In order to develop their practical skills, students should be encouraged to carry out a range of practical experiments related to this topic. Possible
experiments include investigating the reactions of different functional groups, identifying a carbonyl compound by preparing and finding the
melting point of its 2,4-DNPH derivative, preparing and hydrolysing esters. Mathematical skills that could be developed in this topic include
representing chiral molecules with appropriate diagrams, calculating percentage yields and experimental errors. Within this topic, students can
consider how organic synthesis can produce a variety of important materials, such as esters for solvents, flavourings and perfumes. They will also
continue their study of reaction mechanisms, and see the ways in which different mechanisms act as a pattern to describe a range of organic
reactions.
Topic 17A: Chirality
1. know that optical isomerism is a result of chirality in molecules with a single chiral centre
What is the definition of stereoisomerism?
Stereoisomers are isomers with the same structural formula, but a different spatial arrangement of the atoms.
Explain what differentiates a chiral carbon from other carbons?
Chiral carbons have four different atoms or groups bonded to it. Think of it like functional groups in NMR.
2. understand that optical isomerism results from chiral centre(s) in a molecule with asymmetric carbon
atom(s) and that optical isomers are [REDACTED]
Give the definition of an optical isomer.
Optical isomers / enantiomers are the two mirror-image forms of a chiral molecule. These are nonsuperimposable
mirror images of each other.
3. know that optical activity is [REDACTED]
State the definition of ‘optical activity’.
Optical activity is the ability of a single optical isomer to rotate the plane of polarisation of plane-polarised
monochromatic light in molecules containing a single chiral centre.
4. understand the nature of a racemic mixture
What is the definition of a racemic mixture?
A racemic mixture is one that has a 50:50 mixture of two enantiomers.
5. be able to use data on optical activity of reactants and products as evidence for S N1 and SN2 mechanisms
Explain how SN1 and SN2 create optically different products.
SN1 creates new molecules via a carbocation intermediary, which is a trigonal planar
shape. Therefore, the nucleophile can attack from either above or below the plane.
This creates two different enantiomers that are produced in a 1:1 ratio – hence the
mixture is racemic and optically inactive.
SN2 occurs via ‘backside attack’, where the nucleophile can only
attack from one direction, creating one enantiomer. This ‘inverts’ the
, stereochemistry of the reactant and creates a pure optical isomer (assuming the reactant was pure), so it forms
optically active products.
Topic 17B: Carbonyl compounds
6. be able to identify the aldehyde and ketone functional groups
Give the main difference between an aldehyde and a ketone.
Aldehydes occur when a carbonyl group (C=O) is present at
the end of an alkyl chain.
Ketones occur when the carbonyl is inside the alkyl chain.
7. understand that aldehydes and ketones:
i do not form intermolecular hydrogen bonds and this affects their physical properties
Compare the boiling points of the three molecules in the table.
Although all three have similar Mrs, so similar strength of London forces, ethanal also experiences Molecule Type Mr B.P. (oC)
permanent dipole – permanent dipole forces as it is a polar molecule (propane is non-polar). Ethanal Propane Alkane 44 -42
also experiences London forces. Ethanal Aldehyde 44 +21
Ethanol Alcohol 46 +78
Furthermore, ethanol has an even higher boiling point as it contains a carboxyl group (O-H) that can take
part in hydrogen bonding, as well as pd/pd and London force. As hydrogen bonding is stronger than
pd/pd, which is stronger than London forces, ethanol has the highest boiling point, then ethanal, then
propane, since stronger forces require more energy to overcome.
ii can form hydrogen bonds with water and this affects their solubility
Explain why the solubility of carbonyl compounds decreases with chain length.
As chain length increases, the strength of London forces increases between
carbonyl molecules since more electrons are present. Hence, it is less
energetically favourable for the carbonyl to overcome these stronger London
forces and form a hydrogen bond with water, reducing their solubility.
8. understand the reactions of carbonyl compounds with:
i Fehling’s or Benedict’s solution, Tollens’ reagent and acidified dichromate(VI) ions
In equations, the oxidising agent can be represented as [O].
Explain the reaction between Fehling’s / Benedict’s solution and carbonyl compounds.
In both cases, aldehydes reduce blue copper (II) ions to copper (I) ions, which come out of solution to
give a red precipitate of copper (I) oxide (Cu2O). Bear in mind ketones do not react in this way.
Give the full ionic equation for the reaction between Fehling’s solution and ethanal.
CH3CHO + 3OH- -----> CH3COO- + 2H2O + 2e- 2e- + 2OH- + 2Cu2+ -----> Cu2O + H2O
CH3CHO + 2Cu2+ + 5OH- -----> Cu2O + CH3COO- + 3H2O
Explain the reaction between Tollens’s reagent and carbonyl compounds. Give the full ionic equation
for the reaction between Tollens’s and ethanal.
Aldehydes reduce colourless silver (I) ions in solution to silver atoms, which form a silver mirror. Bear in
mind ketones do not react in this way.
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