Class 12 Chemistry | Chapter 7
Alcohols, Phenols and Ethers
Structure • Preparation • Acidity • Mechanisms • Key Reactions
1. Introduction and Classification
Alcohols (R-OH) and Phenols (Ar-OH) contain a hydroxyl (-OH) group. Ethers (R-O-R') contain an oxygen atom bonded to two alkyl/aryl groups.
1.1 Classification of Alcohols
- Allylic alcohols: -OH is attached to an sp3 hybridized carbon next to a carbon-carbon double bond. Example: CH2=CH–CH2OH
- Benzylic alcohols: -OH is attached to an sp3 hybridized carbon next to an aromatic ring. Example: C6H5–CH2OH
- Vinylic alcohols: -OH is attached directly to an sp2 carbon of a C=C double bond (these are usually unstable and tautomerize).
1.2 Physical Properties
- Boiling Points: Alcohols and phenols have much higher boiling points than alkanes, haloalkanes, and ethers of comparable molecular masses due to intermolecular hydrogen bonding. Boiling point increases with carbon chain length and decreases with branching.
- Solubility: Lower alcohols are highly soluble in water due to their ability to form hydrogen bonds with water molecules. Solubility decreases as the hydrophobic alkyl part gets larger.
2. Preparation of Alcohols
2.1 From Alkenes
- Acid Catalysed Hydration: Alkenes react with water in the presence of an acid
(H2SO4) to yield alcohols. Follows Markovnikov's rule.CH3–CH=CH2 + H2O(H+) → CH3–CH(OH)–CH3 (Propan-2-ol Major)
- Hydroboration-Oxidation: Alkenes react with diborane (B2H6)
followed by oxidation with H2O2/OH− to give alcohols. Follows
Anti-Markovnikov's rule (gives primary alcohol from terminal alkene).CH3–CH=CH2 → 1. B2H6 | 2. H2O2/OH− → CH3–CH2–CH2OH (Propan-1-ol)
2.2 From Carbonyl Compounds
- Reduction of Aldehydes and Ketones:
Aldehydes → (H2/Pd or NaBH4 or LiAlH4) → 1° Alcohols.
Ketones → (NaBH4 or LiAlH4) → 2° Alcohols. - Reduction of Carboxylic Acids and Esters: LiAlH4 reduces carboxylic acids directly to 1° alcohols (strong reducing agent needed).
- From Grignard Reagents (RMgX):
HCHO (Formaldehyde) + RMgX → 1° Alcohol.
Other Aldehydes + RMgX → 2° Alcohol.
Ketones + RMgX → 3° Alcohol.
3. Preparation of Phenols
Phenols can be prepared from benzene derivatives:
- From Haloarenes (Dow Process): Chlorobenzene is fused with NaOH at 623 K and 300 atm to form sodium phenoxide, which on acidification yields phenol.
- From Benzene Sulphonic Acid: Benzene + Oleum → Benzene sulphonic acid. Fused with NaOH at high temp, then acidified.
- From Diazonium Salts: Aniline + NaNO2/HCl (0-5°C) → Benzene diazonium chloride. Warming this salt with water yields phenol.
- From Cumene (Isopropylbenzene): (Most important commercial method). Cumene is oxidized to cumene hydroperoxide, which is then cleaved with dilute acid to yield Phenol and Acetone (a valuable byproduct).
4. Chemical Reactions of Alcohols and Phenols
4.1 Acidity of Alcohols and Phenols
- Alcohols: Weak acids (weaker than water). Acidity decreases in order: 1° > 2° > 3°. Electron-donating alkyl groups increase electron density on oxygen, destabilizing the alkoxide ion.
- Phenols: Much stronger acids than alcohols. The phenoxide ion formed after losing H+ is highly resonance stabilized due to delocalization of negative charge over the benzene ring.
- Effect of Substituents on Phenol Acidity:
Electron Withdrawing Groups (EWG like -NO2, -CN, -X) increase acidity by stabilizing the phenoxide ion (especially at ortho/para positions).
Electron Releasing Groups (ERG like -CH3, -OCH3) decrease acidity. Thus, p-nitrophenol is more acidic than phenol, while p-cresol is less acidic.
4.2 Reactions involving cleavage of C–O bond in Alcohols
- Reaction with HX (Lucas Test): Used to distinguish 1°, 2°, 3° alcohols
based on reactivity with conc. HCl + ZnCl2.
3° alcohol: Cloudiness appears immediately.
2° alcohol: Cloudiness appears in ~5 minutes.
1° alcohol: No cloudiness at room temp (only on heating). - Dehydration (forming alkenes): Requires conc. H2SO4 and heat.
Follows Saytzeff's rule.
Ease of dehydration: 3° > 2° > 1° (depends on stability of carbocation intermediate). - Oxidation:
1° Alcohols → (mild, PCC/CrO3) → Aldehydes.
1° Alcohols → (strong, KMnO4/K2Cr2O7) → Carboxylic acids.
2° Alcohols → (CrO3) → Ketones.
3° Alcohols do not undergo oxidation easily (require drastic conditions to break C-C bonds). - Catalytic Dehydrogenation (Cu at 573 K):
1° Alcohol → Aldehyde.
2° Alcohol → Ketone.
3° Alcohol → Alkene (undergoes dehydration instead of dehydrogenation).
4.3 Electrophilic Substitution in Phenols
The -OH group activates the benzene ring towards electrophilic substitution and directs incoming groups to ortho and para positions.
- Nitration: Dilute HNO3 gives a mixture of o- and p-nitrophenol. Conc. HNO3 gives 2,4,6-trinitrophenol (Picric acid).
- Halogenation: Br2 in water gives a white precipitate of 2,4,6-tribromophenol. Br2 in CS2 (low polarity solvent, 273 K) gives mono-bromophenols (mostly para).
💡 Important Name Reactions for Phenol:
1. Kolbe's Reaction: Phenol → 1. NaOH, 2. CO2, then H+ → Salicylic acid (2-Hydroxybenzoic acid).
2. Reimer-Tiemann Reaction: Phenol + CHCl3 + aq. NaOH → Salicylaldehyde (2-Hydroxybenzaldehyde).
1. Kolbe's Reaction: Phenol → 1. NaOH, 2. CO2, then H+ → Salicylic acid (2-Hydroxybenzoic acid).
2. Reimer-Tiemann Reaction: Phenol + CHCl3 + aq. NaOH → Salicylaldehyde (2-Hydroxybenzaldehyde).
5. Ethers
5.1 Preparation of Ethers
- Dehydration of Alcohols (Symmetrical Ethers): 2C2H5OH → (H2SO4, 413 K) → C2H5-O-C2H5. (Note: at 443 K, alkene is formed instead). This is an SN2 mechanism, best for 1° alcohols.
- Williamson Synthesis (Symmetrical & Unsymmetrical): Reacting an alkyl halide with a
sodium alkoxide.R-X + R'-O−Na+ → R-O-R' + NaXCritical Condition: Alkyl halide must be primary (1°) because the alkoxide is a strong base and will cause elimination (forming an alkene) if a 2° or 3° halide is used. To make tert-butyl methyl ether, use tert-butoxide + methyl bromide.
5.2 Chemical Reactions of Ethers
Ethers are highly unreactive, acting mostly as solvents. However, they are cleaved by conc. acids (HI or HBr) at high temperatures.
R-O-R' + HX (excess) → R-X + R'-X + H2O
- Site of cleavage with unsymmetrical ethers:
If alkyl groups are 1° or 2°: The halide I− attacks the smaller alkyl group (SN2 mechanism). E.g., CH3-O-CH2CH3 + HI → CH3I + CH3CH2OH.
If one group is tertiary (3°): The cleavage proceeds via SN1, forming the stable 3° carbocation. The halide I− attacks the tertiary alkyl group. E.g., (CH3)3C-O-CH3 + HI → (CH3)3C-I + CH3OH.
If it's an Alkyl Aryl Ether (Anisole): The O-R bond breaks because the O-Ar bond has partial double bond character due to resonance. Products are always Phenol + Alkyl halide.
Electrophilic Substitution in Aromatic Ethers: The alkoxy group (-OR) is ortho/para directing. Important reactions include halogenation, Friedel-Crafts alkylation/acylation, and nitration of anisole.
🎓 NEET Previous Year Questions
Q1. [NEET 2022] Compound X on reaction with O3 followed by Zn/H2O gives formaldehyde
and 2-methylpropanal as products. The compound X is:
Answer This is ozonolysis. Products are HCHO and
(CH3)2CHCHO. Recombine them by putting a double bond where the oxygens are: CH2=CH-CH(CH3)2.
3-Methylbut-1-ene.
Q2. [NEET 2021] The major product formed in dehydrohalogenation... (Wait, let's use an
alcohol PYQ). Which of the following alcohols will yield the corresponding alkyl chloride on reaction
with concentrated HCl at room temperature?
Answer Reaction with conc. HCl (Lucas test) at room
temp without catalyst is only possible for highly reactive alcohols. Tertiary alcohols
form stable carbocations and react immediately at room temperature.
Q3. [NEET 2020] Anisole on cleavage with HI yields:
Answer Anisole is Methoxybenzene (C6H5-O-CH3). The
O-CH3 bond is weaker than the O-Phenyl bond (which has partial double bond character). Cleavage gives
Phenol (C6H5OH) and Methyl Iodide (CH3I).
Q4. [NEET 2019] The structure of intermediate A in the following reaction is: Cumene +
O2 → A → (H+) Phenol + Acetone.
Answer The intermediate formed by air oxidation of
cumene is Cumene hydroperoxide. C6H5-C(CH3)2-O-OH.
Q5. [NEET 2018] Heating of 2-chloro-1-phenylbutane with EtOK/EtOH gives X as the major
product. Reaction of X with Hg(OAc)2/H2O followed by NaBH4 gives Y as the major product. Y is:
Answer (1) Elimination with strong base gives alkene
conjugated with phenyl ring: 1-phenylbut-1-ene (Zaitsev/Conjugation). (2) Oxymercuration-demercuration
causes Markovnikov addition of water without rearrangement. OH goes to C1 (benzylic carbocation
character stabilizing TS). 1-phenylbutan-1-ol.
💡 Rapid Revision
- Hydroboration-oxidation gives Anti-Markovnikov primary alcohols. Acid-catalyzed hydration gives Markovnikov secondary/tertiary alcohols.
- Phenol is more acidic than alcohol due to resonance stabilization of the phenoxide ion. EWG (-NO2) increases phenol acidity; ERG (-CH3) decreases it.
- Williamson Synthesis trick: The ALKYL HALIDE must be 1°. If halide is 3°, elimination completely dominates over substitution, forming an alkene.
- Ether cleavage with HI: For 1°/2° groups, iodine attacks smaller group (SN2). For 3° groups, iodine attacks the 3° group (SN1). For Alkyl-Aryl (e.g., Anisole), you ALWAYS get Phenol + Alkyl Iodide.
CLASS 12 CHEMISTRY | NCERT SOLUTIONS
Chapter 7 — Alcohols, Phenols & Ethers
22 Solved Questions — Mechanisms, Acidity & Conversions
Note: Mathematical numericals are rare in this chapter. Focus on Arranging by boiling
point/acidity, organic mechanisms (hydration/dehydration), distinguishing tests (Lucas), and step-by-step
conversions.
📝 Physical Properties & Acidity Reasoning (Q1 – Q7)
2 MarksQ1. Explain why propanol has a higher boiling
point than the hydrocarbon, butane, of corresponding molecular mass.
✓ Solution
1. The molecules of butane are held together by weak van der Waals dispersion forces.
2. Propanol contains a highly polar -OH group which leads to strong intermolecular hydrogen bonding between propanol molecules.
Because breaking hydrogen bonds requires much more energy than overcoming van der Waals forces, propanol has a significantly higher boiling point than butane.
1. The molecules of butane are held together by weak van der Waals dispersion forces.
2. Propanol contains a highly polar -OH group which leads to strong intermolecular hydrogen bonding between propanol molecules.
Because breaking hydrogen bonds requires much more energy than overcoming van der Waals forces, propanol has a significantly higher boiling point than butane.
3 MarksQ2. Arrange the following sets of compounds in
order of their increasing boiling points: Pentan-1-ol, butan-1-ol, butan-2-ol, ethanol, propan-1-ol,
methanol.
✓ Solution
Rule 1: Boiling point increases with an increase in the number of carbon atoms (surface area increases, so van der Waals forces increase).
Rule 2: Among isomeric alcohols, branching decreases boiling point by decreasing surface area.
Order taking both rules into account:
Methanol < Ethanol < Propan-1-ol < Butan-2-ol (branched isomer) < Butan-1-ol (straight chain isomer) < Pentan-1-ol.
Rule 1: Boiling point increases with an increase in the number of carbon atoms (surface area increases, so van der Waals forces increase).
Rule 2: Among isomeric alcohols, branching decreases boiling point by decreasing surface area.
Order taking both rules into account:
Methanol < Ethanol < Propan-1-ol < Butan-2-ol (branched isomer) < Butan-1-ol (straight chain isomer) < Pentan-1-ol.
3 MarksQ3. Arrange the following compounds in
increasing order of their acid strength: Propan-1-ol, 2,4,6-trinitrophenol, 3-nitrophenol,
3,5-dinitrophenol, phenol, 4-methylphenol.
✓ Solution
1. Alcohols vs Phenols: Phenols are much more acidic than alcohols due to resonance stabilization of the phenoxide ion.
2. Effect of Substituents on Phenol: Electron-withdrawing groups (EWG) like -NO₂ increase acidity. Electron-releasing groups (ERG) like -CH₃ decrease acidity.
3. Number of EWGs: More -NO₂ groups = more acidic.
Therefore, increasing order is:
Propan-1-ol < 4-methylphenol (p-cresol) < phenol < 3-nitrophenol < 3,5-dinitrophenol < 2,4,6-trinitrophenol (Picric acid).
1. Alcohols vs Phenols: Phenols are much more acidic than alcohols due to resonance stabilization of the phenoxide ion.
2. Effect of Substituents on Phenol: Electron-withdrawing groups (EWG) like -NO₂ increase acidity. Electron-releasing groups (ERG) like -CH₃ decrease acidity.
3. Number of EWGs: More -NO₂ groups = more acidic.
Therefore, increasing order is:
Propan-1-ol < 4-methylphenol (p-cresol) < phenol < 3-nitrophenol < 3,5-dinitrophenol < 2,4,6-trinitrophenol (Picric acid).
2 MarksQ4. Explain why is ortho nitrophenol more
acidic than ortho methoxyphenol?
✓ Solution
The nitro group (-NO₂) is an electron-withdrawing group. It withdraws electrons via resonance (-R) and inductive (-I) effects, decreasing electron density on oxygen and vastly stabilizing the phenoxide ion, making it a stronger acid.
The methoxy group (-OCH₃) is an electron-releasing group via resonance (+R effect). It increases electron density on oxygen, destabilizing the phenoxide ion, making it a weaker acid.
The nitro group (-NO₂) is an electron-withdrawing group. It withdraws electrons via resonance (-R) and inductive (-I) effects, decreasing electron density on oxygen and vastly stabilizing the phenoxide ion, making it a stronger acid.
The methoxy group (-OCH₃) is an electron-releasing group via resonance (+R effect). It increases electron density on oxygen, destabilizing the phenoxide ion, making it a weaker acid.
3 MarksQ5. While separating a mixture of ortho and
para nitrophenols by steam distillation, name the isomer which will be steam volatile. Give reason.
✓ Solution
Ortho-nitrophenol will be steam volatile.
Reason: In o-nitrophenol, the -OH and -NO₂ groups are adjacent to each other, allowing the formation of intramolecular hydrogen bonding (chelation). This prevents association between different molecules.
In p-nitrophenol, the groups are far apart, leading to intermolecular hydrogen bonding between different molecules, causing them to associate (form a network), increasing the boiling point and making it non-volatile in steam.
Ortho-nitrophenol will be steam volatile.
Reason: In o-nitrophenol, the -OH and -NO₂ groups are adjacent to each other, allowing the formation of intramolecular hydrogen bonding (chelation). This prevents association between different molecules.
In p-nitrophenol, the groups are far apart, leading to intermolecular hydrogen bonding between different molecules, causing them to associate (form a network), increasing the boiling point and making it non-volatile in steam.
2 MarksQ6. Why are alcohols comparatively more
soluble in water than hydrocarbons of comparable molecular masses?
✓ Solution
Alcohols contain a polar -OH group. This group is capable of forming hydrogen bonds with the polar molecules of water.
Hydrocarbons are non-polar and cannot form hydrogen bonds with water. Instead, they disrupt the existing H-bonds in water without replacing them with favorable interactions. Thus, alcohols are miscible, while hydrocarbons are essentially insoluble in water.
Alcohols contain a polar -OH group. This group is capable of forming hydrogen bonds with the polar molecules of water.
Hydrocarbons are non-polar and cannot form hydrogen bonds with water. Instead, they disrupt the existing H-bonds in water without replacing them with favorable interactions. Thus, alcohols are miscible, while hydrocarbons are essentially insoluble in water.
2 MarksQ7. Name the reagents used in the
distinguishing test between primary, secondary, and tertiary alcohols (Lucas Test) and state their
observations.
✓ Solution
Reagent: Lucas Reagent (Concentrated HCl and anhydrous ZnCl₂).
Observations:
1. Tertiary (3°) Alcohol: Turbidity (cloudiness due to alkyl chloride formation) appears immediately at room temperature.
2. Secondary (2°) Alcohol: Turbidity appears after about 5 minutes at room temperature.
3. Primary (1°) Alcohol: No turbidity at room temperature. It only appears upon heating.
Reagent: Lucas Reagent (Concentrated HCl and anhydrous ZnCl₂).
Observations:
1. Tertiary (3°) Alcohol: Turbidity (cloudiness due to alkyl chloride formation) appears immediately at room temperature.
2. Secondary (2°) Alcohol: Turbidity appears after about 5 minutes at room temperature.
3. Primary (1°) Alcohol: No turbidity at room temperature. It only appears upon heating.
💡 Mechanisms and Name Reactions (Q8 – Q14)
3 MarksQ8. Give the mechanism of hydration of ethene
to yield ethanol.
✓ Solution
It is an electrophilic addition reaction involving 3 steps:
Step 1: Protonation of alkene to form carbocation by electrophilic attack of H₃O⁺.
CH₂=CH₂ + H₃O⁺ → CH₃-CH₂⁺ + H₂O
Step 2: Nucleophilic attack of water on carbocation.
CH₃-CH₂⁺ + H₂O → CH₃-CH₂-O⁺H₂ (protonated alcohol)
Step 3: Deprotonation to form an alcohol.
CH₃-CH₂-O⁺H₂ + H₂O → CH₃-CH₂-OH + H₃O⁺
It is an electrophilic addition reaction involving 3 steps:
Step 1: Protonation of alkene to form carbocation by electrophilic attack of H₃O⁺.
CH₂=CH₂ + H₃O⁺ → CH₃-CH₂⁺ + H₂O
Step 2: Nucleophilic attack of water on carbocation.
CH₃-CH₂⁺ + H₂O → CH₃-CH₂-O⁺H₂ (protonated alcohol)
Step 3: Deprotonation to form an alcohol.
CH₃-CH₂-O⁺H₂ + H₂O → CH₃-CH₂-OH + H₃O⁺
3 MarksQ9. Write the mechanism of dehydration of
ethanol to yield ethene.
✓ Solution
Dehydration requires concentrated H₂SO₄ at 443 K. The mechanism involves 3 steps:
Step 1: Formation of protonated alcohol.
CH₃-CH₂-OH + H⁺ → CH₃-CH₂-O⁺H₂
Step 2: Formation of carbocation (Slowest, Rate determining step).
CH₃-CH₂-O⁺H₂ → CH₃-CH₂⁺ + H₂O (elimination of water)
Step 3: Formation of ethene by elimination of a proton.
CH₃-CH₂⁺ → CH₂=CH₂ + H⁺
Note: The acid catalyst H⁺ consumed in Step 1 is regenerated in Step 3.
Dehydration requires concentrated H₂SO₄ at 443 K. The mechanism involves 3 steps:
Step 1: Formation of protonated alcohol.
CH₃-CH₂-OH + H⁺ → CH₃-CH₂-O⁺H₂
Step 2: Formation of carbocation (Slowest, Rate determining step).
CH₃-CH₂-O⁺H₂ → CH₃-CH₂⁺ + H₂O (elimination of water)
Step 3: Formation of ethene by elimination of a proton.
CH₃-CH₂⁺ → CH₂=CH₂ + H⁺
Note: The acid catalyst H⁺ consumed in Step 1 is regenerated in Step 3.
3 MarksQ10. Explain Williamson's synthesis. Why is it
not suitable for preparing di-tert-butyl ether?
✓ Solution
Williamson synthesis is an SN2 reaction involving an alkyl halide (R-X) and sodium alkoxide (R'-O⁻Na⁺) to form an ether: R-X + R'-ONa → R-O-R' + NaX.
Limitation: To get maximum yield, the alkyl halide MUST be primary (1°) because the alkoxide ion is not only a good nucleophile but also a strong base.
If a tertiary alkyl halide (e.g., tert-butyl bromide) is used to prepare di-tert-butyl ether, the strong alkoxide base causes an elimination (E2) reaction rather than substitution, yielding an alkene (isobutylene) as the major product.
Williamson synthesis is an SN2 reaction involving an alkyl halide (R-X) and sodium alkoxide (R'-O⁻Na⁺) to form an ether: R-X + R'-ONa → R-O-R' + NaX.
Limitation: To get maximum yield, the alkyl halide MUST be primary (1°) because the alkoxide ion is not only a good nucleophile but also a strong base.
If a tertiary alkyl halide (e.g., tert-butyl bromide) is used to prepare di-tert-butyl ether, the strong alkoxide base causes an elimination (E2) reaction rather than substitution, yielding an alkene (isobutylene) as the major product.
3 MarksQ11. Predict the major product of acid
catalyzed dehydration of 1-methylcyclohexanol.
✓ Solution
Dehydration follows Zaitsev's Rule (more substituted alkene is the major product).
1. Protonation of -OH followed by loss of water generates a tertiary carbocation at C1 (where methyl is attached).
2. A proton (H⁺) can be lost from either the methyl group (yielding 1-methylenecyclohexane) or from the adjacent ring carbon C2/C6 (yielding 1-methylcyclohexene).
The double bond inside the ring (1-methylcyclohexene) is more stable as it is a trisubstituted alkene (3 alkyl groups attached to C=C), whereas the exocyclic one is disubstituted.
Major product: 1-methylcyclohexene.
Dehydration follows Zaitsev's Rule (more substituted alkene is the major product).
1. Protonation of -OH followed by loss of water generates a tertiary carbocation at C1 (where methyl is attached).
2. A proton (H⁺) can be lost from either the methyl group (yielding 1-methylenecyclohexane) or from the adjacent ring carbon C2/C6 (yielding 1-methylcyclohexene).
The double bond inside the ring (1-methylcyclohexene) is more stable as it is a trisubstituted alkene (3 alkyl groups attached to C=C), whereas the exocyclic one is disubstituted.
Major product: 1-methylcyclohexene.
2 MarksQ12. What happens when phenol reacts with
bromine water? Write the equation.
✓ Solution
Phenol acts as a strongly activating group due to the +R effect of the -OH group. In a polar solvent like water, the phenoxide ion is formed, which highly activates the ring towards electrophilic substitution.
Bromination occurs at all available ortho and para positions simultaneously.
Phenol acts as a strongly activating group due to the +R effect of the -OH group. In a polar solvent like water, the phenoxide ion is formed, which highly activates the ring towards electrophilic substitution.
Bromination occurs at all available ortho and para positions simultaneously.
Phenol + 3Br₂(aq) → 2,4,6-Tribromophenol + 3HBr
Observation: A white precipitate of 2,4,6-tribromophenol is formed.2 MarksQ13. Write equations for Reimer-Tiemann
reaction.
✓ Solution
When phenol is treated with chloroform (CHCl₃) in the presence of aqueous sodium hydroxide at 340 K, a formyl group (-CHO) is introduced at the ortho position of the benzene ring.
The intermediate is a substituted benzal chloride which is hydrolyzed by alkali to yield salicylaldehyde.
When phenol is treated with chloroform (CHCl₃) in the presence of aqueous sodium hydroxide at 340 K, a formyl group (-CHO) is introduced at the ortho position of the benzene ring.
The intermediate is a substituted benzal chloride which is hydrolyzed by alkali to yield salicylaldehyde.
1. Phenol + CHCl₃ + 3NaOH(aq) → Sodium salicylaldehyde + 3NaCl +
2H₂O
2. Acidification (H⁺) yields Salicylaldehyde.
2. Acidification (H⁺) yields Salicylaldehyde.
2 MarksQ14. Write equations for Kolbe's reaction.
✓ Solution
Phenol is treated with NaOH to form sodium phenoxide, which is even more reactive than phenol towards electrophilic aromatic substitution.
It is subjected to electrophilic substitution with carbon dioxide (CO₂), a weak electrophile, at 400 K and 4-7 atm pressure.
Phenol is treated with NaOH to form sodium phenoxide, which is even more reactive than phenol towards electrophilic aromatic substitution.
It is subjected to electrophilic substitution with carbon dioxide (CO₂), a weak electrophile, at 400 K and 4-7 atm pressure.
1. Sodium phenoxide + CO₂ → (Heat/Pressure) → Sodium
salicylate
2. Acidification (H⁺) yields Salicylic acid (2-Hydroxybenzoic acid).
2. Acidification (H⁺) yields Salicylic acid (2-Hydroxybenzoic acid).
📈 Ether Cleavage and Organic Conversions (Q15 – Q22)
3 MarksQ15. Write the products when Anisole reacts
with HI.
✓ Solution
Anisole is Methoxybenzene (C₆H₅-O-CH₃).
In alkyl-aryl ethers, the O-R bond is weaker than the O-Ar bond because the oxygen lone pair is in resonance with the benzene ring, imparting partial double bond character to the O-Ar bond.
Therefore, cleavage by HI always occurs at the O-CH₃ bond.
Anisole is Methoxybenzene (C₆H₅-O-CH₃).
In alkyl-aryl ethers, the O-R bond is weaker than the O-Ar bond because the oxygen lone pair is in resonance with the benzene ring, imparting partial double bond character to the O-Ar bond.
Therefore, cleavage by HI always occurs at the O-CH₃ bond.
C₆H₅-O-CH₃ + HI (excess) → C₆H₅OH (Phenol)
+ CH₃I (Methyl iodide)
Note: Phenol does not react further with HI because the O-Ar bond is too strong to break.3 MarksQ16. Write the equation for the reaction of
tert-butyl methyl ether with HI.
✓ Solution
Ether consists of a tertiary group (tert-butyl) and a primary group (methyl).
When one of the alkyl groups is tertiary, the cleavage proceeds via an SN1 mechanism because the tertiary carbocation formed is highly stable.
The nucleophile (I⁻) will attack the stable tertiary carbocation.
Ether consists of a tertiary group (tert-butyl) and a primary group (methyl).
When one of the alkyl groups is tertiary, the cleavage proceeds via an SN1 mechanism because the tertiary carbocation formed is highly stable.
The nucleophile (I⁻) will attack the stable tertiary carbocation.
(CH₃)₃C-O-CH₃ + HI → (CH₃)₃C-I (tert-Butyl
iodide) + CH₃OH (Methanol)
2 MarksQ17. Show how will you synthesize
1-phenylethanol from a suitable alkene.
✓ Solution
1-phenylethanol is C₆H₅-CH(OH)-CH₃.
By acid-catalysed hydration of styrene (phenylethene), we can get this according to Markovnikov's addition.
1-phenylethanol is C₆H₅-CH(OH)-CH₃.
By acid-catalysed hydration of styrene (phenylethene), we can get this according to Markovnikov's addition.
C₆H₅-CH=CH₂ + H₂O(H⁺) →
C₆H₅-CH(OH)-CH₃
3 MarksQ18. Synthesize Cyclohexylmethanol using an
alkyl halide by an SN2 reaction.
✓ Solution
We need to replace the halide with an -OH group using an aqueous base (SN2).
The alkyl halide must be primary to favor SN2 over E2.
We need to replace the halide with an -OH group using an aqueous base (SN2).
The alkyl halide must be primary to favor SN2 over E2.
Alkyl halide: (Bromomethyl)cyclohexane
Mechanism: Cyclohexyl-CH₂Br + NaOH(aq) → (Heat) → Cyclohexyl-CH₂OH + NaBr
Mechanism: Cyclohexyl-CH₂Br + NaOH(aq) → (Heat) → Cyclohexyl-CH₂OH + NaBr
2 MarksQ19. Synthesize Pentan-1-ol using a suitable
alkyl halide.
✓ Solution
Use aqueous KOH to replace the halide via SN2.
Use aqueous KOH to replace the halide via SN2.
1-Bromopentane + KOH(aqueous, boil) → Pentan-1-ol + KBr
CH₃CH₂CH₂CH₂CH₂Br + aq. KOH → CH₃CH₂CH₂CH₂CH₂OH
CH₃CH₂CH₂CH₂CH₂Br + aq. KOH → CH₃CH₂CH₂CH₂CH₂OH
2 MarksQ20. Prepare Phenol from Chlorobenzene.
✓ Solution
This is the Dow Process.
This is the Dow Process.
1. C₆H₅-Cl + NaOH(aq) → (623 K, 300 atm) →
C₆H₅-O⁻Na⁺ (Sodium phenoxide)
2. C₆H₅-O⁻Na⁺ + HCl(dil.) → C₆H₅-OH (Phenol) + NaCl
2. C₆H₅-O⁻Na⁺ + HCl(dil.) → C₆H₅-OH (Phenol) + NaCl
3 MarksQ21. How are the following conversions carried
out? (i) Benzyl chloride to Benzyl alcohol (ii) Ethyl magnesium chloride to Propan-1-ol.
✓ Solution
(i) Benzyl chloride to Benzyl alcohol: React the halide with boiling aqueous NaOH (SN1/SN2 mix).
C₆H₅-CH₂Cl + NaOH(aq) → C₆H₅-CH₂OH + NaCl.
(ii) Ethyl magnesium chloride to Propan-1-ol: A Grignard reaction creating a primary alcohol with one extra carbon requires Formaldehyde (Methanal).
C₂H₅MgCl + HCHO → [C₂H₅-CH₂-OMgCl] adduct.
Adduct + H₂O/H⁺ → C₂H₅-CH₂-OH (Propan-1-ol) + Mg(OH)Cl.
(i) Benzyl chloride to Benzyl alcohol: React the halide with boiling aqueous NaOH (SN1/SN2 mix).
C₆H₅-CH₂Cl + NaOH(aq) → C₆H₅-CH₂OH + NaCl.
(ii) Ethyl magnesium chloride to Propan-1-ol: A Grignard reaction creating a primary alcohol with one extra carbon requires Formaldehyde (Methanal).
C₂H₅MgCl + HCHO → [C₂H₅-CH₂-OMgCl] adduct.
Adduct + H₂O/H⁺ → C₂H₅-CH₂-OH (Propan-1-ol) + Mg(OH)Cl.
2 MarksQ22. Convert Phenol to Aspirin.
✓ Solution
Aspirin is acetylsalicylic acid. We first make salicylic acid via Kolbe's reaction, then acetylate it.
1. Phenol → [NaOH, CO₂ then H⁺] → Salicylic acid (2-Hydroxybenzoic acid).
2. Acetylation: Salicylic acid + Acetic anhydride [(CH₃CO)₂O] in presence of a few drops of conc. H₂SO₄ → Aspirin (2-Acetoxybenzoic acid) + CH₃COOH.
Aspirin is acetylsalicylic acid. We first make salicylic acid via Kolbe's reaction, then acetylate it.
1. Phenol → [NaOH, CO₂ then H⁺] → Salicylic acid (2-Hydroxybenzoic acid).
2. Acetylation: Salicylic acid + Acetic anhydride [(CH₃CO)₂O] in presence of a few drops of conc. H₂SO₄ → Aspirin (2-Acetoxybenzoic acid) + CH₃COOH.
✍ Score Guide — 22 Questions
All questions from NCERT Exercises covering Acidity ordering, Boiling Point logic, Hydration/Dehydration mechanisms, Ether cleavage rules, and step-by-step conversions.
All questions from NCERT Exercises covering Acidity ordering, Boiling Point logic, Hydration/Dehydration mechanisms, Ether cleavage rules, and step-by-step conversions.
High-Yield Facts & Formulas: Alcohols, Phenols & Ethers
Alcohols Solubility
Lower alcohols are miscible with water due to hydrogen bonding. Solubility decreases with increase in molecular mass.
Lower alcohols are miscible with water due to hydrogen bonding. Solubility decreases with increase in molecular mass.
Boiling Point trend
Alcohols and phenols have higher boiling points than alkanes/ethers of same mass due to intermolecular H-bonding.
Alcohols and phenols have higher boiling points than alkanes/ethers of same mass due to intermolecular H-bonding.
Acidic Character of Alcohols
Order of acidity: 1° > 2° > 3°. Alkyl groups (ERG) decrease acidity.
Order of acidity: 1° > 2° > 3°. Alkyl groups (ERG) decrease acidity.
Phenol Acidity
Phenols are stronger acids than alcohols. Phenoxide ion is resonance stabilized.
Phenols are stronger acids than alcohols. Phenoxide ion is resonance stabilized.
Effect of Substituents on Phenol
EWG (-NO2, -CN) increases acidity; ERG (-CH3, -OCH3) decreases it.
EWG (-NO2, -CN) increases acidity; ERG (-CH3, -OCH3) decreases it.
Hydroboration-Oxidation
Alkenes → Alcohols (regioselectivity: Anti-Markovnikov; stereoselectivity: Syn addition).
Alkenes → Alcohols (regioselectivity: Anti-Markovnikov; stereoselectivity: Syn addition).
Grignard Reagent Addition
Formaldehyde → 1°; Other Aldehydes → 2°; Ketones → 3° alcohols.
Formaldehyde → 1°; Other Aldehydes → 2°; Ketones → 3° alcohols.
Lucas Test
3° alcohols (immediate turbidity), 2° (5-10 min), 1° (no turbidity at room temp).
3° alcohols (immediate turbidity), 2° (5-10 min), 1° (no turbidity at room temp).
Victor Meyer's Test
1° (Red), 2° (Blue), 3° (Colorless) colors obtained with specific reagents.
1° (Red), 2° (Blue), 3° (Colorless) colors obtained with specific reagents.
Esterification
R-OH + R'COOH ↔ R'COOR + H2O (Acid catalyzed).
R-OH + R'COOH ↔ R'COOR + H2O (Acid catalyzed).
Dehydration
Alcohols + Conc. H2SO4 → Alkenes (170°C) or Ethers (140°C).
Alcohols + Conc. H2SO4 → Alkenes (170°C) or Ethers (140°C).
Oxidation of 1° Alcohols
To Aldehydes (using PCC) or to Carboxylic acids (using acidified KMnO4).
To Aldehydes (using PCC) or to Carboxylic acids (using acidified KMnO4).
Oxidation of 2° Alcohols
To Ketones using CrO3 or PCC.
To Ketones using CrO3 or PCC.
Kolbe's Reaction
Phenol → Salicylic acid (using NaOH and CO2).
Phenol → Salicylic acid (using NaOH and CO2).
Reimer-Tiemann Reaction
Phenol → Salicylaldehyde (using CHCl3 and NaOH).
Phenol → Salicylaldehyde (using CHCl3 and NaOH).
Coupling Reaction
Phenol + Benzene diazonium chloride → p-hydroxyazobenzene (Orange dye).
Phenol + Benzene diazonium chloride → p-hydroxyazobenzene (Orange dye).
Williamson Synthesis
R-ONa + R'-X → R-O-R' + NaX. (R'X should be 1° to avoid elimination).
R-ONa + R'-X → R-O-R' + NaX. (R'X should be 1° to avoid elimination).
Cleavage of Ethers by HI
Methyl tert-butyl ether + HI → tert-butyl iodide + Methanol (SN1 path).
Methyl tert-butyl ether + HI → tert-butyl iodide + Methanol (SN1 path).
Picric Acid
2,4,6-Trinitrophenol. Strongest organic acid among common phenols.
2,4,6-Trinitrophenol. Strongest organic acid among common phenols.
Cumene Process
Industrial method: Benzene + Propene → Cumene → Phenol + Acetone.
Industrial method: Benzene + Propene → Cumene → Phenol + Acetone.
Aspirin preparation
Acetylation of salicylic acid with acetic anhydride.
Acetylation of salicylic acid with acetic anhydride.
Hydrolysis of Ethers
Ethers are fairly stable but can hydrate back to alcohols with dilute acid under pressure.
Ethers are fairly stable but can hydrate back to alcohols with dilute acid under pressure.
Nature of Ethers
C-O-C bond angle is approx 110° (near tetrahedral). Dipole moment is non-zero.
C-O-C bond angle is approx 110° (near tetrahedral). Dipole moment is non-zero.
Hydrogen bonding in Ethers
Ethers don't form H-bonds with themselves, but can form H-bonds with water molecules.
Ethers don't form H-bonds with themselves, but can form H-bonds with water molecules.
Oxidation of Phenols with
Na2Cr2O7
Gives Benzoquinone (p-quinone).
Gives Benzoquinone (p-quinone).
Dehydrogenation of Alcohols
Using heated Cu catalyst: 1°→Aldehyde, 2°→Ketone, 3°→Alkene.
Using heated Cu catalyst: 1°→Aldehyde, 2°→Ketone, 3°→Alkene.
Reaction with PCl5
R-OH + PCl5 → R-Cl + POCl3 + HCl.
R-OH + PCl5 → R-Cl + POCl3 + HCl.
Symmetry of Ethers
Simple ethers (R=R') vs Mixed ethers (R≠R').
Simple ethers (R=R') vs Mixed ethers (R≠R').
Acidity of Ortho substituted phenols
Ortho effect varies; Intramolecular H-bonding in o-nitrophenol reduces acidity compared to p-isomer.
Ortho effect varies; Intramolecular H-bonding in o-nitrophenol reduces acidity compared to p-isomer.
Methanol toxicity
Oxidized to formaldehyde and formic acid in body; causes blindness and death.
Oxidized to formaldehyde and formic acid in body; causes blindness and death.
Wood Spirit
Another name for Methanol (obtained by destructive distillation of wood).
Another name for Methanol (obtained by destructive distillation of wood).
Grain Alcohol
Another name for Ethanol (obtained by fermentation of grains/molasses).
Another name for Ethanol (obtained by fermentation of grains/molasses).
Glycerol
Propane-1,2,3-triol. A trihydric alcohol.
Propane-1,2,3-triol. A trihydric alcohol.
Glycol
Ethane-1,2-diol. A dihydric alcohol used as antifreeze.
Ethane-1,2-diol. A dihydric alcohol used as antifreeze.
Oxidation of Phenol in air
Phenols turn pink/red in air due to slow oxidation to quinones.
Phenols turn pink/red in air due to slow oxidation to quinones.
Nitration of Phenol (dilute HNO3)
Mixture of o- and p-nitrophenols. Separated by steam distillation (o- is steam volatile).
Mixture of o- and p-nitrophenols. Separated by steam distillation (o- is steam volatile).
Bromination of Phenol in CS2
Monobromination (o- and p- isomers) occurs at low temperature.
Monobromination (o- and p- isomers) occurs at low temperature.
Formaldehyde + Grignard
Gives Primary alcohol always.
Gives Primary alcohol always.
Higher Aldehydes + Grignard
Give Secondary alcohols.
Give Secondary alcohols.
Ketones + Grignard
Give Tertiary alcohols.
Give Tertiary alcohols.
Zeisel's Method
Quantitative estimation of alkoxy groups (-OR) using HI.
Quantitative estimation of alkoxy groups (-OR) using HI.
Protic solvents and SN1
Favor carbocation formation, hence good for SN1 reactions of ethers.
Favor carbocation formation, hence good for SN1 reactions of ethers.
Phenol with FeCl3
Gives Violet/Purple color due to formation of a complex.
Gives Violet/Purple color due to formation of a complex.
Crown Ethers
Cyclic polyethers that complex selectively with metal cations.
Cyclic polyethers that complex selectively with metal cations.
Haloform Reaction in Alcohols
Ethanol shows iodoform test; Methanol does not.
Ethanol shows iodoform test; Methanol does not.
Friedel-Crafts on Ethers
Anisole undergoes alkylation/acetylation at ortho and para positions.
Anisole undergoes alkylation/acetylation at ortho and para positions.
Bakelite precursor
Phenol + Formaldehyde (Polymerization).
Phenol + Formaldehyde (Polymerization).
Phenol solubility
Sparingly soluble in water, but completely soluble in organic solvents.
Sparingly soluble in water, but completely soluble in organic solvents.
Intramolecular H-bonding
In o-nitrophenol, slows down its boiling point compared to p-isomer (Intermolecular).
In o-nitrophenol, slows down its boiling point compared to p-isomer (Intermolecular).
Ether as solvent
Ideal for Grignard reagents as they coordinate and stabilize the Mg center.
Ideal for Grignard reagents as they coordinate and stabilize the Mg center.
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