[University_Of_Toronto]

Which of the following statements is true?

 a. Acid-catalyzed hydration of propene will yield 2-propanol.

 b. Hydroboration of propene followed by oxidation with HOOH under basic conditions will yield 1-propanol.

c. Both A and B

 d. Neither A nor B

I choose C, can someone please verify if I am correct.

[Rico]

I agree statement c. is correct. Acid catalysed hydration occurs over a secondary carbocation giving after addition of water and loss of a proton, isopropanol (2-propanol). Hydroboration occurs by addition of boron to the least hindred end of the alkene, giving after oxidation with hydrogenperoxide, n-propanol (1-propanol).

[University_Of_Toronto]

Thanks for your confirmation!!!
I would love to see more like you...others  beat around the Bush..it a sign that they either pretend to be smart or they have no clue but would like attention

[djay]

Acid catalysed hydration respects the Markovnikov rule -oh goes on the carbon of the double bond with the least hydrogens.

The hydroboration puts the -oh on the carbon with the easiest access because the Rxn forms this little molecule before the NaOOH is added.
    R
    |
R--B--R

[sjb]

Acid catalysed hydration respects the Markovnikov rule -oh goes on the carbon of the double bond with the least hydrogens.

The hydroboration puts the -oh on the carbon with the easiest access because the Rxn forms this little molecule before the NaOOH is added.
    R
    |
R--B--R

Well, you've said that "Acid catalysed hydration respects the Markovnikov rule", so (strictly speaking) does hydroboration, it's just that the attacking species is different so the more electropositive half of the reagent (the BR₂ part) attaches to the carbon with the least hydrogens etc...

S

[ultrashogun]

Acid catalysed hydration respects the Markovnikov rule -oh goes on the carbon of the double bond with the least hydrogens.

The hydroboration puts the -oh on the carbon with the easiest access because the Rxn forms this little molecule before the NaOOH is added.
    R
    |
R--B--R

Well, you've said that "Acid catalysed hydration respects the Markovnikov rule", so (strictly speaking) does hydroboration, it's just that the attacking species is different so the more electropositive half of the reagent (the BR₂ part) attaches to the carbon with the least hydrogens etc...

S

You are correct, but it always depends on how primitive the formulation of Markovnikovs rule is that you are applying.

"The carbon that has the most H accepts the proton from the electrophile" is the way the rule was originally formulated, so with borane this works the other way. But if you generalize the rule and understand that the double bond always attacks the electrophile so that it forms the more stable charge distribution then the terms "Markovnikov" and "anti-Markovnikov" dont really mean much anymore.

[djay]

Both respects the markovnikov rules its just I didnt want to go step by step for hydroboration to prove it. (maybe I should have since it's probably in my final exam need some practice)