Add to the least substituted side of an Epoxide

[panmit]

Destroyer OCHEM # 42 2014 says you add to the most substituted side of an epoxide. However, Chad says you add to the least substituted side of an epoxide. You only add to the most substituted side if the epoxide has a charge on it (i.e. Bromonium ion).

I think Chad is right, but I'm not 100% positive.

---

Members don't see this ad.

 

[BrazilianRider]

Depends. If it's acidic, you attack the most substituted side. If it's basic, you attack the least substituted side.

Also, don't forget to invert stereochemistry but ONLY on the carbon you attack!

 

[disciple155]

BrazilianRider is right! Chad does say the nucleophile attacks the least substituted side in basic conditions, but attacks the more substituted side in acidic conditions just as brazilianrider noted.

 

[Daneosaurus]

I see it as acidic: opens to the LEAST substituted side, which means you add to the most substituted side. Basic: opposite.

This is because:
In acid, you protonate the oxygen first and a carbocation forms, which is where the nucleophile is added. Also, you don't get inversion in acid catalyzed because it proceeds by sn1. Base proceeds by sn2, and therefore you will get inversion on the electrophilic carbon.

 

[skyguy1]

Dane, im not orgo wiz and im not sure if im a 100% right .. but i think there is inversion in acid catalyzed as well on the C u attack.... not 100% sure tho if someone can add on to this it would be great

I see it as acidic: opens to the LEAST substituted side, which means you add to the most substituted side. Basic: opposite.

This is because:
In acid, you protonate the oxygen first and a carbocation forms, which is where the nucleophile is added. Also, you don't get inversion in acid catalyzed because it proceeds by sn1. Base proceeds by sn2, and therefore you will get inversion on the electrophilic carbon.

 

[BrazilianRider]

Dane, im not orgo wiz and im not sure if im a 100% right .. but i think there is inversion in acid catalyzed as well on the C u attack.... not 100% sure tho if someone can add on to this it would be great

Yeah, there's an inversion regardless of acid/base, but only on the carbon you're attacking! All other stereochemistry stays the same!

 

[mpatel24]

Thanks for all the replies. Makes sense now. Also in chads notes I have that an acetylide ion and grigmard add to an epoxide on the least substituted side. Is this correct because these are under basic conditions?

 

[BrazilianRider]

The Hydrogen atom on a terminal alkyne is pretty acidic. Since Grignards are ridiculously basic, it takes the acidic hydrogen and "leaves" the alkyl group behind. The alkyl group then adds to where the hydrogen was.

Not 100% correct, but an easy way to think about it.

 

[Daneosaurus]

I'm fairly certain the acid catalyzed epoxide opening is SN1 and loses it's stereochemistry due to being trigonal planar.

 

[Thanhn]

This is the trick I used.

Acidic = Sn1
Basic = Sn2

A/B = 1/2
So if you it's under acidic conditions, a nucleophile will attack like an SN1 reaction -- the most stable carbocation, meaning the most substituted side.
Basic conditions - B is the second letter so SN2 like -- the easiest/least hindered location.

And yes, all nucleophiles attack an epoxide from the back so if it's chiral, there's an inversion of stereochemistry.

 

[elhammy]

I'm fairly certain the acid catalyzed epoxide opening is SN1 and loses it's stereochemistry due to being trigonal planar.

SN1 reactions are going to attack the most substituted side of an epoxide because SN1 reactions prefer tertiary carbocation intermediates. They are usually non-Grignard reagents (weak nucleophiles) SN2 reactions require a great nucleophile (grignard) and therefore will attack very fast, with a short transition state and they require little to no hindrance from other alkyl groups on the epoxide ring.

 

[derbear]

Also be careful about something that has:
1. OCH3
2. H3O+

This reaction is not done under acidic conditions despite there being acid added AFTER the attack...hence the 1. then 2.