CBSE Class 12 Chemistry Alcohols Phenols and Ethers Questions

Unit - 11

ALCOHOLS, PHENOLS AND ETHERS

Question 1. Write IUPAC names of the following compounds :
(i) \(\text{CH}_3-\text{CH}_2-\text{CH}-\text{CH}-\text{CH}-\text{CH}_3\) with \(-\text{CH}_2\text{OH}\) at C-3 and \(-\text{CH}_2\text{Cl}\), \(-\text{CH}_3\) substituents.
(ii) 3-bromocyclohexan-1-ol
(iii) \(\text{CH}\equiv\text{C}-\text{CH}_2\text{OH}\)
(iv) 4-hydroxy-3-methylbenzyl alcohol (or similar substituted benzyl system)
(v) \(\text{CH}_2=\text{CH}-\text{CH}=\text{CH}-\text{CH}_2\text{OH}\)
(vi) Benzene-1,3-diol
(vii) 2-bromo-4-hydroxybenzonitrile
(viii) 2-ethoxy-1-nitrobenzene
(ix) \(\text{C}_6\text{H}_5\text{OC}_3\text{H}_7\)
(x) \(\text{CH}_3\text{CH}_2\text{OCH}_2\text{CH}_2\text{CH}_2\text{Cl}\)
Answer:
(i) 2-(chloromethyl)-3-methylpentan-1-ol
(ii) 3-bromocyclohexan-1-ol
(iii) prop-2-yn-1-ol
(iv) (4-hydroxy-3-methylphenyl)methanol
(v) pent-2,4-dien-1-ol
(vi) benzene-1,3-diol (Resorcinol)
(vii) 2-bromo-4-hydroxybenzonitrile
(viii) 1-ethoxy-2-nitrobenzene
(ix) 1-propoxybenzene (Propoxybenzene)
(x) 1-chloro-3-ethoxypropane
In simple words: Follow IUPAC priority rules, giving lowest locants to principal functional groups like \(-\text{OH}\).

Exam Tip: Functional group priority order for suffixes: Carboxylic acid > sulfonic acid > ester > acid halide > amide > nitrile > aldehyde > ketone > alcohol > amine > alkene > alkyne > ether > halo.

Question 2. Write the structures of the compounds whose names are given below :
(i) 3, 5-dimethoxyhexane-1, 3, 5-triol
(ii) cyclohexylmethanol
(iii) 2-ethoxy-3-methylpentane
(iv) 3-chloromethylpentan-2-ol
(v) p-nitroanisole
Answer:
(i) \(\text{HO}-\text{CH}_2-\text{CH}_2-\text{C}(\text{OCH}_3)(\text{OH})-\text{CH}_2-\text{C}(\text{OCH}_3)(\text{OH})-\text{CH}_3\)
(ii) Cyclohexane ring attached to \(-\text{CH}_2\text{OH}\).
(iii) \(\text{CH}_3-\text{CH}_2-\text{CH}(\text{CH}_3)-\text{CH}(\text{OC}_2\text{H}_5)-\text{CH}_3\)
(iv) \(\text{CH}_3-\text{CH}(\text{OH})-\text{CH}(\text{CH}_2\text{Cl})-\text{CH}_2-\text{CH}_3\)
(v) Benzene ring substituted with \(-\text{OCH}_3\) at position 1 and \(-\text{NO}_2\) at position 4.
In simple words: Construct the carbon skeleton first, then add functional groups and substituents at specified positions.

Exam Tip: Always number the parent chain from the end that gives lower locants to the principal functional group.

Question 3. Describe the following reactions with example :
(i) Hydroboration oxidation of alkenes
(ii) Acid catalysed dehydration of alcohols at 443K.
(iii) Williamson synthesis
(iv) Reimer-Tiemann reaction.
(v) Kolbe’s reaction
(vi) Friedel-Crafts acylation of Anisole.
Answer:
(i) Alkenes react with diborane (\(\text{B}_2\text{H}_6\)) to form trialkylboranes, which on oxidation with alkaline hydrogen peroxide (\(\text{H}_2\text{O}_2/\text{OH}^-\)) yield alcohols with anti-Markovnikov addition of water. Example: \(\text{CH}_3\text{CH}=\text{CH}_2 \xrightarrow{(1) \text{B}_2\text{H}_6, (2) \text{H}_2\text{O}_2/OH^-} \text{CH}_3\text{CH}_2\text{CH}_2\text{OH}\).
(ii) Alcohols heated with concentrated \(\text{H}_2\text{SO}_4\) at 443 K undergo dehydration to form alkenes via an E1 mechanism involving carbocation intermediates. Example: \(\text{CH}_3\text{CH}_2\text{OH} \xrightarrow{\text{conc. H}_2\text{SO}_4, 443\text{K}} \text{CH}_2=\text{CH}_2 + \text{H}_2\text{O}\).
(iii) Alkyl halide reacts with sodium alkoxide to form ether via an \(\text{S}_\text{N}2\) mechanism. Example: \(\text{CH}_3\text{CH}_2\text{ONa} + \text{CH}_3\text{Cl} \rightarrow \text{CH}_3\text{CH}_2\text{OCH}_3 + \text{NaCl}\).
(iv) Treatment of phenol with chloroform in the presence of aqueous sodium hydroxide at 340 K introduces an aldehyde group at the ortho position, forming salicylaldehyde. Followed by acidification, dichloromethyl intermediate forms the product.
(v) Phenol is treated with sodium hydroxide to form sodium phenoxide, which then reacts with carbon dioxide under pressure (400 K, 4-7 atm) followed by acidification to produce salicylic acid (2-hydroxybenzoic acid).
(vi) Anisole reacts with an acyl chloride (e.g., acetyl chloride) in the presence of anhydrous aluminum chloride (\(\text{AlCl}_3\)) to yield a mixture of 2-methoxyacetophenone and 4-methoxyacetophenone.
In simple words: These are standard named and mechanistic organic reactions used to synthesize functional groups.

Exam Tip: Learn reaction conditions, reagents, and major/minor products for all named reactions as they frequently appear in board exams.

Question 4. Complete the following reactions :
(i) \(\text{CH}_3\text{CH}_2\text{CH}_2\text{CHO} \xrightarrow{\text{Pd/H}_2}\)
(ii) \(\text{CH}_3\text{CHO} \xrightarrow{\text{(i) CH}_3\text{MgBr, (ii) H}_2\text{O/H}^+}\)
(iii) \(\text{CH}_3\text{CH}_2\text{OH} \xrightarrow{\text{Cu/573K}}\)
(iv) \(\text{C}_6\text{H}_5\text{OH} + \text{Br}_2 \xrightarrow{\text{H}_2\text{O}}\)
(v) Salicylic acid + \((\text{CH}_3\text{CO})_2\text{O} \xrightarrow{\text{H}^+}\)
(vi) Sodium phenoxide + \(\text{CH}_3\text{Br} \rightarrow\)
(vii) \(\text{CH}_3\text{CH}_2\text{CH}_2\text{O}-\text{CH}_3 + \text{HBr} \rightarrow\)
(viii) Phenetole (\(\text{C}_6\text{H}_5\text{OC}_2\text{H}_5\)) + \(\text{HBr} \rightarrow\)
(ix) \((\text{CH}_3)_3\text{C}-\text{O}-\text{C}_2\text{H}_5 + \text{HI} \rightarrow\)
(x) Anisole + conc. \(\text{HNO}_3\) / conc. \(\text{H}_2\text{SO}_4 \rightarrow\)
(xi) 3-oxocyclohexanecarboxylic acid + \(\text{NaBH}_4 \rightarrow\)
(xiii) 3-methylbutanal + \(\text{NaBH}_4 \rightarrow\)
(xiv) Benzene sulfonic acid + (i) \(\text{NaOH}, \Delta\), (ii) \(\text{H}^+ \rightarrow\)
Answer:
(i) \(\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{OH}\)
(ii) \((\text{CH}_3)_2\text{CH}-\text{OH}\) (Propan-2-ol)
(iii) \(\text{CH}_3\text{CHO} + \text{H}_2\) (Acetaldehyde)
(iv) 2,4,6-tribromophenol (white precipitate)
(v) Acetylsalicylic acid (Aspirin) + \(\text{CH}_3\text{COOH}\)
(vi) Methoxybenzene (Anisole) + \(\text{NaBr}\)
(vii) \(\text{CH}_3\text{CH}_2\text{CH}_2\text{OH} + \text{CH}_3\text{Br}\) (Cleavage yields smaller alkyl group as halide)
(viii) Phenol + \(\text{C}_2\text{H}_5\text{Br}\)
(ix) \((\text{CH}_3)_3\text{C}-\text{I} + \text{C}_2\text{H}_5\text{OH}\) (Tertiary alkyl halide formed due to stable carbocation)
(x) Mixture of 2-nitroanisole and 4-nitroanisole
(xi) Reduction of carbonyl to secondary alcohol: 3-hydroxycyclohexanecarboxylic acid
(xiii) 3-methylbutan-1-ol
(xiv) Phenol (via sodium phenoxide intermediate)
In simple words: Reagents determine functional group transformations like oxidation, reduction, cleavage, and electrophilic aromatic substitution.

Exam Tip: In ether cleavage with HX, tertiary alkyl ethers yield tertiary alkyl halides via an \(\text{S}_\text{N}1\) mechanism involving a stable carbocation.

Question 5. What happens when :
(i) aluminium reacts with tert-butyl alcohol
(ii) phenol is oxidised with chromic acid
(iii) cumene is oxidised in the presence of air and the product formed is treated with dilute acid.
(iv) phenol is treated with conc. \(\text{HNO}_3\).
(v) phenol is treated with chloroform in presence of dilute NaOH.
Answer:
(i) Aluminium reacts with tert-butyl alcohol to form aluminium tert-butoxide and hydrogen gas.
(ii) Phenol is oxidized by chromic acid (\(\text{H}_2\text{CrO}_4\)) to form benzoquinone (a conjugated dicarbonyl compound).
(iii) Cumene (isopropylbenzene) undergoes aerial oxidation to form cumene hydroperoxide, which upon treatment with dilute acid yields phenol and acetone.
(iv) Phenol reacts with concentrated nitric acid to yield 2,4,6-trinitrophenol (picric acid) in low yield due to strong oxidation.
(v) Phenol reacts with chloroform and aqueous sodium hydroxide (Reimer-Tiemann reaction) to form salicylaldehyde.
In simple words: These reactions demonstrate specific chemical properties and industrial synthesis routes for phenols.

Exam Tip: Cumene oxidation is the principal industrial method for manufacturing phenol.

Question 6. How will you convert
(i) propene to propan-l-ol.
(ii) anisole to phenol
(iii) butan-2-one to butan-2-ol
(iv) ethanal to ethanol
(v) phenol to ethoxybenzene
(vi) 1-phenylethene to 1-phenylethanol
(vii) formaldehyde to cyclohexylmethanol
(viii) butyl bromide to pentan-1-ol.
(ix) toluene to benzyl alcohol
(x) 1-propoxypropane to propyl iodide
(xi) ethyl bromide to 1-ethoxyethane
(xii) methyl bromide to 2-methoxy-2-methylpropane
(xiii) ethyl bromide to ethoxybenzene
(xiv) ethanol to benzyl ethyl ether.
Answer:
(i) Hydroboration-oxidation of propene using \(\text{B}_2\text{H}_6\) followed by alkaline \(\text{H}_2\text{O}_2\).
(ii) Heating anisole with anhydrous \(\text{HI}\) or hydrobromic acid.
(iii) Reduction of butan-2-one using \(\text{NaBH}_4\) or \(\text{LiAlH}_4\).
(iv) Reduction of ethanal using \(\text{H}_2/\text{Ni}\) or \(\text{NaBH}_4\).
(v) Convert phenol to sodium phenoxide with \(\text{NaOH}\), then treat with bromoethane (Williamson synthesis).
(vi) Acid-catalyzed hydration of 1-phenylethene.
(vii) React formaldehyde with cyclohexylmagnesium bromide (Grignard reagent) followed by acidic hydrolysis.
(viii) Convert butyl bromide to pentanenitrile using \(\text{KCN}\), followed by acid hydrolysis to pentanoic acid and reduction with \(\text{LiAlH}_4\).
(ix) Oxidation of toluene with alkaline \(\text{KMnO}_4\) to benzoic acid, followed by reduction with \(\text{LiAlH}_4\), or direct chlorination/hydrolysis.
(x) React 1-propoxypropane with excess hot HI.
(xi) Treat sodium ethoxide with ethyl bromide (Williamson synthesis).
(xii) Convert methyl bromide to tert-butoxide via Grignard reaction or use tert-butyl alcohol and sodium metal, then react with methyl bromide.
(xiii) Treat sodium phenoxide with ethyl bromide.
(xiv) Convert ethanol to sodium ethoxide, react with benzyl chloride (Williamson synthesis).
In simple words: Organic conversions require choosing appropriate reagents to lengthen carbon chains, alter functional groups, or change positions.

Exam Tip: For chain extension by one carbon, use the cyanide (\(-\text{CN}\)) route followed by hydrolysis or reduction.

Question 7. Identify the missing reactant or product A to D in the following equations:
(i) (A) + \(\text{HNO}_3 + \text{H}_2\text{SO}_4 \rightarrow 2,4,6\)-trinitrophenol
(ii) Methylcyclohexane + dil. \(\text{H}_2\text{SO}_4 \rightarrow\) (B)
(iii) (C) + \( \text{H}_2\text{O} \xrightarrow{H^+} (\text{CH}_3)_2\text{C}(\text{OH})(\text{CH}_2)_2\text{CH}_3
(iv) \(\text{CH}_3\text{OC}_6\text{H}_5 + \text{HI} \rightarrow\) (D)
Answer:
(i) (A) is Phenol (\(\text{C}_6\text{H}_5\text{OH}\)).
(ii) (B) is 1-methylcyclohexanol (hydration of alkene intermediate / acid-catalyzed addition).
(iii) (C) is 2-methylpent-2-ene (\((\text{CH}_3)_2\text{C}=\text{CHCH}_2\text{CH}_3\)).
(iv) (D) is Phenol and iodomethane (\(\text{C}_6\text{H}_5\text{OH} + \text{CH}_3\text{I}\)).
In simple words: Work backwards from products using known reaction mechanisms to find reactants.

Exam Tip: Acid-catalyzed hydration of alkenes follows Markovnikov's rule and forms the most stable carbocation.

Question 8. Identify X, Y and Z in the following sequence of reactions :
(i) Phenol \(\xrightarrow{\text{Zn dust}}\) X \(\xrightarrow[\text{Anhyd. AlCl}_3]{\text{CH}_3\text{Cl}}\) Y \(\xrightarrow[\text{OH}^-]{\text{KMnO}_4}\) Z
(ii) Ethanol \(\xrightarrow{\text{PBr}_3}\) X \(\xrightarrow{\text{alc. KOH}}\) Y \(\xrightarrow{\text{dil. H}_2\text{SO}_4}\) Z
(iii) Anisole \(\xrightarrow{\text{HI}}\) X + \(\text{CH}_3\text{I}\)
X + conc. \(\text{HNO}_3 \rightarrow\) Y (a dinitro compound)
X + \(\text{Br}_2(\text{aq}) \rightarrow\) Z (a tribromo product)
Answer:
(i) X = Benzene, Y = Toluene, Z = Benzoic acid.
(ii) X = Bromoethane, Y = Ethene, Z = Ethanol.
(iii) X = Phenol, Y = 2,4-dinitrophenol, Z = 2,4,6-tribromophenol.
In simple words: Trace each step sequentially to find intermediate compounds.

Exam Tip: Zinc dust reduces phenol to benzene efficiently by removing the oxygen atom.

Question 10. Write the mechanism for following reactions :
(i) Acid-catalysed hydration of alkenes
(ii) Acid catalysed dehydration of alcohols
(iii) Acid catalysed nucleophilic substitution reaction (ether formation)
(iv) \(\text{CH}_3\text{OCH}_3 + \text{HI} \rightarrow \text{CH}_3\text{OH} + \text{CH}_3\text{I}\)
(v) \((\text{CH}_3)_3\text{C}-\text{O}-\text{CH}_3 + \text{HI} \rightarrow \text{CH}_3\text{OH} + (\text{CH}_3)_3\text{CI}\)
Answer:
(i) Step 1: Protonation of alkene to form a carbocation. Step 2: Nucleophilic attack of water on the carbocation. Step 3: Deprotonation to form alcohol.
(ii) Step 1: Protonation of alcohol to form protonated alcohol. Step 2: Loss of water molecule to form a carbocation. Step 3: Elimination of a proton to form an alkene.
(iii) Step 1: Protonation of alcohol. Step 2: Nucleophilic attack of a second alcohol molecule on protonated alcohol with displacement of water. Step 3: Deprotonation to yield ether.
(iv) Protonation of ether oxygen, followed by iodide attack on methyl carbon to yield methanol and iodomethane via \(\text{S}_\text{N}2\).
(v) Protonation of ether oxygen, followed by cleavage to form stable tert-butyl carbocation and methanol, subsequently yielding tert-butyl iodide via \(\text{S}_\text{N}1\).
In simple words: Organic mechanisms involve protonation, formation of reactive intermediates (carbocations), nucleophilic attack, and deprotonation.

Exam Tip: Always show formal charges and curved arrows pointing from electron-rich to electron-deficient centers in mechanisms.

Question 11. Give reason for the following :
(i) The C-O-C bond angle in dimethyl ether is \(111.7^\circ\)
(ii) Alcohols have higher boiling points than ethers of comparable molecular masses.
(iii) Phenols are more acidic than alcohols.
(iv) Nitrophenol is more acidic than o-methoxyphenol.
(v) Phenol is more reactive towards electrophilic substitution reaction than benzene.
(vii) The given Williamson synthesis for t-butyl ethyl ether yields elimination products instead of substitution.
(viii) Reaction of sodium methoxide with 1-bromo-4-nitrobenzene does not give 1-methoxy-4-nitrobenzene.
(ix) o-nitrophenol is steam volatile but p-nitrophenol is not.
(x) phenol is less polar than ethanol.
(xi) Phenyl methyl ether reacts with HI to form phenol and iodomethane and not iodobenzene and methanol.
(xii) methanol is less acidic than water.
(xiii) alcohols can act as weak base as well as weak acids.
(xiv) phenols do not give protonation reaction readily.
(xvi) absolute ethanol can not be obtained by fractional distillation of ethanol and water mixture.
Answer:
(i) Due to steric repulsion between bulky methyl groups attached to oxygen.
(ii) Alcohols form intermolecular hydrogen bonding, whereas ethers do not.
(iii) Phenoxide ion is more stable due to resonance stabilization compared to alkoxide ion.
(iv) Nitro group is electron-withdrawing (\(-I\) and \(-M\)), increasing acidity, whereas methoxy group is electron-donating (\(+M\)), decreasing acidity.
(v) The hydroxyl group in phenol is strongly activating and ortho/para-directing due to \(+M\) effect.
(vii) Tertiary alkyl halides undergo elimination rather than substitution when treated with strong alkoxides due to steric hindrance.
(viii) Aryl halides are unreactive toward nucleophilic substitution under normal conditions due to resonance and partial double bond character of the C-X bond.
(ix) o-Nitrophenol exhibits intramolecular hydrogen bonding, whereas p-nitrophenol forms intermolecular hydrogen bonding and associates into chains.
(x) The benzene ring in phenol exerts an electron-withdrawing inductive effect, reducing the dipole moment relative to aliphatic alcohols.
(xi) The phenyl-oxygen bond has partial double bond character due to resonance, making cleavage of the methyl-oxygen bond kinetically favored.
(xii) The methyl group in methanol is electron-donating (\(+I\) effect), intensifying negative charge on oxygen and destabilizing methoxide ion relative to hydroxide ion.
(xiii) Alcohols have a lone pair on oxygen (acting as a Lewis base) and a polar O-H bond (acting as a Brønsted acid).
(xiv) The lone pairs on oxygen in phenol participate in resonance with the benzene ring, reducing electron density on oxygen.
(xvi) Ethanol and water form an azeotropic mixture with a constant boiling point composition.
In simple words: Properties of organic molecules depend on hydrogen bonding, resonance, inductive effects, and steric factors.

Exam Tip: Memorize resonance structures of phenoxide ion and nitroaromatic compounds to explain acidity and reactivity trends.

Question 12. Arrange the following in the increasing order of property shown :
(i) methanol, ethanol, diethylether, ethyleneglycol. (Boiling points)
(ii) phenol, o-nitrophenol, m-nitrophenol, p-nitrophenol. (Acid strength)
(iii) dimethylether, ethanol, phenol. (Solubility in water)
(iv) n-butanol, 2-methylpropan-1-ol, 2-methylpropan-2-ol. (Acid strength)
Answer:
(i) Diethyl ether < methanol < ethanol < ethylene glycol.
(ii) Phenol < o-nitrophenol < m-nitrophenol < p-nitrophenol.
(iii) Phenol < ethanol < dimethyl ether (or based on hydrogen bonding capability: dimethyl ether < phenol < ethanol).
(iv) 2-methylpropan-2-ol < 2-methylpropan-1-ol < n-butanol.
In simple words: Boiling points increase with molecular mass and hydrogen bonding; acid strength increases with electron-withdrawing groups and decreases with branching.

Exam Tip: Branching in alkyl chains decreases surface area, lowering van der Waals forces and boiling points, while increasing electron density and reducing acidity.

Question 13. Give a chemical test to distinguish between the following pair of compounds.
(i) n-propyl alcohol and isopropylalcohol
(ii) methanol and ethanol
(iii) cyclohexanol and phenol.
(iv) propan-2-ol and 2-methylpropan-2-ol.
(v) phenol and anisole
(vi) ethanol and diethyl ether
Answer:
(i) Lucas test: Isopropyl alcohol reacts faster (within 5 minutes) than n-propyl alcohol (reacts only on heating).
(ii) Iodoform test: Ethanol gives a yellow precipitate with \(\text{I}_2/\text{NaOH}\), whereas methanol does not.
(iii) Neutral \(\text{FeCl}_3\) test: Phenol gives a violet coloration, whereas cyclohexanol does not.
(iv) Lucas test: 2-methylpropan-2-ol reacts instantaneously, whereas propan-2-ol takes 5 minutes.
(v) Neutral \(\text{FeCl}_3\) test: Phenol gives a violet color, whereas anisole does not.
(vi) Sodium metal test: Ethanol reacts with sodium metal to evolve hydrogen gas, whereas diethyl ether does not.
In simple words: Chemical tests rely on distinct functional group reactivities that produce color changes, precipitates, or gas evolution.

Question 14. Which of the following compounds gives fastest reaction with HBr and why?
(i) \((\text{CH}_3)_3\text{COH }\)
(ii) \(\text{CH}_3\text{CH}_2\text{CH}_2\text{OH}\)
Answer:
(i) \((\text{CH}_3)_3\text{COH }\) reacts the fastest.
In simple words: Tertiary alcohols react faster because they form stable tertiary carbocation intermediates via an \(\text{S}_\text{N}1\) pathway.

Exam Tip: Alcohol reactivity order with halogen acids: Tertiary > Secondary > Primary.

Question 15. What is the function of \(\text{ZnCl}_2\) (anhyd) in Lucas test for distinction between \(1^\circ, 2^\circ\) and \(3^\circ\) alcohols.
Answer:
Anhydrous \(\text{ZnCl}_2\) acts as a Lewis acid that coordinates with the oxygen atom of the hydroxyl group, polarizing the C-O bond and facilitating the departure of the water molecule to generate a carbocation.
In simple words: It helps convert the poor leaving group \((-\text{OH})\) into a good leaving group \((-\text{OH}_2^+)\).

Exam Tip: Lucas reagent is a mixture of concentrated \(\text{HCl}\) and anhydrous \(\text{ZnCl}_2\).

Question 16. An alcohol A (\(\text{C}_4\text{H}_{10}\text{O}\)) on oxidation with acidified potassium dichromate gives carboxylic acid B (\(\text{C}_4\text{H}_8\text{O}_2\)). Compound A when dehydrated with conc. \(\text{H}_2\text{SO}_4\) at 443 K gives compound C. Treatment of C with aqueous \(\text{H}_2\text{SO}_4\) gives compound D (\(\text{C}_4\text{H}_{10}\text{O}\)) which is an isomer of A. Compound D is resistant to oxidation but compound A can be easily oxidised. Identify A, B, C and D and write their structures.
Answer:
[A] : \((\text{CH}_3)_2\text{CHCH}_2\text{OH}\) (2-methylpropan-1-ol)
[B] : \(\text{CH}_3\text{CH}(\text{CH}_3)\text{COOH}\) (2-methylpropanoic acid)
[C] : \((\text{CH}_3)_2\text{C}=\text{CH}_2\) (2-methylpropene)
[D] : \((\text{CH}_3)_3\text{C}-\text{OH}\) (2-methylpropan-2-ol, a tertiary alcohol resistant to oxidation)
In simple words: Stepwise deduction links oxidation and dehydration products to constitutional isomers.

Exam Tip: Primary alcohols oxidize to carboxylic acids, secondary to ketones, and tertiary alcohols do not oxidize under normal conditions.

Question 17. An organic compound A having molecular formula \(\text{C}_6\text{H}_6\text{O}\) gives a characteristic colour with aqueous \(\text{FeCl}_3\). When A is treated with NaOH and \(\text{CO}_2\) at 400 K under pressure, compound B is obtained. Compound B on acidification gives compound C which reacts with acetyl chloride to form D which is a popular pain killer. Deduce the structure of A, B, C and D. What is the common name of Drug D?
Answer:
A = Phenol (\(\text{C}_6\text{H}_5\text{OH}\))
B = Sodium salicylate
C = Salicylic acid (2-hydroxybenzoic acid)
D = Acetylsalicylic acid
Common name of Drug D: Aspirin.
In simple words: Phenol undergoes Kolbe's reaction to yield salicylic acid, which is acetylated to form aspirin.

Exam Tip: Aspirin is widely used as an analgesic and antipyretic drug.

Question 19. An ether A (\(\text{C}_5\text{H}_{12}\text{O}\)) when heated with excess of hot concentrated HI produced two alkyl halides which on hydrolysis form compounds B and C. Oxidation of B gives an acid D whereas oxidation of C gave a ketone E. Deduce the structures of A, B, C, D and E.
Answer:
(A) = \(\text{CH}_3\text{CH}_2\text{OCH}(\text{CH}_3)_2\) (2-ethoxypropane)
(B) = \(\text{CH}_3\text{CH}_2\text{OH}\) (Ethanol)
(C) = \(\text{CH}_3\text{CHOHCH}_3\) (Propan-2-ol)
(D) = \(\text{CH}_3\text{COOH}\) (Acetic acid)
(E) = \(\text{CH}_3\text{COCH}_3\) (Acetone)
In simple words: Cleavage of ether yields alcohols/halides that oxidize according to their primary or secondary nature.

Exam Tip: Primary alcohols oxidize to acids, while secondary alcohols oxidize to ketones.

Question 20. Phenol, \(\text{C}_6\text{H}_5\text{OH}\) when it first reacts with concentrated sulphuric acid, forms Y. Y is reacted with concentrated nitric acid to form Z. Identify Y and Z and explain why phenol is not converted commercially to Z by reacting it with conc. \(\text{HNO}_3\).
Answer:
Y = Phenol-2,4-disulfonic acid (or 4-hydroxybenzenesulfonic acid derivative)
Z = 2,4,6-trinitrophenol (Picric acid)
Reason: Direct nitration of phenol with conc. \(\text{HNO}_3\) gives a very poor yield due to strong oxidation of the activated benzene ring, leading to tarry by-products. Sulfonation deactivates the ring slightly, allowing controlled nitration.
In simple words: Sulfonation protects and controls subsequent nitration to obtain picric acid in better yields.

Question 21. Synthesise the following alcohols from suitable alkenes.
(a) 1-methylcyclohexanol
(b) 3-methylpentan-3-ol
Answer:
(a) Acid-catalyzed hydration of 1-methylcyclohexene.
(b) Acid-catalyzed hydration of 3-methylpent-2-ene or 3-methylpent-3-ene.
In simple words: Add water across the double bond following Markovnikov's rule.

Question 22. How are the following ethers prepared by Williamson synthesis?
(a) Ethoxybenzene
(b) 2-methoxy-2-methylpropane
Answer:
(a) Treatment of sodium phenoxide with bromoethane (\(\text{C}_6\text{H}_5\text{ONa} + \text{C}_2\text{H}_5\text{Br} \rightarrow \text{C}_6\text{H}_5\text{OC}_2\text{H}_5 + \text{NaBr}\)).
(b) Treatment of sodium tert-butoxide with chloromethane or iodomethane (\((\text{CH}_3)_3\text{CONa} + \text{CH}_3\text{I} \rightarrow (\text{CH}_3)_3\text{C}-\text{O}-\text{CH}_3 + \text{NaI}\)).
In simple words: Williamson synthesis involves reacting a sodium alkoxide/phenoxide with a primary alkyl halide.

Exam Tip: Always use a primary alkyl halide and tertiary alkoxide to prevent competing elimination reactions.