Name: Unit 3 Packet: Activation Energy, Free Radical Chain Reactions, Alkane Preparations, S N 2, E 2

Name: Unit 3 Packet: Activation Energy, Free Radical Chain Reactions, Alkane Preparations, S N 2, E 2

Key Terms For Unit 3 Free Radical Chain Reaction Homolytic Cleavage Free Radical Initiation Propagation Termination Endothermic Exothermic Activation Energy Bond Energies Selectivity Alkane Preparation Hydrogenation Organometallics Nucleophile Electrophile Grignard Reaction Wurtz Reaction Corey-House Reaction 2 S N E 2 Saytzeff Product Hoffmann Product 62

Making an Alkane More Useful: Free Radical Chain Reaction As you ve already learned, alkanes are not very useful in the world of organic chemistry. They are involved in relatively few reactions of interest, capable of doing very little other than burning. There is only really one simple way of taking an alkane, and converting it to something more useful. The pathway used is called a chain reaction. Consider methane, the simplest alkane, with diatomic chlorine. While methane and chlorine are indefinitely stable when combined in the dark, upon irradiation with light, a mixture of the following products are formed. Chlorination: CH 4 + Cl 2 hv CH 3 Cl + CH 2 Cl 2 + CHCl 3 + CCl 4 Methane Chlorine Methyl chloride Methylene chloride Chloroform Carbon Tetrachloride The composition (relative abundances) of the products for this chlorination reaction depends upon the amount of chlorine available and the amount of time the reaction is allowed to run. If sufficient chlorine and sufficient time were allowed, all of the methane would be converted to carbon tetrachloride. Through a composition vs. time analysis, it has been determined that methyl chloride is formed first, and the products are made sequentially from there as shown below. CH 4 + Cl 2 hv CH 3 Cl + Cl 2 hv CH 2 Cl 2 + Cl 2 hv CHCl 3 + Cl 2 hv CCl 4 Reaction 1 Reaction 2 Reaction 3 Reaction 4 Let s focus in for a minute on the initial reaction between methane and chlorine to form methyl chloride. H 3 C H + Cl Cl hv H 3 C Cl + H Cl 105 kcal/mol 59 kcal/mol 85 kcal/mol 103 kcal/mol Breaking bonds costs energy (endothermic), but forming new bonds gives off energy (exothermic). So to calculate the overall energy of reaction, the following calculation is performed. Bonds Made = -85-103 = -188 kcal/mol Bonds Broken = 105 + 59 = +164 kcal/mol -24 kcal/mol The reaction is exothermic! This reaction, like many others in organic chemistry, is not explained with a single step, but rather a combination of several steps. The set of specific steps in a reaction, that display which bonds are being formed and broken, is called a reaction mechanism. 63

Mechanism Free Radical Chlorination of Methane The mechanism for the free radical chlorination of methane (or any alkane, for that matter) involves an initiation step, two propagation steps, and one of any number of termination steps....... Initiation: :Cl Cl: hv 2 :Cl. Propagation:.... H 3 C H +. Cl: H 3 C. + H Cl:........ H 3 C. + :Cl Cl: H 3 C Cl: +. Cl: H 3 C. + H 3 C. H 3 C CH 3 or.... Termination: H 3 C. +. Cl: H 3 C Cl: or........ :Cl. +. Cl: :Cl Cl: Let s take a closer look at the thermochemistry of each of the propagation steps. Although the second step is exothermic by 26 kcal/mol, and thus is very favorable, the first propagation step is endothermic by about 2 kcal/mol. First Propagation Step: Second Propagation Step:............ H 3 C H. Cl: H 3 C. + H Cl: H 3 C. :Cl Cl: H 3 C Cl: +.Cl: 105 kcal/mol 103 kcal/mol 59 kcal/mol 85 kcal/mol This reaction is endothermic by 2 kcal/mol This reaction is exothermic by 26 kcal/mol The carbon-hydrogen bond in methyl chloride is weaker than the one in methane, so methyl chloride that is formed converts quickly to form methylene chloride. Methylene chloride and chloroform are more reactive still, and the reaction will speed up as it progresses towards carbon tetrachloride. 64

Chlorination of n-butane: Halogenation of Other Alkanes In methane, all of the hydrogens are equivalent and primary. In n-butane, there are hydrogens on the terminal carbons and hydrogens on carbons in the middle of the chain. By analyzing the products of monochlorination of n-butane, we can see that chlorine has a preference towards adding to the 2º position to form sec-butyl chloride. CH 3 CH 2 CH 2 CH 3 + Cl 2 hv CH 3 CH 2 CHCH 3 + CH 3 CH 2 CH 2 CH 2 Cl Cl sec-butyl chloride n-butyl chloride (72.2%) (27.8%) This analysis would seem to show that chlorination prefers secondary to primary by a ratio of 72.2 : 27.8, or a factor of 2.6. This however, does not take into consideration that there are 6 primary hydrogens and 4 secondary hydrogens, so to correct for the statistical advantage we must multiply the factor by 6/4 to produce the true factor of 3.9. Thus, chlorination prefers 2º to 1º by a factor of 3.9. Bromination of n-butane: Bromine is significantly more than chlorine in its free radical halogenation: CH 3 CH 2 CH 2 CH 3 + Br 2 hv CH 3 CH 2 CHCH 3 + CH 3 CH 2 CH 2 CH 2 Br Br sec-butyl bromide n-butyl bromide (98.2%) (1.8%) The above statistical adjustment would be the same for bromine, and would indicate that bromination prefers 2º to 1º by a factor of 82. Chlorination of isobutane: CH 3 CH 3 CH 3 CH hv 3 CHCH 3 + Cl 2 CH 3 CHCH 2 Cl + CH 3 CCH 3 Cl 65 isobutyl chloride tert-butyl chloride (64%) (36%) Isobutane has nine primary hydrogens and one tertiary hydrogen chlorination prefers 3º to 1º by a factor of 5.1.

Bromination of isobutane: Halogenation of Other Alkanes (cont.) CH 3 CH 3 CH 3 CH 3 CHCH 3 + Br 2 hv CH 3 CHCH 2 Br + CH 3 CCH 3 Br isobutyl bromide tert-butyl bromide (~0%) (~100%) ***Bromination, being much more selective than chlorination, prefers 3º to 1º by a factor of 1600.*** Selectivities of Halogen atoms in Carbon-Hydrogen Bond Abstraction Halogen Primary Secondary Tertiary F 1 1.2 1.4 Cl 1 3.9 5.1 Br 1 82 1600 Energy Diagrams Exothermic Reaction Endothermic Reaction For Chlorination: 66

1. Formation of Alkyl Halide: Alkane Preparation Reactions CH 3 CH 3 + X 2 CH 3 CH 2 X + H X Alkane diatomic halide alkyl halide binary acid Cl 2 or Br 2 2. Catalytic Hydrogenation of Alkenes: CH 3 CH=CH 2 + H 2 CH 3 CH 2 CH 3 Alkene Hydrogen Alkane 3. Reduction of Alkyl Halide: 67 2 CH 3 CHCH 3 + 2 Zn + 2 HC 2 H 3 O 2 2 CH 3 CH 2 CH 3 + ZnX 2 + Zn(C 2 H 3 O 2 ) 2 X Alkyl halide Zinc Acetic acid Alkane Zinc halide Zinc acetate Alkane Preparation Reactions #4 - #8 - Organometallics ***Organic compound with Carbon directly attached to a metal; highly reactive and extremely usefel. Strong bases, good nucleophiles. Mg, Li, Na, Cu.*** 4. Wurtz Reaction: Organosodiums very reactive, cannot stop CH 3 CH 2 X + 2 Na CH 3 CH 2 :Na + NaX 1º Alkyl Halide Sodium 1º Organosodium Sodium Salt CH 3 CH 2 :Na + CH 3 CH 2 X CH 3 CH 2 CH 2 CH 3 + NaX 1º Organosodium 1º Alkyl Halide Alkane Sodium Salt **Doubles chain length, good for symmetrical alkanes, Alkyl Halide must be primary** 5. Organolithiums 1 st step to Corey-House: CH 3 CH 2 X + 2 Li CH 3 CH 2 :Li + LiX 1º or 2 º Alkyl Halide Lithium 1º Organolithium Lithium Salt 6. Corey-House Reaction: hv Pt, Pd, Ni Pr essure reflux ether, hexane ether, hexane ether, hexane ether, hexane 2 CH 3 CH 2 :Li + CuI [(CH 3 CH 2 :) 2 Cu]Li + LiI 1º 2 º Organolithium Cuprous Iodide Lithium dialkyl cuprate Lithium Iodide **The lithium dialkyl cuprate can then be used to perform S N 2 on a different 1º alkyl halide **

7. Grignard Reaction: CH 3 CH 2 Cl + Mg ether CH 3 CH 2 MgX Alkyl halide Magnesium Alkane:Mg halide 8. Hydrolysis of Grignard: H 2 O Mg(OH)X CH 3 CH 2 MgX + or CH 3 CH 3 + or CH 3 OH Mg(CH 3 O)X Grignard Reagent Acidic Solvent Alkane Magnesium salt 9. S N 2 Substitution Nucleophilic Bimolecular: Backside Attack Involves only one step the rate of reaction is dependant on the concentration of the two molecules in the rate determining step. Substrate must be 1º or 2º. Competes with E 2. **Inversion of Configuration** H H H H H Et H.. H CH 3 CH 2 : - + I CH 3 CH 2 : - ------- ------ :I: - and H.. H :I: - **Transition State** 10. E 2 Elimination Bimolecular: One Step Competes with S N 2 Preferred when the carbon to be attacked is sterically hindered (2º or 3º). CH 3 CH 3 CH 2 = C CH 2 CH 3 CH 3 CH 2 : - + CH 3 CCH 2 CH 3 and X CH 3 CH 3 C = CH CH 3 **Saytzeff s Rule** major product is determined by the greater # of alkyl groups attached to double bonded carbons Hoffmann Product (Minor) Saytzeff Product (Major) 68

S N 2 vs. E 2 - Continued 11. Ethers: **Ethers are chosen as the solvent for these reactions because they are un-reactive in both substitution and elimination**.. CH 3 CH 2 : - + CH 3 O CH 3 No Reaction When determining whether S N 2, E 2 or hydrolysis will occur in a reaction, consider the following: 1. Look for a metal Is one of your reactants a base (anion)? Recognize acid/base rxns 2. Look for an easy (acidic) proton for a base to grab 3. Can one of your reactants act as a nuc:? 4. 2 1º S N 5. 3º E 2 6. 2º S 2 N + E 2 CH 3 CH 2 MgX Organometallics Used as Nucleophiles: CH 3 CH 2 :Li + CH 3 CH 2 CH 2 Br CH 3 CH 2 CH 2 CH 2 CH 3 [(CH 3 CH 2 :) 2 Cu]Li 2 SN Organometallics Used as Bases: CH 3 CH 2 MgX CH 3 CH 3 + Mg(OCH 3 )Cl Hydrolysis CH 3 CH 2 :Li + CH 3 CH 2 O H CH 3 CH 3 + Li(OCH 3 ) [(CH 3 CH 2 :) 2 Cu]Li CH 3 CH 3 + Li(OCH 3 ) + CH 3 CH 2 Cl More Uses as Bases: 69 CH 3 CH 3 CH 2 = C CH 2 CH 3 [(CH 3 CH 2 :) 2 Cu]Li + CH 3 CCH 2 CH 3 and X CH 3 CH 3 C = CH CH 3 Hoffmann Product (Minor) Saytzeff Product (Major)

S N 2 and E 2 Mechanisms A Summary **Correlation of structure and reactivity for substitution and elimination reactions** Halide Type S N 2 Methyl Primary Secondary Tertiary Highly favored with most nuc: Highly favored with strong nuc: Favored by good nuc: in polar, aprotic solvents Does not occur E 2 Does not occur Favored with strong, hindered bases Favored when strong bases are used Highly favored when bases are used S N 2 Substitution, Nucleophilic, Bimolecular - Rate = k[substrate][:nuc] - One-step, concerted mechanism - 100% Stereochemical Inversion - Can compete with E 2 E 2 Elimination, Bimolecular - Rate = k[substrate][base] - One-step, concerted mechanism - Competes with S N 2 70

Alkane Reactions and Properties Worksheet #1 1. Which of the following is most soluble in water? a. Pentan 1 ol b. Hexane c. Diethyl ether d. Ethan 1,2 diol 2. Complete and balance the following (assume complete oxidation): a. Isohexane + oxygen b. Hentriacontane + oxygen c. Tetradecane + oxygen 3. Determine the oxidation number of the indicated atoms in each of the following molecules a. CH 3 CH 2 CH 3 e. Br CH CH 2 CH 3 Br b. CH 3 CH CH 3 OH f. CH 3 CH = CH 2 c. CH 3 C CH 3 :O: g. CH 3 CH CH 3 :NH 2 d. CH 3 CH 2 C OH :O: h. CH3 C NH 2 :O: i. CH 3 CH 2 :Mg:Cl 71

Alkane Properties and Preparation Worksheet #2 1. What isomer of hexane has two (and only two) monobromo derivatives? 2. Without referring to tables, list the following hydrocarbons in order of decreasing boiling points: a. 3,3 dimethylpentane b. n heptane c. 2 methylheptane d. n pentane e. 2 methylhexane 3. Write balanced equations, naming all organic products, for the following reactions: a. Isobutyl bromide + Mg ether b. Tert-butyl bromide + Mg ether c. Product of a + water d. Product of b + water e. Product of a + CH 3 OH f. Sec-butyl chloride + Li, then CuI g. Product of f + ethyl bromide 4. Write Equations for the preparation of n-butane from: a. n butyl bromide b. sec butyl bromide c. ethyl chloride d. but 1 ene (CH 3 CH 2 CH = CH 2 ) 72

Alkane Reactions Worksheet #3 1. If the following names are correct, say so. If not, give the correct name. a. 3 isopropylhexane b. 3 n-propylhexane 2. Arrange the following in order of increasing boiling point. Explain your choices: a. Pentane b. 2,2 dimethylpropane c. Hexane d. Ethane 3. Complete the following with proper equations: a. The production of tert-butylmagnesium bromide from 2 bromo 2 methylpropane b. Product of a plus water c. Product of a plus aqueous HCl d. Isopentyl chloride + zinc + acetic acid e. Complete combustion of hexadecane with oxygen f. Isobutane + chlorine hv g. Show all steps in the free-radical chain reaction for the formation of one of the products in f h. Starting with propane, outline a synthesis of 2,3 dimethylbutane i. Starting with isobutane and propane, how could you prepare 2,2 dimethylpentane? j. Consider the following picture: i. What is the activation energy? ii. What is the sign and magnitude of ΔH? iii. Endothermic or Exothermic? 73

Alkane Reactions Worksheet #4 I. Outline all steps in the free-radical chain reaction representing the monobromination of isobutane. Show 3 possible termination steps. II. Give the IUPAC names for: a. CH 3 (CH 2 ) 55 CH 3 c. b. III. IV. Give the common names for: a. 1 bromo 2,2 dimethylpropane b. 2 methylbutane c. 2 bromo 2 methylbutane d. 1 bromo 4 methylpentane Complete and balance (name the products): a. Ethane and I 2 hv, heat b. 2 chloropentane Zn, HAc c. 4 methylpent 2 ene + H 2 Pd d. Propane + Cl 2 hv e. CH 3 CH 2 CH 2 CH 2 CH 2 Cl + Na ether f. Isopropyl chloride + Mg ether V. Starting with only ethane and any needed inorganic reagents, outline a potential synthesis of 3 methylpentane 74

Material Covered on the Unit 3 Exam 1. Write and balance combustion reactions for alkanes. 2. Be able to assign oxidation numbers to carbon and other elements in organic compounds. 3. Define oxidation and reduction, carbanion and free radical. 4. Write the mechanism for free-radical chain reaction, including termination steps. 5. Define Activation Energy. 6. Draw and label a reaction energy profile for exothermic and endothermic steps in a free radical chain reaction. 7. Use bond energy tables to find ΔH rxn. 8. Explain why chlorine is more reactive than bromine with methane. 9. Explain why Iodine does not react with methane. 10. Use probability and success rate factors to predict % yields of isomers, assuming monohalogenation. 11. Explain what is meant by the statement Bromine is more selective than Chlorine. 12. Explain the stabilities of 3, 2 and 1 free radicals. 13. Define hyperconjugation and the role it plays in the stability of a charged/radical species. 14. Diagram mechanisms for S N 2 and E 2 reactions. 15. Describe the competition between S N 2 and E 2, and identify the factors that influence which mechanism is preferred in a given situation. 16. Predict the products of various alkane preparation reactions. 17. Use alkane preparation reactions in synthesis problems to make more complicated products from simpler reactants. 75