Something to think about

The chemistry covered in this part of sub-topic 10.2 is really very simple. It can be summarized by saying that alcohols burn in oxygen to give carbon dioxide and water, primary alcohols are oxidised first to aldehydes then carboxylic acids, secondary alcohols are oxidized to ketones and tertiary alcohols cannot be oxidized whilst still retaining the carbons chain. All alcohols can react with carboxylic acids to form esters. This condensation reaction is also known as esterification. Be careful to remember that although the class of compounds is correctly called alcohols, the IB calls the -OH functional group the hydroxyl group, although probably it is more correct to call it the hydroxy group so that, for example, the systematic (IUPAC) name of lactic acid, CH₃CH(OH)COOH is 2-hydroxypropanoic acid, not 2-hydroxylpropanoic acid.

In fact the chemistry of alcohols, and ethanol in particular, can provide a useful respite from the purely organic chemistry approach. Ethanol has many attributes similar to water – or put another way the simplest 'alcohol' of all is water, H-O-H. It could be worth considering using ethanol (and the other alcohols) to re-enforce your understanding of  chemistry of some of the other core topics.

For example:

Topic 1. (1.1) Deduction of chemical equations when reactants and products are specified.
You should be able to deduce the general equation for the combustion of any alcohol C_xH_yO.
C_xH_yO + (x + y/4 -½)O₂ → xCO₂ + y/2H₂O

Topic 3. (3.2) Group trends should include the treatment of the reactions of alkali metals with water.
You could deduce what might happen if a very small piece of sodium metal is placed in ethanol rather than water. From the equation with water you should be able to work out that hydrogen gas will be evolved.

Na(s) + H-O-H(l) → Na⁺(aq) + OH^–(aq) + ½H₂(g)
Na(s) + R-O-H(l) → Na⁺(aq) + OR^–(aq) + ½H₂(g)

The reaction of alcohols with sodium is not actually on the IB programme but it does help to re-enforce the understanding of the reaction of sodium with water.

Topic 4. (4.3) Prediction of molecular polarity from bond polarity and molecular geometry.
By substituting just one of the hydrogen atoms in water with a –C₂H₅ group you can easily see that ethanol is polar. You should also be able to deduce that it is not as polar as water as there is only one hydrogen atom with δ+, not two . An interesting investigation is to put a beaker of water (about 10 cm³) in a household microwave oven for ten seconds or so and record the temperature rise. Predict what will happen if you repeat it with 10 cm³ of ethanol. You will probably predict the temperature will not rise as much as it is less polar. In fact it will rapidly boil as the specific heat capacity of ethanol is much lower than that of water.

Topic 5. (5.1) A calorimetry experiment for an enthalpy of reaction should be covered and the results evaluated.
Combusting alcohols using a spirit lamp under a calorimeter and then determining the enthalpy change is a classic experiment.

Topic 6. (6.1) Explanation of the effects of temperature on rate of reaction and (6.1.) Investigation of rates of reaction experimentally and evaluation of the results.
The obvious link with Topic 6 is that heat is required to bring about the oxidation of primary and secondary alcohols with an acidified solution of potassium dichromate(VI). The heat, of course increases the energy of the reacting particles so that more will have the necessary activation energy for the reaction to proceed. However if  deduced what will happen when sodium metal is reacted with ethanol (see Topic 1 above) then you could design or investigate an experiment to measure the rate when different alcohols are used (e.g. CH₃OH, C₂H₅OH, C₃H₇OH and C₄H₉OH) to see what happens to the reaction rate when the carbon chain length increases.

Topic 7. (7.1) Deduction of the equilibrium constant expression (K_c) from an equation for a homogeneous reaction.
One of the classic reactions used to illustrate homogeneous equilibrium is the esterification reaction between ethanol and ethanoic acid. You could even bring in a bit of Topic 2 here and ask yourself how you could determine whether it is the oxygen atom from the alcohol or the carboxylic acid that ends up in the water. The answer of course is to use isotopic labelling with ¹⁸O.

Topic 8. (8.1.) Deduction of the conjugate acid or conjugate base in a chemical reaction.
You know that hydroxide ions are a strong base. This is because hydroxide ions are the conjugate base of water which is a very weak acid (K_w = 1.00 x 10^-14). Ethanol is an even weaker acid (K_a = approximately 10^-16) so you should be able to deduce that the ethoxide ion, C₂H₅O^–, is an even stronger base than the hydroxide ion, OH^–.

Topic 9. (9.1) Deduction of redox reactions using half-equations in acidic or neutral solutions.
Often when writing the alcohol oxidation equations in Topic 10 [O] is used to represent the oxygen from the oxidizing agent. But under topic 9 you could be asked to write the full equation and even in Topic 10 this is still true as it  states under 'Applications and skills', "Writing equations for the oxidation reactions of primary and secondary alcohols(using acidified potassium dichromate(VI) or potassium manganate(VII)  as oxidizing agents).". So use the half-equations for the reduction of dichromate(VI) ions and manganate(VII) ions in acidic solution to deduce the balanced redox equations for the oxidation of ethanol to ethanal and then to ethanoic acid. It is all good practice.

Topic 11. (11.1) Uncertainties and errors in measurement and results.
The molar molecular mass of ethanol is 46.07 g mol⁻¹. If you take 46.07 g of ethanol from the bottle marked ‘ethanol’ in the laboratory you may think you have taken 1.00 mol of ethanol. However there will be about a 4% error in this assumption as laboratory ethanol obtained by distillation is a 96% ethanol:4% water mixture. It is possible to remove the water chemically to obtain absolute ethanol, but absolute ethanol is very expensive and absorbs water when it is exposed to air as it is hygroscopic.