Alcohols (R-OH) and Alkyl halides R-X (X = F, Cl, Br, I) Alcohols. Suffix:

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1 Alcohols (R-O) and Alkyl halides R-X (X = F, Cl,, I) Alcohols Alcohols contain a R-O functional group. Primary, secondary or tertiary? O O O Nomenclature: Suffix: numbering scheme: The number locating it is placed in front of the root name. Other functional groups present the # for the -O is placed before the ol. Prefix: With higher priority functional groups use: Number the longest chain of carbons using higher priority functional group 79

2 Naming examples: Compound Name Line Drawing O O Compound Name Line Drawing O O O O O N 2 O S O O O O O O O O 80

3 Physical Properties: The atom attached to the O Volatililty: Melting points: Boiling points: Aqueous solubility: Reactivity: Alcohols as bases or nucleophiles Nucleophiles are electron rich species that react with electron poor ones. Alcohol oxygen atoms are bases. The C-O is difficult to break, it needs to be assisted Alcohols are very weakly acidic O - O Alcohols react with Na (or K) 2 O: RO: Alkoxide: Alkoxides are important bases in organic chemistry. Replace the with another R' group Reaction with a carboxylic acid Reaction with hydrogen halides 81

4 Nomenclature: Functional group suffix = Functional group prefix = Primary, secondary or tertiary? C 3 Cl Alkyl halides Physical Properties: The carbon halogen bonds are typically quite polar Volatililty: Melting points: Boiling points: Aqueous solubility: Reactivity: Polarity makes the C atom electrophilic. The halogens (Cl, and I) are good leaving groups. Nu CR 3 -X CR 3 -Nu X Bases can remove β-hydrogens and cause 1,2-elimination NaOR Insertion of a metal (esp. Mg) creates an organometallic species. M CR 3 -X CR 3 -MX 82

5 Carbocations Structure: R R C R Stability: Order of simple alkyl carbocations is: C 3 C 3 C C 3 C C C 3 C 3 C 3 C Alkyl groups are weakly electron donating a) Inductive effects. C 3 C 2 C C 3 C C b) yperconjugation c) Resonance effects C C C 2 C 2 C 2 C 3 83

6 Reactivity: Act as electrophiles Loss of can generate a π bond. Rearrangements: generates a more stable carbocation a) 1,2-hyride shifts b) 1,2-alkyl shifts Reactions involving carbocations: a) Substitutions via S N 1 b) Eliminations via E1 c) Additions to alkenes and alkynes (X, 3 O ) Mechanism: This pathway is a multi-step process with the following characteristics: Step 1: Step 2: Multi-step reactions have intermediates (I) and several transition states (TS). TS 1 I TS 2 SM = starting material TS 1 = Transition State 1 I = Intermediate TS 2 = Transition State 2 P P = Products SM 84

7 Free Radicals Stability: Order: C 3 C 3 C C 3 C C C 3 C 3 C 3 C Alkyl groups are weakly electron donating: a) b) c) Structure: R R C R 85

8 Reactivity: Free radicals act as electrophiles: Alternatively, loss of. can generate a pi bond Rearrangements: Free radicals rarely rearrange Radical Substitution: X X C Substitution of R- by X: Alkane R- relative reactivity order : alogen reactivity: Mechanism: 86

9 Radical Chain Mechanism Three distinct type of steps: Step 1 (Initiation) Step 2 (Propagation) (a) A halogen radical removes hydrogen atom (b) The alkyl radical removes bromine atom from another 2 C 3 C 3 C 3 C 3 Step 3 (Termination): Reactions between pairs of radicals, C 3 C 3 C 3 C 3 C 3 C 3 87

10 Selectivity Selectivity of radical halogenations of alkanes is based on two factors: 1) Reactivity of the alkyl (R-) system The bond strength of the R- depends on as 1 º, 2 º or 3 º. R- R Bond dissociation energies for methyl 1 º, 2 º or 3 º hydrogens Type R- kj/mol kcal/mol Me C C 3 C (C 3 ) 2 C (C 3 ) 3 C ) Reactivity of the halogen (X ) F Cl I Reactivity determines selectivity: reactive radicals are more selective Selectivity of the radical reactions can be calculated using a) A reactivity factor, R i, b) A statistical factor, n i. 88

11 ow to calculate the ratio of products; Step 1: What are the possible monosubstituted products? Step 2: ow many hydrogens that yield the same product? (n i ) Step 3: What is the reactivity (R i ) for the hydrogen and halogen type? Step 4: Using the reactivity factors, what are the total possible structures? Step 5: What is the percent of each hydrogen type per total hydrogens? %P i % yield of product "i" R i at 25 C Cl n i number of of type "i" 1 R i type "i" reactivity factor 2 Σ i sum for all types 3 What does the reactivity factors indicate? 89

12 You try: the chlorination of propane: C 3 C 2 C 3 Step 1: Step 2: Step 3: Step 4: Step 5: What about bromination of propane? Conclusions from this data: 90

13 Radical alogenation of Alkanes X X C C Reaction type: Products: Alkane R- relative reactivity order : alogen reactivity: Selectivity: omination Mechanism: Chlorination More highly brominated by-products are possible if methyl bromide reacts with a bromine radical in the same fashion as methane does. Draw the cycle that leads to the formation of dibromomethane. 91

14 Try this: Radical alogenation 1. Calculate the predicted yields for all of the monohalogenated products produced by chlorination and bromination of butane. For chlorination: omination: 92

15 ow to identify the original compound from the product ratios. Step 1: What are the possible starting materials (isomers)? Step 2: Which ones give the correct number of products? Step 3: Calculate which has the correct ratios. Try this: On radical chlorination, an isomer of hexane gave a mixture of monochlorides in the ratio 46: 39: 15. Identify the isomer. Step 1: Step 2: Step 3: 1) 2) Answer: 93

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