Hydroxy Compounds

Hydroxy Compounds (Chapter 34)

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

OH

Three tendencies of reactions ••

3. Attack other substrates

OδR Nu: 1. Nucleophiles attack alkyl group

δ+

H

δ+

:B 2. Bases that attack the hydrogen atom

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)

Halide Formation Bubbling HX(g) OH

HBr

HX is produced ‘in situ’ NaBr + H2SO4 NaHSO4 + HBr HBr + C4H9OH C4H9Br + H2O

Br + H2O

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

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 •Mechanism R-OH + ZnCl2 R-O+H-Zn-Cl2 → R+ + Cl- → RCl

Elimination Mechanism(E1): OH OH2+ CH3CHCH3 + H+ → CH3CHCH3 → CH3C+HCH3 + H2O → CH2=CHCH3 + H+ •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)

Intramolecular Dehydration excess c.H2SO4,170oC

CH3CH2CHCH3 → CH3CH2CH=CH2 OH or Al2O3,350oC + CH3CH=CHCH3 (major)

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.)

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

Intermolecular Dehydration Mechanism (SN2) c. H2SO4

CH3CH2OH 

140oC

CH3CH2OH

CH3CH2O+H2 

CH3CH2O+HCH2CH3 + H2O  CH3CH2OCH2CH3 + H+

As Acids CH3-O-H + H2O

Ka  CH3O:- + H3O+

pKa values: HCl CH3COOH CH3OH H2O CH3CH2OH (CH3)2CHOH (CH3)3COH

-7 14.8 15.5 15.7 15.9 17 18

Strength increase ?

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

As Nucleophiles Esterification: Alkanol + Acid • •

c.H2SO4  Ester + water reflux

Excess acid or alkanol is used to drive the eqm. to the formation of ester. c.H2SO4 is used to 1. Catalyse the reaction 2. Shift the equilibrium position to the product side by removing H2O

Mechansium of esterification R’

O C

H+

R’

OH H shift +

O+H

R :O H

C OH

R’ H-O C O+H2 O R

-H2O

R’ H-O+= C O R

R’ HO C OH O+ R H -H+

R’COOR

Oxidation Oxidizing Agent: K2Cr2O7/H+ 1o alkanol [O] [O] RCH2OH → RCHO → RCOOH aldehyde alkanoic acid 2o alkanol [O] R2COH →

3o alkanol R2C=O ketone

Cannot be oxidized

Mechanism of Oxidation 2o alkanol

R

H +

C R R

H C

R

:O

O Cr OH O

O HO Cr OH

OH

+ H2O

O R C R

O + H2CrO3

Mechanism of Oxidation 1 alkanol o

R

H C

H

R

[O]

C

OH

H

H

R

R

O

H

C

C

.. HO Cr OH

O

:O

:O- O Cr OH

O

O H+

O

R C HO

O

Triiodomethane Formation Substrate: Reagent: e.g.

Alkanol with CH3C(OH)I2 in NaOH(aq) , a mold O.A.

OH I2/NaOH CH3CHC2H5 → C2H5COO-Na+ + CHI3 (a yellow ppt.) Serve as a qualitative test to identify compound with the above structure.

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 π e- system. ∴ weakened O-H bond

.. OH

Phenol .. O-

is stabilized by delocalization of the negative charge into the benzene ring. ..-

O:-

O

O

-..

Reaction of phenols 1. Reaction with sodium C6H5OH + Na →C6H5O-Na+ + ½ H2 (more vigorous than aliphatic alkanol) 2. Reaction with NaOH C6H5OH + NaOH→ C6H5O-Na+ +H2O

Reaction of phenols O OH

NaOH

-OH takes part in πe- system, NOT a good Nu:

O

O

O-Na+ R-C-O-C-R O R-C-O-Cl

O-C-R O

O-C-R