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Question 63

63

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Step 1: Debromination with ($$Zn$$, $$\Delta$$)The reaction of a vicinal dibromide (1,2-dibromo compound) with zinc dust and heat undergoes an elimination reaction. The zinc removes the two bromine atoms to form a carbon-carbon double bond.

  • Intermediate: 3,5-dimethylcyclopent-1-ene.

Step 2: Hydrobromination with ($$HBr$$)The intermediate alkene then reacts with ($$HBr$$). This is an electrophilic addition reaction following Markovnikov's rule.

  1. Protonation: The double bond attacks the ($$H^{+}$$) from ($$HBr$$), forming the most stable carbocation.
  2. Carbocation Formation: In this symmetrical-looking substituted cyclopentene, the carbocation forms at one of the substituted positions to benefit from secondary/tertiary stability (though here they are equivalent).
  3. Nucleophilic Attack: The bromide ion ($$(Br^{-}$$)) attacks the carbocation.

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