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Alcohols, Phenols and Ethers

Alcohols, Phenols and EthersNEET Chemistry · Class 12 · NCERT Chapter 7

Classification and IUPAC Naming

Alcohols

Alcohols contain the hydroxyl group (-OH) attached to a saturated sp3 carbon. They are classified based on the nature of the carbon bearing -OH:

TypeCarbon typeExample
Primary (1°)-OH on C bonded to 1 other C (or CH3OH)Methanol (CH3OH), ethanol (C2H5OH)
Secondary (2°)-OH on C bonded to 2 other C atomsPropan-2-ol (isopropanol)
Tertiary (3°)-OH on C bonded to 3 other C atoms2-methylpropan-2-ol (tert-butanol)

Polyhydric alcohols: diols (glycols, e.g., ethylene glycol HO-CH2-CH2-OH), triols (glycerol HO-CH2-CHOH-CH2-OH), etc.

IUPAC Naming of Alcohols

  • Select the longest chain containing the -OH group as the parent chain.
  • Number from the end nearer to -OH to give it the lowest locant.
  • Replace the terminal "-e" of the alkane name with "-ol". E.g., ethane → ethanol.
  • If -OH gets the lowest locant, write it before the suffix: butan-1-ol, butan-2-ol.
  • Cyclic alcohols: cyclopentanol, cyclohexanol (OH on C-1 by convention).

Phenols

Phenols have -OH directly attached to a benzene ring. The simplest is phenol (C6H5OH). The -OH in phenols is more acidic than in alcohols due to resonance stabilisation of the phenoxide ion (C6H5O-).

Ethers

Ethers have two organic groups (alkyl or aryl) attached to an oxygen: R-O-R'. Named as alkoxy derivatives of the larger alkane: CH3-O-CH2CH3 = methoxyethane (common: diethyl ether). If both groups are same, use "di-": CH3-O-CH3 = dimethyl ether (methoxymethane).

Physical Properties and Hydrogen Bonding

Alcohols and phenols have much higher boiling points than alkanes or ethers of similar molecular mass. The reason is intermolecular hydrogen bonding.

In alcohols, the O-H bond is highly polar. The slightly positive H of one molecule is attracted to the slightly negative O of another molecule. This electrostatic attraction (hydrogen bond) requires extra energy to break, raising the boiling point significantly.

Boiling Point Comparison

CompoundFormulaMol. massBoiling PointH-bonding
EthanolC2H5OH4678°CYes (O-H ... O)
Dimethyl etherCH3OCH346-24°CVery weak (no O-H)
PropaneC3H844-42°CNone

Ethers cannot form strong H-bonds with themselves (no O-H group), so their boiling points are much lower than alcohols of similar mass. Ethers can, however, form weak H-bonds with water, explaining why lower ethers are soluble in water.

Solubility in water: lower alcohols (C1-C4) are miscible with water because they can form H-bonds with water. As chain length increases, the hydrophobic alkyl part dominates and solubility decreases.

Preparation of Alcohols

1. From Alkenes

  • Acid-catalysed hydration (Markovnikov): alkene + H2O (H2SO4 catalyst) → alcohol following Markovnikov's rule. CH2=CH2 + H2O → CH3CH2OH (ethanol).
  • Hydroboration-oxidation (anti-Markovnikov): alkene + B2H6, then H2O2/NaOH → primary alcohol. Syn addition, anti-Markovnikov.
  • Oxymercuration-demercuration (Markovnikov): alkene + Hg(OAc)2/H2O, then NaBH4 → Markovnikov alcohol with no rearrangement.

2. From Haloalkanes

  • Aqueous KOH (SN2): R-X + KOH (aq) → R-OH. Best for 1° haloalkanes.
  • Silver oxide/water: R-X + Ag2O + H2O → R-OH (milder, avoids elimination).

3. Using Grignard Reagents

  • Formaldehyde (HCHO) + RMgX → hydrolysis → 1° alcohol (with one extra C compared to Grignard alkyl group).
  • Aldehyde (RCHO) + R'MgX → hydrolysis → 2° alcohol.
  • Ketone (RCOR') + R''MgX → hydrolysis → 3° alcohol.
  • CO2 + RMgX → hydrolysis → carboxylic acid (not alcohol — included here for completeness).

4. From Carbonyl Compounds (Reduction)

  • Aldehyde + H2 (Ni/Pd catalyst) → 1° alcohol.
  • Ketone + H2 → 2° alcohol.
  • NaBH4 (mild reducing agent): reduces C=O to C-OH; does not reduce C=C, ester, or COOH.
  • LiAlH4 (strong reducing agent): reduces all carbonyl-containing groups including esters, COOH.

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Chemical Reactions of Alcohols

1. Acidic Character

Alcohols are weak acids (Ka ~ 10-16). They react with active metals (Na, K) to liberate hydrogen: 2 R-OH + 2Na → 2 R-ONa + H2. The reaction is much slower and less vigorous than Na with water. Acidity order: water >>> 3° (because alkyl groups donate electrons, destabilising the alkoxide anion).

2. Dehydration (Elimination)

Alcohols undergo dehydration (loss of water) with concentrated H2SO4 at different temperatures:

  • 413 K (140°C): intermolecular dehydration → ether (R-O-R'). Two alcohol molecules react: R-OH + HO-R → R-O-R' + H2O.
  • 443 K (170°C): intramolecular dehydration → alkene (E2). Follows Zaitsev's rule: more substituted alkene is the major product.
  • Reactivity:>> 1° in dehydration (easier to form more stable carbocation).

3. Oxidation Reactions

This is the most important reaction type for NEET. The product depends on the type of alcohol and the oxidising agent used.

SubstrateMild oxidant (PCC, MnO2)Strong oxidant (KMnO4/K2Cr2O7)
1° AlcoholAldehyde (stopped here)Carboxylic acid (further oxidation)
2° AlcoholKetoneKetone (stable)
3° AlcoholNo reactionNo simple product (C-C bond breaks in harsh conditions)

PCC (pyridinium chlorochromate): mild oxidant that converts 1° alcohol to aldehyde without further oxidation to acid.

Oxidation in biology: ethanol is oxidised by liver enzymes (alcohol dehydrogenase) in two steps: ethanol → acetaldehyde → acetic acid.

4. Reaction with HX (Substitution)

R-OH + HX → R-X + H2O. Rate increases with HI > HBr > HCl, and 3° >> 1°. The -OH must be protonated first (by H+ from HX) to form a good leaving group (H2O).

5. Esterification

Alcohols react with carboxylic acids (H2SO4 catalyst) to form esters: R-OH + R'COOH ⇌ R'COOR + H2O. Reversible reaction; Fischer esterification.

6. Lucas Test (ZnCl2/conc. HCl)

  • 3° alcohol: turbidity appears immediately (reacts via SN1).
  • 2° alcohol: turbidity appears in 5 minutes at room temperature.
  • 1° alcohol: no turbidity at room temperature (requires heating).
  • Only useful for alcohols up to C5 (must be soluble in Lucas reagent).

Alcohol Oxidation Simulator

Select the type of alcohol and the oxidising agent to see what product forms. Understand why PCC stops at aldehyde while KMnO₄ goes all the way to carboxylic acid.

Alcohol Type
Example: CH₃CH₂OH (ethanol), CH₃OH (methanol) | α-Hydrogen: Yes
Reagent
Pyridinium chlorochromate in CH₂Cl₂. Stops at aldehyde.
Product
Aldehyde (R-CHO)
Why this product?

PCC and MnO₂ are mild oxidants. They oxidise primary alcohols to aldehydes but CANNOT oxidise the aldehyde further to a carboxylic acid. The mild oxidant removes H from both the α-carbon and the -OH, giving R-CHO. The reaction stops at the aldehyde stage. This is the KEY advantage of PCC: you can make aldehydes from primary alcohols without over-oxidation.

Further oxidation?

Stopped at aldehyde. PCC does not have enough oxidising power to oxidise R-CHO to R-COOH.

Oxidation Summary Table
PCC / MnO₂
KMnO₄ / K₂Cr₂O₇
1° Alcohol
Aldehyde
Carboxylic Acid
2° Alcohol
Ketone
Ketone
3° Alcohol
No reaction
No reaction

Phenols: Structure, Acidity, and Reactions

Why Phenols Are More Acidic Than Alcohols

Phenol (C6H5OH) is a much stronger acid than alcohols (pKa 10 vs pKa 16-18 for alcohols). When phenol loses a proton, it forms the phenoxide ion (C6H5O-). The negative charge on oxygen is delocalised into the benzene ring by resonance through five resonance structures. This resonance stabilisation makes the phenoxide ion much more stable, making phenol more willing to donate a proton (more acidic).

In alkoxide ions (RO-), no such resonance stabilisation is possible, so they are less stable and alcohols are less acidic.

Effect of Substituents on Phenol Acidity

Groups that stabilise the phenoxide ion (withdraw electrons) increase acidity. Groups that destabilise it (donate electrons) decrease acidity.

SubstituentEffect on phenoxide ionEffect on acidity vs phenolExample (pKa)
-NO2 at ortho/paraWithdraws electrons (resonance), stabilises O-More acidic4-nitrophenol (pKa 7.1)
-Cl, -BrMild electron withdrawal (-I dominates)Slightly more acidic4-chlorophenol (pKa 9.4)
-CH3, -OCH3Donates electrons, destabilises O-Less acidic4-methylphenol (pKa 10.2)

Reactions of Phenol

  • FeCl3 test: violet/purple colour. Used to identify phenol and enols.
  • Bromination: phenol + Br2/water → 2,4,6-tribromophenol (white ppt) immediately (no catalyst needed). Phenol ring is highly activated.
  • Nitration: phenol + dilute HNO3 → mixture of o-nitrophenol and p-nitrophenol. o-isomer is more volatile (intramolecular H-bond) and separable by steam distillation.
  • Kolbe reaction: sodium phenoxide + CO2 (high pressure, 125°C) → sodium salicylate → salicylic acid (2-hydroxybenzoic acid). Basis of aspirin synthesis.
  • Reimer-Tiemann reaction: phenol + CHCl3 + NaOH → o-hydroxybenzaldehyde (salicylaldehyde) as major product. The reactive intermediate is dichlorocarbene (:CCl2).
  • Coupling with diazonium salt: C6H5OH + C6H5N2+Cl- (alkaline medium, 0-5°C) → p-hydroxyazobenzene (orange azo dye) by electrophilic aromatic substitution.
  • Reaction with acetic anhydride: phenol → phenyl acetate (ester). Used in synthesis of aspirin (acetylsalicylic acid).

Acidity Comparator: Alcohols, Phenols, and Substituted Phenols

See how substituents on the phenol ring affect acidity. Click any compound to understand the resonance and electronic effects behind the pKa value.

pKa Comparison (lower pKa = stronger acid)

Ethanol

alcohol
pKa = 15.9
C₂H₅OH

p-Methylphenol (p-Cresol)

phenol
pKa = 10.2
4-CH₃-C₆H₄-OH

Phenol

phenol
pKa = 10
C₆H₅OH

p-Chlorophenol

phenol
pKa = 9.4
4-Cl-C₆H₄-OH

p-Nitrophenol

activated phenol
pKa = 7.1
4-NO₂-C₆H₄-OH

2,4-Dinitrophenol

activated phenol
pKa = 4.1
2,4-(NO₂)₂-C₆H₃-OH
Phenol

C₆H₅OH

Substituent Effect

No substituent — reference

Resonance Explanation

Phenoxide ion (C₆H₅O⁻) is stabilised by resonance with the benzene ring through 5 resonance structures. Negative charge delocalised over ring at ortho and para positions. This stabilisation dramatically increases acidity vs alcohols (pKa drops from ~16 to ~10). Reference compound for all phenol acidity comparisons.

pKa
10
vs Phenol (10.0)

Reference

NEET Quick Rule

Acid strength order: carboxylic acids > activated phenols (with -NO₂) > phenol > water > alcohols<br/>For phenol derivatives: EWG at ortho/para → more acidic; EDG at ortho/para → less acidic. Always ask: does the substituent stabilise the phenoxide ion (O⁻) by withdrawing electrons? If yes, more acidic.

Ethers: Preparation and Reactions

Preparation

  • Williamson ether synthesis: R-ONa (sodium alkoxide) + R'-X → R-O-R' + NaX. An SN2 reaction, so R'-X must be primary (3° R'-X gives elimination with alkoxide base instead). Used for both symmetric and unsymmetric ethers.
  • Intermolecular dehydration of alcohol: 2 R-OH + H2SO4 (413 K) → R-O-R + H2O. Only works well for simple symmetric ethers from 1° alcohols.

Reactions of Ethers

  • Cleavage by HI/HBr: R-O-R' + 2HI → R-I + R'-I + H2O. HI > HBr > HCl in reactivity. For unsymmetric ethers (R-O-Ar), C-O bond of the alkyl group breaks (aryl-oxygen bond is strong due to resonance). Product: R-I + Ar-OH (phenol).
  • Mechanism: Protonation of O first (→ oxonium ion), then SN2 by iodide on the less hindered carbon. If one group is 3°, SN1 mechanism operates.
  • Peroxide formation: ethers form explosive peroxides on prolonged exposure to air and light. This is why ether must be tested for peroxides before distillation.
  • Friedel-Crafts (aryl ethers): anisole (CH3-O-C6H5) reacts with CH3COCl + AlCl3 at para position (electron-donating OCH3 activates ring, ortho-para directing).

Epoxides (Oxiranes)

Cyclic ethers with three-membered rings. Very reactive due to ring strain. Undergoes ring-opening with nucleophiles under acid (anti addition, attack at more substituted C in SN1-like fashion) or base conditions (anti addition, attack at less substituted C in SN2 fashion).

Distinction Tests

TestReagentPositive resultDetects
Lucas testZnCl2 + conc. HClTurbidity (immediate for 3°, 5 min for 2°, no reaction for 1°)Distinguishes 1°, 2°, 3° alcohols (C3-C6)
FeCl3 testFeCl3 (aqueous)Violet/purple colourPhenols and enols
Iodoform testI2 + NaOH (NaOI)Yellow CHI3 precipitateMethyl ketones (CH3CO-) and CH3CHOH-
Cerric ammonium nitrate (CAN)Ce(NH4)2(NO3)6Red colour (all alcohols)Presence of -OH group
Sodium metalNaH2 gas evolutionAcidic H (alcohol, phenol, acid)
Bromine waterBr2/H2ODecolourises immediately, white ppt (if phenol)Phenol (tribromophenol ppt) vs alcohol (just decolourise)

Worked NEET Problems

1

NEET-style problem · Acidity comparison

Question

Arrange in increasing order of acidity: ethanol, water, phenol, p-nitrophenol. (A) Ethanol < water < phenol < p-nitrophenol (B) Phenol < water < ethanol < p-nitrophenol (C) Water < ethanol < phenol < p-nitrophenol (D) Ethanol < phenol < water < p-nitrophenol

Solution

Answer: (A) Ethanol < water < phenol < p-nitrophenol. Acidity means ability to donate H+. Ethanol is the weakest acid (pKa ~16-18) — no stabilisation of ethoxide ion. Water (pKa 15.7) is slightly more acidic than ethanol. Phenol (pKa 10) — phenoxide ion is stabilised by resonance with benzene ring. p-Nitrophenol (pKa 7.1) — the -NO2 group at para position withdraws electron density from the ring and specifically delocalises the negative charge of phenoxide through resonance onto the oxygen atoms of NO2, providing extra stabilisation. So the order is: ethanol < water < phenol < p-nitrophenol.
2

NEET-style problem · Oxidation of alcohols

Question

Which of the following reagents converts a primary alcohol directly to an aldehyde without further oxidation to carboxylic acid? (A) KMnO4/H2SO4 (B) K2Cr2O7/H2SO4 (C) PCC (pyridinium chlorochromate) (D) CrO3

Solution

Answer: (C) PCC (pyridinium chlorochromate). KMnO4 and K2Cr2O7 are strong oxidising agents that will oxidise a primary alcohol all the way to a carboxylic acid (aldehyde → acid is fast with these agents). PCC is a mild oxidising agent — it oxidises primary alcohols to aldehydes but stops there; it does not have enough oxidising power to convert the aldehyde to a carboxylic acid. PCC also oxidises secondary alcohols to ketones. Tertiary alcohols are not oxidised by PCC. CrO3 in acetic acid (Jones reagent) oxidises primary alcohols to acids.
3

NEET-style problem · Ether cleavage

Question

When methyl tert-butyl ether (CH3-O-C(CH3)3) reacts with HI, the products are: (A) CH3I + (CH3)3COH (B) CH3OH + (CH3)3CI (C) 2-methylpropan-2-ol + CH3I (D) Isobutylene + CH3OH + HI

Solution

Answer: (B) CH3OH + (CH3)3CI. Ether cleavage with HI proceeds by: first protonation of O to give oxonium ion, then attack of I-. The I- will attack preferentially at the carbon that forms the most stable carbocation — the tert-butyl group gives a 3° carbocation (SN1 pathway). I- then attacks this 3° carbocation to give (CH3)3CI (tert-butyl iodide), while the methyl group is released as methanol (CH3OH). Note: the more substituted C-O bond breaks (SN1 at 3° C), not the methyl C-O bond.

Summary Cheat Sheet

Boiling Point Order (same MW)

Alcohol > Ether Alkane. Reason: H-bonding in alcohol. Ethers have only weak van der Waals, similar to alkanes.

Acidity Order

p-nitrophenol > phenol > water > cyclohexanol > 3° alcohol > 2° alcohol > 1° alcohol

Rule: EWG on phenol ring → more acidic. EDG on phenol ring → less acidic. Alkyl groups on C-bearing OH → less acidic (via alkoxide stability).

Oxidation Quick Reference

  • 1° alcohol + PCC → aldehyde (stopped)
  • 1° alcohol + KMnO4 → carboxylic acid
  • 2° alcohol + any oxidant → ketone
  • 3° alcohol + mild/moderate oxidant → no reaction

Williamson Synthesis Rule

Use alkoxide (strong base) + primary alkyl halide (SN2). Never use alkoxide + tertiary alkyl halide (E2 elimination occurs instead).

Key Named Reactions

  • Kolbe: phenoxide + CO2 (125°C, pressure) → salicylic acid (2-OH benzoic acid)
  • Reimer-Tiemann: phenol + CHCl3 + NaOH → salicylaldehyde (via dichlorocarbene)
  • Williamson: R-ONa + R'X → R-O-R' (ether synthesis, SN2)

Frequently asked questions

Why is the boiling point of ethanol (78°C) so much higher than dimethyl ether (-24°C) even though both have the same molecular formula C2H6O?

Both compounds are isomers with the same molecular mass (46 g/mol). The difference is due to intermolecular hydrogen bonding. Ethanol (CH3CH2OH) has an O-H bond — the hydrogen is partially positive and the oxygen of a neighbouring molecule is partially negative. This O-H...O hydrogen bond is a relatively strong intermolecular force (20-40 kJ/mol) that must be broken when ethanol evaporates, requiring much more energy and therefore a higher boiling point. Dimethyl ether (CH3-O-CH3) has no O-H bond. It cannot form hydrogen bonds with itself — only weak van der Waals forces hold its molecules together. So it boils at -24°C, much lower than ethanol. This principle applies to all alcohol vs ether comparisons.

Why is phenol more acidic than ethanol? What do electron-withdrawing groups do to phenol acidity?

When phenol (C6H5OH) donates a proton, it forms the phenoxide ion (C6H5O-). The negative charge on oxygen is delocalised into the benzene ring through resonance — five resonance structures spread the charge, stabilising the phenoxide ion. This stabilisation makes it easier to form the phenoxide (more acidic, pKa ~10). Ethanol forms an ethoxide ion (C2H5O-) where the negative charge stays on oxygen with no resonance stabilisation — it is less stable, so ethanol is less acidic (pKa ~16). Electron-withdrawing groups (like -NO2) at ortho or para positions on the phenol ring further delocalise the negative charge of the phenoxide ion through resonance onto the EWG oxygens. This extra stabilisation makes phenol even more acidic. For example, 2,4,6-trinitrophenol (picric acid) is as strong as mineral acids!

What is the Williamson ether synthesis and why must the alkyl halide be primary?

Williamson ether synthesis makes an ether by reacting a sodium alkoxide (R-O-Na) with an alkyl halide (R'X): R-ONa + R'X → R-O-R' + NaX. The alkoxide is a strong base/nucleophile. The alkyl halide must be primary (or methyl) because the reaction proceeds by SN2 — the alkoxide attacks the back face of the alkyl carbon. If R'X is secondary or tertiary, the alkoxide will instead act as a base and cause E2 elimination to give an alkene, not an ether. So for making CH3-O-C(CH3)3 (methyl tert-butyl ether, MTBE): use (CH3)3C-ONa (tert-butoxide) + CH3I (methyl iodide), NOT CH3ONa + (CH3)3CBr (which would give isobutylene).

How does the Lucas test distinguish between primary, secondary, and tertiary alcohols?

The Lucas test uses a mixture of anhydrous zinc chloride (ZnCl2) and concentrated hydrochloric acid. The alcohol reacts to form an alkyl chloride (insoluble in the reagent), which appears as turbidity (cloudiness). The test works by SN1: ZnCl2 coordinates with the -OH of the alcohol, making it a better leaving group (as Zn-OH complex). Then Cl- attacks. Tertiary alcohols form stable 3° carbocations immediately — turbidity appears in seconds. Secondary alcohols form 2° carbocations more slowly — turbidity appears in about 5 minutes. Primary alcohols form very unstable 1° carbocations — no turbidity at room temperature (reaction requires heating). The test is only useful for alcohols up to about C5 because larger alcohols are insoluble in water and cannot be tested this way.

What is the difference between Kolbe synthesis and Reimer-Tiemann reaction?

Both reactions start with phenol (phenoxide) and introduce a group onto the ring, but they are completely different. Kolbe synthesis (Kolbe-Schmitt): sodium phenoxide + CO2 gas (125°C, 5 atm) → sodium salicylate (sodium 2-hydroxybenzoate) → salicylic acid on acidification. The product is a carboxylic acid at the ortho position. Salicylic acid is the starting material for aspirin. Reimer-Tiemann reaction: phenol + chloroform (CHCl3) + NaOH (warm) → 2-hydroxybenzaldehyde (salicylaldehyde) as major product, plus some 4-hydroxybenzaldehyde. The reactive intermediate is dichlorocarbene (:CCl2), generated from CHCl3 + NaOH. It reacts with the activated phenoxide ring (highly nucleophilic at ortho/para) to introduce a -CHO group. The product is an aldehyde at the ortho position.

When a mixed ether like methyl tert-butyl ether reacts with HI, which C-O bond breaks and why?

When methyl tert-butyl ether (CH3-O-C(CH3)3) reacts with HI, the tert-butyl C-O bond breaks, not the methyl C-O bond. The mechanism is: (1) O is protonated by HI to give an oxonium ion. (2) I- now acts as nucleophile or the C-O bond breaks. The deciding factor is which carbocation is more stable: CH3+ (methyl, very unstable) vs (CH3)3C+ (tertiary, very stable). The system chooses to break the tert-butyl C-O bond and form the stable 3° carbocation (SN1). I- then attacks this carbocation to give (CH3)3CI (tert-butyl iodide), and CH3OH (methanol) is released. General rule for ether cleavage: I- attacks the more substituted carbon if it can form a stable carbocation (SN1), OR attacks the less hindered carbon if the substrate is primary (SN2).

How many questions from Alcohols, Phenols and Ethers come in NEET, and which topics are tested most?

You can expect 2-4 questions from this chapter each year in NEET. The most frequently tested topics are: (1) Acidity comparison — ordering phenol, substituted phenols, alcohols, water. Know that EWG increases acidity and EDG decreases it. (2) Oxidation of alcohols — what product forms with PCC vs KMnO4, and 1° vs 2° vs 3° alcohols. (3) Named reactions — Kolbe, Reimer-Tiemann, Williamson synthesis. Know products and conditions clearly. (4) Lucas test — which alcohol reacts when. (5) Ether cleavage by HI — which C-O bond breaks and why. Practise these specific question types and you will cover the bulk of what NEET tests.

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