Reactions of Dienes: 1,2 and 1,4 Addition

by James

in Dienes and MO Theory

In today’s post we’ll discuss  1,2- and 1,4- addition to dienes – specifically, the addition of strong acid such as HBr. Here’s the summary of the key points. 

Addition To Alkenes, Revisited

Waaay back in the day we talked about addition reactions of alkenes. Remember Markovnikov’s rule in the addition of electrophiles to alkenes?

For example, take an alkene like 1-butene, and add HBr. What happens?

An addition reaction occurs! (Break C-C pi, and form adjacent C-H and C-Br bonds).

Bromine adds to the most substituted carbon of the alkene, and hydrogen adds to the least substituted carbon. That’s due to carbocation stability: the reaction proceeds through the most stable carbocation, which happens to be the most substituted.

Butadiene Gives “1,2-Addition” And “1,4-Addition” Products

All well and good. But since we’ve been discussing conjugation and resonance, let’s throw in an additional wrinkle.

What happens when we try the same reaction on a diene? Butadiene, for example.

I’ll tell you. You get two products! (But the second one might not be what you think it is).

Here’s what they look like:

Product #1 is essentially the same product that we saw in the 1-butene example:  H and Br are added across two consecutive carbons of a double bond. Note that since butadiene is symmetric,  the same product is formed regardless of which double bond participates! (we get 3-bromo-1-butene either way).

Since addition occurs across two consecutive carbons, we often call this “1,2 addition”.

In contrast, Product #2 shows the result of adding H and Br across four conjugated carbons. All four carbons participate in the reaction.  A new C-H single bond has formed on one end of the diene (C-1), and C-Br formed on the other end (C-4). Note that the C1-C2 and C3-C4 pi bonds are broken, and we’ve formed a new pi bond between C2 and C3.

We call this “1,4 addition”.

[Note: to simplify the discussion here, I’m choosing to ignore double bond isomers (i.e. E and Z) in this analysis.  In the lab, the 1,4- example above will exist as a mixture of (mostly) E product with a small amount of the Z. ]

So why does this “1,4 addition” happen in the first place? What’s different about butadiene, as opposed to 1-butene where only “1,2-addition” was possible?

Protonation Of Butadiene Gives A Resonance-Stabilized Carbocation

The first thing to notice is that the initial protonation of butadiene gives a resonance-stabilized carbocation. See how protonation of C-1 gives a carbocation that has two important resonance forms?

In the case of butadiene,

  • the major contributor to the resonance hybrid will be the resonance form where the carbocation is on the more substituted carbocation (C2)
  • the minor contributor will be the resonance form where the carbocation is on the less substituted carbocation (C4).

Attack of the nucleophile (Br) at the C2 position of the hybrid will lead to the 1,2-product.

Attack of Br– at the C4 position of the resonance hybrid will lead to the 1,4-product.

We can draw it up like this:

[Note: never forget that resonance forms are not in equilibrium with each other. It’s understandable to casually say that “Br(–) will attack the top resonance form with the more stable carbocation” so long as you remember that the true structure of the molecule is a hybrid of the two resonance forms.] In the footnote, I’m including perhaps a more correct way to show formation of the bottom resonance form.

The effect of temperature

Here’s the big question: which one of the two products will be major? The 1,2-addition product or the 1,4-addition product?

We’re so used to seeing “Markovnikov addition” in alkenes (where addition occurs to the most stable carbocation) that it seems intuitive that the 1,2-addition product would be dominant.

And indeed, at low temperature, 1,2 addition to butadiene is favoured. Here’s a literature example.

Interestingly, however, as the temperature is increased, the amount of 1,4 product increases.

At room temperature, the ratio of 1,2- and 1,4- addition is 45:55 .

At 40 degrees Celsius the 1,4- product is dominant (about 80%).

What’s going on here? What structural features are present that could possibly make formation of the 1,4 product more favourable than the 1,2 product, even though it goes through a “less stable” carbocation?

The answer lies in the substitution pattern of the double bond.

Remember Zaitsev’s rule? Same deal. The 1,4 product is more stable because it is a more substituted double bond.

The 1,4 product has a di-substituted double bond, whereas the 1,2-product has a  mono-substituted double bond. Generally speaking, double bond stability increases as the number of carbons directly attached to the double bond is increased.

So why would 1,4 be more favoured under conditions of higher temperature, and the 1,2 be favoured under conditions of lower temperature?

“Kinetic Control” vs “Thermodynamic Control”

At low temperatures, the differentiating factor is the relative energies of the transition states leading to the products. The 1,2 product has a lower-energy transition state, owing to the fact that charge is more stable on the more substituted carbon. The difference between the energies of these transition states will determine the product ratio.

A quick analogy. Imagine you’re hungry, and you only have $5 in your pocket. In a choice between McDonalds and Applebee’s, the only accessible option is McDonalds – even if you like Applebee’s a lot more.

Let’s sketch this out. The carbocation intermediate can pass through the two different transition states  that lead to the 1,2- and 1,4- products, respectively. If we choose a temperature low enough, then the product distribution will reflect the difference in energy between the two activation energies Ea (1,2) and Ea (1,4). So long as the reaction is not reversible, the product with the lower energy transition state will dominate. This is called “kinetic control”.

At higher temperatures, the reaction has the potential to be reversible. [note 2] In this case, this means is that both the 1,2- and 1,4- products can ionize (think of the first step in the SN1 reaction), reforming the carbocation intermediate.

This sets up an equilibriumThe product ratio will now reflect the relative stabilities of the 1,2- and 1,4- products, not the transition states leading to their formation. In the case of butadiene, since the 1,4- product is more stable (it has a disubstituted double bond) it will be the dominant product at higher temperatures.

This is referred to as “thermodynamic control”.

Math interlude:  recall that ΔG = –RT lnK

A difference of 1 kcal/mol in stability doesn’t sound like much, but it translates into a 83:17 ratio of products at equilibrium at room temperature.

Continuing our analogy: with enough money in your pocket, the decision where to eat lunch is more a function of how much you like the overall McDonalds vs. Applebee’s experiences, not how much they cost. 

Using An Energy Diagram To Understand Thermodynamic Versus Kinetic Control

Drawing up the reaction energy diagram can be helpful to understand kinetic and thermodynamic control. This post here goes into more detail, but we’ll repeat the basics here.

The reaction energy diagram for the addition of HBr to butadiene looks like this:

Point A is the starting butadiene, and point B is the transition state for addition of H to butadiene.

The important part to pay attention to is the “local minimum” C, the resonance-stabilized carbocation.

As we mentioned previously, going from intermediate C to transition states D and D represent the energy pathways for 1,2- and 1,4- addition, respectively.

Hence the activation energies for the forward reactions is equal to the difference in energy between D and C (for 1,2-addition) and  and C (for 1,4-addition). We saw that 1,2-addition has a lower activation energy. 

Now let’s look at the reverse reaction.

Points E and E represent the energies of the 1,2- and 1,4- products.

The activation energy for the reverse reaction is the difference in energy between E and transition state D, and E and transition state D, respectively.

It should be clear from this diagram that the activation energies for the forward reaction  (going from C through transition states D and give E and E, respectively)  are much lower than the activation energies going in the reverse reaction (i.e. from E and through transition states D and D back to carbocation C ).

  • So if we keep the temperature low, we favor the forward reaction and hinder the reverse, and obtain the kinetic product. 
  • If the temperature is raised, the reverse reaction becomes energetically accessible, and equilibrium is established. We will then obtain the thermodynamic product. 


A Parting Word Of Warning: The “1,4-Product” Is Not Always The “Thermodynamic” Product

Let’s summarize.  Understanding the following two factors is key to correctly answering exam problems.

  1. The relative importance of the carbocation resonance forms
  2. The relative stabilities of the double bond products.

In the case of butadiene, it is true that the 1,2 product was formed through a more stable carbocation (kinetic product), and the 1,4 product had a more stable double bond (thermodynamic product).

But it will not always be true for all dienes!!

Here are three examples that will help you think through the main issues (answers in the next post).

Other Reactions Can Give 1,2- and 1,4- Additions As Well

This post is long enough, but I would be remiss if I failed to note that 1,2 and 1,4 additions to dienes are also possible for a few other classes of reaction.

  • Addition of HBr to dienes under free-radical conditions (e.g. HBr + peroxides)
  • Addition of Br2 (and Cl2) to dienes.

Note that similar issues will arise (i.e. stability of a reactive intermediate versus stability of the final product). We’ll go into more detail when the time comes.

Many thanks to Tom Struble for help with preparing this post. 


All the textbooks I consulted showed diagrams similar to what I drew above, with Br(-) attacking different resonance forms.  Perhaps a more correct way to draw the formation of the carbocation at the 4-position is to show it like this.

Further Reading

Kharasch, M. S. ; Kritchevsky, J.; Mayo, F.R. “The Addition of Hydrogen Chloride To Butadiene”, J. Org. Chem19372 (5), 489-496 Direct Link

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