The Three Oxygen-Containing Brothers
Alcohols, phenols, and ethers all carry an oxygen atom, but their personalities — and their chemistry — couldn’t be more different. Understanding why they behave differently is what separates a 60% student from a 90% student in this chapter.
The key distinction: alcohols have —OH attached to an sp³ carbon (saturated), phenols have —OH attached directly to a benzene ring, and ethers have oxygen sandwiched between two carbon groups (R—O—R’). That one structural difference cascades into completely different acidity, reactivity, and exam questions.
This chapter carries solid weightage in JEE Main (1-2 questions almost every year) and NEET (2-3 questions), and CBSE gives it a full 8-10 marks in Class 12 boards. Worth investing time here.
Key Terms & Definitions
Alcohol — Compound with —OH group attached to an aliphatic (sp³) carbon. General formula: R—OH.
Phenol — Compound with —OH group directly bonded to an aromatic ring. The benchmark compound is C₆H₅OH (carbolic acid, pKa ≈ 10).
Ether — Compound with the linkage R—O—R’. Can be symmetrical (diethyl ether) or unsymmetrical (methyl ethyl ether).
Primary, Secondary, Tertiary Alcohols:
- 1° alcohol: —OH carbon bears one alkyl group (e.g., ethanol, CH₃CH₂OH)
- 2° alcohol: —OH carbon bears two alkyl groups (e.g., isopropanol)
- 3° alcohol: —OH carbon bears three alkyl groups (e.g., tert-butanol)
The 1°/2°/3° classification is based on the carbon bearing —OH, not the total number of —OH groups. Students mix this up constantly.
Glycol — Diol (two —OH groups), e.g., ethylene glycol (HOCH₂CH₂OH), used as antifreeze.
Cleavage of Ether — Ethers resist most reagents but get cleaved by HI and HBr. This is a favourite exam reaction.
Preparation Methods
Preparation of Alcohols
From Alkenes (Hydration):
Markovnikov’s rule applies — the —OH goes to the more substituted carbon. This gives 2° and 3° alcohols from alkenes easily.
From Carbonyl Compounds (Reduction):
- Aldehyde + [H] → 1° alcohol
- Ketone + [H] → 2° alcohol
- LiAlH₄ in dry ether is the standard reducing agent for CBSE. NaBH₄ also works and is milder.
Grignard Synthesis — This is the heavy hitter for JEE:
JEE Main 2023 asked about the product when methylmagnesium bromide reacts with formaldehyde. Know which carbonyl compound gives which class of alcohol from Grignard — it’s asked almost every alternate year.
Preparation of Phenol
The industrial route from benzene goes: Benzene → Chlorobenzene → Phenol (via Dow process, using NaOH at high temperature and pressure). Academically, the important routes are:
From Diazonium Salt:
This is a high-yield CBSE question. The diazonium salt route is also used to introduce —OH into benzene rings in multi-step synthesis.
Preparation of Ethers (Williamson Synthesis)
For best results, use a 1° alkyl halide. With 3° alkyl halides, elimination (E2) dominates over substitution.
Students write Williamson synthesis with a 3° alkyl halide and expect an ether. Wrong — you get an alkene. The sodium alkoxide (RO⁻Na⁺) is a strong base, and 3° substrates eliminate. Always use 1° halide in Williamson.
Key Reactions
Reactions of Alcohols
With conc. H₂SO₄:
- At 443 K (170°C): Gives alkene (intramolecular dehydration)
- At 413 K (140°C): Gives ether (intermolecular dehydration)
Temperature controls the product — lower temperature, ether; higher temperature, alkene.
Esterification:
This is reversible (Fischer esterification). To push it forward, remove water or use excess of one reagent.
Lucas Test (distinguishing 1°, 2°, 3° alcohols):
- Reagent: ZnCl₂ + conc. HCl (Lucas reagent)
- 3° alcohol: Immediate turbidity (cloudiness)
- 2° alcohol: Turbidity in 5 minutes
- 1° alcohol: No turbidity at room temperature
The turbidity is due to formation of the insoluble alkyl chloride. 3° carbocations form fastest (most stable), hence immediate reaction.
Oxidation:
- 1° alcohol → Aldehyde → Carboxylic acid (using KMnO₄ or K₂Cr₂O₇)
- 2° alcohol → Ketone
- 3° alcohol: Resistant to mild oxidation (no H on the —OH carbon)
Reactions of Phenol
Acidity of Phenol vs Alcohol:
Phenol is far more acidic than alcohol. pKa of phenol ≈ 10, while ethanol ≈ 16. Why? The phenoxide ion (C₆H₅O⁻) is stabilised by resonance — the negative charge delocalises into the ring. Ethoxide has no such stabilisation.
Phenol reacts with NaOH (a weak base won’t work for alcohols). But phenol does NOT react with Na₂CO₃ or NaHCO₃ — it’s not acidic enough to liberate CO₂. Carboxylic acids react with NaHCO₃; phenols don’t. This is a classic differentiation question.
Electrophilic Aromatic Substitution (EAS):
The —OH group is a strong ortho/para director and ring activator. Phenol undergoes:
- Bromination: With Br₂ water (no catalyst needed), gives 2,4,6-tribromophenol immediately (white precipitate). Used to test for phenol.
- Nitration: With dilute HNO₃, gives a mixture of ortho and para nitrophenols.
- Kolbe’s Reaction: Phenol + CO₂ under pressure with NaOH gives sodium salicylate → salicylic acid (aspirin precursor).
- Reimer-Tiemann Reaction: Phenol + CHCl₃ + NaOH → salicylaldehyde (2-hydroxybenzaldehyde).
Kolbe’s and Reimer-Tiemann are NEET and CBSE favourites. JEE Main 2024 Shift 2 had a question on the product of Reimer-Tiemann reaction. Both reactions are specific to phenols, not alcohols.
Reactions of Ethers
Ethers are relatively unreactive — they don’t react with metals, bases, or most oxidising agents. What breaks them:
Cleavage with HI (or HBr):
With excess HI:
For unsymmetrical ethers (R—O—R’ where R ≠ R’), the halide goes to the smaller/less hindered alkyl group in SN2 conditions, or the more stable carbocation in SN1 (for 3° ethers).
Solved Examples
Example 1 — Easy (CBSE Level)
Q: Arrange in increasing order of acidity: ethanol, phenol, water, carbonic acid.
Solution: Acidity increases with the stability of the conjugate base (anion formed after losing H⁺).
- Ethanol → Ethoxide (no resonance stabilisation) — least acidic
- Water → Hydroxide (slightly better due to smaller size)
- Phenol → Phenoxide (resonance stabilised by ring) — more acidic than water
- Carbonic acid → HCO₃⁻ (stabilised, and inductive effect of two oxygens) — most acidic
Increasing order: Ethanol < Water < Phenol < Carbonic acid
Example 2 — Medium (JEE Main Level)
Q: What is the major product when isopropyl alcohol is treated with conc. H₂SO₄ at 443 K?
Solution: At high temperature (443 K), dehydration favours alkene formation.
Isopropyl alcohol (2° alcohol) loses water. The —OH leaves and a proton is lost from the adjacent carbon.
Product: Propene (CH₃—CH=CH₂)
This follows Zaitsev’s rule — the more substituted alkene is the major product. Since propene is the only possible alkene here, it’s straightforward.
If the question said 413 K, the answer would be di-isopropyl ether, not propene. Temperature changes the entire answer. Read carefully.
Example 3 — Hard (JEE Advanced Level)
Q: A Grignard reagent X reacts with formaldehyde to give compound Y. Y on oxidation with PCC gives Z, which gives a positive Tollens test. Identify X, Y, Z.
Solution: Work backwards.
Z gives positive Tollens test → Z is an aldehyde.
Z is obtained by oxidising Y with PCC (mild oxidant) → Y is a 1° alcohol (PCC converts 1° alcohol to aldehyde, stops there).
Y is a 1° alcohol made from Grignard + HCHO →
So Y = RCH₂OH, Z = RCHO.
The Grignard X = RMgBr (any Grignard reagent). If specific molecular formula is given, solve accordingly.
This multi-step reasoning — working backward from product properties — is exactly how JEE Advanced structures organic questions.
Exam-Specific Tips
CBSE Boards (8-10 marks): Focus on named reactions — Williamson synthesis, Kolbe’s, Reimer-Tiemann, Lucas test, Victor Meyer test. Write them with proper conditions. The 5-mark questions usually ask for mechanism OR preparation + 2 reactions. Don’t skip mechanisms.
JEE Main (1-2 questions): Acidity comparisons, Grignard products, ether cleavage by HI, and dehydration temperature effects are the highest-frequency topics. From 2019-2024, at least 6 questions came from ether cleavage and phenol EAS reactions alone.
NEET (2-3 questions): Kolbe’s reaction, Reimer-Tiemann, bromination of phenol (tribromophenol test), and Lucas test are NEET’s favourites. These are more reaction-identification based than mechanism-heavy.
Common Mistakes to Avoid
Mistake 1: Phenol reacts with NaHCO₃. It doesn’t. Only carboxylic acids are strong enough to displace CO₂ from NaHCO₃. Phenol reacts with NaOH but not NaHCO₃. This distinction appears in almost every CBSE paper.
Mistake 2: Using 3° alkyl halide in Williamson synthesis. Strong base + 3° substrate = elimination, not substitution. Always pair Williamson synthesis with 1° halides.
Mistake 3: Oxidising 3° alcohol with KMnO₄ expecting a product. Tertiary alcohols have no hydrogen on the carbon bearing —OH, so mild oxidation fails. With very strong oxidants under vigorous conditions, C—C bond breaks — but that’s not what your exam expects.
Mistake 4: Lucas test — confusing which gives immediate turbidity. 3° → immediate (within seconds), 2° → 5 minutes, 1° → no reaction at room temperature. Students flip 1° and 3°. Remember: tertiary carbocations form fastest.
Mistake 5: Dehydration temperature — swapping 413 K and 443 K. Lower temperature (413 K) → ether; higher temperature (443 K) → alkene. A memory trick: “Higher temperature, higher energy, bond-breaking to give alkene.”
Practice Questions
Q1. Give the IUPAC name of: CH₃—CH(OH)—CH₂—CH₃
Answer: Butan-2-ol
The longest chain with the —OH group has 4 carbons. Number from the end closest to —OH, which gives position 2.
Q2. Which alcohol gives immediate turbidity with Lucas reagent at room temperature?
Answer: Tertiary (3°) alcohol.
Lucas reagent (ZnCl₂ + conc. HCl) works by converting —OH to —Cl. Tertiary carbocations form readily via SN1, giving insoluble alkyl chloride (turbidity) immediately.
Q3. Why is phenol more acidic than cyclohexanol even though both have —OH groups?
Answer: In phenol, after losing H⁺, the phenoxide ion (C₆H₅O⁻) is stabilised by resonance — the negative charge delocalises over the ring through 4 resonance structures. Cyclohexoxide has no such stabilisation. Greater stability of the conjugate base = greater acidity of the parent compound.
Q4. What is the product of phenol with CHCl₃ and NaOH on heating, followed by acidification?
Answer: Salicylaldehyde (2-hydroxybenzaldehyde) — this is the Reimer-Tiemann reaction.
The CHO group gets introduced at the ortho position of phenol. The —OH group is an ortho/para director, and Reimer-Tiemann preferentially gives the ortho product.
Q5. Give one chemical test to distinguish between phenol and ethanol.
Answer: Add FeCl₃ solution.
Phenol gives a characteristic violet/purple colour with neutral FeCl₃. Ethanol gives no colour (or faint yellow). This is the standard CBSE answer. Alternatively: NaHCO₃ — phenol won’t react, carboxylic acids will (but this differentiates carboxylic acids from phenol, not ethanol).
Q6. Predict the product: methyl phenyl ether + excess HI →
Answer: Iodobenzene + methanol (first), then methanol + HI → iodomethane + water.
Wait — actually: methyl phenyl ether cleavage gives phenol + CH₃I. The C—O bond adjacent to the aryl group doesn’t break (aryl halides don’t form easily via SN2). So the alkyl—oxygen bond breaks: product is phenol + methyl iodide.
With excess HI, phenol does NOT convert to iodobenzene (Ar—OH doesn’t react with HI like alkyl alcohols do). So final products: phenol + iodomethane.
Q7. What happens when ethanol is heated with conc. H₂SO₄ at 170°C?
Answer: Ethene (ethylene) is produced.
At 170°C (443 K), intramolecular dehydration occurs — the alcohol molecule loses one water molecule to give an alkene.
Q8. A compound X (mol. formula C₄H₁₀O) does not react with sodium metal. What functional group is present?
Answer: Ether (—O—).
Alcohols react with sodium to give H₂ gas. If there’s no reaction with Na, it cannot be an alcohol or phenol. C₄H₁₀O with no reaction to Na → it’s diethyl ether (CH₃CH₂—O—CH₂CH₃).
FAQs
Q: What is the difference between phenol and carbolic acid?
They’re the same compound — C₆H₅OH. “Carbolic acid” is the older, common name for phenol. You’ll see both terms in older NCERT and reference books.
Q: Why can’t tertiary alcohols be oxidised easily?
The carbon bearing —OH in a 3° alcohol has no C—H bond available for oxidation. Common oxidants (KMnO₄, K₂Cr₂O₇) need to abstract a hydrogen from that carbon to form the carbonyl. No H, no easy oxidation.
Q: Is phenol a weak acid or strong acid?
Phenol is a weak acid (pKa ≈ 10). It’s stronger than alcohols and water, but much weaker than carboxylic acids (pKa ≈ 4-5) and carbonic acid (pKa ≈ 6.4). It partially ionises in water.
Q: Why is diethyl ether used as a solvent in Grignard reactions?
Diethyl ether is aprotic (no O—H or N—H bond to donate protons) and coordinates to the Mg through lone pairs, stabilising the Grignard reagent. Protic solvents would destroy the Grignard immediately by protonation.
Q: What does PCC do, and why is it preferred over KMnO₄ for oxidising 1° alcohols to aldehydes?
PCC (Pyridinium Chlorochromate) is a mild oxidant. It converts 1° alcohols to aldehydes and stops there — it cannot further oxidise the aldehyde to a carboxylic acid. KMnO₄ is too strong and takes the oxidation all the way to carboxylic acid. When the exam wants an aldehyde from a 1° alcohol, PCC is the answer.
Q: How do we distinguish between o-nitrophenol and p-nitrophenol using steam distillation?
Ortho-nitrophenol undergoes intramolecular hydrogen bonding (the —OH and —NO₂ are adjacent), so it doesn’t associate with water molecules much. It has a lower boiling point and is steam volatile. Para-nitrophenol forms intermolecular H-bonds with water, has a higher boiling point, and is not steam volatile. This is a common CBSE 3-mark explanation question.
Q: Which is more acidic — o-cresol or p-cresol?
Both have a methyl group (electron-donating) on the ring, which slightly decreases acidity compared to phenol. For comparison between ortho and para: ortho-cresol has a slightly different effect due to steric and field effects, making this comparison nuanced. For board exams, both are less acidic than phenol — the exact ortho vs para comparison isn’t typically tested at this level.