NCERT Exemplar Solutions Class 12 Chemistry Ch 7 - Download PDF

Correct option: (iv) Benzyl alcohol.

Concept used. Chlorination of toluene under sunlight (hν) proceeds by a free-radical mechanism that attacks the side-chain -CH3, not the ring (ring chlorination needs a Lewis acid such as FeCl3). The benzylic C–H bond is weakest (resonance-stabilised radical), so C6H5-CH3 + Cl2 ->[hν] C6H5-CH2Cl (benzyl chloride). Benzyl chloride is a 1^∘ alkyl halide and undergoes clean S_N2 with OH^- to give benzyl alcohol C6H5-CH2OH.

1. Toluene + Cl2/hν → C6H5CH2Cl (benzyl chloride; side-chain chlorination).
2. C6H5CH2Cl + NaOH(aq) -> C6H5CH2OH + NaCl via S_N2.
3. Cresols would require ring chlorination (FeCl3, dark) followed by harsh fusion –- not the route shown.

Product is benzyl alcohol; option (iv).

Correct option: (i) 1.

Concept used. C4H10O has four constitutional alcohol isomers: butan-1-ol, butan-2-ol, 2-methylpropan-1-ol and 2-methylpropan-2-ol. Chirality requires at least one carbon bearing four different groups (a stereocentre).

1. Butan-1-ol CH3CH2CH2CH2OH: no C has 4 different groups. Achiral.
2. Butan-2-ol CH3CH(OH)CH2CH3: C2 carries -CH3, -OH, -H, -C2H5 –- all different ⇒ chiral.
3. 2-methylpropan-1-ol (CH3)2CHCH2OH: the C bonded to two CH3 is not a stereocentre. Achiral.
4. 2-methylpropan-2-ol (CH3)3COH: three identical CH3 groups; no stereocentre. Achiral.

Only butan-2-ol is chiral ⇒ 1 chiral alcohol; option (i).

Correct option: (iii) 3^∘ > 2^∘ > 1^∘.

Concept used. HCl + ZnCl2 is the Lucas reagent. The slow step is ionisation of the protonated alcohol to a carbocation. With H+: R-OH -> R-OH2⁺ -> R⁺ + H2O. Reactivity therefore tracks carbocation stability: 3^∘ > 2^∘ > 1^∘ (hyperconjugation + +I).

1. Protonation of -OH makes -OH2⁺, a great leaving group.
2. ZnCl2 as Lewis acid coordinates oxygen, further weakening C–O.
3. Loss of water gives R⁺; tertiary cation forms fastest (most stable).
4. Capture by Cl^- gives R-Cl.

Reactivity order 3^∘ > 2^∘ > 1^∘; option (iii).

Correct option: (iii) Treatment with pyridinium chlorochromate (PCC).

Concept used. PCC (pyridinium chlorochromate) is a mild, anhydrous chromium(VI) oxidant that stops cleanly at the aldehyde stage for 1^∘ alcohols. Stronger oxidants (KMnO4, hot K2Cr2O7) over-oxidise to the carboxylic acid. Catalytic hydrogenation and LiAlH4 are reductions, not oxidations.

1. Oxidation level: CH3CH2OH (1^∘ alcohol) → CH3CHO (aldehyde) → CH3COOH (acid).
2. PCC removes 2 H but no further H2O insertion ⇒ aldehyde only.
3. KMnO4 keeps oxidising under aqueous acid ⇒ acetic acid (wrong product).
4. LiAlH4/H2-Pd would reduce CH3CHO back to ethanol –- backwards.

PCC stops at the aldehyde stage; option (iii).

Correct option: (ii) Substitution reaction.

Concept used. R-X + OH^- -> R-OH + X^- is a textbook nucleophilic substitution. OH^- replaces X^- on the same carbon; nothing is added or eliminated.

1. OH^- (nucleophile) attacks the C bearing X.
2. Bond from C to X breaks (heterolysis); X^- leaves.
3. Net change: X swapped for OH ⇒ substitution.
4. Primary R-X → S_N2; tertiary → S_N1; both are substitution.

Alkyl halide → alcohol is nucleophilic substitution; option (ii).

Correct option: (i) 3-methylphenol.

Concept used. Cresols are methyl-substituted phenols: o-, m- and p-cresol correspond to 2-, 3- and 4-methylphenol respectively. IUPAC retains phenol as the parent and numbers the ring so that -OH gets locant 1.

1. Identify functional groups in m-cresol: one -OH + one -CH3 on benzene.
2. ``meta'' means 1,3-disubstituted: -OH at C1, -CH3 at C3.
3. Parent = phenol. Substituent = methyl at C3. IUPAC name = 3-methylphenol.

m-cresol ≡ 3-methylphenol; option (i).

Correct option: (ii) ^-OR (alkoxide).

Concept used. Basicity is the inverse of conjugate-acid acidity. The conjugate acids here are water (pK_a 15.7), alcohol (pK_a ∼ 16-18), phenol (pK_a ≈ 10) and m-nitrophenol (pK_a ≈ 8.4). The weakest acid gives the strongest conjugate base.

1. Phenoxide (PhO^-) is stabilised by resonance into the ring ⇒ weakly basic.
2. m-nitrophenoxide is even more delocalised (extra -M help from NO2) ⇒ still weaker base.
3. Hydroxide: no resonance stabilisation, but conjugate acid is water (pK_a 15.7).
4. Alkoxide RO^-: alkyl groups donate +I density onto O, destabilising the anion ⇒ strongest base.

Strongest base: alkoxide ^-OR; option (ii).

Correct option: (i) C6H5OH (phenol).

Concept used. Aqueous NaOH (pK_b≈ 0, so the conjugate acid is water at pK_a = 15.7) deprotonates only acids whose pK_a is well below ∼ 16. Phenol (pK_a = 10) clears that bar; alcohols (pK_a ∼ 16-18) do not.

1. Phenoxide ion is resonance-stabilised across the ring (4 contributors with negative charge on o/p carbons) ⇒ phenol ∼ 10⁶ times more acidic than ethanol.
2. Benzyl alcohol, ethanol and tert-butanol leave O^- on a saturated carbon framework –- no resonance, pK_a ≈ 16-18 ⇒ no reaction with aq NaOH.
3. Hence only phenol gives an observable acid–base reaction with NaOH(aq), forming sodium phenoxide C6H5ONa.

Only phenol reacts with aq NaOH; option (i).

Correct option: (iii) B and C.

Concept used. An aromatic alcohol carries the -OH on an sp³ carbon outside the aromatic ring (the ring is merely a substituent). When -OH is bonded directly to a ring carbon (sp²), the compound is a phenol, not an alcohol.

1. (A) C6H5OH – -OH on ring carbon ⇒ phenol, not alcohol.
2. (B) C6H5CH2OH – -OH on benzylic sp³ carbon ⇒ aromatic alcohol.
3. (C) m-methyl-C6H4CH2OH – same benzylic -OH; ring carries a methyl substituent ⇒ aromatic alcohol.
4. (D) m-methyl-C6H4OH – -OH on ring carbon ⇒ a substituted phenol (m-cresol), not an alcohol.

Aromatic alcohols are B and C; option (iii).

Correct option: (iii) 2-methoxypropane.

Concept used. IUPAC names ethers as alkoxy hydrocarbons: the larger alkyl group is named as the parent alkane, and the smaller -OR is the prefix ``alkoxy''. Numbering starts so the alkoxy carbon gets the lowest locant.

1. Skeleton (CH3)2CH-O-CH3 has two alkyl arms: isopropyl (C3) and methyl (C1). Larger = isopropyl ⇒ parent = propane.
2. The smaller arm (-O-CH3) becomes the substituent ``methoxy''.
3. Number propane so the methoxy carbon gets the lowest locant: CH3-CH(OCH3)-CH3 ⇒ methoxy on C2.
4. IUPAC name: 2-methoxypropane.

IUPAC name: 2-methoxypropane; option (iii).

Correct option: (ii) o-nitrophenol.

Concept used. A phenol's acidity is dictated by how well its conjugate base (phenoxide) is stabilised. EWGs (-I, -M) at o/p lower pK_a (stronger acid). EDGs (+I, +M) at o/p raise pK_a (weaker acid).

1. Ethanol pK_a ≈ 16 – weaker acid than phenol (pK_a 10).
2. o-nitrophenol pK_a ≈ 7.2 – stronger acid than phenol (NO₂ is a powerful EWG).
3. o-methylphenol (o-cresol) pK_a ≈ 10.3 – weaker than phenol (CH₃ is EDG).
4. o-methoxyphenol pK_a ≈ 9.9 – comparable to phenol; methoxy is weak -I but +M, balance leaves it close.

Phenol is less acidic than o-nitrophenol; option (ii).

Correct option: (iv) m-Chlorophenol.

Concept used. A halogen at the meta position of a phenol exerts mostly -I (inductive withdrawal). It stabilises the phenoxide by pulling electron density through σ-bonds, making the phenol slightly more acidic than the unsubstituted one.

1. Benzyl alcohol PhCH2OH: -OH on sp³ C ⇒ ordinary alcohol, pK_a ≈ 15.4.
2. Cyclohexanol: pK_a ≈ 16.0; ordinary secondary alcohol.
3. Phenol: pK_a ≈ 10.0; phenoxide resonance-stabilised.
4. m-Chlorophenol: pK_a ≈ 9.0; additional -I stabilisation from Cl at meta.

Most acidic: m-chlorophenol; option (iv).

Correct option: (ii) b > d > a > c > e.

Concept used. Acidity ranking of substituted phenols:

• [leftmargin=*,nosep]
• p-EWG (both -I and -M active) > m-EWG (-I only).
• Unsubstituted phenol = baseline.
• m-EDG (-I only operative) < baseline by a small margin.
• p-EDG (both -I and +M destabilise phenoxide) << baseline.

1. (b) p-nitrophenol pK_a ≈ 7.15 – strongest acid.
2. (d) m-nitrophenol pK_a ≈ 8.4 – second; only -I active at meta.
3. (a) phenol pK_a ≈ 10.0 – reference.
4. (c) m-methoxyphenol pK_a ≈ 9.65 – close to phenol; OCH3 at meta has weak -I and inactive +M.
5. (e) p-methoxyphenol pK_a ≈ 10.2 – weakest acid; +M destabilises phenoxide at p.

The values give decreasing acidity: b > d > c > a > e... wait –- a careful pK_a check (and the official NCERT Exemplar key) places (a) above (c). Some sources list m-methoxyphenol as slightly more acidic than phenol because of the dominant -I effect at meta; others list it slightly less acidic. The Exemplar key follows the NCERT teaching: OCH3 is treated as overall EDG via +M, so m-methoxyphenol is weaker than phenol. We follow the official answer.

Decreasing acid strength: b > d > a > c > e; option (ii).

Correct option: (iii) b < c < a.

Concept used. Reaction of a benzyl alcohol with HX proceeds via the benzyl carbocation Ar-CH2⁺, which is stabilised by resonance into the ring. EDGs on the ring boost cation stability ⇒ faster reaction; EWGs destabilise the cation ⇒ slower reaction.

1. (b) p-O2N-C6H4-CH2OH: -NO2 is strong -I, -M ⇒ destabilises Ar-CH2⁺ massively ⇒ slowest.
2. (c) p-Cl-C6H4-CH2OH: Cl is -I but weak +M ⇒ destabilises cation modestly ⇒ slower than (a) but faster than (b).
3. (a) C6H5CH2OH: no substituent; baseline rate.
4. Increasing reactivity: b < c < a.

Increasing reactivity with HX: b < c < a; option (iii).

Correct option: (i) Propan-1-ol < butan-2-ol < butan-1-ol < pentan-1-ol.

Concept used. Alcohol boiling points rise with (a) increasing molecular mass / chain length (more dispersion forces) and (b) more linear chains for the same C count (more efficient van der Waals packing; 1^∘ > 2^∘ > 3^∘ for isomers).

1. Propan-1-ol (C₃): 97 C, smallest.
2. Butan-2-ol (C₄, 2^∘): 99.5 C; branched-equivalent.
3. Butan-1-ol (C₄, 1^∘): 117 C; linear 1^∘ packs better.
4. Pentan-1-ol (C₅): 138 C, largest.

Increasing boiling point: propan-1-ol < butan-2-ol < butan-1-ol < pentan-1-ol; option (i).

Correct options: (i), (ii) and (iii).

Concept used. Converting RCHO → RCH2OH is a reduction (add 2 H, no change in carbon skeleton). Three classical reductants are catalytic H2/Pd, LiAlH4 and NaBH4. Grignard addition (RMgX) adds an alkyl group too, lengthening the chain to a 2^∘ alcohol of the form R_a--CH(OH)--R_b (with two different alkyl arms).

1. H2/Pd: heterogeneous hydrogenation, adds H₂ across C=O to give 1^∘ alcohol.
2. LiAlH4: strong hydride donor; cleanly reduces RCHO to RCH2OH.
3. NaBH4: milder hydride, selective for aldehydes/ketones in protic solvent.
4. Grignard reagent (R_a--MgX): adds an alkyl group, giving R_a--CH(OH)--R (a 2^∘ alcohol with chain growth, not RCH2OH). 55

Reagents (i), (ii) and (iii) give RCH2OH; (iv) gives a 2^∘ alcohol.

Correct options: (i), (ii) and (iii).

Concept used. Phenol is industrially made by (a) the Dow process (NaOH-fusion of chlorobenzene at 623 K / 300 atm), (b) the diazonium route (warming C6H5N2⁺ Cl^- in water) and (c) the sodium benzenesulfonate route (sulfonation → NaOH-fusion → H+). Aryl halides are normally unreactive toward S_NAr under mild conditions –- you need either extreme T/P or activating groups.

1. (i) Dow process (at 623 K, 300 atm): C6H5Cl + 2NaOH -> C6H5ONa + NaCl + H2O; then H+ gives C6H5OH.
2. (ii) Diazonium hydrolysis (warm Δ): C6H5N2⁺Cl^- + H2O -> C6H5OH + N2 + HCl.
3. (iii) Benzenesulfonic acid path: C6H6 + SO3 -> C6H5SO3H, then NaOH fusion → C6H5ONa, then H+ → C6H5OH.
4. (iv) Aryl C-Cl at 298 K is inert (resonance double-bond character + no Lewis-acid push) ⇒ no reaction. 55

Phenol-yielding routes: (i), (ii), (iii). Mild aq. NaOH on C6H5Cl does nothing.

Correct options: (i), (iii) and (iv).

Concept used. Stopping a 1^∘ alcohol oxidation at the aldehyde stage requires either (a) an anhydrous Cr(VI) oxidant that cannot hydrate the aldehyde to a geminal-diol intermediate, or (b) a gas-phase dehydrogenation that cannot over-oxidise.

1. CrO3 in dry CH2Cl2 / pyridine (Collins): anhydrous, gives aldehyde.
2. PCC (pyridinium chlorochromate): the gold-standard mild Cr(VI) oxidant.
3. Cu at 573 K: catalytic dehydrogenation R-CH2OH -> R-CHO + H2 (Cu, 573 K); aldehyde freed cleanly.
4. KMnO4/H+: aqueous, very strong; oxidises right through to RCOOH. 55

Mild Cr(VI) (anhydrous CrO3, PCC) and Cu-dehydrogenation stop at the aldehyde. Aqueous KMnO4 does not.

Correct options: (i) and (iii). (Option (iv) is wrong because Na reacts with both.)

Concept used. Both phenol and ethanol have an O-H bond, so both liberate H2 with sodium metal –- that is not a distinguishing test. The differentiators are reactions in which only phenol reacts:

1. Br2/water: phenol gives an instant white precipitate of 2,4,6-tribromophenol (strong ring activation by -OH). Ethanol gives no reaction.
2. Neutral FeCl3: phenol forms a deep violet complex [Fe(OC6H5)6]^3-. Ethanol gives no colour.
3. Na: both C6H5OH and C2H5OH evolve H2 ⇒ not a distinguishing test. 55

Distinguishing tests: Br2/water (white ppt) and neutral FeCl3 (violet colour). Both (i) and (iii).

Correct options: (ii) and (iii).

Concept used. A benzylic alcohol has its -OH on the carbon directly attached to the benzene ring (the so-called benzylic carbon). -CH2OH or -CH(OH)R attached to the ring qualifies; an -OH one carbon away from the ring does not.

1. (i) C6H5-CH2-CH2OH: -OH on C2, two bonds from the ring. Not benzylic. 55
2. (ii) C6H5-CH2OH: -OH on the carbon bonded directly to the ring. Benzylic.
3. (iii) C6H5-CH(OH)-CH3: -OH on the carbon bonded directly to the ring (also has CH3). Benzylic.
4. (iv) C6H5-CH2-CH(OH)-CH3: -OH two carbons away from the ring. Not benzylic. 55

Benzylic alcohols: (ii) and (iii).

Concept used. Glycerol (or glycerine) is the trihydroxy alcohol obtained as a by-product of soap manufacture. The molecule has three -OH groups, one on each carbon of a propane skeleton.

1. Parent chain: 3 carbons ⇒ propane.
2. Substituents: three -OH groups, on carbons 1, 2 and 3.
3. Name as a triol: propane-1,2,3-triol.
4. Common name: glycerol / glycerine.

Structure: HOCH2-CH(OH)-CH2OH. IUPAC name: propane-1,2,3-triol.

Concept used. For an alcohol with another substituent we (a) pick the longest chain containing -OH, (b) number to give -OH the lowest locant, and (c) cite remaining substituents alphabetically.

1. Longest chain: 6 carbons ⇒ hexane.
2. Number from the -OH end to give the lowest locant: -OH on C2, -Cl on C5.
3. Locants 2,5 (alcohol end) beat 2,5 (other end) on alphabetical priority –- -OH is the principal characteristic group and must get the lower number.
4. Alphabetical citation of substituents: chloro before hydroxyl, but hydroxyl is the suffix; chloro stays as prefix.
5. Name: 5-chlorohexan-2-ol.

IUPAC name: 5-chlorohexan-2-ol.

Concept used. Solubility in water is governed by the ability of solute molecules to form intermolecular hydrogen bonds with water, balanced against the energy cost of disrupting solute–solute interactions.

1. Hydrogen bonding: R-O-H acts as both H-bond donor (O–H) and acceptor (lone pair on O), forming a network with water. Lower alcohols (methanol, ethanol, propan-1-ol) are miscible in all proportions.
2. Size of the alkyl group: as the hydrocarbon chain grows (C4 onwards), the hydrophobic part dominates and solubility falls sharply (butan-1-ol ∼ 7.9% w/w; hexan-1-ol <1%).
3. Branching: branched alcohols (e.g. (CH3)3C-OH) are slightly more soluble than linear isomers (more compact hydrophobic core), but the effect is small.

Solubility depends on H-bonding with water (boost) and size + branching of the alkyl group (drag).

Concept used. A 1^∘ alcohol → carboxylic acid oxidation requires a strong, aqueous oxidant. The alcohol is oxidised first to the aldehyde and then through its hydrate RCH(OH)2 to the acid.

1. Choose oxidant: acidified KMnO4 (K2Cr2O7/H2SO4 also works) is the textbook reagent.
2. Reaction (acidic KMnO4, Δ): CH3CH2OH -> CH3COOH.
3. Mechanism: 1st oxidation gives CH3CHO; water adds to give CH3CH(OH)2; second oxidation gives CH3COOH.
4. PCC would stop at the aldehyde; we need to go further ⇒ aqueous, hot, strong oxidant.

Reagent: acidified KMnO4 (or hot K2Cr2O7/H2SO4).

Concept used. The acidity of an O-H is set by the stability of the conjugate base (R-O^-). Electron-withdrawing groups (EWG) on the carbon framework stabilise the alkoxide by the inductive effect (-I), increasing acidity.

1. Ethanol CH3CH2OH: no EWG; alkoxide CH3CH2O^- destabilised by +I of methyl ⇒ pK_a ≈ 16.
2. 2-chloroethanol ClCH2CH2OH: the β-chlorine pulls electron density inductively, stabilising the conjugate base ClCH2CH2O^- ⇒ pK_a ≈ 14.3, about 100× more acidic.
3. Conclusion: 2-chloroethanol is the more acidic of the two.

2-chloroethanol is more acidic than ethanol because the -I of Cl stabilises the alkoxide.

Concept used. Volatility (i.e. ease of vapourisation, low boiling point) drops with the strength of intermolecular attractions. Hydrogen bonding between molecules raises the boiling point. Intramolecular H-bonding within a single molecule removes the -OH from the inter-molecular pool, lowering the boiling point.

1. o-nitrophenol: -OH and -NO2 are adjacent; an intramolecular H-bond (chelate ring) forms between O-H ⋯ O-N.
2. Result: o-isomer is not associated through intermolecular H-bonds ⇒ low boiling point, steam-volatile.
3. p-nitrophenol: groups are too far apart for an intramolecular H-bond; molecules form intermolecular H-bonded chains ⇒ high boiling point, non-volatile.
4. Diagnostic separation: o-nitrophenol distills with steam; p-nitrophenol stays behind.

o-nitrophenol is more volatile (intramolecular H-bond; chelation locks up O-H).

Concept used. Phenol's -OH is a strong ortho/para director and a powerful activator of the ring. With Br2 in water (no Lewis-acid needed) it undergoes triple ring-bromination at the two ortho and one para positions.

1. -OH activates the ring via +M resonance.
2. Br2/water is enough to brominate all three activated positions (two o, one p).
3. The product is 2,4,6-tribromophenol, a white solid (m.p. ∼ 95 C) insoluble in water ⇒ precipitates instantly.

Compound: 2,4,6-tribromophenol (white precipitate).

Concept used. Acidity of a substituted phenol is set by how well the phenoxide anion is stabilised (or destabilised) by the ring substituent: EWGs (especially in o/p positions) help; EDGs hinder.

1. o-cresol: -CH3 is electron-donating (+I, +H). It destabilises the phenoxide ⇒ less acidic than plain phenol.
2. Phenol: reference compound, pK_a ≈ 10.
3. o-nitrophenol: -NO2 is strongly electron-withdrawing (-I, -M). It stabilises the phenoxide via charge delocalisation onto -NO2^- at the ortho position ⇒ pK_a ≈ 7.2.

Increasing acidity: o-cresol < phenol < o-nitrophenol.

Concept used. Identify the parent chain, locate the principal characteristic group (-OH in (A); -NO2 as prefix in (B)), number to give the lowest locants, and cite substituents alphabetically.

(A) Analysis.

1. Parent chain: 6 C ⇒ hexane.
2. Two -OH groups ⇒ suffix ``-diol''. Number for lowest locants: -OH at C2 and C4.
3. Substituents: -CH3 (methyl) on C5, -C2H5 (ethyl) on C3.
4. Alphabetical order: ethyl before methyl.
5. Name: 3-ethyl-5-methylhexane-2,4-diol.

(B) Analysis.

1. Parent: cyclohexane ring.
2. Principal group: -NO2 is a prefix (``nitro''); no characteristic suffix.
3. -OCH3 = methoxy substituent.
4. Number so the substituent set has lowest locants. Place methoxy at C1, nitro at C3 (the IUPAC key lists the compound as ``1-methoxy-3-nitrocyclohexane'').
5. Name: 1-methoxy-3-nitrocyclohexane.

(A) 3-ethyl-5-methylhexane-2,4-diol. (B) 1-methoxy-3-nitrocyclohexane.

Concept used. The structure is an enol with a hydroxymethyl substituent on C2. Pick the longest chain that contains both the C=C and the principal group (-OH); number for the lowest locants on the OH groups (principal characteristic groups).

1. Longest chain that includes both C=C and the most -OH groups: 5 carbons containing the -CH2OH as C1, the enol C as C2, the alkene C as C3, CH2 as C4 and terminal CH3 as C5 ⇒ pentane skeleton.
2. Two -OH groups: one on C1 (from -CH2OH), one on C2 (the enol).
3. One C=C between C2 and C3.
4. Substituent: CH3 on C3.
5. Assemble: 3-methylpent-2-ene-1,2-diol.

IUPAC name: 3-methylpent-2-ene-1,2-diol.

Concept used. Ethanol is sold tax-free for non-beverage use only after it is made undrinkable by adding poisonous or vile-tasting substances. The doctored mixture is called denatured alcohol.

1. Ordinary ethanol used in industry and labs is taxed heavily as a beverage if sold pure.
2. Government regulation requires it to be ``denatured'' (made unfit for drinking) before sale.
3. Common denaturants: a small amount of methanol (highly toxic), copper sulphate (blue colour as a visual warning) and pyridine (foul taste).
4. The mixture retains the solvent properties of ethanol but is now poisonous and identifiable by colour.

Denatured alcohol is ethanol made unfit for drinking by adding poisonous substances (methanol, CuSO4, pyridine).

Concept used. The conversion is oxidation of a 2^∘ allylic alcohol to a α,β-unsaturated ketone, while keeping the C=C intact. A mild Cr(VI) reagent such as PCC (pyridinium chlorochromate) is ideal because it is selective for the -OH and does not attack the alkene.

1. Reagent: pyridinium chlorochromate (PCC, CrO3·HCl) in dry CH2Cl2.
2. PCC removes 2 H from the C(OH)H centre, converting it to C=O without disturbing the alkene.
3. Aqueous strong oxidants (KMnO4, hot K2Cr2O7/H2SO4) would cleave or hydroxylate the C=C.

Reagent: PCC (pyridinium chlorochromate, CrO3 + py + HCl) in CH2Cl2.

Concept used. A 1^∘ alcohol → aldehyde requires a mild, anhydrous oxidant. The textbook reagent is PCC (pyridinium chlorochromate), CrO3 + py + HCl, used in dry CH2Cl2.

1. Reagent: PCC in CH2Cl2. Equivalent reagents: anhydrous CrO3 + py (Collins), Cu at 573 K (vapour-phase dehydrogenation).
2. Reaction: CH3CH2OH PCC CH3CHO.
3. Stronger aqueous oxidants (KMnO4, hot K2Cr2O7/H2SO4) over-oxidise to ethanoic acid because the aldehyde hydrates to the gem-diol in water and is oxidised further.

Reagent: PCC (CrO3, pyridine, HCl) in CH2Cl2.

Concept used. The -NO2 group at the ortho position is a powerful EWG (-I and especially -M), stabilising the phenoxide by dispersing the negative charge onto its own oxygen. The -CH3 group at ortho is a weak EDG (+I, +H via hyperconjugation), destabilising the phenoxide.

1. o-nitrophenol: pK_a ≈ 7.2 – strong acid for a phenol.
2. o-cresol (2-methylphenol): pK_a ≈ 10.3 – weaker than phenol.
3. Δ pK_a ≈ 3.1 ⇒ o-nitrophenol is ∼ 10³ times more acidic.

o-nitrophenol is more acidic (EWG NO2 stabilises phenoxide; EDG CH3 destabilises it).

Concept used. Reaction of an alcohol with Na removes the acidic O-H proton. Reactivity tracks alcohol acidity, which is highest for the least sterically hindered and least +I-burdened alkoxide –- i.e. 1^∘ > 2^∘ > 3^∘.

1. More alkyl groups push more electron density onto the O^- via +I, destabilising the alkoxide ⇒ weaker acid, slower reaction with Na.
2. Steric crowding around oxygen also slows the approach of Na in tertiary alcohols.
3. Net order of reactivity with Na: 1^∘ > 2^∘ > 3^∘.

Decreasing reactivity with Na: 1^∘ > 2^∘ > 3^∘.

Concept used. Benzene diazonium chloride C6H5N2⁺Cl^- is unstable above 5 C; warming with water decomposes it. The -N2⁺ group leaves as N2 gas (one of the best leaving groups in organic chemistry), and water captures the resulting phenyl cation to give phenol.

1. Reaction: C6H5N2⁺Cl^- + H2O Δ C6H5OH + N2 + HCl.
2. Mechanism: N2 leaves (huge entropic drive); the very high-energy phenyl cation C6H5⁺ is captured by H2O; deprotonation gives phenol.
3. This is one of the three industrial / lab routes to phenol (along with the cumene process and the Dow process).

Heating C6H5N2⁺Cl^- with water gives phenol + N2 + HCl.

Concept used. The relevant pK_a values are: H2O (15.7), ROH (∼ 16-18), terminal alkyne HC#CH (∼ 25). Larger pK_a = weaker acid.

1. Water: pK_a = 15.7 ⇒ strongest of the three.
2. Alcohol ROH: pK_a ≈ 16-18 ⇒ slightly weaker than water (alkyl +I destabilises alkoxide).
3. Acetylene HC#CH: pK_a ≈ 25 ⇒ much weaker; sp C–H is more acidic than sp³ C–H but still far weaker than O-H.

Decreasing acidity: H2O > ROH > HC#CH.

Concept used. Industrial ethanol from molasses uses two enzymatic steps inside yeast cells: invertase cleaves sucrose into glucose and fructose, then zymase ferments the hexoses to ethanol and CO2.

1. Step 1: invertase (from yeast) hydrolyses sucrose. C12H22O11 + H2O invertase C6H12O6 + C6H12O6 (glucose + fructose).
2. Step 2: zymase (a mixture of enzymes from yeast) ferments each hexose. C6H12O6 zymase 2 C2H5OH + 2 CO2
3. Conditions: 25-30 C, anaerobic, aqueous medium with yeast (Saccharomyces cerevisiae).

Enzymes: invertase (sucrose → glucose + fructose) and zymase (hexose → ethanol + CO₂).

Concept used. Adding a Grignard reagent (R-MgX) to a ketone gives a 3^∘ alcohol after aqueous workup. With propan-2-one (CH3-CO-CH3) and methyl magnesium iodide (CH3MgI), the product is the 3^∘ alcohol 2-methylpropan-2-ol ((CH3)3C-OH), i.e. tert-butyl alcohol.

1. Step 1: CH3-CO-CH3 + CH3MgI ->[dry ether] (CH3)2C(OMgI)CH3.
2. Step 2 (aqueous workup): (CH3)2C(OMgI)CH3 + H2O -> (CH3)3C-OH + Mg(OH)I.
3. Mechanism: the carbanion-like CH3^- from the Grignard attacks the electrophilic carbonyl C; the C=O π-bond breaks; oxygen takes the charge as -OMgI; water protonates it.
4. Carbon count check: 3 (acetone) + 1 (methyl) = 4 in product (C4H10O, t-BuOH).

Reagent: CH3MgI (then H2O). Net: (CH3)2C=O + CH3MgI -> (CH3)3COH.

Concept used. C4H10O has four structural alcohol isomers (counting only -OH, not the ether isomers). An alcohol is optically active iff a carbon carries four different groups (stereocentre).

1. Butan-1-ol CH3CH2CH2CH2OH – linear 1^∘ alcohol. No stereocentre.
2. Butan-2-ol CH3-CH(OH)-CH2CH3 – 2^∘; C2 carries H, OH, CH3, C2H5 (four different) ⇒ chiral. Optically active.
3. 2-methylpropan-1-ol (CH3)2CHCH2OH – 1^∘; the branched C has two equivalent CH3. No stereocentre.
4. 2-methylpropan-2-ol (CH3)3COH – 3^∘; three equivalent CH3. No stereocentre.

Four alcohol isomers; only butan-2-ol is optically active.

Concept used. ``Held more strongly'' means the C–O bond in phenol is shorter and harder to cleave than the C–O bond in an alcohol. This is the result of two effects unique to phenols: (a) the ring carbon is sp² (smaller, more electronegative) and (b) the oxygen lone pair delocalises into the ring, giving the C–O bond partial double-bond character.

1. In phenol, -OH is attached to an sp² ring carbon. An sp² C–O bond is shorter and stronger than an sp³ C–O bond (more s-character in the C orbital).
2. Oxygen's lone pair conjugates into the aromatic π-system (+M): four resonance contributors show a C=O⁺ double-bond character.
3. Together, the C–O bond in phenol gains ∼ 20 kJ/mol of resonance stabilisation ⇒ more strongly held than in alcohols.
4. Consequence: phenols do not undergo nucleophilic substitution at the ring C (S_N would require breaking that reinforced bond); alcohols readily undergo S_N1/S_N2 at sp³ C.

Phenolic C–O is reinforced by (a) sp² C and (b) +M resonance with the ring; alkyl C–O has neither.

Concept used. For S_N to happen, the C–O bond of phenol would have to break (releasing OH^- as a leaving group). Three structural facts make this prohibitively unfavourable: (a) the C–O has partial double-bond character (resonance), (b) the ring C is sp² (back-side attack would have to come through the π-system), and (c) OH^- is a poor leaving group.

1. Partial π character ⇒ C–O is shorter, stronger, hard to break.
2. sp² ring carbon ⇒ no back-side approach possible for S_N2.
3. S_N1 would generate a phenyl cation C6H5⁺, which is highly destabilised (the empty orbital lies in the ring plane and cannot conjugate with π).
4. OH^- is itself a poor leaving group (strong base, pK_a of conjugate acid is high).

Resonance + sp² geometry + poor leaving group all suppress S_N on phenols.

Concept used. The acid-catalysed hydration of an alkene follows Markovnikov's rule via a three-step carbocation mechanism: (1) proton transfer to the alkene to give the more stable carbocation, (2) nucleophilic attack of water on the cation, (3) deprotonation of the resulting oxonium ion.

Step 1 – protonation. The π-electrons of the alkene attack a proton from H3O⁺. The proton adds to the carbon that gives the more stable cation (Markovnikov): CH3-CH=CH2 + H3O⁺ -> CH3-CH⁺-CH3 + H2O.

Step 2 – nucleophilic addition of water. A water molecule attacks the carbocation through its oxygen lone pair: CH3-CH⁺-CH3 + H2O -> CH3-CH(OH2⁺)-CH3.

Step 3 – deprotonation. A second water molecule removes the proton from the oxonium ion, regenerating H3O⁺ and giving the alcohol: CH3-CH(OH2⁺)-CH3 + H2O -> CH3-CH(OH)-CH3 + H3O⁺.

1. Net reaction: CH3-CH=CH2 + H2O H⁺ CH3-CH(OH)-CH3 (propan-2-ol, Markovnikov product).
2. H+ is a catalyst (consumed in step 1, regenerated in step 3).

Three-step mechanism: protonation → water attack → deprotonation; Markovnikov product (more substituted alcohol).

Concept used. Net molecular polarity is the vector sum of all bond dipoles. If the geometry causes the bond dipoles to cancel, the molecule is non-polar even though individual bonds are polar.

1. CO2 is linear (O=C=O, 180). The two C=O bond dipoles are equal in magnitude and point in opposite directions ⇒ they cancel. Net μ = 0 ⇒ non-polar.
2. R-O-R has a bent geometry at oxygen (∼ 111; sp³ O with two lone pairs). The two C–O bond dipoles do not cancel; they add to give a net dipole pointing along the bisector of the C–O–C angle.
3. Typical μ(R-O-R) ≈ 1.3 D (e.g. dimethyl ether μ = 1.30 D).

CO2 is linear ⇒ bond dipoles cancel (non-polar). Ether is bent ⇒ dipoles add (polar).

Concept used. Lucas test converts R-OH to R-Cl through a carbocation intermediate. The rate-determining step is ionisation, so the rate tracks carbocation stability: 3^∘ > 2^∘ > 1^∘.

1. Protonation: R-OH + H+ -> R-OH2⁺ (water is now a good leaving group).
2. ZnCl2 coordinates the oxygen, further weakening C–O.
3. Slow step (cation formation): R-OH2⁺ -> R⁺ + H2O. Rate depends on cation stability.
4. 3^∘ cation (CH3)3C⁺: nine α C–H bonds for hyperconjugation + three +I methyls ⇒ very stable, fast.
5. 2^∘ cation (CH3)2CH⁺: six α C–H + two +I methyls ⇒ moderately stable.
6. 1^∘ cation CH3CH2⁺: three α C–H + one +I methyl ⇒ poor, slow.
7. Observable: 3^∘ cloudy in <1 min; 2^∘ in 5-10 min; 1^∘ only on heating.

Different rates because cation stability differs: 3^∘ > 2^∘ > 1^∘.

Concept used. The conversion uses Kolbe's reaction to put a -COOH group ortho to phenol (giving salicylic acid), followed by O-acetylation with acetic anhydride to give aspirin (acetylsalicylic acid).

Step 1 – form sodium phenoxide. C6H5OH + NaOH -> C6H5ONa + H2O.

Step 2 – Kolbe-Schmitt carboxylation. Heat C6H5ONa with CO2 under pressure (125 C, 4-7 atm), then acidify: C6H5ONa + CO2 ΔH⁺ o-HO-C6H4-COOH giving salicylic acid (2-hydroxybenzoic acid).

Step 3 – O-acetylation. React salicylic acid with acetic anhydride in the presence of conc. H2SO4: o-HO-C6H4-COOH + (CH3CO)2O H⁺ o-CH3COO-C6H4-COOH + CH3COOH giving aspirin (acetylsalicylic acid).

1. Phenol → sodium phenoxide (NaOH).
2. Phenoxide → salicylic acid (Kolbe with CO2, then H+).
3. Salicylic acid → aspirin (acetic anhydride, H2SO4).

Three steps: NaOH → Kolbe (CO₂, Δ, H⁺) → acetic anhydride / H2SO4 to give aspirin.

Concept used. Electrophilic aromatic substitution (EAS) is faster when the ring is more electron-rich. -OH donates a lone pair into the ring via +M resonance, raising the HOMO of the ring and making it a stronger nucleophile toward the nitronium ion NO2⁺.

1. Benzene: no activating group; reacts only with conc. HNO3 + conc. H2SO4 at ∼ 50 C.
2. Phenol: -OH activates by +M. Even dilute HNO3 at room temperature gives a mixture of o- and p-nitrophenol.
3. With conc. HNO3/H2SO4, phenol gives 2,4,6-trinitrophenol (picric acid).

Phenol is nitrated more easily because -OH activates the ring (+M) and raises electron density at o/p.

Concept used. CO2 is a weak electrophile. Phenol's ring is activated but not enough for CO2 to attack. The phenoxide ion C6H5O^- is much more reactive than phenol toward EAS because (a) the negatively charged -O^- is an even stronger +M donor than -OH, and (b) the formal negative charge raises the ring's HOMO further.

1. Convert phenol to phenoxide first: C6H5OH + NaOH -> C6H5O^-Na⁺.
2. Phenoxide's resonance structures place full negative charge at o, p ring carbons (vs only partial δ^- in phenol).
3. These electron-rich o, p carbons are nucleophilic enough to attack CO2.
4. Net result: sodium salicylate, then H+ gives salicylic acid.

Phenoxide is far more nucleophilic than phenol; it is needed because CO2 is too weak an electrophile to react with neutral phenol.

Concept used. The net dipole of an R-OH molecule comes from the C–O and O–H bond dipoles. In phenol, the ring's -I effect on the sp² carbon makes the C–O bond less polar than in methanol, where the sp³ C–O has full alkyl-to-O polarisation. The oxygen lone pair also delocalises into the ring, further reducing the C–O bond moment in phenol.

1. Methanol μ = 1.71 D: C–O bond fully polarised, sp³ C, no resonance back-donation.
2. Phenol μ = 1.55 D: C–O is partly π-double-bond (resonance with ring) and C(sp²) is more electronegative, so the C–O bond moment is smaller and the O lone pair is partly tied up in the ring.
3. The benzene ring's electron-withdrawing effect (from the perspective of the substituent) reduces the electron density at oxygen.

Phenol μ = 1.55 D < methanol μ = 1.71 D. Cause: resonance and the sp² ring carbon reduce the C–O bond moment in phenol.

Concept used. Williamson ether synthesis proceeds by S_N2. With a 3^∘ alkyl halide and a bulky alkoxide (t-BuO^-), S_N2 is blocked by steric crowding. The base then promotes E2 elimination instead, giving an alkene rather than an ether.

1. Williamson recipe (general): R-X + R-ONa -> R-O-R + NaX (back-side attack of alkoxide on C of R-X).
2. For di-t-butyl ether you would need (CH3)3C-Br + (CH3)3C-O^-Na⁺.
3. (CH3)3C-Br is a 3^∘ alkyl halide – back-side attack is sterically impossible (the three methyls block approach).
4. (CH3)3C-O^- is a strong, bulky base. Instead of S_N2 it abstracts a β-H, giving E2 elimination.

Actual outcome (E2): (CH3)3C-Br + (CH3)3C-ONa -> (CH3)2C=CH2 + (CH3)3C-OH + NaBr.

Steric blocking of S_N2 + strong-bulky-base ⇒ E2 elimination dominates; alkene is obtained, not the ether.

Concept used. The oxygen in both alcohols and ethers is sp³ hybridised with two bond pairs and two lone pairs. VSEPR predicts ≈ 109.5, but two corrections apply: (i) lone pair-bond pair repulsion compresses bond angles, (ii) bulk of substituents pushes them apart.

1. Alcohols (e.g. methanol, CH3-O-H): the H on oxygen is tiny; lone-pair repulsion dominates and compresses C–O–H to ∼ 108.9 (slightly less than tetrahedral).
2. Ethers (e.g. dimethyl ether, CH3-O-CH3): two bulky alkyl groups repel each other strongly; the C–O–C angle is forced open to ∼ 111 (slightly greater than tetrahedral).
3. Lone-pair effect is the same in both; only the bond-pair sizes differ.

C–O–H ∼ 108.9 < 109.5 (small H, lone-pair repulsion wins). C–O–C ∼ 111 > 109.5 (bulky alkyl groups push each other apart).

Concept used. Solubility in water depends on the ability to form intermolecular hydrogen bonds with water, balanced against the cost of disrupting solute-solute interactions and the hydrophobic penalty of forcing alkyl chains into the water network.

1. Low MW alcohols (CH3OH, C2H5OH, C3H7OH) have small alkyl groups and one -OH each.
2. -OH is both an H-bond donor (O–H) and an acceptor (lone pair on O); each alcohol molecule can form ∼ 2 H-bonds with water (∼ 21 kJ/mol each).
3. The hydrophobic penalty of a small alkyl chain (∼ 3.5 kJ/mol per -CH2-) is easily paid for by the H-bond enthalpy.
4. Methanol, ethanol and propan-1-ol are infinitely miscible. From butan-1-ol onward the hydrophobic penalty starts to dominate and solubility drops.

H-bonding with water dominates for small alcohols (C₁-C₃ are infinitely miscible). Solubility falls sharply from C₄ onward.

Concept used. The -NO2 group at the para position strongly stabilises the p-nitrophenoxide anion through both -I (inductive electron-pull) and -M (resonance delocalisation onto the nitro oxygens). The phenoxide of unsubstituted phenol has only ring-carbon resonance contributors.

1. p-nitrophenoxide resonance: in addition to the four ring contributors of plain phenoxide, an extra contributor places the negative charge directly on a nitro oxygen.
2. This dispersal of charge onto an electronegative O atom is energetically very favourable ⇒ much more stable anion.
3. Lower anion energy ⇒ lower pK_a ⇒ stronger acid.
4. Values: phenol pK_a = 10.0; p-nitrophenol pK_a = 7.15 (about 1000 times more acidic).

-NO2 at para delocalises the phenoxide charge onto its own oxygen (-M). The resulting anion is far more stable, so p-nitrophenol is ∼ 1000× more acidic than phenol.

Concept used. Boiling point depends on the strength of intermolecular forces. Alcohols form strong intermolecular hydrogen bonds between O-H of one molecule and the oxygen lone pair of a neighbour. Ethers have oxygen lone pairs but no O-H donor, so they cannot self-H-bond; they are held together only by dispersion and weak dipole-dipole forces.

1. Example pair: ethanol (C2H5OH, MW 46) b.p. 78 C; dimethyl ether (CH3OCH3, MW 46) b.p. -24 C. Difference > 100 C.
2. H-bond energy ∼ 21 kJ/mol; each ethanol molecule donates ∼ 1 and accepts ∼ 2 H-bonds in liquid.
3. Total H-bond contribution to ethanol's Δ H_vap: ∼ 30-40 kJ/mol on top of dispersion ⇒ much higher b.p.

Alcohols self-H-bond (donor + acceptor) ⇒ high b.p. Ethers cannot self-H-bond (acceptor only) ⇒ low b.p.

Concept used. Two effects combine to reinforce the C–O bond in phenol relative to methanol: (a) the ring carbon is sp² (more s-character, shorter and stronger σ-bond than sp³); (b) the oxygen lone pair delocalises into the aromatic ring, giving the C–O bond partial double-bond character (+M resonance).

1. In methanol CH3-OH: C is sp³; C–O bond length ≈ 1.43 ; bond energy ≈ 339 kJ/mol; no resonance.
2. In phenol C6H5-OH: C is sp²; C–O bond length ≈ 1.36 ; bond energy ≈ 358 kJ/mol; reinforced by ∼ 20 kJ/mol of resonance stabilisation.
3. Net: phenol's C–O is shorter and stronger by ∼ 20-40 kJ/mol.

Phenol C–O is reinforced by (a) sp² C and (b) +M resonance; methanol's is not.

Concept used. Acidity is governed by the stability of the conjugate base. Phenoxide is resonance-stabilised across the ring (four contributors), so phenol is the strongest acid of the three. Ethoxide is destabilised by the +I effect of C2H5 (alkyl group pushes electron density onto O^-), so ethanol is the weakest acid. Water sits in between (no resonance, no +I).

1. Phenol (pK_a = 10.0): phenoxide resonance-stabilised by 4 contributors with charge on o, p ring carbons ⇒ strongest acid.
2. Water (pK_a = 15.7): hydroxide has no resonance and no destabilising +I ⇒ moderate acid.
3. Ethanol (pK_a ≈ 16): ethoxide destabilised by +I of C2H5 (alkyl pushes density onto O^-) ⇒ weakest acid.

Increasing acidity: ethanol < water < phenol.

Concept used. Map each common use/role to the right oxygen compound from everyday chemistry.

1. (i) Antifreeze: ethylene glycol (HOCH2CH2OH, low-melting, lowers water's freezing point) ⇒ (e).
2. (ii) Perfume solvent: ethanol (volatile, mixes with aroma oils) ⇒ (f).
3. (iii) Picric acid (2,4,6-trinitrophenol) is made by nitrating phenol ⇒ (d).
4. (iv) ``Wood spirit'' is the trivial name for methanol (originally from destructive distillation of wood) ⇒ (c).
5. (v) Phenolic group is detected with neutral FeCl3 (violet colour) ⇒ (a).
6. (vi) Soap-industry by-product used as cosmetic humectant: glycerol ⇒ (b).

(i)→(e); (ii)→(f); (iii)→(d); (iv)→(c); (v)→(a); (vi)→(b).

Concept used. Map each IUPAC structure to its widely-used trivial name from the phenol / ether dictionary.

1. (i) 2-methylphenol = o-cresol ⇒ (d).
2. (ii) Benzene-1,2-diol = catechol ⇒ (c).
3. (iii) Benzene-1,3-diol = resorcinol ⇒ (f).
4. (iv) Benzene-1,4-diol = hydroquinone ⇒ (a).
5. (v) Methoxybenzene (C6H5-O-CH3) = anisole ⇒ (g).
6. (vi) Ethoxybenzene (C6H5-O-C2H5) = phenetole ⇒ (b).
7. ``Quinone'' (e) is oxidised hydroquinone (1,4-benzoquinone) and is not in Column I.

(i)→(d), (ii)→(c), (iii)→(f), (iv)→(a), (v)→(g), (vi)→(b).

Concept used. Cleavage of an ether by HI follows two rules:

• [leftmargin=*,nosep]
• Aryl alkyl ether (Ar-O-R) always gives phenol + R-I (aryl side is reinforced by resonance and never breaks).
• Two alkyl groups: when one is 3^∘, the cleavage proceeds by S_N1 at that side (most stable carbocation); when both are ≤ 2^∘, S_N2 takes place at the less hindered carbon, giving the bulkier alcohol + smaller alkyl iodide.

1. (i) CH3-O-CH3 + HI -> CH3OH + CH3I ⇒ (d).
2. (ii) (CH3)2CH-O-CH3 + HI -> (CH3)2CH-OH + CH3I (S_N2 on the less hindered methyl) ⇒ (e).
3. (iii) (CH3)3C-O-CH3 + HI -> (CH3)3C-I + CH3OH (S_N1; tertiary carbocation is the stable intermediate) ⇒ (b).
4. (iv) C6H5-O-CH3 + HI -> C6H5OH + CH3I (aryl-O never breaks) ⇒ (a).

(i)→(d), (ii)→(e), (iii)→(b), (iv)→(a).

Concept used. Match each label to its defining synonym / reagent / product.

1. (i) Methanol = wood spirit (from destructive distillation of wood) ⇒ (d).
2. (ii) Kolbe's reaction = phenoxide + CO2 → salicylic acid (o-hydroxybenzoic acid; here called ``o-hydroxysalicylic acid'') ⇒ (a).
3. (iii) Williamson's synthesis = R-O^-Na⁺ + R_a-X → R-O-R_a + NaX ⇒ (f).
4. (iv) 2^∘ alcohol → ketone uses heated Cu at 573 K (dehydrogenation) ⇒ (e).
5. (v) Reimer-Tiemann reaction = phenol + CHCl3/NaOH → salicylaldehyde ⇒ (c).
6. (vi) Fermentation of sucrose → ethanol (ethyl alcohol) ⇒ (b).

(i)→(d), (ii)→(a), (iii)→(f), (iv)→(e), (v)→(c), (vi)→(b).

Correct option: (ii) Both A and R are wrong statements.

Concept used. Acid-catalysed hydration of an alkene follows Markovnikov's rule via the most stable carbocation. With but-1-ene (CH3CH2CH=CH2), the proton attaches to C1 (terminal CH₂), generating a 2^∘ carbocation at C2; water then attacks to give butan-2-ol, NOT butan-1-ol. The reaction also does NOT proceed via a primary carbocation, which would be very unstable.

1. Check A: but-1-ene + H+/H2O → butan-2-ol (Markovnikov), not butan-1-ol. A is false.
2. Check R: a primary carbocation CH3CH2CH2CH2⁺ is too unstable; mechanism proceeds via the 2^∘ cation CH3CH2CH⁺CH3. R is false.
3. Both A and R false ⇒ option (ii).

Option (ii): Both A and R are wrong (gives butan-2-ol via 2^∘ cation).

Correct option: (i) Both A and R are correct and R is the correct explanation of A.

Concept used. The acidity of a substituted phenol is set by stabilisation of its phenoxide. A p-NO2 group stabilises the phenoxide through both -I and especially -M resonance, which delocalises the negative charge onto a nitro oxygen.

1. Check A: pK_a of phenol = 10.0; pK_a of p-nitrophenol = 7.15. p-nitrophenol is ∼ 1000× more acidic. A is true.
2. Check R: in the p-nitrophenoxide anion an additional resonance contributor places the negative charge onto a nitro O, dispersing it through the conjugated π-system. This dispersal stabilises the anion. R is true and is the correct explanation of A.
3. Both correct, R explains A ⇒ option (i).

Option (i): Both true, R correctly explains A.

Correct option: (iv) A is wrong but R is correct.

Concept used. Apply the IUPAC ``alkoxy-alkane'' rule: the larger alkyl arm is the parent alkane; the smaller -OR is named as an alkoxy prefix.

1. Skeleton: (CH3)2CH-O-CH2CH2CH3. Left arm = isopropyl (C3, sec). Right arm = n-propyl (C3). Both have 3 C; the propyl side is selected as parent (longest straight chain) ⇒ propane.
2. Substituent on the parent: the isopropoxy group from the other side – -O-CH(CH3)2 = ``propan-2-yloxy'' or commonly written ``2-propoxy'' or ``isopropoxy''. Locant on propane: position 1.
3. Correct IUPAC name: 1-(propan-2-yloxy)propane (commonly: isopropyl n-propyl ether).
4. The proposed name ``2-ethoxy-2-methylethane'' is nonsensical (ethane has only 2 C; ``2-methylethane'' is not valid; no ethoxy group is present in this molecule). A is false.
5. Check R: the IUPAC rule for ethers does describe them as a hydrocarbon with H replaced by -OR or -OAr. R is true.

Option (iv): A false (name is wrong), R true (rule is correctly stated).

Correct option: (iv) A is wrong but R is correct.

Concept used. VSEPR for ethers (R-O-R): oxygen has two bond pairs (to the two R groups) and two lone pairs. Lone-pair repulsion alone would compress C–O–C below 109.5, but the bulky alkyl groups push each other apart, dominating the balance and opening the angle to slightly greater than 109.5.

1. Check A: ∠C-O-C in dimethyl ether ≈ 111.7, slightly greater (not less) than 109.5. A is false.
2. Check R: there genuinely is steric repulsion between the two bulky alkyl groups, and this is the reason the angle widens. R is true.
3. A false, R true ⇒ option (iv).

Option (iv): A false (angle is greater, not less); R true (reason itself is correct).

Correct option: (iii) A is true but R is wrong.

Concept used. Alcohols self-hydrogen-bond (donor + acceptor) and have higher boiling points than comparable hydrocarbons. Ethers have no O-H donor; they cannot self-H-bond. Their boiling points are therefore close to those of alkanes of similar molecular mass.

1. Check A: alcohols genuinely have high b.p. relative to alkanes (ethanol 78 C vs ethane -88 C). Ethers have b.p. only modestly above alkanes (dimethyl ether -24 C vs ethane -88 C; b.p. of propane -42 C is in the same range). The Exemplar key accepts this generalisation; ``high'' is interpreted loosely. A is true as the textbook generalisation.
2. Check R: alcohols H-bond but ethers do NOT self-H-bond (no O-H donor). So the reason as stated applies only to alcohols, not to both. R is false as a blanket explanation.
3. A true, R false ⇒ option (iii).

Option (iii): A true, R false (ethers cannot self-H-bond).

Correct option: (iv) A is wrong but R is correct.

Concept used. Benzene's ring is unactivated and needs a Lewis acid (e.g. FeBr3) to polarise Br2 before substitution. Phenol's ring is strongly activated by +M donation of -OH; it brominates without any Lewis acid – in fact even Br2/water alone gives 2,4,6-tribromophenol immediately.

1. Check A: phenol does NOT need a Lewis acid. Br2 alone in water gives instant tribromination; in CS2/273 K, Br2 alone gives monobromination. A is false.
2. Check R: it is true that Lewis acids (FeBr3, AlBr3) polarise Br2 to generate the electrophile Br⁺, but the reaction in phenol does not need that boost. The statement about Lewis acid behaviour is correct in itself. R is true.
3. A false, R true ⇒ option (iv).

Option (iv): A false (no Lewis acid needed for phenol), R true (general statement about Lewis acid polarisation is correct).

Correct option: (v) Both A and R are correct but R is not the correct explanation of A.

Concept used. The lower water-solubility of o-nitrophenol is caused by intramolecular H-bonding (chelation between -OH and the adjacent -NO2 group), which removes its -OH from the pool available for H-bonding with water. The m- and p-isomers cannot chelate (groups too far apart) and so their -OH is available for both inter-solute H-bonding (``associated molecules'') and H-bonding with water – so they dissolve more.

1. Check A: solubility (per 100 g H2O at 20 C): o-nitrophenol ∼ 0.2 g; m-nitrophenol ∼ 1.4 g; p-nitrophenol ∼ 1.6 g. o- is indeed least soluble. A is true.
2. Check R: m- and p-nitrophenols genuinely do form intermolecular H-bonded associated structures in the solid state. R is true.
3. But the real reason A is true is intramolecular chelation in o-nitrophenol (which locks up its -OH), not the association behaviour of the other isomers. So R, while true, is NOT the correct explanation.

Option (v): Both true; but R (association of m, p isomers) is not the correct cause of A (which is intramolecular H-bond in o).

Correct option: (ii) Both A and R are wrong statements.

Concept used. The solvent controls how strongly the ring is brominated. In a non-polar medium (CS2, CHCl3 at low T) only monobromination occurs, giving mainly p-bromophenol (with some o-bromophenol). The tri-brominated product 2,4,6-tribromophenol forms only when phenol meets Br2/water (polar, ionising medium). The genuine cause of phenol's facile bromination is the strong +M activation of the ring by -OH, not any special polarisation of Br2 by CS2.

1. Examine A: In CS2 at 273 K, the actual product is p-bromophenol (+ a little o-isomer), not 2,4,6-tribromophenol. So A is false.
2. Examine R: CS2 is essentially non-polar (ε ≈ 2.6) and does not polarise Br2 meaningfully. What actually activates the ring is the +M donation of the -OH group (or of the phenoxide ion in water); Br2 polarisation by phenoxide (not by CS2) is the operative cause. So R, as stated, is also false.
3. Combined verdict: A false, R false ⇒ option (ii).

Option (ii): CS2 gives mono-bromophenol (mostly p-), not the tribromo product, and CS2 is not what polarises Br2 –- phenol's +M activation is.

Correct option: (iii) A is true but R is false.

Concept used. Acidity of R-OH vs Ar-OH is set by conjugate-base stability. Phenoxide is resonance-stabilised across the ring; ethoxide is not. Hence phenol is a much stronger acid (pK_a ≈ 10) than ethanol (pK_a ≈ 16).

1. Verify A: phenol (pK_a 10) is more acidic than ethanol (pK_a 16). So ethanol is the weaker acid. A is true.
2. Examine R: aq. NaOH (conjugate acid water, pK_a 15.7) cannot deprotonate ethanol (pK_a 16) because the equilibrium constant K ≈ 10^15.7-16≈ 0.5 already disfavours it, and the equilibrium is further shifted to the left by the huge excess of water. Industrially, NaOEt is made from Na metal + ethanol, not from aq. NaOH.
3. Therefore R is false; the experimental claim that aq. NaOH converts ethanol to NaOEt is wrong.

Option (iii): A is true (phenol > ethanol in acidity); R is false (aq. NaOH does not deprotonate ethanol meaningfully).

Correct option: (iv) A is wrong but R is correct.

Concept used. Conc. HNO3 + H2SO4 is a powerful nitrating mixture that generates NO2⁺ in large quantity. On phenol's highly activated ring, this leads to trinitration at 2, 4, and 6 (giving picric acid), NOT a simple mono-nitration. Mono-nitration of phenol to give o- and p-nitrophenol uses dilute HNO3 instead.

1. Check A: conc. HNO3/H2SO4 gives 2,4,6-trinitrophenol (picric acid), not the mono o/p products. A is false.
2. Check R: -OH is genuinely o-, p-directing because +M resonance places electron density at o, p ring carbons. R is true.
3. A false, R true ⇒ option (iv).

Option (iv): A false (conc. acid gives picric acid, not mono-nitro), R true (-OH is o-, p-director).

Concept used. The cleavage of an alkyl aryl ether (Ar-O-R) by HI always gives phenol + alkyl iodide, never aryl iodide. This is because the C(sp²)–O bond of aryl ethers has partial double-bond character (resonance with the ring) and is much harder to break than the C(sp³)–O bond of the alkyl side.

Step 1: Protonation. HI first protonates the ether oxygen, converting -O- into the much better leaving group -O+H-R: C6H5-O-CH3 + HI -> C6H5-O⁺(H)-CH3 + I^-

Step 2: S_N2 attack of I^- on the alkyl carbon. Iodide, an excellent nucleophile, attacks the sp³ carbon (CH3). The C(sp²)–O bond resists breaking (loss of aryl-O resonance), so cleavage occurs at the methyl side: C6H5-O⁺(H)-CH3 + I^- -> C6H5-OH + CH3I

1. Protonation of O activates the ether.
2. I^- attacks the methyl carbon (back-side, S_N2).
3. Bond cleavage at CH3-O liberates C6H5-OH.
4. Products: phenol + methyl iodide.

C6H5-O-CH3 + HI -> C6H5-OH + CH3I. Mechanism: protonation, then S_N2 at -CH3.

Concept used. Modern phenol industry uses the cumene process; phenol's ring is strongly activated by -OH via +M resonance, so it brominates without a Lewis acid, but the solvent picks how many bromines go on.

(a) Starting material. The industrial route is the cumene (isopropylbenzene) hydroperoxide process, in three steps:

1. [leftmargin=*,nosep]
2. Friedel–Crafts alkylation (H+ catalyst): C6H6 + CH3-CH=CH2 -> C6H5-CH(CH3)2.
3. Air oxidation (O2): C6H5-CH(CH3)2 -> C6H5-C(CH3)2-OOH.
4. Acid-catalysed rearrangement (H+): C6H5-C(CH3)2-OOH -> C6H5OH + (CH3)2C=O.

So cumene (or equivalently benzene + propene) is the starting material. Acetone is obtained as a valuable by-product.

(b) Bromination in two media.

Aqueous medium (Br2/water): polar solvent; the ring is hugely activated, giving the tribromo product: C6H5OH + 3 Br2 -> (2,4,6-Br3)C6H2OH + 3 HBr (2,4,6-tribromophenol precipitates as a white solid).

Non-aqueous medium (Br2 in CS2 or CHCl3 at 273 K): only one Br enters the ring: C6H5OH + Br2 -> C6H4(OH)(Br) + HBr The major product is p-bromophenol with a small amount of o-bromophenol.

(c) Why no Lewis acid is needed. Phenol's -OH donates a lone pair into the ring by +M resonance. This raises the ring's electron density at the ortho/para positions to such an extent that Br2 alone (with H2O helping to polarise it to Br⁺ ⋯ Br^-) is electrophilic enough. Adding FeBr3 would over-activate the ring and lead to poly-bromination of any remaining product.

1. Industrial starting material: cumene (from benzene + propene).
2. Aq. Br2: triple substitution to 2,4,6-tribromophenol.
3. CS2/CHCl3, 273 K: monobromination, mainly p.
4. -OH activates so strongly via +M that Br2 alone reacts; no Lewis acid required.

(a) Cumene. (b) Aq: 2,4,6-tribromophenol; CS2/273 K: p-bromophenol. (c) -OH is so strongly activating (+M) that ring electron density is already sufficient for unassisted bromination.

Concept used. The conversion proceeds in three steps: deprotonate phenol with NaOH, carry out Kolbe's reaction with CO2 to give salicylic acid (2-hydroxybenzoic acid), then O-acetylate the phenolic -OH with acetic anhydride to give aspirin (acetylsalicylic acid).

Step 1 – form sodium phenoxide. C6H5-OH + NaOH -> C6H5-O^-Na⁺ + H2O

Step 2 – Kolbe-Schmitt carboxylation. Sodium phenoxide reacts with CO2 (125 C, 4-7 atm) to put a -COO^-Na⁺ ortho to the OH; acidification then gives salicylic acid. C6H5O^-Na⁺ + CO2 ΔH⁺ HO-C6H4-COOH (salicylic acid)

Step 3 – O-acetylation with acetic anhydride. The phenolic -OH is converted to -OCOCH3 using acetic anhydride ((CH3CO)2O) and a catalytic amount of conc. H2SO4. The carboxylic acid is left untouched (under-reactive vs the phenolic OH). HO-C6H4-COOH + (CH3CO)2O H⁺ CH3COO-C6H4-COOH + CH3COOH The product CH3COO-C6H4-COOH is aspirin (acetylsalicylic acid).

1. Phenol → sodium phenoxide (deprotonation with NaOH).
2. Sodium phenoxide → salicylic acid (Kolbe with CO2 at Δ, then H+).
3. Salicylic acid → aspirin (acetic anhydride, H2SO4 catalyst).

Three steps: NaOH → Kolbe (CO₂, Δ, H⁺) → acetic anhydride / H2SO4. Final product: acetylsalicylic acid (aspirin).

Concept used. A biocatalyst is a catalyst of biological origin – typically an enzyme or a whole-cell microbial preparation. The classic example in alcohol chemistry is fermentation of carbohydrates by yeast enzymes to produce ethanol. Yeast supplies two enzymes: invertase (hydrolyses sucrose to glucose + fructose) and zymase (ferments hexose to ethanol + CO2).

Step 1 – invertase. Sucrose is the disaccharide feedstock (from molasses or sugar cane juice). Invertase from yeast hydrolyses the glycosidic bond: C12H22O11 + H2O invertase C6H12O6 + C6H12O6 giving glucose + fructose (both reducing hexoses).

Step 2 – zymase. Zymase (actually a complex of ∼ 12 enzymes including alcohol dehydrogenase) ferments each hexose to ethanol + carbon dioxide: C6H12O6 zymase 2 C2H5OH + 2 CO2

Industrial conditions.

• [leftmargin=*,nosep]
• Temperature: 25-30 C (yeast optimal).
• Atmosphere: anaerobic (oxygen would push the process to CO2 and water).
• pH: slightly acidic (∼ 5).
• Time: 3-5 days, until ethanol concentration reaches ∼ 14% (yeast stops at higher alcohol levels because ethanol itself becomes toxic to yeast).

Workup. The fermented broth contains ∼ 14% ethanol. Fractional distillation gives rectified spirit (95% ethanol + 5% water; this is the constant-boiling azeotrope). Removing the last 5% of water requires anhydrous benzene azeotrope or molecular-sieve drying to give absolute alcohol.

1. Sucrose → glucose + fructose (invertase).
2. Hexose → ethanol + CO2 (zymase).
3. Conditions: yeast, 25-30 C, anaerobic, ∼5 days.
4. Fractional distillation → 95% ethanol (rectified spirit) → absolute ethanol via azeotrope removal.

Ethanol from sucrose: invertase + zymase (yeast) catalyse the two enzymatic steps at 25-30 C anaerobically.