AQA A-Level Organic Chemistry 3.3.5 Alcohols Classification and Reactions
3.3.5

Classification and Reactions of Alcohols

Understand how to classify alcohols, write combustion reactions, and compare industrial pathways for ethanol production.

Alcohols are a family of organic compounds containing the hydroxyl group, \( -\text{OH} \), bonded to a saturated carbon atom. They are widely used as solvents, chemical intermediates, and fuels. In this lesson, we cover alcohol classification, their complete combustion, and the two major industrial processes used to manufacture ethanol.

🔑 Key Principle

The chemical behaviour of an alcohol is heavily influenced by its structure, specifically the number of alkyl groups attached to the carbon holding the hydroxyl group. This structural classification directly dictates their reactivity, particularly towards oxidising agents.

Classification of Alcohols

Alcohols are classified as primary, secondary, or tertiary based on the environment of the carbon atom that is directly bonded to the \( -\text{OH} \) group:

Primary (1°) Alcohol

An alcohol in which the carbon atom bonded to the \( -\text{OH} \) group is attached to only one other carbon atom (or zero other carbon atoms, as in methanol, \( \text{CH}_3\text{OH} \)).

Secondary (2°) Alcohol

An alcohol in which the carbon atom bonded to the \( -\text{OH} \) group is attached to exactly two other carbon atoms (alkyl groups).

Tertiary (3°) Alcohol

An alcohol in which the carbon atom bonded to the \( -\text{OH} \) group is attached to three other carbon atoms (alkyl groups).

📝 AQA Examiner Tip

When identifying the classification of an alcohol, locate the \( \text{C-OH} \) carbon first. Count the number of alkyl groups attached directly to that carbon. Do not get confused by long chains: only the carbon atoms directly bonded to the \( \text{C-OH} \) carbon determine the classification.

Classification of Alcohols (1°, 2°, 3°) Alcohol Classification and Structures Primary (1°) One R group on C-OH C OH H H R Example: Ethanol CH₃CH₂OH Secondary (2°) Two R groups on C-OH C OH H R₂ R₁ Example: Propan-2-ol CH₃CH(OH)CH₃ Tertiary (3°) Three R groups on C-OH C OH R₃ R₂ R₁ Example: 2-Methylpropan-2-ol (CH₃)₃COH
✏️ Worked Example: Classifying Alcohols
Classify the following alcohols as primary, secondary, or tertiary:
  1. Pentan-3-ol
  2. 2-methylbutan-2-ol
  3. 2,2-dimethylpropan-1-ol

Solution:

  1. Pentan-3-ol: The structural formula is \( \text{CH}_3\text{CH}_2\text{CH(OH)CH}_2\text{CH}_3 \). The carbon bonded to the \( -\text{OH} \) group is attached to two other carbon atoms (an ethyl group on either side). Therefore, it is a secondary (2°) alcohol.
  2. 2-methylbutan-2-ol: The structural formula is \( \text{CH}_3\text{CH}_2\text{C(OH)(CH}_3)_2 \). The carbon holding the \( -\text{OH} \) group is attached directly to three other carbon atoms (an ethyl group and two methyl groups). Therefore, it is a tertiary (3°) alcohol.
  3. 2,2-dimethylpropan-1-ol: The structural formula is \( (\text{CH}_3)_3\text{C-CH}_2\text{-OH} \). The carbon atom holding the \( -\text{OH} \) group is attached directly to only one other carbon atom (the quaternary carbon). Therefore, it is a primary (1°) alcohol.

Combustion of Alcohols

Alcohols make clean-burning fuels because they contain oxygen within their structures, which aids complete combustion. When burned in a plentiful supply of oxygen, they undergo complete combustion to produce carbon dioxide and water:

\[ \text{C}_2\text{H}_5\text{OH(l)} + 3\text{O}_2\text{(g)} \rightarrow 2\text{CO}_2\text{(g)} + 3\text{H}_2\text{O(l)} \quad \Delta H = -1367\text{ kJ mol}^{-1} \]

This reaction is highly exothermic, which explains why ethanol is commonly blended with petrol (as E10 fuel) to reduce reliance on pure fossil fuels.

Industrial Ethanol Production

There are two key chemical processes used to produce ethanol on a commercial scale: the hydration of ethene (a synthetic pathway derived from crude oil) and fermentation (a biochemical pathway using crops). You must know the conditions, equations, and comparative metrics for both.

Pathway 1: Hydration of Ethene

Ethene is obtained by cracking fractions of crude oil. It is reacted with steam at high temperature and pressure in the presence of an acid catalyst to produce ethanol:

\[ \text{C}_2\text{H}_4\text{(g)} + \text{H}_2\text{O(g)} \rightleftharpoons \text{C}_2\text{H}_5\text{OH(g)} \]

Pathway 2: Fermentation of Glucose

Glucose is obtained from crops containing starch or sugars (such as sugarcane, maize, or sugar beet). Yeast is added, which produces the zymase enzyme to catalyse the breakdown of glucose into ethanol and carbon dioxide:

\[ \text{C}_6\text{H}_{12}\text{O}_6\text{(aq)} \rightarrow 2\text{C}_2\text{H}_5\text{OH(aq)} + 2\text{CO}_2\text{(g)} \]
📝 AQA Examiner Tip

Remember the temperature constraint for fermentation: if the temperature exceeds \( 40^\circ\text{C} \), the zymase enzymes in the yeast will denature, permanently stopping the reaction. If the temperature is too low (e.g. below \( 25^\circ\text{C} \)), the reaction rate becomes too slow due to low kinetic energy, although yeast is not killed.

Comparing the Industrial Pathways

You must be able to contrast these two methods across several criteria. Use this comparative table for your revision:

Criteria Hydration of Ethene Fermentation of Glucose
Raw Materials Crude oil (ethene) and steam (non-renewable) Sugars/starch from crops (renewable)
Reaction Rate Fast (takes seconds) Slow (takes several days)
Product Purity High (essentially 100% pure) Low (produces 10-15% ethanol; fractional distillation needed to concentrate)
Operating Conditions High temp (\(300^\circ\text{C}\)) and pressure (\(65\text{ atm}\)); high energy cost Low temp (\(35^\circ\text{C}\)) and atmospheric pressure; low energy cost
Process Type Continuous (highly automated, low labour costs) Batch (requires shutting down, cleaning, and restarting; high labour costs)
Atom Economy 100% (addition reaction; no waste products) 51.1% (produces waste \( \text{CO}_2 \) as a co-product)
✏️ Worked Example: Calculating Atom Economy
Calculate and compare the atom economy for the production of ethanol via:
a) Hydration of ethene: \( \text{C}_2\text{H}_4 + \text{H}_2\text{O} \rightarrow \text{C}_2\text{H}_5\text{OH} \)
b) Fermentation of glucose: \( \text{C}_6\text{H}_{12}\text{O}_6 \rightarrow 2\text{C}_2\text{H}_5\text{OH} + 2\text{CO}_2 \)
Use the relative formula masses: \( M_r(\text{C}_2\text{H}_4) = 28.0 \), \( M_r(\text{H}_2\text{O}) = 18.0 \), \( M_r(\text{C}_2\text{H}_5\text{OH}) = 46.0 \), \( M_r(\text{C}_6\text{H}_{12}\text{O}_6) = 180.0 \), \( M_r(\text{CO}_2) = 44.0 \).

Solution:

Atom economy is calculated using the formula:

\[ \text{Atom Economy} = \frac{\text{Total } M_r \text{ of desired product}}{\text{Total } M_r \text{ of all reactants}} \times 100 \]

a) Hydration of Ethene:

The desired product is ethanol (\( M_r = 46.0 \)). The reactants are ethene (\( 28.0 \)) and water (\( 18.0 \)).

\[ \text{Atom Economy} = \frac{46.0}{28.0 + 18.0} \times 100 = \frac{46.0}{46.0} \times 100 = 100\% \]

This is an addition reaction, so all reactant atoms end up in the desired product, giving a 100% atom economy.

b) Fermentation of Glucose:

The desired product is 2 moles of ethanol (\( 2 \times 46.0 = 92.0 \)). The reactant is glucose (\( M_r = 180.0 \)).

\[ \text{Atom Economy} = \frac{2 \times 46.0}{180.0} \times 100 = \frac{92.0}{180.0} \times 100 = 51.1\% \]

Even though fermentation is carbon-neutral when crops grow, its chemical atom economy is low because carbon dioxide is a waste co-product.

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