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: mechanism of electrophilic addition
- 3 things to know
- AS Unit 2
Overview
Recall questions on Markovnikov's rule.
What you need to know
Answer each one before you reveal it.
Why is a secondary carbocation more stable than a primary one?
Identify the major products when hydrogen bromide adds to propene and to 2-methylbut-2-ene, and explain why the more stable carbocation gives the major product.
Questions from this subtopic
Each one with the model answer underneath.
- 1
Bromine is a non-polar molecule, so how can it act as an electrophile towards ethene?
Model answer
As the bromine molecule approaches, the high electron density of the π-bond repels the electron pair in the Br–Br bond towards the far bromine atom, so a dipole is induced with the nearer atom becoming δ+ and the far atom δ−, and it is that δ+ atom that is attacked.
- 2
Describe the mechanism for the reaction between ethene and bromine, saying where each curly arrow starts and ends.
Model answer
The first curly arrow starts at the C=C π-bond and ends at the δ+ bromine atom, forming a C–Br bond; at the same time a second curly arrow starts at the Br–Br bonding pair and ends on the δ− bromine atom, so the Br–Br bond breaks heterolytically to give a bromide ion and a carbocation. A third curly arrow then starts at a lone pair on the bromide ion and ends at the positively charged carbon, forming the second C–Br bond and giving 1,2-dibromoethane.
- 3
What kind of bond fission occurs in the Br–Br bond during the addition of bromine to ethene?
Model answer
It is heterolytic fission, because both of the bonding electrons go to the bromine atom furthest from the double bond, which leaves as a bromide ion while the other bromine becomes attached to carbon.
- 4
Write the equation for the addition of bromine to ethene and name the product.
Model answer
The equation is C₂H₄ + Br₂ → CH₂BrCH₂Br, and the product is 1,2-dibromoethane.
- 5
Name the intermediate formed halfway through the electrophilic addition of bromine to ethene and say why it is so reactive.
Model answer
The intermediate is a carbocation, in this case CH₂BrCH₂⁺; it is highly reactive because the positively charged carbon has only six electrons in its outer shell, so it readily accepts a lone pair from the bromide ion.
- 6
Why is the addition of bromine to an alkene classified as electrophilic addition rather than nucleophilic addition?
Model answer
The reaction is started by the species that accepts electrons from the double bond, that is, the δ+ bromine atom acting as an electrophile, and the overall result is that both atoms of the bromine molecule are added across the C=C bond.
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