Comments on: Hydration and Oxymercuration of Alkynes

By: James Ashenhurst

James Ashenhurst — Mon, 07 Oct 2019 17:11:07 +0000

By definition a chiral center must have four different groups attached. Two hydrogens are delivered to the alpha carbon of the ketone. Therefore formation of enantiomers is not possible in this reaction.

By: Sina riahi

Sina riahi — Fri, 27 Sep 2019 06:28:41 +0000

In part 4(mechanism of addition of water),in step2(tautomerization)
i want to ask about hydrogen that h3O donates.can that H be added by any side so we will have enantiomers or it is stereospecific?

By: James Ashenhurst

James Ashenhurst — Mon, 15 Apr 2019 19:53:12 +0000

In reply to Sara. No, there's no free carbocation.

By: James Ashenhurst

James Ashenhurst — Mon, 15 Apr 2019 19:41:30 +0000

In reply to charles.

Thanks for this comment. The problem is not regioselectivity but that 1,1-dibora products form, which is why bulky boranes must be used. See JACS 1967 vol. 89 p. 291. https://pubs.acs.org/doi/10.1021/ja00978a022

By: Shama Tanveer

Shama Tanveer — Sun, 25 Nov 2018 16:33:23 +0000

In reply to Emily. it will be a ketone

By: Meghan

Meghan — Fri, 17 Aug 2018 06:47:29 +0000

What an amazing page ! It cleared all my doubts regarding this topic . My Chemistry teacher couldn’t have taught it better :) Please keep up the good work .

By: Ptachia Bar - On

Ptachia Bar - On — Mon, 14 May 2018 05:28:23 +0000

If water are added via oxymercurium path, don’t a second stage, of using hydroboran to reduce oxymercurium inermidiate, is needed?

By: charles

charles — Wed, 18 Apr 2018 23:35:15 +0000

In reply to james. Is the regioselectivity of BH3 any good? Seems like hydroboration via 9-BBN or some other bulky R2BH reagent greatly enhances the anti-Markovnikov product via sterics. While the electronegativity difference is real, you're hoping for a collision that leads to the π bond attacking boron for the syn transfer of the H to the more substituted side of the alkene (or alkyne)–is there any way something as small as BH3 can reliably select for the anti-M product?

By: Andre

Andre — Wed, 06 Dec 2017 22:41:34 +0000

In reply to Emily. Hydration of a terminal alkyne will first form an enol following Markovnikov's rule. The enol will then undergo Keto-Enol Tautomerization and rearrange to a methyl ketone.

By: Deesier

Deesier — Sun, 01 May 2016 16:54:20 +0000

In reply to Emily. For H2so4 reactions its always a ketone for Bh3 reaction it can be ketone or aldehyde