CBSE Class 12 Chemistry Alcohols Phenols and Ethers Important Questions Set 06

Very Short Answer Questions (1 Mark)

Question 1. Cumene is a better starting material for the preparation of phenol. Explain. (CBSE Sample Paper 2016)
Answer:
Cumene is oxidised by air to cumene hydroperoxide, which on treatment with dilute acid gives phenol and acetone:
C₆H₅–CH(CH₃)₂ (Cumene) \( \xrightarrow{O_2\ (\text{air})} \) C₆H₅–C(CH₃)₂–O–O–H (Cumene hydroperoxide) \( \xrightarrow{H^+/H_2O} \) C₆H₅OH (Phenol) + CH₃COCH₃ (Acetone)
Cumene is a better starting material for the preparation of phenol for the following reasons :
(i) Cheap reagents such as O₂ is used for oxidation of cumene to cumene hydroperoxide and dil. H₂SO₄ for conversion of cumene hydroperoxide to phenol and acetone.
(ii) Acetone is a by product of this reaction which is a widely used laboratory and industrial solvent.

Teacher's Note:
1. Cumene itself is made cheaply from benzene and propene, which is why this is the main industrial route to phenol.

Question 2. How is that alcohol and water are miscible in all proportions ? (CBSE 2007)
Answer:
Due to intermolecular H-bonding (as shown below) alcohol and water are miscible in all proportions.
R–O–H ···· O(H)–H  and  H–O–H ···· O(R)–H, i.e., the O of the alcohol accepts H-bonds from water and its O–H hydrogen forms H-bonds with the O of water.

Teacher's Note:
1. Only lower alcohols are miscible in all proportions; as the alkyl part grows, the non-polar portion dominates and solubility falls.

Question 3. Arrange the four isomeric butyl alcohols in order of their decreasing boiling points.
Answer:
Amongst isomeric alcohols, boiling points decrease as branching increases on the carbon atom carrying the OH group due to a corresponding decrease in surface area. Among isomeric 1° alcohols, the branched chain alcohol has lower boiling point than the corresponding n-alcohols. Thus, the boiling points of the four isomeric alcohols decrease in the order :
butan-1-ol > 2-methylpropan-1-ol > butan-2-ol > 2-methylpropan-2-ol

Teacher's Note:
1. Boiling point and solubility in water run in opposite orders for these isomers; learn one and reverse it for the other.

Question 4. Arrange the four isomeric butyl alcohols in order of their increasing solubility in water.
Answer:
As branching increases, the surface area of the non-polar hydrocarbon part decreases and hence the solubility in water increases with branching. Thus, the solubility of the four isomeric butyl alcohols in water increases in the order :
butan-1-ol < 2-methylpropan-1-ol < butan-2-ol < 2-methylpropan-2-ol

Teacher's Note:
1. A compact (branched) alkyl group disturbs the hydrogen-bonded network of water less, so the more branched isomer dissolves better.

Question 5. Why do alcohols have higher boiling points than haloalkanes of comparable molecular mass ?
Answer:
Alcohols undergo extensive intermolecular H-bonding. As a result, alcohols exist as associated molecules as shown below :
H–O(R)····H–O(R)····H–O(R)····H–O(R)····
Quite a large amount of energy is needed to break these H-bonds. Therefore, the boiling points of alcohols are much higher than haloalkanes of comparable molecular mass because the molecules of haloalkanes are held together only by weak dipole-dipole and van der Waals forces of attraction. For example, the b.p. of methyl chloride is 249 K while that of methanol is 337·5.

Teacher's Note:
1. Methanol (M = 32) and methyl chloride (M = 50.5) make a good pair to quote: the lighter alcohol still boils almost 90 K higher.

Question 6. Anhydrous CaCl₂ is not recommended as a drying agent for alcohols and amines. Why so ?
Answer:
Alcohols and amines combine with anhyd. CaCl₂ to form complexes. For example, with C₂H₅OH, it gives a complex of molecular formula, CaCl₂.3C₂H₅OH.

Teacher's Note:
1. Alcohols are usually dried with quicklime (CaO) instead, which does not form such complexes.

Question 7. What is the order of acidic character of primary, secondary and tertiary alcohols ?
Answer:
The 1°, 2° and 3° alcohols have one, two and three alkyl groups on the carbon atom attached to the OH group as shown below :
R→CH₂→O→H (1° Alcohol),   R₂CH→O→H (2° Alcohol),   R₃C→O→H (3° Alcohol)
Since alkyl groups have +I-effect, therefore, electron density in O–H bond increases from 1° to 3° alcohols. In other words, the O–H bond in 3° alcohols is the strongest and hence most difficult to break followed by O–H bond in 2° alcohols while the O–H bond in 1° alcohols is the weakest and hence most easy to break. Thus, the acid strength of alcohols follows the order :
primary > secondary > tertiary

Teacher's Note:
1. The same +I argument also explains why the alkoxide ion of a 3° alcohol is the least stable and the strongest base.

Question 8. Why is sulphuric acid not used during the reaction of alcohols with KI in the conversion of an alcohol to the alkyl iodide ? (CBSE 2010 S)
Answer:
Alkyl iodides are formed when alcohols are heated with constant boiling hydriodic acid (57%)
R–OH + HI \( \xrightarrow{\text{Heat}} \) R–I + H₂O
The HI needed for the purpose, however, cannot be prepared in situ by heating KI with H₂SO₄. The reason being that H₂SO₄ being an oxidising agent, oxidises HI to I₂ before it is able to react with alcohols to form alkyl iodides.
KI + H₂SO₄ \( \xrightarrow{\text{Heat}} \) KHSO₄ + HI
2 HI + H₂SO₄ \( \xrightarrow{\text{Heat}} \) I₂ + SO₂ + 2 H₂O.

Teacher's Note:
1. A non-oxidising acid such as phosphoric acid (H₃PO₄) is used with KI instead.

Question 9. What is the order of reactivity of HCl, HBr and HI with alcohols ?
Answer:
Reactivity decreases as the strength of the H–X bond increases, i.e., HI > HBr > HCl.

Teacher's Note:
1. HCl is so unreactive that it needs anhydrous ZnCl₂ (Lucas reagent) to react with 1° and 2° alcohols.

Question 10. How will you convert ethanol into ethene ? (CBSE 2011)
Answer:
Dehydration of ethanol with conc. H₂SO₄ at 443 K gives ethene.
CH₃CH₂OH (Ethanol) \( \xrightarrow[(\text{Dehydration})]{\text{Conc. }H_2SO_4,\ 443\text{ K}} \) CH₂=CH₂ (Ethene) + H₂O

Teacher's Note:
1. Temperature decides the product: at 413 K excess ethanol with conc. H₂SO₄ gives diethyl ether instead of ethene.

Question 11. The relative ease of dehydration of alcohols is : tertiary > secondary > primary.
Answer:
The dehydration of alcohols occurs through carbocation intermediates. Since the stability of carbocations decreases in the order :
3° > 2° > 1°. Therefore, dehydration of alcohols follows the same order, i.e., tertiary > secondary > primary.

Teacher's Note:
1. Note the milder conditions this allows: 3° alcohols dehydrate with 20% H₃PO₄ at about 358 K, while ethanol needs conc. H₂SO₄ at 443 K.

Question 12. Name a reagent which converts primary alcohols exclusively to corresponding aldehydes ?
Answer:
Pyridinium chlorochromate
RCH₂OH (1° Alcohol) \( \xrightarrow[CH_2Cl_2]{C_5H_5NH^+CrO_3Cl^-} \) R–CHO (Aldehyde)

Teacher's Note:
1. PCC works in anhydrous CH₂Cl₂; without water the aldehyde cannot form a hydrate, so it is not oxidised further to the acid.

Question 13. What is the main product obtained when vapours of t-butyl alcohol are passed over copper at 573 K ?
Answer:
Dehydration occurs to give isobutylene (2-methylpropene).
(CH₃)₃C–OH (tert-Butyl alcohol) \( \xrightarrow{Cu,\ 573\text{ K}} \) (CH₃)₂C=CH₂ (2-Methylpropene) + H₂O

Teacher's Note:
1. A 3° alcohol has no H on the carbinol carbon, so hot copper cannot dehydrogenate it; it is dehydrated instead (1° and 2° alcohols give aldehydes and ketones).

Question 14. Name one reagent which is used for the distinction of primary, secondary and tertiary alcohols. (Raj. Board 2011)
or What is Lucas reagent ? (Raj. Board 2011)
Answer:
Lucas reagent (anhyd. ZnCl₂ + conc. HCl).

Teacher's Note:
1. With Lucas reagent, 3° alcohols give turbidity at once, 2° in about 5 minutes, and 1° alcohols not at room temperature.

Question 15. Arrange the following alcohols in order of increasing reactivity towards Lucas reagent :
2-butanol, 1-butanol, 2-methyl-2-propanol
Answer:
The order of reactivity of alcohols towards Lucas reagent follows the sequence : 3° > 2° > 1°. Thus, the correct sequence of reactivity is : 2-methyl-2-propanol (3°) > 2-butanol (2°) > 1-butanol (1°)
Hence, in increasing order of reactivity : 1-butanol < 2-butanol < 2-methyl-2-propanol.
Reason : the reaction with Lucas reagent proceeds through carbocations, and the stability of carbocations is 3° > 2° > 1°; the more stable the carbocation, the faster the alkyl chloride (turbidity) is formed.

Teacher's Note:
1. Read the question carefully: it asks for increasing reactivity, so write the order starting from the 1° alcohol.

Short Answer Questions (2 Marks)

Question 16. Account for the following : The order of reactivity of halogen acids towards alcohols is : HI > HBr > HCl.
Answer:
The reaction R–OH + HX → R–X + H₂O involves breaking of the H–X bond. The bond dissociation enthalpy of H–X decreases in the order H–Cl > H–Br > H–I as the size of the halogen increases. Thus, HI is the strongest acid and gives the protonated alcohol most readily, and I⁻ is also the best nucleophile among the halide ions. Hence, the reactivity order is HI > HBr > HCl.

Teacher's Note:
1. Because HCl is the least reactive, it needs anhydrous ZnCl₂ as catalyst (Lucas reagent).

Question 17. “Formation of products by the reaction of ethanol and sulphuric acid depends on experimental conditions.” Justify the statement.
Answer:
Ethanol and conc. H₂SO₄ give different products under different conditions :
(i) At 273 K (cold) : ethyl hydrogen sulphate (via oxonium salt) : C₂H₅OH + H₂SO₄ \( \xrightarrow{273\text{ K}} \) C₂H₅HSO₄ + H₂O
(ii) With excess ethanol at 413 K : diethyl ether : 2C₂H₅OH \( \xrightarrow{H_2SO_4,\ 413\text{ K}} \) C₂H₅OC₂H₅ + H₂O
(iii) With excess acid at 443 K : ethene : C₂H₅OH \( \xrightarrow{H_2SO_4,\ 443\text{ K}} \) CH₂=CH₂ + H₂O
Thus, the product depends on temperature and on which reactant is in excess.

Teacher's Note:
1. Lower temperature and excess alcohol favour substitution (ether); higher temperature and excess acid favour elimination (alkene).

Question 18. Give the IUPAC name of iodoform and give its preparation using acetone, sodium hydroxide and iodine. (Hr. Board 2011)
Answer:
IUPAC name of iodoform (CHI₃) : Triiodomethane.
Acetone is warmed with iodine and sodium hydroxide; yellow crystals of iodoform separate out :
CH₃COCH₃ + 3I₂ + 4NaOH → CHI₃ (Iodoform) + CH₃COONa + 3NaI + 3H₂O

Teacher's Note:
1. Here NaOH and I₂ form NaOI (sodium hypoiodite), which iodinates the CH₃ group next to C=O.

Question 19. What is iodoform test ? (HP Board 2013)
Answer:
When an alcohol containing the CH₃–CH(OH)– group (or a carbonyl compound containing the CH₃CO– group) is warmed with iodine and sodium hydroxide solution, a yellow precipitate of iodoform is formed. This is called iodoform test. For example,
CH₃CH(OH)CH₃ + 4I₂ + 6NaOH → CHI₃ (yellow ppt.) + CH₃COONa + 5NaI + 5H₂O
Ethanol and all 2° alcohols of type CH₃CH(OH)R give this test, while methanol, other 1° alcohols and 3° alcohols do not.

Teacher's Note:
1. The test identifies a structural unit, not a class of alcohol: ethanol (1°) gives it, but propan-1-ol (1°) does not.

Question 20. Write the arrow head equations for the manufacture of ethyl alcohol from canesugar by fermentation. (West Bengal Board 2013)
Answer:
C₁₂H₂₂O₁₁ (Cane sugar) + H₂O \( \xrightarrow{\text{Invertase}} \) C₆H₁₂O₆ (Glucose) + C₆H₁₂O₆ (Fructose)
C₆H₁₂O₆ (Glucose or fructose) \( \xrightarrow{\text{Zymase}} \) 2C₂H₅OH (Ethyl alcohol) + 2CO₂

Teacher's Note:
1. Write the enzyme above each arrow; that is what “arrow head equations” asks for.

Question 21. What is meant by (i) absolute alcohol (ii) methylated spirit (iii) power alcohol (iv) rectified spirit ?
Answer:
(i) Absolute alcohol : 100% ethyl alcohol, obtained by azeotropic distillation of rectified spirit.
(ii) Methylated spirit : ethyl alcohol made unfit for drinking (denatured) by adding a little methanol (and some pyridine/copper sulphate for smell and colour); it is used for industrial purposes.
(iii) Power alcohol : a mixture of absolute alcohol (20%) and petrol (80%) with some benzene as cosolvent, used as a motor fuel.
(iv) Rectified spirit : a mixture of 95% ethyl alcohol and 5% water, obtained by fractional distillation of the fermented liquor.

Teacher's Note:
1. Methanol is added to denature alcohol because it is poisonous; drinking methylated spirit can cause blindness.

Question 22. (a) Give two uses of ethyl alcohol. (Hr. Board 2012)
(b) Mention one use each of (i) carbolic acid (ii) oil of winter green.
Answer:
(a) (i) As a solvent for paints, varnishes, dyes, perfumes and medicines (tinctures). (ii) As a fuel (power alcohol) and in alcoholic beverages; also as an antiseptic and for the manufacture of ether, chloroform, etc.
(b) (i) Carbolic acid (phenol) is used as an antiseptic and disinfectant (and in the manufacture of bakelite). (ii) Oil of winter green (methyl salicylate) is used in pain-relieving ointments for muscular pain and as a flavouring agent.

Teacher's Note:
1. Carbolic acid is simply phenol, and oil of winter green is methyl salicylate; learn the common names with the formulas.

Question 23. Identify P, Q, R and S in the following conversions :
CH₃OH \( \xrightarrow{PCl_5} \) P \( \xrightarrow{\text{Alc. KCN}} \) Q \( \xrightarrow{LiAlH_4} \) R \( \xrightarrow{HNO_2,\ 273\text{ K}} \) S (Karnataka Board 2012)
Answer:
P = CH₃Cl (Chloromethane) ; Q = CH₃CN (Ethanenitrile) ; R = CH₃CH₂NH₂ (Ethanamine) ; S = CH₃CH₂OH (Ethanol)
CH₃OH + PCl₅ → CH₃Cl + POCl₃ + HCl ; CH₃Cl + KCN → CH₃CN + KCl ; CH₃CN + 4[H] \( \xrightarrow{LiAlH_4} \) CH₃CH₂NH₂ ; CH₃CH₂NH₂ + HONO → CH₃CH₂OH + N₂ + H₂O

Teacher's Note:
1. This sequence is a standard way of stepping up an alcohol by one carbon atom (methanol to ethanol).

Question 24. What are phenols? How do they differ structurally from aromatic alcohols. (J & K Board 2012)
Answer:
Phenols are compounds in which one or more –OH groups are attached directly to a carbon atom of a benzene (aromatic) ring, e.g., phenol, C₆H₅OH; cresols, CH₃C₆H₄OH.
In aromatic alcohols, the –OH group is attached to an sp³-hybridised carbon of a side chain and not directly to the ring, e.g., benzyl alcohol, C₆H₅CH₂OH. In phenols the –OH group is attached to an sp²-hybridised carbon of the ring.

Teacher's Note:
1. C₆H₅CH₂OH and o-cresol are isomers (C₇H₈O), but only cresol is a phenol; check where the –OH is attached.

Question 25. Both methanol and phenol have an –OH group but the dipole moment of methanol (1.71 D) is higher than that of phenol (1.54 D). Why ?
Answer:
In methanol, the C–O bond is polar with the negative end on oxygen, and the electron-releasing CH₃ group further increases the electron density on oxygen. In phenol, the C–O bond is less polar because (i) the carbon attached to O is sp²-hybridised and more electronegative, and (ii) due to resonance, the lone pair of oxygen is delocalised into the benzene ring, which reduces the electron density on oxygen. Hence, the dipole moment of phenol is lower than that of methanol.

Teacher's Note:
1. In phenol the resonance moment acts from O towards the ring, opposite to the inductive C–O moment, so they partly cancel.

Question 26. Explain the cause of acidity of phenol.
Answer:
(i) In phenol, the –OH group is attached to an sp²-hybridised carbon, which is more electronegative; this makes the O–H bond more polar. Further, due to resonance, oxygen acquires a partial positive charge, which weakens the O–H bond so phenol releases H⁺ easily.
(ii) The phenoxide ion formed is stabilised by resonance, the negative charge being delocalised over the ortho and para carbons of the ring : C₆H₅OH + H₂O ⇄ C₆H₅O⁻ + H₃O⁺. Resonance in phenoxide ion is more effective than in phenol (no charge separation), so the equilibrium lies to the right to some extent; hence phenol is acidic.

Teacher's Note:
1. Phenol itself has charge-separated resonance forms; phenoxide does not, which is why phenoxide gains more stability by resonance.

Short Answer Questions (3 Marks)

Question 27. (a) Write the reaction and state the condition for each of the following conversions :
(i) Propanone to 2-propanol
(ii) Propene to 1-propanol (CBSE 2013)
(iii) Propene to propan-2-ol (CBSE 2013)
(iv) Ethanal to 2-propanol. (CBSE 2013)
(v) Propanone to 2-methyl-2-propanol
(vi) Methanal to ethanol (Hr. Board 2011)
(vii) Ethanoic acid to ethanol. (Hr. Board 2011)
Answer:
(i) CH₃COCH₃ (Propanone) \( \xrightarrow{H_2/Ni\ \text{or}\ NaBH_4} \) CH₃CH(OH)CH₃ (2-Propanol)
(ii) 6 CH₃CH=CH₂ \( \xrightarrow{B_2H_6\ (\text{in THF})} \) 2 (CH₃CH₂CH₂)₃B \( \xrightarrow{H_2O_2/OH^-} \) 6 CH₃CH₂CH₂OH (1-Propanol)
(iii) CH₃CH=CH₂ \( \xrightarrow{\text{Conc. }H_2SO_4} \) CH₃CH(OSO₃H)CH₃ \( \xrightarrow{H_2O,\ \text{boil}} \) CH₃CH(OH)CH₃ (Propan-2-ol) ; or directly CH₃CH=CH₂ + H₂O \( \xrightarrow{H^+} \) CH₃CH(OH)CH₃
(iv) CH₃CHO (Ethanal) \( \xrightarrow[\text{dry ether}]{CH_3MgBr} \) CH₃CH(OMgBr)CH₃ \( \xrightarrow{H^+/H_2O} \) CH₃CH(OH)CH₃ (2-Propanol)
(v) CH₃COCH₃ (Propanone) \( \xrightarrow[\text{dry ether}]{CH_3MgBr} \) (CH₃)₃C–OMgBr \( \xrightarrow{H^+/H_2O} \) (CH₃)₃C–OH (2-Methyl-2-propanol)
(vi) HCHO (Methanal) \( \xrightarrow[\text{dry ether}]{CH_3MgBr} \) CH₃CH₂–OMgBr \( \xrightarrow{H^+/H_2O} \) CH₃CH₂OH (Ethanol)
(vii) CH₃COOH (Ethanoic acid) \( \xrightarrow[(ii)\ H_2O]{(i)\ LiAlH_4,\ \text{ether}} \) CH₃CH₂OH (Ethanol)

Teacher's Note:
1. NaBH₄ reduces aldehydes and ketones but not carboxylic acids; for (vii) only LiAlH₄ (or conversion to an ester followed by catalytic hydrogenation) works.

Question 28. How are the following conversions carried out ?
(i) Benzyl chloride to benzyl alcohol. (CBSE 2010)
(ii) Ethylmagnesium chloride to propan-1-ol. (CBSE 2010)
(iii) Methylmagnesium bromide to 2-methylpropan-2-ol (CBSE 2010, 2011)
(iv) Ethene to ethanol (Manipur Board 2012)
Answer:
(i) By hydrolysis with aqueous NaOH (or KOH) : C₆H₅CH₂Cl (Benzyl chloride) + NaOH(aq) \( \xrightarrow{\Delta} \) C₆H₅CH₂OH (Benzyl alcohol) + NaCl
(ii) By reaction with formaldehyde followed by hydrolysis : HCHO + CH₃CH₂MgCl \( \xrightarrow{\text{Dry ether}} \) CH₃CH₂CH₂–OMgCl \( \xrightarrow{H^+/H_2O} \) CH₃CH₂CH₂OH (Propan-1-ol) + Mg(OH)Cl
(iii) By reaction with acetone followed by hydrolysis : CH₃COCH₃ + CH₃MgBr \( \xrightarrow{\text{Dry ether}} \) (CH₃)₃C–OMgBr \( \xrightarrow{H^+/H_2O} \) (CH₃)₃C–OH (2-Methylpropan-2-ol) + Mg(OH)Br
(iv) By acid-catalysed hydration : CH₂=CH₂ + H₂O \( \xrightarrow{H^+} \) CH₃CH₂OH, or by passing ethene into conc. H₂SO₄ and boiling the ethyl hydrogen sulphate with water : CH₂=CH₂ + H₂SO₄ → CH₃CH₂OSO₃H \( \xrightarrow{H_2O,\ \Delta} \) CH₃CH₂OH + H₂SO₄

Teacher's Note:
1. Write the name of every intermediate under its formula; examiners award marks for the adduct step in Grignard conversions.

Question 29. Explain giving reasons :
(i) Alcohols are more soluble in water than hydrocarbons of comparable molecular masses. (CBSE 2012)
(ii) Boiling point of propanol is greater than that of butane. (Raj. Board 2012)
(iii) Alcohols are freely soluble in water but alkyl halides are not.
(iv) The solubility of alcohols in water decreases with increase in molecular mass.
Answer:
(i) Alcohols form H-bonds with water molecules (R–O–H····O(H)–H), and the energy released compensates for breaking the H-bonds of water. Hydrocarbons cannot form H-bonds with water and hence are insoluble.
(ii) Propanol (M = 60) molecules are associated by intermolecular H-bonding, which needs extra energy to break, while butane (M = 58) molecules are held only by weak van der Waals forces. Hence, b.p. of propanol (370 K) is much higher than that of butane (273 K).
(iii) Alcohols can form H-bonds with water but alkyl halides, though polar, cannot form H-bonds with water. The energy needed to break the H-bonds of water is not compensated by the weak dipole-dipole attractions between alkyl halides and water. Hence alkyl halides are insoluble in water.
(iv) With increase in molecular mass, the size of the alkyl group (hydrophobic part) increases. It hinders the formation of H-bonds between the –OH group and water molecules, so solubility decreases.

Teacher's Note:
1. One idea, hydrogen bonding, explains all four parts; state it with a small labelled H-bond sketch in the exam.

Question 30. What happens when primary, secondary and tertiary alcohols are :
(i) Oxidised using alkaline potassium permanganate
(ii) Passed over heated copper at 573 K.
Answer:
(i) Primary alcohols are oxidised first to aldehydes and then to carboxylic acids (containing the same number of C atoms) : RCH₂OH \( \xrightarrow{[O]} \) RCHO \( \xrightarrow{[O]} \) RCOOH
Secondary alcohols are oxidised to ketones (same number of C atoms) : R₂CHOH \( \xrightarrow{[O]} \) R₂C=O
Tertiary alcohols are resistant to oxidation under mild conditions; under drastic conditions they give a mixture of ketones and acids, each containing fewer carbon atoms than the alcohol.
(ii) Primary alcohols undergo dehydrogenation to give aldehydes : RCH₂OH \( \xrightarrow{Cu,\ 573\text{ K}} \) RCHO + H₂
Secondary alcohols undergo dehydrogenation to give ketones : R₂CHOH \( \xrightarrow{Cu,\ 573\text{ K}} \) R₂CO + H₂
Tertiary alcohols undergo dehydration to give alkenes : (CH₃)₃COH \( \xrightarrow{Cu,\ 573\text{ K}} \) (CH₃)₂C=CH₂ + H₂O

Teacher's Note:
1. The difference comes from the H atoms on the carbinol carbon: 3° alcohols have none, so they cannot lose H₂ from there.

Question 31. What happens when : (i) ethanol is oxidised with acidified KMnO₄ solution
(ii) ethanol is treated with PCl₅. (Bihar Board 2011)
(iii) Vapours of ethyl alcohols are passed over red hot alumina at 623 K. (UP Board 2012)
(iv) Ethyl methyl ketone is treated with sodium amalgam and water. (Maharashtra Board 2012)
(v) (CH₃)₃C–OH is treated with Cu at 573 K. (CBSE (Foreign) 2017)
Answer:
(i) Ethanol is oxidised first to ethanal and then to ethanoic acid : CH₃CH₂OH \( \xrightarrow{KMnO_4/H^+\ [O]} \) CH₃CHO \( \xrightarrow{[O]} \) CH₃COOH (Ethanoic acid)
(ii) Ethyl chloride is formed : C₂H₅OH + PCl₅ → C₂H₅Cl + POCl₃ + HCl
(iii) Dehydration takes place to form ethene : CH₃CH₂OH \( \xrightarrow{Al_2O_3,\ 623\text{ K}} \) CH₂=CH₂ + H₂O
(iv) Reduction occurs to give butan-2-ol : CH₃COCH₂CH₃ + 2[H] \( \xrightarrow{Na(Hg)/H_2O} \) CH₃CH(OH)CH₂CH₃ (Butan-2-ol)
(v) Being a 3° alcohol, it undergoes dehydration to give 2-methylpropene : (CH₃)₃C–OH \( \xrightarrow{Cu,\ 573\text{ K}} \) (CH₃)₂C=CH₂ + H₂O

Teacher's Note:
1. Both Al₂O₃ at 623 K and conc. H₂SO₄ at 443 K dehydrate ethanol to ethene; know both conditions.

Question 32. What happens when : (i) Ethanol is mixed with conc. H₂SO₄ at 273 K
(ii) Excess of ethyl alcohol is heated with conc. H₂SO₄ at 413 K.
(iii) Ethanol is heated with excess of conc. H₂SO₄ at 443 K. (Hr. Board 2012)
Answer:
(i) Due to the presence of two lone pairs of electrons on the oxygen atom, alcohol acts as a Lewis base. Therefore, at 273 K, ethanol reacts with H₂SO₄ to form oxonium salts (protonated alcohols).
CH₃CH₂–O–H + H₂SO₄ \( \overset{273\text{ K}}{\rightleftharpoons} \) [CH₃CH₂–O⁺H₂] HSO₄⁻ (Oxonium salt)
(ii) Diethyl ether is formed : 2 CH₃CH₂OH \( \xrightarrow{\text{Conc. }H_2SO_4,\ 413\text{ K}} \) CH₃CH₂–O–CH₂CH₃ + H₂O
(iii) Ethene is formed by dehydration : CH₃CH₂OH \( \xrightarrow{\text{Conc. }H_2SO_4,\ 443\text{ K}} \) CH₂=CH₂ + H₂O

Teacher's Note:
1. Excess of ethanol in (ii) is essential; it provides the second molecule that acts as the nucleophile.

Question 33. Explain the mechanism of the following reaction :
CH₃CH₂OH \( \xrightarrow[443\text{K}]{H^+} \) CH₂=CH₂ + H₂O (CBSE 2013, 2018 S, 2019)
or Discuss the mechanism of dehydration of ethanol to form ethene. (Hr. Board 2011 ; Pb. Board 2012 ; CBSE Sample Paper 2018-19)
Answer:
The mechanism involves the following three steps :
Step 1. Formation of protonated alcohol (fast) : CH₃–CH₂–O–H + H⁺ ⇄ CH₃–CH₂–O⁺H₂
Step 2. Formation of carbocation by loss of water (slow, rate determining step) : CH₃–CH₂–O⁺H₂ ⇄ CH₃–CH₂⁺ + H₂O
Step 3. Formation of ethene by elimination of a proton (fast) : H–CH₂–CH₂⁺ ⇄ CH₂=CH₂ + H⁺
The acid used in Step 1 is released in Step 3. To drive the equilibrium to the right, ethene is removed as it is formed.

Teacher's Note:
1. Draw the curved arrows: lone pair of O to H⁺ in Step 1, C–O bond pair to O in Step 2, and C–H bond pair to the C–C bond in Step 3.

Question 34. Convert the following :
(i) Ethanol into ethyl ethanoate.
(ii) Ethanol into ethoxyethane.
(iii) Methanol to ethanol. (CBSE 2007 ; HP Board 2011)
(iv) Propan-2-ol to propanone (CBSE 2013)
(v) Propan-2-ol to 2-methylpropan-2-ol (CBSE 2015)
Answer:
(i) CH₃CH₂OH \( \xrightarrow{KMnO_4/H^+} \) CH₃COOH ; then CH₃COOH + CH₃CH₂OH \( \xrightarrow{\text{Conc. }H_2SO_4} \) CH₃COOC₂H₅ (Ethyl ethanoate) + H₂O
(ii) 2 CH₃CH₂OH \( \xrightarrow{\text{Conc. }H_2SO_4,\ 413\text{ K}} \) CH₃CH₂OCH₂CH₃ (Ethoxyethane) + H₂O
(iii) CH₃OH \( \xrightarrow{PCl_5} \) CH₃Cl \( \xrightarrow{Mg/\text{dry ether}} \) CH₃MgCl \( \xrightarrow[(ii)\ H^+/H_2O]{(i)\ HCHO} \) CH₃CH₂OH (Ethanol)
(iv) CH₃CH(OH)CH₃ \( \xrightarrow{Cu,\ 573\text{ K}\ \text{or}\ K_2Cr_2O_7/H^+} \) CH₃COCH₃ (Propanone)
(v) CH₃CH(OH)CH₃ \( \xrightarrow{Cu,\ 573\text{ K}} \) CH₃COCH₃ \( \xrightarrow[(ii)\ H^+/H_2O]{(i)\ CH_3MgBr} \) (CH₃)₃COH (2-Methylpropan-2-ol)

Teacher's Note:
1. To step up a carbon chain by one carbon and keep a 1° alcohol, use a Grignard reagent with HCHO; the HCHO itself can be made by oxidising methanol.

Question 35. How does Lucas reagent help in the distinction of primary, secondary and tertiary alcohols ? Discuss the reactions involved. (HP Board 2011 ; UP Board 2012 ; MP Board 2013)
Answer:
Lucas reagent is a mixture of conc. HCl and anhydrous ZnCl₂. Alcohols are soluble in it but the alkyl chlorides formed are insoluble, so their formation is shown by turbidity in the solution.
(i) Tertiary alcohols produce turbidity immediately at room temperature : (CH₃)₃COH + HCl \( \xrightarrow{ZnCl_2} \) (CH₃)₃CCl + H₂O
(ii) Secondary alcohols produce turbidity within 5 minutes : (CH₃)₂CHOH + HCl \( \xrightarrow{ZnCl_2} \) (CH₃)₂CHCl + H₂O
(iii) Primary alcohols do not produce turbidity at room temperature; turbidity appears only on heating : CH₃CH₂CH₂OH + HCl \( \xrightarrow{ZnCl_2,\ \Delta} \) CH₃CH₂CH₂Cl + H₂O
The reactivity order 3° > 2° > 1° is due to the stability of carbocation intermediates.

Teacher's Note:
1. Turbidity appears because the alkyl chloride is insoluble in the aqueous reagent; that is what you actually observe.

Question 36. How will you distinguish between primary, secondary and tertiary alcohols by Victor Meyer’s test ? Give chemical reactions also.
Answer:
The alcohol is converted into alkyl iodide (red P + I₂), which is treated with silver nitrite to form nitroalkane. The nitroalkane is treated with nitrous acid (NaNO₂ + HCl) and the solution is made alkaline with NaOH.
1° alcohol : RCH₂OH \( \xrightarrow{P/I_2} \) RCH₂I \( \xrightarrow{AgNO_2} \) RCH₂NO₂ \( \xrightarrow{HONO} \) RC(=NOH)NO₂ (Nitrolic acid) \( \xrightarrow{NaOH} \) Blood red colour
2° alcohol : R₂CHOH \( \xrightarrow{P/I_2} \) R₂CHI \( \xrightarrow{AgNO_2} \) R₂CHNO₂ \( \xrightarrow{HONO} \) R₂C(NO)NO₂ (Pseudonitrole) \( \xrightarrow{NaOH} \) Blue colour
3° alcohol : R₃COH \( \xrightarrow{P/I_2} \) R₃CI \( \xrightarrow{AgNO_2} \) R₃CNO₂ \( \xrightarrow{HONO} \) No reaction (no α-H) \( \xrightarrow{NaOH} \) Colourless

Teacher's Note:
1. Remember the colours as “RBC”: Red (1°), Blue (2°), Colourless (3°).

Question 37. Complete the following equations and name the products formed :
(i) C₂H₅OH \( \xrightarrow{\text{Conc. }H_2SO_4/443\text{ K}} \)
(ii) CH₃CH₂OH \( \xrightarrow{Cu/573\text{ K}} \)
(iii) CH₃CH₂OH + PCl₅ → (HP Board 2013)
(iv) (CH₃)₂CHOH \( \xrightarrow{Cu/573\text{ K}} \) (Hr. Board 2012)
(v) CH₃CH₂CH₂OH + SOCl₂ → ? (Assam Board 2013)
(vi) CH₃CH₂CH(OH)CH₃ \( \xrightarrow{Cu/573\text{ K}} \) (CBSE 2016)
(vii) (CH₃)₂C=O \( \xrightarrow{LiAlH_4} \)
(viii) Cyclohexanol + SOCl₂ → (Assam Board 2012 ; CBSE 2014)
(ix) CH₃–CH=CH–CH₂OH \( \xrightarrow{PCC} \) (CBSE 2017)
Answer:
(i) C₂H₅OH \( \xrightarrow{\text{Conc. }H_2SO_4/443\text{ K}} \) CH₂=CH₂ (Ethene) + H₂O
(ii) CH₃CH₂OH \( \xrightarrow{Cu/573\text{ K}} \) CH₃CHO (Ethanal) + H₂
(iii) CH₃CH₂OH + PCl₅ → CH₃CH₂Cl (Chloroethane) + POCl₃ + HCl
(iv) (CH₃)₂CHOH \( \xrightarrow{Cu/573\text{ K}} \) (CH₃)₂C=O (Propanone) + H₂
(v) CH₃CH₂CH₂OH + SOCl₂ → CH₃CH₂CH₂Cl (1-Chloropropane) + SO₂ + HCl
(vi) CH₃CH₂CH(OH)CH₃ \( \xrightarrow{Cu/573\text{ K}} \) CH₃CH₂COCH₃ (Butan-2-one) + H₂
(vii) (CH₃)₂C=O \( \xrightarrow{LiAlH_4} \) (CH₃)₂CHOH (Propan-2-ol)
(viii) C₆H₁₁–OH (Cyclohexanol) + SOCl₂ → C₆H₁₁–Cl (Chlorocyclohexane) + SO₂ + HCl
(ix) CH₃–CH=CH–CH₂OH \( \xrightarrow{PCC} \) CH₃–CH=CH–CHO (But-2-enal)

Teacher's Note:
1. PCC oxidises only the –CH₂OH group; it does not touch the C=C double bond, as (ix) shows.

Question 38. How will you distinguish between the following pairs by chemical reactions ?
(i) CH₃OH and C₂H₅OH
(ii) 1-Propanol (n-propyl alcohol) and 2-propanol (isopropyl alcohol)
(iii) 1-Butanol and 2-butanol
(iv) sec-Butanol and tert-butanol
(v) 1-Propanol and 2-methyl-2-propanol
(vi) Ethanol and 1-propanol
(vii) Ethyl alcohol and 2-propanol
Answer:
(i) Iodoform test : Ethanol on warming with I₂ and NaOH gives a yellow ppt. of iodoform while methanol does not. C₂H₅OH + 4I₂ + 6NaOH → CHI₃↓ + HCOONa + 5NaI + 5H₂O
(ii) Iodoform test : 2-Propanol (CH₃CH(OH)CH₃) gives a yellow ppt. of iodoform while 1-propanol does not. (Lucas test also distinguishes : 2-propanol gives turbidity within 5 minutes, 1-propanol gives no turbidity at room temperature.)
(iii) Iodoform test : 2-Butanol (CH₃CH(OH)CH₂CH₃) gives a yellow ppt. of iodoform while 1-butanol does not.
(iv) Lucas test : tert-Butanol gives turbidity immediately while sec-butanol gives turbidity within 5 minutes. (Also, sec-butanol gives the iodoform test while tert-butanol does not.)
(v) Lucas test : 2-Methyl-2-propanol (3°) gives turbidity immediately while 1-propanol (1°) gives no turbidity at room temperature.
(vi) Iodoform test : Ethanol gives a yellow ppt. of iodoform while 1-propanol does not.
(vii) Lucas test : 2-Propanol (2°) gives turbidity within 5 minutes while ethyl alcohol (1°) does not give turbidity at room temperature. (Both give the iodoform test, so it cannot be used here.)

Teacher's Note:
1. Before choosing a test, check whether both compounds give it; in (vii) both have the CH₃CH(OH)– unit, so only the Lucas test works.

Long Answer Questions (5 Marks)

Question 39. Phenol, C₆H₅OH, when it first reacts with concentrated sulphuric acid, forms Y. The compound Y is reacted with concentrated nitric acid to form Z. Identify Y and Z and explain why phenol is not converted commercially to Z by reacting it with conc. HNO₃.
Answer:
Y: phenol-2,4-disulphonic acid  
Z: 2,4,6-trinitrophenol (picric acid)  
Phenol is not reacted directly with conc. HNO₃ because the yield of picric acid is very poor; conc. HNO₃ oxidises much of the phenol (tarry products).

Teacher's Note:
1. Sulphonation first protects the ring from oxidation; the –SO₃H groups are then replaced by –NO₂ groups.

Question 40. What are primary, secondary and tertiary alcohols ? Starting with a Grignard reagent, how will you obtain each of these ? What happens when these three classes of alcohols are
(i) passed over heated copper
(ii) oxidised with K₂Cr₂O₇ and H₂SO₄ ?
Answer:
Alcohols in which the –OH group is attached to a primary, secondary or tertiary carbon atom are called primary (1°, –CH₂OH), secondary (2°, >CHOH) and tertiary (3°, R₃C–OH) alcohols respectively, e.g., CH₃CH₂OH (1°), (CH₃)₂CHOH (2°), (CH₃)₃COH (3°).
Preparation from Grignard reagent : Grignard reagents add to carbonyl compounds and the adducts on hydrolysis give alcohols.
(a) 1° alcohol (with formaldehyde) : HCHO + RMgX → RCH₂OMgX \( \xrightarrow{H^+/H_2O} \) RCH₂OH + Mg(OH)X ; e.g., CH₃MgBr gives ethanol.
(b) 2° alcohol (with any other aldehyde) : R′CHO + RMgX → R′CH(R)OMgX \( \xrightarrow{H^+/H_2O} \) R′CH(OH)R ; e.g., CH₃CHO + CH₃MgBr gives propan-2-ol.
(c) 3° alcohol (with a ketone) : R′2CO + RMgX → R′2C(R)OMgX \( \xrightarrow{H^+/H_2O} \) R′2C(OH)R ; e.g., CH₃COCH₃ + CH₃MgBr gives 2-methylpropan-2-ol.
(i) Heated copper (573 K) : 1° alcohols are dehydrogenated to aldehydes, RCH₂OH \( \xrightarrow{Cu,\ 573\text{ K}} \) RCHO + H₂ ; 2° alcohols are dehydrogenated to ketones, R₂CHOH \( \xrightarrow{Cu,\ 573\text{ K}} \) R₂CO + H₂ ; 3° alcohols are dehydrated to alkenes, (CH₃)₃COH \( \xrightarrow{Cu,\ 573\text{ K}} \) (CH₃)₂C=CH₂ + H₂O.
(ii) K₂Cr₂O₇/H₂SO₄ : 1° alcohols are oxidised first to aldehydes and then to carboxylic acids with the same number of carbon atoms, RCH₂OH \( \xrightarrow{[O]} \) RCHO \( \xrightarrow{[O]} \) RCOOH ; 2° alcohols are oxidised to ketones with the same number of carbon atoms, R₂CHOH \( \xrightarrow{[O]} \) R₂CO ; 3° alcohols are not oxidised under these conditions (under drastic conditions they give a mixture of ketones and acids, each with fewer carbon atoms). In each positive case the orange colour of dichromate turns green (Cr³⁺).

Teacher's Note:
1. The carbonyl compound chosen decides the class of alcohol: HCHO → 1°, RCHO → 2°, R₂CO → 3°.