Comments on: Alkene Addition Pattern #2: The “Three-Membered Ring” Pathway

By: EDWIN K A

EDWIN K A — Wed, 07 Apr 2021 07:32:19 +0000

Really great work sir

By: James Ashenhurst

James Ashenhurst — Mon, 22 Mar 2021 15:17:33 +0000

Yes it will! It will form a cyclic ester known as a “lactone”. This reaction is specifically called “halolactonization”. Iodolactonization is a prominent example. https://en.wikipedia.org/wiki/Iodolactonization

By: cheyenne Darrow

cheyenne Darrow — Wed, 17 Mar 2021 03:47:22 +0000

Hi im having a hard time with Br2 (or any halogenation reaction) followed by an INTRAmolecular nucleophilic reaction. Such as Br2 added to a alkene that has a carboxylic acid attached? i know it forms a cyclic product but will it be a an ester type product?

By: George

George — Tue, 12 May 2020 18:30:50 +0000

… unless because it is in acid… maybe the alcohol on the more substituted carbon then gets protonated again (it is in acid) and leaves forming a carbocation, so the epoxide reforms and is now scrambled in stereochemistry (because it can attack the carbocation from below or above the ring) and the reaction with alcohol then occurs again so you get enantiomers… I dunno…

By: George

George — Tue, 12 May 2020 18:23:35 +0000

Karin,
I’m with you — I don’t think you get enantiomers with epoxide + acid… also I think James agrees (unless I’m missing something) in his post here
https://www.masterorganicchemistry.com/2015/02/02/opening-of-epoxides-with-acid/

By: Ha

Ha — Sun, 20 Oct 2019 04:26:59 +0000

Thank you for your post; it was very helpful!
I’m just a bit confused on how to draw anti vs. syn in an opened chain. Why would it differ if it was anti or syn if the bond is in free rotation? how can we depict it in the drawing?

Again, thank you!

By: James Ashenhurst

James Ashenhurst — Sun, 07 Jul 2019 04:06:06 +0000

In reply to Sam.

It’s a valid question. I’d start by asking, “what’s the purpose of dashes and wedges” ? And I’d say that ultimately it’s to resolve ambiguity. If there’s more than one way to arrange the groups on a carbon, then dashes and wedges are required to unambiguously depict which configuration we’re dealing with.
On the example you mention, the carbon is attached to two methyl groups and an OH, and there’s only one way to draw that configuration. It would not be incorrect to continue to draw that CH3 as a dash, but it would be more work, and since it’s not necessary, I just didn’t do it.
The carbon is tetrahedral, yes, but that’s always assumed for an sp3 carbon.

By: James Ashenhurst

James Ashenhurst — Sat, 02 Mar 2019 17:57:34 +0000

In reply to Sam. Very interesting, did not know that. Thank you.

By: An

An — Tue, 01 Jan 2019 10:04:32 +0000

In the last reaction, why would the final product have Cl bonded with the tertiary carbon and not with either the tertiary or the secondary one?

By: Sam

Sam — Tue, 27 Nov 2018 16:27:16 +0000

In reply to James. With ICl, you actually get close to a 50:50 mixture on 1-methylcyclohexene derivatives. With BrCl, you get good selectivity for the bromine at the more substituted carbon, which suggests that the chloronium forms, then bromide acts as the nucleophile. This is explained by the fact that the Br- is a more stable leaving group, even though it is less electronegative.