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Organic Chemistry - Alcohols

Grade 12IGCSEChemistry

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

🔑Concepts

Alcohols are a homologous series containing the hydroxyl functional group (OH-OH) bonded to a saturated carbon atom.

The general formula for aliphatic alcohols is CnH2n+1OHC_nH_{2n+1}OH.

Alcohols are classified as primary (11^\circ), secondary (22^\circ), or tertiary (33^\circ) based on the number of carbon atoms attached to the carbon bearing the OH-OH group.

Ethanol is manufactured by two main methods: the fermentation of glucose (using yeast at 37C37^\circ C in anaerobic conditions) and the catalytic hydration of ethene (C2H4C_2H_4) using H3PO4H_3PO_4 catalyst at 300C300^\circ C and 6060 atm pressure.

Primary alcohols can be oxidized to aldehydes (RCHORCHO) and further to carboxylic acids (RCOOHRCOOH) using acidified potassium dichromate(VI) (K2Cr2O7/H+K_2Cr_2O_7/H^+). The color change observed is from orange (Cr2O72Cr_2O_7^{2-}) to green (Cr3+Cr^{3+}).

Secondary alcohols are oxidized to ketones (RCORRCOR) using acidified K2Cr2O7K_2Cr_2O_7, whereas tertiary alcohols are resistant to oxidation under these conditions.

Elimination of water (dehydration) from alcohols using a catalyst like concentrated H2SO4H_2SO_4 or hot Al2O3Al_2O_3 yields alkenes.

Esterification occurs when an alcohol reacts with a carboxylic acid in the presence of an acid catalyst (H2SO4H_2SO_4), forming an ester and water (e.g., CH3CH2OH+CH3COOHCH3COOCH2CH3+H2OCH_3CH_2OH + CH_3COOH \rightleftharpoons CH_3COOCH_2CH_3 + H_2O).

📐Formulae

CnH2n+1OHC_nH_{2n+1}OH

C6H12O6yeast, 37C2C2H5OH+2CO2C_6H_{12}O_6 \xrightarrow{\text{yeast, } 37^\circ C} 2C_2H_5OH + 2CO_2

C2H4+H2O(g)H3PO4,300C,60 atmC2H5OHC_2H_4 + H_2O(g) \xrightarrow{H_3PO_4, 300^\circ C, 60 \text{ atm}} C_2H_5OH

RCH2OH+[O]distilRCHO+H2ORCH_2OH + [O] \xrightarrow{\text{distil}} RCHO + H_2O

RCH2OH+2[O]refluxRCOOH+H2ORCH_2OH + 2[O] \xrightarrow{\text{reflux}} RCOOH + H_2O

CH3CH(OH)CH3+[O]CH3COCH3+H2OCH_3CH(OH)CH_3 + [O] \rightarrow CH_3COCH_3 + H_2O

C2H5OHconc. H2SO4,170CC2H4+H2OC_2H_5OH \xrightarrow{\text{conc. } H_2SO_4, 170^\circ C} C_2H_4 + H_2O

💡Examples

Problem 1:

Identify the products formed when Propan-1-ol (CH3CH2CH2OHCH_3CH_2CH_2OH) is heated under reflux with excess acidified potassium dichromate(VI). Write the chemical equation.

Solution:

CH3CH2CH2OH+2[O]CH3CH2COOH+H2OCH_3CH_2CH_2OH + 2[O] \rightarrow CH_3CH_2COOH + H_2O

Explanation:

Propan-1-ol is a primary alcohol. Under reflux with excess oxidizing agent (K2Cr2O7/H+K_2Cr_2O_7/H^+), it undergoes complete oxidation to form a carboxylic acid, which in this case is propanoic acid (CH3CH2COOHCH_3CH_2COOH).

Problem 2:

Explain why ethanol (C2H5OHC_2H_5OH) has a much higher boiling point (78C78^\circ C) than propane (C3H8C_3H_8, 42C-42^\circ C), despite having a similar relative molecular mass.

Solution:

Ethanol contains a highly electronegative oxygen atom bonded to hydrogen, allowing for the formation of intermolecular hydrogen bonds.

Explanation:

Propane is a non-polar hydrocarbon with only weak London dispersion forces (Van der Waals forces). Ethanol molecules are held together by much stronger hydrogen bonds between the OH-OH groups, requiring significantly more thermal energy to overcome during boiling.

Problem 3:

Write the balanced equation for the combustion of ethanol in excess oxygen.

Solution:

C2H5OH+3O22CO2+3H2OC_2H_5OH + 3O_2 \rightarrow 2CO_2 + 3H_2O

Explanation:

In excess oxygen, alcohols undergo complete combustion to produce carbon dioxide (CO2CO_2) and water (H2OH_2O). This reaction is highly exothermic, making ethanol a useful fuel.