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(B) The reaction proceeds via the electrophilic addition of $HI$ to the alkene. $1$. Protonation of the double bond occurs to form the most stable car

Vedclass · Accepted Answer

$1-$iodo$-1,2-$dimethylcyclohexane

Vedclass · Answer

(B) The reaction proceeds via the electrophilic addition of $HI$ to the alkene. $1$. Protonation of the double bond occurs to form the most stable carbocation. The initial carbocation formed is a secondary carbocation adjacent to a cyclopentyl ring. $2$. To relieve ring strain and form a more stable tertiary carbocation,a ring expansion occurs,converting the five-membered ring into a six-membered ring. $3$. The resulting tertiary carbocation is then attacked by the iodide ion $(I^-)$ to form the final product,which is $1$-iodo-$1,2$-dimethylcyclohexane.

Taking into account the stability of various cycloalkanes and carbocations,as well as the rules governing mechanisms of carbocation rearrangements,what is the most likely product of this reaction?

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