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Problem:

enter image description here

$\ce{A}$ can be?

  1. concentrated sulphuric acid
  2. alcoholic potassium hydroxide
  3. triethylamine
  4. potassium tert-butoxide

My comments:

Option (A): Conc. $\ce{H2SO4}$, is an acid, it gives $\ce{H+}$ ions which must act like electrophile and add to the double bond, which must produce a carbocation and then, $\ce{HSO4^-}$ ion must act like a nucleophile and form a bond with the carbocation. Hence, the reaction must be addition of $\ce{H+}$ and $\ce{HSO4^-}$

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 is A. How is this so?

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    $\begingroup$ 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? $\endgroup$
    – Nick
    Commented Jan 14, 2013 at 15:36

1 Answer 1

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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?

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    $\begingroup$ Answer to extra credit: Because of the hydride migration, it forms a tertiary carbocation which is very much stable than secondary carbocation $\endgroup$
    – claws
    Commented Jan 18, 2013 at 20:41
  • $\begingroup$ 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. $\endgroup$
    – claws
    Commented Jan 18, 2013 at 20:44
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    $\begingroup$ @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. $\endgroup$
    – F'x
    Commented Jan 18, 2013 at 20:50

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