Asked Jul 12, 2019

Draw both resonance structures of the most stable carbocation intermediate in the reaction shown.


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Expert Answer

Step 1

The given compound is substituted diene. The reaction of diene with hydrochloric acid results in the formation of 1,2 product and 1,4 product.

Step 2

In the first step, the double bond at 1st and 2nd position in a reactant abstracts a proton from the hydrochloric acid to yield tertiary and secondary carbocation intermediates as mentioned as I and II.


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CHз CH3 CH3 or Н—СІ CHз .CH3 .CHз II о- е

Step 3

II carbocation intermediate is more stable because it is in conjugation with the double bond. The double bond shifts form 3rd and 4th position to 2nd and 3rd position a...


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CH3 CHз CHз 2 4CH3 .CH3 .CH3


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