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Hydroxy Compounds
(Chapter 34)
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Hydroxy compounds
Aliphatic Monohydric Alcohols
1o Primary RCH2OH (one –R)2o Secondary R2CHOH (two –R)3o Tertiary R3COH (three –R)
Phenol OH
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Three tendencies of reactions
R+
O-
H+
Nu:1. Nucleophiles attack alkyl group
:B2. Bases that attack the hydrogen atom
3. Attack other substrates
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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)
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Halide Formation
OHHBr Br + H2O
Bubbling HX(g)
HX is produced ‘in situ’
NaBr + H2SO4 NaHSO4 + HBrHBr + C4H9OH C4H9Br + H2O
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Halide Formation
PX3 ( P + X2) or SOCl2
PCl3 + 3 C2H5OH 3 C2H5Cl + P(OH)3
SOCl2 + 2 C2H5OH 2 C2H5Cl + SO2 + H2O
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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
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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+
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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.)
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Intermolecular Dehydration
c. H2SO4
2CH3CH2OH CH3CH2OCH2CH3
140oC
•For 1o alkanol (2o,3o Alkenes form)•Not suitable for unsymmetrical ether•SN2 mechanism
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Intermolecular Dehydration
CH3CH2OH CH3CH2O+H2
CH3CH2O+HCH2CH3 + H2O CH3CH2OCH2CH3 + H+
c. H2SO4
140oC
CH3CH2OH
Mechanism (SN2)
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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 ?
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Reaction with sodium
e.g. 2CH3OH + 2Na 2CH3O- Na+ + H2
CH3O- Methoxide ion
A stronger base than OH-. Why?
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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
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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
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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
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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
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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
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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.
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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..
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Phenol
O-..
is stabilized by delocalizationof the negative charge into thebenzene ring.
O:- O..-
O
-..
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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
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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: