Chapter 11: Alcohols, Phenols, and Ethers
Highly Detailed Mark-wise Solutions for Board Exam Preparation
1 Mark Questions (Very Short Answer)
Answer: The IUPAC name of anisole ($C_6H_5-O-CH_3$) is Methoxybenzene.
Answer: Cobalt naphthenate is used as the catalyst for the oxidation of cumene to cumene hydroperoxide.
Answer: Phenol is more acidic than ethanol because the phenoxide ion formed after losing a proton is stabilized by resonance, whereas the ethoxide ion has no resonance stabilization.
Answer: Phenol reacts rapidly with bromine water to give a white precipitate of 2,4,6-tribromophenol.
2 Mark Questions (Short Answer-I)
Answer:
Williamson Synthesis: It is a laboratory method to prepare symmetrical and unsymmetrical ethers. It involves the reaction of an alkyl halide with sodium alkoxide (or sodium phenoxide) via an $S_N2$ mechanism.
$CH_3-CH_2-Br + CH_3-O^-Na^+ \rightarrow CH_3-CH_2-O-CH_3 + NaBr$
(Ethyl bromide) (Sod. methoxide) (Ethyl methyl ether)
Condition for unsymmetrical ethers:
To get a good yield, the alkyl halide used must be primary ($1^\circ$). If secondary or tertiary alkyl halides are used, the strong base (alkoxide) causes elimination ($\beta$-elimination) to dominate over substitution, forming an alkene instead of an ether.
Answer:
Chlorobenzene is highly unreactive towards nucleophilic substitution. However, under drastic conditions (623 K and 300 atm), it fuses with aqueous sodium hydroxide ($NaOH$) to form sodium phenoxide. This salt upon acidification with dilute HCl yields phenol.
$C_6H_5-Cl + 2NaOH \xrightarrow{623 \text{ K}, 300 \text{ atm}} C_6H_5-O^-Na^+ + NaCl + H_2O$
$C_6H_5-O^-Na^+ + dil. HCl \rightarrow C_6H_5-OH \text{ (Phenol)} + NaCl$
Answer:
When sodium phenoxide (prepared from phenol and NaOH) is heated with carbon dioxide ($CO_2$) at 400 K and 4-7 atm pressure, it undergoes electrophilic substitution to form sodium salicylate. Subsequent acidification with dilute acid yields Salicylic acid (2-hydroxybenzoic acid).
$C_6H_5ONa + CO_2 \xrightarrow{400 \text{ K}, 4-7 \text{ atm}} \text{Sodium salicylate} \xrightarrow{H_3O^+} \text{Salicylic acid}$
This reaction is specifically used to introduce a carboxyl group ($-COOH$) at the ortho position of the phenol ring.
3 Mark Questions (Short Answer-II / Mechanisms)
Answer:
Step 1: Oxidation of Cumene
Cumene (isopropylbenzene) is oxidized in the presence of air or pure oxygen at 423 K using a cobalt naphthenate catalyst. This produces cumene hydroperoxide.
$C_6H_5-CH(CH_3)_2 + O_2 \xrightarrow{\text{Co-naphthenate, 423 K}} C_6H_5-C(OOH)(CH_3)_2$
Step 2: Hydrolysis
The cumene hydroperoxide is then decomposed by treating it with dilute acid (like dilute $H_2SO_4$). This cleaves the molecule to yield Phenol and Acetone.
$C_6H_5-C(OOH)(CH_3)_2 \xrightarrow{H^+, \text{ } H_2O} C_6H_5OH \text{ (Phenol)} + CH_3COCH_3 \text{ (Acetone)}$
Why it is an excellent commercial method:
This process is highly preferred industrially because a very valuable byproduct, Acetone, is produced in large quantities. The sale of acetone significantly offsets the cost of production, making the entire process highly economical.
Answer:
Acidic Nature of Phenol:
Phenol is weakly acidic because the oxygen of the $-OH$ group is attached to an $sp^2$ hybridized carbon of the benzene ring. After losing a proton ($H^+$), it forms the phenoxide ion ($C_6H_5O^-$). The phenoxide ion is highly stabilized by resonance, as the negative charge is delocalized over the ortho and para positions of the benzene ring. This stabilization drives the equilibrium forward, making phenol acidic.
Effect of Substituents:
- p-Nitrophenol is more acidic: The nitro group ($-NO_2$) is a powerful Electron Withdrawing Group (EWG). It exhibits both -I (inductive) and -R (resonance) effects. It pulls electron density away from the oxygen atom, making the release of $H^+$ easier. Furthermore, it highly stabilizes the resulting phenoxide ion by dispersing the negative charge. Hence, acidity increases.
- p-Cresol is less acidic: The methyl group ($-CH_3$) is an Electron Donating Group (EDG) showing a +I effect and hyperconjugation. It pumps electron density into the ring and towards the oxygen atom, making the release of $H^+$ difficult and destabilizing the phenoxide ion by intensifying the negative charge. Hence, acidity decreases.
4 Mark Questions (Long Answer / Combined)
Note: In the HSC Board Exam, 4-mark questions in Chemistry are frequently split into sub-questions (e.g., a 2-mark name reaction combined with a 2-mark reasoning question).
(b) Why are primary alkyl halides preferred in Williamson synthesis? Give the reaction to prepare ethyl methyl ether. [2 Marks] March 2018, March 2022
Answer (a): Reimer-Tiemann Reaction
When phenol is treated with chloroform ($CHCl_3$) in the presence of an aqueous base like sodium hydroxide ($NaOH$) at 340 K, an intermediate substituted benzal chloride is formed. This undergoes hydrolysis in the alkaline medium, followed by acidification to yield Salicylaldehyde (2-hydroxybenzaldehyde) as the major product.
$\text{Phenol} + CHCl_3 + 3NaOH(aq) \xrightarrow{340 \text{ K}} \text{Intermediate} \xrightarrow{H_3O^+} \text{Salicylaldehyde} + 3NaCl + 2H_2O$
Answer (b): Williamson Synthesis Preference
Williamson synthesis proceeds via an $S_N2$ mechanism where an alkoxide ion (a strong base and nucleophile) attacks the alkyl halide. Primary alkyl halides are preferred because they have minimal steric hindrance, allowing smooth backside attack by the nucleophile to form the ether. If bulky secondary or tertiary halides are used, the strong basic nature of the alkoxide causes elimination to overpower substitution, resulting in an alkene instead of an ether.
Preparation of Ethyl methyl ether:
We use a primary halide (Methyl bromide) and an alkoxide (Sodium ethoxide).
$CH_3-Br + CH_3-CH_2-O^-Na^+ \rightarrow CH_3-O-CH_2-CH_3 \text{ (Ethyl methyl ether)} + NaBr$
(b) How are primary alcohols prepared from aldehydes? Write the equation. [2 Marks] Oct 2013, March 2020
Answer (a): Action of conc. $HNO_3$ on Phenol
Phenol reacts violently with a nitrating mixture (concentrated $HNO_3$ in the presence of concentrated $H_2SO_4$) to yield a tri-substituted product, 2,4,6-trinitrophenol, which is commonly known as Picric acid. Note: The yield is often poor due to the oxidative cleavage of the phenol ring by conc. nitric acid.
$C_6H_5OH + 3HNO_3(\text{conc}) \xrightarrow{\text{conc. } H_2SO_4, \Delta} C_6H_2(NO_2)_3OH \text{ (Picric acid)} + 3H_2O$
Answer (b): Preparation of $1^\circ$ Alcohols from Aldehydes
Primary alcohols are prepared by the reduction of aldehydes. This can be achieved by catalytic hydrogenation using Hydrogen gas in the presence of a finely divided metal catalyst like Nickel (Ni), Platinum (Pt), or Palladium (Pd). Alternatively, strong chemical reducing agents like Lithium aluminium hydride ($LiAlH_4$) or Sodium borohydride ($NaBH_4$) can be used.
$R-CHO + H_2 \xrightarrow{Ni/Pt/Pd \text{ or } LiAlH_4} R-CH_2-OH$
Example: $CH_3CHO \text{ (Acetaldehyde)} + H_2 \xrightarrow{Ni} CH_3CH_2OH \text{ (Ethanol)}$
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