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Q.Write the IUPAC name of the following compound: CH2=CH−CO−CH3CH_2=CH-CO-CH_3 (where −CO−-CO- denotes a C=OC=O carbonyl group)

CBSECBSE Class XII Board 2019Subjective· 1mImportance★★★★★
✓ Free question

The compound CH2=CH−CO−CH3CH_2=CH-CO-CH_3 contains both an alkene and a ketone group. Prioritizing the ketone group for numbering, the longest carbon chain is identified, leading to the IUPAC name but-3-en-2-one.

When naming organic compounds with multiple functional groups, the IUPAC system follows a hierarchy to determine the principal functional group. This principal group dictates the suffix of the name and receives the lowest possible locant (number) in the carbon chain. Other functional groups are then treated as prefixes or indicated by their own locants within the name.

Here's how to systematically name the given compound:

  1. Identify all functional groups present.

    The compound is CH2=CH−CO−CH3CH_2=CH-CO-CH_3.

    • We have a carbon-carbon double bond (C=CC=C), which indicates an alkene.
    • We have a carbonyl group (C=OC=O) where the carbon is bonded to two other carbons (R−CO−R′R-CO-R'), which indicates a ketone.
  2. Determine the principal functional group.

    According to IUPAC priority rules, ketones have higher priority than alkenes. Therefore, the ketone group will be the principal functional group, and its suffix ('-one') will be used in the name. The double bond will be indicated by the '-en-' infix.

  3. Identify the longest continuous carbon chain that includes the principal functional group and the multiple bond.

    The carbon chain is: C−C−C−CC-C-C-C.

    CH2=CH−CO−CH3CH_2=CH-CO-CH_3

    This is a 4-carbon chain. The parent alkane is butane.

  4. Number the carbon chain.

    Number the chain from the end that gives the principal functional group (ketone) the lowest possible locant. If there's a tie, then give the multiple bond the lowest possible locant.

    Let's consider the two possible numbering directions:

    • Option A (Left to Right):

      1CH2=2CH−3CO−4CH3^1CH_2=^2CH-^3CO-^4CH_3

      Here, the ketone group is at carbon 3 (C3C^3). The double bond starts at carbon 1 (C1C^1).

    • Option B (Right to Left):

      4CH2=3CH−2CO−1CH3^4CH_2=^3CH-^2CO-^1CH_3

      Here, the ketone group is at carbon 2 (C2C^2). The double bond starts at carbon 3 (C3C^3).

    Comparing the locants for the principal functional group (ketone), Option B gives the ketone a lower number (2) than Option A (3). Therefore, Option B is the correct numbering.

    The correctly numbered chain is:

    CH3−CO−CH=CH2CH_3-CO-CH=CH_2

    1CH3−2CO−3CH=4CH2^1CH_3-^2CO-^3CH=^4CH_2

  5. Construct the IUPAC name.

    • The parent chain has 4 carbons, so the base name is 'but'.
    • There is a double bond starting at carbon 3, so we use the infix '-3-en-'.
    • There is a ketone group at carbon 2, so we use the suffix '-2-one'.

    Combining these parts, the name is but-3-en-2-one.

Watch out

A common mistake is to number the chain to give the double bond the lowest number, ignoring the higher priority of the ketone group. Always prioritize the principal functional group first.

✓Final answer

The IUPAC name of the compound CH2=CH−CO−CH3CH_2=CH-CO-CH_3 is but-3-en-2-one\boxed{\text{but-3-en-2-one}}.

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