Ace Your Next Organic Chemistry
Exam.

With these Downloadable PDF Study Guides

Our Study Guides

Dienes and MO Theory

By James Ashenhurst

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

Last updated: August 31st, 2020 |

Kinetic Versus Thermodynamic Control In Addition of HBr to Dienes: 1,2- and 1,4- Addition

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. 
summary of 12 and 14 addition with addition of hbr to butadiene thermodynamic versus kinetic product varies with temperature kinetic versus thermodynamic control

Table of Contents

  1. Addition To Alkenes, Revisited
  2. Reaction Of Butadiene With Acid Gives “1,2-Addition” And “1,4-Addition” Products
  3. The “1,4 Addition” Product Of Acids Adding To Butadiene
  4. Protonation Of Butadiene Gives A Resonance-Stabilized Carbocation
  5. The Effect Of Temperature: Low Temperatures Give “1,2” Addition To Butadiene
  6. Higher Temperatures Give More Of The “1,4” Product
  7. With Butadiene, The “1,4” Product Is More Stable Because It Has A More Substituted Double Bond
  8. “Kinetic Control” vs “Thermodynamic Control”: At Low Temperatures The Reaction Is Irreversible And Products Are Determined By Relative Rates
  9. Thermodynamic Control: At Higher Temperatures The Reaction Is Reversible And Product Distribution Is Determined By Stability
  10. Using An Energy Diagram To Understand Thermodynamic Versus Kinetic Control
  11. A Parting Word Of Warning: The “1,4-Product” Is Not Always The “Thermodynamic” Product
  12. Other Reactions Can Give 1,2- and 1,4- Additions As Well
  13. Notes
  14. (Advanced) References and Further Reading

1. 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?

review the addition of acids to alkenes with hbr giving alkyl bromides also works for hi and h2o follows markovnikovs rule

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.

example of markovnikov rule with 1-butene step 1 attack hydrogen of hbr resulting in more substituted carbocation followed by attack of bromide ion to give alkyl bromide

2. Reaction Of Butadiene With Acid 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.

question what happens when butadiene is treated with hbr what product do you get

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:

when butadiene is treated with hbr two products result we call these 12 and 14 addition the 12 product is formed at lower temperature the 14 product formed at higher temperature

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”.

definition of 12 addition in addition of hbr to butadiene the hydrogen ends up on carbon 1 and the bromine ends up on carbon 2 adds to adjacent carbons

3. The “1,4 Addition” Product Of Acids Adding To Butadiene

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”.

definition of 14 addition in addition of hbr to butadiene the h ends up attached to carbon 1 and the bromine ends up attached to carbon 4 hence 14 addition internal alkene

[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?

4. 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?

protonatoin of butadiene gives carbocation that is resonance stabilized with major and minor resonance contributors depending on substitution of carbocation recall resonance hybrid

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:
intermediate from protonation of butadiene with hcl

[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.

5. The Effect Of Temperature: Low Temperatures Give “1,2” Addition To Butadiene

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.

low temperature addition of hbr to butadiene gives 12 addition about 80 per cent at minus 80 degrees

6. Higher Temperatures Give More Of The “1,4” Product

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%).

at higher temperature addition of hbr to butadiene gives 14 addition at 40 degrees celsius addition of hbr to butadiene 80 per cent 14

7. With Butadiene, The “1,4” Product Is More Stable Because It Has A More Substituted Double Bond

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.

recall that stability of alkenes increases as number of attached carbons increases 12 has monosubstituted double bond 14 has disubstituted 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?

8. “Kinetic Control” vs “Thermodynamic Control”: At Low Temperatures The Reaction Is Irreversible And Products Are Determined By Relative Rates

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”.

reaction energy diagram of kinetic control low temperature 12 transition state and 14 transition state activation energy less for 12 addition

9. Thermodynamic Control: At Higher Temperatures The Reaction Is Reversible And Product Distribution Is Determined By Stability

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.

at high enough temperatures 12 product is reversible can go back to resonance stabilized carbocaiton 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”.

at high temperatures there is equilibrium between 12 product and 14 product which is called 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. 

10. 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:

reaction energy diagram for 12 versus 14 addition of hbr to butadiene showing resonance stabilized carbocations and reversibility to 12 and 14 products

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. 

11.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).

real life exam problems to think through 12 and 14 addition to various dienes cyclopentadiene 25 dimethyl 2 4 hexadiene and also 1 methyl cyclohexadiene

12. 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. 


Notes

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.

equivalent way to draw 14 addition to butadiene showing multiple arrows

 


(Advanced) References and Further Reading

  1. THE ADDITION OF HYDROGEN CHLORIDE TO BUTADIENE
    M. S. KHARASCH, J. KRITCHEVSKY, and F. R. MAYO
    The Journal of Organic Chemistry 1937, 02 (5), 489-496
    DOI: 10.1021/jo01228a010
    An early paper by the esteemed chemist M. S. Kharasch on the addition of HCl to butadiene, seeing whether the ratio of 1,2 to 1,4 addition varied with different reaction conditions.
  2. THE PEROXIDE EFFECT IN THE ADDITION OF REAGENTS TO UNSATURATED COMPOUNDS. XIII. THE ADDITION OF HYDROGEN BROMIDE TO BUTADIENE
    M. S. KHARASCH, ELLY T. MARGOLIS, and FRANK R. MAYO
    The Journal of Organic Chemistry 1936, 01 (4), 393-404
    DOI: 10.1021/jo01233a008
    Another early study by M. S. Kharasch in which he studies the addition of HBr to butadiene, in which he attempts to rigorously separate the two modes of addition – electrophilic vs. radical.
  3. —Mobile-anion tautomerism. Part I. A preliminary study of the conditions of activation of the three-carbon system, and a discussion of the results in relation to the modes of addition to conjugated systems
    Harold Burton and Christopher Kelk Ingold
    J. Chem. Soc. 1928, 904-921
    DOI: 10.1039/JR9280000904
    A classic paper from a pioneer of physical organic chemistry, Prof. C. K. Ingold. He suggests the general idea that the 1,2 and 1,4 addition products are in equilibrium and that the 1,2-product can reverse reaction and form the 1,4 product. This is the basis of what we now call “kinetic” and “thermodynamic” control.
  4. The modes of addition to conjugated unsaturated systems. Part IX. A discussion of mechanism and equilibrium, with a note on three-carbon prototropy
    P. B. D. de la Mare, E. D. Hughes and C. K. Ingold
    J. Chem. Soc. 1948, 17
    DOI: 10.1039/JR9480000017
    Here, C. K. Ingold has reviewed all work (including Kharasch’s) done up to that point in the study of electrophilic additions to conjugated systems.
  5. Properties of conjugated compounds. Part XI. Addition of hydrogen bromide to βγ- and αδ-dimethylbutadiene
    Ernest Harold Farmer and Frederick G. B. Marshall
    J. Chem. Soc. 1931, 129
    DOI: 10.1039/JR9310000129
    When methyl groups are added to butadiene, the 1,2 and 1,4 addition ratios can change significantly!  This is because the stability of the carbocation intermediates will change.

 

Comments

Comment section

8 thoughts on “Reactions of Dienes: 1,2 and 1,4 Addition

  1. Difficult Concepts made so easy
    Do you offer courses from scratch to proficient level
    Do get back on my mail id
    I am interested in a regular course that compels you to study
    Am a tutor
    Duration should be around 6 to 9 months

  2. Why does the more stable intermediate have a lower barrier to its transition state? Shouldn’t the opposite be true?

Leave a Reply

Your email address will not be published. Required fields are marked *

This site uses Akismet to reduce spam. Learn how your comment data is processed.