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
Halogenoalkanes
Reactivity of the carbon–halogen bond and the silver nitrate test
- 3 things to know
- AS Unit 2
Overview
28 recall questions on reactivity of the carbon–halogen bond and the silver nitrate test.
What you need to know
Answer each one before you reveal it.
Why is ethanol added to the mixture in the silver nitrate test?
Why must sodium hydroxide not be used in place of water in the silver nitrate test?
Questions from this subtopic
Each one with the model answer underneath.
- 1
Place the C–F, C–Cl, C–Br and C–I bonds in order of decreasing polarity, and explain the order.
Model answer
The order is C–F, then C–Cl, then C–Br, then C–I, because electronegativity falls down the group from fluorine at 4.0 through chlorine at 3.0 and bromine at 2.8 to iodine at 2.5, so the difference from carbon at 2.5 gets smaller and the C–I bond is essentially non-polar.
- 2
Place the C–F, C–Cl, C–Br and C–I bonds in order of decreasing bond enthalpy.
Model answer
The order is C–F at about 467 kJ mol⁻¹, then C–Cl at about 346 kJ mol⁻¹, then C–Br at about 290 kJ mol⁻¹, then C–I at about 228 kJ mol⁻¹, since the halogen atom gets larger down the group so the shared pair is further from its nucleus and the overlap is poorer.
- 3
If bond polarity alone decided the rate of hydrolysis, which halogenoalkane would react fastest?
Model answer
A fluoroalkane would react fastest, because the C–F bond is by far the most polar and so has the most strongly δ+ carbon for a nucleophile to attack.
- 4
Fluoroalkanes are in fact the least reactive halogenoalkanes towards hydroxide ions. Explain this apparent contradiction.
Model answer
Bond polarity governs how strongly the carbon attracts the nucleophile, but the slow step of the reaction requires the carbon–halogen bond to break, and the C–F bond is by far the strongest of the four at about 467 kJ mol⁻¹, so bond enthalpy rather than polarity is the factor that controls the rate.
- 5
Arrange 1-chlorobutane, 1-bromobutane and 1-iodobutane in order of increasing rate of hydrolysis and justify the order.
Model answer
1-Chlorobutane is slowest, 1-bromobutane is intermediate and 1-iodobutane is fastest, because the carbon–halogen bond enthalpy falls from about 346 kJ mol⁻¹ for C–Cl to about 290 kJ mol⁻¹ for C–Br and about 228 kJ mol⁻¹ for C–I, so less energy is needed to break the bond and the activation energy is lower.
- 6
Why does the carbon–halogen bond get weaker as you go down the halogen group?
Model answer
The halogen atom becomes larger, so its bonding electrons lie further from the nucleus and are shielded by more inner shells; the orbitals overlapping with carbon are also more diffuse, giving poorer overlap and a longer, weaker bond.
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