JEE Main & Advanced

Alcohols, Phenols and Ethers

Alcohols, Phenols and Ethers for JEE Main & Advanced

1
Module 1

Alcohols

Nomenclature, Preparation, and Properties of AlcoholsTopic 1

Alcohols: Organic compounds with $-OH$ group on $sp^3$ carbon. General formula $R-OH$.

Classification:

TypeDescriptionExample
Primary (1°)$-OH$ on C bonded to 1 other CMethanol, ethanol
Secondary (2°)$-OH$ on C bonded to 2 other C'sIsopropanol
Tertiary (3°)$-OH$ on C bonded to 3 other C'stert-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)
Worked Examples
1

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.

2

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.

✎ Self-Check — 5 questions0 / 5
Q1.

Methanol is:

Q2.

Glycerol has:

Q3.

Hydroboration-oxidation gives:

Q4.

Most reduce-able to alcohol:

Q5.

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)
Worked Examples
1

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).

2

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.

✎ Self-Check — 5 questions0 / 5
Q1.

Order of acidity of alcohols:

Q2.

$3°$ alcohol on $K_2Cr_2O_7/H_2SO_4$:

Q3.

Esterification needs:

Q4.

Dehydration of $2°$ alcohol gives:

Q5.

$CH_3OH$ on heating with $Cu$ at $573$ K:

2
Module 2

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:

  1. Resonance stabilization of phenoxide ion ($C_6H_5O^-$): negative charge delocalized over 4 ring positions (ipso, ortho, para).
  2. The conjugate base (phenoxide) is more stable than alkoxide.
  3. $-OH$ 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).

Worked Examples
1

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.

2

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).

✎ Self-Check — 5 questions0 / 5
Q1.

Phenol is:

Q2.

With Na metal, both phenol and ethanol release $H_2$. Which is more acidic in water?

Q3.

Phenol + FeCl₃:

Q4.

Reimer-Tiemann reaction gives:

Q5.

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). $C_6H_5OCH_3 + CH_3COCl \xrightarrow{AlCl_3} p-CH_3C(O)C_6H_4OCH_3$

Worked Examples
1

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).

2

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.

✎ Self-Check — 5 questions0 / 5
Q1.

Williamson synthesis:

Q2.

Best alkyl halide for Williamson:

Q3.

Cleavage of $CH_3OC_2H_5$ by HI:

Q4.

Ethers form peroxides on:

Q5.

Anisole is:

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