આ MCQ મોડ્યુલ આના પર આધારિત છે: Properties Reactions Alcohols
Properties Reactions Alcohols
આ મૂલ્યાંકન આના પર આધારિત હશે: Properties Reactions Alcohols
મૂલ્યાંકન બનાવવામાં તેમની સામગ્રી સામેલ કરવા ચિત્રો, PDF અથવા Word દસ્તાવેજ અપલોડ કરો.
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.
| Compound | Molar mass (g mol⁻¹) | Boiling point (°C) | H-bonding? |
|---|---|---|---|
| n-Butane CH₃(CH₂)₂CH₃ | 58 | –0.5 | No |
| Diethyl ether C₂H₅OC₂H₅ | 74 | 34.6 | Acceptor only |
| Propan-1-ol CH₃CH₂CH₂OH | 60 | 97 | Donor + acceptor |
| Ethane-1,2-diol HOCH₂CH₂OH | 62 | 197 | Two –OH → extensive H-bonds |
| Phenol C₆H₅OH | 94 | 182 | Yes |
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:
(b) Reaction with active metals
Sodium, potassium and aluminium liberate hydrogen gas from an alcohol, giving the corresponding alkoxide:
(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.
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:
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:
(b) With PCl₅, PCl₃, SOCl₂
(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.
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.
Dehydrogenation (passing the alcohol vapour over hot copper, 573 K) removes two hydrogens without adding oxygen:
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.
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:
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.
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
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.
(b) Halogenation
(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.
(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.
7.8.4 Oxidation
Mild oxidation of phenol with Na₂Cr₂O₇/H₂SO₄ or chromic acid gives the bright-yellow dicarbonyl benzoquinone:
Worked Examples
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.
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).
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.
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).
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.
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.
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.
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.
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.
- Mix and note the time at which each mixture becomes cloudy.
- Record: immediate (< 1 min), slow (5–10 min) or no cloudiness at RT.
- 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.
Competency-Based Questions — Properties & Reactions
1. Which of the following is compound X?
2. Arrange in decreasing boiling point: diethyl ether, n-butane, butan-1-ol, propan-1-ol.
3. True/False: 2,4,6-trinitrophenol (picric acid) is a weaker acid than ethanol.
4. Fill in the blank: The aromatic aldehyde produced when phenol reacts with CHCl₃ in aq. NaOH is ________ , formed by the ________ reaction.
5. Why does phenol dissolve in aqueous NaOH but not in aqueous NaHCO₃, whereas a carboxylic acid dissolves 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.
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.
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.
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