Reactions Involving Alcohols


This is part of the HSC Chemistry course under the topic Alcohols.

HSC Chemistry Syllabus

  • investigate the production of alcohols, including: hydration of alkenes, substitution reactions of halogenated organic compounds and fermentation

  • conduct a practical investigation to measure and reliably compare the enthalpy of combustion for a range of alcohols

  • write equations, state conditions and predict products to represent the reactions of alcohols, including but not limited to (ACSCH128, ACSCH136):

– combustion
– dehydration
– substitution with HX
– oxidation


    • investigate the products of the oxidation of primary and secondary alcohols

    Production of Alcohols

    Hydration of Alkene

    • An alkane can react with water in the presence of an acid catalyst (H+/H2SO4 (aq)) to form an alcohol.


    • Alkene hydration follows Markovnikov’s rule – the major alcohol product will be the one where the –OH group is connected to the carbon atom that has the most alkyl groups attached.


    • It is difficult to produce primary alcohols using acid-catalysed hydration of alkenes. This is because primary alcohol requires the formation of a primary carbocation which is the least stable.


    Substitution of Halogenated Organic Compounds

    • Reaction between a halogenated alkane with hydroxide ions e.g. NaOH produces an alcohol as the halogen is substituted (replaced) by the OH
    • Substitution of halogenated alkane to produce alcohol can also be performed using water as the reagent. However, this is less effective as water is a poorer nucleophile compared to hydroxide ion.



    • Substitution of halogenated alkane does not produce more than one organic product.
    • The position of alcohol (–OH) depends on position of the halogen atom(s).


    Fermentation of Carbohydrates

    • Fermentation of glucose (C6H12O6) by yeast produces ethanol and carbon dioxide.
    • Fermentation is a renewable method of producing alcohol as carbohydrates, such as glucose, are derived from plants. For example, sugar canes are rich in glucose and therefore, are regarded as good raw materials for producing ethanol via fermentation.



    • Fermentation requires specific reaction conditions:





    Enzyme (biological catalyst) is required for fermentation.

    Anaerobic environment (absence of oxygen)


    • Yeasts only catalyse fermentation when oxygen supply is insufficient.
    • Ethanol can be oxidised to produce acetic acid in the presence of oxygen


    Dilute solution of carbohydrate

    If starting carbohydrate concentration is too high, large amount of alcohol is produced which damages and kills yeast cells. Ideally, the alcohol content is kept < 15%.

    Temperature of 37ºC

    This is an optimal temperature for yeast’s enzymatic activity. If the temperature is lower than 37ºC, the reaction rate is lowered. If the temperature is greater than 37ºC, yeast cells will die.

    Combustion of Alcohols and Enthalpy of Combustion 


    • All alcohols undergo combustion with oxygen to produce carbon dioxide and water (complete combustion). For example, the combustion of ethanol:


    $$2C_2H_5OH(l) + 2O_2(g) \rightarrow 2CO_2(g) + 4H_2O(l)$$


    • Combustion of alcohol is exothermic which means the energy released during bond formation of products is greater than the energy absorbed during bond breaking of reactants. In alcohol combustion, all of the energy comes from the formation of C=O bonds in CO2 and H–O bonds in water.


    Bond type

    Energy absorbed or released

    (kJ mol–1)

    Number of extra bond broken/formed for each new carbon atom

    Change in enthalpy for each new carbon atom (kJ mol–1)




    1 × 347 = 347




    2 × 413 = 826




    2 × –746 = –1492




    2 × –464 = –928


    • Molar ∆H (in kJ mol–1) increases with alcohol size.

    For each additional carbon atom added to an alcohol molecule, two extra C–H bonds and one extra C–H bond are broken during combustion. This is followed by the formation of two extra C=O bonds in a carbon dioxide molecule and two extra O–H bonds in a water molecule. Thus, as shown in table above, adding carbon atoms to an alcohol molecule causes its combustion to produce more energy i.e. ∆H to become more negative.


    Table: heat of combustion and molecular weight of first eight straight-chained alcohols.


    • ∆H (in kJ g–1) increases with alcohol size.

    For each additional carbon atom added to an alcohol molecule, its molecular mass increases constantly (1 carbon atom and 2 hydrogen atoms). Since O–H bonds in alcohol molecules are not broken during combustion, mass of oxygen atom(s) (16.00 g mol–1) does not contribute to ∆H of combustion. As carbon atoms are added to increase the size of alcohol, oxygen’s contribution to molecular mass becomes smaller, leading to an increase in ∆H (in kJ g–1).

    • Position and chain isomers have the same heat of combustion (kJ mol–1 and kJ g–1) because the bonds broken and formed are identical despite their differences in structure. For example, 1-propanol and 2-propanol are position isomers.


    Number of bonds broken

    C–H = 7

    C–C = 2

    C–O = 1


    Number of bonds broken

    C–H = 7

    C–C = 2

    C–O = 1



    Investigating ∆cH of Alcohol

    • Heat of combustion (∆cH) of an alcohol is the amount of heat energy it releases per mole or gram of alcohol consumed.
    • An example of a first-hand investigation to measure a 1-propanol’s heat of combustion:


    1.     1-propanol is added to a spirit burner and the total mass is weighed on an electronic balance.

    2.     A known quantity of water is added to a 500 mL beaker.

    3.     The initial temperature of water is recorded with a thermometer.

    4.     The spirit burner is lit with a match.

    5.     The water is gently stirred with a thermometer and the change is temperature is monitored.



    1. The spirit burner is extinguished after a substantial increase in temperature is observed.
    2. The final temperature (highest) temperature of water is recorded.
    3. The mass of spirit burner and 1-propanol is re-weighed on an electronic balance and recorded.
    4. The heat absorbed by the water is calculated:



    1. The heat of combustion is calculated by dividing by the mole of alcohol consumed.


    $$\Delta{H} = \frac{-q}{n}$$ 

    n(alcohol) = mass change of spirit burner molecular weight of alcohol used


    • The experimental results of this first-hand investigation are invalid because
    • Incomplete combustion may occur due to an insufficient supply of oxygen to the spirit burner. This can be easily identified by the formation of soot underneath the beaker. Incomplete combustions have a lower heat of combustion than complete combustions.
    • It is unlikely that all the heat produced by combustion will be absorbed by the water in the beaker. The increase in temperature only represents a fraction of what’s released.


    • To improve experimental validity, two main actions can be taken:
    1. Minimise heat loss by
      • Decreasing the distance between the flame and the beaker
      • Eliminating sources of air current which may carry away the heat that is produced
      • Minimise heat dissipated from the beaker by surrounding it with insulating material e.g. polystyrene.


    1. Prevent incomplete combustion by
      • Using another combustion apparatus as it is difficult to increase oxygen to fuel supply with a spirit burner.


    • If the procedure is repeated in the same manner, its results would be reliable i.e consistent.


    • The results could be made more accurate if numerical values are measured correctly. This includes temperature change of water, mass of water used, mass of alcohol used. However, the invalid experimental set-up would significantly reduce the accuracy of results. In other words, improving the validity will also improve accuracy.



    Reactions Involving Alcohols: Dehydration, Substitution with HX and Oxidation


    Dehydration of Alcohol

    • In the presence of concentrated acid e.g. concentrated H2SO4, an alcohol (hydroxyl) functional group can be eliminated to form an alkene and water.


    Dehydration of ethanol produces ethene and water


    • Dehydration is considered as an elimination reaction, opposite to hydration of an alkene (addition reaction). The two reactions are reversible, with dehydration being endothermic and hydration exothermic.
    • Reagents and reaction condition:
      • Concentrated H2SO4 (catalyst, increases reaction rate)
      • Heat (Increases yield of alkene as dehydration is endothermic)


    • Position isomers are produced during dehydration reaction of either a secondary (2º) or tertiary (3º) alcohol. The major product is always more substituted – the resultant C=C bond prefers to be formed between non-terminal carbon atoms. This is known as Zaitsev’s rule.


    Zaitsev’s rule states that:


     In an elimination reaction, the major product will be one where the resultant double carbon-carbon bond is formed in a more substituted position.


    Substitution of Alcohol with HX

    • When halogen halide is supplied in large concentration, the –OH group can be directly substituted by a halogen. This transforms an alcohol into a mono-substituted haloalkane.
    • Reagents:
      • Concentrated hydrogen halide e.g. HCl
      • ZnCl2 (catalyst) 


    • Substitution of tertiary and secondary alcohols with HX typically does not require ZnCl2 as a catalyst as the reaction proceeds through a favourable, stable tertiary and secondary carbocation respectively. In contrast, Substitution of primary alcohol with HX always requires a catalyst.


    • This method of forming a haloalkane is slightly different to hydrohalogenation of an alkene.
      • In the reaction between an alkene and a hydrogen halide, the halogen is selectively connected to the carbon with more alkyl substituents.
      • In contrast, the position of the halogen in a haloalkane (produced from substitution with alcohol) will always depend on the position of its former alcohol.


    For example,


    Oxidation of Alcohol

    • Oxidation states of carbon atoms in organic molecules
      • Every bond between C and another C does not alter the oxidation state
      • Every bond between C and H will decrease the oxidation state by 1
      • Every bond between C and a more electronegative element will increase the oxidation state by 1



    • Oxidation of primary (1º) alcohol yields aldehyde as an intermediate and carboxylic acid as the final product.



    • Oxidation of secondary (2º) alcohol produces ketone. Ketone cannot be oxidised further as this would break a C–C bond which requires too much energy.





    • Tertiary (3º) alcohols cannot be oxidised because this would otherwise involve breaking a C–C bond (sigma bond) which requires too much energy.




    Oxidation state change

    –1 to +1


    +1 to +3

    0 to +2

    No oxidation occurs