Anti-Markovnikov Addition of H-Br (How does the placement of Br differ in radical vs. polar addition of HBr to an alkene?)

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1 hemactivity 21 Anti-Markovnikov Addition of 1 hemactivity 21 Anti-Markovnikov Addition of - (ow does the placement of differ in radical vs. polar addition of to an alkene?) Model 1: Radical Reactions of Alkenes Figure 1a: Radical hain Reactions with 2 and Rxn I excess Rxn II excess ritical Thinking Questions 1. Review: Use curved arrows to show propagation steps that lead to the products shown in Rxn I. Do not show any termination steps.

) Model 1: Radical Reactions of Alkenes Figure 1a: Radical hain Reactions with 2 and Rxn I excess Rxn II

2 hemactivity 21 Anti-Markovnikov Addition of 2 Rxn I excess Figure 1b: Reaction of a p Bond with a Radical R X X R Will form the lowest V.E. radical product 2. Devise a series of radical chain reaction propagation steps that lead to the products shown in Rxn II (redrawn for you below). Do not show any termination steps. Rxn II excess 3. Explain why the following radical is not produced during Rxn II.

Devise a series of radical chain reaction propagation steps that lead to the products shown in Rxn

3 hemactivity 21 Anti-Markovnikov Addition of 3 A primary radical is higher in P.E. than a secondary radical, and the lowest P.E. radical is most likely to form. Model 2: Radical hain Reaction Initiators Previously we have seen that a strong light source can be used to initiate a radical chain reaction by breaking a bond. hν owever, this method can only be used when there is a high concentration of 2 in solution. Rxn II (above) has a high concentration of - instead of -.

Model 2: Radical hain Reaction Initiators Previously we have seen that a strong light source can be used to initiate a

4 hemactivity 21 Anti-Markovnikov Addition of 4 In cases where concentration is low, peroxides are used as radical initiators because the - bond is easy to break. Benzoyl peroxide is a common radical initiator. heat ( ) or light (hν) benzoyl peroxide Note: is an abbreviation indicating the addition of heat to a reaction. Bond dissociation energy (BDE) is the amount of energy required to break a bond in a non-polar fashion ("hemolytic bond cleavage"), with one electron going to each of the two atoms involved in the bond. The BDE of the bond of hydrogen peroxide ( ) = 51 kcal/mole (kcal = a measure of energy) benzoyl peroxide = 18 kcal/mole ritical Thinking Questions 3. Which is easier to break, the bond of hydrogen peroxide or the bond of benzoyl peroxide? ite data from Model 2 to support your answer. The BDE of benzoyl peroxide is only 18 kcal/mole as compared to 51 kcal/mole for hydrogen peroxide. This means that it takes much less energy to break the - bond of benzoyl peroxide. 4. onstruct a reasonable mechanism for the following reaction. Show... i. an initiation step. ii. iii. propagation steps that lead to the product shown. two possible termination steps. excess a tiny amount of benzoyl peroxide (R R) heat ( ) where R = other products initiation step (net production of radicals) R R R R propogation steps R R (one radical consumed one radical produced) one possible termination step R (net consumption of radicals) R another possible termination step

Bond dissociation energy (BDE) is the amount of energy required to break a bond in a non-polar fashion ("hemolytic bond cleavage"), with one electron going to each of the two atoms involved in the

5 hemactivity 21 Anti-Markovnikov Addition of 5 Model 3: Radical and Polar Addition of to an Alkene Rxn IIIa dark, no peroxides Markovnikov propene (propylene) Rxn IIIb benzoyl peroxide (R R) heat ( ) or light (hν) Anti-Markovnikov + other products styrene Rxn IVa Rxn IVb dark, no peroxides benzoyl peroxide (R R) heat ( ) or light (hν) Markovnikov Anti-Markovnikov ritical Thinking Questions 5. onsider the regiochemistry of reactions IIIa and IIIb and label each product Markovnikov or anti-markovnikov, as appropriate. a) Show the mechanism of Rxn IVa in the space below. Rxn IVa styrene dark, no peroxides Markovnikov b) Draw the major carbon containing product of Reactions IVa at the top of the page. 6. Summarize the pattern you see regarding which reagents/conditions produce Markovnikov products and which produce anti-markovnikov products. In the dark, with no peroxides and no possibility of radical formation, a Markovnikov addition product is observed. With a peroxide and light or heat to initiate a radical reaction, an anti- Markovnikov product is produced. 7. Without going through the mechanism, draw the product of Rxn IVb, above, and label it as Mark. or anti-mark., as appropriate.

onsider the regiochemistry of reactions IIIa and IIIb and label each product Markovnikov or anti-markovnikov, as appropriate. a) Show the mechanism of Rxn IVa in the space below.

6 hemactivity 21 Anti-Markovnikov Addition of 6 8. A student proposes the following multi-step synthesis of the starting from 2-methylpropene. (By convention, in a synthesis, reagents are placed above each arrow, and non-carbon containing products are not shown.) dark, no peroxides Na S S use, benzoyl peroxide and heat a) ne step in this synthesis is wrong. ross out the reagents for this step. b) orrect the student by writing in reagents that will accomplish this step. 9. To each box below add reagents that will accomplish the transformation shown. 2 hν tbutyl, heat benzoyl peroxide N N radical bromination E2 anti-mark. add'n of S N Write the appropriate reaction name below each reaction arrow in TQ 7: hoose from: S N 2, E2, radical bromination, anti-markov. addition of. 11. Add reagents to the box to complete the multi-step synthesis below. R dilute l NEW RXN + enantiomer + enantiomer Note: The first step is a new reaction found on reaction sheet R4. You do not need to know the mechanism of this new reaction.

) dark, no peroxides Na S S use, benzoyl peroxide and heat a) ne step in this synthesis is wrong. ross out the reagents for this step.

7 hemactivity 21 Anti-Markovnikov Addition of 7 Exercises 1. Design a synthesis of the following s beginning with the starting material given. Write the reagents needed for each transformation above the reaction arrows for that step in the synthesis. a) + enantiomer b) c)

8 hemactivity 21 Anti-Markovnikov Addition of 8 2. ver each reaction arrow, write in reagents that would give the product shown? and 3. Write the mechanism of each reaction in Exercise #2. 4. xygen oxygen single bonds are relatively easy to break. onstruct an explanation for why benzoyl peroxide has an especially small B.D.E. compared to other peroxides? (see Model 2) 5. Read the assigned pages in the text and do the assigned problems.

Write the mechanism of each reaction in Exercise #2. 4.

9 hemactivity 21 Anti-Markovnikov Addition of 9

CH 3 Addition to an alkene with Br 2. No reaction when an aromatic molecule is mixed with Br 2. No Reaction. + H Br

CH 3 Addition to an alkene with Br 2. No reaction when an aromatic molecule is mixed with Br 2. No Reaction. + H Br RADIALS Reactions with 2 : 2 3 Addition to an alkene with 2 2 No reaction when an aromatic molecule is mixed with 2 2 (in the dark) No Reaction 2 h (in the light) During a demonstration by Dr., the reactants