AQA · A-Level · Chemistry

    Course version: 7405 · Version from Sept 2016

    Aldehydes and unsymmetrical ketones form mixtures of enantiomers when they…

    Full specification wording

    Aldehydes and ketones (A-level only): Aldehydes and unsymmetrical ketones form mixtures of enantiomers when they react with KCN followed by dilute acid.

    The carbonyl carbon is trigonal planar, so the cyanide ion can approach from above or below the plane with equal probability.

    Checked against the AQA 7405 specification (opens in a new tab) on .

    • Point 5 of 7
    • 4 key terms

    Study notes

    From the checked teaching notes for this specification point.

    The idea

    The carbonyl carbon is trigonal planar, so the cyanide ion can approach from above or below the plane with equal probability. In the product, that carbon carries four groups: hydroxyl, nitrile, and the two original groups. If these two are different, the carbon is chiral. Equal approach from both faces produces equal amounts of the two enantiomers, forming an optically inactive racemate. However, if the original groups are identical, no chiral centre forms. Methanal (two hydrogens) and symmetrical ketones like propanone (two methyls) are key exceptions; their reaction with KCN/H+ yields achiral products that do not exhibit optical isomerism.

    What earns the marks

    • Stating that the carbonyl carbon is planar.
    • Saying the cyanide ion attacks from above or from below with equal probability.
    • Identifying the new chiral centre as a carbon with four different groups.
    • Stating that equal amounts of the two enantiomers are formed, so the product is a racemate and optically inactive.

    Common mistakes and how to put them right

    • Saying the product is chiral without explaining that attack occurs equally from both faces.

    • Claiming methanal or propanone gives a racemate, when their products lack a chiral centre due to identical groups.

    • Arguing that one enantiomer forms in excess because it is more stable.

    • Describing the planar species as a carbocation, when here it is the neutral carbonyl carbon.

    Exam tips

    • Check the four groups on the new carbon before claiming enantiomers; two identical groups (e.g., in methanal) means none.
    • Use the three-step chain: planar carbon, attack from either side, equal amounts of enantiomers.
    • Finish by saying the mixture is optically inactive, since that is often the final mark.

    Try the lesson

    The start of the Aldehydes and ketones (A-level only): Aldehydes and unsymmetrical ketones form mixtures of enantiomers when they react with KCN followed by dilute acid. lesson, made from checked teaching notes for this specification point.

    Have a go first

    Which compound forms a racemic mixture of enantiomers when it reacts with KCN followed by dilute acid?

    Listen

    The carbonyl carbon is trigonal planar, so the cyanide ion can approach from above or below the plane with equal probability. In the product, that carbon carries four groups: hydroxyl, nitrile, and the two original groups. If these two are different, the carbon is chiral. Equal approach from both faces produces equal amounts of the two enantiomers, forming an optically inactive racemate. However, if the original groups are identical, no chiral centre forms. Methanal (two hydrogens) and symmetrical ketones like propanone (two methyls) are key exceptions; their reaction with KCN/H+ yields achiral products that do not exhibit optical isomerism. A strong answer earns marks for these points. stating that the carbonyl carbon is planar. saying the cyanide ion attacks from above or from below with equal probability. identifying the new chiral centre as a carbon with four different groups. stating that equal amounts of the two enantiomers are formed, so the product is a racemate and optically inactive. Exam tip: Check the four groups on the new carbon before claiming enantiomers; two identical groups (e.g., in methanal) means none.

    See it

    A top-down flowchart. The first box states that the carbonyl carbon is trigonal planar, leading to a second box saying the cyanide ion attacks from above or below with equal probability. This leads to a third box: the new carbon carries four groups — hydroxyl, nitrile, and the two original groups. A decision box then asks whether the two original groups are different. On the Yes branch, the carbon is chiral, leading to equal amounts of the two enantiomers forming, and then to a racemate that is optically inactive. On the No branch, no chiral centre forms, leading to an achiral product with no optical isomerism.
    It shows how attack of cyanide on a planar carbonyl carbon leads either to a racemate when the two original groups differ, or to an achiral product when they are identical.

    Key words

    Say what each word means before you turn the card.

    planar carbonyl carbon

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    new chiral centre

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    racemic product

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    achiral exceptions

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    See every definition
    planar carbonyl carbon
    The carbonyl carbon is trigonal planar, so the cyanide ion can approach from above or below the plane with equal probability.
    new chiral centre
    In the product, that carbon carries four groups: hydroxyl, nitrile, and the two original groups, and if these two are different, the carbon is chiral.
    racemic product
    Equal approach from both faces produces equal amounts of the two enantiomers, forming an optically inactive racemate.
    achiral exceptions
    If the original groups are identical, no chiral centre forms, so methanal and symmetrical ketones like propanone yield achiral products that do not exhibit optical isomerism.

    Label the diagram

    Some boxes have lost their names. Pick the right name for each number.

    The diagram again, with 3 boxes numbered instead of named. A top-down flowchart. The first box states that the carbonyl carbon is trigonal planar, leading to a second box saying the cyanide ion attacks from above or below with equal probability. This leads to a third box: the new carbon carries four groups — hydroxyl, nitrile, and the two original groups. A decision box then asks whether the two original groups are different. On the Yes branch, the carbon is chiral, leading to equal amounts of the two enantiomers forming, and then to a racemate that is optically inactive. On the No branch, no chiral centre forms, leading to an achiral product with no optical isomerism.

    • 1Box 1

    • 2Box 2

    • 3Box 3

    0 of 3 labelled

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