Alcohols, Phenols and Ethers

What to remember

• Hydrogen bonding controls the physical properties. Alcohols and phenols boil much higher than ethers, haloalkanes and hydrocarbons of similar mass. Boiling point rises with chain length and falls with branching.
• Alcohols react in two ways: O–H cleavage (acidity, esterification) and C–O cleavage (with HX, PX₃, dehydration). Reactivity in C–O cleavage with HX: 3° > 2° > 1°. Oxidation: 1° gives aldehyde then acid, 2° gives ketone, 3° resists.
• Phenol is more acidic than alcohols because the phenoxide ion is stabilised by resonance. Electron-withdrawing groups (–NO₂) at ortho and para positions increase acidity; electron-releasing groups (–CH₃) decrease it. The –OH group is a strong ortho-para director and activates the ring.
• Know the named reactions: Kolbe (salicylic acid), Reimer-Tiemann (salicylaldehyde), Williamson (ethers), cumene process (phenol and acetone), Lucas test, and the ether cleavage rules with HI.

Classification and nomenclature

Alcohols have –OH on an sp³ carbon; phenols have –OH on an sp² carbon of a benzene ring; ethers have the group R–O–R′.

• By number of –OH groups: monohydric, dihydric (diols), trihydric and polyhydric.
• By the carbon bearing –OH: primary (1°), secondary (2°), tertiary (3°). Allylic alcohols have –OH on an sp³ carbon next to C=C; benzylic alcohols have it next to a benzene ring.
• Ethers are symmetrical (R = R′) or unsymmetrical (R ≠ R′).

In ether names, the smaller alkyl part with oxygen is the alkoxy group; the longer chain is the parent. Example: CH₃OCH₂CH₃ is methoxyethane.

Structures of functional groups

In an alcohol the oxygen is sp³ hybridised. In methanol the C–O–H angle is about 109° (108.9°), slightly less than the tetrahedral angle because of lone pair repulsion. The C–O bond is about 142 pm. In phenol the C–O bond is shorter, about 136 pm, because of partial double bond character from resonance and because the carbon is sp². In an ether the C–O–C angle is larger than the tetrahedral angle (about 112° in methoxymethane) because of repulsion between the two alkyl groups; the C–O bond is about 141 pm.

Preparation of alcohols

Preparation of phenols

• 1. From haloarenes: chlorobenzene + NaOH at 623 K and 300 atm → sodium phenoxide; acidification with HCl gives phenol.
• 2. From benzenesulphonic acid: fuse with NaOH to give sodium phenoxide; acidify to get phenol.
• 3. From diazonium salts: benzenediazonium chloride warmed with water → phenol + N₂ + HCl.
• 4. From cumene (isopropylbenzene): cumene + O₂ → cumene hydroperoxide; treatment with dilute acid gives phenol and acetone. This is the main industrial route, and it gives a valuable by-product.

Physical properties

• Boiling point. Alcohols and phenols form intermolecular hydrogen bonds, so they boil much higher than hydrocarbons, haloalkanes and ethers of comparable mass. Ethers cannot form hydrogen bonds with each other.
• Boiling point rises with the number of carbons (more van der Waals forces) and falls with branching (smaller surface area). Among isomeric alcohols: 1° > 2° > 3°.
• Solubility. Lower alcohols and phenols dissolve in water by hydrogen bonding. Solubility falls as the hydrocarbon part grows. Ethers can accept hydrogen bonds from water through the oxygen lone pairs, so ethoxyethane and butan-1-ol have similar solubility, about 7.5 g per 100 mL.

Acidity of alcohols and phenols

Alcohols react with sodium metal: 2 R–OH + 2 Na → 2 R–ONa + H₂. They are weaker acids than water (pKa about 16) and do not react with NaOH. Acidic strength: water > primary > secondary > tertiary alcohol, because alkyl groups release electrons (+I effect) and make the alkoxide ion less stable.

Phenol is much more acidic (pKa about 10). It reacts with NaOH to give sodium phenoxide, but not with NaHCO₃ (it is a weaker acid than carbonic acid). The phenoxide ion is stabilised by delocalisation of the negative charge into the ring, while an alkoxide ion has no such stabilisation.

Effect of substituents. Electron-withdrawing groups (–NO₂) at the ortho and para positions stabilise the phenoxide ion and increase acidity. Electron-releasing groups (–CH₃) destabilise it and decrease acidity.

Order of acidity: 2,4,6-trinitrophenol (picric acid) > p-nitrophenol > m-nitrophenol > phenol > p-cresol > ethanol.

Esterification

R–OH + R′COOH ⇌ R–O–COR′ + H₂O (conc. H₂SO₄ catalyst). Alcohols and phenols also react with acid chlorides or anhydrides (in the presence of a base such as pyridine). Example: salicylic acid + ethanoic anhydride → 2-acetoxybenzoic acid (acetylsalicylic acid, aspirin), a pain-reliever.

Reactions involving cleavage of the C–O bond in alcohols

With HX. R–OH + HX → R–X + H₂O. Reactivity of HX: HI > HBr > HCl. Reactivity of alcohols: 3° > 2° > 1°.

Lucas test. The reagent is concentrated HCl with anhydrous ZnCl₂, and it distinguishes 1°, 2° and 3° alcohols (lower alcohols that dissolve in it).

The turbidity is due to the insoluble alkyl chloride.

With phosphorus halides. 3 R–OH + PBr₃ → 3 R–Br + H₃PO₃.

Dehydration. Heating with concentrated H₂SO₄ or H₃PO₄ removes water.

• Ethanol at 443 K gives ethene. Ethanol at 413 K with excess alcohol gives ethoxyethane.
• Typical temperatures for forming alkenes: primary about 443 K, secondary about 440 K, tertiary about 358 K. The ease of dehydration is 3° > 2° > 1°.
• Mechanism: (1) protonation of the –OH group to R–OH₂⁺; (2) loss of water to form a carbocation (slow step); (3) loss of H⁺ from the adjacent carbon to give the alkene. The Saytzeff rule applies: butan-2-ol gives but-2-ene as the major product.

Oxidation.

Example: CH₃CH₂OH + PCC → CH₃CHO; with acidified K₂Cr₂O₇ it goes on to CH₃COOH.

Reactions of phenol

• Electrophilic substitution. The –OH group activates the ring and directs to ortho and para positions.
• Nitration with dilute HNO₃ at low temperature (about 298 K) gives a mixture of o-nitrophenol and p-nitrophenol. o-Nitrophenol has intramolecular hydrogen bonding and is steam volatile; p-nitrophenol has intermolecular hydrogen bonding and is not, so steam distillation separates them. Concentrated HNO₃ gives 2,4,6-trinitrophenol (picric acid).
• Halogenation with Br₂ in CS₂ or CHCl₃ at low temperature gives a mixture of o-bromophenol and p-bromophenol (para major). With bromine water the product is a white precipitate of 2,4,6-tribromophenol.
• Kolbe's reaction. Phenol + NaOH → sodium phenoxide; heat with CO₂ at about 400 K and 4–7 atm; acidify → salicylic acid (2-hydroxybenzoic acid).
• Reimer-Tiemann reaction. Phenol + CHCl₃ + aqueous NaOH at about 340 K, then acidification → salicylaldehyde (2-hydroxybenzaldehyde). The electrophile is dichlorocarbene, :CCl₂. With CCl₄ in place of CHCl₃ the product is salicylic acid.
• With zinc dust. Phenol + Zn → benzene + ZnO.
• Oxidation. Phenol + Na₂Cr₂O₇/H₂SO₄ → benzoquinone (a conjugated cyclic diketone).

Commercially important alcohols

• Methanol (wood spirit). Made by hydrogenation of carbon monoxide: CO + 2 H₂ → CH₃OH at 200–300 atm and 573–673 K over a ZnO–Cr₂O₃ catalyst. It is a good solvent and fuel but is poisonous: even small amounts cause blindness, and larger amounts cause death.
• Ethanol. Made by fermentation of sugars (molasses): sucrose → glucose + fructose (invertase), then glucose and fructose → ethanol + CO₂ (zymase). It is also made by acid-catalysed hydration of ethene. Commercial alcohol is made unfit for drinking (denatured) by adding copper sulphate (colour) and pyridine (bad smell) so that it is not misused.

Ethers: preparation

• 1. Dehydration of alcohols. 2 R–OH → R–O–R + H₂O with conc. H₂SO₄ at 413 K and excess of primary alcohol. It suits only symmetrical ethers from primary alcohols; secondary and tertiary alcohols give alkenes.
• 2. Williamson synthesis. R–X + R′–ONa → R–O–R′ + NaX. This is an SN2 reaction, so the halide should be primary. A tertiary halide gives an alkene through elimination. To make tert-butyl ethyl ether use sodium tert-butoxide + bromoethane, not sodium ethoxide + tert-butyl bromide. Phenols give aryl alkyl ethers: sodium phenoxide + CH₃I → anisole. An aryl halide cannot be used as the halide.

Ethers: physical properties and cleavage

Ethers are slightly polar, with a C–O–C bent shape. Their boiling points are close to those of alkanes of similar mass and much lower than alcohols.

Cleavage with HX. The order of reactivity is HI > HBr > HCl.

• R–O–R′ + HX → R–X + R′–OH; with excess HX both parts become halides.
• If both alkyl groups are primary or secondary (SN2), the halide ion attacks the smaller alkyl group. CH₃OCH₂CH₃ + HI → CH₃I + CH₃CH₂OH.
• If one group is tertiary (SN1), the tertiary halide forms. (CH₃)₃COCH₃ + HI → (CH₃)₃CI + CH₃OH.
• Anisole + HI → phenol + CH₃I. Phenol is not converted further, and iodobenzene is not formed, because the C(sp²)–O bond has partial double bond character and is difficult to break.

Electrophilic substitution in anisole. The –OCH₃ group activates the ring (+R effect) and directs to ortho and para positions.

NEET traps

• 1. Phenol is acidic enough to react with NaOH but not with NaHCO₃; alcohols react with neither.
• 2. Among alcohols, tertiary is the least acidic but the most reactive towards HX and in dehydration.
• 3. In the Lucas test, tertiary alcohols give immediate turbidity; primary alcohols give none at room temperature.
• 4. PCC stops at the aldehyde; acidified K₂Cr₂O₇ or KMnO₄ goes on to the acid.
• 5. Hydration of an alkene follows Markovnikov; hydroboration-oxidation gives the anti-Markovnikov alcohol.
• 6. Kolbe gives salicylic acid (CO₂); Reimer-Tiemann gives salicylaldehyde (CHCl₃).
• 7. In Williamson synthesis the alkyl halide must be primary; a tertiary halide gives an alkene.
• 8. Anisole with HI gives phenol and methyl iodide, not iodobenzene and methanol.
• 9. o-Nitrophenol is steam volatile because of intramolecular hydrogen bonding; p-nitrophenol is not.
• 10. Ethers boil lower than isomeric alcohols, but dissolve in water about as well as alcohols of the same mass.

One-liners

• 1. Alcohols boil higher than ethers of similar mass because of hydrogen bonding.
• 2. Boiling point of isomeric alcohols: 1° > 2° > 3°.
• 3. Acidity order: water > 1° alcohol > 2° alcohol > 3° alcohol.
• 4. Phenol is more acidic than ethanol because of resonance stabilisation of the phenoxide ion.
• 5. p-Nitrophenol is more acidic than phenol; p-cresol is less acidic.
• 6. Reduction of a ketone gives a secondary alcohol; reduction of an aldehyde gives a primary alcohol.
• 7. HCHO with a Grignard reagent gives a primary alcohol; a ketone gives a tertiary alcohol.
• 8. Cumene process gives phenol and acetone.
• 9. Bromine water converts phenol to 2,4,6-tribromophenol.
• 10. Concentrated HNO₃ converts phenol to picric acid.
• 11. Zinc dust converts phenol to benzene.
• 12. Methanol is made from CO and H₂ over ZnO–Cr₂O₃ at 573–673 K.
• 13. Copper at 573 K dehydrogenates 1° alcohols to aldehydes and 2° alcohols to ketones.
• 14. Williamson synthesis is an SN2 reaction of an alkoxide with a primary alkyl halide.
• 15. Ether cleavage with HI gives an alkyl iodide from the smaller group, unless one group is tertiary.

Practice questions

1. Hydroboration-oxidation of propene (B₂H₆, then H₂O₂/OH⁻) gives

   1. propan-1-ol
   2. propanal
   3. propane-1,2-diol
   4. propan-2-ol
   Answer

   A. propan-1-ol

   The addition is anti-Markovnikov, so the –OH goes to the terminal carbon: propan-1-ol.

2. Acid-catalysed hydration of but-1-ene gives mainly

   1. 2-methylpropan-1-ol
   2. 2-methylpropan-2-ol
   3. butan-2-ol
   4. butan-1-ol
   Answer

   C. butan-2-ol

   Markovnikov addition places –OH on the more substituted carbon: butan-2-ol.

3. Hydrogenation of carbon monoxide at 200–300 atm and 573–673 K over a ZnO–Cr₂O₃ catalyst gives

   1. methanoic acid
   2. ethanol
   3. methanal
   4. methanol
   Answer

   D. methanol

   CO + 2 H₂ → CH₃OH is the industrial route to methanol.

4. Methanal reacts with CH₃MgBr, followed by hydrolysis, to give

   1. propan-1-ol
   2. ethanol
   3. methanol
   4. propan-2-ol
   Answer

   B. ethanol

   HCHO gives a primary alcohol with one extra carbon: CH₃CH₂OH.

5. In the cumene process, cumene is oxidised by air and the hydroperoxide is treated with dilute acid. The products are

   1. benzaldehyde and propan-2-ol
   2. benzoic acid and methanol
   3. phenol and propanone
   4. phenol and ethanal
   Answer

   C. phenol and propanone

   Cumene hydroperoxide rearranges to phenol and acetone.

6. Which alcohol gives turbidity immediately with concentrated HCl and anhydrous ZnCl₂ at room temperature?

   1. 2-methylpropan-2-ol
   2. butan-1-ol
   3. butan-2-ol
   4. propan-2-ol
   Answer

   A. 2-methylpropan-2-ol

   A tertiary alcohol forms the carbocation fastest, so the chloride separates at once.

7. Which compound has the highest boiling point?

   1. methoxymethane
   2. ethane
   3. propane
   4. ethanol
   Answer

   D. ethanol

   Ethanol forms intermolecular hydrogen bonds; the others do not.

8. Propan-1-ol is oxidised with PCC in CH₂Cl₂. The product is

   1. propanoic acid
   2. propanal
   3. propanone
   4. propene
   Answer

   B. propanal

   PCC is a mild oxidant and stops at the aldehyde stage: propanal.

9. Phenol reacts with bromine water to give

   1. 4-bromophenol only
   2. 2,4,6-tribromophenol
   3. 2-bromophenol only
   4. 3,5-dibromophenol
   Answer

   B. 2,4,6-tribromophenol

   The strongly activating –OH group gives a white precipitate of the tribromo compound.

10. Heating sodium phenoxide with CO₂ at about 400 K and 4–7 atm, then acidifying, gives

    1. benzoic acid
    2. salicylaldehyde
    3. salicylic acid
    4. catechol
    Answer

    C. salicylic acid

    This is Kolbe's reaction; the product is 2-hydroxybenzoic acid.

11. Phenol heated with CHCl₃ and aqueous NaOH at 340 K, followed by acidification, gives

    1. salicylaldehyde
    2. salicylic acid
    3. benzaldehyde
    4. phenyl chloride
    Answer

    A. salicylaldehyde

    This is the Reimer-Tiemann reaction; the product is 2-hydroxybenzaldehyde.

12. Sodium ethoxide reacts with bromomethane to give

    1. methoxymethane
    2. ethoxyethane
    3. ethanol
    4. methoxyethane
    Answer

    D. methoxyethane

    Williamson synthesis: C₂H₅O⁻ replaces Br⁻ to give CH₃OCH₂CH₃, methoxyethane.

13. Anisole is heated with HI. The products are

    1. benzene and methyl alcohol
    2. iodobenzene and methanol
    3. phenol and iodomethane
    4. iodobenzene and iodomethane
    Answer

    C. phenol and iodomethane

    The halide attacks the CH₃ group; the C(sp²)–O bond stays. Products: C₆H₅OH + CH₃I.

14. Heating ethanol with excess concentrated H₂SO₄ at 443 K gives mainly

    1. ethane
    2. ethanal
    3. ethoxyethane
    4. ethene
    Answer

    D. ethene

    At 443 K dehydration to the alkene occurs: ethene.

15. Phenol is heated with zinc dust. The organic product is

    1. benzene
    2. cyclohexane
    3. benzoquinone
    4. toluene
    Answer

    A. benzene

    Zn removes the oxygen as ZnO: C₆H₅OH + Zn → C₆H₆ + ZnO.

16. Oxidation of phenol with Na₂Cr₂O₇ in H₂SO₄ gives

    1. catechol
    2. benzoquinone
    3. cyclohexanone
    4. benzoic acid
    Answer

    B. benzoquinone

    The product is the conjugated cyclic diketone, benzoquinone.

17. Given below are two statements. Statement I: Phenol is more acidic than ethanol. Statement II: The phenoxide ion is stabilised by resonance, whereas the ethoxide ion is not.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    A. Both Statement I and Statement II are correct

    Resonance delocalisation of charge makes the phenoxide ion stable. Both correct.

18. Given below are two statements. Statement I: Tertiary alcohols are more acidic than primary alcohols. Statement II: Alkyl groups release electrons and make the alkoxide ion less stable.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    D. Statement I is incorrect but Statement II is correct

    Alkyl groups destabilise the alkoxide, so tertiary alcohols are less acidic; I is wrong, II is correct.

19. Given below are two statements. Statement I: Ethers boil at higher temperatures than alcohols of comparable molar mass. Statement II: Alcohols form intermolecular hydrogen bonds.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    D. Statement I is incorrect but Statement II is correct

    Ethers cannot form hydrogen bonds with each other and boil lower; I is wrong, II is correct.

20. Given below are two statements. Statement I: Anisole undergoes Friedel-Crafts acylation to give mainly p-methoxyacetophenone. Statement II: Anisole undergoes electrophilic substitution more slowly than benzene.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    C. Statement I is correct but Statement II is incorrect

    –OCH₃ activates the ring so it reacts faster than benzene; I is correct, II is wrong.

21. Which of the following statements are correct? 1. o-Nitrophenol is steam volatile. 2. p-Nitrophenol has intermolecular hydrogen bonding. 3. o-Nitrophenol has strong intermolecular hydrogen bonding.

    1. 1 and 3 only
    2. 1 and 2 only
    3. 2 and 3 only
    4. 1, 2 and 3
    Answer

    B. 1 and 2 only

    o-Nitrophenol has intramolecular hydrogen bonding, so 3 is wrong; 1 and 2 are correct.

22. Given below are two statements. Statement I: A primary alkyl halide and an alkoxide give an ether in good yield by the Williamson method. Statement II: A tertiary alkyl halide with an alkoxide gives mainly an alkene.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    A. Both Statement I and Statement II are correct

    Primary halides undergo SN2; tertiary halides eliminate. Both correct.

23. Given below are two statements. Statement I: (CH₃)₃C–O–CH₃ reacts with HI to give (CH₃)₃C–I and CH₃OH. Statement II: Anisole reacts with HI to give iodobenzene and methanol.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    C. Statement I is correct but Statement II is incorrect

    Tertiary ether cleaves by SN1 to give the tertiary iodide; anisole gives phenol and CH₃I. I correct, II wrong.

24. Given below are two statements. Statement I: PCC oxidises primary alcohols to carboxylic acids. Statement II: Copper at 573 K converts tertiary alcohols into ketones.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    B. Both Statement I and Statement II are incorrect

    PCC stops at the aldehyde; tertiary alcohols give alkenes with hot copper. Both incorrect.

25. Given below are two statements. Statement I: Ethanol heated with conc. H₂SO₄ at 413 K with excess alcohol gives ethoxyethane. Statement II: Ethanol heated with conc. H₂SO₄ at 443 K gives ethene.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    A. Both Statement I and Statement II are correct

    Both are correct as per the temperature-dependent outcomes.

26. Which of the following statements are correct? 1. Phenol reacts with NaHCO₃ to release CO₂. 2. Phenol reacts with aqueous NaOH to form sodium phenoxide. 3. Ethanol does not react with aqueous NaOH.

    1. 1 and 2 only
    2. 1 and 3 only
    3. 1, 2 and 3
    4. 2 and 3 only
    Answer

    D. 2 and 3 only

    Phenol is a weaker acid than carbonic acid, so 1 is wrong; 2 and 3 are correct.

27. Assertion (A): Alcohols have higher boiling points than ethers of comparable molar mass. Reason (R): Alcohols form intermolecular hydrogen bonds.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    A. Both A and R are true and R is the correct explanation of A

    Hydrogen bonding raises the boiling point of alcohols; R explains A.

28. Assertion (A): Phenol reacts with bromine water to give 2,4,6-tribromophenol. Reason (R): Phenol is acidic.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    B. Both A and R are true but R is not the correct explanation of A

    Both true; the reaction occurs because –OH activates the ring, not because phenol is acidic.

29. Assertion (A): Tertiary alcohols react fastest with Lucas reagent. Reason (R): The tertiary carbocation is the least stable carbocation.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    C. A is true but R is false

    A is true; a tertiary carbocation is the most stable, so R is false.

30. Assertion (A): Anisole reacts with HI to give iodobenzene and methanol. Reason (R): The C(sp²)–O bond in anisole has partial double bond character.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    D. A is false but R is true

    Anisole gives phenol and CH₃I, so A is false; R is true and explains why the aryl C–O bond is not cleaved.

31. Assertion (A): Alcohols are weaker acids than water. Reason (R): The +I effect of the alkyl group increases electron density on oxygen and makes the release of H⁺ difficult.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    A. Both A and R are true and R is the correct explanation of A

    Both true and R explains A.

32. Assertion (A): Ethanol is miscible with water in all proportions. Reason (R): Ethanol contains a hydrocarbon part.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    B. Both A and R are true but R is not the correct explanation of A

    A is true because of hydrogen bonding; the hydrocarbon part does not explain miscibility, so R is true but not an explanation.

33. Match the reaction in List I with the product in List II. List I: a. Kolbe's reaction b. Reimer-Tiemann reaction c. Cumene process d. Phenol with zinc dust List II: i. salicylaldehyde ii. phenol and acetone iii. salicylic acid iv. benzene

    1. a-i, b-iii, c-ii, d-iv
    2. a-iii, b-ii, c-i, d-iv
    3. a-iii, b-i, c-ii, d-iv
    4. a-iii, b-i, c-iv, d-ii
    Answer

    C. a-iii, b-i, c-ii, d-iv

    Kolbe-salicylic acid (iii); Reimer-Tiemann-salicylaldehyde (i); cumene-phenol and acetone (ii); Zn-benzene (iv).

34. Match each alcohol heated with Cu at 573 K with the product. List I: a. propan-1-ol b. propan-2-ol c. 2-methylpropan-2-ol d. methanol List II: i. propanone ii. propanal iii. 2-methylpropene iv. methanal

    1. a-iv, b-i, c-iii, d-ii
    2. a-ii, b-i, c-iv, d-iii
    3. a-ii, b-i, c-iii, d-iv
    4. a-i, b-ii, c-iii, d-iv
    Answer

    C. a-ii, b-i, c-iii, d-iv

    Primary alcohols give aldehydes, secondary give ketones, tertiary give alkenes.

35. Match the reagent in List I with its function in List II. List I: a. Lucas reagent b. PCC c. SOCl₂ d. NaBH₄ List II: i. reduces aldehydes to primary alcohols ii. stops oxidation of primary alcohols at the aldehyde iii. distinguishes 1°, 2° and 3° alcohols iv. converts an alcohol to an alkyl chloride

    1. a-iii, b-iv, c-ii, d-i
    2. a-ii, b-iii, c-iv, d-i
    3. a-iii, b-i, c-iv, d-ii
    4. a-iii, b-ii, c-iv, d-i
    Answer

    D. a-iii, b-ii, c-iv, d-i

    Lucas-distinguishes classes (iii); PCC-aldehyde (ii); SOCl₂-alkyl chloride (iv); NaBH₄-reduction (i).

36. Match the ether in List I (heated with HI) with the products in List II. List I: a. CH₃OCH₂CH₃ (1 equivalent HI) b. (CH₃)₃COCH₃ c. C₆H₅OCH₃ d. CH₃OCH₃ (excess HI) List II: i. CH₃I and C₆H₅OH ii. two molecules of CH₃I iii. CH₃I and CH₃CH₂OH iv. (CH₃)₃CI and CH₃OH

    1. a-i, b-iv, c-iii, d-ii
    2. a-iii, b-iv, c-i, d-ii
    3. a-iv, b-iii, c-i, d-ii
    4. a-iii, b-iv, c-ii, d-i
    Answer

    B. a-iii, b-iv, c-i, d-ii

    Primary groups: I⁻ attacks the smaller CH₃ (iii); tertiary group: SN1 (iv); anisole gives phenol (i); excess HI gives 2 CH₃I (ii).

37. Match the phenol reaction in List I with the product in List II. List I: a. dilute HNO₃, 298 K b. concentrated HNO₃ c. Br₂ in CS₂ at low temperature d. bromine water List II: i. 2,4,6-trinitrophenol ii. 2,4,6-tribromophenol iii. o- and p-nitrophenol iv. o- and p-bromophenol

    1. a-iii, b-i, c-iv, d-ii
    2. a-iii, b-i, c-ii, d-iv
    3. a-iii, b-iv, c-i, d-ii
    4. a-i, b-iii, c-iv, d-ii
    Answer

    A. a-iii, b-i, c-iv, d-ii

    Dilute HNO₃-mono-nitro (iii); conc. HNO₃-picric acid (i); Br₂/CS₂-mono-bromo (iv); bromine water-tribromo (ii).

38. Match the method in List I with the alcohol obtained in List II. List I: a. hydration of propene with H⁺ b. hydroboration-oxidation of propene c. CH₃MgBr with HCHO, then H₃O⁺ d. CH₃MgBr with propanone, then H₃O⁺ List II: i. propan-1-ol ii. propan-2-ol iii. ethanol iv. 2-methylpropan-2-ol

    1. a-i, b-ii, c-iii, d-iv
    2. a-ii, b-i, c-iv, d-iii
    3. a-ii, b-i, c-iii, d-iv
    4. a-iii, b-i, c-ii, d-iv
    Answer

    C. a-ii, b-i, c-iii, d-iv

    Markovnikov hydration-propan-2-ol (ii); hydroboration-propan-1-ol (i); HCHO-ethanol (iii); ketone-tertiary alcohol (iv).

39. Which pair of reactants is best for preparing tert-butyl ethyl ether by Williamson synthesis?

    1. sodium ethoxide and tert-butyl bromide
    2. tert-butyl alcohol and ethanol with conc. H₂SO₄ at 413 K
    3. sodium ethoxide and chlorobenzene
    4. sodium tert-butoxide and bromoethane
    Answer

    D. sodium tert-butoxide and bromoethane

    The primary halide must be the alkyl halide; a tertiary halide would give 2-methylpropene.

40. A compound C₃H₈O gives no turbidity with Lucas reagent at room temperature and gives propanal with PCC. The compound is

    1. methoxyethane
    2. propan-1-ol
    3. propan-2-ol
    4. propan-1,2-diol
    Answer

    B. propan-1-ol

    A primary alcohol gives an aldehyde on mild oxidation: propan-1-ol.

41. A mixture of o-nitrophenol and p-nitrophenol is steam distilled. The compound present in the distillate is

    1. o-nitrophenol
    2. p-nitrophenol
    3. both equally
    4. neither
    Answer

    A. o-nitrophenol

    Intramolecular hydrogen bonding makes the ortho isomer volatile.

42. Butan-2-ol is heated with concentrated H₂SO₄. The major alkene formed is

    1. butane
    2. but-2-ene
    3. 2-methylpropene
    4. but-1-ene
    Answer

    B. but-2-ene

    Saytzeff rule: the more substituted alkene but-2-ene is major.

43. Sodium phenoxide is heated with CO₂ at 400 K and 4–7 atm, the product is acidified, and the acid is then treated with ethanoic anhydride. The final compound is

    1. methyl 2-hydroxybenzoate
    2. phenyl ethanoate
    3. 2-hydroxybenzaldehyde
    4. 2-acetoxybenzoic acid
    Answer

    D. 2-acetoxybenzoic acid

    Kolbe's reaction gives salicylic acid; acetylation of its –OH gives 2-acetoxybenzoic acid (aspirin).

44. Anisole is treated with CH₃Cl and anhydrous AlCl₃. The major products are

    1. 3-methylanisole
    2. phenol and CH₃Cl
    3. 2-methylanisole and 4-methylanisole
    4. toluene and methanol
    Answer

    C. 2-methylanisole and 4-methylanisole

    –OCH₃ is ortho-para directing, so Friedel-Crafts alkylation gives o- and p-methylanisole.

45. Which reagent gives a white precipitate with phenol but no precipitate with ethanol?

    1. bromine water
    2. dilute HCl
    3. dilute H₂SO₄
    4. aqueous NaCl
    Answer

    A. bromine water

    Bromine water forms insoluble 2,4,6-tribromophenol with phenol only.