In which of these cases would the addition of $\ce{Br2/H2O}$ be highly regioselective?

$$\ce{CH3CH=CHCH2CH3}$$ or $$\ce{(CH3)2C=CH2}$$

since both are asymmetric alkenes, how do I answer the question?

  • $\begingroup$ Which one will give the most stable cation? $\endgroup$
    – Waylander
    May 22, 2019 at 18:05
  • $\begingroup$ @Waylander Comparing the stability of bromonium intermediate, I think the second compound will be more stable due to the presence of a 3-degree carbon which has a stronger inductive effect than those with degree 2 as in the first compound. But does the carbon number matter (because the first compound has a greater number of carbons on both sides of the double bond)? $\endgroup$
    – Tapi
    May 22, 2019 at 18:21
  • $\begingroup$ Tertiary cations are always more favoured over secondary $\endgroup$
    – Waylander
    May 22, 2019 at 18:51
  • $\begingroup$ @Waylander, okay, so, how is stability related to regioselectivity? $\endgroup$
    – Tapi
    May 22, 2019 at 18:53

1 Answer 1


With molecule (2), you have two possible cations from addition of Br+; one is tertiary, one is secondary. Tertiary is much more stable so you will get very little product resulting from water reacting the secondary cation = high selectivity observed.

With molecule (1), Br+ addition gives 2 possible secondary cations; nothing much to choose between them = poor selectivity, ~1:1 mixture observed.


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