1. Chapter 8 Reactions of Alcohols I.Oxidation and Reduction of Alcohols A.Inorganic Oxidation and Reduction 1)Oxidation = loss of electrons: Cu + Cu 3+ 2)Reduction = gain of electrons: Zn 2+ Zn 0 B.Organic Oxidation and Reduction 1)Oxidation = addition of electronegative atoms (O, Cl, N) = removal of H 2)Reduction = removal of electronegative atoms (O, Cl, N) = addition of H 3)Relationship between alcohols and carbonyl groups 1)C==O = carbonyl -2 e- +2 e-

2. 2)Aldehyde is an oxidized Primary Alcohol (reduced aldehyde) 3)Ketone is an oxidized Secondary Alcohol (reduced ketone) C.Alcohol Synthesis by Reduction of Aldehydes and Ketones 1)Hydrogenation = adding H 2 to a double bond 2)Catalyst = reactant that doesn’t get destroyed; it speeds up the reaction by lowering the E a for the reaction a)Heterogeneous Catalyst = insoluble, reaction occurs at its surface i.Fine powder increases surface area (put cat. on fine charcoal) ii.Pd/C or Pt metal or Ni metal work as hydrogenation cat.’s b)Homogeneous Catalyst = soluble, reaction occurs in solution 3)Hydrogenation of Carbonyls gives Alcohols aldehyde1 o ROHketone2 o ROH

3. D.Alcohol Synthesis by Hydride Reduction of Carbonyl’s 1)Carbonyl groups are polar 2)Hydride = has several efficient sources for Organic Synthesis a)Na + H - or Li + H - M + + H - (not very organic soluble) b)NaBH 4 Na + + B(solvent) + 4 H - c)LiAlH 4 Li + + Al(solvent) + 4 H - 3)Reduction of Carbonyl’s to Alcohols 4)NaBH 4 Mechanism

4. 5)LiAlH 4 is too reactive to use in Protic Solvents 6)LiAlH 4 in Aprotic Organic Solvent E.Oxidation of Alcohols to Carbonyls 1)Reduction reactions can be reversed to give the aldehydes or ketones 2)Oxidizing Reagent is Cr(VI) a)(Na 2 Cr 2 O 7 or K 2 Cr 2 O 7 or CrO 3 ) and H 2 SO 4 and H 2 O b)Primary alcohols can be overoxidized to carboxylic acids

5. 3)Do Primary alcohol oxidation without water a)PCC = pyridinium chlorochromate = b)Anhydrous conditions and PCC don’t overoxidize primary alcohol 4)Mechanism involves Chromic Ester

6. II.Organometallic Reagents = carbon-metal bonds A.Nucleophilic Carbon 1)H- is a nucleophilic hydrogen that gives new C—H bonds 2)Formation of new C—C bonds is the key requirement in organic synthesis B.Alkylmetal Reagents 1)Haloalkanes can be transformed into organometallic compounds a)Alkyllithium Synthesis (I > Br > Cl reactivity) b)Alkylmagnesium Synthesis (Grignard Reagent)

7. 2)Alkylmetals are very strong Nucleophiles a)Use them as soon as you make them (…made in situ…) b)Air and water sensitive, must do reaction under N 2 or Ar 3)True structure involves coordinated solvent—require ether or THF 4)Very polar bond, metal is very electropositive a)Opposite of usual situation for carbon, as in Haloalkanes b)Treat the molecule like R - = Carbanion c)Resonance forms 5)Alkylmetals are very basic a)Basicity: RCH 2 - > RNH - > RO - (electronegativity) b)Acidity: RCH 3 < H 2 NR < ROH c)Leads to fast hydrolysis (see above) in protic solvents Hydrolysis

8. 6)Making Alkanes from Haloalkanes a)Use alkyl metal hydrolysis b)Use hydride nucleophile c)Useful for Deuterating compounds (D = deuterium = 2 H) C.Alkylmetal reagents in Alkanol Synthesis 1)Formation of new C—C bond by using C Nucleophile 2)Nucleophilic attack of Haloalkane by alkylmetal reagent is too slow

9. 3)Must use Carbonyl electrophile (C  ==O  ) a)Ketone gives tertiary alcohol product b)Aldehyde gives a secondary alcohol product c)Formaldehyde (CH 2 O) gives primary alcohol product

10. III.Synthetic Strategies A.Use Mechanisms to predict the products of a reaction 1) 2) B.Know all of the tools (reactions) you can use

11. C.Synthesis of Complex Alcohols Using Alkylmetal Reagents 1. Useful Reaction Sequence 2.Retrosynthetic Analysis = Work Backwards from Target a)Complex molecules are made from simple parts b)Look for C—C bonds to break (form)

12. 3)Avoid Synthetic Pitfalls a)Use fewest possible steps i.2 steps at 90% yield = (0.9)(0.9) = 81% total yield ii.4 steps at 95% yield = (0.95)(0.95)(0.95)(0.95)= 81% total yield b)Convergent Synthesis is better than Linear Synthesis A B C D (50% yield at each step) E F D G H c)Don’t use functional groups that would interfere in the reaction d)Remember mechanistic limitations i.Br 2 is very selective form tertiary H ii.S N 2 can’t happen for tertiary electrophile 80g40g20g 10g 20g 10g