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16.8 Electrophilic Addition: 1,2- Versus 1,4-Addition

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  • Addition reactions of isolated dienes proceed more or less as expected from the behavior of simple alkenes. Thus, if one molar equivalent of 1,5-hexadiene is treated with one equivalent of bromine a mixture of 5,6-dibromo-1-hexene, 1,2,5,6-tetrabromohexane and unreacted diene is obtained, with the dibromo compound being the major product (about 50%)

    CH2=CH(CH2)2CH=CH2 + Br2 arrow2.gif BrCH2CHBr(CH2)2CH=CH2 + BrCH2CHBr(CH2)2CHBrCH2Br + CH2=CH(CH2)2CH=CH2
        5,6-dibromo-1-hexene 1,2,5,6-tetrabromohexane 1,5-hexadiene

    Similar reactions of conjugated dienes, on the other hand, often give unexpected products. The addition of bromine to 1,3-butadiene is an example. As shown below, a roughly 50:50 mixture of 3,4-dibromo-1-butene (the expected product) and 1,4-dibromo-2-butene (chiefly the E-isomer) is obtained. The latter compound is remarkable in that the remaining double bond is found in a location where there was no double bond in the reactant. This interesting relocation requires an explanation.

    CH2=CH-CH=CH2 + Br2 arrow2.gif BrCH2CHBr-CH=CH2 + BrCH2CH=CHCH2Br
        3,4-dibromo-1-butene 1,4-dibromo-2-butene

    The expected addition product from reactions of this kind is the result of 1,2-addition, i.e. bonding to the adjacent carbons of a double bond. The unexpected product comes from 1,4-addition, i.e. bonding at the terminal carbon atoms of a conjugated diene with a shift of the remaining double bond to the 2,3-location. (These numbers refer to the four carbons of the conjugated diene and are not IUPAC nomenclature numbers.) Product compositions are often temperature dependent: at 40 oC, 85% of the product mixture in the addition reaction above is the 1,4 product, whereas at 0 oC, only about 30% is the 1,4 product.

    CH2=CH-CH=CH2 + HBr arrow2.gif CH3CHBr-CH=CH2 + CH3CH=CHCH2Br
    reaction temperature   1,2 addition yield 1,4 addition yield
    0 ºC   70% 30%

    40 ºC


    Bonding of an electrophilic atom or group to one of the end carbon atoms of a conjugated diene ( carbon #1 in the figure below) generates an allyl cation intermediate. Such cations are stabilized by charge delocalization, and it is this delocalization that accounts for the 1,4-addition product produced in such addition reactions. As shown in the diagram, the positive charge is distributed over carbons #2 and #4 so it is at these sites that the nucleophilic component bonds. Note that resonance stabilization of the allyl cation is greater than comparable stabilization of 1,3-butadiene, because charge is delocalized in the former, but created and separated in the latter.


    An explanation for the temperature influence is shown in the following energy diagram for the addition of HBr to 1,3-butadiene. The initial step in which a proton bonds to carbon #1 is the rate determining step, as indicated by the large activation energy (light gray arrow). The second faster step is the product determining step, and there are two reaction paths (colored blue for 1,2-addition and magenta for 1,4-addition). The 1,2-addition has a smaller activation energy than 1,4-addition - it occurs faster than 1,4 addition, because the bromide nucleophile is closer to carbon #2 then to carbon #4. However, the 1,4-product is more stable than the 1,2-product. At low temperatures, the products are formed irreversibly and reflect the relative rates of the two competing reactions. This is termed kinetic control. At higher temperatures, equilibrium is established between the products, and the thermodynamically favored 1,4-product dominates.


    When a conjugated diene is attacked by an electrophile, the resulting products are a mixture of 1,2 and 1,4 isomers. Kinetics and Thermodynamics control a reaction when there are two products under different reaction conditions. The Kinetic product (Product A) will be formed fast, and the Thermodynamic product (Product B) will be formed more slowly. Usually the first product formed is the more stable favored product, but in this case, the slower product formed is the more stable product; Product B. Section 16.9 will go into more detail on kinetic vs. thermodynamic control of this reaction.


    1. Write out the products of 1,2 addition and 1,4- addition of HBr to 1,3-cyclo-hexadiene.
    2. What is unusual about the products of 1,2- and 1,4- addition of HX to an unsubstituted cyclic 1,3-diene?
    3. Write out the products of 1,2 addition and 1,4- addition of Br2 to 1,3-cyclo-hexadiene.
    4. Is the 1,2-addition product formed more rapidly at higher temperatures, even though it is the 1,4-addition product that predominates under these conditions?
    5. Why is the 1,4-addition product the thermodynamically more stable product?

    6. Addition of 1 equivalent of bromine to 2,4-hexadiene at 0 degrees C gives 4,5-dibromo-2-hexene plus an isomer. Which of the following is that isomer:

    1. 5,5-dibromo-2-hexene
    2. 2,5-dibromo-3-hexene
    3. 2,2-dibromo-3-hexene
    4. 2,3-dibromo-4-hexene


    Problems 7. and 8.

    Practice Problem 10.gif

    7. The kinetically controlled product in the above reaction is:

    1. 3-chloro-1-butene
    2. 1-chloro-2-butene

    8. For the reaction above, which product is the result of 1,4-addition?

    1. 3-chloro-1-butene
    2. 1-chloro-2-butene

    1. The same product will result from 1,2 and 1,4 addition.

    Answer to 1.gif

    2. Addition of the HX to unsubstituted cycloalka-1,3-dienes in either 1,2- or 1,4- manner gives the same product because of symmetry.

    3. Both 1,2 and 1,4 products will form.

    Answer to 1 B.gif

    4. Yes. the Kinetic Product will still form faster but in this case there will be enough energy to form the thermodynamic product because the thermodynamic product is still more stable.

    5. The 1,4- product is more thermodynamically stable because there are two alkyl groups on each side of the double bond and more substituted alkenes are more stable. 

    6. b - this is the 1,4 product, d is an incorrect name and a and c each have both bromines adding to the same carbon.

    8. a

    8. b


    • Natasha Singh
    • Layne Morsch (University of Illinois Springfield)
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