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Organic Chemistry - Alcohols (Fermentation and hydration)

Grade 11IGCSEChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Alcohols are a homologous series of organic compounds containing the hydroxyl functional group −OH-OH. Their general formula is CnH2n+1OHC_n H_{2n+1} OH.

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Ethanol (C2H5OHC_2H_5OH) is the most common alcohol and can be manufactured by two main methods: Fermentation and Catalytic Hydration.

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Fermentation involves the breakdown of glucose (C6H12O6C_6H_{12}O_6) using yeast enzymes (zymase) in anaerobic conditions (no oxygen) at approximately 30∘C30^\circ C to 35∘C35^\circ C.

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Hydration involves the addition reaction of ethene (C2H4C_2H_4) and steam (H2OH_2O) at a high temperature of 300∘C300^\circ C, a pressure of 60−7060-70 atm, and using a phosphoric acid (H3PO4H_3PO_4) catalyst.

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Fermentation is a renewable process (uses plant sugars) but is slow and produces impure ethanol in batches. Hydration is a fast, continuous process producing pure ethanol, but relies on non-renewable crude oil.

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Alcohols undergo complete combustion to produce carbon dioxide (CO2CO_2) and water (H2OH_2O). For example: C2H5OH+3O2→2CO2+3H2OC_2H_5OH + 3O_2 \rightarrow 2CO_2 + 3H_2O.

📐Formulae

CnH2n+1OHC_n H_{2n+1} OH

C6H12O6→yeast2C2H5OH+2CO2C_6H_{12}O_6 \xrightarrow{\text{yeast}} 2C_2H_5OH + 2CO_2

C2H4+H2O→60 atm300∘C,H3PO4C2H5OHC_2H_4 + H_2O \xrightarrow[60 \text{ atm}]{300^\circ C, H_3PO_4} C_2H_5OH

CxHyOH+O2→CO2+H2OC_xH_yOH + O_2 \rightarrow CO_2 + H_2O

💡Examples

Problem 1:

A student performs fermentation at 60∘C60^\circ C. Explain why the production of ethanol will likely stop at this temperature.

Solution:

The production of ethanol stops because the enzymes in the yeast become denatured at 60∘C60^\circ C.

Explanation:

Yeast is a living organism containing protein-based catalysts called enzymes. High temperatures (typically above 40∘C40^\circ C to 45∘C45^\circ C) change the shape of the enzyme's active site, preventing the substrate (glucose) from binding, thus stopping the reaction.

Problem 2:

Calculate the mass of ethene required to produce 230g230g of ethanol via the hydration process, assuming 100%100\% yield.

Solution:

Molar mass of ethanol (C2H5OHC_2H_5OH) = (2×12)+(6×1)+16=46 g/mol(2 \times 12) + (6 \times 1) + 16 = 46\text{ g/mol}. Moles of ethanol needed = 23046=5 mol\frac{230}{46} = 5\text{ mol}. The molar ratio of ethene (C2H4C_2H_4) to ethanol (C2H5OHC_2H_5OH) is 1:11:1. Moles of ethene required = 5 mol5\text{ mol}. Molar mass of ethene (C2H4C_2H_4) = (2×12)+(4×1)=28 g/mol(2 \times 12) + (4 \times 1) = 28\text{ g/mol}. Mass of ethene = 5×28=140g5 \times 28 = 140g.

Explanation:

By using the balanced equation C2H4+H2O→C2H5OHC_2H_4 + H_2O \rightarrow C_2H_5OH, we determine the stoichiometric ratio is 1:11:1. We convert the mass of the product to moles, then use the ratio to find the moles of reactant, and finally convert back to mass.

Problem 3:

Explain why fermentation must be carried out in the absence of air.

Solution:

Fermentation must be anaerobic to prevent the oxidation of ethanol into ethanoic acid (CH3COOHCH_3COOH) and to prevent the yeast from respiring aerobically, which would produce water and carbon dioxide instead of ethanol.

Explanation:

If oxygen is present, aerobic bacteria or oxygen itself can react with the ethanol formed, turning the mixture into vinegar (ethanoic acid). Additionally, yeast prefers aerobic respiration for energy, which does not yield alcohol.

Alcohols (Fermentation and hydration) Revision - Grade 11 Chemistry IGCSE