The reaction of 5-methylenecyclohex-2-enone w | vedclass

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(D) The reaction involves the electrophilic addition of $HBr$ to the two double bonds present in $5-$methylenecyclohex$-2-$enone. First,$HBr$ adds to

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$3-$bromo$-5-$bromo$-5-$methylcyclohexanone

Vedclass · Answer

(D) The reaction involves the electrophilic addition of $HBr$ to the two double bonds present in $5-$methylenecyclohex$-2-$enone. First,$HBr$ adds to the exocyclic double bond. Protonation occurs to form a stable $3^{\circ}$ carbocation at the $5-$position,which is then attacked by the bromide ion to give a $5-$bromo$-5-$methylcyclohex$-2-$enone intermediate. Second,$HBr$ adds to the remaining endocyclic double bond. According to Markovnikov's rule,the proton adds to the carbon with more hydrogens,and the bromide adds to the carbon that forms a more stable carbocation (stabilized by the inductive effect of the carbonyl group). The final product is $3-$bromo$-5-$bromo$-5-$methylcyclohexanone.

The reaction of $5-$methylenecyclohex$-2-$enone with $2$ equivalents of $HBr$ yields which of the following products?

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