NCERT Solutions Class 12 Chemistry Chapter 8 AKCA - Download PDF

Picture-first. Every derivative in this list is a nucleophile adding to the same electrophile, the carbonyl carbon. The carbon is C^δ+; nucleophile forms the new C–Nu bond; oxygen becomes -OH; if the nucleophile carries an NH next to the new bond, that NH kicks the OH out as water and we get a C=N.

Grouping by mechanism:

• Simple addition (stop at the tetrahedral adduct): cyanohydrin (HCN), hemiacetal (ROH, one equiv).
• Addition then substitution by a second nucleophile: acetal, ketal (two equivs of ROH with dry HCl).
• Addition–elimination (C=O → C=N): imine, oxime, semicarbazone, 2,4-DNP, Schiff base (all nitrogen nucleophiles with an N-H).

Alternative approach (the electrophilicity–steric–resonance trio). Three independent factors decide how easily each nucleophile can land on the carbonyl carbon: (a) how positive that carbon already is (electrophilicity), (b) how crowded its two faces are (steric), and (c) whether the carbonyl is part of a π system that gives away positive charge to a neighbour (resonance). For HCHO all three are favourable, so it forms every adduct in this list at room temperature. For a hindered or resonance-stabilised carbonyl (e.g. amide R-CO-NH2, ester R-CO-OR), the same nucleophiles struggle. Reading any new substrate through this trio is faster than memorising case-by-case.

Concept linkage: α-H acidity, tautomerism and oxidation/reduction. Each adduct here keeps the original carbonyl carbon's oxidation state intact (it is still C⁺¹ in an aldehyde-derived adduct). That is the bridge to Q 8.7 (aldol, which needs α-H acidity), to Q 8.20 (oxidation of -CHO to -COOH) and to Q 8.15 (Clemmensen / Wolff–Kishner, which take the carbon two oxidation states down). Recognising ``addition only'' vs ``oxidation'' vs ``reduction'' is the first sort you do on any carbonyl reagent.

1. Cyanohydrin: (CH3)2C=O + HCN - (CH3)2C(OH)CN. An industrial precursor to methyl methacrylate (Plexiglas monomer).
2. Acetal vs hemiacetal: CH3CHO + C2H5OH <=> CH3CH(OH)(OC2H5) (hemiacetal), then + C2H5OH, -H2O → CH3CH(OC2H5)2 (acetal).
3. Semicarbazone, oxime and 2,4-DNP follow the same addition–elimination: R2C=O + H2N-X - R2C=N-X + H2O with X = NHCONH2, OH, NH-C6H3(NO2)2 respectively.
4. Aldol: the α-carbanion of one CH3CHO attacks the carbonyl carbon of a second CH3CHO to give CH3CH(OH)CH2CHO. This is the only entry on the list that builds a new C-C bond.
5. Imine / Schiff: same addition–elimination but with a primary amine. When the amine is aromatic (C6H5NH2, aniline) the imine is called a Schiff's base.

Exam relevance. The CBSE 2024 and 2023 board papers both asked one-mark identifications of cyanohydrin and 2,4-DNP products, and a three-mark question on the differentiation of oxime vs Schiff base. Knowing the nucleophile-suffix pairing (see exam tip above) gives you the answer in under a minute.

Memory hook. One-shot addition for HCN / single ROH. Two-shot substitution for full acetal/ketal. Eliminate water whenever the nucleophile is an amine derivative. Aldol is the odd one out: a carbon nucleophile, so the new bond is C-C.

Definitions plus one balanced equation per term, as in the main solution.

Strategic angle. Skim each structure for the highest-priority suffix-bearing group (-COOH beats -CHO beats C=O). Anchor it at the lowest locant, then read the chain out as a substituent list plus suffix.

Alternative approach (priority table). When several functional groups appear in the same molecule, the IUPAC seniority order is -COOH > -COOR > -CONH₂ > -CN > -CHO > C=O > -OH > -NH₂. Only the highest-ranked group earns the suffix; every other group falls back to a prefix (oxo- for a ketone C=O, formyl- for an aldehyde -CHO, cyano- for -CN, etc.). The first scan of any new compound should be this priority sort.

Concept linkage. Each name fixes the connectivity, which in turn fixes the reactivity. Pentane-2,4-dione (iv) has two α-Hs between the carbonyls (the methylene), so its pK_a is unusually low (∼ 9) - that links straight into the enol–enolate chemistry of Q 8.7. The 3,3,5-trimethylhexan-2-one (v) has no α-H on the C-3 side (quaternary carbon), so the aldol enolate forms only on the methyl side; this matters in later questions.

1. (i) Pentanal with a methyl on C-4: 4-Methylpentanal. Locants 1 (CHO) and 4 (CH₃) sum to 5; numbering the other way gives 1 (CHO) and 2 (CH₃) sum 3, but that puts CHO at the wrong end of the chain - recall CHO is always at C-1 so the only legal numbering is from CHO.
2. (ii) Six-carbon ketone, Cl at C-6, C2H5 branch on C-4: 6-Chloro-4-ethylhexan-3-one. The ketone wins the lowest locant (3); alphabetical order ``chloro'' before ``ethyl'' is read.
3. (iii) Four-carbon α,β-unsaturated aldehyde: (E)-But-2-enal. trans configuration is more stable due to lower steric strain.
4. (iv) Symmetrical 1,3-diketone: Pentane-2,4-dione (acetylacetone). The methylene between two C=O groups is unusually acidic (pK_a ≈ 9).
5. (v) Six-carbon ketone with gem-dimethyl at C-3 and methyl at C-5: 3,3,5-Trimethylhexan-2-one. Locants 2,3,3,5 beat any alternative numbering.
6. (vi) Branched neopentyl acid: 3,3-Dimethylbutanoic acid. Four-carbon parent because -COOH must be C-1.
7. (vii) Two aldehydes para on benzene: Benzene-1,4-dicarbaldehyde. Both CHO groups keep the carbon, so they appear as ``carbaldehyde'', not ``-al''.

Exam relevance. CBSE awards full marks only for IUPAC names with the correct locant set, hyphenation and capitalisation. A common 1-mark trap is asking the name of CH3COCH2COCH3 - many write ``2,4-pentanedione'' but the locants 2,4 must precede a hyphen on the parent (pentane-2,4-dione, not 2,4-pentanedione).

Why this matters. Naming locks in the structure; once you write the name correctly, you know how each reagent will hit the molecule (which α-H is there, where steric crowding is, where the polar group sits).

Same set of seven IUPAC names as above.

Strategic angle. Build each skeleton as parent + decorations. Pick the parent (butanal, pentan-2-one, pentanoic acid, benzaldehyde, benzophenone) and pin the principal group at C-1 or lowest locant. Then drop on substituents.

Alternative approach (skeleton-first). For any name, first sketch the bare carbon chain or ring (the parent), then add the principal group at the locant given, and only after that hang on the substituents. This forces you to keep the carbon count correct. Try it on (viii): hex = 6 carbons + COOH at C-1 + double bond 2–3 + triple bond 4–5, and you get the structure in three pen strokes.

1. (i) Parent butanal C1HO-C2H2-C3H2-C4H3; add CH3 at C-3 to get OHC-CH2-CH(CH3)-CH3, i.e. isovaleraldehyde.
2. (ii) Propiophenone is PhCOC2H5; p-NO2 gives O2N-C6H4-CO-C2H5, a yellow crystalline solid.
3. (iii) Benzaldehyde with p-methyl is p-tolualdehyde, 4-CH3-C6H4-CHO, the precursor to terephthalic acid.
4. (iv) Pentan-2-one with C3=C4 and a methyl on C-4: (CH3)2C=CH-COCH3 (mesityl oxide). This is the α,β-unsaturated ketone from diacetone alcohol.
5. (v) Pentan-2-one + Cl at C-4: CH3COCH2CHClCH3. The C-4 chlorine is at the γ position relative to the carbonyl.
6. (vi) Pentanoic acid with phenyl at C-4 and Br at C-3: CH3CH(C6H5)CHBrCH2COOH.
7. (vii) Benzophenone with OH para on each ring: HO-C6H4-CO-C6H4-OH (used as a UV-absorbing sunscreen ingredient).
8. (viii) Hexanoic acid with C2=C3 and C4≡C5: CH3C#CCH=CHCOOH. A conjugated en-ynoic acid.

Exam relevance. Structure-from-name questions reward neat, unambiguous bond drawings. Always show the double/triple bonds explicitly (don't compress them into ≡ on a typed linear formula in your answer sheet).

Why this matters. Practising structure drawing in both directions (name → structure, structure → name) fixes IUPAC priorities firmly in mind.

Eight structures as listed above.

Quick reading. For mixed cases like (i) and (iii), count from CHO or place the ketone at the lowest locant first. For (iv), note the styryl (PhCH=CH-) group.

Alternative approach (count, then place). (a) count carbons in the principal chain (including the suffix carbon); (b) place the principal group at the lowest possible locant; (c) read remaining substituents alphabetically with their locants. For (i) CH3COC5H11, count = 7, ketone at C-2: heptan-2-one. For (vi) PhCOPh, count = 1 (the carbonyl carbon - the phenyls are substituents on it), suffix -methanone: diphenylmethanone.

Concept linkage. The names anchor reactivity. Heptanal (iii) has α-H, so it does aldol; benzaldehyde-type compounds (no α-H, see Q 8.7) do Cannizzaro instead. Cinnamaldehyde (iv) has both a C=C and a C=O - selective reduction of -CHO (NaBH₄ at low temperature) gives cinnamyl alcohol; selective C=C hydrogenation (H₂/Pd) gives 3-phenylpropanal.

1. (i) Heptan-2-one (methyl pentyl ketone). M_w = 114, used as a fragrance.
2. (ii) 4-Bromo-2-methylhexanal. Two stereocentres at C-2 and C-4 (not asked, but worth noting).
3. (iii) Heptanal (oenanthaldehyde). Six α-H total but only the two on C-2 take part in aldol.
4. (iv) (E)-3-Phenylprop-2-enal (cinnamaldehyde). Conjugated α,β-unsaturated aldehyde; the aroma of cinnamon.
5. (v) Cyclopentanecarbaldehyde. No common name in widespread use.
6. (vi) Diphenylmethanone (benzophenone). No α-H, no C-H that can transfer hydride either: stable to all the standard carbonyl reactions.

Exam relevance. Mixed naming questions test both IUPAC mechanics and recognition of trivial names. Two-mark questions typically ask one IUPAC + one common name; full marks need both.

Why this matters. The IUPAC name fixes the connectivity; the common name (cinnamaldehyde, benzophenone) is the everyday tag. Knowing both helps when reading literature.

Same six pairs of names as the main solution.

Picture-first. Every derivative is a one-swap operation on the parent carbonyl. Draw the parent first, then swap.

Alternative approach (mechanism class shortcut). Three mechanism classes cover all six:

• Addition–elimination (nitrogen nucleophile, C=O → C=N): 2,4-DNP (i), oxime (ii), semicarbazone (iv).
• Two-step substitution (oxygen nucleophile, 2 equiv ROH, C=O → C(OR)2): acetal (iii), ketal (v).
• One-step addition (one equiv ROH stops at the hemiacetal): (vi).

Concept linkage. The C=N stretch of an oxime, semicarbazone or 2,4-DNP appears around 1620cm^-1, distinctly lower than the parent C=O stretch near 1715cm^-1. So IR spectroscopy tells you instantly whether the derivative has formed. For acetals and hemiacetals, C=O disappears completely (replaced by C-O singles around 1100cm^-1) - another diagnostic.

1. (i) PhCH=N-NHC6H3(NO2)2 (orange solid; m.p. 237).
2. (ii) Cyclopropane ring with exocyclic =N-OH. The ring strain leaves the oxime slightly less stable than an acyclic one.
3. (iii) CH3CH(OCH3)2. Used as a solvent and as an anhydrous source of acetaldehyde.
4. (iv) Cyclobutane ring with exocyclic =N-NH-CO-NH2.
5. (v) Hexan-3-one with the C-3 sealed inside a 1,3-dioxolane. Standard C=O protecting strategy for later LiAlH4 steps.
6. (vi) HOCH2OCH3 (methoxymethanol). Unstable in dilute acid - collapses back to HCHO + CH3OH.

Exam relevance. 2-mark CBSE questions ask for the structural formula of one specified derivative given the parent carbonyl and nucleophile. Always show stereochemistry of the C=N (E/Z) if asked, but the default is just the connectivity.

Why this matters. Recognising these on a spectrum (e.g. a strong C=N stretch around 1620cm^-1, no C=O near 1715cm^-1) confirms which derivative you have made.

Six labelled structures, drawn as in the main solution.

Picture-first. The substrate is an aliphatic aldehyde with an unhindered C=O. So everything proceeds smoothly: addition with PhMgBr, oxidation with Tollens', condensation with semicarbazide, acetal formation with ethanol, and Clemmensen reduction.

Alternative approach (reagent-class sort). Sort each reagent by what it does to the carbonyl carbon's oxidation state and hybridisation:

• PhMgBr is a strong carbon nucleophile: adds C, leaves oxidation state unchanged, carbon becomes sp³.
• Tollens' is a mild oxidiser: -CHO → -COOH, oxidation state C⁺¹ → C⁺³, still sp².
• Semicarbazide: addition–elimination, swaps =O for =N-NHCONH2, sp² retained.
• Excess ROH/H+: addition-substitution, swaps =O for two -OR, carbon becomes sp³.
• Clemmensen: full reduction, C⁺¹ → C^-2, carbon becomes sp³.

Concept linkage with carbonyl reactivity. The electrophilicity of an aldehyde carbon depends on (a) the absence of an extra alkyl (+I) donor, and (b) the absence of resonance donors. Here, cyclohexyl is just one +I alkyl –- so the substrate is more reactive than acetone but less than formaldehyde. This matches the ``aldehyde > ketone'' ladder you will use in Q 8.12 (i).

1. (i) C6H11-CH(OH)-Ph: secondary alcohol from Grignard addition + protonation. Two-step mechanism (alkoxide intermediate, then aqueous workup).
2. (ii) C6H11-COOH: silver mirror plus carboxylic acid. The mirror is silver(0) deposited on the inner test-tube wall.
3. (iii) C6H11-CH=N-NH-CO-NH2: cyclohexanecarbaldehyde semicarbazone. A crystalline solid with sharp m.p. used to identify the parent aldehyde.
4. (iv) C6H11-CH(OC2H5)2: the diethyl acetal. Bears two -OEt on the same carbon.
5. (v) C6H11-CH3: methylcyclohexane. The deepest reduction available without breaking any C-C bond.

Exam relevance. CBSE 2022 asked exactly this product-set question on a different aldehyde (propanal). Marks are split: 1 per product, with 1 extra mark for naming the named reaction (Clemmensen, Rosenmund, etc.). Always label the named reaction next to the arrow.

Why this matters. The same carbonyl carbon plays five different roles depending on the reagent: electrophile for Grignard, substrate for oxidation, substrate for addition–elimination, substrate for protection, substrate for reduction. The carbonyl is the most reagent-flexible functional group you meet in Class 12.

Five products as listed; methylcyclohexane from Clemmensen is the deepest reduction.

Decision tree. For each compound: (a) Is it a carbonyl? If no → neither. (b) Does it have α-H? If yes → aldol; if no → Cannizzaro.

Alternative approach (aldol vs Cannizzaro decision). Both reactions need (i) a carbonyl and (ii) base. Aldol additionally needs an α-H so the base can deprotonate it; Cannizzaro additionally needs a transferable hydride (the C-H of the -CHO itself). A ketone like benzophenone has no α-H and no C-H on the carbonyl carbon, so neither reaction runs. Run this two-question test on every compound: ``Is there an α-H?'' then ``Is there a transferable C-H?''

• (i), (iii), (ix): no α-H ⇒ Cannizzaro.
• (ii), (v), (vi), (vii): α-H present ⇒ Aldol (gives β-hydroxy carbonyl, then dehydrates).
• (iv): no α-H but it is a ketone; benzophenone does not undergo Cannizzaro (no hydride donor); neither.
• (viii): alcohol, not a carbonyl; neither.

Concept linkage with α-H acidity and tautomerism. An aldehyde or ketone α-H has pK_a ≈ 20, much more acidic than a simple alkane (pK_a ≈ 50) because the resulting carbanion is delocalised onto the electronegative oxygen (enolate). The same enolate is the nucleophilic species in aldol. The keto–enol tautomerism is the structural underpinning: the enol form R-C(OH)=CH2 is the protonated enolate, and the two forms interconvert in solution.

1. Methanal: 2 HCHO -[OH^-] CH3OH + HCOO^-. Disproportionation is fast; commercial HCOONa is made by exactly this Cannizzaro on industrial scale.
2. Benzaldehyde: 2 PhCHO -[OH^-] PhCH2OH + PhCOO^-. Classic Cannizzaro; product mix is 50:50 alcohol:carboxylate.
3. 2,2-Dimethylbutanal: 2 RCHO -[OH^-] RCH2OH + RCOO^-, R = C(CH3)2C2H5.
4. 2-Methylpentanal aldol (after β-elimination): 2-methyl-3-propylhex-2-enal.
5. Cyclohexanone aldol: 2-(1-cyclohexenyl)cyclohexan-1-one.
6. Phenylacetaldehyde aldol: 2,3-diphenylprop-2-enal.

Exam relevance. This is one of the most-asked CBSE board questions on the chapter. Expect a 5-mark variant where the question lists 4–5 compounds and asks for sorting + product structures + mechanism. Memorise the decision tree as a 30-second sort.

Why this matters. The single rule ``no α-H = no enolate'' tells you every time which test will fire. Aldol is the α-chemistry; Cannizzaro is the carbonyl-only chemistry.

Aldol: (ii), (v), (vi), (vii); Cannizzaro: (i), (iii), (ix); Neither: (iv), (viii).

Quick reading. All three targets sit on the same family tree, with the aldol product CH3CH(OH)CH2CHO as the trunk.

Alternative approach (Hell–Volhard–Zelinsky logic does not apply here, but selective oxidant choice does). For polyfunctional targets always ask: which functional group am I changing, and what do I leave alone? Here the trunk has -OH and -CHO; each branch chooses one selective transformation: reduce -CHO but keep -OH (use NaBH₄), dehydrate but keep -CHO (use Δ), or oxidise -CHO but keep C=C (use Tollens'). Selectivity is the central skill of multifunctional synthesis.

1. Form the aldol: 2 CH3CHO -[dil. NaOH] CH3CH(OH)CH2CHO. Only a catalytic amount of NaOH is needed; too much base drives the system to the α,β-unsaturated product directly (Claisen–Schmidt-like).
2. (i) Reduce both functional groups: NaBH4 first reduces the CHO; the C(OH) already exists. Net product CH3CH(OH)CH2CH2OH. NaBH4 in MeOH at room temperature is the textbook choice.
3. (ii) Heat the aldol to drive off water across C-2,C-3 to give CH3CH=CHCHO. The newly formed C=C is conjugated with C=O - the aldol product was higher in energy than this α,β-unsaturated aldehyde because of the lost conjugation; that is the driving force.
4. (iii) Mild oxidation of (ii) with Tollens' converts -CHO → -COOH without touching the alkene. The [Ag(NH3)2]⁺ complex is a mild Ag(I) oxidiser that stops at the carboxylic acid stage.

Concept linkage –- the aldol family tree. Starting from any aldehyde with α-H, the aldol gives a β-hydroxy carbonyl. Then four standard moves give four common targets: reduce → 1,3-diol; dehydrate → enal; oxidise → enoic acid; hydrogenate → saturated β-hydroxy ketone. Pattern fluency beats route-design.

Exam relevance. CBSE 2024 set a 3-mark version on propanal-to-2-methylpent-2-enal. The marking scheme demands all three steps shown explicitly, with reagents above each arrow.

Why this matters. Recognising one ``branch point'' intermediate (the aldol product here) lets you reach a whole family of products in 2–3 steps instead of designing each route from scratch.

Aldol → (i) butane-1,3-diol; (ii) but-2-enal; (iii) but-2-enoic acid.

Strategic angle. Label one aldehyde the nucleophile (its α-C attacks) and the other the electrophile (its C=O is attacked). With two aldehydes that both have α-H, you get 2 × 2 = 4 combinations.

Alternative approach (combinatorial). Number the aldehydes A and B, then enumerate (nucleophile)+(electrophile) pairs: AA, AB, BA, BB. Two of these are self-aldols and two are crosses. For each pair, the β-hydroxy product has the carbon chain of the electrophile extended at its carbonyl carbon by the α-carbon of the nucleophile.

Concept linkage –- carbanion (enolate) reactivity. The α-H of a carbonyl has pK_a ≈ 20; in dilute NaOH the equilibrium concentration of the enolate is small but enough to react. The aldol product's -OH is then dehydrated on heating to give an α,β-unsaturated aldehyde with conjugation between C=C and C=O, which lowers the molecule's energy. The same enolate logic governs Claisen, Reformatsky and Knoevenagel reactions.

1. P+P aldol: CH3CH2CH(OH)CH(CH3)CHO → 2-methylpent-2-enal. (Self-aldol of propanal.)
2. B+B aldol: CH3CH2CH2CH(OH)CH(C2H5)CHO → 2-ethylhex-2-enal. (Self-aldol of butanal.)
3. P-nuc + B-elec: CH3CH2CH2CH(OH)CH(CH3)CHO → 2-methylhex-2-enal. (Cross product 1.)
4. B-nuc + P-elec: CH3CH2CH(OH)CH(C2H5)CHO → 2-ethylpent-2-enal. (Cross product 2.)

Exam relevance. CBSE 2023 awarded 1 mark per correctly labelled product (4 marks total) plus 1 mark for explicitly identifying nucleophile vs electrophile in each case (5 marks total). Always label, don't just list.

Why this matters. The crossed aldol is a C-C bond-forming reaction that builds complex aldehyde skeletons from simple ones; the selectivity problem here is a classic challenge in modern synthesis (Mukaiyama, Evans aldol).

Four aldols / α,β-unsaturated aldehydes as listed.

Structural observation. Three signposts:

• 2,4-DNP and Tollens' + Cannizzaro ⇒ aryl aldehyde Ar-CHO.
• Vigorous oxidation gives phthalic acid (1,2-diacid) ⇒ two side chains ortho on benzene, both end up as COOH.
• C9H10O leaves C3H6 outside the ring, split as -CHO + -C2H5.

Alternative approach (DoU first, then tests). Degree of unsaturation immediately tells you the ring + carbonyl count, so the skeleton is set before any test result is used. DoU = (2C+2-H)/2 = 5; four count for the aromatic ring (3 π + 1 ring), leaving 1 for a single C=O. Now layer on the test results to choose between possible isomers.

1. DoU = (2 × 9 + 2 - 10)/2 = 5. Four for the benzene ring + one for C=O.
2. Place -CHO at C-1 (Cannizzaro ⇒ no α-H), -C2H5 at C-2 (ortho).
3. Vigorous oxidation: -CHO → -COOH (the existing aldehyde) and -CH2CH3 → -COOH (the ethyl loses both of its carbons except the one bonded to the ring; that ring carbon ends up as a -COOH). Both ortho: phthalic acid.
4. Confirm carbon count of products: ortho-diacid is C8H6O4 - the original 9-carbon compound lost one carbon during ethyl oxidation. Always check loss/gain of carbon in side-chain oxidation.

Concept linkage –- side-chain oxidation, hyperconjugation and Hammett. The reactivity of a benzylic C-H in KMnO4 oxidation tracks the same hyperconjugation that makes toluene more easily oxidised than benzene (which is inert to KMnO4). Electron-donating p-substituents accelerate side-chain oxidation; strong electron-withdrawers (-NO2) slow it down. So the test ``vigorous KMnO₄ cuts every benzylic side chain'' is qualitatively right but rate-sensitive in practice.

Exam relevance. Structure-elucidation questions like this are a CBSE 5-mark staple. Marks split: 1 for molecular formula arithmetic, 1 for each functional-group test interpretation, 1 for the final structure. Show your atom-counting work explicitly.

Why this matters. Side-chain oxidation always cuts back to the ring carbon, no matter how long the alkyl. So an ortho-ethyl becomes an ortho-COOH, locking the substitution pattern.

Compound: 2-Ethylbenzaldehyde.

Quick reading. C8H16O2, DoU = 1 ⇒ one π-bond, so either ester or acid. ``Hydrolyses to acid + alcohol'' fixes it as an ester. The alcohol dehydrates to but-1-ene, fixing C = butan-1-ol. Then B = oxidation product of C = butanoic acid, so A is butyl butanoate.

Alternative approach (Saytzeff rule check). But-1-ene is the less-substituted alkene; if C were sec-butanol (CH3CH(OH)CH2CH3), Saytzeff dehydration would give the more-substituted but-2-ene as the major product, not but-1-ene. So C must be a primary alcohol where the only possible alkene is the terminal one –- butan-1-ol. This Saytzeff cross-check rules out the secondary isomer immediately.

1. C: but-1-ene comes from CH3CH2CH2CH2OH via -H2O (no rearrangement at 443 K).
2. B: CH3CH2CH2CH2OH -[CrO3/H2SO4] CH3CH2CH2COOH. Chromic acid is a strong oxidiser: primary alcohol → carboxylic acid, secondary alcohol → ketone.
3. A: connect B and C as an ester: CH3CH2CH2-COO-CH2CH2CH2CH3 via Fischer esterification.
4. Verify C8H16O2: C3H7COOC4H9 = C8H16O2. DoU = 1 matches the ester C=O.

Concept linkage with oxidation–reduction map. The R-CH2OH → R-CHO → R-COOH ladder is the standard oxidation map for primary alcohols. Mild oxidants (PCC, MnO₂) stop at the aldehyde; strong oxidants (CrO₃/H₂SO₄, hot acidic KMnO4) go to the carboxylic acid. Knowing which oxidant stops where is essential for selective synthesis.

Exam relevance. Three-step structure-elucidation problems are a classic CBSE 5-mark question. Show molecular formula verification (atom-counting) at the end –- it earns a separate mark.

Why this matters. Pattern: acid + primary alcohol of the same carbon count and the alcohol → alkene with terminal double bond fixes the alcohol as the unbranched primary one.

A: butyl butanoate; B: butanoic acid; C: butan-1-ol.

Strategic angle. For (i), +I and steric crowding both reduce HCN addition. For (ii), the closer the -I group to COOH, the stronger the acid; alkyl +I groups weaken the acid in proportion to branching. For (iii), score each substituent: -OCH3 (+M, para) weakens; -NO2 (-I, -M) strengthens.

Alternative approach (numerical pK_a ladder for acid strength). The qualitative orderings here follow from real pK_a values:

• Isobutyric acid pK_a = 4.86; butanoic acid pK_a = 4.82.
• β-bromobutanoic acid pK_a ≈ 4.0; α-bromobutanoic acid pK_a ≈ 2.97.
• 4-Methoxybenzoic pK_a = 4.47; benzoic acid pK_a = 4.20; 4-nitrobenzoic pK_a = 3.41; 3,4-dinitrobenzoic acid pK_a ≈ 2.8.

Lower pK_a = stronger acid. Practise placing real numbers on the ladder and the qualitative ordering becomes obvious.

Concept linkage with carbonyl reactivity. For (i), HCN addition rate ∝ electrophilicity of carbonyl C. Electrophilicity decreases as more +I alkyl groups crowd around it (alkyl donates into the empty π^* of C=O, and also blocks the approach of CN^-). This is the same logic as ``aldehydes are more reactive than ketones towards nucleophilic addition''.

1. (i) Order: Di-t-Bu ketone (most hindered; kinetically dead) < Me-t-Bu ketone < Acetone < Acetaldehyde (least hindered, fastest).
2. (ii) Branched alkyl (+I strongest) < straight chain (+I smaller) < β-bromo (-I at distance) < α-bromo (-I closest). Net: lowest pK_a for the α-bromo compound.
3. (iii) -OCH3 para (+M destabilises anion) < H (benzoic, no substituent) < one -NO2 (-I, -M stabilises) < two -NO2 groups (additive -I, -M).

Exam relevance. Acid-strength ordering is asked almost every CBSE year. Marks split: 1 mark for the correct sequence, 1 mark for naming the dominant effect (± I, ± M, steric, etc.). Always state which effect dominates.

Why this matters. The Hammett substituent effect is one of the few quantitative tools for predicting reactivity from structure. The orderings here are the qualitative form of that idea.

Three orderings as listed above.

Quick reading. Three tools cover the table:

• Tollens' (or Fehling) distinguishes -CHO from no -CHO.
• Iodoform distinguishes CH3CO- / CH3CH(OH)- from not.
• NaHCO3 distinguishes -COOH from -OH / ester.

Alternative approach (functional-group signature card). Build a one-line mental signature for each test:

• Tollens': silver mirror (visible Ag⁰ on inner wall), R-CHO only.
• Fehling: brick-red Cu2O precipitate, aliphatic R-CHO only.
• Iodoform: yellow CHI3 precipitate (m.p. 119 ^∘C), CH3CO-R or CH3CH(OH)-R only.
• NaHCO3: brisk effervescence of CO2, R-COOH only.
• Neutral FeCl3: violet/purple colour, phenols (and enols).

Concept linkage with the oxidation/reduction map. Tollens' and Fehling are mild oxidants that take -CHO → -COOH without touching C=C. Iodoform is a haloform cleavage: a C-C bond breaks, producing a carboxylate and the methyl ketone is shortened by one carbon. NaHCO3 is purely acid-base. Knowing the chemistry (not just the observation) tells you which test to pick.

1. Tollens' fires for: (i) propanal, (vi) benzaldehyde. Aliphatic propanal gives both Tollens' and Fehling; benzaldehyde gives only Tollens'.
2. Iodoform fires for: (ii) acetophenone (has CH3-CO-), (v) pentan-2-one (has CH3-CO-), (vii) ethanal (has CH3-CHO).
3. NaHCO3 fires for: (iii) benzoic acid (vs phenol), (iv) benzoic acid (vs ester).

Exam relevance. CBSE has asked this exact 7-pair question in 2018, 2020 and 2023. Marking: 1 mark per pair (must state the reagent and the visible observation). Write both. Saying ``Tollens' positive'' alone scores 0.5; saying ``Tollens' –- silver mirror with propanal, no change with propanone'' earns full marks.

Why this matters. A small kit of three reagents distinguishes seven different aldehyde/ketone/acid/phenol pairs. Recognising what the test detects is more important than memorising each case.

Tests as above for all seven pairs.

Strategic angle. Pin the central question: what directs the electrophile to the right position? -CH3 on the ring is o/p directing (activator); -COOH is meta directing (deactivator). This single rule decides every sequence.

Alternative approach (work backwards from the target). For each target, identify the last functional-group install:

• Methyl benzoate → esterify COOH → make COOH from CH3 → start from benzene → toluene.
• m-Nitrobenzoic acid → nitrate benzoic acid (meta-direct) → make benzoic acid first.
• p-Nitrobenzoic acid → oxidise p-nitrotoluene (already-installed o/p-directing methyl decides para) → nitrate toluene.
• Phenylacetic acid → hydrolyse -CN → install -CN from -Cl → benzylic chlorination of toluene.
• p-Nitrobenzaldehyde → hydrolyse -CHBr2 or use Etard on p-nitrotoluene → nitrate toluene to install NO2 para.

Concept linkage –- the directing-effect map. Activators (-CH3, -OH, -NH2, -OR) are o/p directors; deactivators except halogens are meta directors. Halogens are an exception: deactivators but o/p directors. For multi-step benzene synthesis the rule is simple: install activators before running another electrophilic substitution; install deactivators last (or use them as meta-directors).

1. (i) Friedel–Crafts methylation, then KMnO4 oxidation (side-chain to -COOH), then Fischer esterification with CH3OH/H+.
2. (ii) Get to benzoic acid first (-COOH meta-directs); nitrate (HNO3/H2SO4) gives m-nitrobenzoic acid.
3. (iii) Nitrate toluene first (-CH3 o/p-directs; para is the major thermodynamic product because of steric reasons); oxidise side chain second.
4. (iv) Toluene → benzyl chloride (Cl2/hν, free-radical) → C6H5CH2CN (KCN, SN2) → C6H5CH2COOH (hydrolysis with aq. H2SO4 or NaOH then acidify).
5. (v) Toluene → p-nitrotoluene (nitration, o/p) → Etard (or double side-chain bromination with Br2/hν then hydrolysis) gives p-nitrobenzaldehyde.

Exam relevance. Multi-step synthesis is a 5-mark CBSE classic. Marking scheme: 1 mark per correctly named reagent; candidates lose marks for forgetting to label ``meta'' or ``para'' explicitly under the structure.

Why this matters. Sequence matters as much as reagents in electrophilic aromatic substitution; one swap of order changes meta to para and the whole synthesis.

Five-step routes as written; choice of order is the key synthetic decision.

Quick reading. Group the conversions by the central operation.

• Reductions: (i), (ii), (ix).
• Aldols: (iii), (vii).
• Grignards: (v), (vi).
• Cyanohydrin: (viii).
• Aromatic sequence: (iv).

Alternative approach (named reactions as one-step compressors). Each line below is a one-step name that compresses what otherwise would take 3-4 elementary steps:

• Rosenmund = R-COCl + H2 → R-CHO (stops at aldehyde because Pd-BaSO₄ is poisoned).
• Clemmensen = R2C=O → R2CH2 in acidic conditions.
• Wolff–Kishner = same in basic conditions.
• Aldol = build C-C bond between α-C and carbonyl-C.
• Friedel–Crafts = install acyl/alkyl on benzene ring.
• Grignard = make secondary/tertiary alcohol from aldehyde or ketone + RMgX.
• Cyanohydrin → acid = HCN adds, then hydrolyse -CN to -COOH; raises carbon count by 1.

Concept linkage –- the reduction selectivity ladder. Strength order: LiAlH4 > NaBH4 ≫ H2/Pd-BaSO4 > NaBH4/CeCl3 > Rosenmund (poisoned). The stronger the reductant, the less selective. For multi-functional substrates pick the gentlest reductant that still does the job.

1. (i) (CH3)2CO -[NaBH4] (CH3)2CHOH -[H+, Δ] CH3CH=CH2.
2. (ii) PhCOOH -[SOCl2] PhCOCl -[H2/Pd-BaSO4] PhCHO. (Rosenmund.)
3. (iii) EtOH Cr2O7^2-/H+ CH3CHO OH^- CH3CH(OH)CH2CHO. (Self-aldol of acetaldehyde.)
4. (iv) C6H6 -[CH3COCl/AlCl3] PhCOCH3 -[HNO3/H2SO4] m-O2N-C6H4-COCH3. The -COCH3 is meta-directing, fixing nitration position.
5. (v) PhCHO + PhMgBr -[H3O+] Ph2CHOH -[K2Cr2O7] Ph2C=O. Grignard + oxidation.
6. (vi) PhBr -[Mg] PhMgBr -[CH3CHO, H3O+] PhCH(OH)CH3. Grignard route to a secondary alcohol.
7. (vii) PhCHO + CH3CHO -[OH^-] PhCH=CHCHO -[H2/Ni] PhCH2CH2CH2OH. Cross-aldol + reduction.
8. (viii) PhCHO -[HCN] PhCH(OH)CN -[H3O+] PhCH(OH)COOH. Cyanohydrin route to mandelic acid.
9. (ix) PhCOOH -[HNO3/H2SO4] m-O2N-PhCOOH -[LiAlH4] m-O2N-PhCH2OH. Nitrate first, then reduce the acid.

Exam relevance. The CBSE pattern: 1 mark per conversion (reagents + product), bonus 1-2 marks for naming the reaction (``Rosenmund'', ``Clemmensen'', etc.). Always label.

Why this matters. Routine syntheses use a tiny toolbox: Rosenmund, Clemmensen, aldol, Grignard, cyanohydrin, Friedel-Crafts. Knowing what each step does and what it leaves alone makes 2-step problems easy.

Same nine sequences as in the main solution.

Quick reading. Four small ideas with a shared theme: building or breaking C-C and C-O bonds at the carbonyl carbon.

Alternative approach (functional-group matrix). Sort each ``describe'' by what bond changes and where:

• Acetylation: new C(O)-O or C(O)-N bond on a nucleophile (OH/NH₂); no change at the carbonyl carbon of the acetyl group.
• Cannizzaro: new C-H bond (hydride transfer to carbonyl) and new C-O (in carboxylate); breaks the C-H of the donor's -CHO.
• Cross aldol: new C-C bond between α-C of one and carbonyl-C of the other; can be followed by β-elimination.
• Decarboxylation: breaks C-C between the carboxylate carbon and the next.

Concept linkage –- acetylation and amide/ester chemistry. Acetylation of an alcohol gives an ester, of an amine gives an amide. Both are protective: an amide doesn't react with electrophiles the way an amine does; an ester doesn't react with bases the way an alcohol does. This protective use links into the synthesis strategy in Q 8.14–8.17.

1. Phenol → phenyl acetate by acetic anhydride/pyridine. Aniline → acetanilide. Acetanilide deactivates the ring enough to allow controlled mono-bromination.
2. HCHO disproportionates to CH3OH + HCOO^-. Industrial source of sodium formate.
3. Benzaldehyde + ethanal → cinnamaldehyde (clean cross aldol because PhCHO has no α-H, so only ethanal can be the nucleophile).
4. Sodium benzoate + soda lime → benzene + Na2CO3. The classical lab prep of benzene from benzoic acid.

Exam relevance. Definition-style questions are 4 marks (1 × 4). Marks split: 1 mark per definition with 1 example each. Definitions without examples score only half marks.

Why this matters. Each of these is a one-step transformation that you can drop into a multi-step synthesis. Acetylation protects; Cannizzaro makes both an alcohol and an acid at once; cross aldol builds a C-C bond; decarboxylation strips an acid group.

Definitions plus one example each, as above.

Strategic angle. Identify the named transformation behind each blank: (A) Perkin synthesis of cinnamic acids; (B) chemoselective hydrogenation of C=C over C=O; (C) nitrile-to-acid hydrolysis.

Alternative approach (functional-group bookkeeping). For each blank, write down the starting functional groups and the ending functional groups; the reagent must accomplish exactly the difference and no more.

• (A) Start: ArCHO + anhydride. End: ArCH=CH-COOH. So we built a C-C bond (aldol-like) and lost water –- that's Perkin.
• (B) Start: ArCH=CH-CO-CH3. End: ArCH2CH2-CO-CH3. So we reduced C=C but not C=O. Choose mild H2/Pd-C.
• (C) Start: R-CH(OH)CN. End: R-CH(OH)COOH. So we hydrolysed -CN to -COOH; aqueous mineral acid or aqueous NaOH/heat works.

Concept linkage –- condensation + chemoselective reduction + hydrolysis. The three blanks span three of the most-asked exam reaction types. Recognising each as a single named transformation is the trick.

1. Perkin: PhCHO + Ac2O / NaOAc → E-cinnamic acid + acetic acid by-product.
2. H2/Pd-C adds across C=C first; ketone is untouched.
3. Aqueous mineral acid converts -CN to -COOH through an amide intermediate; the product here is 2-hydroxypropanoic acid (lactic acid).

Exam relevance. CBSE 2019 and 2024 set fill-in-the-blank synthesis questions modelled on this. Marking: 1 mark for the correct reagent/product, 1 mark for naming the named reaction.

Why this matters. The three transformations span condensation, chemoselective reduction and hydrolysis, three of the most often tested classes of carbonyl chemistry.

Three blanks filled as above.

Strategic angle. Three problems, three different principles: steric hindrance, resonance donation, and equilibrium control.

Alternative approach (apply the electrophilicity–steric–resonance trio). For each item, decide which member of the trio explains the observation:

• (i) Steric dominates: methyls at C-2, C-2, C-6 cage the C=O from both diastereotopic faces. Electronically the C=O is unchanged from cyclohexanone, but kinetics fail.
• (ii) Resonance dominates: the carbonyl-adjacent -NH2 is conjugated into the amide π system, losing its lone-pair availability. Inductive effects play a minor role.
• (iii) Neither steric nor resonance –- a thermodynamic equilibrium control (Le Chatelier).

1. (i) Steric: methyls at C-2, C-2, C-6 wall off the C=O face; CN^- cannot get within bonding distance. The electronic environment is similar to cyclohexanone but the approach trajectory is blocked. This is the same idea that makes acetone > pinacolone > di-t-butyl ketone in HCN reactivity (Q 8.12).
2. (ii) Resonance: the -NH2 next to C=O in semicarbazide donates into the amide π system and is no longer nucleophilic; only the -NH2 on the hydrazine nitrogen (the one bonded to -NH-) reacts with a carbonyl. Drawing the resonance structure helps make this clear.
3. (iii) Le Chatelier: pull out water (or ester) and the equilibrium shifts toward more ester. This is the basis of azeotropic esterification (Dean–Stark trap) and the reason molecular sieves are used in modern Fischer esterification.

Concept linkage with carbonyl reactivity. Every nucleophilic addition to a carbonyl is gated by exactly these three factors (electrophilicity of C, steric access, resonance donation by substituents). Knowing the trio lets you predict every related reactivity ordering in this chapter.

Exam relevance. CBSE 2017, 2021 and 2024 set very similar ``explanation'' questions. Marking is consistent: name the principle, then explain with a structural sketch where possible.

Why this matters. Each item is a model problem: ``steric vs electronic'' (i), ``resonance switches off nucleophilicity'' (ii), ``shift the equilibrium by removing a product'' (iii). These three ideas recur all over organic chemistry.

Three explanations as in the main solution.

Strategic angle. Run the arithmetic, then read the clues.

Alternative approach (combine tests before arithmetic). A faster strategy: read the tests first.

• ``Forms NaHSO3 adduct'' ⇒ methyl ketone or aldehyde with limited steric bulk.
• ``No Tollens'' ⇒ not an aldehyde ⇒ methyl ketone.
• ``Iodoform positive'' ⇒ CH3CO-R confirmed.
• ``Vigorous oxidation gives CH3COOH + CH3CH2COOH'' ⇒ structure is CH3-CO-CH2CH2CH3 (the methyl-side gives CH3COOH, the C₃ side gives CH3CH2COOH).

Concept linkage with iodoform and oxidative cleavage. The methyl-ketone signature recurs through this chapter: iodoform ( 8.13), NaHSO3 adduct ( 8.19), α-bromination via HVZ-like or direct Br2/H+ (more in Q 8.20), and oxidative cleavage (this Q). Recognising it on first sight saves you 5 minutes per problem.

1. %O = 18.60. Ratio C:H:O = 5.81:11.63:1.16 = 5:10:1. Empirical C5H10O, mass 86 = given M_w. Molecular formula C5H10O.
2. DoU = (2 × 5 + 2 - 10)/2 = 1; one C=O, consistent with all tests.
3. Tests ⇒ methyl ketone (iodoform +; NaHSO3 +; Tollens' -).
4. Cleavage to ethanoic + propanoic acid pins the methyl on a C2 neighbour: CH3-CO-CH2-CH2-CH3 (pentan-2-one). Other C5H10O isomers (e.g. 3-pentanone, C2H5COC2H5, gives propanoic acid only) are ruled out.

Exam relevance. 5-mark CBSE structure-elucidation problems follow this exact template. Marking: 1 for empirical, 1 for molecular, 1 for DoU and tests, 1 for the structure, 1 for ruling out isomers. Always write out the structural isomer rule-out.

Why this matters. Mixed quantitative-and-qualitative problems are a CBSE staple. The algorithm is always: empirical from %, molecular from M_w, then layer the tests to fix the isomer.

Pentan-2-one, CH3COC3H7, M_w = 86.

Picture-first. Count resonance structures, but weigh them. Carboxylate's two structures place the charge symmetrically on the two oxygens; phenoxide's four structures spread the charge over O, ortho-C, para-C, with the C-charged structures dominating numerically but contributing weakly.

Alternative approach (charge-on-oxygen weighting). Two factors determine resonance quality:

• Electronegativity of the atom carrying the negative charge: O (χ = 3.5) is much better than C (χ = 2.5). A structure with negative charge on O is about 10⁴ times more stable than one on C.
• Equivalence: two equivalent structures (like the two C-O bonds of carboxylate) give complete delocalisation; non-equivalent structures contribute unequally.

Concept linkage with the pK_a map. Acidity scales:

• Strong acids (pK_a < 0): HCl, HNO₃, RSO₃H.
• Carboxylic acids (pK_a 3-5): RCOOH, ArCOOH.
• Phenols (pK_a 9-11): ArOH.
• Alcohols (pK_a 15-19): ROH.
• α-H of carbonyls (pK_a 18-25).
• Hydrocarbons (pK_a > 40).

1. Carboxylate (RCOO^-): two equivalent structures, both O^-. High-quality stabilisation. The C-O bond order is 1.5, charges -1/2 on each oxygen.
2. Phenoxide (PhO^-): four structures, only one with O^-; three put the negative on a ring carbon, which is less stable. Also, the C-charged structures break the aromatic sextet partially.
3. Effective stabilisation: carboxylate > phenoxide. Therefore RCOOH (pK_a ≈ 4.7) is more acidic than PhOH (pK_a ≈ 10).
4. Inductive support: the carboxylate's C=O also inductively withdraws electron density (the σ^* of C-O is -I), reinforcing resonance stabilisation. Phenoxide sits on an electron-rich benzene ring with no such reinforcement.

Exam relevance. The carboxylic-acid-vs-phenol question is asked in some form almost every CBSE year (1-mark to 3-mark variants). Key phrases for full marks: ``equivalent resonance structures'', ``charge on electronegative oxygen'', ``inductive reinforcement''.

Why this matters. Acid-strength explanations in organic chemistry almost always boil down to ``stability of the conjugate base''. Do not count resonance forms blindly; ask whether the charge sits on the better atom.

Carboxylic acid is stronger because its conjugate base puts the charge on equivalent, electronegative oxygens; phenoxide's extra resonance structures place the charge on less stable ring carbons.