Alcohols, Phenols and Ethers
Alcohols, Phenols and Ethers for JEE Main & Advanced
Alcohols
Nomenclature, Preparation, and Properties of AlcoholsTopic 1
Alcohols: Organic compounds with $-OH$ group on $sp^3$ carbon. General formula $R-OH$.
Classification:
| Type | Description | Example |
|---|---|---|
| Primary (1°) | $-OH$ on C bonded to 1 other C | Methanol, ethanol |
| Secondary (2°) | $-OH$ on C bonded to 2 other C's | Isopropanol |
| Tertiary (3°) | $-OH$ on C bonded to 3 other C's | tert-butanol |
By number of -OH groups:
- Monohydric: 1 -OH (ethanol)
- Dihydric: 2 -OH (ethylene glycol, $HOCH_2CH_2OH$)
- Trihydric: 3 -OH (glycerol, $HOCH_2CH(OH)CH_2OH$)
- Polyhydric: Many -OH (sorbitol)
Nomenclature:
- Common: methyl alcohol, ethyl alcohol
- IUPAC: methanol, ethanol, propan-1-ol, propan-2-ol
Examples:
- $CH_3OH$: methanol
- $CH_3CH_2OH$: ethanol
- $(CH_3)_2CHOH$: 2-propanol (isopropanol)
- $(CH_3)_3COH$: 2-methyl-2-propanol (tert-butanol)
Preparation of Alcohols:
1. From Alkenes (Hydration):
Acid-catalyzed hydration: $CH_2=CH_2 + H_2O \xrightarrow{H^+} CH_3CH_2OH$ (Markovnikov)
Oxymercuration-Demercuration: No rearrangement, Markovnikov. $CH_2=CH_2 + Hg(OAc)_2 + H_2O \to CH_3CH_2HgOAc \xrightarrow{NaBH_4} CH_3CH_2OH$
Hydroboration-Oxidation (Brown's reaction): Anti-Markovnikov. $3CH_3CH=CH_2 + BH_3 \to (CH_3CH_2CH_2)_3B \xrightarrow{H_2O_2, OH^-} 3CH_3CH_2CH_2OH$
2. From Alkyl Halides: $RX + KOH(aq) \xrightarrow{\Delta} R-OH + KX$ (substitution; $S_N$).
3. From Carbonyl Compounds (Reduction):
- Aldehydes → 1° alcohols: $RCHO + 2[H] \to RCH_2OH$
- Ketones → 2° alcohols: $RCOR' + 2[H] \to RCH(OH)R'$
- Carboxylic acids → 1° alcohols: $RCOOH + 4[H] \to RCH_2OH$
- Esters → 1° alcohols: $RCOOR' + 4[H] \to RCH_2OH + R'OH$
Common reducing agents: $LiAlH_4$ (strong, reduces all); $NaBH_4$ (milder, reduces aldehydes and ketones but not esters/acids); $H_2/Ni$ (catalytic).
4. From Grignard Reagents: $RMgX + HCHO \xrightarrow{H_3O^+} RCH_2OH$ ($1°$) $RMgX + R'CHO \to R'(R)CH(OH)$ ($2°$) $RMgX + R'COR'' \to R'(R)(R'')COH$ ($3°$)
5. Industrial:
- Methanol: $CO + 2H_2 \xrightarrow{ZnO, Cr_2O_3, 600K, 200atm} CH_3OH$
- Ethanol: Fermentation of sugars by yeast: $C_6H_{12}O_6 \xrightarrow{zymase} 2C_2H_5OH + 2CO_2$
Physical Properties:
- Liquid (small alcohols); higher BP than corresponding alkanes/ethers/aldehydes due to H-bonding
- Methanol, ethanol, propanol miscible with water; solubility decreases with chain length
- BP increases with chain length; $1° > 2° > 3°$ for same C number (steric, H-bonding)
- BP order: HCOOH > $H_2O$ > alcohol > ether > aldehyde > alkane (similar carbon)
Predict products of: (a) $CH_3CH_2MgBr + HCHO$ then $H_3O^+$ (b) $CH_3CH_2MgBr + CH_3CHO$ then $H_3O^+$
Show solution
(a) $CH_3CH_2MgBr + HCHO \to CH_3CH_2CH_2OMgBr \xrightarrow{H_3O^+} CH_3CH_2CH_2OH$ (n-propanol, 1°)
(b) $CH_3CH_2MgBr + CH_3CHO \to CH_3CH_2CH(CH_3)OMgBr \xrightarrow{H_3O^+} CH_3CH_2CH(OH)CH_3$ (2-butanol, 2°)
Final Answer: (a) n-propanol; (b) sec-butanol.
Why does $CH_3OH$ have higher BP than $CH_3OCH_3$?
Show solution
$CH_3OH$ has $-OH$ group → forms intermolecular H-bonds; needs much energy to break, hence high BP ($65°$C). $CH_3OCH_3$ has no $-OH$; only dipole-dipole forces → much lower BP ($-23°$C).
Final Answer: H-bonding in alcohol; absent in ether.
Methanol is:
Glycerol has:
Hydroboration-oxidation gives:
Most reduce-able to alcohol:
RMgX + HCHO → ?
Reactions of AlcoholsTopic 2
Reactions Based on -OH Group:
1. Reactions with Active Metals (Acidic character):
Alcohols are weak acids; acidity: water > 1° > 2° > 3° (alkyl groups donate electrons, destabilize alkoxide).
$2R-OH + 2Na \to 2RO^-Na^+ + H_2$ (sodium alkoxide; e.g., sodium ethoxide)
2. With Hydrogen Halides (Substitution of -OH by X):
$R-OH + HX \to R-X + H_2O$
Reactivity: $HI > HBr > HCl$; among alcohols: $3° > 2° > 1°$.
Lucas Test: As before (distinguishes 1°, 2°, 3°).
3. With Phosphorus Halides: $R-OH + PCl_5 \to R-Cl + POCl_3 + HCl$ $3R-OH + PCl_3 \to 3R-Cl + H_3PO_3$ $R-OH + SOCl_2 \to R-Cl + SO_2 + HCl$ (clean reaction)
4. Dehydration (Elimination):
Intra-molecular (gives alkene): Heat with conc. $H_2SO_4$ or $Al_2O_3$: $CH_3CH_2OH \xrightarrow{443K, H_2SO_4} CH_2=CH_2 + H_2O$
Order of reactivity: $3° > 2° > 1°$ (carbocation stability). Saytzeff's rule applies.
Inter-molecular (gives ether): Lower T with $H_2SO_4$: $2CH_3CH_2OH \xrightarrow{413K, H_2SO_4} CH_3CH_2OCH_2CH_3 + H_2O$ (diethyl ether)
5. Oxidation:
1° alcohols: oxidized to aldehydes (mild) or carboxylic acids (strong). $RCH_2OH \xrightarrow{[O]} RCHO \xrightarrow{[O]} RCOOH$
2° alcohols: oxidized to ketones. $R-CHOH-R' \xrightarrow{[O]} R-CO-R'$
3° alcohols: No easy oxidation (no $\alpha$-H on -OH carbon for removal).
Oxidizing agents:
- $KMnO_4/H_2SO_4$: strong; gives -COOH from 1°
- $K_2Cr_2O_7/H_2SO_4$: strong (similar)
- $Cu/573K$: dehydrogenation; gives aldehyde from 1°, ketone from 2°
- PCC (Pyridinium chlorochromate): Mild oxidant; gives aldehyde (stops there)
6. Esterification: $RCOOH + R'OH \rightleftharpoons RCOOR' + H_2O$ (with H⁺ catalyst)
- Equilibrium; conc. $H_2SO_4$ shifts toward ester (removes water).
7. Reactions with $Cu$ at 573K (Dehydrogenation):
- 1° alcohol → aldehyde
- 2° alcohol → ketone
- 3° alcohol → alkene (dehydration only; no $\alpha$-H on OH-C)
Predict products on oxidation of: (a) $CH_3CH_2OH$ with $K_2Cr_2O_7/H_2SO_4$ (b) $(CH_3)_2CHOH$ with $K_2Cr_2O_7/H_2SO_4$ (c) $(CH_3)_3COH$ with $K_2Cr_2O_7/H_2SO_4$
Show solution
(a) $1°$: $CH_3CH_2OH \to CH_3CHO \to CH_3COOH$ (acetic acid, end product). (b) $2°$: $(CH_3)_2CHOH \to (CH_3)_2C=O$ (acetone). (c) $3°$: No easy oxidation (no $\alpha$-H on $-OH$ carbon).
Final Answer: (a) acetic acid; (b) acetone; (c) no reaction (or slow at extreme conditions).
Predict major product of $CH_3CH_2OH \xrightarrow{H_2SO_4} ?$ at: (i) $413$ K (ii) $443$ K
Show solution
(i) At lower T ($413$ K): intermolecular dehydration → diethyl ether: $CH_3CH_2OCH_2CH_3 + H_2O$. (ii) At higher T ($443$ K): intramolecular dehydration → ethene: $CH_2=CH_2 + H_2O$.
Final Answer: Lower T → ether; higher T → alkene.
Order of acidity of alcohols:
$3°$ alcohol on $K_2Cr_2O_7/H_2SO_4$:
Esterification needs:
Dehydration of $2°$ alcohol gives:
$CH_3OH$ on heating with $Cu$ at $573$ K:
Phenols and Ethers
Phenols — Acidity, Preparation, and ReactionsTopic 1
Phenols: Aromatic compounds with $-OH$ directly attached to aromatic ring. Simplest: phenol ($C_6H_5OH$).
Nomenclature:
- Phenol ($C_6H_5OH$); cresols ($CH_3$-substituted phenol: o-, m-, p-cresol); naphthols (1- and 2-naphthol).
- IUPAC: benzenol (or accept "phenol").
Preparation:
1. From Cumene (industrial):
- Cumene ($C_6H_5CH(CH_3)_2$) air-oxidized → cumene hydroperoxide → cleaved by H⁺ → phenol + acetone:
$C_6H_5CH(CH_3)_2 + O_2 \to C_6H_5C(OOH)(CH_3)_2 \xrightarrow{H^+} C_6H_5OH + (CH_3)_2CO$
2. From Benzene Diazonium Salts: $C_6H_5N_2^+Cl^- + H_2O \xrightarrow{\Delta} C_6H_5OH + N_2 + HCl$
3. From Chlorobenzene (Dow's Process): $C_6H_5Cl + NaOH \xrightarrow{623K, 300\,atm} C_6H_5ONa \xrightarrow{H^+} C_6H_5OH$
4. From Sulfonic Acids (Alkali Fusion): $C_6H_5SO_3Na + 2NaOH \xrightarrow{fusion} C_6H_5ONa + Na_2SO_3 + H_2O$ $C_6H_5ONa + H^+ \to C_6H_5OH + Na^+$
Acidity of Phenol:
Phenol is weakly acidic ($pK_a \approx 10$). Much more acidic than alcohols ($pK_a \approx 16-18$).
Reasons for higher acidity vs alcohols:
- Resonance stabilization of phenoxide ion ($C_6H_5O^-$): negative charge delocalized over 4 ring positions (ipso, ortho, para).
- The conjugate base (phenoxide) is more stable than alkoxide.
- $-OH$ on $sp^2$ C (more EN) — withdrawal of e⁻ from H makes it more acidic.
Effect of Substituents on Phenol Acidity:
- EW groups (-NO₂, -CN, -COR, -CHO, -X) at o/p positions: Increase acidity (stabilize phenoxide further)
- ED groups (-CH₃, -NH₂, -OR, -OH) at o/p positions: Decrease acidity
- m-position effects are less (no resonance, only inductive)
Order of acidity (typical examples): $2,4,6$-trinitrophenol (picric acid, $pK_a = 0.4$) > p-nitrophenol > o-nitrophenol > m-nitrophenol > phenol > p-methylphenol > p-methoxyphenol > p-aminophenol
Reactions of Phenol:
1. With NaOH: $C_6H_5OH + NaOH \to C_6H_5ONa + H_2O$ (acidic enough to react with strong base, unlike alcohols which need Na metal).
2. With Na metal: $2C_6H_5OH + 2Na \to 2C_6H_5O^-Na^+ + H_2$.
3. Electrophilic Aromatic Substitution (-OH is o/p activator):
- Bromination: $C_6H_5OH + 3Br_2 \xrightarrow{Br_2 \text{ in water}} 2,4,6$-tribromophenol (white ppt) — very rapid; no catalyst needed (-OH activates strongly).
- Nitration:
- Dilute HNO₃ at low T: o- and p-nitrophenols (separable by steam distillation since o- has intra-molecular H-bond; p- has inter-molecular)
- Conc. HNO₃: 2,4,6-trinitrophenol (picric acid)
- Sulfonation: o-/p-phenolsulfonic acid (temperature controls preference).
4. Reimer-Tiemann Reaction: Phenol + CHCl₃ + NaOH → salicylaldehyde (o-hydroxybenzaldehyde). $C_6H_5OH + CHCl_3 + 4NaOH \to o-HOC_6H_4CHO + 3NaCl + 3H_2O$
- Electrophile: $:CCl_2$ (dichlorocarbene) generated from CHCl₃ + OH⁻
- Major: ortho product (resonance and H-bonding stability)
5. Kolbe-Schmitt Reaction: Phenol + CO₂ + NaOH → salicylic acid. $C_6H_5ONa + CO_2 \xrightarrow{Na, 4atm} o-HOC_6H_4COONa \xrightarrow{H^+} o-HOC_6H_4COOH$
- Used in aspirin manufacture (salicylic acid + acetic anhydride → aspirin).
6. Reduction (Zn dust): $C_6H_5OH + Zn \to C_6H_6 + ZnO$.
7. Oxidation: With $Na_2Cr_2O_7$ → para-benzoquinone (purple → yellow).
8. With FeCl₃: Violet color (test for phenol).
Arrange in increasing acidity: phenol, $p$-nitrophenol, $p$-cresol, picric acid, $p$-methoxyphenol.
Show solution
EW groups at o/p increase acidity. Picric acid has 3 NO₂'s → most acidic. Order (least to most acidic): $p$-methoxyphenol < $p$-cresol < phenol < $p$-nitrophenol < picric acid.
Final Answer: p-methoxyphenol < p-cresol < phenol < p-nitrophenol < picric acid.
Predict major product of Reimer-Tiemann reaction with phenol.
Show solution
Phenol + CHCl₃ + NaOH (heat) → ortho-hydroxybenzaldehyde (salicylaldehyde) as major + small para isomer.
- Mechanism: $:CCl_2$ attacks o-position (activation by phenoxide O⁻ resonance).
- Salicylaldehyde isolated by steam distillation.
Final Answer: Salicylaldehyde (o-hydroxybenzaldehyde).
Phenol is:
With Na metal, both phenol and ethanol release $H_2$. Which is more acidic in water?
Phenol + FeCl₃:
Reimer-Tiemann reaction gives:
Kolbe-Schmitt reaction gives:
Ethers — Williamson Synthesis and ReactionsTopic 2
Ethers: Compounds with -O- linking two carbon groups: $R-O-R'$.
Types:
- Symmetrical: $R = R'$ (e.g., dimethyl ether, diethyl ether)
- Unsymmetrical (mixed): $R \neq R'$ (e.g., methyl ethyl ether)
Nomenclature:
- Common: alkyl alkyl ether (e.g., methyl ethyl ether, diethyl ether)
- IUPAC: alkoxy-alkane (smaller alkyl as alkoxy)
- $CH_3OCH_3$: methoxymethane
- $CH_3CH_2OCH_3$: methoxyethane
- $CH_3OCH_2CH_2CH_3$: 1-methoxypropane
Preparation:
1. Williamson's Synthesis (most important): $R-O^-Na^+ + R'-X \to R-O-R' + NaX$
- Sodium alkoxide + alkyl halide
- Best for primary alkyl halides (S_N2)
- With tertiary halides → elimination (alkene) dominates
For unsymmetrical ether: $CH_3O^-Na^+ + CH_3CH_2Br \to CH_3OCH_2CH_3 + NaBr$
For aromatic ethers (anisole, $C_6H_5OCH_3$): $C_6H_5O^-Na^+ + CH_3I \to C_6H_5OCH_3$
2. Dehydration of Alcohols (only for symmetrical ethers): $2C_2H_5OH \xrightarrow{H_2SO_4, 413K} C_2H_5OC_2H_5 + H_2O$ (intermolecular)
Properties:
- Volatile liquids; characteristic smell
- $R-O-R'$: O is $sp^3$ hybridized; lone pair on O
- No H-bonding between ether molecules → low BP (compared to alcohols)
- BP order: alcohol > ether > alkane (same C)
- Slight solubility in water (O can accept H-bond from water)
- Diethyl ether is good general anesthetic (historical)
Reactions of Ethers:
1. Cleavage by HX: $R-O-R' + HX \to R-X + R'-OH$ or both as R-X (excess HX)
Mechanism: $S_N2$ (primary R) or $S_N1$ (tertiary R). Reactivity: $HI > HBr > HCl$. Cleavage at: ether O is protonated → C-O bond breaks; less hindered side gets attacked by nucleophile in $S_N2$ (small Nu).
Example: $CH_3OC_6H_5 + HI \to CH_3I + C_6H_5OH$ (Phenyl group never attacked; methyl side cleaved).
2. Reaction with Cold HCl at higher pressure: similar but slower.
3. Reactions with $O_2$ (Hazardous): Ethers form peroxides on exposure to air/light: $R-OCH_2-R' + O_2 \to R-O-CH(OOH)-R'$ (auto-oxidation) Peroxides are explosive. Test for peroxides: $KI/H_2SO_4$ — turns brown if peroxides present.
4. Combustion: $C_4H_{10}O + 6O_2 \to 4CO_2 + 5H_2O$ (highly exothermic; ether vapors are flammable)
5. Friedel-Crafts on Aryl Alkyl Ethers: Anisole undergoes electrophilic substitution at ortho/para positions (-OR is o/p activator). $C_6H_5OCH_3 + CH_3COCl \xrightarrow{AlCl_3} p-CH_3C(O)C_6H_4OCH_3$
Predict product of $CH_3OC_2H_5 + HI$ (excess).
Show solution
HI (1 equiv): $CH_3OC_2H_5 + HI \to CH_3I + C_2H_5OH$ (methyl side cleaves; $S_N2$ on less hindered C is faster).
With excess HI: $C_2H_5OH$ further reacts → $C_2H_5I$. Total: $CH_3OC_2H_5 + 2HI \to CH_3I + C_2H_5I + H_2O$.
Final Answer: $CH_3I + C_2H_5OH$ (with 1 equiv); $CH_3I + C_2H_5I + H_2O$ (excess).
Why isn't Williamson's synthesis used with tert-butyl halide?
Show solution
With $3°$ alkyl halide, alkoxide ($RO^-$) acts more as a base than nucleophile; promotes E2 elimination instead of $S_N2$. $RO^- + (CH_3)_3CBr \to ROH + (CH_3)_2C=CH_2 + Br^-$ (alkene + alcohol — no ether formed).
Strategy: To make tert-butyl methyl ether, use methyl halide ($1°$) with tert-butoxide ($3°$ alkoxide): $(CH_3)_3CO^-Na^+ + CH_3I \to (CH_3)_3COCH_3$ (works; $S_N2$ on $CH_3I$ is good).
Final Answer: Use $1°$ alkyl halide and the bulkier alkoxide; bulky halide gives elimination.
Williamson synthesis:
Best alkyl halide for Williamson:
Cleavage of $CH_3OC_2H_5$ by HI:
Ethers form peroxides on:
Anisole is:
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