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Alkene Reactions

By James Ashenhurst

Stereoselectivity In Alkene Addition Reactions: Syn vs Anti Addition

Last updated: July 2nd, 2019 |

Stereoselectivity In Alkene Addition Reactions: “Syn” vs “Anti”

In the last post on alkene addition reactions, we discussed one of the two key themes to look for in addition reactions: regiochemistry (in other words – what is the favored direction in which the pi-bond breaks). This post is about the second key theme in addition reactions of alkenes: stereochemistry. 

We’re going to look at three key reactions of alkenes and see how they each demonstrate a different pattern of stereochemistry in addition reactions.

Table of Contents

  1. A Collection of Observations On Reaction Alkene Addition Stereochemistry That Nobody Predicted Ahead Of Time
  2. Addition Of H-Br To Alkenes Is Not Stereoselective, And Gives A Roughly Equal Mixture Of “Syn” And “Anti” Products
  3. Addition Of Bromine To Alkenes Is Stereoseletvie, Giving “Anti” Addition Stereochemistry
  4. Hydrogenation Of Alkenes With Pd-C And H2 Is Selective For Addition Stereochemistry
  5. Summary: Stereoselectivity In Alkene Addition Reactions

1. A Collection Of Observations That Nobody Predicted Ahead Of Time Until They Did The Experiment

Those three reactions we’ll look at today are addition of HBr, bromination with Br2, and hydrogenation with Pd-C and H2. However, later we’ll see that each of these reactions is characteristic of a particular “family” of reactivity for addition reactions. (The carbocation pathway, the “3-membered ring” pathway, and the “concerted” pathway)

Remember, these are results from experiment. They are observations. Without prior knowledge, it isn’t possible to predict from first principles how they proceed – those who discovered these reactions in the late 1800’s and early 1900’s didn’t know what we know now.

Later on, we will use this evidence to make hypotheses about how these reactions work. Let’s have a look.

2. Addition Of H-Br To Alkenes Is Not Stereoselective, And Gives A Roughly Equal Mixture Of “Syn” And “Anti” Products

First example: let’s take a cyclic molecule like 1,2-dimethylcyclohexene and treat it with hydrobromic acid (HBr). Here’s what we get.

 In the previous post I said to ignore all the dashes and wedges, because we’d deal with them later. “Later” is now!  Examine the placement of the H and the Br that are added. Notice how in the left hand product, the H and Br are on opposite sides of the ring, whereas in the right hand product, they are on the same side? These two compounds are not the same – they are “stereoisomers“. The “connectivity” of each molecule is the same, but they differ in their orientation in space!

In the product on the left, the Br and H are on opposite sides of the ring. In other words, H and Br added to opposite faces of the starting alkene. Our term for this relationship is “anti” *. In the product on the right, the Br and H are on the same side of the ring (and therefore have added to the same face of the alkene). Our term for this relationship is “syn“.

So a feature of this reaction is that it produces a mixture of syn and anti products. They exist in roughly equal amounts in this example, but the point is that the mechanism does not selectively deliver either the syn or the anti product. Any mechanism we propose for this reaction will have to be able to explain why we end up with a mixture of these two products.

3. Addition Of Bromine (Br2) To Alkenes Is Stereoselective, Giving “Anti” Addition Stereochemistry 

Let’s look at a different reaction next. When we treat an alkene with a halogen such as Br2, (often in a halogenated solvent such as CH2Cl2 or CCl4) we obtain the following product using 1,2-dimethylcyclohexene.

Again, pay attention to the dashes and wedges. Here, notice that we observe only the “anti” product and none of the “syn” product. In other words, the reaction is highly selective for one stereoisomer over the other. We could go even further and say that because of the complete absence of the “syn” product, the reaction is stereospecific for the “anti”. Only one type of stereoisomer is formed.

We’ll see that this pattern is observed for other reactants similar to Br2. Again, any mechanism we propose will have to account for the fact that we only get the “anti” product and none of the “syn”.

4. Hydrogenation Of Alkenes With Pd-C and H2 Is Selective For “Syn” Addition Stereochemistry

Finally, let’s look at a third category of addition reaction. When 1,2-dimethylcyclohexane is treated with hydrogen gas and palladium catalyst (Pd-C), the result is as follows:

Notice how the only product of this reaction is the one where two hydrogens have added to the same face of the alkene (“syn” stereoselectivity). The product where hydrogens add to opposite faces is not observed. Again, this is an example of a highly stereoselective reaction. The mechanism will once again have to explain why we only obtain the syn product of this reaction and none of the anti product.

5. Summary: Stereoselectivity For Syn vs Anti Products In Alkene Addition Reactions

The key point from this post is to pay close attention to the stereochemistry of addition reactions.

There are three key categories of alkene reaction pathways:

  • an non-stereoselective mixture of syn vs anti products (e.g. H-Br to alkene)
  • reactions that are stereoselective for the anti product (e.g. Br2 to alkene)
  • reactions that are stereoselective for the syn product (e.g. Pd-C/H2 to alkene)

Different reagents can lead to very different stereochemical results – a point which is often tested for by organic chemistry instructors. Stereochemistry really is the key theme of Org 1!

In the next post we’ll start looking at some of these reactions in more detail.

NEXT POST – Markovnikov’s Rule 

 

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Comment section

21 thoughts on “Stereoselectivity In Alkene Addition Reactions: Syn vs Anti Addition

  1. For the third reaction, I get that it is a syn addition….however, does it matter if it is a wedge or a dash for the syn addition?

  2. Is there any way to add two hydrogens across a double bond in an anti fashion? I saw you said hydrogenation occurs only in syn.

  3. For the name of 1,2-dimethylcyclohexene, why there is no position number before -ene-(indicates where the double bond is)

    1. I probably should have put it in. But generally if you don’t see a number before the -ene it’s safe to assume that the double bond is between C-1 and C-2.

  4. How do we know if the reaction proceeds via syn or anti addition when predicting the major product of the reaction?

    1. Through experiment. And reading about the results of these experiments in your textbook. You’ll learn that each reagent is either not stereoselective, stereoselective for “syn” , or stereoselective for “anti”.

  5. But how do you decide which compound undergoes syn and which anti secondly suppose there is reaction of 1 methyl cyclo hex2ene in which MCPBA is used and then H2O is used where O18 is radioactive oxygen then would the addition be decided according to MCPBA or H20 . And please tell how to decide whether its syn or anti.
    THANKS AND REGARDS
    SHIVAM SAHIL

    1. Attack will be first of MCPBA, converting diene to epoxide. H2O will will then react with the epoxide formed. It will be anti addition, for reason refer to mechanism (preferably from Solomons) from a book or from the internet.

  6. when writing the products of the rxn, how do you determine what atom gets the dash, the wedge or none? It seems that the atoms like to switch their orientation after the rxn…why? Please advise. Thank you.

    1. It depends. That’s a very vague question. If you are reacting an alkene with an X2 reaction, the products will be anti only. Draw out the mechanism for the X2 reaction (ex: Br2) and you will be able to see that the X in X2 can attach from the top and bottom of the carbon due to its p orbitals. As long as your wedges and dashes make chemical sense, you are fine.

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