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 OH OH OH OH
CH32 – CH OH CH22 – CH CH22 – CH – CH Ethyl alcohol Ethylene glycol Glycerol (monohydric) (dihydric) (trihydric)
Example 1: Classify the following into primary, secondary and tertiary alcohols:
(a) CH3 (b) OH (c) HO OH H3C
Solution: (a) Tertiary (b) Secondary (c) Tertiary
GENERAL METHODS OF PREPARATION
1. From Alkenes :
(i) By direct hydrolysis : OH H24 SO CH32 – CH = CH2 + H O CH33 – CH – CH (ii) Oxymercuration demercuration : OH OH NaBH4 Hg (OAc)2 CH – CH = CH + H O CH2 – CH – CH2 – CH33 – CH – CH 322 THF OH Hg(OAc) (iii) Hydroboration oxidation : OH BH26 H22 O /OH 6CH3 – CH = CH2 2(CH32 – CH – CH23 –) B 6CH3 – CH2 – CH 2 + 2H 3 BO 3
Overall result of above reaction is anti Markwonikoff addition of water and with no rearrangement.
(iv) Oxo process followed by hydrogenation : O [Co(CO) ] 42 CH – CH – CH – C – H CH3 – CH = CH22 + CO + H high temperature 3 22 and high pressure H2 /Pd CH3 – CH 2 – CH 2 – CH 2 – OH 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. – R X OHR OH X 3. Reduction of Carbonyl Compounds, Carboxylic Acids and their Derivatives : O red. agent R – C – H R – CH2 OH O OH red. agent R – C – R O red. agent R – C – OR R – CH2 OH + R – OH O red. agent R – C – X R – CH2 OH O O red. agent R – C – O – C – R 2R – CH OH 2
4. Using Grignard Reagent : (i) From aldehydes or ketones : O OMgX OH HO/H+ OH – C – + R – MgX – C – R 2 – C – R + Mg X
In this reaction Formaldehyde gives 1°-alcohol Other aldehyde gives 2°-alcohol Ketones give 3°-alcohol (ii) From carboxylic acid and their derivatives : O O Z R – C – Z + R – MgX R – C – R + Mg X (iii) From epoxides :
O OMgX OH + OH HO/H2 CH32 – CH — CH + R – MgX CH32 – CH – CH CH32 – CH – CH + Mg X R R 5. Hydrolysis of Ether :
Ether undergo acid hydrolysis with dilute H2SO4 under pressure to give corresponding alcohols.
dil. H2SO4 R O R H2O R OH R OH Example 2: (a) Find A, B, C, D, E. O
Mg H A B PhBr dry ether C i CH3 MgBr (b) CH3 COOC 2 H 5 D E ii H2 O / H Solution: (a) O OH A = PhMgBr B = C = Ph Ph (b) CH3
H3C C OH C2H5OH
CH3 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.
2ROH 2Na 2R ONa H2 (iii) Esterification : When carboxylic acid is treated with alcohols in the presence of acid as catalyst, esters are formed. O O H+ R – C – OH + H – O – R 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.
Anhyd. ZnCl2 (a) R OH HCl(g) R Cl H2 O
Note : HCl + anhyd. ZnCl2 is called Lucas reagent. H /H24 SO (b) R OH HBr conc. R Br H2 O
H /H24 SO (c) R OH HI conc. R I H2 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. R OH Zn dust R H ZnO 4. Oxidation : O O [O] [O] [O] CH OH H – C – H H – C – OH CO 3 2 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 : O Cu CH – CH – OH CH – C – H + H (Dehydrogenation) (i) 3 2 573K 32 OH O Cu (ii) CH3 – CH – CH3 573K CH3 – C – CH 3 + H 2 (Dehydrogenation) (iii) Tertiary alcohols undergo dehydration to give alkene under similar condition.
CH3 CH2 Cu CH3 – C – OH 573 K CH32 – C + H O CH CH 3 3 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 :
P + I2 AgNO2 –HO2 R – CH2 OH R – CH2 I R – CH2 – NO 2 + HNO 2 R – C – NO2 (1°-alcohol) N OH (Nitrolic acid) NaOH
blood red colour
P + I2 AgNO2 –HO2 R2 CH – OH R2 – CH – I R2 CH – NO 2 + HNO 2 R2 C – NO2 (2°-alcohol) NO Pseudonitrol NaOH blue colour
P + I2 AgNO2 HNO2 NaOH R3 – C – OH R3 C – I R32 C – NO No reaction Colourless (3°-alcohol) 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?
CH3
CH CH CH CH OH , CH CH CHCH , CH CH CH OH , 3 2 2 2 3 2 3 3 2 H3C C CH3
OH OH CH3
Solution : CH COH, it being a tertiary alcohol. 3 3
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 CH3
HO2 H3C C CH CH2 H H3C C CHCH3
CH3 CH3 OH Hydroboration
CH3 CH3
i BH3 H C C CH CH H C C CH CH OH 3 2 ii H22 O , OH 3 2 2
CH 3 CH3 Oxymercuration – demercuration
CH3 CH3
i Hg OAC , THF, H O H C C CH CH 2 2 H C C CHCH 3 2 ii NaBH4 3 3
CH3 CH3 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. –2H H–O–H R–O–R HO R–OH HO–R Water 2R Ether 2 Alcohol (2 moles) 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.
Conc. H24 SO CH–OH2 5 HO–CH 2 5 CH–O–CH 2 5 2 5 HO 2 Ethanol (2 molecules)413K Diethyl ether 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) :
CHI2 5 AgO 2 ICH 2 5 CHOCH 2 5 2 5 2AgI Dry
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 OCH3 + NaBr
Sod. phenoxide Methoxybenzene (Anisole) However CH CH3 3 | | + CH –Cl + NaO–C–CH CH3 –O–C–CH3 + NaCl 33| | 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.
Cl CH3 | | + CH3 –CH 2 –CH–CH 3 + NaO–C–CH 3 CH32 –CH –CH=CH2 + HCl | (Major) CH3 4. Methyl ethers can be prepared by treating primary or secondary alcohol or phenol with diazomethane in presence of BF3.
BF3 C2 H 5 OH CH 2 N 2 C 2 H 5 OCH 3 N 2 Ethylmethyl ether 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 CH3 –CH 2 –O–CH 2 –CH 3 Cl 2 CH.CHCl–O–CH.CH 3 2 3 –Chlorodiethyl ether
Cl2
CH2 Cl.CHCl – O – CH 2 .CH 3 CH 3 CHCl – O – CHCl.CH 3 , –Dichlorodiethyl ether , ´–Dichlorodiethyl ether
light CH3 CH 2 – O – CH 2 .CH 3 10Cl 2 CCl 3 .CCl 2 – O – CCl 2 .CCl 3 Perchlorodiethyl ether 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.
– (C2 H 5 ) 2 O H 2 SO 4 [(C 2 H 5 ) 2 OH] HSO 4 Diethyl ether Diethyloxonium hydrogen sulphate 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.
R23 O + BF RO2 BF3
RO2 R
2R2 O + RMg X Mg RO X 2 It is for this reason that ethers are used as solvent for Grignard reagents. C. Properties due to carbon-oxygen bond : 1. Hydrolysis :
H24 SO CH–O–CH2 5 2 5 HO 2 2CHOH 2 5 Ethyl alcohol 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 :
C25 H – O – C 2524 H H SO (conc.) C 25 H OH C 254 H HSO Ethyl alcohol Ethyl hydrogen sulphate
C2 H 5 OH H 2 SO 4 (conc.) C 2 H 5 HSO 4 H 2 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.
CH–O–CH2 5 2 5 HI CHI 2 5 CHOH 2 5 Ethyl ether Ethyl iodide Ethyl alcohol 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 OH + HI + CH3 I
CH3 | CH3 OH | O–C–CH3 + HI | + 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 : OCH OCH 3 3 OCH3
NO2 conc. HNO3 + conc. H24 SO
NO2 Methylphenyl ether Methyl 2-nitrophenyl Methyl 4-nitrophenyl (Anisole) ether or o-Nitroanisole ether or p-Nitroanisole (ii) Bromination :
OCH3 OCH3 Br Br Br2 /Fe
Br Anisole 2, 4, 6, Tribromoanisole
OCH3 OCH3 OCH3 Br CS2 + Br2 +
Br Anisole 2-Bromoanisole 4-Bromoanisole (iii) Sulphonation : OCH OCH OCH3 3 3
SO3 H H24 SO
SO3
SO3 H
Anisole p-Methoxybenzene o-Methoxybenzene sulphonic acid sulphonic acid
Example 5 : What are crow ethers? How can the following reaction be made to proceed?
CH2F CH2Br KF 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.
O O
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 OH
(ii) dihydric : Catechol OH
(iii) trihydric : HO OH 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. .
...... H O 109° H H O . .. H H–C H O.. H–C C 108.5° 111.7° H H H 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.
SOH3 SO3 Na ONa OH
+ H/HO2 + NaOH
2. From Haloarenes : (Dow's process) + Cl ONa OH
623 K H+ + NaOH (aq.) 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 N2 Cl OH NaNO + HCl 2 HO2 + N + H–Cl 0 – 5°C H SO 2 2 4
Note : In the absence of H2SO4 diazocoupling will also take place.
+ – N2 Cl OH OH
+ + HCl
N=N
4. Cumene Process :
CH3 CH3 CH C H CH C–O–O–H OH H 3 3 O + H PO light H /H2 O 34 + O +CH3 – C = CH 2 2 + CH3— C—CH3 Propene Benzene Cumene Cumene hydroperoxide 5. Grignard's Synthesis : OH 1 HO/H2 Mg C6H5MgBr + O2 C6H5OMgBr C6H5OH + 2 Br 6. From Salicylic Acid : OH OH OH
COOH COONa CaO, + NaOH + NaOH –Na CO 23 Chemical Properties 1. Acidic Nature : (i) Phenol behave as a weak acid forming phenoxide ion with strong alkalies.
C65 H OH + NaOH C65 H ONa + H2O Phenol Sodium phenoxide (ii) It also reacts with sodium metal to form sodium phenoxide and hydrogen is evolved 1 C65 H OH+ Na C6 H 5 ONa H 2 Phenol 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 –9 –9 –10 Ka = 7.7 × 10 1.6 × 10 6.3 × 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. NaOH CH3 I C6 H 5 OH C6 H 5 ONa C 6 H 5 OCH 3 NaI PhenolHO2 Sodium Methyl phenylether phenoxide (Anisole)
NaOH C25 H Br C6 H 5 OH C 6 H 5 ONa C 6 H 5 OC 2 H 5 NaI Phenol Sodium Ethyl phenyl phenoxide ether (Phenetole) 3. Claisen rearrangement :
C6H5ONa + BrCH2 – CH = CH2 C6 H 5 O CH 2 CH CH 2 NaBr Allyl phenyl ether 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 = CH OH 2 2 CH –CH = CH* 475K 2 2
o-Allyl phenol Note : Carbon attached with oxygen is not attached with the carbon of benzene ring in the product.
4. Acylation and benzoylation : O O Pyridine CH – C – Cl C6 H 5 – O – C – CH 3 C65 H OH + 3 Phenol Acetyl chloride 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. O OH
O – C – CH3 OH O
C – CH3
O = C – CH3 Phenyl acetate Heat o-hydroxyacetophenone + p-Hydroxyacetophenone AlCl3 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 P(OC6 H 5 ) 3 3HCl Triphenyl phosphate
The yields of C6H5Cl is very poor due to the formation of triaryl phosphate. (ii) Reaction with Ammonia :
ZnCl2 CHOHNH6 5 3 CHNH 6 5 2 HO 2 Phenol573 K Aniline (iii) Reaction with Zinc Dust : Δ C6 H 5 OH Zn C 6 H 6 + ZnO Phenol Benzene
(iv) Reaction with Neutral FeCl3: ( Test for phenol)
3C6H5OH + FeCl3 (C6 H 5 O) 3 Fe 3HCl Ferric phenoxide (Violet) 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 OH Br Br
HO2 + 3Br2 + 3HBr Phenol Br 2,4, 6-Tribromophenol (yellow ppt.) OH OH Br Br
+ 3Br2 (aq.) + 3HBr + H2SO4 Br SO3 H p-Phenolsulphonic acid 2,4, 6-Tribromophenol (yellow ppt.) (ii) Nitration : OH OH OH NO2 (a) + dil HNO 293 K + 3 NO2 o-Nitrophenol p-Nitrophenol Phenol (40% yield) (13% yield) (b) With concentrated nitric acid and sulphuric acid, it forms 2, 4, 6-trinitrophenol (Picric acid). OH
OH ON2 NO2
H24 SO conc. + 3HNO3 (conc.) NO2 2,4, 6-trinitrophenol (Picric acid) (iii) Sulphonation : When heated with conc. sulphuric acid, phenol forms hydroxy benzene sulphonic acid. OH OH
SO3 H H24 SO , 298 K –H O 2 373 K OH
H24 SO , 373 K
–H2 O
SO3 H p-Hydroxy benzene sulphonic acid
(iv) Friedel-Crafts Alkylation and Acylation : Phenol undergo both these reaction to form mainly p-isomer. OH
OH OH
CH3 CH3 (Major product) Phenol Anhyd. AlCl3 p–Cresol o-Cresol + CH3Cl + OH OH
COCH3 + OH COCH3 o- p-
Anhyd. AlCl3 Hydroxy acetophenone + CH3COCl 8. Kolbe’s reaction OH ONa COONa OH COOH
Sodium salicylate Sodium phenoxide 398 K, 4– 7 atm (Main product) H Salicylic acid + CO2
Note : (i) Methylsalicylate with methanol.
OH O
OH C – OCH3 COOH
+ CH OH H24 SO Methyl Salicylate 3 few drop (ii) Salol (with phenol) : OH OH O COOH HSO24 C – O – C65 H +C65 H OH
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 OH CHO + CHCl + 3NaOH 333 343 + 3NaCl + 2H O 3 (aq.) H 2 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. ONa ONa OH OH COONa COOH NaOH CCl3 3NaOH(aq.) Dil. H24 SO –3NaCl + CCl4 340K
In this reaction CCl3 is formed as intermediate which attack on benzene ring as electrophile. 10. Coupling with Diazonium Salts : C H N Cl + C H OH 0 5 C N = N OH 6 5 2 6 5 pH 9 10 p-Hydroxy azobenzene (An orange dye) 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 O H H
N O C O H O o - hydroxy benzaldehydes o - nitro phenol