# How can I illustrate Markovnikov's rule by the reaction of propene with hydrobromic acid?

Sep 13, 2015

Treat the acid as $D - B r$, i.e. $D$ = deuterium. Consider the intemediate after the olefin has reacted with ${D}^{+}$.

#### Explanation:

${H}_{2} C = C H C {H}_{3} + D - B r \rightarrow {H}_{2} D C - {C}^{+} H C {H}_{3} + B {r}^{-}$
${H}_{2} C = C H C {H}_{3} + D - B r \rightarrow {H}_{2} {C}^{+} \left(- C H D C {H}_{3}\right) + B {r}^{-}$
As you know, the secondary carbocation is more stable than the primary carbocation. Because the carbocation intermediate is then going to react with the electrophile, $B {r}^{-}$, we should see more ${H}_{2} \left(D\right) C - C H B r C {H}_{3}$ than the anti-Markownikow product, ${H}_{2} \left(B r\right) C - C H D C {H}_{3}$.