alcohols, ethers and phenols - s3-ap-southeast … · r–o–r. they are regarded as dialkyl...
TRANSCRIPT
ALCOHOLS, ETHERS AND PHENOLS
ALCOHOLS
Molecules containing –OH group are termed as alcohols. Classification of alcohols they are classified as primary, secondary or tertiary alcohol according to the carbon that is bonded with –OH. Again when any molecule contains 1, 2 or 3 –OH groups then it is called mono, di or tri hydric alcohols respectively. (as in case of alkyl halides)
OH
CH – CH2 2
OH OH
CH – CH – CH2 2
OH OH
CH – CH OH3 2
Ethyl alcohol(monohydric)
Ethylene glycol(dihydric)
Glycerol(trihydric)
Example 1: Classify the following into primary, secondary and tertiary alcohols:
(a)
OH
CH3
(b) CH3
OH
(c) OH
Solution: (a) Tertiary (b) Secondary (c) Tertiary
GENERAL METHODS OF PREPARATION
1. From Alkenes :
(i) By direct hydrolysis :
CH – CH = CH + H O23 2 CH – CH – CH3 3
OHH SO2 4
(ii) Oxymercuration demercuration :
CH – CH = CH + H O23 2
Hg (OAc)2
THF
CH – CH – CH22
OH
CH – CH – CH3 3
OHNaBH4
OH–
Hg(OAc) (iii) Hydroboration oxidation :
6CH – CH = CH3 2 2(CH – CH – CH –) B3 2 2 3
B H2 6 H O /OH2 2
6CH – CH – CH3 + 2H BO2 2 3 3
OH
Overall result of above reaction is anti Markwonikoff addition of water and with no rearrangement.
(iv) Oxo process followed by hydrogenation :
CH – CH = CH + CO + H3 2 2
[Co(CO) ]4 2
high temperatureand high pressure
CH – CH – CH – C – H3 2 2
H /Pd2
CH – CH – CH – CH – OH3 2 2 2
O
Product has one more carbon. 2. From Alkyl Halides : When alkyl halides are treated with aq. KOH or aq. NaOH or moist Ag2O, alcohols are formed.
–XOHRHOXR
3. Reduction of Carbonyl Compounds, Carboxylic Acids and their Derivatives :
O
R – C – Hred. agent
R – CH OH2
O
R – C – Rred. agent
OH
O
R – C – ORred. agent
red. agent
red. agent
R – CH OH + R – OH2
O
R – C – X R – CH OH2
O
R – C – O – C – R 2R – CH OH2
O
4. Using Grignard Reagent :
(i) From aldehydes or ketones :
– C – + R – MgX – C – R – C – R + MgH O/H2
+
O OMgX OHOH
X
In this reaction
Formaldehyde gives 1°-alcohol
Other aldehyde gives 2°-alcohol
Ketones give 3°-alcohol
(ii) From carboxylic acid and their derivatives :
R – C – Z + R – MgX
OZ
XR – C – R + Mg
O
(iii) From epoxides :
CH – CH — CH + R – MgX3 2
OH
XCH – CH – CH3 2
OMgXH O/H2
+
CH – CH – CH + Mg3 2
OH
R
O
R
5. Hydrolysis of Ether :
Ether undergo acid hydrolysis with dilute H2SO4 under pressure to give corresponding alcohols.
OHROHROHROR 42SOH.dil2
Example 2: (a) Find A, B, C, D, E.
PhBrMg
dry ether A
O
B
O
HC
(b)
3
2
i CH MgBr
3 2 5 ii H O / HCH COOC H D E
Solution: (a) A = PhMgBr B = Ph
OC =
Ph
OH
(b)
CH3 C
CH3
CH3
OH C2H5OH
Physical Properties
(i) Physical state : At ordinary temperature, lower members are colourless liquids with burning taste and a pleasant smell.
(ii) Boiling Point : The boiling point of alcohols rise regularly with the rise in the molecular mass. Amongst isomeric alcohols, the boiling point decrease in the order.
1° > 2° > 3°
(iii) Solubility : The extent of solubility of any alcohol in water depends upon the capability of its molecules to form hydrogen bonds with water molecule.
(iv) Alcohols are lighter than water however, the density increases with the increase in molecular mass.
Chemical Properties
1. Reactions involving cleavage of O – H Bond
Alcohols are acidic in nature but they are less acidic than water hence they do not give H+ in aqueous solution. They do not change the colour of litmus paper.
Their acidic strength increases by increasing–I strength of the groups attached and decreases by increasing +I strength of the groups.
(i) Alcohols do not react with aqueous alkali, as it does not give H+ in aqueous solution.
(ii) Action of active metal : When alcohols are treated with active metal they form alkoxides with the liberation of H2 gas.
22ROH 2Na 2R ONa H
(iii) Esterification : When carboxylic acid is treated with alcohols in the presence of acid as catalyst, esters are formed.
O
R – C – OH + H – O – RH
+O
R – C – OR + HOH
(iv) Reaction with Grignard Reagent : When Grignard reagents are treated with alcohol (or any proton donor) they form alkanes.
R – MgX + H – OR R – H + R – OMgX
Other proton donors can be carboxylic acids, phenols, alkynes, H2O, Amines, NH3 etc.
2. Reactions Involving Cleavage of C– O Bond
(i) Reaction with HX : Most alcohols undergo SN1.
(a) Anhyd. ZnCl
2R OH HCl(g) R Cl H O 2
Note : HCl + anhyd. ZnCl2 is called Lucas reagent.
(b) 2 4H /H SO
2conc.R OH HBr R Br H O
(c) 2 4H /H SO
conc. 2R OH HI R I H O
Reactivity order of HX is HI > HBr > HCl
(ii) Dehydration : Alkyl chlorides can also be prepared by following methods :
R – OH + PCl5 R – Cl + POCl3 + HCl
3R – OH + PCl3 3R – Cl + H3PO3
R – OH + SOCl2 R – Cl + SO2 + HCl (Darzen's process)
Darzen’s process is the best method as the other products are gases.
3. Reduction :
Alcohols are reduced to alkanes when they are treated with Zn-dust or red P + HI.
Zn dustR OH R H ZnO
4. Oxidation :
O
CH OH3 H – C – H CO2
[O] [O]
O
H – C – OH[O]
3°-alcohols can’t be oxidised.
(i) Strong oxidising agent like KMnO4 or K2Cr2O7 cause maximum oxidation as above.
(ii) If 1°-alcohol has to be converted into aldehyde PCC + CH2Cl2 or CrO3 should be used
among which PCC + CH2Cl2 is the best.
(iii) 2°-alcohol can converted to ketone best by PCC + CH2Cl2 or CrO3 or H2CrO4 in aq. acetone
(Jones reagent). (iv) MnO2 selectively oxidises the –OH group of allylic and benzylic 1° and 2° alcohols to
aldehydes and ketones respectively.
5. Action of Heated Copper :
(i)
O
CH – 3 CH – OH2 CH – C – H + H (Dehydrogenation)3 2
Cu
573K
(ii)
O
CH 3 – CH – CH3 CH – C – CH + H (Dehydrogenation)3 3 2
Cu
573K
OH
(iii) Tertiary alcohols undergo dehydration to give alkene under similar condition.
CH – C – OH3
CH3
CH3
Cu
573 KCH – C + H O3 2
CH2
CH3
Distinction Between 1°, 2° and 3° Alcohols
1. Lucas Test :
Any alcohol is treated with Lucas reagent (HCl + an hyd. ZnCl2) at room temperature if
(i) Solution becomes cloudy immediately, alcohol is 3°.
(ii) Solution becomes cloudy after 5-min, alcohol is 2°.
(iii) In solution cloud does not form at room temperature, alcohol is 1°.
2. Victor Meyer’s Method :
R – CH OH2
P + I2R – CH I2
AgNO2R – CH – NO + HNO2 2 2
–H O2R – C – NO2
N
OH(Nitrolic acid)
NaOH
blood red colour
(1°-alcohol)
R CH – OH2
P + I2R – CH – I2
AgNO2R CH – NO + HNO2 2 2
–H O2R C – NO2 2
NOPseudonitrol
NaOH
blue colour
(2°-alcohol)
R – C – OH3
P + I2R C – I3
AgNO2R C – NO3 2
HNO2No reaction
(3°-alcohol)
NaOHColourless
Note :
Rectified Spirit : Azeotropic mixture of 95% C2H5OH and 5% H2O is called rectified spirit.
Denatured Spirit : Azeotropic mixture of C2H5OH and CH3OH is called methylated spirit.
Example 3 : Which of the following alcohols would react faster with Lucas reagent?
CH3CH2CH2CH2OH , CH3CH2CH
OH
CH3 , CH3CH
OH
CH2OH , CH3 C
CH3
CH3
CH3
Solution : 3 3CH COH , it being a tertiary alcohol.
Example 4 : List three methods with pertinet chemical equations for the preparation of alcohols from
alkenes.
Solution: Hydration, hydroboration and oxymercuration – demercuration of alkenes. Hydration:
CH3 C
CH3
CH3
CH CH22H O
H CH3 C
CH3
CH3
CHCH3
OH
Hydroboration
CH3 C
CH3
CH3
CH CH2
3
2 2
i BH
ii H O , OH CH3 C
CH3
CH3
CH2CH2OH
Oxymercuration – demercuration
CH3 C
CH3
CH3
CH CH2
22
4
i Hg OAC , THF, H O
ii NaBH CH3 C
CH3
CH3
CHCH3
OH
ETHERS
Ethers are those organic compounds which contain two alkyl groups attached to an oxygen atom, i.e., R–O–R. They are regarded as dialkyl derivatives of water or anhydrides of alcohols.
2
–2H
H OWater 2R Ether Alcohol (2 moles)
H – O – H R – O – R R – OH HO – R
Ethers may be of two types : (i) Symmetrical or simple ether are those in which both the alkyl groups are identical and (ii) unsymmetrical or mixed ethers are those in which the two alkyl groups are different.
CH3–O–CH3; C6H5–O–C6H5 CH3–O–C2H5; CH3–O–C6H5
Symmetrical (simple) ethers Unsymmetrical (mixed)ethers
Like water, ether has two unshared pair of electrons on oxygen atom, yet its angle is greater than normal tetrahedral (109°28´) and different from that in water (105°). This is because of the fact that in ethers the repulsion between lone pairs of electrons is overcome by the repulsion between the bulky alkyl groups.
Preparation of Ethers :
1. By dehydrating excess of alcohols : Simple ethers can be prepared by heating an excess of primary alcohols with conc. H2SO4 at 413K. Alcohol should be taken in excess so as to avoid its
dehydration to alkenes.
2 4Conc. H SO
2 5 2 5 2 5 2 5 2413KEthanol (2 molecules) Diethyl ether
C H – OH HO – C H C H – O – C H H O
Dehydration may also be done by passing alcohol vapours over heated catalyst like alumina under high pressure and temperature of 200 – 250°C.
2. By heating alkyl halide with dry silver oxide (only for simple ethers) :
2 5 2 2 5 2 5 2 5Dry
C H I Ag O IC H C H OC H 2AgI
Remember that reaction of alkyl halides with moist silver oxide (Ag2O + 2H2O = 2AgOH) gives
alcohols
C2H5I + Ag2O (moist) C2H5OH + AgI
3. By heating alkyl halide with sod. or pot. alkoxides (Williamson synthesis):
C2H5ONa + ICH3 C2H5OCH3 + NaI
ONa + BrCH3 OCH + NaBr3
Sod. phenoxide Methoxybenzene(Anisole)
However
CH –Cl + NaO–C–CH3 3
+
CH|
3
|CH3
CH –O– + NaCl3 C–CH3
CH|
3
|CH3
If alkyl halide is other than methyl halide and it is treated with tertiary alkoxide ion, Hoffmann elimination takes place instead of Williamson's ether synthesis.
CH –CH –CH–CH + NaO–C–CH3 2 3 3
CH|
3
|CH3
+
Cl|
CH –CH –CH= + HCl3 2 CH2
(Major)
4. Methyl ethers can be prepared by treating primary or secondary alcohol or phenol with diazomethane in presence of BF3.
3BF
2 5 2 2 2 5 3 2Ethylmethyl ether
C H OH CH N C H OCH N
Chemical Properties :
A. Properties due to Alkyl Groups :
1. Halogenation : When ethers are treated with chlorine or bromine in the dark, substitution
occurs at the -carbon atom. The extent of substitution depends upon the reaction conditions.
´´dark
3 2 2 3 2 3 2 3–Chlorodiethyl ether
2
2 2 3 3 3, –Dichlorodiethyl ether , ´–Dichlorodiethyl ether
CH – CH – O – CH – CH Cl CH .CHCl – O – CH .CH
Cl
CH Cl.CHCl – O – CH .CH CH CHCl – O – CHCl.CH
light
3 2 2 3 2 3 2 2 3Perchlorodiethyl ether
CH CH – O – CH .CH 10Cl CCl .CCl – O – CCl .CCl
2. Combustion :
C2H5.O.C2H5 + 6O2 4CO2 + 5H2O
B. Properties due to Ethereal Oxygen :
1. Chemical inertness : Since ethers do not have an active group, in their molecules, these do not react with active metals like Na, strong bases like NaOH, reducing or oxidising agents.
2. Formation of peroxide (Autoxidation) : On standing in contact with air and light ethers are
converted into unstable peroxides (R2O O) which are highly explosive even in low
concentrations.
3. Basic nature : Owing to the presence of unshared electron pairs on oxygen, ether behave as Lewis bases. Hence they dissolve in strong acids (e.g. conc. HCl, conc. H2SO4) at low
temperature to form oxonium salts.
–
2 5 2 2 4 2 5 2 4Diethyl ether Diethyloxonium hydrogen sulphate
(C H ) O H SO [(C H ) OH] HSO
On account of this property, ether is removed from ethyl bromide by shaking with conc. H2SO4.
Being Lewis bases, ethers also form coordination complexes with Lewis acids like BF3,
AlCl3, RMgX, etc.
R O + BF2 3 R O2 BF3
2R O + RM X2 g
R O2
R O2
R
X
Mg
It is for this reason that ethers are used as solvent for Grignard reagents.
C. Properties due to carbon-oxygen bond :
1. Hydrolysis :
2 4H SO
2 5 2 5 2 2 5Ethyl alcohol
C H – O – C H H O 2C H OH
The hydrolysis may also be effected by boiling the ether with water or by treating it with steam.
Note : Ethers can never be hydrolysed in alkaline medium.
2. Action of conc. sulphuric acid :
2 5 2 5 2 4 2 5 2 5 4
Ethyl alcohol Ethyl hydrogen sulphate
C H – O – C H H SO (conc.) C H OH C H HSO
2 5 2 4 2 5 4 2C H OH H SO (conc.) C H HSO H O
3. Action of hydroiodic or hydrobromic acid :
In cold, ether react with HI or HBr to give the corresponding alkyl halide and alcohol. In case of mixed ethers, the halogen atom attaches itself to the smaller alkyl group.
2 5 2 5 2 5 2 5
Ethyl ether Ethyl iodide Ethyl alcohol
C H – O – C H HI C H I C H OH
The order of reactivity of halogen acids is :
HI > HBr > HCl
If one of the the group around oxygen is aryl group then I– will always attack on the group other than aryl group.
O–CH3+ HI
OH+ ICH3
O–C–CH3|CH3
CH3
|+ HI OH
+ I–C–CH3|CH3
CH3
|
D. Properties Due to Benzene Nucleus :
Alkoxy group, being o-, p- directing, anisole undergoes substitution in o- and p- positions. However, –OR group is less activating than the phenolic group.
(i) Nitration :
OCH3 OCH3
conc. HNO3
conc. H SO2 4
NO2
NO2
+
Methylphenyl ether
(Anisole)
Methyl 2-nitrophenyl
ether or o-Nitroanisole
Methyl 4-nitrophenyl
ether or p-Nitroanisole
OCH3
(ii) Bromination :
OCH3
Br /Fe2
OCH3
BrBr
Br
Anisole 2, 4, 6, Tribromoanisole
OCH3
+ Br2
CS2
OCH3
Br+
OCH3
Br
Anisole 2-Bromoanisole 4-Bromoanisole
(iii) Sulphonation :
SO3
Anisole p-Methoxybenzenesulphonic acid
o-Methoxybenzenesulphonic acid
OCH3OCH3 OCH3
SO H3
H SO2 4
SO H3
Example 5 : What are crow ethers? How can the following reaction be made to proceed?
CH2BrKF
CH2F
KBr
Solution: Crown ethers are large ring polyethers and are basically cyclic oligomers of oxirane
which may have annulated rings. They are designated according to ring size and the
number of complexing oxygen atoms, thus 18-crown – 6 denotes an 18-membered ring
with 6-oxygens. The molecule is shaped like a “doughnut”, and has a hole in the middle.
OO
O
O O
O
These are phase transfer catalysts. This is a unique example of “host-guest relationship”.
The crown ether is the host, the cation is the guest. The cavity is well suited to fit a K+ or
Rb+ which is held as a complex. Interaction between host and guest in all these
complexes are mainly through electrostatic forces and hydrogen bonds.
The reaction can be made to process by using catalytic amount of crown ether, 18-
crown-6.
PHENOLS
Phenols are compounds with an OH group attached directly to a carbon of an aromatic ring. The compound phenol is C6H5OH.
OH
If OH group is present on the side chain, then it is called aliphatic alcohol. For example
CH2OH
Phenols may be
(i) monohydric :
OH
(ii) dihydric :
OHOH
Catechol
(iii) trihydric :
OH
OHOH
Phloroglucinol
In phenol R– of alcohol is replaced by aryl ring.
Comparison of bond Angles in Phenols, Alcohols and Ethers :
Bond angle increases with the increase in hindrance.
..
..
H109°
O
O.. ..
H–C
H
H
C
H
HH
111.7°
O.. ..
H–C
H
H108.5°
H
Bond angle increases with the increase in hindrance.
Method of Preparation
1. From Aryl Sulphonic Acids :
When aryl sulphonic acids are fused with NaOH at 570 – 620 K followed by hydrolysis phenols are formed.
+ NaOH
SO H3
SO Na3 ONa OH
H /H O+
2
2. From Haloarenes : (Dow's process)
Cl
+ NaOH
ONa OH
H+
(aq.) 623 K
320 atm
+
Note : Condition of reaction become less vigorous when –M groups are present at ortho or para or both position to the chlorine atom.
3. From Benzene Diazonium Salts :
NH2 OH
H O2NaNO + HCl2
0 – 5°C
N Cl2
+
H SO42
+ N + H–Cl 2
Note : In the absence of H2SO4 diazocoupling will also take place.
N Cl2
–+
+
OH OH
N=N
+ HCl
4. Cumene Process :
+CH – C = CH3 2
H
BenzenePropene
H PO3 4
CH3
CCH3 H
+ O2
CH3
C–O–O–HCH3
H /H O+
2
OH
+ CH C—CH3— 3
O
Cumene Cumene hydroperoxide
light
5. Grignard's Synthesis :
C6H5MgBr + 2
1O
2 C6H5OMgBr 2H O/H
C6H5OH + MgOH
Br
6. From Salicylic Acid :
OH
COOH+ NaOH
OH
COONa+ NaOH
CaO,
–Na CO2 3
OH
Chemical Properties
1. Acidic Nature :
(i) Phenol behave as a weak acid forming phenoxide ion with strong alkalies.
Phenol
C H OH6 5 + NaOH
Sodium phenoxide
C H ONa6 5 + H2O
(ii) It also reacts with sodium metal to form sodium phenoxide and hydrogen is evolved
Phenol
C H OH6 5+ Na 6 5 2
1C H ONa H
2
(a) Effect of substituents on the acidity of phenols : It should be noted that the presence of electron withdrawing groups like –NO2, – CN, –CHO,–X, –COOH, etc. increases the acidic
strength (because of the greater polarity of O–H bond the greater stability of the phenoxide ion by the dispersal of –ve charge, by –R effect). On the other hand, electron-releasing groups like –CH3, –NH2, –OH, etc., tend to destabilize the phenoxide ion by intensifying its –
ve charge by +R effect and hence decreases the acidic strength.
o - chlorophenol > m - chlorophenol > p - chlorophenol
Ka = 7.7 × 10–9 1.6 × 10–9 6.3 × 10–10
In case of haloarenes –I effect of halogens dominates over it's +M effect. (except for fluorine)
p-nitrophenol > o-nitrophenol > m - nitrophenol > phenol
(b) Steric Effect : 3, 5 -dimethyl-4-nitrophenol is weaker acid than the isomeric 2, 6-dimethyl - 4 - nitrophenol.
2. Alkylation or Etherification : When sodium phenoxide is treated with alkyl halides (but not with aryl halides as they are inert) form phenolic ethers.
CH INaOH
6H OPhenol Sodium Methyl phenylether
phenoxide (Anisole)
C H OH C H ONa C H OCH NaI
3
26 5 5 6 5 3
C H BrNaOH
Phenol Sodium Ethyl phenylphenoxide ether (Phenetole)
C H OH C H ONa C H OC H NaI 2 5
6 5 6 5 6 5 2 5
3. Claisen rearrangement :
C6H5ONa + BrCH2 – CH = CH2 Allyl phenyl ether
C H O CH CH CH NaBr 6 5 2 2
When aryl allyl ether is heated to 475 K, the allyl group of the ether migrates from ethereal oxygen to the ring carbon at ortho position.
O–CH –CH = CH2 2*
475K CH –CH = CH2 2
OH
o-Allyl phenol
*
Note : Carbon attached with oxygen is not attached with the carbon of benzene ring in the product.
4. Acylation and benzoylation :
Phenol
C H OH6 5 + CH – C – Cl3
O
Acetyl chloride
Pyridine C H – O – C – CH6 5 3
O
Phenyl acetate
5. Fries Rearrangement : When heated with anhydrous aluminium chloride, phenyl esters undergo Fries rearrangement forming a mixture of o- and p-hydroxy ketones.
Phenyl acetate
O – C – CH3
O
Heat
AlCl
3
o-hydroxyacetophenone
C – CH3
OOH
+ O = C – CH3
OH
p-Hydroxyacetophenone
The para isomer is formed predominantly at low temperature while at higher temperatures o - isomer is predominant.
6. Reactions due to C–O Bond :
(i) Reaction with PCl5 :
C6H5OH + PCl5 C6H5Cl + POCl3 + HCl
3C6H5OH + PCl3 Triphenyl phosphate
P(OC H ) 3HCl6 5 3
The yields of C6H5Cl is very poor due to the formation of triaryl phosphate.
(ii) Reaction with Ammonia :
ZnCl
573 KPhenol Aniline
C H OH NH C H NH H O 2
6 5 3 6 5 2 2
(iii) Reaction with Zinc Dust :
Δ
Phenol Benzene
C H OH Zn C H 6 5 6 6+ ZnO
(iv) Reaction with Neutral FeCl3: ( Test for phenol)
3C6H5OH + FeCl3 Ferric phenoxide(Violet)
(C H O) Fe 3HCl6 5 3
7. Electrophilic Substitution Reaction on the Benzene Ring :
From the contributing structure of phenol, it is clear that ortho- and para-position on it become
rich in electron density. Thus the electrophilic attack at these positions is facilitated. Again ....
present on the benzene ring is the very powerful ring activator towards electrophilic aromatic substitution.
(i) Bromination :
OH
Phenol
+ 3Br2 H O
2 Br
OHBr Br
2,4, 6-Tribromophenol(yellow ppt.)
+ 3HBr
SO H3
OH
p-Phenolsulphonic acid
+ 3Br2 (aq.)
OH
Br
Br Br
2,4, 6-Tribromophenol(yellow ppt.)
+ 3HBr + H2SO4
(ii) Nitration :
(a)
OH
Phenol
+ dil HNO3 293 K
OH
NO2
o-Nitrophenol(40% yield)
+ NO2
OH
p-Nitrophenol(13% yield)
(b) With concentrated nitric acid and sulphuric acid, it forms 2, 4, 6-trinitrophenol (Picric acid).
OH
+ H SO conc.
(conc.)
3HNO 2 4
3
OH
NO2
O N2 NO2
2,4, 6-trinitrophenol(Picric acid)
(iii) Sulphonation : When heated with conc. sulphuric acid, phenol forms hydroxy benzene sulphonic acid.
OH
SO H3
OH
p-Hydroxy benzene sulphonic acid
OH
SO H3
373 K
H SO , 298 K2 4
–H O2
H SO , 373 K2 4
–H O2
(iv) Friedel-Crafts Alkylation and Acylation : Phenol undergo both these reaction to form mainly p-isomer.
OH
Phenol + CH3Cl Anhyd. AlCl
3
CH3
OH
(Major product)p–Cresol +
OH
CH3
o-Cresol
OH
+ CH3COCl Anhyd. AlCl
3
OH
COCH3
COCH3
OH
+
o- p-
Hydroxy acetophenone
8. Kolbe’s reaction
ONa
Sodium phenoxide+ CO2 398 K, 4– 7 atm
OH
COONa
Sodium salicylate(Main product) H
OH
COOH
Salicylic acid
Note : (i) Methylsalicylate with methanol.
OHCOOH
+ CH3OH H SO
few drop2 4
OH
C – OCH3
O
Methyl Salicylate
(ii) Salol (with phenol) :
OH
COOH+C H OH6 5
H SO2 4
OH
C – O – C H6 5
O
Phenyl salicylate(Salol)
Methyl salicylate is known as oil of winter green. Phenyl salicylate is known as salol. Salol is an intestinal antiseptic.
9. Riemer Tiemann Reaction :
(a) On heating with chloroform and alkali phenols are converted to phenolic aldehydes
OH
+ CHCl3 + 3NaOH 333 343
(aq.) H
OHCHO
+ 3NaCl + 2H2O
In this reaction dichloro carbene is formed as intermediate which attack on benzene ring as electrophile.
(b) If instead of chloroform, carbon tetrachloride is used, salicylic acid is formed. Some para isomers is also formed.
OH
+ CCl4
NaOH
340K
ONaCCl3 3NaOH(aq.)
–3NaCl
ONa
COONaDil. H SO
2 4
OH
COOH
In this reaction 3CCl is formed as intermediate which attack on benzene ring as
electrophile.
10. Coupling with Diazonium Salts :
C6H5N2Cl + C6H5OH 0 5 C
pH 9 10
p-Hydroxy azobenzene(An orange dye)
N = N OH
11. Test for phenol
(i) Neutral FeCl3 test Aqueous solution of phenol gives a violet colouration with FeCl3.
(ii) Br2 water test Aqueous solution of phenol gives a yellow precipitate of 2, 4, 6-
tribromophenol with bromine water.
(iii) Phenol gives Liebermann's nitroso reaction.
Example 6: Explain the law a boiling point and decreased water solubility by o – nitro phenol and o –
hydroxy benzoyldehydes as compared with this m and p – isomers.
Solution : Intramolecular H-bonding (chelation) in the o-isomers inhibits intermolecular attraction,
lowering the boiling point and reduces H-bonding with H2O, decreasing water solubility.
Intramolecular chelation can not occurs in m – and p – isomers.
O
N O
H
O
o - nitro phenol
O
C O
H
H
o - hydroxy benzaldehydes