Alcohols, Phenols & Ethers
Complete Organic Chemistry Revision Guide for CBSE
1. Introduction & Classification
Alcohols and phenols are formed when a hydrogen atom in a hydrocarbon (aliphatic and aromatic respectively) is replaced by an -OH (hydroxyl) group. Ethers are formed when a hydrogen atom in a hydrocarbon is replaced by an -OR or -OAr (alkoxy or aryloxy) group.
- Alcohols ($R-OH$): Classified as primary (1°), secondary (2°), or tertiary (3°) based on whether the -OH group is attached to a primary, secondary, or tertiary carbon atom.
- Phenols ($Ar-OH$): The -OH group is directly attached to an $sp^2$ hybridized carbon of an aromatic ring. (Note: Benzyl alcohol, $C_6H_5CH_2OH$, is an alcohol, not a phenol, because the -OH is attached to an $sp^3$ carbon).
- Ethers ($R-O-R'$): Classified as symmetrical (simple) if $R = R'$ (e.g., diethyl ether), and unsymmetrical (mixed) if $R \neq R'$ (e.g., ethyl methyl ether).
2. Master Cheat Sheet: Name Reactions & Key Mechanisms
A. Golden Name Reactions (Phenols)
- 1. Reimer-Tiemann Reaction: Treatment of phenol with chloroform in the presence of aqueous NaOH at 340 K, followed by hydrolysis, gives Salicylaldehyde (2-hydroxybenzaldehyde).
$$C_6H_5OH + CHCl_3 + 3NaOH \xrightarrow{340\text{ K}} C_6H_4(OH)CHO + 3NaCl + 2H_2O$$
- 2. Kolbe's Reaction: Phenol reacts with NaOH to form sodium phenoxide, which undergoes electrophilic substitution with $CO_2$ under pressure (4-7 atm) at 400 K to form Salicylic acid (2-hydroxybenzoic acid).
$$C_6H_5ONa + CO_2 \xrightarrow[H^+]{400\text{ K, 4-7 atm}} C_6H_4(OH)COOH$$
- 3. Cumene Process (Industrial prep of Phenol): Cumene (isopropylbenzene) is oxidized in air to cumene hydroperoxide, which on treatment with dilute acid yields Phenol and Acetone (a valuable byproduct).
B. Golden Reactions (Alcohols & Ethers)
- 4. Hydroboration-Oxidation: Alkenes react with diborane ($B_2H_6$) followed by oxidation with $H_2O_2$ in alkaline medium to give alcohols. Addition follows Anti-Markovnikov's rule.
$$CH_3-CH=CH_2 \xrightarrow[\text{2. } H_2O_2, OH^-]{\text{1. } B_2H_6} CH_3-CH_2-CH_2-OH \quad (\text{Propan-1-ol})$$
- 5. Williamson Ether Synthesis: Reaction of an alkyl halide with sodium alkoxide to form an ether. Follows $S_N2$ mechanism.
$$R-X + R'-O^-Na^+ \rightarrow R-O-R' + NaX$$Rule: Alkyl halide MUST be primary (1°). If tertiary (3°) alkyl halide is used, elimination dominates to form an alkene.
3. Exhaustive Theoretical Concepts
3.1 Preparation of Alcohols
From Alkenes:
- Acid Catalyzed Hydration: Follows Markovnikov's rule.
Mechanism (Crucial for CBSE):- Protonation of alkene to form a carbocation by electrophilic attack of $H_3O^+$.
- Nucleophilic attack of water on the carbocation.
- Deprotonation to form an alcohol.
- Hydroboration-Oxidation: Yields primary alcohols (Anti-Markovnikov product).
From Carbonyl Compounds:
- Reduction: Aldehydes reduce to 1° alcohols, ketones to 2° alcohols (using $LiAlH_4$, $NaBH_4$, or catalytic hydrogenation $H_2/Ni$).
- From Grignard Reagents ($RMgX$):
Formaldehyde ($HCHO$) + $RMgX \rightarrow$ 1° Alcohol.
Any other Aldehyde ($RCHO$) + $RMgX \rightarrow$ 2° Alcohol.
Ketone ($RCOR$) + $RMgX \rightarrow$ 3° Alcohol.
3.2 Acidity of Alcohols and Phenols
Why is Phenol more acidic than Alcohol?
Reason: In phenol, the -OH group is attached to an $sp^2$ hybridized carbon, which is more electronegative than the $sp^3$ carbon in alcohols, aiding the release of the $H^+$ ion. More importantly, after losing a proton, phenol forms the phenoxide ion. The phenoxide ion is highly stabilized by resonance (delocalization of the negative charge over the benzene ring). In contrast, the alkoxide ion ($R-O^-$) from alcohols is localized and destabilized by the +I (electron-donating) effect of the alkyl group.
Effect of Substituents on Acidity of Phenol:
- Electron Withdrawing Groups (EWG): Groups like $-NO_2$, $-CN$, $-X$ increase acidic strength by stabilizing the phenoxide ion through -R and -I effects. The effect is most pronounced at ortho and para positions. E.g., p-Nitrophenol is more acidic than phenol.
- Electron Donating Groups (EDG): Groups like $-CH_3$, $-OCH_3$, $-NH_2$ decrease acidic strength by destabilizing the phenoxide ion (+I, +R effects). E.g., Cresols are less acidic than phenol.
3.3 Chemical Reactions of Alcohols
- Oxidation:
1° Alcohols $\xrightarrow{PCC}$ Aldehydes $\xrightarrow{KMnO_4}$ Carboxylic Acids.
2° Alcohols $\xrightarrow{CrO_3}$ Ketones.
3° Alcohols generally do not undergo oxidation under normal conditions (require strong oxidizing agents at elevated temps to cleave C-C bonds). - Dehydration (Heating with conc. $H_2SO_4$ at 443 K): Forms alkenes. Ease of dehydration: 3° > 2° > 1° (due to stability of carbocation intermediate).
- Lucas Test (To distinguish 1°, 2°, 3° alcohols): Reagent is conc. $HCl + \text{anhydrous } ZnCl_2$.
3° alcohol $\rightarrow$ immediate turbidity (cloudiness).
2° alcohol $\rightarrow$ turbidity after 5 minutes.
1° alcohol $\rightarrow$ no turbidity at room temperature.
3.4 Reactions of Ethers
Ethers are highly unreactive, but they undergo cleavage of the C-O bond under drastic conditions with excess of hydrogen halides ($HI$ or $HBr$).
Cleavage of Ethers by HI (CBSE Favorite Rule)
Rule 1 (For 1° and 2° alkyl groups): The reaction follows $S_N2$ mechanism. The nucleophile ($I^-$) attacks the smaller (less sterically hindered) alkyl group.
$CH_3-O-CH_2CH_3 + HI \rightarrow CH_3I + CH_3CH_2OH$
Rule 2 (If one group is 3° tertiary): The reaction follows $S_N1$ mechanism. The halide attaches to the tertiary group due to the formation of a highly stable 3° carbocation.
$(CH_3)_3C-O-CH_3 + HI \rightarrow (CH_3)_3C-I + CH_3OH$
Rule 3 (Alkyl Aryl Ethers e.g., Anisole): The $O-Ar$ bond has partial double bond character due to resonance and is stronger. Cleavage always happens at the $O-R$ bond. Thus, it always yields Phenol and an Alkyl Halide.
$C_6H_5-O-CH_3 + HI \rightarrow C_6H_5OH (\text{Phenol}) + CH_3I$
4. Most Important CBSE Board Questions
Strictly aligned with the latest CBSE sample papers and marking scheme.
Section A: 1 Mark Questions (MCQs & A-R)
Q1. Which of the following reacts fastest with Lucas reagent? 1 Mark
- Butan-1-ol
- Butan-2-ol
- 2-Methylpropan-1-ol
- 2-Methylpropan-2-ol
Explanation: Lucas test involves the formation of a carbocation. 2-Methylpropan-2-ol is a tertiary (3°) alcohol, which forms a highly stable 3° carbocation, reacting almost instantaneously to give turbidity.
Q2. The correct order of acidic strength is: 1 Mark
- Phenol < p-cresol < p-nitrophenol
- p-cresol < Phenol < p-nitrophenol
- p-nitrophenol < Phenol < p-cresol
- Phenol < p-nitrophenol < p-cresol
Explanation: Electron Withdrawing Groups ($-NO_2$) increase acidity by stabilizing the phenoxide ion. Electron Donating Groups ($-CH_3$ in cresol) decrease acidity by destabilizing the phenoxide ion via +I effect.
Assertion-Reasoning
Q3. Assertion (A): Ether formation by Williamson synthesis using a tertiary alkyl halide fails.
Reason (R): Alkoxides are strong bases and they favor elimination over substitution with 3° alkyl halides. 1 Mark
Explanation: Williamson synthesis is an $S_N2$ reaction. 3° alkyl halides are extremely sterically hindered. Alkoxides (like $CH_3O^-$) act as strong bases instead of nucleophiles, extracting a proton and causing $\beta$-elimination to form an alkene instead of an ether.
Section B: 2 Mark Questions (Very Short Answer)
Q4. How will you convert:
(i) Propene to Propan-2-ol
(ii) Phenol to 2,4,6-tribromophenol 2 Marks
$$CH_3-CH=CH_2 + H_2O \xrightarrow{H^+} CH_3-CH(OH)-CH_3$$
(ii) Treat phenol with bromine water ($Br_2(aq)$). The highly activating -OH group causes trisubstitution immediately, forming a white precipitate.
$$C_6H_5OH + 3Br_2(aq) \rightarrow 2,4,6\text{-tribromophenol} \downarrow + 3HBr$$
Q5. Write the mechanism of hydration of ethene to yield ethanol. 2 Marks
Step 1: Protonation of ethene to form a carbocation by electrophilic attack of $H_3O^+$.
$$CH_2=CH_2 + H_3O^+ \rightleftharpoons CH_3-CH_2^+ + H_2O$$
Step 2: Nucleophilic attack of water on the carbocation.
$$CH_3-CH_2^+ + H_2O \rightleftharpoons CH_3-CH_2-O^+H_2$$
Step 3: Deprotonation to form ethanol.
$$CH_3-CH_2-O^+H_2 + H_2O \rightleftharpoons CH_3-CH_2-OH + H_3O^+$$
Section C: 3 Mark Questions (Short Answer)
Q6. Give simple chemical tests to distinguish between the following pairs of compounds:
(i) Phenol and Ethanol
(ii) Propan-1-ol and Propan-2-ol 3 Marks
Phenol reacts with neutral $FeCl_3$ to give a characteristic violet/purple color due to the formation of an iron-phenol complex. Ethanol does not give this test.
(ii) Propan-1-ol (1°) and Propan-2-ol (2°): Use the Iodoform test (or Lucas test).
Propan-2-ol contains the $CH_3-CH(OH)-$ group. When treated with $I_2$ and $NaOH$ (or $NaOI$), it gives a yellow precipitate of Iodoform ($CHI_3$). Propan-1-ol does not contain this group and gives no precipitate.
*(Alternatively, Lucas test: 2° gives turbidity in 5 mins, 1° gives no turbidity at room temp).*
Q7. Write the major product(s) in the following reactions:
(i) $CH_3-CH_2-O-CH_3 + HI \rightarrow$
(ii) $(CH_3)_3C-O-CH_3 + HI \rightarrow$
(iii) Anisole ($C_6H_5OCH_3$) + $HI \rightarrow$ 3 Marks
Product: $CH_3I$ (Iodomethane) + $CH_3CH_2OH$ (Ethanol).
(ii) Contains a tertiary alkyl group. Reaction follows $S_N1$. Halide attaches to the 3° carbocation.
Product: $(CH_3)_3C-I$ (tert-Butyl iodide) + $CH_3OH$ (Methanol).
(iii) Alkyl aryl ether. The $O-Ar$ bond is exceptionally strong due to resonance. Cleavage occurs at the $O-CH_3$ bond.
Product: $C_6H_5OH$ (Phenol) + $CH_3I$ (Iodomethane).
Section D: 4 Mark Case-Based Questions
Q8. Read the passage given below and answer the following questions:
(i) Why do ethers have lower boiling points than isomeric alcohols? (1 Mark)
(ii) Arrange the following in increasing order of their boiling points: Pentan-1-ol, n-butane, Pentanal, Ethoxyethane. (1 Mark)
(iii) Explain why o-nitrophenol is steam volatile but p-nitrophenol is not. (2 Marks)
(ii) Increasing order of boiling point: n-butane (alkane, weak Van der Waals) < Ethoxyethane (ether, dipole-dipole) < Pentanal (aldehyde, stronger dipole) < Pentan-1-ol (alcohol, strong H-bonding).
(iii) In o-nitrophenol, the -OH and -NO₂ groups are close to each other, forming intramolecular hydrogen bonding (within the same molecule). This prevents association with other molecules, making it steam volatile. In p-nitrophenol, the groups are far apart, leading to intermolecular hydrogen bonding (between different molecules), causing molecular association and a much higher boiling point, hence it is not steam volatile.
Section E: 5 Mark Long Answer Questions
Q9. (a) Write the chemical equations for the following name reactions:
(i) Reimer-Tiemann reaction
(ii) Kolbe's reaction
(iii) Williamson ether synthesis (3 Marks)
(b) An organic compound (A) with molecular formula $C_6H_6O$ gives a characteristic colour with neutral $FeCl_3$. When (A) is treated with $CO_2$ and $NaOH$ at 400 K under pressure, it gives compound (B), which on acidification forms compound (C). Compound (C) reacts with acetyl chloride to form a popular painkiller (D). Identify A, B, C, and D. (2 Marks)
(i) Reimer-Tiemann:
$$C_6H_5OH + CHCl_3 + 3NaOH \xrightarrow{340\text{ K}} C_6H_4(OH)CHO (\text{Salicylaldehyde}) + 3NaCl + 2H_2O$$
(ii) Kolbe's Reaction:
$$C_6H_5OH \xrightarrow{NaOH} C_6H_5ONa \xrightarrow[\text{400K, 4-7 atm}]{\text{1. } CO_2, \text{ 2. } H^+} C_6H_4(OH)COOH (\text{Salicylic acid})$$
(iii) Williamson ether synthesis:
$$CH_3CH_2Br + CH_3CH_2O^-Na^+ \rightarrow CH_3CH_2-O-CH_2CH_3 (\text{Diethyl ether}) + NaBr$$
(b) Identification of Compounds:
1. Compound (A) ($C_6H_6O$) gives a color with $FeCl_3$, meaning it is a phenol. (A) = Phenol.
2. Phenol reacting with $CO_2/NaOH$ is Kolbe's reaction. It first forms Sodium salicylate (B).
3. Acidification of (B) yields Salicylic acid (C).
4. Salicylic acid reacts with acetyl chloride (acetylation) to form Acetylsalicylic acid, which is famously known as Aspirin (D).
A = Phenol
B = Sodium salicylate
C = Salicylic acid (2-Hydroxybenzoic acid)
D = Aspirin (Acetylsalicylic acid)
Mastering 'Alcohols, Phenols & Ethers'
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The "Rule of Ethers" Trap: When asked to write the product of an ether with $HI$, stop and check the alkyl groups. If one is 3° (tertiary), the Iodine goes to the 3° group ($S_N1$). For 1° and 2°, Iodine goes to the smaller group ($S_N2$). If it's Anisole, ALWAYS form Phenol!
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Learn the Hydration Mechanism: The 3-step mechanism of Acid-catalyzed hydration of ethene to ethanol is one of the most frequently asked mechanism questions in the entire organic chemistry syllabus. Memorize the arrow pushing.
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Acidity Reasoning is Crucial: Be prepared to explain why phenol is more acidic than ethanol (resonance of phenoxide) and how $-NO_2$ increases this acidity (ortho/para).
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Name Reaction Chain: The preparation of Aspirin from Phenol (via Kolbe's reaction) is a standard 5-mark chain reaction. Ensure you know the structures of Salicylic acid and Aspirin.
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