Understanding Syn and Anti Elimination in E2 and Sn2 Reactions

[chops369]

Can someone please help me understand the difference between syn and anti elimination in terms of what they favor? My [extremely] limited understanding leads me to believe that syn elimination favors E2 since it mimics frontside attack, which will not work for Sn2 since there is no net-bonding when the nucleophile attacks the empty sigma antibonding orbital from this angle. I would also venture that anti elimination favors Sn2 since it mimics backside attack. Is this the only difference between the two arrangements? Please correct if I'm wrong.

 

[DDTea]

This barely makes sense and I'm really not quite sure what you're asking; could you clarify what you mean by, "what they favor?" Do you mean what substrates favor syn or anti-elimination? Are you trying to discuss which reactions (E2, E1, Sn2, Sn1) can compete with one another?

 

[chops369]

This barely makes sense and I'm really not quite sure what you're asking; could you clarify what you mean by, "what they favor?" Do you mean what substrates favor syn or anti-elimination? Are you trying to discuss which reactions (E2, E1, Sn2, Sn1) can compete with one another?

My bad. I meant to ask what the syn and anti arrangements favor. My organic textbook says that syn and anti are favored in the E2 reaction over both a 60^o dihedral angle and a 120^o dihedral angle. This makes sense to me, since these arrangements would require further rotation to make a planar alkene. Then it goes off on some tangent about how the syn arrangement, when attacked by a nucleophile/base, resembles what is seen in an SN2 backside attack, whereas the anti arrangement appears similar to frontside attack. The book is not very clear about this, hence my confusion and my badly written question.

 

[DDTea]

Well, E2 won't happen with a syn arrangement--so I don't know what your book is talking about there! If you look at the transition state for E2, it requires that the base and the leaving group to be be anti-periplanar to one another. In linear molecules, syn and anti in this case aren't a problem because free rotation of the relevant C-C bond can bring it into the proper geometry for E2 reaction.

Free rotation is not going to be a major obstacle to any reaction (keep in mind, bonds rotate at ~10^12 times/second!). If the rotation is hindered somehow (e.g., in a cyclopentane ring), then it might not be possible to reach the transition state for a particular reaction.

In general though, if you're confused, Sn1 reactions are going to compete with E1 reactions (since both involve an intermediate carbocation); Sn2 reactions are going to compete with E2 (since both are concerted mechanisms that occur under the same reaction conditions). In reality, few reactions are "pure" Sn1 or Sn2 reactions. They tend to be some shade of grey in between them (again, look at the transition states! notice the similarity between Sn1 and Sn2 in that light).

 

[AsuraSky]

I don't know if I have the same book as the topic starter (Organic Chemistry 7th edition by L.G. Wade Jr) but my book also mentions the E2 reaction occurring when there's a syn-coplanar relationship between the leaving group and an H when there is no free rotation (like in a cycloalkane). The book says:

The transition state for the anti-coplanar arrangement is a staggered conformation, with the base far away from the leaving group. In most cases, this transition state is lower in energy than that for the syn-coplanar elimination

Ok, makes sense but then the book goes on to say

Some molecules are rigidly held in eclipsed (or nearly eclipsed) conformations, with a hydrogen atom and a leaving group in a syn-coplanar arrangement. Such compounds are likely to undergo E2 elimination by a concerted syn-coplanar mechanism. Deuterium labeling (using D, the hydrogen isotope with mass number 2) is used in the following reaction to show which atom is abstracted by the base. Only the hydrogen atom is abstracted, because it is held in a syn-coplanar position with the bromine atom. Remember that syn-coplanar eliminations are unusual, however, and anti-coplanar eliminations are more common

The book then goes on to show the following picture

I'm confused about this too. I've missed many a test question thinking that there's a syn-coplanar elimination when the correct answer is some other reaction mechanism. Is syn-coplanar elimination that unusual so that it almost never occurs and follows some other reaction mechanism instead (even in ring compounds with no free rotation)?

 

[DDTea]

I love when I learn something new. After checking out the wiki page for the E2 reaction, it turns out it *does* happen in a synperiplanar geometry. However, it is energetically less favorable than the antiperiplanar geometry. Now I'm going to have to look this up in my physical organic book. Cool thread!

 

[chemisttree]

The book then goes on to show the following picture

I'm confused about this too. I've missed many a test question thinking that there's a syn-coplanar elimination when the correct answer is some other reaction mechanism. Is syn-coplanar elimination that unusual so that it almost never occurs and follows some other reaction mechanism instead (even in ring compounds with no free rotation)?

So, in this reaction, the rate determing step involves the alkoxide. Can you posit a mechanism for this E2 elimination? Your example is a very special case. The deuterium and the bromine should eliminate in an anti fashion except that there is a steric interference somewhere in this case.