Chemistry Stack Exchange is a question and answer site for scientists, academics, teachers and students. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm tutoring few Class 12 students for their chemistry exams. I'm stuck with few problems. I don't know if the key is wrong or I am wrong.


enter image description here

My comments:

Option (A): Conc. $\ce{H2SO4}$, is an acid, it gives $H^+$ ions which must act like electrophile and add to the double bond, which must produce a carbocation and then, $\ce{HSO_4^-}$ ion must act like a nucleophile and form a bond with the carbocation. Hence, the reaction must be addition of $\ce{H+}$ and $\ce{HSO_4^-}$

Options (B),(C),(D): are bases, with basicity order: (D) > (B) > (C). But what would they do to an alkene?

Answer given in the book:


How is this so?

share|improve this question
Under (A) you have heat added. What is to stop a second molecule reacting with the first under electrophilic addition to remove the H2SO4 as a good leaving group and give the product shown in the question? – Nick Jan 14 '13 at 15:36
up vote 10 down vote accepted

Bottom line: $\ce{HSO4-}$ is a really bad nucleophile, and it doesn't enter in the picture.

enter image description here

(I realize after scanning that I'm missing the "+ H+" alongside the final product.)

Question for extra credit: after the second step, why is the third step (carbocation migration) favorable rather than directly reforming the double bond on the right-hand side ring?

share|improve this answer
Answer to extra credit: Because of the hydride migration, it forms a tertiary carbocation which is very much stable than secondary carbocation – claws Jan 18 '13 at 20:41
In you second last step. You've removed a proton. I strongly advice my students not to just protons getting away. Instead, I ask them to show a base picking up a proton. Who took that proton? Last step is like an Elimination 1 reaction. Which usually needs a strong base, but I don't see a strong base here. – claws Jan 18 '13 at 20:44
@claws the last step has nothing to do with an elimination. The carbocation is the conjugate acid of the alkene, so under acidic conditions it is formed in small amounts (equilibrium with alkene). The proton is not “picked up” by the base: this is an acidic solution, protons come and go all the time. – F'x Jan 18 '13 at 20:50

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.