Reactivity of Alkenes with HBr [closed]

Question

I am new in the Organic Chemistry (World of Reactions). I had got one question in my test. But, when I had seen it, I got no proper logic to tick the correct answer, and then when I saw solution, then there is only written, "More Substituted Alkene, is more Reactive." But, how they are more reactive ? I mean, that if they are substituted with ALKYL GROUPS, then they will get more stable (due to + Inductive Effect) and will not React. I think, I am wrong somewhere, and when I searched on the internet, there is No Proper Logic, for this question. So, I just want to know the Correct Logic and Mechanism of this question. Please help.

Answer 1

What you should be looking at is the stability of the carbocation intermediate formed when the strong acid $\ce{HBr}$ protonates the alkene. This stability is more sensitive to alkyl substitution than the stability of the uncharged alkene, so the net effect is that alkyl substitution lowers the activation energy versus having no substitution. Ergo a faster reaction, e.g.

2,3-dimethyl-but-2-ene (two methyl groups at each double-bonded carbon, tertiary cation when protonated)

$>$ but-2-ene (one methyl group, secondary cation)

$>$ ethene (no substitution, primary cation).