WJEC A-Level Chemistry 1410QS
Formulae and Equations
Basic Ideas About Atoms
Chemical Calculations
Bonding
Solid Structures
The Periodic Table
Simple Equilibria and Acid-Base Reactions
Thermochemistry
Rates of Reaction
The Wider Impact of Chemistry
Organic Compounds
Hydrocarbons
Halogenoalkanes
Alcohols and Carboxylic Acids
Instrumental Analysis
Redox and Standard Electrode Potential
Redox Reactions
Chemistry of the P-Block
Chemistry of the D-Block Transition Metals
Chemical Kinetics
Enthalpy Changes for Solids and Solutions
Entropy and Feasibility of Reactions
Equilibrium Constants
Acid-Base Equilibria
Stereoisomerism
Aromaticity
Alcohols and Phenols
Aldehydes and Ketones
Carboxylic Acids and Their Derivatives
Amines
Amino Acids, Peptides and Proteins
Organic Synthesis and Analysis
Practical Skills and Techniques
Planning, Analysis and Evaluation
Core Experiments and Their Chemistry
Mathematical Skills in Chemistry
Hydrocarbons
Electrophilic addition, Markovnikov's rule and testing for alkenes: testing for a C=C bond
- 3 things to know
- AS Unit 2
Overview
Eleven recall questions on electrophilic addition, Markovnikov's rule and testing for alkenes: testing for a C=C bond.
What you need to know
Answer each one before you reveal it.
How would you show experimentally that a colourless liquid is an alkene rather than an alkane?
Chlorine also adds to ethene. Write the equation and name the product.
Questions from this subtopic
Each one with the model answer underneath.
- 1
A sample of 0.560 g of an alkene containing one double bond reacts exactly with 8.00 cm³ of a 1.25 mol dm⁻³ solution of bromine. Determine the molecular formula of the alkene.
Model answer
The amount of bromine used is 0.00800 dm³ × 1.25 mol dm⁻³ = 0.0100 mol, and since one C=C bond adds one Br₂ molecule the alkene is also 0.0100 mol. Its molar mass is therefore 0.560 ÷ 0.0100 = 56.0 g mol⁻¹, and as an alkene has the general formula CnH2n with 14.0 g mol⁻¹ per unit, n = 56.0 ÷ 14.0 = 4, so the alkene is C₄H₈.
- 2
Burning 0.100 g of an alkene produces 0.314 g of carbon dioxide and 0.129 g of water. Determine its empirical formula, and its molecular formula given that its relative molecular mass is 56.0.
Model answer
The amount of carbon dioxide is 0.314 ÷ 44.0 = 7.14 × 10⁻³ mol, giving 7.14 × 10⁻³ mol of carbon atoms. The amount of water is 0.129 ÷ 18.0 = 7.17 × 10⁻³ mol, giving twice as many hydrogen atoms, 1.43 × 10⁻² mol. The ratio of carbon to hydrogen is 7.14 × 10⁻³ : 1.43 × 10⁻², which is 1 : 2, so the empirical formula is CH₂ with a mass of 14.0. Since 56.0 ÷ 14.0 = 4, the molecular formula is C₄H₈.
- 3
An unknown hydrocarbon contains 85.7% carbon and 14.3% hydrogen by mass and has a relative molecular mass of 56.0. Determine its molecular formula and identify the class of compound.
Model answer
In 100 g there are 85.7 ÷ 12.0 = 7.14 mol of carbon and 14.3 ÷ 1.0 = 14.3 mol of hydrogen, a ratio of 1 : 2, so the empirical formula is CH₂. As the empirical formula mass is 14.0 and 56.0 ÷ 14.0 = 4, the molecular formula is C₄H₈, which fits the general formula CnH2n and so is an alkene.
- 4
Why is the first step of an electrophilic addition the slow, rate-determining step?
Model answer
Forming the carbocation requires the π-bond and the bond in the attacking molecule both to break, so it has a high activation energy, whereas the second step simply brings together an anion and a positively charged carbon, which is fast.
- 5
Chlorine also adds to ethene. Write the equation and name the product.
Model answer
The equation is C₂H₄ + Cl₂ → CH₂ClCH₂Cl, and the product is 1,2-dichloroethane.
- 6
Explain why the bromide ion attacks the carbocation rather than the carbocation attacking the bromide ion.
Model answer
The bromide ion is the electron-rich species, with lone pairs and a full negative charge, so it acts as the nucleophile and donates a pair of electrons to the electron-deficient positive carbon, which is the electrophile in that step.
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