Alcohols, Phenols, and Ethers
Comprehensive Reactions, Cumene Process, Mechanisms & Solved PYQs
1. Introduction and 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 by substituting the hydrogen atom of a hydrocarbon by an alkoxy or aryloxy group ($R-O-R'$).
Classification
- Monohydric, Dihydric, Trihydric: Based on the number of $-OH$ groups (one, two, or three).
- Primary ($1^\circ$), Secondary ($2^\circ$), Tertiary ($3^\circ$) Alcohols: Based on the nature of the $sp^3$ carbon atom attached to the $-OH$ group.
- Allylic Alcohols: $-OH$ is attached to an $sp^3$ hybridized carbon next to a $C=C$ double bond ($CH_2=CH-CH_2-OH$).
- Benzylic Alcohols: $-OH$ is attached to an $sp^3$ hybridized carbon next to an aromatic ring ($C_6H_5-CH_2-OH$).
- Vinylic Alcohols: $-OH$ is attached directly to an $sp^2$ hybridized carbon of a double bond ($CH_2=CH-OH$).
2. Preparation of Alcohols
A. From Alkenes
- Acid-Catalyzed Hydration: Follows Markovnikov's rule.
$CH_3-CH=CH_2 + H_2O \xrightarrow{H^+} CH_3-CH(OH)-CH_3$ (Propan-2-ol) - Hydroboration-Oxidation: Yields anti-Markovnikov product.
$6CH_3-CH=CH_2 + B_2H_6 \rightarrow 2(CH_3-CH_2-CH_2)_3B$ (Tripropylborane)
$(CH_3-CH_2-CH_2)_3B + 3H_2O_2 \xrightarrow{OH^-} 3CH_3-CH_2-CH_2-OH + B(OH)_3$ (Propan-1-ol)
B. By Reduction of Carbonyl Compounds
- Aldehydes reduce to Primary ($1^\circ$) alcohols using $H_2/Ni$, $LiAlH_4$, or $NaBH_4$.
- Ketones reduce to Secondary ($2^\circ$) alcohols.
- Carboxylic Acids and Esters are reduced to Primary alcohols using strong reducing agents like $LiAlH_4$.
C. From Grignard Reagent
Grignard reagents ($RMgX$) react with carbonyl compounds to form adducts which on acid hydrolysis yield alcohols.
- Formaldehyde (HCHO) $\rightarrow$ Primary Alcohol
- Any other Aldehyde (RCHO) $\rightarrow$ Secondary Alcohol
- Ketone (RCOR') $\rightarrow$ Tertiary Alcohol
3. Preparation of Phenols
1. From Cumene (Commercial Method - Very Important)
Cumene (Isopropylbenzene) is oxidized in the presence of air to cumene hydroperoxide. It is then treated with dilute acid to form Phenol and Acetone (a valuable byproduct).
$C_6H_5-CH(CH_3)_2 + O_2 \xrightarrow{423 \text{ K}, \text{Co-naphthenate}} C_6H_5-C(OOH)(CH_3)_2$ (Cumene hydroperoxide)
$C_6H_5-C(OOH)(CH_3)_2 \xrightarrow{dil. H^+} C_6H_5OH \text{ (Phenol)} + CH_3COCH_3 \text{ (Acetone)}$
2. From Chlorobenzene (Dow Process)
Chlorobenzene is fused with $NaOH$ at 623 K and 300 atm pressure to form sodium phenoxide, which is acidified to yield phenol.
$C_6H_5Cl + 2NaOH \xrightarrow{623 \text{ K, } 300 \text{ atm}} C_6H_5ONa + NaCl + H_2O$
$C_6H_5ONa + HCl \rightarrow C_6H_5OH + NaCl$
3. From Benzene Sulfonic Acid
Benzene sulfonic acid is neutralized with aqueous $NaOH$ to sodium benzenesulfonate, which is fused with solid $NaOH$ at 573 K to form sodium phenoxide. Acidification gives phenol.
4. From Aniline (Diazotization)
Aniline reacts with nitrous acid ($NaNO_2 + HCl$) at 273-278 K to form benzene diazonium chloride. Warming it with water hydrolyzes it to phenol.
$C_6H_5NH_2 \xrightarrow{NaNO_2 + HCl, \text{ 273 K}} C_6H_5N_2^+Cl^-$
$C_6H_5N_2^+Cl^- + H_2O \xrightarrow{\text{Warm}} C_6H_5OH + N_2 \uparrow + HCl$
4. Preparation of Ethers
A. Williamson Synthesis (Symmetrical and Unsymmetrical Ethers)
It involves the reaction of an alkyl halide with sodium alkoxide or sodium phenoxide. This reaction follows the $S_N2$ mechanism.
$R-X + R'-O^-Na^+ \rightarrow R-O-R' + NaX$
B. Dehydration of Alcohols
Primary alcohols undergo intermolecular dehydration in the presence of protic acids ($H_2SO_4$ or $H_3PO_4$) at 413 K to form symmetrical ethers.
$2CH_3CH_2OH \xrightarrow{H_2SO_4, 413 \text{ K}} CH_3CH_2-O-CH_2CH_3 + H_2O$
Note: If the temperature is 443 K, intramolecular dehydration occurs to form an alkene (ethene).
5. Physical Properties
- Boiling Points: Alcohols and phenols have much higher boiling points than corresponding alkanes, haloalkanes, and ethers of similar molecular mass. Reason: Presence of strong intermolecular hydrogen bonding in alcohols and phenols.
- Solubility: Lower alcohols and phenols are soluble in water because they can form hydrogen bonds with water molecules. Solubility decreases with an increase in the size of the hydrophobic alkyl/aryl group.
6. Chemical Reactions of Alcohols
A. Acidic Nature of Alcohols
Alcohols react with active metals like Na, K, Al to liberate hydrogen gas, showing weak acidic nature. Acidity decreases in the order: $1^\circ > 2^\circ > 3^\circ$.
B. Oxidation
Oxidation involves the cleavage of $O-H$ and $C-H$ bonds to form a $C=O$ double bond.
- Primary Alcohols: Oxidized to aldehydes (using mild oxidizing agents like PCC) or to carboxylic acids (using strong agents like acidic $KMnO_4$ or $K_2Cr_2O_7$).
- Secondary Alcohols: Oxidized to ketones using $CrO_3$ or acidic $K_2Cr_2O_7$.
- Tertiary Alcohols: Resist oxidation under normal conditions. Under vigorous conditions, carbon-carbon bond cleavage occurs to give a mixture of carboxylic acids with fewer carbon atoms.
C. Action of Heated Copper (Dehydrogenation)
- $1^\circ$ alcohol $\xrightarrow{Cu, 573 K}$ Aldehyde + $H_2$
- $2^\circ$ alcohol $\xrightarrow{Cu, 573 K}$ Ketone + $H_2$
- $3^\circ$ alcohol $\xrightarrow{Cu, 573 K}$ Alkene + $H_2O$ (Undergoes dehydration instead of dehydrogenation).
7. Chemical Reactions of Phenols
Acidity of Phenols
Phenols are significantly more acidic than alcohols. They react with aqueous $NaOH$ to form sodium phenoxide, whereas alcohols do not.
Reason: The phenoxide ion formed after the loss of $H^+$ is highly stabilized by resonance (delocalization of negative charge over the ortho and para positions of the benzene ring). In contrast, the alkoxide ion from an alcohol has no resonance stabilization and the alkyl group destabilizes it via the +I effect.
Effect of substituents: Electron-withdrawing groups (EWG like $-NO_2$) increase acidity. Electron-donating groups (EDG like $-CH_3$) decrease acidity.
Electrophilic Aromatic Substitution
The $-OH$ group is strongly activating and ortho, para-directing.
- Nitration:
- With dilute $HNO_3$ at low temp (298 K): Yields mixture of o-nitrophenol and p-nitrophenol.
- With concentrated $HNO_3$ and conc. $H_2SO_4$: Yields 2,4,6-trinitrophenol (Picric Acid).
- Halogenation (Bromination):
- With $Br_2$ in $CS_2$ or $CHCl_3$ at low temp: Yields o-bromophenol and p-bromophenol.
- With Bromine water ($Br_2/H_2O$): Yields a white precipitate of 2,4,6-tribromophenol.
Kolbe's Reaction
Sodium phenoxide is heated with $CO_2$ at 400 K and 4-7 atm pressure, followed by acidification to yield Salicylic acid (2-Hydroxybenzoic acid).
Reimer-Tiemann Reaction
Phenol is treated with chloroform ($CHCl_3$) and aqueous $NaOH$ at 340 K, followed by acid hydrolysis to yield Salicylaldehyde (2-Hydroxybenzaldehyde).
8. Chemical Reactions of Ethers
Ethers are highly unreactive compounds. They do not react with bases, reducing agents, oxidizing agents, or active metals under ordinary conditions.
Cleavage of C-O Bond by Hydrogen Halides ($HX$)
Ethers are cleaved by concentrated $HI$ or $HBr$ at high temperatures.
$R-O-R' + HX \xrightarrow{\Delta} R-X + R'-OH$
If an excess of $HX$ is used, the alcohol formed further reacts to yield another alkyl halide.
Rule for Cleavage of Unsymmetrical Ethers:
- If both alkyl groups are primary or secondary, the halide ion ($I^-$ or $Br^-$) attacks the smaller alkyl group (due to less steric hindrance, $S_N2$ mechanism).
- If one of the alkyl groups is tertiary, the halide ion attacks the tertiary group (due to the formation of a stable tertiary carbocation, $S_N1$ mechanism).
- In alkyl aryl ethers (e.g., Anisole), the bond between Oxygen and Alkyl group breaks because the Aryl-Oxygen bond is stronger due to partial double bond character from resonance. Yields Phenol + Alkyl halide.
9. Solved Textbook Reasoning Questions
Question 1: Acidic Strength Comparison
Arrange the following in increasing order of acidic strength: Phenol, ethanol, p-nitrophenol.
Answer:
Ethanol < Phenol < p-Nitrophenol
Reason: Ethanol is weakly acidic because the alkoxide ion has no resonance stabilization, and the ethyl group exhibits a +I (electron-donating) effect, destabilizing the negative charge. Phenol is more acidic than ethanol because the phenoxide ion is resonance stabilized. p-Nitrophenol is the most acidic because the $-NO_2$ group is a strong electron-withdrawing group (-I and -R effect), which strongly stabilizes the phenoxide ion by withdrawing electron density, making it easier to lose $H^+$.
Question 2: Preparation of t-Butyl methyl ether
How will you synthesize t-butyl methyl ether using Williamson synthesis?
Answer:
To prepare an unsymmetrical ether where one group is tertiary, the alkyl halide must be primary and the alkoxide must be tertiary. If a tertiary halide is used, elimination will occur to form an alkene instead.
Correct Reactants: Sodium t-butoxide and Methyl chloride.
$(CH_3)_3C-O^-Na^+ + CH_3-Cl \rightarrow (CH_3)_3C-O-CH_3 + NaCl$
10. Board PYQs with Complete Answers
Verified previous year questions from the Maharashtra State Board HSC Chemistry exams.
1 Mark Questions (VSA)
Q1. Write the IUPAC name of cumene. (March 2013, Oct 2015)
Answer: Isopropylbenzene (or 2-Phenylpropane).
Q2. Name the catalyst used in the dehydration of primary alcohols to ethers at 413 K. (March 2016)
Answer: Concentrated Sulfuric acid ($H_2SO_4$) or Phosphoric acid ($H_3PO_4$).
2 Mark Questions (SA-I)
Q3. Write the chemical equation for the preparation of phenol from chlorobenzene (Dow process). (March 2014, March 2019, Oct 2021)
$C_6H_5Cl + 2NaOH \xrightarrow{623 \text{ K, } 300 \text{ atm}} C_6H_5O^-Na^+ (\text{Sodium phenoxide}) + NaCl + H_2O$
$C_6H_5O^-Na^+ + dil. HCl \rightarrow C_6H_5OH (\text{Phenol}) + NaCl$
Q4. What is the action of bromine water on phenol? Give the chemical equation. (March 2017, March 2022)
Phenol reacts with bromine water rapidly to form a white precipitate of 2,4,6-tribromophenol.
$C_6H_5OH + 3Br_2(aq) \rightarrow C_6H_2Br_3OH \downarrow (\text{white ppt}) + 3HBr$
3 Mark Questions (SA-II)
Q5. Write a short note on Reimer-Tiemann reaction and Kolbe's reaction. (Oct 2014, March 2018, March 2023)
Reimer-Tiemann Reaction: When phenol is treated with chloroform ($CHCl_3$) in the presence of aqueous sodium hydroxide ($NaOH$) at 340 K, followed by hydrolysis with dilute acid, Salicylaldehyde (2-hydroxybenzaldehyde) is formed. This reaction introduces an aldehyde group at the ortho position of the benzene ring.
Kolbe's Reaction: Sodium phenoxide (formed by treating phenol with NaOH) is heated with carbon dioxide ($CO_2$) gas at 400 K and 4-7 atm pressure to form sodium salicylate, which upon acidification yields Salicylic acid (2-hydroxybenzoic acid). This reaction introduces a carboxyl group at the ortho position.
Q6. How is phenol prepared from cumene? Write the chemical equations. Why is this considered the best commercial method? (March 2015, Oct 2020)
Preparation: Cumene (isopropylbenzene) is oxidized in the presence of air or oxygen at 423 K using cobalt naphthenate catalyst to form cumene hydroperoxide. This is then decomposed by treating with dilute acid to yield phenol and acetone.
1. $C_6H_5-CH(CH_3)_2 + O_2 \xrightarrow{423 \text{K}} C_6H_5-C(OOH)(CH_3)_2$
2. $C_6H_5-C(OOH)(CH_3)_2 \xrightarrow{dil. H^+} C_6H_5OH (\text{Phenol}) + CH_3COCH_3 (\text{Acetone})$
Why it is the best commercial method:
Because a very valuable byproduct, Acetone, is obtained in large quantities. The sale of acetone significantly offsets the cost of production, making the process highly economical.
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