohol. The carbon atom bonded to the -OH group (C2) is attached to two other carbon atoms (C1 and C3). Therefore, it is a secondary (2°) 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 '-ol' and a number to show the position of the -OH group.
  • Primary (1°) alcohols have the -OH on a carbon attached to one other carbon.
  • Secondary (2°) alcohols have the -OH on a carbon attached to two other carbons.
  • Tertiary (3°) alcohols have the -OH on a carbon attached to three other carbons.

Quick check

  1. Name the compound CH3C(CH3)(OH)CH2CH3 and classify it.2 marks

2. Physical Properties: Boiling Points and Solubility

Alcohols have significantly higher boiling points than alkanes with similar relative molecular mass. This is because alcohol molecules can form strong intermolecular hydrogen bonds with each other, which require a large amount of energy to overcome. Alkanes only have weak van der Waals forces between molecules. Similarly, short-chain alcohols (e.g., methanol, ethanol) are completely miscible with water. This is because they can form hydrogen bonds with water molecules, allowing them to mix. As the hydrocarbon chain length increases, the non-polar alkyl part of the molecule becomes more significant, disrupting the hydrogen bonding with water and making the alcohol less soluble.

Key term

Hydrogen Bond: A strong intermolecular dipole-dipole attraction between a hydrogen atom covalently bonded to a highly electronegative atom (like oxygen) and another nearby electronegative atom.

Fun fact

The 'tears of wine' seen on the inside of a wine glass are caused by the Marangoni effect, where alcohol evaporates faster than water, increasing the surface tension of the remaining liquid and causing it to climb the glass before falling back as 'tears'.

Worked example 14 marks

Explain why the boiling point of ethanol (78 °C) is much higher than that of propane (-42 °C), even though they have very similar relative molecular masses (Mr of ethanol = 46.0; Mr of propane = 44.0).

  1. 1

    Step 1: Identify the intermolecular forces in each substance. Propane (C3H8) 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 ethanol. Ethanol (CH3CH2OH) has a polar O-H bond, allowing it to form strong hydrogen bonds between its molecules, in addition to van der Waals forces.

  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 force strength to boiling point. More energy is required to overcome the strong hydrogen bonds in ethanol compared to the weak van der Waals forces in propane. Therefore, ethanol has a much higher boiling point.

Recap

  • Alcohols have high boiling points due to intermolecular hydrogen bonding.
  • Hydrogen bonds are stronger than van der Waals forces.
  • Short-chain alcohols are soluble in water because they can form hydrogen bonds with water.
  • Solubility in water decreases as the hydrocarbon chain length increases.

Quick check

  1. Which would you expect to have a higher boiling point, propan-1-ol or butan-1-ol? Explain your reasoning.2 marks

3. Oxidation of Alcohols

The oxidation of alcohols is a key reaction used to distinguish between the different classes. The usual oxidising agent is a solution of potassium dichromate(VI), K2Cr2O7, acidified with dilute sulfuric acid, H2SO4. During the reaction, the orange dichromate(VI) ion (Cr2O7^2-) is reduced to the green chromium(III) ion (Cr^3+).

  • Primary (1°) Alcohols: Can be oxidised twice. Gentle heating with the oxidising agent, with the product distilled off as it forms, yields an aldehyde. Strong heating under reflux with excess oxidising agent yields a carboxylic acid.
  • Secondary (2°) Alcohols: Can be oxidised once. Heating under reflux with the oxidising agent yields a ketone.
  • Tertiary (3°) Alcohols: Are resistant to oxidation by acidified potassium dichromate(VI). The orange solution remains unchanged. This is because oxidation would require breaking a strong C-C bond.

Primary Alcohol to Aldehyde: RCH2OH + [O] → RCHO + H2O (distil)

Primary Alcohol to Carboxylic Acid: RCH2OH + 2[O] → RCOOH + H2O (reflux)

Secondary Alcohol to Ketone: R2CHOH + [O] → R2CO + H2O (reflux)

Tertiary Alcohol: R3COH + [O] → No reaction

Key term

Reflux: The process of heating a chemical reaction for an extended period while continually cooling the vapour produced back into liquid form, preventing the loss of volatile reactants and products.

Examiner insight

Marks are frequently awarded for correctly stating the conditions (distillation vs. reflux) needed to isolate an aldehyde or a carboxylic acid from a primary alcohol.

Worked example 13 marks

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

  1. 1

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

  2. 2

    Step 2: Identify the reaction conditions. 'Heated under reflux with excess' indicates complete oxidation.

  3. 3

    Step 3: Determine the product of complete oxidation of a primary alcohol. This is a carboxylic acid.

  4. 4

    Step 4: Name the product. The three-carbon carboxylic acid is propanoic acid.

  5. 5

    Step 5: Write the equation. The alcohol is oxidised to the carboxylic acid, requiring two oxidising equivalents. CH3CH2CH2OH + 2[O] → CH3CH2COOH + H2O.

Worked example 21 mark

Describe what you would observe when propan-2-ol is warmed with acidified potassium dichromate(VI) solution.

  1. 1

    Step 1: Identify the class of alcohol. Propan-2-ol is a secondary alcohol.

  2. 2

    Step 2: Recall the reaction of secondary alcohols with this reagent. Secondary alcohols are oxidised to ketones.

  3. 3

    Step 3: Recall the colour change of the reagent. The oxidising agent, potassium dichromate(VI), is reduced.

  4. 4

    Step 4: State the observation. The solution would change colour from orange (Cr2O7^2-) to green (Cr^3+).

Recap

  • Primary alcohols oxidise to aldehydes (distil) or carboxylic acids (reflux).
  • Secondary alcohols oxidise to ketones (reflux).
  • Tertiary alcohols do not oxidise with K2Cr2O7/H2SO4.
  • The colour change for a positive oxidation test is from orange to green.
  • Distillation separates volatile products, while reflux ensures complete reaction.

Quick check

  1. What organic product is formed when 2-methylpropan-2-ol is warmed with acidified potassium dichromate(VI)?1 mark

4. The Tri-iodomethane (Iodoform) Test

The tri-iodomethane (or iodoform) test is a specific chemical test used to identify the presence of a particular structural feature in an organic molecule. A positive result is given by any alcohol containing the CH3CH(OH)- group (a methyl group and a hydrogen on the carbon attached to the -OH group). This means all primary alcohols except ethanol, and all tertiary alcohols, will give a negative result. Ethanol is the only primary alcohol that gives a positive test. All secondary alcohols with the structure CH3CH(OH)R will give a positive test. The test is carried out by warming the substance with an alkaline solution of iodine (iodine in sodium hydroxide solution). A positive result is the formation of a pale yellow precipitate of tri-iodomethane (CHI3), which has a characteristic 'antiseptic' or 'hospital' smell.

General reaction: RCH(OH)CH3 + 4I2 + 6OH- → RCOO- + CHI3 + 5I- + 5H2O

Key term

Tri-iodomethane: A pale yellow solid (CHI3), also known as iodoform, whose formation indicates a positive result in the iodoform test.

Examiner insight

This test is a classic tool for distinguishing between positional isomers, such as propan-1-ol (negative) and propan-2-ol (positive). Be prepared to use it in structural identification problems.

Worked example 12 marks

An unknown alcohol C4H10O is found to give a positive result in the tri-iodomethane test. Draw the structure and give the name of this alcohol.

  1. 1

    Step 1: Understand the requirement for a positive test. The alcohol must contain the CH3CH(OH)- group.

  2. 2

    Step 2: Consider the molecular formula C4H10O. This means there are 4 carbon atoms in total.

  3. 3

    Step 3: Construct the structure. Start with the required group: -CH(OH)CH3. This uses 2 carbons, 1 oxygen, and 4 hydrogens.

  4. 4

    Step 4: Add the remaining atoms (C2H6) to form a valid structure. The remaining part is an ethyl group (C2H5). Attaching this to the carbon with the -OH group gives CH3CH2CH(OH)CH3.

  5. 5

    Step 5: Name the structure. It is a four-carbon chain with the -OH group on the second carbon. The name is butan-2-ol.

  6. 6

    Step 6: Check other isomers. Butan-1-ol, 2-methylpropan-1-ol and 2-methylpropan-2-ol do not have the required CH3CH(OH)- group.

Recap

  • The iodoform test identifies the CH3CH(OH)- group in an alcohol.
  • Reagents are aqueous iodine and sodium hydroxide, with warming.
  • A positive result is a pale yellow precipitate of tri-iodomethane (CHI3).
  • Ethanol is the only primary alcohol to give a positive test.
  • Any secondary alcohol with a methyl group next to the C-OH carbon will give a positive test.

Quick check

  1. Would pentan-3-ol give a positive result with the tri-iodomethane test? Explain your answer.2 marks

5. Dehydration of Alcohols

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

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 resulting in the formation of a double or triple bond.

Common pitfall

Forgetting that dehydration of unsymmetrical alcohols like butan-2-ol can form more than one structural isomer of the alkene product.

Worked example 13 marks

Butan-2-ol is heated with concentrated sulfuric acid. Identify the possible organic products and state the type of reaction.

  1. 1

    Step 1: Identify the reactant and reaction type. Butan-2-ol is a secondary alcohol, and heating with concentrated acid causes dehydration (an elimination reaction).

  2. 2

    Step 2: Locate the -OH group and adjacent carbons with hydrogen. The -OH is on C2. Adjacent carbons are C1 and C3.

  3. 3

    Step 3: Consider elimination of H from C1. Removing H from C1 and OH from C2 forms a double bond between C1 and C2. Product: CH2=CHCH2CH3 (But-1-ene).

  4. 4

    Step 4: Consider elimination of H from C3. Removing H from C3 and OH from C2 forms a double bond between C2 and C3. Product: CH3CH=CHCH3 (But-2-ene).

  5. 5

    Step 5: Identify isomers of the second product. But-2-ene exists as E/Z (cis/trans) stereoisomers. So, both E-but-2-ene and Z-but-2-ene can be formed.

  6. 6

    Step 6: List all possible organic products. The products are but-1-ene, E-but-2-ene, and Z-but-2-ene.

Recap

  • Dehydration of an alcohol forms an alkene and water.
  • This is an elimination reaction.
  • Conditions are heating with a concentrated acid catalyst (H2SO4 or H3PO4).
  • Alternatively, pass alcohol vapour over hot aluminium oxide.
  • Dehydration of unsymmetrical alcohols can lead to a mixture of isomeric alkenes.

Quick check

  1. What is the single organic product formed from the dehydration of propan-1-ol?1 mark

6. Other Key Reactions

Alcohols undergo several other important reactions.

1. Combustion: Like most organic compounds, alcohols burn readily in a plentiful supply of oxygen to produce carbon dioxide and water. This is a highly exothermic reaction, which is why ethanol is used as a biofuel.

2. Reaction with Sodium: Alcohols react with reactive metals like sodium in a redox reaction. The alcohol acts as a very weak acid, and the O-H bond breaks. A salt called a sodium alkoxide is formed, and hydrogen gas is evolved. The observation is effervescence (fizzing). For example, ethanol reacts with sodium to form sodium ethoxide and hydrogen.

3. Substitution to form Halogenoalkanes: The hydroxyl group can be replaced by a halogen atom (a substitution reaction). This can be achieved using several different reagents, for example, reacting the alcohol with a hydrogen halide (e.g., HBr, made in situ from NaBr and H2SO4) or with phosphorus(V) chloride (PCl5). With PCl5, the reaction is vigorous at room temperature, producing misty fumes of HCl gas.

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

Reaction with Sodium: 2CH3CH2OH + 2Na → 2CH3CH2ONa + H2

Substitution with HBr: CH3CH2OH + HBr → CH3CH2Br + H2O

Substitution with PCl5: CH3CH2OH + PCl5 → CH3CH2Cl + POCl3 + HCl

Key term

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

Worked example 13 marks

A small piece of sodium is added to a sample of dry ethanol in a test tube. State two observations and write a balanced equation for the reaction.

  1. 1

    Step 1: Recall the reaction between an alcohol and sodium. It's a redox reaction producing an alkoxide and hydrogen gas.

  2. 2

    Step 2: Predict the observations. The production of hydrogen gas will cause effervescence (fizzing). The reaction is exothermic, so the test tube will feel warm. The sodium metal will dissolve as it reacts.

  3. 3

    Step 3: State two observations. Observation 1: Effervescence / fizzing / bubbles are produced. Observation 2: The sodium solid dissolves/disappears.

  4. 4

    Step 4: Write the balanced equation. The reactants are ethanol (C2H5OH) and sodium (Na). The products are sodium ethoxide (C2H5ONa) and hydrogen (H2). Balancing gives: 2C2H5OH + 2Na → 2C2H5ONa + H2.

Recap

  • Complete combustion of alcohols produces CO2 and H2O.
  • Alcohols react with sodium to produce a sodium alkoxide and hydrogen gas.
  • The -OH group can be substituted by a halogen using reagents like hydrogen halides or PCl5.
  • The reaction with sodium shows that alcohols can act as very weak acids.

Quick check

  1. What gaseous products are formed when ethanol reacts with phosphorus(V) chloride, PCl5?1 mark

End-of-chapter exercise

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

  1. Draw the displayed formula for 2-methylpropan-2-ol and classify it as a primary, secondary or tertiary alcohol.2 marks
  2. Ethanol (C2H5OH) has a boiling point of 78 °C, while methoxymethane (CH3OCH3) has a boiling point of -24 °C. Both have the same molecular formula C2H6O. Explain this difference in boiling points in terms of intermolecular forces.3 marks
  3. An alcohol, X, with molecular formula C3H8O, is warmed with acidified potassium dichromate(VI). The solution turns from orange to green, and the organic product formed is Y. Product Y does not react with Fehling's solution. Identify X and Y.3 marks
  4. Write a balanced chemical equation for the complete combustion of propan-1-ol (C3H7OH).2 marks
  5. State the reagents and conditions required to convert ethanol into ethene.2 marks
  6. There are four structural isomers of C4H9OH. One of these isomers, butan-2-ol, gives a positive tri-iodomethane (iodoform) test. Explain why butan-2-ol gives a positive test, while its isomer butan-1-ol does not.3 marks
  7. Describe how you could use a chemical test to distinguish between pentan-2-ol and pentan-3-ol. State the reagents you would use and the expected observation for each alcohol.3 marks
  8. Propan-1-ol can be converted into propanoic acid. State the reagents and conditions for this conversion and write an equation for the reaction, using [O] for the oxidising agent.3 marks
  9. An unknown alcohol Z is a liquid at room temperature. It does not react when warmed with acidified potassium dichromate(VI) solution. When a small piece of sodium is added to Z, effervescence is observed. Deduce the class of alcohol (primary, secondary or tertiary) to which Z belongs, explaining your reasoning.3 marks
  10. Devise a two-step synthesis to convert propan-2-ol into propanone, and then propanone back into propan-2-ol. For each step, state the reagents, conditions and type of reaction.6 marks

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