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JEE Main levelSingle correct optionDifficulty 2 of 3nm253

What is the von Baeyer name of the compound shown? Give the double bond the lowest locant.

  1. Abicyclo[2.2.1]hept-2-ene
  2. Bbicyclo[2.2.2]oct-2-ene
  3. Cbicyclo[3.1.1]hept-2-ene
  4. Dbicyclo[2.2.1]hept-5-ene

Answer: A

The compound is bicyclo[2.2.1]hept-2-ene (norbornene). The ring system is bicyclo[2.2.1]heptane; the double bond is placed in a two-carbon bridge and numbered from bridgehead C1 to get the lowest locant, 2. The locant 5 describes the same bond numbered the long way round.
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