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
Organic Synthesis and Analysis
Planning a synthesis as a sequence of reactions
- 3 things to know
- A2 Unit 4
Overview
24 recall questions on planning a synthesis as a sequence of reactions.
What you need to know
Answer each one before you reveal it.
Suggest a three-step route from propan-1-ol to butanoic acid, giving reagents and conditions for each step.
What reagents and conditions convert propan-2-ol into propanone?
Questions from this subtopic
Each one with the model answer underneath.
- 1
Why can a target molecule often not be made from the available starting material in a single step?
Model answer
Because the target usually differs from the starting material in two ways at once, in the functional group present and in the number of carbon atoms, and no single reagent makes both changes. Each reagent performs one specific conversion, so a sequence of steps is needed, each one moving the molecule closer to the target.
- 2
Name the two general ways of lengthening a carbon chain by one atom in organic synthesis, and state what each produces.
Model answer
A halogenoalkane can be heated under reflux with potassium cyanide dissolved in ethanol, giving a nitrile with one more carbon atom. Alternatively an aldehyde or ketone can be treated with hydrogen cyanide, generated from potassium cyanide and dilute acid, giving a hydroxynitrile with one more carbon atom. In both cases the new carbon arrives as the −CN group.
- 3
Suggest a three-step route from propan-1-ol to butanoic acid, giving reagents and conditions for each step.
Model answer
Step 1: heat propan-1-ol under reflux with sodium bromide and concentrated sulfuric acid to give 1-bromopropane. Step 2: heat 1-bromopropane under reflux with potassium cyanide in ethanol to give butanenitrile, adding the fourth carbon atom. Step 3: heat butanenitrile under reflux with dilute hydrochloric acid to hydrolyse the nitrile to butanoic acid.
- 4
How would you convert bromoethane into propan-1-amine?
Model answer
First heat the bromoethane under reflux with potassium cyanide in ethanol. The cyanide ion substitutes the bromine to give propanenitrile, CH₃CH₂CN, which has three carbon atoms. Then reduce the nitrile, either with lithium tetrahydridoaluminate in dry ether or with hydrogen over a nickel catalyst, to give propan-1-amine, CH₃CH₂CH₂NH₂.
- 5
Ethanal can be converted into 2-hydroxypropanoic acid in two steps. Give the reagents, and explain why the product is not optically active.
Model answer
Ethanal is treated with hydrogen cyanide, made in situ from potassium cyanide and dilute sulfuric acid, giving 2-hydroxypropanenitrile, CH₃CH(OH)CN. This is then heated under reflux with dilute hydrochloric acid, which hydrolyses the nitrile group to a carboxylic acid, giving CH₃CH(OH)COOH. The product has a chiral centre, but the carbonyl carbon of ethanal is trigonal planar, so the cyanide ion attacks with equal probability from either face. Equal amounts of the two enantiomers are formed, and their rotations cancel, so the mixture is not optically active.
- 6
What reagents and conditions convert propan-2-ol into propanone?
Model answer
Heat the propan-2-ol under reflux with potassium dichromate(VI) acidified with dilute sulfuric acid. The secondary alcohol is oxidised to the ketone propanone, and the orange dichromate(VI) turns green as it is reduced to chromium(III). No further oxidation occurs, because a ketone has no hydrogen atom on the carbonyl carbon.
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