Hydroxy compounds
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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.

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Hydroxy Compounds

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Hydroxy compounds

Hydroxy Compounds

(Chapter 34)


Hydroxy compounds1

Phenol

OH

Hydroxy compounds

Aliphatic Monohydric Alcohols

1o Primary RCH2OH (one –R)

2o Secondary R2CHOH (two –R)

3o Tertiary R3COH (three –R)


Three tendencies of reactions



O-

H+

R+

Three tendencies of reactions

3. Attack other

substrates

Nu:

:B

  • Nucleophiles attack

  • alkyl group

2. Bases that attack

the hydrogen atom


Nucleophilic substitution

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)

HBr

Br

+ H2O

OH

Halide Formation

HX is produced ‘in situ’

NaBr + H2SO4 NaHSO4 + HBr

HBr + C4H9OH  C4H9Br + H2O


Halide formation1

Halide Formation

PX3 ( P + X2) or SOCl2

PCl3 + 3 C2H5OH 3 C2H5Cl + P(OH)3

SOCl2 + 2 C2H5OH 2 C2H5Cl + SO2 + H2O


Lucas reaction

  • 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


Elimination

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)


Intramolecular dehydration

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.)


Intermolecular dehydration

Intermolecular Dehydration

  • c. H2SO4

  • 2CH3CH2OH  CH3CH2OCH2CH3

  • 140oC

  • For 1o alkanol (2o,3o Alkenes form)

  • Not suitable for unsymmetrical ether

  • SN2 mechanism


Intermolecular dehydration1

Mechanism (SN2)

c. H2SO4

CH3CH2OH

CH3CH2OH  CH3CH2O+H2 

CH3CH2O+HCH2CH3 + H2O  CH3CH2OCH2CH3 + H+

140oC

Intermolecular Dehydration


As acids

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)2CHOH17

(CH3)3COH18


Reaction with sodium

Reaction with sodium

e.g. 2CH3OH + 2Na  2CH3O- Na+ + H2

CH3O- Methoxide ion

A stronger base than OH-. Why?


As nucleophiles

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


Mechansium of esterification

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


Oxidation

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


Mechanism of oxidation

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


Mechanism of oxidation1

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


Triiodomethane formation

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.


Phenol

..

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


Phenol1

is stabilized by delocalization

of the negative charge into the

benzene ring.

..-

O:-

O

O

..

-..

O-

Phenol


Reaction of phenols

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


Reaction of phenols1

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:


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