CBSE Class 12 Chemistry Alcohols Phenols and Ethers Important Questions Set 07

Very Short Answer Questions (1 Mark)

Question 1. Name the only primary alcohol which gives iodoform test.
Answer:
Ethanol.
CH₃CH₂OH + 4I₂ + 6NaOH → CHI₃ (Iodoform, yellow ppt.) + HCOONa + 5NaI + 5H₂O

Teacher's Note:
1. The iodoform test needs a CH₃–CH(OH)– group; ethanol is the only 1° alcohol that has it.

Question 2. How will you distinguish between 1-phenylethanol and 2- phenylethanol ?
Answer:
By iodoform test. 1-Phenylethanol contains the grouping –CHOH–CH₃ linked to carbon and hence gives iodoform test while 2- phenylethanol does not contain the grouping –CHOH–CH₃ and hence does not give iodoform test.
C₆H₅–CHOH–CH₃ (1-Phenylethanol) \( \xrightarrow{I_2/NaOH} \) CHI₃ (Iodoform) ;
C₆H₅CH₂CH₂OH (2-Phenylethanol) \( \xrightarrow{I_2/NaOH} \) No reaction

Teacher's Note:
1. In a distinction question, always state what each compound does: the yellow precipitate for one and no precipitate for the other.

Question 3. In the process of wine making, ripened grapes are crushed so that sugar and enzyme should come in contact with each other and fermentation should start. What will happen if anaeropic conditions are not maintained during this process. (CBSE Sample Paper 2017)
Answer:
Fermentation occurs under anaerobic (absence of air) conditions. If anaerobic conditions are not maintained then ethanol present in wine will be oxidised to ethanoic acid.
CH₃CH₂OH (Ethanol, Ethyl alcohol) \( \xrightarrow[\text{or oxygen}]{\text{Air}} \) CH₃COOH (Ethanoic acid, Acetic acid)

Teacher's Note:
1. This is why wine left open to air turns sour (vinegar-like).

Question 4. What is rectified spirit ?
Answer:
A mixture of 95% ethyl alcohol and 5% water is called rectified spirit. It is obtained by distillation of fermentation mixture obtained from molasses.

Teacher's Note:
1. Rectified spirit cannot be made purer than 95% by simple distillation because it is an azeotrope.

Question 5. What is absolute alcohol ?
Answer:
100% ethyl alcohol is called absolute alcohol. It is obtained by azeotropic distillation of rectified spirit.

Teacher's Note:
1. Remember the chain: wash (fermented liquor) → rectified spirit (95%) → absolute alcohol (100%).

Question 6. What is ‘Power alcohol’ ?
Answer:
A mixture of absolute alcohol (20%), petrol (80%) and some benzene as a cosolvent is called power alcohol.

Teacher's Note:
1. Benzene is added because absolute alcohol and petrol do not mix well on their own.

Question 7. Why can’t rectified spirit be converted into absolute alcohol by simple distillation ?
Answer:
Rectified spirit containing 95% ethyl alcohol and 5% water forms an azeotropic mixture which distils at a constant temperature of 351.1 K.

Teacher's Note:
1. An azeotrope boils with the same composition in liquid and vapour, so distillation cannot separate its components.

Question 8. Why boiling points of phenols are higher than those of the corresponding aromatic hydrocarbons and alkyl halides ?
Answer:
Due to intermolecular H-bonding (···O(C₆H₅)–H···O(C₆H₅)–H···) phenols have higher boiling point than the corresponding aromatic hydrocarbons and alkyl halides which have only weak van der Waals forces of attraction.

Teacher's Note:
1. Toluene (M = 92, b.p. 384 K) and phenol (M = 94, b.p. 455 K) make a good pair to quote.

Question 9. Why are phenols more acidic than alcohols. Give two reactions to show that phenols are acidic in nature. (Hr. Board 2012)
Answer:
Since the phenoxide ion left after the removal of a proton is stabilized by resonance (the negative charge is delocalised over the ortho and para positions of the benzene ring) whereas alkoxide ion (left after the removal of a proton from alcohol) is not. Therefore, phenols are more acidic than alcohols.
The two reactions of phenol showing its acidic nature are :
(i) Phenols dissolve in NaOH forming sodium phenoxide.
C₆H₅OH + NaOH → C₆H₅ONa + H₂O
(ii) Phenols react with metallic sodium to evolve H₂ gas
2 C₆H₅OH + 2 Na → 2 C₆H₅ONa + H₂

Teacher's Note:
1. Alcohols react with Na but not with aqueous NaOH; reaction with NaOH is the one that shows phenol is the stronger acid.

Question 10. Arrange the following in order of decreasing acid strength : CH₃OH, H₂O, C₆H₅OH
Answer:
Due to +I-effect of the CH₃ group, the electron density in the O–H bond of CH₃OH is more than that in water, H–OH. Therefore, O–H bond in CH₃OH is stronger and hence more difficult to break than that in H₂O and hence CH₃OH is a weaker acid than H₂O.
In contrast, phenoxide ion (left after the removal of a proton) is stabilized by resonance, whereas methoxide ion (left after the removal of a proton from CH₃OH) is not. Therefore, C₆H₅OH is a stronger acid than CH₃OH. Thus, the overall acidic strength decreases in the order :
C₆H₅OH > H₂O > CH₃OH.

Teacher's Note:
1. Two separate effects decide this order: the +I effect of CH₃ (methanol vs water) and resonance stabilisation of the anion (phenol vs both).

Question 11. Give one reaction to show that phenol is acidic in character.
Answer:
Turns blue litmus red or dissolves in NaOH (C₆H₅OH + NaOH → C₆H₅ONa + H₂O).

Teacher's Note:
1. Phenol is too weak an acid to react with NaHCO₃; use NaOH or Na in your example.

Question 12. How will you know whether a given OH group is alcoholic or phenolic in nature ?
Answer:
Phenolic OH group gives blue or violet colouration with neutral FeCl₃ while alcoholic OH group does not.

Teacher's Note:
1. The ferric chloride test is the standard one-line distinction between phenols and alcohols.

Question 13. Write the formula of the reagent used in bromination of phenol to 2, 4, 6-tribromophenol.
Answer:
Br₂/H₂O (bromine-water).

Teacher's Note:
1. With Br₂ in a non-polar solvent like CS₂ at low temperature, only mono-bromination (o- and p-bromophenol) occurs.

Question 14. Phenol reacts with benzoyl chloride in presence of NaOH. What is the product formed ? What is the name of this reaction ?
Answer:
Phenyl benzoate. Schotten-Baumann reaction.
C₆H₅OH + C₆H₅COCl \( \xrightarrow{NaOH} \) C₆H₅COOC₆H₅ (Phenyl benzoate) + HCl

Teacher's Note:
1. NaOH converts phenol into the more nucleophilic phenoxide ion and also neutralises the HCl formed.

Question 15. Explain why phenols do not undergo substitution of the –OH group like alcohols?
Answer:
The C–O bond in phenols has some double bond character due to resonance and hence cannot be easily cleaved by a nucleophile. In contrast, the C–O bond in alcohols is a pure single bond and hence can be easily cleaved by a nucleophile.

Teacher's Note:
1. The carbon of the C–O bond in phenol is also sp² hybridised, which makes the bond shorter and stronger.

Short Answer Questions (2 Marks)

Question 16. Ethyl alcohol and phenol both contain OH group. Why phenol is acidic and alcohol is neutral in nature. Give reason. (MP Board 2013)
Answer:
The phenoxide ion formed by loss of a proton from phenol is stabilised by resonance (negative charge delocalised over the ring), and the O–H bond in phenol is weakened because the oxygen lone pair is involved in resonance with the ring. In ethanol, the electron-releasing ethyl group (+I) strengthens the O–H bond and destabilises the ethoxide ion, which has no resonance. Hence, phenol loses H⁺ and is acidic, while ethyl alcohol is practically neutral.

Teacher's Note:
1. Ethanol still reacts with sodium metal, but it does not turn blue litmus red or react with NaOH, while phenol does.

Question 17. Phenols are more acidic than alcohols. (CBSE 2011 ; Raj. Board 2012)
Answer:
In phenol, the –OH group is attached to an sp²-hybridised carbon and the lone pair of oxygen is in resonance with the ring, so the O–H bond is weakened. The phenoxide ion formed is stabilised by resonance as the negative charge is delocalised over the benzene ring.
In alcohols, the –OH group is attached to an sp³ carbon, the alkyl group (+I) increases the electron density on oxygen, and the alkoxide ion is not stabilised by resonance (in fact destabilised by +I effect). Hence, phenols are more acidic than alcohols.

Teacher's Note:
1. Phenol (pK_a 10) reacts with NaOH, but ethanol (pK_a about 16) does not; quote this as evidence.

Question 18. Arrange the following in decreasing order of their acidic character :
(i) p-CH₃O–C₆H₄OH, (ii) p-NO₂–C₆H₄OH, (iii) C₆H₅OH
Answer:
The electron-withdrawing NO₂ group stabilises the phenoxide ion and increases acidity, while the electron-releasing OCH₃ group (+R effect at the para position) destabilises the phenoxide ion and decreases acidity. Hence, the order is :
(ii) > (iii) > (i), i.e., p-nitrophenol > phenol > p-methoxyphenol

Teacher's Note:
1. At the para position, the +R effect of –OCH₃ outweighs its −I effect; at the meta position the −I effect wins and m-methoxyphenol is slightly more acidic than phenol.

Question 19. Give the structural formulae and names of the products of the following reactions :
(i) Phenol is treated with excess of bromine water (ii) Phenol is treated with Br₂/CS₂ at 278 K.
Answer:
(i) A white precipitate of 2,4,6-tribromophenol is formed : C₆H₅OH + 3Br₂ \( \xrightarrow{H_2O} \) 2,4,6-Br₃C₆H₂OH (2,4,6-Tribromophenol) + 3HBr
(ii) Monobromination occurs giving a mixture of 4-bromophenol (major) and 2-bromophenol (minor) : C₆H₅OH + Br₂ \( \xrightarrow{CS_2,\ 278\text{ K}} \) p-Br–C₆H₄–OH (4-Bromophenol) + o-Br–C₆H₄–OH (2-Bromophenol) + HBr

Teacher's Note:
1. In water phenol ionises partly to the highly reactive phenoxide ion, which is why bromine water gives tri-substitution.

Question 20. Carry out conversion of phenol to benzoquinone. (CBSE 2011; CBSE Sample Paper 2016)
Answer:
Phenol on oxidation with chromic acid (sodium dichromate and sulphuric acid) gives benzoquinone :
C₆H₅OH \( \xrightarrow{Na_2Cr_2O_7/H_2SO_4} \) p-O=C₆H₄=O (p-Benzoquinone, cyclohexa-2,5-diene-1,4-dione)

Teacher's Note:
1. Phenol slowly turns pink-red in air for the same reason: it is oxidised to coloured quinone-type products.

Question 21. Why phenol undergoes electrophilic substitution more easily than benzene ? (CBSE 2019)
Answer:
In phenol, the lone pairs of electrons on the oxygen atom of the –OH group are delocalised into the benzene ring (+R effect). This increases the electron density in the ring, especially at the ortho and para positions. Hence, the ring in phenol is activated and is attacked by electrophiles more easily than benzene. Also, the carbocation (arenium ion) formed in the attack is stabilised by the –OH group.

Teacher's Note:
1. That is why phenol reacts with bromine water without any Lewis acid catalyst, while benzene needs FeBr₃.

Question 22. How will you convert (write chemical equations)
(i) Phenol to 4-bromophenol
(ii) Phenol to 2-acetoxybenzoic acid.
Answer:
(i) C₆H₅OH + Br₂ \( \xrightarrow{CS_2,\ 273\text{ K}} \) p-Br–C₆H₄–OH (4-Bromophenol, major) + o-Br–C₆H₄–OH (minor) + HBr
(ii) C₆H₅OH \( \xrightarrow[(iii)\ H^+]{(i)\ NaOH,\ (ii)\ CO_2,\ 400\text{ K},\ 4\text{-}7\text{ atm}} \) o-HO–C₆H₄–COOH (Salicylic acid) \( \xrightarrow{(CH_3CO)_2O/H^+} \) o-CH₃COO–C₆H₄–COOH (2-Acetoxybenzoic acid)

Teacher's Note:
1. 2-Acetoxybenzoic acid is the IUPAC-style name of aspirin.

Question 23. Distinguish between aromatic and aliphatic ethers. (Tamil Nadu Board 2012)
Answer:
Aliphatic ethers : both groups attached to the oxygen atom are alkyl groups, e.g., CH₃–O–CH₃ (methoxymethane), C₂H₅–O–C₂H₅ (ethoxyethane).
Aromatic ethers : one or both groups attached to the oxygen atom are aryl groups, e.g., C₆H₅–O–CH₃ (anisole, an alkyl aryl ether) and C₆H₅–O–C₆H₅ (diphenyl ether, a diaryl ether).
In aromatic ethers the aryl C–O bond has partial double bond character due to resonance; so on cleavage with HI, aryl alkyl ethers give phenol and alkyl iodide, and the ring undergoes electrophilic substitution at o- and p-positions.

Teacher's Note:
1. Anisole is the standard example of an aromatic ether; diethyl ether is the standard aliphatic one.

Question 24. Write Williamson synthesis reaction. (Pb. Board 2012 ; AP Board 2012 ; Hr. Board 2011)
Answer:
When an alkyl halide is heated with sodium alkoxide (or sodium phenoxide), an ether is formed. This is called Williamson synthesis. It is an S_N2 reaction.
R–X + R′–ONa \( \xrightarrow{\Delta} \) R–O–R′ + NaX
Example : CH₃CH₂Br + CH₃CH₂ONa → CH₃CH₂–O–CH₂CH₃ (Ethoxyethane) + NaBr

Teacher's Note:
1. Both symmetrical and unsymmetrical ethers can be made by this method; it is the most important laboratory method.

Question 25. Write Williamson synthesis. Illustrate its limitations. (Hr. Board 2012)
Answer:
Williamson synthesis : R–X + R′ONa → R–O–R′ + NaX, e.g., CH₃I + C₂H₅ONa → CH₃–O–C₂H₅ + NaI.
Limitations : (i) It works best with primary alkyl halides. With secondary and especially tertiary alkyl halides, elimination competes with substitution because alkoxides are strong bases, and alkenes are formed. For example, (CH₃)₃C–Br + CH₃ONa → (CH₃)₂C=CH₂ (2-Methylpropene) + CH₃OH + NaBr, not tert-butyl methyl ether.
(ii) Aryl halides cannot be used (they do not undergo S_N2 reaction); so aryl alkyl ethers are made from sodium phenoxide and alkyl halide, not from aryl halide and sodium alkoxide.

Teacher's Note:
1. When planning an unsymmetrical ether, put the bulky group on the alkoxide and the small (primary) group on the halide.

Question 26. How will you convert CH₃CH₂OH → (CH₃CH₂)₂O ? (West Bengal Board 2013)
Answer:
By heating excess of ethanol with conc. H₂SO₄ at 413 K :
2 CH₃CH₂OH \( \xrightarrow{\text{Conc. }H_2SO_4,\ 413\text{ K}} \) (CH₃CH₂)₂O + H₂O
Alternatively, by Williamson synthesis : CH₃CH₂OH \( \xrightarrow{Na} \) CH₃CH₂ONa \( \xrightarrow{CH_3CH_2Br} \) (CH₃CH₂)₂O + NaBr

Teacher's Note:
1. Keep alcohol in excess and the temperature at 413 K; at 443 K the main product would be ethene.

Short Answer Questions (3 Marks)

Question 27. What is fermentation? How is ethyl alcohol obtained from molasses? Explain with chemical equations. (Uttarakhand Board 2012)
Answer:
Fermentation is the slow decomposition of complex organic compounds into simpler compounds by the action of enzymes produced by micro-organisms such as yeast, usually with evolution of CO₂.
From molasses : Molasses (the dark syrupy mother liquor left after crystallisation of sugar from cane juice) contains about 40% sucrose. It is diluted with water to about 10% sugar, a little ammonium sulphate/phosphate is added as food for yeast, the pH is kept at about 4-5 and yeast is added. The mixture is kept at about 303 K for 2-3 days.
C₁₂H₂₂O₁₁ (Sucrose) + H₂O \( \xrightarrow{\text{Invertase}} \) C₆H₁₂O₆ (Glucose) + C₆H₁₂O₆ (Fructose)
C₆H₁₂O₆ \( \xrightarrow{\text{Zymase}} \) 2C₂H₅OH + 2CO₂
The fermented liquor (wash, about 8-10% alcohol) is fractionally distilled to give rectified spirit (95% ethanol).

Teacher's Note:
1. Both enzymes (invertase and zymase) are supplied by yeast; air is kept out so that ethanol is not oxidised to acetic acid.

Question 28. Briefly describe the manufacture of ethanol from starch by fermentation process, mentioning the role of enzymes in this process. Give necessary chemical equations. (Assam Board 2012)
Answer:
Starchy materials (potatoes, rice, maize, barley) are mashed and heated with steam to get a starch solution. It is treated with malt (germinated barley) at about 323 K. The enzyme diastase present in malt converts starch into maltose :
2(C₆H₁₀O₅)n (Starch) + nH₂O \( \xrightarrow{\text{Diastase}} \) nC₁₂H₂₂O₁₁ (Maltose)
Yeast is then added at about 303 K. The enzyme maltase of yeast converts maltose into glucose, and the enzyme zymase converts glucose into ethanol :
C₁₂H₂₂O₁₁ (Maltose) + H₂O \( \xrightarrow{\text{Maltase}} \) 2C₆H₁₂O₆ (Glucose)
C₆H₁₂O₆ \( \xrightarrow{\text{Zymase}} \) 2C₂H₅OH + 2CO₂
The fermented liquor is fractionally distilled to obtain rectified spirit.

Teacher's Note:
1. Each enzyme catalyses one specific step: diastase (starch → maltose), maltase (maltose → glucose), zymase (glucose → ethanol).

Question 29. Give the structures of A, B and C
(i) CH₃COOH \( \xrightarrow{LiAlH_4} \) A \( \xrightarrow{Cu/573\text{K}} \) B (Hr. Board 2011)
(ii) CH₃CHO \( \xrightarrow{CH_3MgBr} \) A \( \xrightarrow{B} \) CH₃CHOHCH₃ (UP Board 2012)
(iii) CH₃COOH \( \xrightarrow{LiAlH_4} \) A \( \xrightarrow{SOCl_2} \) B (Assam Board 2012)
(iv) CH₃Br \( \xrightarrow{KCN} \) A \( \xrightarrow{LiAlH_4} \) B \( \xrightarrow{HNO_2,\ 273\text{ K}} \) C (CBSE 2014)
Answer:
(i) A = CH₃CH₂OH (Ethanol) ; B = CH₃CHO (Ethanal)
(ii) A = (CH₃)₂CH–OMgBr (Adduct) ; B = H₂O/H⁺ (acid hydrolysis)
(iii) A = CH₃CH₂OH (Ethanol) ; B = CH₃CH₂Cl (Chloroethane)
(iv) A = CH₃CN (Ethanenitrile) ; B = CH₃CH₂NH₂ (Ethanamine) ; C = CH₃CH₂OH (Ethanol)

Teacher's Note:
1. In (iv) the cyanide adds one carbon, so the final alcohol has one carbon more than the starting bromide.

Question 30. (a) How is phenol obtained from
(i) chlorobenzene (Manipur Board 2012)
(ii) aniline
(iii) benzene ? (Hr. Board 2012)
Give equations for the reactions involved.
(b) Give an equation of reaction for the preparation of phenol from curmene. (CBSE Sample Paper 2016)
Answer:
(a) (i) From chlorobenzene (Dow’s process) : C₆H₅Cl + 2NaOH \( \xrightarrow{623\text{ K},\ 300\text{ atm}} \) C₆H₅ONa + NaCl + H₂O ; C₆H₅ONa + HCl → C₆H₅OH + NaCl
(ii) From aniline (via diazonium salt) : C₆H₅NH₂ + NaNO₂ + 2HCl \( \xrightarrow{273\text{-}278\text{ K}} \) C₆H₅N₂⁺Cl⁻ + NaCl + 2H₂O ; C₆H₅N₂⁺Cl⁻ + H₂O \( \xrightarrow{\text{Warm}} \) C₆H₅OH + N₂ + HCl
(iii) From benzene (via benzenesulphonic acid) : C₆H₆ + H₂SO₄ (oleum) → C₆H₅SO₃H + H₂O ; C₆H₅SO₃H \( \xrightarrow[(ii)\ H^+]{(i)\ NaOH\ (\text{fuse})} \) C₆H₅OH
(b) From cumene : C₆H₅CH(CH₃)₂ (Cumene) + O₂ (air) → C₆H₅C(CH₃)₂OOH (Cumene hydroperoxide) \( \xrightarrow{H^+/H_2O} \) C₆H₅OH + CH₃COCH₃

Teacher's Note:
1. Chlorobenzene needs drastic conditions because the C–Cl bond of an aryl halide has partial double bond character.

Question 31. Explain giving reasons :
(i) Why alcohols are higher boiling compounds than hydrocarbons of corresponding molecular masses but have lower boiling points than corresponding phenols.
(ii) Phenols boil at higher temperature than arenes of comparable molecular masses.
(iii) The lower homologues of alcohols are soluble in water whereas phenols are slightly soluble in water.
(iv) Why are boiling points of phenols higher than boiling points of alcohols of comparable molecular masses ?
Answer:
(i) Alcohols are associated by intermolecular H-bonding while hydrocarbons have only weak van der Waals forces; hence alcohols boil higher. In phenols, the O–H bond is more polar (due to resonance) and the molecules are larger with greater van der Waals forces, so the H-bonds are stronger; hence phenols boil higher than corresponding alcohols.
(ii) Phenols form intermolecular H-bonds while arenes have only weak van der Waals forces. Extra energy is needed to break the H-bonds; hence phenols boil higher.
(iii) Lower alcohols form H-bonds with water and the small alkyl group does not hinder it. In phenols, the large hydrophobic benzene ring hinders H-bonding with water; hence phenols are only slightly soluble (phenol about 8 g/100 g water).
(iv) The O–H bond of phenols is more polar than that of alcohols because the lone pair on O is delocalised into the ring; hence phenols form stronger intermolecular H-bonds. Also their larger surface area gives greater van der Waals forces. Thus phenols boil higher.

Teacher's Note:
1. Comparable molecular mass is the key condition; always compare compounds of nearly equal mass, e.g. phenol (94) with toluene (92).

Question 32. Explain the following : (i) Phenol is more acidic than ethanol. (CBSE 2011)
(ii) o- and p-Nitrophenols are more acidic than phenol. (CBSE 2009)
(iii) o-Nitrophenol is steam volatile while p-Nitrophenol is less volatile ? (CBSE 2018 S, 2019)
Answer:
(i) The phenoxide ion is stabilised by resonance (negative charge delocalised over the ring) while the ethoxide ion is not, and is in fact destabilised by +I effect of the ethyl group. Hence, phenol loses H⁺ more easily and is more acidic than ethanol.
(ii) The NO₂ group is electron-withdrawing (−I and −R effects). At o- and p-positions it withdraws electron density from the O–H bond and delocalises the negative charge of the phenoxide ion onto the oxygen atoms of the NO₂ group, further stabilising it. Hence, o- and p-nitrophenols are more acidic than phenol.
(iii) o-Nitrophenol has intramolecular H-bonding (chelation) and exists as discrete molecules, so it is steam volatile. p-Nitrophenol has intermolecular H-bonding and exists as associated molecules, so it is less volatile.

Teacher's Note:
1. Electron-withdrawing groups increase and electron-releasing groups decrease the acidity of phenol; the effect is strongest from o- and p-positions.

Question 33. Account for the following : The OH group in phenols is ortho and para-directing.
Or Nitration of phenol gives ortho and para products.
Answer:
The lone pairs of electrons on the oxygen atom of the –OH group are in conjugation with the benzene ring (+R effect). The resonance structures of phenol show that a negative charge (higher electron density) develops at the ortho (C-2, C-6) and para (C-4) positions :
In these resonance structures the oxygen carries a positive charge (=O⁺H) and the negative charge appears in turn at C-2, C-4 and C-6 of the ring.
Therefore, an electrophile such as NO₂⁺ attacks preferably at the ortho and para positions. Hence, the –OH group is o, p-directing and nitration of phenol (with dil. HNO₃ at 298 K) gives a mixture of o-nitrophenol and p-nitrophenol.
C₆H₅OH \( \xrightarrow{\text{Dil. }HNO_3,\ 298\text{ K}} \) o-O₂N–C₆H₄–OH + p-O₂N–C₆H₄–OH

Teacher's Note:
1. The same +R effect also makes the ring more reactive than benzene, so phenol is nitrated even by dilute nitric acid.

Question 34. Convert phenol into
(i) Salicylaldehyde (Assam Board 2012 ; Hr. Board 2012)
(ii) Salicylic acid (Hr. Board 2012 ; Pb. Board 2012)
(iii) Benzene (Hr. Board 2012)
(iv) Picric acid or 2, 4, 6-trinitrophenol (Hr. Board 2012 ; Karnataka Board 2012 ; CBSE 2013)
Answer:
(i) Reimer-Tiemann reaction : C₆H₅OH \( \xrightarrow[(ii)\ H^+]{(i)\ CHCl_3 + aq.\ NaOH,\ 340\text{ K}} \) o-HO–C₆H₄–CHO (Salicylaldehyde)
(ii) Kolbe’s reaction : C₆H₅OH \( \xrightarrow{NaOH} \) C₆H₅ONa \( \xrightarrow[(ii)\ H^+]{(i)\ CO_2,\ 400\text{ K},\ 4\text{-}7\text{ atm}} \) o-HO–C₆H₄–COOH (Salicylic acid)
(iii) Distillation with zinc dust : C₆H₅OH + Zn \( \xrightarrow{\Delta} \) C₆H₆ + ZnO
(iv) Nitration with conc. HNO₃ : C₆H₅OH \( \xrightarrow{\text{Conc. }HNO_3\ (+\ \text{conc. }H_2SO_4)} \) 2,4,6-(O₂N)₃C₆H₂OH (Picric acid) + 3H₂O. (Nowadays phenol is first sulphonated with conc. H₂SO₄ to phenol-2,4-disulphonic acid, which is then nitrated with conc. HNO₃ to give picric acid in good yield.)

Teacher's Note:
1. Direct nitration of phenol with conc. HNO₃ gives poor yields because phenol is also oxidised; sulphonation first protects the ring.

Question 35. Convert salicylic acid to
(i) aspirin, (ii) methyl salicylate
(iii) phenolphthalein (Tamil Nadu Board 2012)
Answer:
(i) o-HO–C₆H₄–COOH + (CH₃CO)₂O \( \xrightarrow{H^+} \) o-CH₃COO–C₆H₄–COOH (Aspirin) + CH₃COOH
(ii) o-HO–C₆H₄–COOH + CH₃OH \( \xrightarrow{\text{Conc. }H_2SO_4} \) o-HO–C₆H₄–COOCH₃ (Methyl salicylate) + H₂O
(iii) Salicylic acid is first decarboxylated to phenol by heating its sodium salt with soda lime; phenol is then heated with phthalic anhydride and conc. H₂SO₄ :
o-HO–C₆H₄–COONa \( \xrightarrow{NaOH + CaO,\ \Delta} \) C₆H₅OH (after acidification) ; Phthalic anhydride + 2C₆H₅OH \( \xrightarrow{\text{Conc. }H_2SO_4,\ \Delta} \) Phenolphthalein + H₂O

Teacher's Note:
1. In aspirin the phenolic –OH is acetylated; in methyl salicylate the –COOH is esterified. Do not mix the two.

Question 36. What happens when :
(i) Phenol is treated with neutral FeCl₃ (HP Board 2013)
(ii) Phenol is warmed with CCl₄ in presence of aqueous NaOH.
(iii) Ethanoyl chloride is treated with phenol.
(iv) Phenol is heated with zinc dust. (Assam Board 2013 ; CBSE Foreign 2017)
Answer:
(i) A violet colouration is produced due to the formation of a complex : 6C₆H₅OH + FeCl₃ → [Fe(OC₆H₅)₆]³⁻ (Violet) + 3H⁺ + 3HCl
(ii) Salicylic acid is formed (Reimer-Tiemann reaction) : C₆H₅OH + CCl₄ + aq. NaOH \( \xrightarrow{\Delta} \) o-NaO–C₆H₄–COONa \( \xrightarrow{H^+} \) o-HO–C₆H₄–COOH
(iii) Phenyl ethanoate is formed (acetylation) : C₆H₅OH + CH₃COCl \( \xrightarrow{\text{Pyridine}} \) CH₃COOC₆H₅ (Phenyl ethanoate) + HCl
(iv) Benzene is formed : C₆H₅OH + Zn \( \xrightarrow{\Delta} \) C₆H₆ + ZnO

Teacher's Note:
1. Pyridine is used in acetylation to remove the HCl formed, which shifts the equilibrium to the right.

Question 37. How will you convert phenol into
(i) p-hydroxyazobenzene (ii) phenetole (iii) phenyl ethanoate (iv) cyclohexanol
(v) acetophenone (CBSE 2011, 2012)
(vi) 2,4,6-tribromophenol. (Raj. Board 2011 ; CBSE 2011)
Answer:
(i) By coupling with benzenediazonium chloride in alkaline medium : C₆H₅N₂⁺Cl⁻ + C₆H₅OH \( \xrightarrow{OH^-,\ 273\text{-}278\text{ K}} \) p-HO–C₆H₄–N=N–C₆H₅ (p-Hydroxyazobenzene, orange dye) + HCl
(ii) C₆H₅OH \( \xrightarrow{NaOH} \) C₆H₅ONa \( \xrightarrow{C_2H_5I} \) C₆H₅–O–C₂H₅ (Phenetole) + NaI
(iii) C₆H₅OH + CH₃COCl \( \xrightarrow{\text{Pyridine}} \) CH₃COOC₆H₅ (Phenyl ethanoate) + HCl
(iv) C₆H₅OH + 3H₂ \( \xrightarrow{Ni,\ 433\text{ K}} \) C₆H₁₁OH (Cyclohexanol)
(v) C₆H₅OH \( \xrightarrow{Zn\ \text{dust},\ \Delta} \) C₆H₆ \( \xrightarrow{CH_3COCl/\text{anhyd. }AlCl_3} \) C₆H₅COCH₃ (Acetophenone) + HCl
(vi) C₆H₅OH + 3Br₂ \( \xrightarrow{H_2O} \) 2,4,6-Br₃C₆H₂OH (2,4,6-Tribromophenol) + 3HBr

Teacher's Note:
1. For (v), Friedel-Crafts acylation must be done on benzene, not on phenol, because the –OH group reacts with AlCl₃.

Question 38. Complete the following equations and name the products formed :
(i) C₆H₅OH + Br₂ \( \xrightarrow{H_2O} \)
(ii) C₆H₅OH + CHCl₃ \( \xrightarrow[(ii)\ H^+]{(i)\ CHCl_3 + aq.\ NaOH,\ 340\text{ K}} \) (CBSE 2016)
(iii) C₆H₅COCl + C₆H₅OH \( \xrightarrow{NaOH} \) (HP Board 2011)
(iv) C₆H₆ (Benzene) + CH₃–CH=CH₂ \( \xrightarrow{H_3PO_4} \)
(v) o-HO–C₆H₄–COOH \( \xrightarrow[H^+]{(CH_3CO)_2O} \) (CBSE 2017)
(vi) C₆H₅OH + CO₂ \( \xrightarrow[400\text{ K},\ 4\text{-}7\text{ atm}]{NaOH} \) (HP Board 2011)
or C₆H₅OH \( \xrightarrow[(ii)\ CO_2,\ H^+]{(i)\ aq.\ NaOH} \) (CBSE 2016)
(vii) C₆H₅OH + H₂SO₄ \( \xrightarrow{373\text{K}} \)
Answer:
(i) C₆H₅OH + 3Br₂ \( \xrightarrow{H_2O} \) 2,4,6-Br₃C₆H₂OH (2,4,6-Tribromophenol, white ppt.) + 3HBr
(ii) o-HO–C₆H₄–CHO (Salicylaldehyde or 2-hydroxybenzaldehyde)
(iii) C₆H₅COOC₆H₅ (Phenyl benzoate) + HCl (Schotten-Baumann reaction)
(iv) C₆H₅CH(CH₃)₂ (Cumene or isopropylbenzene)
(v) o-CH₃COO–C₆H₄–COOH (Acetylsalicylic acid or aspirin) + CH₃COOH
(vi) o-HO–C₆H₄–COONa (Sodium salicylate), which on acidification gives o-HO–C₆H₄–COOH (Salicylic acid) (Kolbe’s reaction)
(vii) p-HO–C₆H₄–SO₃H (p-Hydroxybenzenesulphonic acid, major) + H₂O

Teacher's Note:
1. Sulphonation of phenol is temperature-controlled: about 290 K gives mainly the ortho acid and 373 K mainly the para acid.

Long Answer Questions (5 Marks)

Question 39. (a) Explain the preparation of ethyl alcohol from molasses.
(b) How does ethyl alcohol react with the following?
(i) Metallic Na (ii) CH₃COOH
(iii) CH₃MgI (iv) conc. H₂SO₄ at 443 K (AP Board 2012)
Answer:
(a) Molasses (the mother liquor left after crystallisation of cane sugar, containing about 40% sugar) is diluted with water to about 8-10% sugar. Ammonium sulphate or phosphate is added as food for yeast and the pH is adjusted to about 4-5. Yeast is added and the mixture is kept at about 303 K for 2-3 days in the absence of air. The enzyme invertase converts sucrose into glucose and fructose, and the enzyme zymase converts these into ethanol :
C₁₂H₂₂O₁₁ + H₂O \( \xrightarrow{\text{Invertase}} \) C₆H₁₂O₆ (Glucose) + C₆H₁₂O₆ (Fructose) ;   C₆H₁₂O₆ \( \xrightarrow{\text{Zymase}} \) 2C₂H₅OH + 2CO₂
The fermented liquor (wash, 8-10% alcohol) on fractional distillation gives rectified spirit (95.6% ethanol), from which absolute alcohol is obtained by azeotropic distillation.
(b) (i) Sodium ethoxide and hydrogen are formed : 2C₂H₅OH + 2Na → 2C₂H₅ONa + H₂↑
(ii) Ethyl acetate is formed (esterification, in presence of conc. H₂SO₄) : CH₃COOH + C₂H₅OH \( \overset{\text{Conc. }H_2SO_4}{\rightleftharpoons} \) CH₃COOC₂H₅ + H₂O
(iii) Methane is evolved (the acidic H of the alcohol destroys the Grignard reagent) : C₂H₅OH + CH₃MgI → CH₄↑ + C₂H₅OMgI
(iv) Ethene is formed by dehydration : C₂H₅OH \( \xrightarrow{\text{Conc. }H_2SO_4,\ 443\text{ K}} \) CH₂=CH₂ + H₂O

Teacher's Note:
1. Reaction (iii) is why Grignard reagents must be kept away from water and alcohols during their preparation.

Question 40. List the action of following reagents on ethanol?
(i) Lucas reagent,
(ii) Phosphorus pentachloride,
(iii) Red phosphorus and bromine,
(iv) Acidified potassium dichromate. (Maharashtra Board 2012)
Answer:
(i) Ethanol, being a 1° alcohol, does not give turbidity with Lucas reagent (conc. HCl + anhyd. ZnCl₂) at room temperature; on heating, ethyl chloride is formed : C₂H₅OH + HCl \( \xrightarrow{ZnCl_2,\ \Delta} \) C₂H₅Cl + H₂O
(ii) Ethyl chloride is formed : C₂H₅OH + PCl₅ → C₂H₅Cl + POCl₃ + HCl
(iii) Red phosphorus and bromine form PBr₃, which converts ethanol into ethyl bromide : 2P + 3Br₂ → 2PBr₃ ; 3C₂H₅OH + PBr₃ → 3C₂H₅Br + H₃PO₃
(iv) Ethanol is oxidised first to ethanal and then to ethanoic acid; the orange colour of dichromate changes to green : CH₃CH₂OH \( \xrightarrow{K_2Cr_2O_7/H_2SO_4\ [O]} \) CH₃CHO \( \xrightarrow{[O]} \) CH₃COOH ; K₂Cr₂O₇ + 4H₂SO₄ → K₂SO₄ + Cr₂(SO₄)₃ + 4H₂O + 3[O]

Teacher's Note:
1. The orange-to-green colour change in (iv) is the basis of the old breath-analyser test for alcohol.