Hydroxy compounds
This presentation is the property of its rightful owner.
Sponsored Links
1 / 23

Hydroxy Compounds PowerPoint PPT Presentation


  • 154 Views
  • Uploaded on
  • Presentation posted in: General

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.

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.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Presentation Transcript


Hydroxy Compounds

(Chapter 34)


Phenol

OH

Hydroxy compounds

Aliphatic Monohydric Alcohols

1o Primary RCH2OH (one –R)

2o Secondary R2CHOH (two –R)

3o Tertiary R3COH (three –R)




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

  • 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)


Bubbling HX(g)

HBr

Br

+ H2O

OH

Halide Formation

HX is produced ‘in situ’

NaBr + H2SO4 NaHSO4 + HBr

HBr + 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


  • 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


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)


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

  • c. H2SO4

  • 2CH3CH2OH  CH3CH2OCH2CH3

  • 140oC

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

  • Not suitable for unsymmetrical ether

  • SN2 mechanism


Mechanism (SN2)

c. H2SO4

CH3CH2OH

CH3CH2OH  CH3CH2O+H2 

CH3CH2O+HCH2CH3 + H2O  CH3CH2OCH2CH3 + H+

140oC

Intermolecular Dehydration


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

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

    • Catalyse the reaction

    • Shift the equilibrium position to the product side

    • by removing H2O


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

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


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


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


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.


..

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


is stabilized by delocalization

of the negative charge into the

benzene ring.

..-

O:-

O

O

..

-..

O-

Phenol


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


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:


  • Login