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
Aromaticity
Electrophilic substitution: nitration, halogenation and alkylation
- 3 things to know
- A2 Unit 4
Overview
Recall questions on nitration, halogenation and alkylation.
What you need to know
Answer each one before you reveal it.
Write equations for the alkylation of benzene by chloromethane, showing the electrophile.
Why can a Friedel-Crafts alkylation give a mixture containing polysubstituted products?
Questions from this subtopic
Each one with the model answer underneath.
- 1
Why does an electrophile attack the ring of methylbenzene rather than its methyl side chain?
Model answer
The delocalised π system of the ring is a region of high electron density that attracts an electron-pair acceptor, whereas the methyl group is a saturated alkyl group with no such region and no lone pair, so it offers the electrophile nothing to attack.
- 2
Why is benzene attacked by electrophiles rather than by nucleophiles?
Model answer
The ring carries six delocalised pi electrons in clouds above and below its plane, so it is a region of high electron density. That attracts electron-deficient species, while nucleophiles, which are themselves electron rich, are repelled.
- 3
Outline the general pattern of electrophilic substitution in benzene.
Model answer
The electrophile is attracted to the delocalised ring, and two of the pi electrons form a bond to it, giving a positively charged intermediate in which the delocalisation is broken and spread over only five carbons. A base then removes the hydrogen from the carbon that was attacked, and the electron pair from that bond returns to the ring, restoring the full delocalised system.
- 4
Why is the intermediate formed during electrophilic substitution unstable?
Model answer
Forming the bond to the electrophile uses two of the six delocalised electrons, so the intermediate is left with only four spread over five carbon atoms and carries a positive charge. It has lost the aromatic stabilisation of the parent ring, so it is high in energy and reverts to an aromatic structure as fast as it can.
- 5
Why does the intermediate lose a proton rather than gain a nucleophile?
Model answer
Losing the proton returns the electron pair to the ring and restores the delocalised system, recovering the aromatic stabilisation energy. Adding a nucleophile would give a saturated product with no aromatic ring, which is much higher in energy.
- 6
State the reagents and conditions for the nitration of benzene.
Model answer
Concentrated nitric acid with concentrated sulfuric acid as catalyst, warmed under reflux to about 50 to 55 °C. Keeping the temperature below about 55 °C limits the reaction to a single substitution, since a hotter mixture goes on to give dinitrobenzene.
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