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Alkene Reactions

By James Ashenhurst

Addition Reactions: Elimination’s Opposite

Last updated: July 1st, 2019 |

Addition Reactions of Alkenes

In previous series, we’ve discussed acid-base reactions, nucleophilic substitution reactions, and elimination reactions. These represent three of the four most important reaction types in a typical Org 1 course.

What each of these reactions have had in common so far is that each of them begin with a Lewis base (which we call a “base” or “nucleophile” depending on whether it is attacking hydrogen or carbon) donating a lone pair to an electrophile (either hydrogen or carbon).

In this series of posts we’ll cover the fourth major class of reactions, addition reactions. As we’ll see, what makes this class of reactions different is that a double bond (a π-bond, to be more specific) that will act as the electron-pair donor. In other words, we’ll see that π-bonds can be nucleophiles too!

Furthermore, since by definition π-bonds span two carbons, we shall see that this class of reactions will have interesting ramifications concerning the identity of the products (“regioselectivity” and “stereoselectivity”, respectively) that we will cover in subsequent posts.

So let’s start, shall we?

Table of Contents

  1. Three Reactions of Alkenes – Don’t Worry About “Why”, Yet, Just Focus On The Bonds That Form And Break
  2. The Key Pattern Of All These Reactions Is That They Break A C–C Pi Bond And Form Two Single Bonds On Adjacent Carbons  (The Exact Reverse Of Elimination)
  3. The General Pattern For Addition Reactions
  4. Summary: Addition Reactions

1. Three Reactions of Alkenes – Don’t Worry About “Why”, Yet, Just Focus On The Bonds That Form And Break

I always think it’s important to describe the “what” before we get to the “why” or “how”. Before we can understand how or why something happens, it’s important to be able to just recognize the essential pattern. And as always, this will come from what experimental observations tell us.

Let’s look at an experimental observation that dates back well over 140 years. In the late 1860’s, the Russian chemist Vladimir Markovnikov made the following observation: alkenes treated with hydrobromic acid formed alkyl bromides.

1-addition

Note the pattern of bond-forming and bond-breaking here: we’re breaking a C-C π bond and forming a C-Br and C-H bond on adjacent carbons.

Here’s another example. In the late 1800’s it was discovered by French chemist Paul Sabatier that when alkenes are treated with hydrogen gas in the presence of finely divided nickel, the following reaction occurs:

Sabatier won the 1912 Nobel Prize in chemistry for the development of this reaction, which was subsequently found to occur with many different varieties of metal catalysts besides nickel,  including palladium, platinum, and many other “late” metals. Again, note the pattern: breaking a C–C  π bond and forming two C–H bonds on adjacent carbons. [Don’t worry so much about the dashes and wedges for now – we’ll get there in a later post].

Here’s one last example. If you take an alkene (like cyclohexane, for instance) and add elemental (liquid) bromine, the following reaction occurs:

2. The Key Pattern Of All These Reactions Is That They Break A C–C Pi Bond And Form Two Single Bonds On Adjacent Carbons  (The Exact Reverse Of Elimination)

Again, note the pattern – break C–C  π [and Br-Br]  and form  C–Br bonds on adjacent carbons. [We’ll deal with the dashes and wedges in subsequent posts – it’s OK to just ignore them for now].

If you’ve got a really good memory, you might notice that this pattern is strangely familiar. If we go waaay back into the archives, we’ve seen a reaction that fits this pattern exactly…. but in reverse! It’s our old friend the elimination reaction!

[I’ve left the “strong base” here as generic, but a typical example would be NaOCH3 or NaOCH2CH3]

As we’ve previously seen, elimination reactions involve breaking two single bonds on adjacent carbons and forming a new C–C π bond. Notice how these two reactions (addition and elimination) achieve the exact opposite results here.

  • In the addition reaction [the first reaction at the top of the page] , we’re forming C-Br and C-H, and breaking C–C π  [we’re also breaking H-Br]
  • In the elimination reaction, we’re breaking C–Br and C–H, and forming C–C π [and forming a bond between the base and hydrogen]

3. The General Pattern For Addition Reactions

We can even generalize these patterns beyond this specific example of H-Br.  Likewise, for addition reactions, the general pattern looks like this:

As we’ll see , there are many, many more examples of addition reactions we’ll see beyond the 3 examples we’ve seen here. But they all follow the same essential pattern. We’ll always break a C-C π bond and we’ll always be forming two new single bonds to carbon.

4. Summary: The General Pattern For Addition Reactions

Lots of mysteries remain for us, however: for instance, did you notice in the first example that Br added to the most substituted carbon? And how in the second, the two hydrogens added to the same side of the alkene, but in the third, the bromines added to opposite sides? We’ll go through these patterns in more detail in subsequent posts.

NEXT POST: Addition Reactions – Regioselectivity 

Comments

Comment section

3 thoughts on “Addition Reactions: Elimination’s Opposite

  1. What I dont get is why you can do an elimination reaction of water to get a double bond (catalyzed by H2SO4, E1 mechanism) but you can also get the addition of water to an alkene in acidic conditions. I don’t see how you get 2 different results using the same conditions.

    1. It’s an equilibrium. It depends on whether you have a high concentration of water relative to your alkene (addition) or whether you are actively sequestering water as it’s being formed, taking it out of equilibrium (elimination). Follows Le Chatelier.

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