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Properties Reactions Alcohols

🎓 Class 12 Chemistry CBSE Theory Ch 7 – Alcohols, Phenols and Ethers ⏱ ~14 min
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Properties Reactions Alcohols

7.6 Physical Properties

All three families — alcohols, phenols and ethers — contain a C–O bond, yet their physical behaviour splits cleanly into two camps depending on whether a hydrogen sits on oxygen. Alcohols and phenols carry an –OH group and are powerful hydrogen-bond donors and acceptors; ethers carry no O–H and can only accept hydrogen bonds, not donate them.

7.6.1 Boiling Points

Intermolecular hydrogen bonding between molecules of an alcohol creates large clusters that must be separated before boiling can occur. The resulting boiling points are far higher than those of alkanes of comparable molar mass, and also much higher than those of the analogous ethers which have no O–H.

CompoundMolar mass (g mol⁻¹)Boiling point (°C)H-bonding?
n-Butane CH₃(CH₂)₂CH₃58–0.5No
Diethyl ether C₂H₅OC₂H₅7434.6Acceptor only
Propan-1-ol CH₃CH₂CH₂OH6097Donor + acceptor
Ethane-1,2-diol HOCH₂CH₂OH62197Two –OH → extensive H-bonds
Phenol C₆H₅OH94182Yes
CH₃–O–H O–H | CH₃ O–H | CH₃ CH₃–O–H Hydrogen bonds (≈ 20 kJ mol⁻¹)
Fig 7.6: Inter-molecular hydrogen bonding in liquid methanol. Breaking the network absorbs additional energy and raises the boiling point.

7.6.2 Solubility in Water

Lower alcohols (methanol, ethanol, propan-1-ol, tert-butanol) are miscible with water because their –OH inserts into the water H-bond network. As the hydrocarbon chain lengthens the hydrophobic part dominates and solubility falls — decan-1-ol is practically insoluble. Phenol is only moderately soluble (≈ 8.3 g per 100 g H₂O at 20 °C); the large aromatic ring repels water. Phenol dissolves freely in most organic solvents.

7.7 Chemical Properties of Alcohols

The reactions of an alcohol fall into three groups sorted by which bond breaks: the O–H bond, the C–O bond, or the C–H bond α to the oxygen (oxidation/dehydrogenation).

7.7.1 Reactions Involving Cleavage of the O–H Bond

(a) Acidity

Alcohols are very weak acids, roughly as acidic as water (pKa ≈ 16–18). The equilibrium R–OH ⇌ R–O⁻ + H⁺ lies far to the left. Adding an electron-donating alkyl group destabilises the alkoxide anion, so acidity decreases along the series:

Acidity order: H₂O > 1° R–OH > 2° R–OH > 3° R–OH.

(b) Reaction with active metals

Sodium, potassium and aluminium liberate hydrogen gas from an alcohol, giving the corresponding alkoxide:

2 R–OH + 2 Na → 2 R–O⁻Na⁺ + H₂ ↑
6 C₂H₅OH + 2 Al → 2 Al(OC₂H₅)₃ + 3 H₂ ↑

(c) Esterification

In the presence of concentrated H₂SO₄, an alcohol and a carboxylic acid give an ester and water in an equilibrium that can be driven forward by removing water.

R–OH + R'–COOH  conc. H₂SO₄, Δ⇌ R'–COO–R + H₂O

Mechanism: protonation of the carbonyl activates it towards nucleophilic addition by the alcohol; a tetrahedral intermediate then loses water and a proton to deliver the ester. With an acid chloride or anhydride the reaction is much faster and essentially irreversible:

R–OH + R'COCl → R'COOR + HCl   (pyridine absorbs HCl)

7.7.2 Reactions Involving Cleavage of the C–O Bond

(a) With HX — substitution

Protonation converts the poor leaving group –OH into water (the best leaving group). The halide then displaces water: 3° alcohols go by SN1 (carbocation), 1° alcohols by SN2. The classic Lucas test exploits this rate difference:

Lucas test (conc. HCl + anhydrous ZnCl₂ at room temperature): 3° alcohol produces an immediate cloudy solution; 2° alcohol turns cloudy in 5–10 min; 1° alcohol does not react noticeably at room temperature. Reactivity order: 3° > 2° > 1°.

(b) With PCl₅, PCl₃, SOCl₂

R–OH + PCl₅ → R–Cl + POCl₃ + HCl
3 R–OH + PCl₃ → 3 R–Cl + H₃PO₃
R–OH + SOCl₂ → R–Cl + SO₂↑ + HCl↑   (cleanest — both by-products are gases)

(c) Dehydration to alkenes

Concentrated H₂SO₄ at 443 K, or passage over alumina (Al₂O₃) at 623 K, converts an alcohol into an alkene by loss of water. Reactivity order of alcohols toward dehydration follows carbocation stability: 3° > 2° > 1°. When more than one alkene is possible, the more substituted (Saytzeff) product dominates.

CH₃CH₂–OH + H⁺ CH₃CH₂–O⁺H₂ –H₂O CH₃CH₂⁺ (1°, poor; only with primary acid-catalysed) CH₃CH₂⁺ (or via concerted E2) –H⁺ from β-C CH₂=CH₂ + H₃O⁺ Sequence: protonation → loss of water → loss of β-proton → alkene. For 1° alcohols the route is essentially E2 (concerted), for 3° it is E1 through a carbocation.
Fig 7.7: Acid-catalysed dehydration of ethanol. Conc. H₂SO₄ (443 K) gives ethene.

7.7.3 Oxidation and Dehydrogenation

Primary alcohols climb the "oxidation ladder" in two steps: first to an aldehyde (lose 2 H), then to a carboxylic acid (gain O / lose 2 H). Secondary alcohols stop at the ketone. Tertiary alcohols have no α-hydrogen to remove and therefore resist oxidation under ordinary conditions — strong reagents will cleave a C–C bond and break the carbon skeleton.

R–CH₂OH (1°) PCC R–CHO KMnO₄ / K₂Cr₂O₇ R–COOH R₂CH–OH (2°) K₂Cr₂O₇ / CrO₃ R₂C=O (ketone) stops here (no α-H gone) R₃C–OH (3°) no easy oxidation strong oxidant breaks C–C bond
Fig 7.8: Oxidation ladder. PCC (pyridinium chlorochromate) is the cleanest way to stop at the aldehyde; KMnO₄/K₂Cr₂O₇ drives on to the acid.

Dehydrogenation (passing the alcohol vapour over hot copper, 573 K) removes two hydrogens without adding oxygen:

R–CH₂OH  Cu, 573K→ R–CHO + H₂   (1° → aldehyde)
R₂CHOH  Cu, 573K→ R₂C=O + H₂   (2° → ketone)
R₃COH  Cu, 573K→ R₂C=CR' + H₂O   (3° → alkene, not carbonyl)

7.8 Chemical Properties of Phenols

7.8.1 Acidity — The Defining Feature

Phenol (pKa ≈ 10) is a million times more acidic than ethanol (pKa ≈ 16). The phenoxide ion is stabilised by resonance: the negative charge is delocalised onto the ortho and para carbons of the ring, spreading it over four atoms instead of localising it on one oxygen.

OI (O⁻ localised) OII (ortho, top) OIII (ortho, bottom) OIV (para) OV (O⁻ again) Negative charge delocalised over O, ortho × 2 and para carbons → stable phenoxide.
Fig 7.9: Five resonance structures of the phenoxide ion. In an alkoxide the negative charge is trapped on oxygen with no such delocalisation.

Substituent effects on phenol acidity:

  • Electron-withdrawing groups (–NO₂, –Cl, –CHO, –CN) especially at the ortho and para positions increase acidity by further stabilising the phenoxide. 4-nitrophenol pKa = 7.15; picric acid (2,4,6-trinitrophenol) pKa ≈ 0.4 — as strong as a mineral acid.
  • Electron-donating groups (–CH₃, –OCH₃, –OH, –NH₂) decrease acidity; p-cresol pKa ≈ 10.3.

Because phenol is considerably stronger than carbonic acid, aqueous NaOH — but not aqueous NaHCO₃ — deprotonates it to sodium phenoxide:

C₆H₅OH + NaOH → C₆H₅O⁻Na⁺ + H₂O   (alcohols need Na metal, not NaOH)

7.8.2 Esterification

Phenols react faster with acid chlorides or anhydrides than with carboxylic acids. Acetylation of salicylic acid gives aspirin — one of the biggest-selling pharmaceuticals of the twentieth century.

C₆H₅OH + (CH₃CO)₂O → C₆H₅OCOCH₃ + CH₃COOH   (phenyl ethanoate)

7.8.3 Electrophilic Aromatic Substitution on the Ring

The –OH lone pair donates electron density into the ring, activating it strongly and directing incoming electrophiles to the ortho and para positions.

(a) Nitration

C₆H₅OH + dil. HNO₃ → o-nitrophenol + p-nitrophenol

The two isomers are separated by steam distillation: the ortho isomer forms an intramolecular hydrogen bond (chelation between –OH and –NO₂) so it is volatile and distils over; the para isomer relies on intermolecular H-bonds, forms an associated polymer-like network, and stays behind in the flask.

C₆H₅OH + conc. HNO₃ → 2,4,6-trinitrophenol (picric acid) — yellow, explosive

(b) Halogenation

C₆H₅OH + Br₂ / CS₂ (low polar solvent, cold) → o- + p-bromophenol
C₆H₅OH + Br₂ (aq.) → 2,4,6-tribromophenol (white ppt) — qualitative test for phenol

(c) Kolbe's reaction

Sodium phenoxide — a stronger nucleophile than phenol — attacks CO₂ at the ortho position; workup gives salicylic acid, the starting point for aspirin.

C₆H₅ONa + CO₂  400K, 7 atm→ o-HOC₆H₄COONa  H⁺→ salicylic acid

(d) Reimer–Tiemann reaction

Phenol + chloroform in aq. NaOH produces a dichlorocarbene electrophile (:CCl₂). The carbene attacks the ortho position; hydrolysis of the intermediate benzal chloride gives an aldehyde — salicylaldehyde.

C₆H₅OH + CHCl₃ + 3 NaOH → o-HOC₆H₄CHO (salicylaldehyde) + 3 NaCl + 2 H₂O
C₆H₅–ONa sodium phenoxide CO₂ / 400K o-HO–C₆H₄–COOH salicylic acid (→ aspirin) Kolbe C₆H₅–OH CHCl₃ / NaOH (: CCl₂ generated) o-HO–C₆H₄–CHO salicylaldehyde (Reimer-Tiemann)
Fig 7.10: The Kolbe and Reimer–Tiemann reactions both install a carbon substituent ortho to the phenolic –OH.

7.8.4 Oxidation

Mild oxidation of phenol with Na₂Cr₂O₇/H₂SO₄ or chromic acid gives the bright-yellow dicarbonyl benzoquinone:

C₆H₅OH  Na₂Cr₂O₇ / H₂SO₄→ 1,4-benzoquinone (C₆H₄O₂)

Worked Examples

Example 7.8 — Acidity order

Arrange the following in decreasing order of acidity: ethanol, water, phenol, 4-nitrophenol, 2-methylpropan-2-ol.

Phenol pKa 10; 4-nitrophenol pKa 7.15; water 15.7; ethanol 15.9; tert-butanol 18. Hence:

4-nitrophenol > phenol > water > ethanol > 2-methylpropan-2-ol.

Example 7.9 — Lucas test prediction

Three unlabelled bottles contain propan-1-ol, propan-2-ol and 2-methylpropan-2-ol. How would you identify them using the Lucas reagent?

Add a few drops of each alcohol to Lucas reagent (conc. HCl + anhydrous ZnCl₂). The 3° alcohol (2-methylpropan-2-ol) turns cloudy immediately. The 2° alcohol (propan-2-ol) turns cloudy within 5–10 min. The 1° alcohol (propan-1-ol) shows no immediate change (needs warming).

Example 7.10 — Choose the reagent for controlled oxidation

How would you oxidise propan-1-ol to propanal without going to propanoic acid?

Use PCC (pyridinium chlorochromate) in anhydrous CH₂Cl₂. It stops cleanly at the aldehyde. KMnO₄ or K₂Cr₂O₇ in aqueous acid would oxidise further to propanoic acid.

Example 7.11 — Dehydration regiochemistry

Predict the major product when butan-2-ol is heated with conc. H₂SO₄ at 443 K.

The 2° cation CH₃–⁺CH–CH₂CH₃ can lose a β-H from either the methyl on C-1 or the methylene on C-3. Loss from C-3 gives the more substituted but-2-ene (Saytzeff product); loss from C-1 gives but-1-ene. Major: but-2-ene (both E and Z).

Example 7.12 — Why is o-nitrophenol steam-volatile?

The ortho isomer forms an intramolecular H-bond between the –OH and the adjacent –NO₂, so each molecule exists as a compact chelate and does not associate with its neighbours. The para isomer has no such chelation; instead it forms intermolecular H-bonds with other p-nitrophenol molecules, giving a higher effective molar mass in the liquid and a lower vapour pressure. The ortho isomer therefore distils over with steam and the para isomer remains.

Example 7.13 — Draw the Kolbe product

Phenol is treated with NaOH, then with CO₂ at 400 K / 7 atm, then acidified. Identify the product.

NaOH → sodium phenoxide; CO₂ attack at the ortho carbon of the phenoxide; acidification regenerates –OH and –COOH. Product: 2-hydroxybenzoic acid (salicylic acid). Acetylation of its –OH gives aspirin.

Example 7.14 — Ester mechanism

Write the tetrahedral-intermediate mechanism of the Fischer esterification of ethanoic acid and ethanol.

(i) H⁺ protonates the C=O of CH₃COOH. (ii) C₂H₅OH attacks the electrophilic carbonyl C; a tetrahedral intermediate with +OH₂ leaving group forms. (iii) Proton transfers give a tetrahedral species with –OH₂⁺; water leaves. (iv) Deprotonation delivers ethyl ethanoate CH₃COOC₂H₅. All steps reversible — excess alcohol or removal of water pushes the equilibrium forward.

Example 7.15 — Phenol identification test

Aqueous bromine is added to an unknown. A white precipitate appears immediately. Which functional group is likely present?

A phenol — it reacts with aqueous Br₂ to give a white precipitate of 2,4,6-tribromophenol. Alcohols do not give this reaction.

Activity 7.2 — Alcohol Classification via the Lucas ReagentL4 Analyse

You have three labelled alcohols (1 mL each): propan-1-ol, propan-2-ol, 2-methylpropan-2-ol. Add 3 mL Lucas reagent (conc. HCl + anhydrous ZnCl₂) to each tube.

Predict: Which tube goes cloudy first, which last, and which not at all at room temperature?
  1. Mix and note the time at which each mixture becomes cloudy.
  2. Record: immediate (< 1 min), slow (5–10 min) or no cloudiness at RT.
  3. Relate the rate to the class of alcohol and the stability of the carbocation intermediate.

2-methylpropan-2-ol (3°): immediate cloudiness (SN1 via stable 3° cation).

Propan-2-ol (2°): cloudiness in 5–10 min (SN1, less stable cation).

Propan-1-ol (1°): no cloudiness at RT; needs warming; would proceed by SN2.

Interactive — Alcohol Classification Quiz

Enter an alcohol structure (use SMILES-like shorthand or a common name) and see its class, Lucas-test result and oxidation product.

Result will appear here …

Competency-Based Questions — Properties & Reactions

A chemistry student is told to deduce the structure and reactions of a mystery alcohol X (C₄H₁₀O). She finds that X reacts with Lucas reagent to give an oily layer only after warming, and that K₂Cr₂O₇/H₂SO₄ oxidises it cleanly to a carboxylic acid.

1. Which of the following is compound X?

  • (a) butan-1-ol
  • (b) butan-2-ol
  • (c) 2-methylpropan-2-ol
  • (d) methoxypropane
(a) butan-1-ol. Only a 1° alcohol needs warming with Lucas reagent and is oxidised to a carboxylic acid by K₂Cr₂O₇.

2. Arrange in decreasing boiling point: diethyl ether, n-butane, butan-1-ol, propan-1-ol.

butan-1-ol (118 °C) > propan-1-ol (97 °C) > diethyl ether (35 °C) > n-butane (–0.5 °C). Alcohols H-bond; ether accepts only; butane has only London forces.

3. True/False: 2,4,6-trinitrophenol (picric acid) is a weaker acid than ethanol.

False. Three electron-withdrawing –NO₂ groups pull picric acid's pKa down to ≈ 0.4 — similar to a mineral acid — whereas ethanol pKa ≈ 16.

4. Fill in the blank: The aromatic aldehyde produced when phenol reacts with CHCl₃ in aq. NaOH is ________ , formed by the ________ reaction.

Salicylaldehyde (2-hydroxybenzaldehyde); Reimer–Tiemann reaction.

5. Why does phenol dissolve in aqueous NaOH but not in aqueous NaHCO₃, whereas a carboxylic acid dissolves in both?

Phenol (pKa 10) is stronger than water (pKa 15.7) but weaker than carbonic acid (pKa₁ ≈ 6.3). NaOH (base of very weak H₂O) deprotonates phenol; NaHCO₃ (base of H₂CO₃) cannot. Carboxylic acids (pKa ≈ 4–5) are stronger than H₂CO₃, so they dissolve in both.

Assertion–Reason Questions

Assertion (A): Phenol is more acidic than cyclohexanol.

Reason (R): The phenoxide ion is stabilised by resonance into the aromatic ring; the cyclohexoxide has no such stabilisation.

  • A. Both true, R explains A.
  • B. Both true, R does not explain A.
  • C. A true, R false.
  • D. A false, R true.
Answer: A. Five resonance structures delocalise the phenoxide negative charge; this is the dominant factor that lowers phenol's pKa.

Assertion (A): Tertiary alcohols are very difficult to oxidise under ordinary laboratory conditions.

Reason (R): Oxidation of an alcohol requires removal of an α-hydrogen, which is absent on the carbinol carbon of a 3° alcohol.

  • A. Both true, R explains A.
  • B. Both true, R does not explain A.
  • C. A true, R false.
  • D. A false, R true.
Answer: A. A 3° alcohol has the carbinol C bonded to three other carbons and no H — no dehydrogenation possible without breaking a C–C bond.

Assertion (A): o-nitrophenol can be separated from p-nitrophenol by steam distillation.

Reason (R): o-nitrophenol has stronger intermolecular hydrogen bonds than p-nitrophenol.

  • A. Both true, R explains A.
  • B. Both true, R does not explain A.
  • C. A true, R false.
  • D. A false, R true.
Answer: C. A is correct, but R is the opposite of the truth — o-nitrophenol has intramolecular H-bonds (chelation), which reduce intermolecular association, raising its volatility.

Frequently Asked Questions - Properties Reactions Alcohols

What is the main concept covered in Properties Reactions Alcohols?
In NCERT Class 12 Chemistry Chapter 7 (Alcohols, Phenols and Ethers), "Properties Reactions Alcohols" covers the core chemistry principles and reactions students need for board exam success. The MyAiSchool lesson explains the topic with definitions, structural diagrams, reaction mechanisms, worked examples, and interactive simulations. Key reactions, IUPAC names, and chemical reasoning are highlighted throughout aligned with CBSE 2025-26 syllabus.
How is Properties Reactions Alcohols useful in real-life or applied chemistry?
Real-life applications of "Properties Reactions Alcohols" from NCERT Class 12 Chemistry Chapter 7 include drug design, polymer industry, food chemistry, electrochemical cells, fuel cells, dyes/pigments, agrochemicals, and biochemistry. The MyAiSchool lesson links every concept to a tangible industrial or biological example so students see chemistry as a problem-solving framework for the molecular world.
What are the key reactions students should memorize for Properties Reactions Alcohols?
Key reactions in "Properties Reactions Alcohols" (NCERT Class 12 Chemistry Chapter 7 Alcohols, Phenols and Ethers) are tabulated in the MyAiSchool reaction map. Students should memorize each reaction with its reagent, conditions, mechanism class (SN1/SN2/E1/E2/electrophilic addition/etc), product, and stereochemistry. The Summary section provides a quick-reference reaction chart for last-minute revision.
How does this part connect to other parts of Chapter 7?
NCERT Class 12 Chemistry Chapter 7 (Alcohols, Phenols and Ethers) is structured so each part builds chemical understanding sequentially. "Properties Reactions Alcohols" connects to neighbouring parts via shared functional groups, reaction mechanisms, and structural concepts. The MyAiSchool lesson cross-references related concepts with internal links so students can navigate the whole chapter as one connected story rather than disconnected fragments.
What types of CBSE board questions come from Properties Reactions Alcohols?
CBSE board questions from "Properties Reactions Alcohols" typically include: (1) 1-mark MCQs on definitions and IUPAC naming, (2) 2-mark short-answer reactions/products, (3) 3-mark mechanism questions, (4) 5-mark long-answer combining mechanism + product + stereochemistry + application. The MyAiSchool lesson tags each Competency-Based Question (CBQ) with Bloom level (L1-L6) so students know how to study for each weight.
How can students use the interactive simulation effectively?
The interactive simulation in the "Properties Reactions Alcohols" lesson allows students to explore reaction outcomes, predict products, or compare reaction conditions, with live visual feedback. To use it effectively: (1) try every option/configuration, (2) compare with the analytical reasoning, (3) check IUPAC names and structural correctness, (4) test edge cases from worked examples. The simulation reinforces conceptual intuition that pure mechanism memorisation cannot provide.
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