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

R+

O-

H+

Nu:1. Nucleophiles attack alkyl group

:B2. Bases that attack the hydrogen atom

3. Attack other substrates

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

OHHBr Br + H2O

Bubbling HX(g)

HX is produced ‘in situ’

NaBr + H2SO4 NaHSO4 + HBrHBr + C4H9OH C4H9Br + 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 immediately2o 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

•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)

Mechanism(E1):

CH3CHCH3 + H+ CH3CHCH3 CH3C+HCH3 + H2O CH2=CHCH3 + H+

OH OH2+

Intramolecular Dehydration

excess c.H2SO4,170oC

CH3CH2CHCH3 CH3CH2CH=CH2 OH or Al2O3,350oC + CH3CH=CHCH3

(major)

Saytzeff’s rule: In the elimination reactions, the majorproduct should be the one with greater number of alkyl groups attached to the C=C bond.(higher substitutedalkenes 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

CH3CH2OH CH3CH2O+H2

CH3CH2O+HCH2CH3 + H2O CH3CH2OCH2CH3 + H+

c. H2SO4

140oC

CH3CH2OH

Mechanism (SN2)

As Acids

Ka

CH3-O-H + H2O CH3O:- + H3O+

pKa values: HCl -7CH3COOH 14.8CH3OH 15.5H2O 15.7CH3CH2OH 15.9(CH3)2CHOH 17(CH3)3COH 18

Strengthincrease ?

Reaction with sodium

e.g. 2CH3OH + 2Na 2CH3O- Na+ + H2

CH3O- Methoxide ion

A stronger base than OH-. Why?

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

1. Catalyse the reaction2. Shift the equilibrium position to the product side by removing H2O

Mechansium of esterificationR

COR’

OH

H+C

O+HR’

OH

:OH

C

R’

HO OH

O+

R H

C

R’

H-O O+H2

OR

H+ shiftC

R’

H-O+=

OR

-H2O -H+

R’COOR

Oxidation

1o alkanol [O] [O]RCH2OH RCHO RCOOH aldehyde alkanoic acid

Oxidizing Agent: K2Cr2O7/H+

2o alkanol [O]R2COH R2C=O ketone

3o alkanol

Cannot be oxidized

Mechanism of Oxidation

2o alkanol R HC

R OH

+ HO Cr OH

O

O

R HC

R O Cr OH

O

:O+ H2O

RC

RO + H2CrO3

Mechanism of Oxidation

1o alkanolR H

C

H OH

[O]R

C

HO

HO Cr OH

O

O

HC

RO

..

R HC

:O- O Cr OH

O

:O

H+R

C

HOO

Triiodomethane Formation

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

e.g. OH I2/NaOH CH3CHC2H5 C2H5COO-Na+ + CHI3

(a yellow ppt.)

Serve as a qualitative test to identify compoundwith the above structure.

PhenolAcid strengthC6H5OH(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

OH..

Phenol

O-..

is stabilized by delocalizationof the negative charge into thebenzene 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

OH O-Na+

NaOH R-C-O-C-R

O OO-C-R

O

R-C-O-Cl

O

O-C-R

O-OH takes part ine- system, NOTa good Nu: