chem 201 activity 24: radical chain mechanisms model 1 –...

Chem 201 Activity 24: Radical chain mechanisms

(What do radicals do? What does a radical chain mechanism look like)

Model 1 – Homolysis Reactions are Highly Endothermic

Heterolysis

Y Z Y+

Z Example

C Cl

CH3

CH3

CH3 C

CH3

CH3

CH3 Cl

Homolysis

Y Z Y Z

Examples

Br Br Br Br

CH3 CH3 CH3CH3

'Hrxn (kcal/mol)

46

88

Questions 1. Add lone pairs and curved arrows as needed to the heterolysis example in Model 1.

2. A curved arrow (two barbs on arrowhead) shows the movement of an electron pair.

a. How many electrons does a ‘fish hook’ arrow (one barb on arrowhead) move? b. How many fish hooks are required to show a homolysis reaction?

3. Add lone pairs, unpaired electrons, and fish hooks to the homolysis examples in Model 1.

4. Radical (or free radical) refers to an atom with an unpaired electron. It can also refer to the molecule that contains this atom.

a. Draw the radicals that appear in the examples in Model 1 here: b. What kind of reaction, heterolysis or homolysis (circle one), produces radicals? c. How many radicals does a homolysis produce? d. Draw a reaction coordinate diagram for Br2 ĺ 2Br. Assume the barrier for the reverse reaction,

radical recombination, is very small.

PE

Br2 2 Br

Ch201 Activity 24: Radical chain mechanisms (Ch. 12) 2

e. Why is homolysis of Br-Br and H3C-C nonexistent at room temperature? 1

Model 2 – Halogenation of Alkanes Gives Mixtures Chlorine (Cl2) and bromine (Br2) react with alkanes when irradiated with ultraviolet light (signified by UV or hȣ). The following reactions are typical:

CH3CH C

H2

CH2

CH CH3

CH3

CH3

X2

hQX X

X

X2 = Cl2 32% 42% 26%

X2 = Br2 <1% 9% 91%

Similar halogenations occur with F2, but are very fast, release large quantities of energy, and generate large amounts of corrosive, toxic HF. Halogenation does not occur to a significant extent with I2.

Questions 5. Classify the reactions in Model 2 as substitution, addition, or elimination.

6. C-X and C-O bonds are treated the same way when calculating oxidation numbers. Classify the reactions in Model 2 as reduction, oxidation, or neither.

7. Which substitution of H by X occurs with greater selectivity, H ĺ Cl or H ĺ Br?

8. Which type of H appears to be replaced most quickly by Br: 1°H, 2°H or 3°H? Which type of H is replaced most slowly?

_____ (fastest) >> _____ >> _____ (slowest)

Fact 24.1. The speed of H ĺ Br is negatively correlated with the strength of the alkane CH bond:

3° CH (weakest) < 2° CH < 1° CH < CH4 (strongest)

9. What are the number of 1°, 2° and 3°H in 2,5-dimethylhexane, the starting material in Model 2?

1 Activity 6, Model 5 contained these data: half-life of a molecule ‘protected’ by a barrier of 130 kJ/mol was ‘days’ at 100˚C and ‘forever’ at 25˚C (1 kcal/mol § 4.2 kJ/mol so 46 and 88 kcal/mol § 190 and 370 kJ/mol, respectively).


10. Use the data in Model 2 to construct supporting explanations for, or complete, the following statements:

a. If every H in 2,5-dimethylhexane were equally reactive, the product distribution would be 67% 1°X and 22% 2°X and just 11% 3°X.

b. Equal amounts of 1°X and 2°X would be produced if each 2°H were ______ times as reactive as each 1°H.

c. Equal amounts of 1°X and 3°X would be produced if each 3°H were ______ times as reactive as each 1°H.

11. Use the data in Model 2 to complete these statements:

a. Each 3°H in 2,5-dimethylhexane reacts about _______ times faster with Br2 than each 2° H. b. Each 3°H in 2,5-dimethylhexane reacts _______ (faster, slower, at about the same rate) with Cl2

______ (than, as) each 2° H.

Model 3 – Halogenation Occurs by a Radical Chain Mechanism A chain mechanism is one that consumes a species necessary for the reaction and then produces that species again so that the reaction can continue. The individual steps in a chain mechanism are called propagation steps.

Propagation step #1 Propagation step #2

Questions 12. According to the examples in Model 3, how many fish hooks are needed

a. To show the breaking of a covalent bond: ______ b. To show the forming of a covalent bond: ______

13. Cross out any species that appears both as a reactant and as a product in the “sum” equation in Model 3. Write a “net” equation for the two-step chain in Model 3 based on the remaining species.


14.

a. Draw the species in Model 3 that justifies calling the mechanism a chain mechanism. b. Why can’t we use the consumption of alkane, RH, to justify calling this a chain mechanism?

c. Why can’t we use the production of HX to justify calling this a chain mechanism?

d. T or F. Propagation step #1 consumes and produces the same radical. If T, draw the radical that appears as a reactant and product in step #1.

15. Two commonly observed steps in radical chain mechanisms are hydrogen atom transfer and halogen atom transfer. Label each propagation step in Model 3 with its appropriate name.

Fact 24.2 Fish hook rules: The examples in Model 3 show that every electron that participates in a radical reaction must be guided by a fish hook. Fish hooks start at bonds or unpaired electrons, never at atoms. Fish hooks meet where new bonds are required. A fish hook that points to an atom delivers an unpaired electron to that atom. More rules: Never mix curved arrows and fish hooks in the same drawing. An unpaired electron on one atom never jumps to another atom.

16. According to Fact 24.2, why are 3 fish hooks used in each atom transfer reaction in Model 3?

17. The fish hooks in the following drawings violates one or more rules in Fact 24.2. Find and describe all of the fish hooks violations.

Y Z Y Z

Y Z ZX X Y

Y Z ZX X Y

Y Z ZX X Y


Model 4 – Bu3SnH Reduces Haloalkanes by a Radical Chain Mechanism Bu = CH2CH2CH2CH3 (n-butyl), Bu3SnH = tributyl tin hydride. The same column in the periodic table that contains C also contains Si, Ge (germanium), Sn (tin), and Pb (lead).

Questions 15. Add fish hooks to the reactants of each step in Model 4

16. Write the “net reaction” for Model 4 under the line.

a. What tin-containing species is consumed in the net reaction? b. What tin-containing species is produced by the net reaction? c. What is the tin-containing reagent that reduces haloalkanes?

17. What radical in Model 4 justifies calling this a radical chain mechanism?

18. Label each propagation step in Model 4 as either hydrogen atom transfer or halogen transfer.

Model 5 – NBS Creates Allylic & Benzylic Bromides by a Radical Chain Mechanism

NBr

OO

N-BromosuccinimideNBS


Questions 19. How many propagation steps appear in the mechanism in Model 5?

20. ‘Succ’ stands for succinimide. This compound is related to NBS by replacing Br with H. Draw ‘Succ’.

21. Label propagation steps in Model 5 that involve radicals as hydrogen atom transfer or halogen transfer as needed.

22. Draw fish hooks on the reactants of propagation steps in Model 5 that involve radicals.

23. Label allyl radical wherever it appears in Model 5. Draw fish hooks needed to interconvert the two resonance structures in propagation step #1.

24. Propagation step #2 is very fast, but does not involve radicals. It probably looks something like this:

a. Add lone pairs and curved arrows to the reactants. b. Draw the resonance structure implied by the curved arrows on the conjugate acid of NBS. c. Add lone pairs and curved arrows where needed to the Br’s in the ion pair intermediates. Also

add lone pairs to the products of the Br transfer. d. The HO-C=N structure is an unstable short-lived tautomer of succinimide. You have seen an

analogous pair of tautomers in connection with vinyl alcohols. Draw brackets around the unstable tautomer below (BVA!) and draw the more stable compound that it turns into.

CH3

C CCH3O

H

H


Fact 24.3 Selectivity of allylic/benzylic bromination reflects the fact that the speed of H ĺ Br is negatively correlated with the strength of the alkane CH bond:

Benzylic CH (weakest) < Allylic CH < 3° CH < 2° CH < 1° CH < CH4 (strongest)

Fact 24.4 NBS (reagent recipe shown below) gives exclusively allylic and benzylic bromination. Example:

NBS

(PhCOO)2, '

Br

25. Use Fact 24.4 to predict the products of these reactions:

NBS

(PhCOO)2, '

NBS

(PhCOO)2, 'draw dibromo product

Model 6 – Initiation Steps Create Chain-Carrying Radicals Model 1 provided evidence that covalent bonds will not undergo homolysis spontaneously. Exceptions to this rule-of-thumb are 1) molecules that contain unusually weak bonds (these break when heated) or 2) molecules that undergo homolysis after they absorb UV radiation (this can excite an electron from a BMO to an ABMO).

The chemical agents that produce radicals are called radical initiators and the chemical reactions that lead to chain-carrying radicals are called initiation steps.


Questions 26. Three radical initiators are shown in Model 6. Circle the non-radical reactant (if an abbreviation is provided, circle that instead) and label it as ‘light-sensitive initiator’ or ‘heat-sensitive initiator’.

27. Draw fish hooks for the initiation steps shown in Model 6.

28. The two heat-sensitive initiators produce ‘allyl-like’ radicals in the first step. Draw another resonance structure for each of these radicals.

29. Initiators, like catalysts, can be added in small amounts. However, the justification for this is different.

a. Which substance, an initiator or a catalyst, is not consumed during a reaction? b. Construct an explanation for why initiators can be used successfully in small amounts.

30. Each initiation sequence in Model 6 includes a molecule that will act as the reactant for the desired reaction. This molecule must be present in large amounts. Circle these molecules and label them ‘reactant’.

31. Each initiation sequence in Model 6 produces a radical that can carry a chain mechanism forward. Circle these radicals and label them ‘chain carriers’.

7 – O2 Creates Allylic Peroxides by a Radical Chain Mechanism Triplet O2, is ubiquitous in the natural environment and can create radicals for chain mechanisms. Here are the initiation steps leading to allylic peroxides.


Allyl radical carries a chain mechanism that consumes more O2 and leads to allylic peroxides.

Questions 32. Label all ‘allyl radicals’ that appear in Model 7.

33. What radical ‘carries’ the chain mechanism in Model 7, i.e., is consumed and regenerated?

34. Use Fact 24.2 to explain why O2 reacts much faster with allylic CH (and benzylic CH) than with alkyl CH. What kind of CH bond might be even more O2-sensitive than allylic CH?

28. Combine the initiation steps in Model 7 and draw the net reaction.

29. How many propagation steps are there in the chain mechanism in Model 7?

30. Combine the propagation steps in Model 7 and draw the net reaction.

31. Complete the sentence: According to Model 7, the H and allyl groups bonded to O2 in the allyl peroxide come from ___________ (the same/different) alkene(s).

32. Termination steps are reactions that destroy radicals needed for the chain to continue. Termination steps often involve radical recombination.

a. Draw the product of each radical recombination.

b. Describe how each radical recombination would affect the rate of allyl peroxide formation (raise/lower the rate).


c. A chain reaction can run indefinitely as long as there are available reactants. In practice, a chain might repeat 100’s or 1000’s of times before termination occurs.

o Are the concentrations of short-lived chain-carrying radicals lower than, about the same as, or higher than, the concentrations of the reactants?

o Which kind of collision is likely to occur more often: radical-radical or radical-reactant?

o Which kind of reaction probably occurs more often: propagation step or termination step?

chem 201 activity 24: radical chain mechanisms model 1 –...

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