1 / 23

Hydroxy Compounds

Hydroxy Compounds. (Chapter 34). Phenol. OH. Hydroxy compounds. Aliphatic Monohydric Alcohols 1 o Primary RCH 2 OH (one –R) 2 o Secondary R 2 CHOH (two –R) 3 o Tertiary R 3 COH (three –R). . O -. H +. R +. Three tendencies of reactions. 3. Attack other substrates.

gaetan
Download Presentation

Hydroxy Compounds

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Hydroxy Compounds (Chapter 34)

  2. Phenol OH Hydroxy compounds Aliphatic Monohydric Alcohols 1o Primary RCH2OH (one –R) 2o Secondary R2CHOH (two –R) 3o Tertiary R3COH (three –R)

  3.  O- H+ R+ Three tendencies of reactions 3. Attack other substrates Nu: :B • Nucleophiles attack • alkyl group 2. Bases that attack the hydrogen atom

  4. Nucleophilic Substitution • In acidic medium, -OH is protonated to facilitate C-O bond cleavage (-OH2+ is a better leaving group) • RCH2OH + H+ RCH2-OH2+ • SN1 mainly (down-grading of Nu: in acidic medium)

  5. Bubbling HX(g) HBr Br + H2O OH Halide Formation HX is produced ‘in situ’ NaBr + H2SO4 NaHSO4 + HBr HBr + C4H9OH  C4H9Br + H2O

  6. Halide Formation PX3 ( P + X2) or SOCl2 PCl3 + 3 C2H5OH 3 C2H5Cl + P(OH)3 SOCl2 + 2 C2H5OH 2 C2H5Cl + SO2 + H2O

  7. Mechanism R-O+H-Zn-Cl2  R+ + Cl-  RCl R-OH + ZnCl2 Lucas reaction • Use to distinguish 1o, 2o,3o alkanols • Reagent: ZnCl2(s) in conc.HCl • SN1 mainly, R-Cl is formed • Observation: • 3o Two distinct layers formed immediately • 2o Two distinct layers appear in 10 min. • 1o A cloudy appearance after a few hour

  8. Mechanism(E1): CH3CHCH3 + H+ CH3CHCH3  CH3C+HCH3 + H2O  CH2=CHCH3 + H+ OH OH2+ Elimination • Dehydration, -H2O • Tend to be first order, 2 steps, leaving group led. • 3o alkanols eliminate most readily • Unlike haloalkanes, SN and E do not occur in competition. Each set of reagents do just one job. (PI3 for SN, c.H2SO4/Al2O3 as water grabbers)

  9. excess c.H2SO4,170oC CH3CH2CHCH3 CH3CH2CH=CH2 OH or Al2O3,350oC + CH3CH=CHCH3 (major) Intramolecular Dehydration Saytzeff’s rule: In the elimination reactions, the major product should be the one with greater number of alkyl groups attached to the C=C bond.(higher substituted alkenes are more stable.)

  10. Intermolecular Dehydration • c. H2SO4 • 2CH3CH2OH  CH3CH2OCH2CH3 • 140oC • For 1o alkanol (2o,3o Alkenes form) • Not suitable for unsymmetrical ether • SN2 mechanism

  11. Mechanism (SN2) c. H2SO4 CH3CH2OH CH3CH2OH  CH3CH2O+H2  CH3CH2O+HCH2CH3 + H2O  CH3CH2OCH2CH3 + H+ 140oC Intermolecular Dehydration

  12. Strength increase ? As Acids Ka CH3-O-H + H2O  CH3O:- + H3O+ pKa values: HCl -7 CH3COOH 14.8 CH3OH 15.5 H2O 15.7 CH3CH2OH 15.9 (CH3)2CHOH 17 (CH3)3COH 18

  13. Reaction with sodium e.g. 2CH3OH + 2Na  2CH3O- Na+ + H2 CH3O- Methoxide ion A stronger base than OH-. Why?

  14. As Nucleophiles Esterification: c.H2SO4 Alkanol + Acid  Ester + water reflux • Excess acid or alkanol is used to drive the eqm. to • the formation of ester. • c.H2SO4 is used to • Catalyse the reaction • Shift the equilibrium position to the product side • by removing H2O

  15. R’ :O R’ R’ H O O+H H+ C OH HO C C O+ R H OH OH R’ R’ -H2O -H+ H+ shift R’COOR C O+H2 H-O C H-O+= O O R R Mechansium of esterification R

  16. Oxidation Oxidizing Agent: K2Cr2O7/H+ 1o alkanol [O] [O] RCH2OH  RCHO  RCOOH aldehyde alkanoic acid 2o alkanol [O] R2COH  R2C=O ketone 3o alkanol Cannot be oxidized

  17. O R H OH HO Cr + C R OH O R H :O R C + H2O + H2CrO3 C O R O OH Cr R O Mechanism of Oxidation 2o alkanol

  18. R H R [O] C C O H OH H O R .. H OH HO Cr :O C H :O- O OH O Cr C O R R H+ O C O HO Mechanism of Oxidation 1o alkanol

  19. e.g. OH I2/NaOH CH3CHC2H5 C2H5COO-Na+ + CHI3 (a yellow ppt.) Triiodomethane Formation Substrate: Alkanol with CH3C(OH)- Reagent: I2 in NaOH(aq) , a mold O.A. Serve as a qualitative test to identify compound with the above structure.

  20. .. OH Phenol Acid strength C6H5OH(aq) C6H5O-(aq) + H+(aq) Ka = 1x10-10, much stronger than aliphatic alkanols. • Reason: • Non-bonded e- of oxygen takes part • in the delocalized  e- system. •  weakened O-H bond

  21. is stabilized by delocalization of the negative charge into the benzene ring. ..- O:- O O .. -.. O- Phenol

  22. Reaction of phenols • Reaction with sodium • C6H5OH + Na C6H5O-Na+ + ½ H2 • (more vigorous than aliphatic alkanol) 2. Reaction with NaOH C6H5OH + NaOH C6H5O-Na+ +H2O

  23. O O O O OH O-Na+ O-C-R O-C-R R-C-O-C-R NaOH O R-C-O-Cl Reaction of phenols -OH takes part in e- system, NOT a good Nu:

More Related