NCERT Exemplar Solutions Class 12 Chemistry Ch 8 - Download PDF

Tautomer-first angle. Skip the full mechanism –- whichever option corresponds to the keto form of a Markovnikov-derived enol is the right one. The aqueous Hg²⁺ ion first complexes the π-bond of the terminal alkyne, then water attacks the more substituted carbon (internal) and a proton settles on the terminal one. This delivers the enol CH3CH2-C(OH)=CH2, which immediately tautomerises to the more stable ketone.

Three quick eliminations. (a) Option (i) would require anti-Markovnikov hydration (hydroboration territory, not Hg²⁺). (b) Option (iii) is the trapped enol –- only an intermediate, never the isolated product. (c) Option (iv) is a cleavage product (ozonolysis style), not hydration.

Mnemonic. ``Internal OH, terminal H, then flip to ketone.'' For terminal alkynes R-C#CH the result is always a methyl ketone R-CO-CH3 –- the only exception is HC#CH, which gives CH3CHO.

CH3CH2COCH3 (butan-2-one); option (ii).

Substituent-count angle. Less ``stuff'' attached to the carbonyl carbon ⇒ more reactive. CH3CHO carries just one methyl and one H on the C=O –- the most sterically exposed and most electrophilic carbonyl in the list. Each extra alkyl group hides the δ⁺ carbon behind a cone of +I density and adds steric bulk; each aryl group additionally delocalises the carbonyl π into the ring through resonance, weakening the δ⁺ even further.

Energy-diagram angle. In the rate-limiting step a nucleophile attacks C=O; the transition state already shows partial sp³ character. Bigger groups raise the energy of that sp³-like TS (steric strain), and electron-donating groups raise the LUMO of C=O (less hungry for Nu^-). Both effects penalise (iv) C6H5COCH3 the most –- it sits at the bottom of the reactivity ladder, exactly opposite to (i).

Quick filter. HCHO > aliphatic CHO > aromatic CHO > aliphatic ketone > aromatic ketone. Place the four options on this ladder: CH3CHO ranks highest.

CH3CHO (acetaldehyde); option (i).

pK_a-ladder angle. Memorise four anchor numbers and the problem solves itself: alcohol ∼16, phenol ∼10, aliphatic R-COOH ∼5, α-chloro ClCH2COOH ∼3. The smaller the pK_a, the stronger the acid; arranging upwards from weakest to strongest gives the answer in seconds.

Why each step drops. Going from ethanol (one localised O^-) to phenol, the negative charge spreads over three ring carbons and one oxygen (∼5 resonance forms) –- a 6 pK_a-unit jump. Going from phenol to acetic acid, the charge now splits equally between two oxygens (better acceptors than ring C) –- another 5-unit jump. Adding an α-Cl to acetic acid pulls σ-electron density off the carboxylate by induction (-I), dropping pK_a by another ∼2 units. Each effect is independent and additive.

Trap to avoid. Do not confuse -I (always reduces pK_a) with +M (always raises it). Chlorine is a special case: it is -I but +M. Through a saturated -CH2- linker, only -I reaches the COO^-, so Cl acts purely as an acid-strengthener here.

Ethanol < Phenol < AcOH < ClAcOH; option (iii).

Reagent-pair angle. Phenol esters are almost always made from acyl chlorides, not the parent acid. The acid route is sluggish because phenol is a poor nucleophile (the lone pair is delocalised into the ring), and the equilibrium of acid + phenol sits far to the reactant side. The acid chloride bypasses this issue –- the Cl leaving group makes the carbonyl carbon extremely electrophilic.

Why pyridine, not NaOH? A strong base like NaOH would deprotonate phenol to phenoxide, but it would also hydrolyse PhCOCl to PhCOOH faster than ester formation. Pyridine is just basic enough (pK_aH ∼5) to mop up HCl without destroying the acyl chloride, and it doubles as a nucleophilic catalyst forming a reactive acyl-pyridinium intermediate that phenol attacks easily.

Mechanism cartoon. (1) PhCOCl + pyridine → PhCO-N⁺Py + Cl^-. (2) Phenol's O attacks the acyl carbon, displacing pyridine. (3) Pyridine grabs the proton from phenol's O-H. Net (in pyridine): PhOH + PhCOCl -> PhOCOPh + HCl (the HCl is scavenged as PyH+Cl^-).

PhOH + PhCOCl/pyridine → PhOCOPh; option (ii).

Diagnostic-test angle. Fehling is the textbook diagnostic for aliphatic aldehydes only. The reagent is a deep-blue Cu²⁺-bistartrate complex in NaOH; it relies on a mild two-electron oxidation of R-CHO to R-COO^-, with Cu²⁺ being reduced to brick-red Cu2O. Any carbonyl without the aliphatic-aldehyde combination of (i) an oxidisable C-H on the carbonyl carbon and (ii) no resonance stabilisation flunks the test.

Why each option behaves as it does. (i) Sodium hydrogensulphite NaHSO3 adds to almost every aldehyde and small methyl ketone –- including both acetone and benzaldehyde. (ii) Phenyl hydrazine PhNHNH2 forms a phenylhydrazone with any C=O. (iv) Grignards add to all aldehydes and ketones (and overshoot esters to alcohols). Only Fehling's strikes out twice.

Common pitfall. Tollens' reagent [Ag(NH3)2]⁺ is stronger than Fehling's and oxidises both aliphatic and aromatic aldehydes (silver-mirror test). So benzaldehyde gives Tollens+ but Fehling-. Confusing the two reagents costs marks.

Fehling's reacts with neither CH3COCH3 nor PhCHO; option (iii).

α-H scan angle. For every aldehyde on the list, scan for a C-H on the carbon directly attached to the carbonyl. If even one such H exists, hydroxide will prefer to remove it (forming an enolate) and aldol condensation takes over; Cannizzaro can run only when this competing path is shut down.

Pass/fail per option. (i) Cyclohexyl-1-methyl-1-carbaldehyde: α-C is quaternary (already four substituents), zero α-H ⇒ Cannizzaro-capable. (ii) Benzaldehyde: α-C is an sp² ring carbon with no H to lose for an enolate ⇒ Cannizzaro-capable. (iii) Formaldehyde: no α-C exists at all (the only ``other'' substituent on C=O is another H) ⇒ Cannizzaro-capable. (iv) Acetaldehyde: three α-Hs sit on the methyl ⇒ aldol wins overwhelmingly.

Kinetic reason. The aldol path uses cheap enolate formation (pK_a of α-C-H ∼20 under NaOH) followed by fast addition to a second molecule. Cannizzaro requires hydride transfer from a tetrahedral alkoxide –- a higher-barrier step. Hence Cannizzaro only wins when aldol is impossible.

CH3CHO has α-H, no Cannizzaro; option (iv).

No-α-H angle. Benzaldehyde plus concentrated alkali = Cannizzaro disproportionation. The α-position of PhCHO is an sp² ring carbon –- no α-H to deprotonate, so the aldol pathway is locked out. With aldol blocked, the next best reaction available to HO^- is addition to the carbonyl to form a tetrahedral alkoxide, which then becomes a hydride donor.

Counter-ion bookkeeping. The question carefully specifies KOH, not NaOH. Whichever metal hydroxide is present, its cation appears in the final salt. So KOH ⇒ potassium benzoate, PhCOO^- K⁺. Option (i), which proposes sodium benzoate, is a deliberate distractor.

Stoichiometry sanity-check. Two molecules of PhCHO + one KOH → one PhCOOK + one PhCH2OH. The 1:1 product ratio means a 50% yield ceiling –- exactly half is oxidised and half is reduced. Option (iii) (``benzoic acid + unreacted benzaldehyde'') ignores both the disproportionation symmetry and the fact that excess KOH deprotonates any PhCOOH to its potassium salt.

Potassium benzoate + benzyl alcohol; option (ii).

Naming-the-enol angle. Two pieces of information settle the question: (a) the regiochemistry of acid-catalysed alkyne hydration (Markovnikov, OH on more substituted carbon), and (b) the relationship between an enol and its ketone (tautomerism, never geometrical or metamerism).

Step-by-step. The propyne molecule CH3-C#CH is numbered with C1 at the terminal CH. Markovnikov puts OH on C2 (more substituted) and H on C1, giving CH3-C(OH)=CH2. IUPAC numbering of the resulting alkene: the C=C is between C1 and C2, the OH is on C2, three carbons total ⇒ prop-1-en-2-ol.

Why ``tautomerism'', not ``geometrical isomerism''. The keto form CH3-CO-CH3 and the enol form CH3-C(OH)=CH2 are constitutional isomers –- their atoms are connected differently (a proton has moved from O to C, and a double bond has shifted from C=C to C=O). Geometrical isomerism would require the same constitution with different spatial arrangement; metamerism requires different alkyl groups around the same functional group (e.g. ethers, ketones with different chains). Neither fits here.

A = prop-1-en-2-ol; type of isomerism with acetone = tautomerism. Option (iv).

Skeleton-and-group angle. To classify two molecules as positional / functional / chain isomers, ask two questions in order. (1) Same molecular formula? (2) If yes –- same functional group? If yes again, they are either positional or chain isomers depending on whether only the locant of the functional group has changed, or the carbon skeleton has been rearranged. If the functional group itself changed (e.g. CHO → CO or OH → OR), they are functional isomers.

For this question. Grignard addition → hydrolysis preserves the carbon skeleton (no rearrangement). Oxidation of the resulting alcohol with PCC oxidises a C-OH to C=O on the same carbon, so the position of the carbonyl in C is dictated by where the OH ended up in A/B. Because the original Grignard added to a different carbonyl than the one regenerated, the locant of the carbonyl group differs between A and C while the molecular formula and the functional-group type are unchanged.

Rule out the wrong choices. ``Identical'' fails because A and C have different connectivity around the central carbon. ``Functional isomers'' fails because both are carbonyls (same family). ``Optical isomers'' would require a chiral carbon to be present in both with opposite configurations –- not the case here.

Positional isomers; option (ii).

Reagent-screening angle. Look at the substrate end and the product end. The carbon count goes from C4 to C3, so a C1 fragment is expelled. The fragment ends up as CHI3 (the yellow iodoform) –- a dead giveaway for the haloform reaction with I2 + NaOH.

Why the other reagents fail. (i) Tollens' reagent ([Ag(NH3)2]⁺) only oxidises aldehydes, not secondary alcohols, and cannot cleave a C-C bond. (ii) Benzoyl peroxide is a radical initiator used in polymerisation or benzylic bromination; it does not selectively oxidise alcohols to carboxylic acids with loss of a C1 fragment. (iv) Sn + NaOH is a mild reducing system (used for nitro → amine reductions, for instance) –- the opposite of what we need.

Net transformation. CH3CH(OH)CH2CH3 + 4 I2 + 6 NaOH -> CH3CH2COO^-Na⁺ + CHI3↓ + 5 NaI + 5 H2O The four iodines do double duty: three end up in CHI3, the fourth is expelled as NaI during the first oxidation step.

I2/NaOH (iodoform); option (iii).

Alcohol-class angle. Classify each candidate as primary, secondary or tertiary by counting the carbons attached to the carbinol carbon. Butan-1-ol: HO-CH2-CH2CH2CH3 –- the carbinol carbon has one alkyl, so it is primary. Butan-2-ol: CH3-CH(OH)-CH2CH3 –- the carbinol carbon has two alkyl groups, so it is secondary.

Why 1^∘ overshoots. Primary alcohols are oxidised first to an aldehyde, but the aldehyde C-H is the weakest C-H in the molecule (BDE ∼88 kcal/mol). KMnO4 continues to attack the aldehyde, oxidising it to the carboxylic acid. There is no ``stop'' point at the aldehyde with strong oxidants in aqueous base –- that's why we need PCC, DIBAL, or Rosenmund-style reagents when an aldehyde is the desired product.

Why 2^∘ stops cleanly at the ketone. Once the carbinol carbon has been converted to C=O, there is no C-H on the carbonyl carbon to attack. Further oxidation of a ketone requires C-C bond cleavage, which needs much harsher conditions (hot, concentrated KMnO4/H+ over hours). Under standard KMnO4/OH^- the ketone is the final product.

Why ``both'' is wrong. Butan-1-ol gives butanoic acid (an acid), not butanone (a ketone). Only butan-2-ol delivers the target.

Butan-2-ol; option (ii).

Reagent-recall angle. Memorise the two-word phrase ``zinc amalgam, hydrochloric acid'' –- this is the Clemmensen combo and the rest of the options are deliberate distractors.

Why each wrong option fails. (ii) Sodium amalgam + HCl: Na would react violently with HCl to give hydrogen gas long before any organic substrate could be reduced; the amalgam is also a poor surface for the hydride-shuttle mechanism that Clemmensen relies on. (iii) Zinc amalgam + HNO3: nitric acid is a strong oxidiser –- it would oxidise the substrate (and the amalgam) rather than allowing reduction. (iv) Sodium amalgam + HNO3: combines both errors –- wrong metal and wrong acid.

When to pick Clemmensen vs Wolff–Kishner. Both convert C=O → CH2, but Clemmensen needs strong acid (so use it for acid-stable substrates) while Wolff–Kishner needs strong base (use it for base-stable substrates with NH2NH2/KOH in glycol). If your substrate carries an acid-sensitive group (acetal, tert-OH), choose Wolff–Kishner; if base-sensitive (ester, α,β-unsat ketone prone to Michael), choose Clemmensen.

Zinc amalgam + HCl; option (i).

α-H scan angle. Mark the carbonyl carbon, then box the carbon next to it (this is α). Count its hydrogens. Zero α-H ⇒ no enolate ⇒ no aldol. Apply this rule mechanically to every option:

Walk-through. (i) CH3CHO: the methyl carbon is α and carries three Hs –- aldol works. (ii) C6H5CHO: the α-C is part of the aromatic ring, sp² with no C-H that could be removed without destroying aromaticity. No aldol; this molecule undergoes Cannizzaro instead. (iii) CH3COCH3: two methyl groups, six α-Hs –- aldol enthusiastically yields diacetone alcohol. (iv) (CH3)3CCHO (pivaldehyde): the α-C is the quaternary central carbon, no C-H at all ⇒ no aldol; pivaldehyde does Cannizzaro just like PhCHO.

Pattern lock. The classic ``no-α-H'' aldehydes are HCHO, PhCHO and (CH3)3CCHO –- formaldehyde, benzaldehyde, pivaldehyde. Whenever any one of these appears in a multi-aldehyde question, expect Cannizzaro / crossed-Cannizzaro chemistry.

Options (ii) and (iv) cannot undergo aldol.

Saponification angle. For every ester R-C(=O)-O-R^′ in NaOH, mentally cut the C(=O)-O bond; the acid side picks up a Na as the carboxylate, the alcohol side keeps its -OH. For phenyl benzoate, R=Ph (on the carbonyl) and R^′=Ph (on the oxygen), so the carbonyl side becomes PhCOO^-Na⁺ (sodium benzoate) and the alkyl side becomes phenol.

Why phenol also ends up ionised. Phenol (pK_a ∼10) is acidic enough that excess NaOH converts it to sodium phenoxide PhO^-Na⁺. So under standard saponification conditions the isolated products are both sodium salts –- option (i) phenol would only be observed after acid work-up.

Why benzophenone is impossible. Option (iv) requires formation of a C-C bond between two phenyls –- saponification is a polar O–C bond cleavage and cannot reorganise the carbon skeleton.

Mechanism in two lines. (1) HO^- adds to C=O giving a tetrahedral sp³ intermediate. (2) Phenoxide (PhO^-, the better leaving group because it is resonance-stabilised) departs, regenerating C=O on benzoate. Acid–base in alkali then ties up both products as sodium salts.

Sodium phenoxide + sodium benzoate; options (ii) and (iii).

Reaction-scope angle. Clemmensen reduction Zn(Hg)/HCl has exactly one job: convert C=O (in aldehydes or ketones) all the way down to CH2. It does not stop at the alcohol oxidation state, and it does not reduce other functional groups such as -COCl, -COOH or -COOR. With that scope clearly defined, every distractor is easy to spot.

Per-option check. (i) PhCHO → PhCH2OH stops at the alcohol stage –- Clemmensen would overshoot it to PhCH3 (toluene). So this option mis-states the product. (ii) Cyclohexanone → cyclohexane: classic Clemmensen target, the six-membered ring tolerates strong acid well. (iii) Benzoyl chloride → benzaldehyde is the canonical Rosenmund reaction (H2/Pd-BaSO4, poisoned catalyst); Clemmensen would simply reduce PhCOCl all the way to PhCH3 if it worked, but the Cl is generally lost as HCl first. (iv) Benzophenone → diphenylmethane: textbook Clemmensen on a diaryl ketone –- works.

Two-line reagent map. C=O → CH2 uses Clemmensen (acid) or Wolff–Kishner (base). RCOCl → RCHO uses Rosenmund. RCN → RCHO uses Stephen (Sn-Cl-H) or DIBAL.

Clemmensen-feasible: options (ii) and (iv).

Bond-formation angle. A clean rule of thumb: count whether the two starting molecules end up linked by a fresh C-C bond. If yes, the carbon count of the product chain exceeds either starting material; if no, no chain growth.

Walk-through. (i) Grignard R-MgX delivers a carbanion to the C=O carbon, forming a C-C bond between the R of the Grignard and the carbonyl carbon of the substrate. Net carbon gain ≥ 1. (ii) Cannizzaro disproportionates two aldehyde molecules but never welds them together –- the products PhCH2OH and PhCOO^- each keep the original carbon count. (iii) Aldol couples the α-carbon (nucleophile, after enolate formation) of one carbonyl to the C=O carbon (electrophile) of another; product chain doubles in carbon count. (iv) HVZ swaps an α-H of a carboxylic acid for Br –- no new carbon, no chain growth.

Quick check. Of the four, only Grignard and aldol involve nucleophilic carbon attacking an electrophilic carbon. Cannizzaro is a hydride shuffle (atom transferred is H^-, not C); HVZ is an α-halogenation (atom transferred is Br).

Chain-lengthening routes: options (i) and (iii).

Selectivity angle. The pivotal idea is selectivity –- of the four options, which routes stop cleanly at the ketone without over-reacting? Friedel–Crafts acylation and dialkyl-cadmium addition do; Grignard does not; the Gattermann–Koch reaction makes the wrong product.

Mechanism map for (i). AlCl3 abstracts the chloride from PhCOCl, generating the resonance-stabilised acylium ion Ph-C+=O. Benzene attacks this electrophile via standard EAS; loss of H⁺ regenerates the aromatic ring, leaving Ph-CO-Ph. Because Ph-CO-Ph is deactivated relative to benzene, a second acylation is too slow –- the reaction stops at the mono-ketone.

Why dialkyl-cadmium works for (ii). R2Cd is a soft organometallic –- the C-Cd bond is far less polar than C-Mg. It transfers an R group to the electrophilic C=O of an acyl chloride, but is not nucleophilic enough to attack the resulting ketone. Grignard (R-MgX) is too aggressive –- it ploughs through both stages, ending at the 3^∘ alcohol Ph3COH.

Why (iv) is wrong. Gattermann–Koch is C6H6 + CO/HCl → PhCHO (with AlCl3/CuCl) –- it formylates benzene to benzaldehyde, not benzophenone. The product is an aldehyde with only one ring.

Selective ketone routes: options (i) and (ii).

Two-step polar angle. Every nucleophilic addition to a carbonyl traverses exactly two intermediates: the alkoxide (immediately after Nu^- adds) and the neutral addition product (after the alkoxide picks up H+ from solvent or work-up). Both are tetrahedral at the former carbonyl carbon –- the sp² → sp³ rehybridisation is the geometric signature of nucleophilic addition.

Why options (iii) and (iv) are wrong. (iii) An enolate CH2=C(O^-)-CH3 arises when a base removes an α-H, not when Nu^- adds to the carbonyl carbon. Enolates lead to aldol/Claisen chemistry, an entirely different pathway. (iv) A radical-pair intermediate would imply homolytic cleavage of C=O into two unpaired electrons –- never observed under standard polar nucleophilic-addition conditions (would require photochemistry or a radical initiator).

Mechanism mini-cartoon. The carbonyl carbon is sp² and δ⁺. Nu^- approaches from above or below the C=O plane (Bürgi–Dunitz trajectory, ∼107^∘ to the C=O bond). The π electrons of C=O flow to oxygen as Nu–C σ forms; geometry collapses to tetrahedral sp³. The alkoxide carries a full -1 charge on O –- the σ-bonded Nu is now stable on carbon. Aqueous work-up protonates the alkoxide to deliver the neutral cyanohydrin R2C(CN)(OH).

Alkoxide (i) and the neutral protonated form (ii) are the genuine intermediates.

Donor-vs-acceptor angle. The cleanest framework is to classify each functional group as a hydrogen-bond donor, acceptor, or both. Butan-1-ol's O-H is a strong donor and the same oxygen lone pair is a strong acceptor –- a bi-functional unit that knits a continuous H-bond network through the whole liquid. Butanal's C=O is an acceptor only –- it has no O-H donor of its own, so individual molecules cannot link up except via the weak C-H ⋯ O contact (∼2 kJ/mol, barely a true H-bond).

Energetic accounting. Each genuine H-bond costs ∼20–25 kJ/mol to break. In butan-1-ol every molecule participates in ∼2 H-bonds (one donor, one acceptor), so vaporisation must overcome ∼40 kJ/mol of network energy. In butanal the dipole–dipole pull of C=O is only ∼8 kJ/mol per pair. The Δ H_vap gap directly maps to the ∼42 ^∘C boiling-point gap.

Molar-mass control. The two compounds have nearly identical molar masses (butanal 72 vs butan-1-ol 74). London dispersion is therefore essentially equal, so the boiling-point difference is a clean measure of the H-bonding contribution –- a favourite JEE trick to test whether students confuse mass effects with polar effects.

Butan-1-ol's O-H ⋯ O network gives it a much higher b.p.

Methyl-ketone angle. Draw both structures and scan for the CH3-CO- unit on the carbonyl. Pentan-2-one is CH3-CO-CH2CH2CH3 –- the methyl on the left of the C=O qualifies. Pentan-3-one is CH3CH2-CO-CH2CH3 –- both sides of the carbonyl are ethyl groups, no methyl directly on C=O, so the iodoform pathway cannot start.

Why the iodoform test is so selective. In alkali, the CH3 of a methyl ketone is deprotonated to an enolate; the enolate's α-carbon is then halogenated three times in succession to give R-CO-CI3. The -CI3 is a remarkably good leaving group (three Cl stabilise the carbanion) –- HO^- expels it as CHI3↓ (yellow ppt, m.p. 119 ^∘C), leaving the carboxylate R-COO^-. No other structural motif provides all three pieces: three identical α-H, an electron-sink C=O, and a tri-halogenated good leaving group.

Lab observation. On warming pentan-2-one with I2 in aqueous NaOH (∼60 ^∘C), a fine, pale yellow, sweet-smelling solid of CHI3 precipitates within seconds. Filtering and weighing this gives both qualitative and quantitative proof of a methyl-ketone group. Pentan-3-one under identical conditions shows no precipitate.

Iodoform: pentan-2-one +, pentan-3-one -.

Suffix-pick angle. A three-line decision tree settles every carbonyl-name. (1) If CHO is counted in the principal chain, use suffix -al with the CHO-carbon as C1. (2) If CHO is attached to a ring not counted in the main chain, use -carbaldehyde (and the ring carbon bearing CHO is automatically C1 of the ring). (3) If an extra C=O sits inside the chain, name it with the prefix -oxo- and a locant.

Walk-through. (i) Ph-CH=CH-CHO: chain has three carbons (CHO=C1), a C=C between C2–C3, and a phenyl at C3. Suffix ``-al'' + locant ``-2-en'' + prefix ``3-phenyl'' gives 3-phenylprop-2-enal. (ii) C6H11-CHO: CHO hangs off a saturated cyclohexane ring; ring is principal ⇒ cyclohexanecarbaldehyde. (iii) CH3CH2-CO-CH2-CHO: a 5-carbon chain with CHO at C1 (top priority) and a second C=O at C3 (named ``3-oxo'') ⇒ 3-oxopentanal.

Priority anchor. When both an aldehyde and a ketone appear, the aldehyde wins (gets the suffix, lowest locant); the ketone is relegated to the ``-oxo-'' prefix. This is why entry (iii) is named 3-oxopentanal, not 3-oxo-1-pentanone or 1-formylbutan-2-one.

(i) 3-phenylprop-2-enal; (ii) cyclohexanecarbaldehyde; (iii) 3-oxopentanal.

Parse-the-name angle. For every name in the question, identify (a) the parent skeleton (the last word), (b) the principal functional group that gives the parent its suffix, and (c) any prefix substituents with their locants. Then draw.

(i) 4-nitropropiophenone. Parse: phenone = phenyl ketone, propio = propanoyl (CH3CH2CO-). So the parent is C6H5-CO-CH2CH3. The ``4-nitro'' substituent goes para on the phenyl ring (numbering starts from the carbon attached to the CO). Final: p-O2N-C6H4-CO-CH2CH3.

(ii) 2-hydroxycyclopentane-1-carbaldehyde. Parent = cyclopentane (5-ring); the suffix ``-carbaldehyde'' attaches a CHO to a ring carbon, automatically C1. The 2-hydroxy adds OH to the neighbouring carbon. So: a cyclopentane ring with CHO at C1 and OH at the adjacent C2.

(iii) Phenyl acetaldehyde. Parse: acetaldehyde (CH3CHO) with one H on the methyl replaced by phenyl –- C6H5-CH2-CHO. The IUPAC equivalent is 2-phenylethanal.

Pitfall. ``Propiophenone'' is a semi-systematic legacy name; beginners often draw benzaldehyde + ethyl group instead. The ketonic C=O sits between the phenyl and the propanoyl ethyl group –- not at the end of a chain.

Structures as drawn step-by-step above.

Suffix-and-locant angle. Three IUPAC rules nail every ring–CHO compound. (1) The aldehyde group CHO has the highest priority among the substituents in this question, so it defines the principal characteristic group. (2) When CHO is attached to a ring (and the ring is the parent), use the ``-carbaldehyde'' suffix; when there are two such groups, use ``-dicarbaldehyde'' with appropriate locants. (3) Number the ring so that the suffix carbons get the lowest possible locants first, then minimise the locants of remaining substituents.

(i) Terephthalaldehyde. Two CHO groups sit at opposite ends (para) of a benzene ring. Number C1 and C4 to give the suffix groups: benzene-1,4-dicarbaldehyde. The 1,4-locant set 1,4 is the smallest possible for two para substituents.

(ii) 3-bromobenzaldehyde. A single CHO becomes C1 of the ring (suffix-bearing). The bromine sits meta to it. Going clockwise gives Br at C3; going anticlockwise also gives C3 (meta is symmetric). Name: 3-bromobenzaldehyde.

Why not ``benzene-1-carbaldehyde'' or ``5-bromobenzaldehyde''? When there is only one ring–CHO, IUPAC accepts the shortcut name ``benzaldehyde'' as the parent; you don't need to spell out ``benzene-1-carbaldehyde''. And Br at C5 (also meta) gives the higher locant, so we pick C3 by the lowest-locant rule.

(i) benzene-1,4-dicarbaldehyde; (ii) 3-bromobenzaldehyde.

Two-stage angle. The conversion has two distinct stages, each with its own selectivity rule.

Stage 1 –- side-chain chlorination. Toluene + 2 equivalents of Cl2 under UV light (or peroxide initiator, high temperature) proceeds via radical chain: Cl· abstracts a benzylic H (BDE only ∼88 kcal/mol because the resulting benzyl radical is resonance-stabilised over the ring), and the benzyl radical captures Cl· from Cl2. The second chlorination at the same carbon is even easier –- PhCH2Cl → PhCHCl2 – and controllable stoichiometry stops short of PhCCl3. No Lewis acid: adding FeCl3 would redirect chlorination to the ring (EAS at the ortho/para positions).

Stage 2 –- hydrolysis of the gem-dichloride. In dilute aqueous NaOH each C-Cl undergoes S_N substitution with OH^-, giving the unstable gem-diol PhCH(OH)2. Gem-diols of aromatic aldehydes are unstable (the two OH groups on one sp³ carbon have an unfavourable steric/electronic clash) and lose one molecule of water to give the planar aldehyde PhCHO –- the thermodynamic minimum.

Why this route is preferred industrially. Toluene is cheap, Cl2/hν is one-step gas-phase chemistry, and the subsequent aqueous hydrolysis runs at ∼100 ^∘C in steam. This is in fact one of the historical industrial routes to benzaldehyde, before selective oxidations of toluene took over.

PhCH3  (Cl2, hν)  → PhCHCl2  (NaOH/aq, Δ)  → PhCHO.

Lewis-acid activation angle. AlCl3 is the master chloride-abstractor in Friedel–Crafts chemistry. It coordinates to the Cl of PhCOCl, weakens the C-Cl bond, and finally pulls Cl^- off entirely as AlCl4^-. What is left is the acylium cation PhCO⁺ –- the genuine electrophile that attacks the ring.

Why acylium beats alkyl carbocation. An alkyl carbocation R⁺ (generated by R-Cl + AlCl3) often rearranges to a more stable carbocation before attacking benzene (Friedel–Crafts alkylation suffers from chain shuffles). The acylium ion does not rearrange because the + charge is already heavily stabilised by resonance with the lone pair of oxygen (R-C⁺=O → R-C#O⁺). That extra stability is also why Friedel–Crafts acylation always installs the acyl group cleanly and predictably.

Why monoacylation, not polyacylation. The product PhCOPh has a deactivating C=O on the ring (the carbonyl withdraws electron density via -M); a second EAS is much slower than the first because the ring electron density has dropped. This is why one equivalent of PhCOCl + benzene gives clean benzophenone, not a di-acylated product.

Electrophile is the benzoylium cation PhCO⁺; reaction is the Friedel–Crafts acylation.

Symmetry-of-the-skeleton angle. 2,5-dimethylhexan-3-one is nearly symmetrical about its C=O. The left half is (CH3)2CH- (isopropyl, C3); the right half is -CH2-CH(CH3)2 (C4). Both halves are branched, both α-carbons bear H, so strong oxidation cleaves both C_α-CO bonds.

Tracking the carbons. Left half → (CH3)2CH-COOH directly (isobutyric acid). Right half: the original -CH2-CH(CH3)2 chain undergoes oxidation at the terminal CH2 to give HOOC-CH(CH3)2 (also isobutyric acid). The single carbon that was the carbonyl C in the ketone ends up as CO2 (escapes the flask).

Sanity check on stoichiometry. 2,5-dimethylhexan-3-one is C8H16O (formula weight 128). Two molecules of isobutyric acid (CH3)2CHCOOH are 2 × C4H8O2 = C8H16O4 (weight 176). The carbon and hydrogen balances close cleanly; the extra three oxygens are supplied by KMnO4/H+, and one carbon (formerly the carbonyl C) leaves as CO2.

Why ketones are oxidatively cleaved at all. The α-C-H of a ketone is acidic (pK_a ∼20); under hot oxidising conditions, the enol tautomer is preferentially attacked, and successive oxidation steps break the C=C of the enol –- effectively cleaving the C-CO bond.

Both sides cleave to give 2-methylpropanoic (isobutyric) acid; CO2 is released.

Acidity-ladder angle. Two filters do the heavy lifting: (1) separate the carboxylic acids from the alcohol (carboxylates win because of two-O resonance), and (2) rank the carboxylic acids by the -I pull of the α-substituent.

Filter 1 –- alcohol vs acid. CH3CH2OH has only one oxygen to host the negative charge; ethoxide CH3CH2O^- is poorly stabilised. Every carboxylic acid hosts the charge on two equivalent oxygens (X-ray: identical C-O bond lengths ∼1.27 ). pK_a gap ∼11 units –- a million-million-fold difference in K_a.

Filter 2 –- α-substituent pull. Numerical pK_a values nail the order: FCH2COOH 2.59, ClCH2COOH 2.87, C6H5CH2COOH 4.31, CH3COOH 4.76. F wins on raw electronegativity (3.98 vs 3.16 for Cl); through a saturated -CH2- linker only the σ-frame inductive effect reaches the carboxylate, and -I scales with electronegativity.

Why phenyl beats methyl. A phenyl group attached to CH2 is mildly electron-withdrawing through the σ frame (the sp² ring carbon is more electronegative than sp³). Methyl is electron-donating (+I). So PhCH2COOH is slightly stronger than CH3COOH –- a subtle but examinable point.

FCH2COOH > ClCH2COOH > PhCH2COOH > CH3COOH > CH3CH2OH.

Why four products angle. In a cross-aldol with two α-H-bearing carbonyls, each molecule can play either role –- enolate (nucleophile) or electrophile (C=O). The combinatorial count is 2 × 2 = 4. Two of these are self-condensations; two are genuine crosses.

Tracking the connectivity. The aldol C-C bond forms between the α-carbon of the enolate and the carbonyl carbon of the electrophile. Mark these positions for each pairing and draw out the product. For propanal as enolate (α-C is the central CH2 of CH3CH2CHO) and 2-methylpropanal as electrophile (carbonyl C is the CHO carbon of (CH3)2CHCHO), the bond forms between CH3-CH- and -CH(OH)-CH(CH3)2, giving the cross-aldol A as above.

Dehydration step. In refluxing NaOH, each β-hydroxy aldehyde loses one water molecule (E1cb mechanism: enolate-like H removal at α of CHO followed by collapse of the C-OH). The product is an α,β-unsaturated aldehyde (an ``aldol condensation product''). The push for dehydration comes from conjugation between the new C=C and the CHO.

Real-world consequence. The mixture of four products is generally a synthetic dead end. Modern aldol chemistry uses enolate pre-formation (LDA in THF at -78 ^∘C) to generate only one enolate kinetically, then adds the electrophilic carbonyl in a separate step –- selective cross-aldol.

Cross-aldol condensation; four β-hydroxy aldehydes, dehydrate to four α,β-unsaturated aldehydes.

Functional-group ladder angle. Three reagent signatures identify the three families.

Reagent 1 –- KMnO4/OH^-. Hot alkaline permanganate oxidises a primary alcohol RCH2OH all the way to a carboxylic acid R-COOH (it overshoots the aldehyde stage). So B must be a primary alcohol and A must be the corresponding carboxylic acid.

Reagent 2 –- LiAlH4. Lithium aluminium hydride delivers two hydrides to a carboxylic acid, going through an aldehyde intermediate that gets reduced further to the primary alcohol. The ``A → B'' direction confirms the assignments from Reagent 1.

Reagent 3 –- H2SO4/Δ on acid + alcohol. This is the textbook Fischer esterification. The fruity smell is the diagnostic clue: short-chain esters are responsible for banana, pear, pineapple, and apple aromas in nature.

Worked example. If B = ethanol, then A = acetic acid, and C = ethyl acetate (pear-banana aroma). If B = butan-1-ol, then A = butyric acid, and C = butyl butyrate (pineapple). The question does not specify R, but the families are unambiguous: alcohol, acid, ester.

A = carboxylic acid; B = primary alcohol; C = ester.

-I-strength angle. The order is fixed by how strongly each α-substituent withdraws electron density through the σ-frame. NO2 tops the list because it carries a formal positive charge on its nitrogen (the N⁺ has a permanent σ-acceptor effect that beats even fluorine). F comes second –- pure atomic electronegativity but no σ-frame positive charge. Phenyl (directly on COOH) is mildly electron-withdrawing through the sp² ring carbon, but compared to NO2 or F, the effect is small.

pK_a snapshot. NO2CH2COOH 1.7; FCH2COOH 2.59; C6H5COOH 4.20. Each step is roughly ∼1.5 to 2 pK_a units, consistent with the inductive-pull hierarchy.

Trap to avoid. Some students think benzoic acid should be more acidic than a substituted-acetic acid because ``conjugation on the ring stabilises the anion''. In fact, benzoate's resonance stabilisation is partial –- the -M of COO^- on the ring is opposed by the natural -M of COOH on the ring, so the net resonance gain is small. The dominant effect for ranking these three acids is -I at the α-carbon.

NO2CH2COOH > FCH2COOH > C6H5COOH (rank set by -I pull at α).

Polarity-of-π angle. The π bond is just a region of electron density above and below the σ frame. Whether that density is symmetric or skewed sets the chemistry.

Alkenes –- symmetric π. C=C joins two atoms of identical electronegativity. The π cloud is evenly spread; it behaves like a nucleophile (offers electrons to an electrophile). The first step in alkene chemistry is invariably E⁺ attacking the π bond –- H⁺ in acid hydration, Br⁺ in Br2 addition, Hg²⁺ in oxymercuration.

Carbonyls –- polar π. C=O joins atoms of electronegativity 2.5 (C) and 3.5 (O). The π cloud is pulled toward oxygen; the carbon develops δ⁺. Now the molecule has an electrophilic carbon, and the first move is Nu^- adding to that carbon –- CN^- in cyanohydrin formation, H^- in hydride reduction, R^- in Grignard addition.

Two-flow diagram. For C=C: E⁺ approaches first, the π electrons flow toward E⁺, and a carbocation forms on the other carbon. For C=O: Nu^- approaches the δ⁺ carbon, the π electrons flow onto oxygen, and an alkoxide forms. The opposite arrow direction at step 1 is the reason for the opposite mechanism class –- electrophilic vs nucleophilic addition.

Why both end at sp³. Both reactions saturate a π bond. In both cases the central atoms rehybridise from sp² to sp³. The difference is purely kinetic (the order of arrival of E⁺ vs Nu^-), not thermodynamic.

C=C is symmetric (attacked by E⁺); C=O is polar at δ⁺ C (attacked by Nu^-).

Resonance-deactivation angle. Carboxylic acids and aldehydes both have a C=O, but the environment of the carbonyl carbon differs. In RCOOH, the second oxygen (-OH) is directly attached to the carbonyl carbon and donates a lone pair into the C=O via π overlap (a textbook +M effect). This delocalisation places partial negative charge on the carbonyl oxygen and partial positive charge on the -OH oxygen, but neutralises the δ⁺ on the carbonyl carbon. The result: the carbon is no longer the strong electrophile it is in an aldehyde or ketone.

Proton-protection angle. Even when a nucleophile is strong enough to attempt addition (e.g. LiAlH4, R-MgX, HCN), the O-H is acidic (pK_a ∼5 vs the nucleophile's conjugate-acid pK_a of ∼45 for Grignard, ∼10 for HCN). The nucleophile preferentially picks up the proton and is consumed as its conjugate acid; the carbonyl is bypassed entirely. RMgX + R^′COOH → RH + R^′COO^-MgX⁺ is the canonical illustration –- the Grignard is destroyed in a side-equilibrium before reaching the carbonyl.

What does work. Carboxylic acids can be reduced by very strong hydrides (excess LiAlH4 in dry ether –- the acid is first deprotonated to the carboxylate, then a second hydride adds to the C=O of the salt). They can also be converted to acyl chlorides (SOCl2), esters (H2SO4 + alcohol, Fischer), or amides (heat with amine), all by mechanisms that sidestep classical nucleophilic addition.

Resonance from -OH reduces δ⁺ on carbonyl C; the acidic O-H proton also intercepts nucleophiles –- no classical nucleophilic addition.

Work-backwards angle. The product is tert-butyl alcohol (CH3)3C-OH. Subtract acetic acid's contribution CH3-C(OH)-: two extra methyls are needed on the carbonyl carbon. The only standard reagent that delivers a methyl carbanion is CH3MgBr (or CH3Li). So A must be a methyl Grignard.

Why A also destroys B. Acetic acid has a very acidic O-H (pK_a 4.76). Grignards are super-bases (conjugate-acid pK_a ∼50). The first equivalent of CH3MgBr is therefore protonated as fast as it meets the acid: CH3MgBr + CH3COOH -> CH4 + CH3COO-MgBr. The Grignard is wasted as methane gas; the only organic by-product on the magnesium side is the carboxylate C.

To make the alcohol cleanly. You need to either (i) use three equivalents of CH3MgBr (one to neutralise the acid, two more to add to the carbonyl of the resulting carboxylate), or (ii) convert the acid to the methyl ester first (H2SO4/MeOH) and then use 2 equiv. CH3MgBr. The exemplar sequence is asking you to spot this protection-deprotection wrinkle.

Identification. A = CH3MgBr (methyl-magnesium bromide); B = CH3COOH (acetic acid); C = CH3COO-MgBr (the deprotonation by-product) –- with methane gas CH4 evolved alongside.

A = CH3MgBr, B = CH3COOH, C = CH3COO-MgBr.

Conjugate-base angle. The acidity of an O-H is set entirely by how comfortable the resulting O^- is. The more the negative charge can spread (delocalisation) or be pulled toward electronegative neighbours (induction), the more stable A^- becomes and the easier HA donates a proton.

Three-way comparison. (a) Alkoxide R-O^-: charge is fully localised on a single oxygen; the σ-donating alkyl group (+I) actively destabilises it. (b) Phenoxide PhO^-: charge spreads over the oxygen and three ring carbons via π-conjugation (four resonance forms), but the ring positions are carbon-bearing (less electronegative), so the stabilisation is only modest. (c) Carboxylate R-COO^-: two equivalent resonance forms place identical -12 charges on each of two oxygens (X-ray data confirm identical C-O bond lengths ∼1.27 ). This is the most efficient stabilisation pattern.

pK_a numbers tell the story. Ethanol ∼16, phenol ∼10, acetic acid ∼4.76 –- each step buys ∼5 to 6 pK_a units of acidity, exactly matching the order-of-magnitude jump in conjugate-base stability.

Why the question is conceptually rich. It tests whether students grasp that the identity of O-H alone is not enough –- it's the environment of the oxygen, both before and after deprotonation, that decides acidity. Same atom, very different chemistry.

Carboxylate stabilisation (two-O resonance) > phenoxide (ring delocalisation) > alkoxide (no resonance).

Step-by-step angle. Each step in this sequence follows a single, well-defined rule. Spot the rule, apply it.

Step 1 –- Grignard on a ketone. CH3MgBr delivers a CH3^- equivalent to the electrophilic carbonyl carbon of acetone. The product after aqueous quench is a tertiary alcohol: (CH3)2C(OH)CH3 = (CH3)3C-OH. Net: one new C-C bond, C=O → C-OH, carbon count +1.

Step 2 –- Na as a deprotonating base. Sodium metal is a mild but effective deprotonator of alcohols (analogous to its reaction with water: 2 Na + 2 ROH -> 2 RO^-Na⁺ + H2). The product is the sodium tert-butoxide salt, a strong, hindered base widely used in organic synthesis (especially for E2 eliminations).

Step 3 –- Williamson ether synthesis. The tert-butoxide alkoxide attacks CH3Br in an S_N 2 step. The methyl halide is unhindered (1^∘, no β-H branching), so S_N 2 wins over E2. The product is the mixed ether, 2-methoxy-2-methylpropane –- common name tert-butyl methyl ether (MTBE, the petrol additive used as an oxygenate before being replaced by ethanol).

Why Williamson runs from this direction. You always combine the tertiary alkoxide with the primary halide, never the other way around. The reverse pairing (CH3O^-Na⁺ + (CH3)3CBr) would give 99% elimination (isobutylene + methanol + NaBr) because tertiary halides have a strong E2 preference with strong bases. Choosing the right donor/acceptor pair is the cardinal rule of Williamson ether synthesis.

A = (CH3)3COH; B = (CH3)3CONa; C = (CH3)3C-O-CH3 (MTBE).

Two-flaws-of-alkylation angle. Direct Friedel–Crafts alkylation has two well-known limitations.

Flaw 1 –- carbocation rearrangement. The mechanism of FC alkylation generates a free carbocation R⁺ from R-Cl and AlCl3. Free carbocations are notorious for hydride and methyl shifts that take the cation to a more stable (often secondary or tertiary) species. With C2H5Cl, the ethyl cation CH3CH2⁺ is primary and cannot rearrange (no 1,2-shift gives a more stable cation in just two carbons), so this flaw is mild for ethyl. But with n-propyl chloride, the cation rearranges to isopropyl, and you get cumene contaminating n-propylbenzene.

Flaw 2 –- polyalkylation. An alkyl group on the ring is an activating substituent (donates electron density by +I). The product alkylbenzene is therefore more nucleophilic than the starting benzene, and the second EAS is faster than the first. Result: a statistical mixture of mono-, di-, and tri-alkylated products.

Why acylation avoids both. The acylium ion RCO⁺ is resonance-stabilised (R-C⁺=O ↔ R-C#O⁺) and is remarkably resistant to rearrangement. The acyl product C6H5-CO-R has a -M carbonyl on the ring, deactivating it, so the second acylation is too slow to compete. Pure mono-acyl product is the rule.

Step b –- Clemmensen. Reduction of the acyl group to methylene completes the formal alkylation cleanly: C6H5-CO-R Zn(Hg)/HCl C6H5-CH2-R. Net outcome: ethylbenzene from benzene, no rearrangement, no over-substitution.

Acylation + Clemmensen avoids carbocation rearrangement and the polyalkylation problem of direct FC alkylation.

Mechanistic-twin angle. Both reactions are textbook EAS, both work via an acylium-type cation as the kinetic electrophile, both require AlCl3 as Lewis acid. In this sense Gattermann–Koch is just ``Friedel–Crafts acylation with formyl''.

Where they differ in practice. Friedel–Crafts acylation delivers R-CO- where R is alkyl or aryl –- giving a ketone. The starting acyl chloride RCOCl is a stable, bottleable compound. But for R = H, the would-be reagent ``HCOCl'' (formyl chloride) does not survive at room temperature –- it spontaneously decomposes back to HCl + CO. So you cannot run a standard Friedel–Crafts acylation with formyl chloride. Gattermann–Koch sidesteps this by generating the formyl cation HCO⁺ in situ from HCl + CO + AlCl3 (the CuCl co-catalyst helps stabilise the gaseous CO).

Why this matters for synthesis. Aromatic aldehydes (ArCHO) are key building blocks for fragrances, dyes, and drugs. Direct one-step installation of -CHO onto benzene rings is therefore very useful; Gattermann–Koch is one of the cleanest routes (others: Vilsmeier with HCONMe2/POCl3, Reimer–Tiemann on phenols with CHCl3/KOH, Stephen reduction of nitriles). Each chooses a different electrophilic-equivalent of the formyl group.

Summary verdict. Mechanistically, Gattermann–Koch is the formyl variant of Friedel–Crafts acylation. Practically, they complement each other –- one delivers aldehydes, the other ketones, and together they cover every Ar-CO-R (R = H or alkyl).

Yes –- mechanism, catalyst, and EAS framework match. They differ in the electrophile used (formyl vs acyl) and hence in the product class (aldehyde vs ketone).

Common-name lookup angle. The five common names in Column I each map to a single structure. Once the structure is fixed, the IUPAC name is mechanical.

Structures then names. (i) Cinnamaldehyde C6H5-CH=CH-CHO –- three-carbon chain with CHO at C1, C=C between C2 and C3, phenyl on C3 ⇒ 3-phenylprop-2-enal. (ii) Acetophenone C6H5-CO-CH3 –- two-carbon chain with C=O at C1, phenyl on C1 ⇒ 1-phenylethan-1-one. (iii) Valeraldehyde CH3(CH2)3CHO –- five-carbon chain ending in CHO ⇒ pentanal. (iv) Acrolein CH2=CH-CHO –- three-carbon chain with CHO at C1, C=C between C2 and C3 ⇒ prop-2-enal. (v) Mesityl oxide (CH3)2C=CH-CO-CH3 –- five-carbon chain with C=O at C2, C=C between C3 and C4, methyl branch at C4 ⇒ 4-methylpent-3-en-2-one.

Trap to avoid. ``Acrolein'' and ``cinnamaldehyde'' are both α,β-unsaturated aldehydes; acrolein is the parent (propenal), cinnamaldehyde is the 3-phenyl derivative. Don't swap them.

i–d; ii–e; iii–a; iv–b; v–c.

Carbon-count angle. The trivial dicarboxylic-acid names line up with carbon counts on a simple ladder. Once you know the count, the IUPAC name is just ``n-ane-1,n-dioic acid''.

Walk-through. (i) Phthalic = ortho-C6H4(COOH)2, aromatic ⇒ benzene-1,2-dicarboxylic acid (the ring is the parent, not a six-carbon chain). (ii) Oxalic = (COOH)2, two carbons both as COOH ⇒ ethanedioic. (iii) Succinic = 4 carbons, two at the ends as COOH ⇒ butanedioic. (iv) Adipic = 6 carbons (the nylon-6,6 precursor) ⇒ hexanedioic. (v) Glutaric = 5 carbons (named after gluten in the original isolation) ⇒ pentanedioic.

Trap. Phthalic acid is the aromatic ortho-diacid; do not equate it with hexanedioic acid just because both have 6 C's. The parent skeleton differs (benzene ring vs alkane chain).

i–b; ii–e; iii–d; iv–a; v–c.

Product-shape angle. For every Column I entry, ask: where did the original oxygen end up, and was a new carbon added? The answer immediately pinpoints the reagent class.

(i) Benzophenone → diphenylmethane. Oxygen has vanished entirely; no carbon added. This is full C=O → CH2 reduction, the signature Clemmensen transformation ⇒ (c) Zn(Hg)/HCl. (Wolff–Kishner would do the same job but isn't in the reagent column.)

(ii) Benzaldehyde → 1-phenylethanol PhCH(OH)CH3. A new CH3 carbon and an OH have appeared –- the C=O has been converted to a C(OH)(CH3). Only an organometallic delivers a methyl group; Grignard CH3MgBr ⇒ (d).

(iii) Cyclohexanone → cyclohexanol. Oxygen retained, but the C=O is reduced to C-OH; no carbon added. This is the classic hydride reduction; LiAlH4 does this with universal reliability ⇒ (a).

(iv) Phenyl benzoate → benzaldehyde. Starting material is an ester (PhCOOPh) and product is an aldehyde –- partial reduction stopping at the RCHO stage. Only DIBAL–H (at -78 ^∘C, 1 eq) achieves this selectivity ⇒ (b). LiAlH4 would barrel through to PhCH2OH.

Rule of thumb. ``Oxygen-state delta'' tells you the reagent family; ``carbon-count delta'' tells you whether a C-C bond was made (Grignard/cadmium) or not (hydride/dissolving-metal).

i–c, ii–d, iii–a, iv–b.

Reagent-fingerprint angle. Every named reaction has a characteristic reagent combination –- spot the combination, name the reaction.

(i) C6H6 + RCOCl + AlCl3. The trio aromatic substrate + acyl chloride + Lewis acid is the signature of Friedel–Crafts acylation. Product is an aromatic ketone.

(ii) 2 RCHO + conc. NaOH. An aldehyde with no α-H (here HCHO) disproportionates to the carboxylate + alcohol –- Cannizzaro reaction.

(iii) RCOOH + Br2/P (red phosphorus). The Hell–Volhard–Zelinsky reaction brominates the α-carbon of a carboxylic acid.

(iv) 2 CH3CHO + dil. NaOH. An aldehyde with an α-H undergoes base-catalysed coupling to give a β-hydroxy aldehyde –- aldol condensation (strictly ``aldol addition''; the dehydration step would give the α,β-unsaturated aldehyde).

(v) RCOCl + H2/Pd-BaSO4. Poisoned palladium (deactivated by BaSO4 and quinoline) stops the reduction at the aldehyde stage –- Rosenmund's reduction.

(vi) RCN + SnCl2/HCl, H3O+. Stannous chloride and HCl reduce a nitrile to an imine, which hydrolyses on aqueous work-up to an aldehyde –- Stephen's reaction (the modern substitute is DIBAL at -78 ^∘C).

i–a; ii–d; iii–b; iv–c; v–e; vi–f.

Geometry-from-hybridisation angle. The clearest way to see the assertion-reason link is to trace the bonding around the central carbon of formaldehyde, H2C=O.

Step 1 –- count σ bonds and lone-pair-bearing p orbitals. Carbon makes three σ bonds (two C-H and one C-O) and contributes one π bond to C=O. Three σ frameworks plus one π ⇒ three sp² hybrid orbitals in a plane, one unhybridised p_z perpendicular to that plane.

Step 2 –- map orbitals to geometry. VSEPR says three sp² orbitals point to the corners of a trigonal arrangement, ∼120^∘ apart. So H, H, and O all lie in the same plane as the central carbon. The molecule's atoms all live in a single plane ⇒ planar.

Step 3 –- verify against data. Microwave spectroscopy of H2CO gives ∠HCH = 116^∘ and ∠HCO = 122^∘ –- distorted slightly from 120^∘ by the larger lone pairs on O but unambiguously planar. The Reason (R) is not merely a true statement; it is the causal mechanism for the Assertion (A).

Common trap. Confusing ``trigonal planar'' (three substituents, sp<sup>2</sup>) with ``tetrahedral'' (four substituents, sp³). Tetrahedral sp³ carbons are not planar; only sp and sp² centres are.

Option (i): A true, R true, and R explains A.

Logic-check angle. Assertion-Reason questions test causality, not just isolated facts. To answer correctly you must (a) verify A independently, (b) verify R independently, and (c) ask whether R logically explains why A is true. Failure of step (c) is the most common pitfall.

Step (a) –- is A true? Yes. Aldehydes contain a C-H on the carbonyl carbon, which is unusually weak (BDE ∼88 kcal/mol) because the resulting acyl radical is stabilised by the adjacent C=O. This makes R-CHO readily oxidised by even mild reagents such as Tollens', Fehling's, K2Cr2O7/H+, KMnO4, H2O2, and even atmospheric O2.

Step (b) –- is R true? Yes. LiAlH4 in dry ether reduces R-COOH all the way to R-CH2OH via two hydride additions (carboxylate → aldehyde intermediate → alkoxide → alcohol after work-up). This is a textbook reaction.

Step (c) –- does R explain A? No. R describes a reduction of carboxylic acid going backwards on the redox ladder; A talks about the oxidation of aldehyde going forwards. They are opposite-direction processes on different substrates. The ease of R-CHO → R-COOH has nothing to do with the fact that LiAlH4 also happens to reduce carboxylic acids. Two independent true facts, no causal link.

Verdict. A true, R true, R does not explain A ⇒ option (v).

Option (v): A correct, R correct, R does not explain A.

Pitfall-spotting angle. The Assertion-Reason pair is deliberately set up so that the Reason is correct but the Assertion contains a flipped polarity. Many students rush in and pick option (i) (``both true, R explains A'') without checking the direction of the claim.

Reason check. The Reason states a textbook fact: the α-deprotonated carbanion is delocalised onto the carbonyl oxygen, the resulting enolate is significantly more stable than a plain alkyl carbanion. This explains why α-H is acidic at all.

Assertion check. Is the α-H ``less acidic''? Compared to what? If the comparison is with normal alkyl C-H (CH4, pK_a ∼45), then α-H of a ketone (pK_a ∼20) is much more acidic –- by 25 pK_a units. The Assertion is therefore wrong.

The trick. R would explain A if A were ``α-H is unusually acidic''. The flipped polarity in A is the trap; with the Reason intact, the correct option is (iv) (A wrong, R correct).

pK_a snapshot. Methane 45, alkyl C-H 50, acetaldehyde 17, acetone 20, ethyl acetate 25, acetylacetone 9. The pattern matches the enolate-stabilisation argument: stronger conjugation with C=O (or two C=O groups) ⇒ more acidic α-H.

Option (iv): A is wrong (the α-H is unusually acidic); R is correct.

Independent-truth angle. Two facts to check; the trap is in the Reason.

Assertion truth. Both HCHO and any Ar-CHO (benzaldehyde, furfural, etc.) lack an α-H. In conc. NaOH they cannot form an enolate, so the aldol path is shut. The slower Cannizzaro disproportionation (hydride shuttle between two carbonyl molecules) takes over. So both classes undergo Cannizzaro ⇒ A is true.

Reason check –- the trap. The Reason claims aromatic aldehydes are ``as reactive as formaldehyde''. This is false. Formaldehyde is the most reactive carbonyl known (smallest, no +I substituent, two H's exposing the δ⁺). PhCHO has a large aryl group that donates electrons into the C=O via resonance (+M); this reduces the carbonyl carbon's δ⁺ and slows nucleophilic addition. Experimentally, PhCHO adds HCN many times slower than HCHO does.

Resolution. Both aldehydes can do Cannizzaro (A true) not because they are equally reactive, but because they both lack α-H (the structural prerequisite). Reactivity is a separate axis from α-H availability. So R fails as both a fact and an explanation.

Option (iii): A true; R false.

Diagnostic-test angle. Tollens' reagent is the classic silver-mirror test for aldehydes. The mechanism is an oxidation: R-CHO donates the C-H on the carbonyl carbon (and one of the lone pairs on oxygen) to Ag⁺, becoming the carboxylate R-COO^-; the Ag⁺ accepts two electrons and deposits as Ag↓ on the glass wall.

Why ketones fail. The carbonyl carbon of a ketone has no C-H bond to give up –- both substituents are R (alkyl or aryl). The two-electron oxidation that converts CHO to COOH has no analogue for R-CO-R^′ (you would need to break a C-C bond, which requires harsh conditions outside Tollens' regime). So the silver mirror only forms with aldehydes.

R-check. Both aldehydes and ketones do indeed contain a carbonyl group –- a true statement. But sharing the carbonyl is not sufficient for the diagnostic test; the key is the oxidisable C-H on the carbonyl carbon, which only aldehydes possess.

Verdict. A is wrong (silvers only with aldehydes); R is a true but insufficient statement. Option (iv): A wrong, R correct.

Option (iv): A wrong, R correct.

Test-stack angle. Decode each test to a structural constraint, then intersect.

B passes both tests. (1) Fehling+ ⇒ B is an aliphatic aldehyde R-CHO. (2) Iodoform+ ⇒ B has a CH3-CO- or CH3-CH(OH)- motif. The only aldehyde with CH3-CO- on the same carbon is acetaldehyde CH3-CHO (CHO is the CO part of CH3-CO- in this single molecule). So B = acetaldehyde.

C is iodoform+ but Fehling-. (1) Fehling- rules out aldehyde (so C is a ketone). (2) Iodoform+ on a ketone requires the methyl-ketone motif CH3-CO-R. The smallest methyl ketone is acetone CH3-CO-CH3 (3 carbons). For our C5H10 counting, C must be acetone (so that B + C total C5 together: 2 + 3 = 5).

Recover A. Ozonolysis splits the C=C and puts a C=O on each carbon. Reverse: B's CHO carbon and C's central CO carbon were once linked by C=C. Re-form that bond: (CH3)2C=CH-CH3 = 2-methylbut-2-ene. Carbon count C5; H count 5 × 2 = 10; C5H10 .

Why no other alkene fits. If A were 2-methylbut-1-ene, ozonolysis would give HCHO + CH3CH2COCH3 (butan-2-one) –- HCHO is Fehling+ but is not iodoform+ (no CH3CO-), so B-pattern fails. If A were pent-2-ene, ozonolysis would give CH3CHO + CH3CH2CHO; CH3CH2CHO (propanal) is Fehling+ but is not iodoform+ (no CH3CO either side), so C-pattern fails. The unique fit is 2-methylbut-2-ene.

A = 2-methylbut-2-ene; B = CH3CHO; C = CH3COCH3.

Diagnostic-stack angle. Read off each piece of information as a structural filter and intersect the filters until only one structure survives.

Filter 1 –- 2,4-DNP positive, orange-red. Excludes everything that is not an aldehyde or a ketone. Survivors: R-CHO or R-CO-R^′.

Filter 2 –- Tollens', Fehling's both negative. Excludes aliphatic aldehydes (Fehling+) and even excludes most aromatic aldehydes from being the answer (Tollens+). Survivors: a ketone –- aliphatic or aromatic.

Filter 3 –- Br2/water -, Baeyer's -. No alkene and no primary/secondary alcohol. Survivors: a saturated ketone.

Filter 4 –- iodoform positive. A methyl ketone, CH3-CO-R. Survivors: R-CO-CH3 where R has no C=C or -OH.

Filter 5 –- formula C8H8O. Degree of unsaturation Ω = (2· 8+2-8)/2 = 5. A benzene ring uses 4 of them; the C=O uses one. So the molecule is benzene + one C=O + side chains summing to 2 carbons and 6 Hs. Combined with Filter 4 (methyl ketone), the only fit is Ph-CO-CH3 –- acetophenone.

Filter 6 –- drastic CrO3 gives C7H6O2. Hot chromic acid cleaves a benzylic side chain right back to the ring, converting it to -COOH. Acetophenone's -CO-CH3 is oxidatively chopped off, leaving benzoic acid C6H5COOH (formula C7H6O2).

Why no other formula fits. Phenylacetaldehyde (PhCH2CHO, also C8H8O) would give Tollens+ and Fehling+; ruled out. Styrene oxide (C8H8O epoxide) would decolourise Baeyer's; ruled out. Only acetophenone passes the full diagnostic stack.

A = acetophenone C6H5COCH3; B = benzoic acid C6H5COOH.

Why propanal beats acetone angle. Both molecules have a C=O, but propanal has one alkyl group (ethyl) on the carbonyl carbon, while acetone has two methyls. Two effects favour propanal: (1) Less steric crowding –- CN^- approaches the carbonyl carbon along the B"urgi–Dunitz trajectory at ∼107^∘ to the C=O axis; one alkyl group leaves more room than two. (2) Less +I destabilisation of the carbonyl –- each alkyl donates electron density into the C=O, reducing the carbon's δ⁺; one alkyl makes propanal's carbonyl carbon hungrier for Nu^- than acetone's.

Mechanism in detail. Under mild basic conditions (small NaCN added to give CN^- in equilibrium with HCN), the cyanide ion attacks the carbonyl carbon. The π electrons of C=O relocalise onto oxygen, giving a tetrahedral alkoxide intermediate R-CN|C-O^-H. The alkoxide picks up a proton from HCN or solvent to give the neutral cyanohydrin R-C(OH)(CN)-H, regenerating CN^-.

Why the reaction is reversible. Cyanohydrin formation has a small but measurable reverse rate: HO^- can deprotonate the O-H, and the resulting alkoxide can re-expel CN^- to regenerate C=O. The equilibrium constant is modest, not infinite.

Why strong acid kills the reaction. The active nucleophile is CN^-, not HCN (HCN is too weak to add). Strong acid protonates CN^- to HCN, removing it from the reactive pool. With less cyanide, the rate plummets; with the reverse reaction unaffected, the equilibrium shifts toward reactants, and the product concentration falls.

Propanal is faster (steric + electronic); two-step polar addition; reversible (does not go to completion); strong acid lowers product concentration.

Two-test diagnostic angle. Two tests partition the four options (aldehyde or ketone) into a unique answer.

Tollens+ filter. Only aldehydes (aliphatic or aromatic) give the silver mirror. The mechanism is two-electron oxidation of the C-H on the carbonyl carbon: R-CHO + Ag⁺ -> R-COO^- + Ag. Ketones lack this C-H and fail the test. So A is the aldehyde.

NaHSO3 filter. Bisulphite adds to nearly every aldehyde and most small methyl ketones (acetone, methyl-ethyl-ketone). The test is not aldehyde-specific –- it's a steric filter: only carbonyls with small substituents on the C=O carbon are accessible. Failure of large ketones (e.g. benzophenone) is due to steric crowding around the C=O.

Resolution. A is an aldehyde (passes both); B is a small methyl ketone (passes only the bisulphite test).

Why the bisulphite adduct crystallises. R-C(OH)(SO3Na)-R^′ is a high-melting, water-soluble, but ethanol-insoluble salt. It is often used as a purification handle for aldehydes: precipitate out as the bisulphite adduct, wash, then liberate the aldehyde by treating the crystals with Na2CO3 or dilute mineral acid.

Mechanism note. The bisulphite ion HSO3^- attacks the carbonyl carbon with sulphur as the nucleophile (the S lone pair of the SO3^2- resonance form); the alkoxide is protonated by the bisulphite's own O-H to give the neutral hydroxysulphonate.

A is the aldehyde (Tollens+ and NaHSO3+); B is the small methyl ketone (only NaHSO3+).