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Chemistry - Alkenes

Grade 9IGCSE

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

🔑Concepts

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Alkenes are a homologous series of unsaturated hydrocarbons. 'Unsaturated' means they contain at least one carbon-carbon double bond (C=CC=C).

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The general formula for alkenes is CnH2nC_n H_{2n}. This means they have twice as many hydrogen atoms as carbon atoms.

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Functional Group: The C=CC=C double bond is the functional group, which makes alkenes more reactive than alkanes.

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Cracking: Alkenes are produced by breaking down long-chain alkanes into smaller, more useful molecules using heat and a catalyst (e.g., Al2O3Al_2 O_3 or SiO2SiO_2).

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Addition Reactions: Because of the double bond, alkenes can undergo addition reactions where the double bond 'opens up' to allow new atoms to bond to the carbon atoms.

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Test for Unsaturation: When an alkene is added to orange/brown bromine water (Br2Br_2), the solution becomes colorless. This is because the bromine adds across the double bond to form a colorless dibromoalkane.

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Hydrogenation: The addition of hydrogen (H2H_2) to an alkene to form an alkane, typically using a nickel catalyst at 150∘C150^\circ C.

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Hydration: The addition of steam (H2OH_2 O) to an alkene to produce an alcohol, using a phosphoric acid catalyst (H3PO4H_3 PO_4) at 300∘C300^\circ C and 6060 atm pressure.

📐Formulae

CnH2nC_n H_{2n}

C2H4+3O2→2CO2+2H2OC_2 H_4 + 3O_2 \rightarrow 2CO_2 + 2H_2 O

C2H4+Br2→C2H4Br2C_2 H_4 + Br_2 \rightarrow C_2 H_4 Br_2

C2H4+H2→Ni,150∘CC2H6C_2 H_4 + H_2 \xrightarrow{Ni, 150^\circ C} C_2 H_6

C2H4+H2O→H3PO4,300∘CC2H5OHC_2 H_4 + H_2 O \xrightarrow{H_3 PO_4, 300^\circ C} C_2 H_5 OH

💡Examples

Problem 1:

Identify the molecular formula and name of the alkene with 44 carbon atoms.

Solution:

C4H8C_4 H_8 (Butene)

Explanation:

Using the general formula CnH2nC_n H_{2n}, if n=4n = 4, then the number of hydrogens is 2×4=82 \times 4 = 8. The prefix for 4 carbons is 'but-', and the suffix for alkenes is '-ene'.

Problem 2:

What is observed when propene (C3H6C_3 H_6) is bubbled through a solution of bromine in water?

Solution:

The orange-brown color of the bromine water disappears, and the solution becomes colorless.

Explanation:

Propene is an alkene. It reacts with bromine (Br2Br_2) via an addition reaction to form 1,2−dibromopropane1,2-dibromopropane (C3H6Br2C_3 H_6 Br_2), which is a colorless liquid.

Problem 3:

Calculate the mass of ethene (C2H4C_2 H_4) required to produce 460460 g of ethanol (C2H5OHC_2 H_5 OH) via hydration. (Ar:C=12,H=1,O=16A_r: C=12, H=1, O=16)

Solution:

Mr(C2H5OH)=(2×12)+(6×1)+16=46 g/molM_r(C_2 H_5 OH) = (2 \times 12) + (6 \times 1) + 16 = 46 \text{ g/mol} Moles of ethanol=46046=10 molMoles\ of\ ethanol = \frac{460}{46} = 10 \text{ mol} C2H4+H2O→C2H5OHC_2 H_4 + H_2 O \rightarrow C_2 H_5 OH 1 mole of C2H4 produces 1 mole of C2H5OH1\text{ mole of } C_2 H_4 \text{ produces } 1 \text{ mole of } C_2 H_5 OH Moles of ethene needed=10 molMoles\ of\ ethene\ needed = 10 \text{ mol} Mr(C2H4)=(2×12)+(4×1)=28 g/molM_r(C_2 H_4) = (2 \times 12) + (4 \times 1) = 28 \text{ g/mol} Mass of ethene=10×28=280 gMass\ of\ ethene = 10 \times 28 = 280 \text{ g}

Explanation:

The stoichiometric ratio between ethene and ethanol is 1:11:1. By finding the moles of the product (ethanol), we can determine the moles of reactant (ethene) needed and convert that back into mass.