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Addition to
Conjugated Dienes

Understand how electrophilic addition to conjugated dienes produces 1,2-addition and 1,4-addition products, and learn what controls their ratio.

Organic Chemistry 1

What is a Conjugated Diene?

A conjugated diene has two \( \mathrm{C{=}C} \) double bonds separated by exactly one single bond. The \( \pi \) electrons are delocalized across all four carbons, creating special reactivity not seen in isolated dienes.

\(\mathrm{C1}\)
\(\mathrm{C2}\)
\(\mathrm{C3}\)
\(\mathrm{C4}\)
1,3-Butadiene
delocalized

Types of Diene Systems

Conjugated
\( \mathrm{C{=}C{-}C{=}C} \): double bonds separated by one single bond. Overlap of \( \pi \) orbitals gives extra stability.
Isolated
\( \mathrm{C{=}C{-}C{-}C{=}C} \): separated by two or more single bonds. Each double bond reacts independently.
Cumulated
\( \mathrm{C{=}C{=}C} \): double bonds share a carbon (allene). Orthogonal \( \pi \) systems, no conjugation.

Why Conjugation Matters

When an electrophile (such as \( \mathrm{HBr} \)) attacks a conjugated diene, the intermediate carbocation is allylic: it can be stabilized by resonance across the remaining \( \pi \) bond. This is what makes two different products possible.

Allylic Carbocation (resonance stabilized)
\(\mathrm{C1}\)
\(\mathrm{C2}\)
\(\mathrm{C3}\)
\(\mathrm{C4}\)
\(+\)
\(\leftrightarrow\)
\(\mathrm{C1}\)
\(\mathrm{C2}\)
\(\mathrm{C3}\)
\(\mathrm{C4}\)
\(+\)
Charge on C2
Charge on C4

1,2-Addition vs. 1,4-Addition

1,2-Addition (Direct)
\( \mathrm{H} \) adds to C1, \( \mathrm{Br} \) adds to C2. The remaining double bond stays between C3 and C4. This is the kinetic product.
1,4-Addition (Conjugate)
\( \mathrm{H} \) adds to C1, \( \mathrm{Br} \) adds to C4. A new double bond forms between C2 and C3. This is the thermodynamic product.
1,3-Butadiene \(+\) HBr
1,2-add
1,4-add
1,2-Product
\(\mathrm{H}\)
\(\mathrm{Br}\)
C1
C2
C3
C4
\(\mathrm{C3{=}C4}\) double bond remains
1,4-Product
\(\mathrm{H}\)
\(\mathrm{Br}\)
C1
C2
C3
C4
\(\mathrm{C2{=}C3}\) double bond (new!)

Mechanism: Step by Step

1
Protonation at C1 (Markovnikov): \( \mathrm{H^+} \) adds to the terminal carbon, forming an allylic carbocation with the positive charge delocalized over C2 and C4.
2
Nucleophilic attack by \( \mathrm{Br^-} \): the bromide ion can attack at C2 (1,2-addition) or C4 (1,4-addition), because both carbons share the positive charge.
3
Product formation: attack at C2 gives the 1,2-product; attack at C4 gives the 1,4-product with an internal double bond between C2-C3.

Kinetic vs. Thermodynamic Control

Kinetic Control
Low temperature (\( -80\,^\circ\mathrm{C} \)). Reaction is irreversible. The faster-forming product dominates: 1,2-addition.
Thermodynamic Control
High temperature (\( 40\,^\circ\mathrm{C} \)). Reaction is reversible. The more stable product accumulates: 1,4-addition.
Energy Diagram
Energy
Reaction coordinate
Reactants
Intermediate
TS (1,2)
1,2-product
TS (1,4)
1,4-product
lower barrier
more stable

Why 1,2 is the Kinetic Product

The 1,2-product forms faster because \( \mathrm{Br^-} \) attacks the carbon closest to where the positive charge initially forms. C2 is right next to the original protonation site: the nucleophile doesn't need to "travel" far.

Why 1,4 is the Thermodynamic Product

The 1,4-product has an internal (more substituted) double bond between C2 and C3. Internal alkenes are more stable than terminal alkenes due to hyperconjugation and better orbital overlap.

1
More substituted \( \mathrm{C{=}C} \): the 1,4-product places the double bond between two carbons that each bear substituents, stabilizing it.
2
Lower energy product: sits in a deeper energy well on the energy diagram, so at equilibrium it is thermodynamically favored.
3
Reversibility: high temperature lets the system reach equilibrium, and the more stable product wins.

Key Distinction: Where Does \( \mathrm{Br^-} \) Attack?

1,2-Addition 1,4-Addition
Site of \( \mathrm{Br^-} \) attack C2 C4
Double bond position \(\mathrm{C3{=}C4}\) (terminal) \(\mathrm{C2{=}C3}\) (internal)
Stability Less stable More stable
Formation rate Faster Slower
Favored at Low temperature High temperature
Control type Kinetic Thermodynamic

How to Predict the Product

1
Identify the conjugated diene: confirm you have alternating double-single-double bond pattern.
2
Add \( \mathrm{H^+} \) to a terminal carbon (Markovnikov): this creates the most stable allylic carbocation.
3
Draw both resonance forms of the allylic cation: identify C2 and C4 as sites bearing the positive charge.
4
Attack at C2 gives 1,2-product; attack at C4 gives 1,4-product.
5
Check the temperature: low \( T \) favors kinetic (1,2); high \( T \) favors thermodynamic (1,4).

Example 1: \( \mathrm{HBr} \) + 1,3-Butadiene

Predict both products of the reaction of \( \mathrm{HBr} \) with 1,3-butadiene at \( -80\,^\circ\mathrm{C} \).

Reactant
\(+ \mathrm{HBr}, -80\,^\circ\mathrm{C}\)
1,2-Product (major)
\(\mathrm{H}\)
\(\mathrm{Br}\)
1,4-Product (minor)
\(\mathrm{H}\)
\(\mathrm{Br}\)

Example 2: Same Reaction at High Temperature

Now run the same reaction, \( \mathrm{HBr} \) + 1,3-butadiene, but at \( 40\,^\circ\mathrm{C} \). What changes?

1,4-Product (major)
At \( 40\,^\circ\mathrm{C} \), the reaction is reversible. Equilibrium favors the more stable internal alkene: the 1,4-product dominates.
1,2-Product (minor)
Still forms (faster), but at high \( T \) it can revert back to the allylic cation and re-form as the 1,4-product.

Example 3: \( \mathrm{HCl} \) + 2-Methyl-1,3-butadiene

With an unsymmetrical diene, you must decide which end \( \mathrm{H^+} \) adds to. Apply Markovnikov's rule: protonate the end that gives the more stable allylic carbocation.

1
Protonate C1: gives a tertiary allylic cation (charge delocalized to C2, which is tertiary due to the methyl group, or C4). This is more stable than the alternative.
2
1,2-product: \( \mathrm{Cl} \) at C2 with double bond at \(\mathrm{C3{=}C4}\).
3
1,4-product: \( \mathrm{Cl} \) at C4 with new double bond at \(\mathrm{C2{=}C3}\) (internal, trisubstituted).

Common Mistakes

✕ Mixing up kinetic/thermodynamic
Students often think "thermodynamic = fast" because it sounds powerful. Remember: thermodynamic = most stable, not fastest.
✕ Forgetting to draw both resonance forms
If you only draw one resonance form of the allylic cation, you'll miss one of the two products entirely.
✕ Wrong protonation site
For unsymmetrical dienes, protonating the wrong end gives a less stable (non-allylic) cation. Always form the allylic intermediate.

Summary: The Big Picture

Feature 1,2-Addition 1,4-Addition
Atoms added across Carbons 1 and 2 Carbons 1 and 4
Double bond in product \(\mathrm{C3{=}C4}\) (terminal) \(\mathrm{C2{=}C3}\) (internal)
Relative stability Less stable More stable
Activation energy Lower (faster) Higher (slower)
Favored by Low T (kinetic) High T (thermodynamic)
Key principle Fastest path wins Most stable product wins

Question 1 of 5

When \( \mathrm{HBr} \) reacts with 1,3-butadiene at low temperature, which product is the major product?

Question 2 of 5

Why is the 1,4-addition product more thermodynamically stable than the 1,2-addition product?

Question 3 of 5

Both the 1,2- and 1,4-products form through the same intermediate. What is that intermediate?

Question 4 of 5

If you want to maximize the yield of the 1,4-addition product, what reaction conditions should you use?

Question 5 of 5

In the addition of \( \mathrm{HCl} \) to 2-methyl-1,3-butadiene (isoprene), which carbon does the proton add to first?