NCERT Solutions for Class 11 Chemistry Chapter 8: Organic Chemistry: Some Basic Principles and Techniques

Concept. The hybridisation of a carbon atom is decided by the number of σ bonds plus lone pairs around it (its steric number): four σ bonds (all single bonds) → sp³; three σ bonds (one double bond) → sp²; two σ bonds (a triple bond or two double bonds) → sp. A quick rule: the number of π bonds on carbon tells the story — no π bond = sp³, one π bond = sp², two π bonds = sp.

1. CH₂=C=O (ketene): CH₂ carbon has one C=C double bond → sp²; the central carbon is joined by a double bond to CH₂ and a double bond to O, i.e. two π bonds → sp.
2. CH₃CH=CH₂ (propene): CH₃ (four single bonds) → sp³; CH= → sp²; =CH₂ → sp².
3. (CH₃)₂CO (propanone): the two CH₃ carbons → sp³; the carbonyl carbon (C=O, one π bond) → sp².
4. CH₂=CHCN (acrylonitrile): CH₂ → sp²; CH → sp²; the nitrile carbon (C≡N, two π bonds) → sp.
5. C₆H₆ (benzene): each of the six carbons is bonded to two carbons and one hydrogen by σ bonds and takes part in the delocalised π system → all six are sp².

Extra depth. The geometry follows directly: sp³ → tetrahedral (109.5°), sp² → trigonal planar (120°), sp → linear (180°). So the C=C=O unit of ketene is linear at the central carbon, the CH₂ end is planar, and acrylonitrile has a linear C–C≡N chain attached to a planar vinyl group. The ‘count π bonds’ shortcut works equally for allene CH₂=C=CH₂ (middle carbon sp, the two end carbons sp²) — a favourite follow-up question. Because sp carbon has 50 % s-character, it is more electronegative than sp² (33 %) and sp³ (25 %), which is why the C–H of ethyne is acidic.

Conclusion: Ketene: sp², sp; propene: sp³, sp², sp²; propanone: sp³, sp³, sp²; acrylonitrile: sp², sp², sp; benzene: all sp².

Concept. Every covalent bond contains exactly one σ bond; a double bond has one σ plus one π bond, and a triple bond one σ plus two π bonds. So count all bonds from a fully written structure. (For C₆H₁₂ the textbook means cyclohexane, a saturated ring.)

1. C₆H₆ (benzene): 6 C–C σ + 6 C–H σ = 12 σ; three C=C π bonds = 3 π.
2. C₆H₁₂ (cyclohexane): 6 C–C σ + 12 C–H σ = 18 σ, no π bond.
3. CH₂Cl₂: 2 C–H + 2 C–Cl = 4 σ, no π.
4. CH₂=C=CH₂ (allene): 4 C–H σ + 2 C–C σ (one in each C=C) = 6 σ; two C=C π = 2 π.
5. CH₃NO₂: 3 C–H + 1 C–N + 1 N=O (σ) + 1 N–O (σ, coordinate) = 6 σ; the N=O double bond has 1 π.
6. HCONHCH₃ (N-methylformamide): C–H (formyl) 1 + C=O (σ) 1 + C–N 1 + N–H 1 + N–CH₃ 1 + 3 C–H (methyl) = 8 σ; one C=O π = 1 π.

Extra depth. In benzene the three π bonds are not localised: the six p electrons are delocalised over the ring (resonance), yet for counting purposes the Kekulé picture gives 3 π bonds. In allene the two π bonds lie in perpendicular planes, which makes the molecule twisted (the planes of the two end CH₂ groups are at 90°). In nitromethane the N–O bonds are really equivalent (resonance); counting one N=O and one N–O is only a representation of the two Lewis forms, giving 6 σ and 1 π bond for the formal structure. A total check: all σ + π lines drawn in the structure equals the number of bond-lines (e.g. 9 in N-methylformamide).

Conclusion: σ/π: C₆H₆ 12/3; C₆H₁₂ 18/0; CH₂Cl₂ 4/0; CH₂=C=CH₂ 6/2; CH₃NO₂ 6/1; HCONHCH₃ 8/1.

Concept. In a bond-line formula the carbon chain is drawn as a zig-zag; each line end and each vertex is a carbon atom (CH₃ at the ends unless a functional group is written); hydrogens on carbon are not shown; only heteroatoms and the functional groups (OH, =O) are written. First write the condensed formula from the name, then draw the zig-zag.

1. Isopropyl alcohol = propan-2-ol: three carbons, OH on C-2: CH₃CH(OH)CH₃. Zig-zag of three carbons with OH on the middle vertex.
2. 2,3-Dimethylbutanal: butanal (C₄ chain with CHO as C-1) with methyl groups at C-2 and C-3: OHC–CH(CH₃)–CH(CH₃)–CH₃. Zig-zag of four carbons, =O at the first carbon, a methyl line at C-2 and another at C-3.
3. Heptan-4-one: seven-carbon chain with C=O on C-4 (the middle carbon): CH₃CH₂CH₂COCH₂CH₂CH₃. Zig-zag of seven carbons with =O on the central vertex.

Extra depth. A useful check on a bond-line drawing is to count carbons (vertices + ends) and hydrogens (each carbon is completed to four bonds): for 2,3-dimethylbutanal the carbons are 6 (C₆H₁₂O), for heptan-4-one 7 (C₇H₁₄O). The aldehyde carbon is always drawn at the chain end (CHO written as =O with its H implied), while the ketone carbonyl is drawn inside the chain. Isopropyl alcohol is the common name of propan-2-ol (a secondary alcohol).

Conclusion: Propan-2-ol: a three-carbon zig-zag with OH on the middle carbon; 2,3-dimethylbutanal: four-carbon zig-zag, =O at C-1, CH₃ at C-2 and C-3; heptan-4-one: seven-carbon zig-zag with =O on C-4.

Concept. Procedure: (1) pick the principal functional group (suffix); (2) find the longest chain containing it; (3) number to give the functional group the lowest number, then the substituents the lowest set of locants; (4) write substituents alphabetically with locants; (5) combine. For (a)–(e) first translate the bond-line formula into a condensed formula (each vertex/end = C).

1. (a) A three-carbon chain (CH₃–CH₂–CH₂–) attached to a benzene ring: C₆H₅CH₂CH₂CH₃ → propylbenzene (also called 1-phenylpropane); the ring is the parent.
2. (b) CH₃CH₂CH(CH₃)CH₂CN: the nitrile carbon is C-1 (suffix ‘nitrile’, included in the chain); chain: C-1 (CN), C-2 (CH₂), C-3 (CH, with methyl), C-4, C-5 → pentane chain, methyl at C-3 → 3-methylpentanenitrile.
3. (c) Seven-carbon chain with methyl groups on the second carbon and on the fifth carbon (numbering from the end nearer the first branch gives 2,5 rather than 3,6): 2,5-dimethylheptane.
4. (d) Seven-carbon chain with both Br and Cl on carbon 3 (numbering from the end nearer to the substituents, 3 < 5): substituents in alphabetical order, bromo before chloro → 3-bromo-3-chloroheptane.
5. (e) Cl–CH₂–CH₂–CH₂–CHO: the aldehyde (suffix -al) carbon is C-1 → four-carbon chain butanal, Cl on C-4 → 4-chlorobutanal.
6. (f) Cl₂CHCH₂OH: ethanol parent (OH principal group, on C-1 by default); two chloro groups on C-2 → 2,2-dichloroethanol.

Extra depth. Common pitfalls: in (b) the CN carbon belongs to the chain (so the chain is five, not four, carbons); in (e) the CHO carbon is C-1, so Cl gets 4; in (d) a tie between locants is broken alphabetically but here both substituents sit on the same carbon so only the numbering direction (3 vs 5) matters. In (a) alkyl-substituted benzene rings are named as substituted benzenes, not as phenyl-alkanes, when the alkyl chain is short.

Conclusion: (a) Propylbenzene; (b) 3-methylpentanenitrile; (c) 2,5-dimethylheptane; (d) 3-bromo-3-chloroheptane; (e) 4-chlorobutanal; (f) 2,2-dichloroethanol.

Concept. Each choice tests a specific nomenclature rule: (a) the number of locants must equal the number of identical substituents; (b) lowest set of locants; (c) alphabetical order breaks a tie; (d) the principal functional group (OH, suffix -ol) gets the lowest number, and the name ends with the suffix of the principal group.

1. (a) 2,2-Dimethylpentane is correct. Two methyl groups need two locants (‘2,2-’); ‘2-dimethylpentane’ has one locant for two substituents, so it is wrong.
2. (b) 2,4,7-Trimethyloctane is correct. Numbering from the end that gives the lowest set of locants: {2,4,7} vs {2,5,7}; at the first point of difference 4 < 5, so 2,4,7 is correct.
3. (c) 2-Chloro-4-methylpentane is correct. The locant sets are the same either way {2,4}; the lower number goes to the substituent that comes first alphabetically (chloro before methyl), so Cl gets 2.
4. (d) But-3-yn-1-ol is correct. OH is the principal group (suffix -ol) and must get the lowest locant (1); the triple bond then gets 3. ‘But-4-ol-1-yne’ violates both the lowest-locant rule for OH and the format (the suffix of the principal group comes last).

Extra depth. General rules to remember: locants are separated by commas, numbers from letters by hyphens; the multiplying prefixes (di, tri) are ignored for alphabetical order; a principal functional group (OH > C=C > C≡C) gets the lowest locant and the name ends with its suffix. Test the answer (d): HOCH₂CH₂C≡CH — numbering from the OH carbon gives OH at 1 and the triple bond at 3, but numbering from the alkyne end would give the triple bond 1 and OH 4: priority goes to OH.

Conclusion: (a) 2,2-dimethylpentane; (b) 2,4,7-trimethyloctane; (c) 2-chloro-4-methylpentane; (d) but-3-yn-1-ol.

Concept. A homologous series is a family of compounds with the same functional group, a general formula, and successive members differing by a –CH₂– unit (14 mass units). To write the first five members, start with the given compound and add one CH₂ each time.

1. (a) Carboxylic acids (C_nH_2n+1COOH): HCOOH (methanoic acid), CH₃COOH (ethanoic acid), CH₃CH₂COOH (propanoic acid), CH₃CH₂CH₂COOH (butanoic acid), CH₃CH₂CH₂CH₂COOH (pentanoic acid).
2. (b) Ketones (R–CO–R′), starting with propanone: CH₃COCH₃ (propanone), CH₃COCH₂CH₃ (butanone), CH₃COCH₂CH₂CH₃ (pentan-2-one), CH₃COCH₂CH₂CH₂CH₃ (hexan-2-one), CH₃COCH₂CH₂CH₂CH₂CH₃ (heptan-2-one).
3. (c) Alkenes (C_nH_2n), starting with ethene: CH₂=CH₂ (ethene), CH₃CH=CH₂ (propene), CH₃CH₂CH=CH₂ (but-1-ene), CH₃CH₂CH₂CH=CH₂ (pent-1-ene), CH₃CH₂CH₂CH₂CH=CH₂ (hex-1-ene).

Extra depth. Each pair of consecutive members differs by CH₂: their molecular masses differ by 14 u and their physical properties (boiling point) change gradually, whereas chemical properties are similar because the functional group is the same. Strictly, in series (b) the first ketone is propanone (CH₃COCH₃); in series (c) the first alkene is ethene, since methene (CH₂) does not exist as a stable alkene.

Conclusion: (a) HCOOH, CH₃COOH, C₂H₅COOH, C₃H₇COOH, C₄H₉COOH; (b) CH₃COCH₃ … CH₃CO(CH₂)₄CH₃; (c) ethene, propene, but-1-ene, pent-1-ene, hex-1-ene.

Concept. Convert each name into a skeleton: the parent chain from the root (pent-, propane-, hex-), the suffix gives the principal functional group, and the prefixes give the substituents at the stated positions. Then write the condensed formula and the zig-zag bond-line formula.

1. (a) 2,2,4-Trimethylpentane: pentane chain (C-1 to C-5) with two methyls on C-2 and one on C-4: (CH₃)₃C–CH₂–CH(CH₃)₂ (C₈H₁₈). Functional group: none — it is a saturated hydrocarbon (alkane).
2. (b) 2-Hydroxypropane-1,2,3-tricarboxylic acid (citric acid): propane chain with a COOH on each of C-1, C-2, C-3 and an OH on C-2: HOOC–CH₂–C(OH)(COOH)–CH₂–COOH (C₆H₈O₇). Functional groups: three carboxylic acid groups (–COOH) and one hydroxyl (alcohol) group (–OH).
3. (c) Hexanedial: six-carbon chain with an aldehyde group at each end: OHC–CH₂–CH₂–CH₂–CH₂–CHO (C₆H₁₀O₂). Functional groups: two aldehyde groups (–CHO).

Extra depth. 2,2,4-Trimethylpentane is commercially known as iso-octane, the reference fuel for the octane-number scale (octane number 100). Citric acid is the acid of citrus fruits and an intermediate of the Krebs cycle; its name shows the naming rule in action: the principal group is COOH (three of them, so the chain carbons that carry COOH are counted ‘tricarboxylic acid’ of propane) and OH is a prefix (hydroxy). For hexanedial note that both CHO carbons are part of the six-carbon chain.

Conclusion: (a) (CH₃)₃CCH₂CH(CH₃)₂, no functional group; (b) HOOCCH₂C(OH)(COOH)CH₂COOH, –OH and 3 –COOH; (c) OHC(CH₂)₄CHO, 2 –CHO.

Concept. Look at every atom or group attached to the ring or chain and match it with Table 8.4. Remember that the benzene ring itself is the ‘aromatic’ part and that an ester is a single functional group (–COO–) even though it contains C=O and C–O.

1. (a) Aldehyde (–CHO), ether (the methoxy group, –OCH₃ written as OMe) and a phenolic hydroxyl group (–OH attached directly to the aromatic ring), on a benzene ring. (This is vanillin: 4-hydroxy-3-methoxybenzaldehyde.)
2. (b) A primary aromatic amino group (–NH₂) on the ring, an ester group (–COO–), and a tertiary amino group (–N(C₂H₅)₂) in the side chain. (The compound is 2-(diethylamino)ethyl 4-aminobenzoate, the local anaesthetic procaine.)
3. (c) A nitro group (–NO₂), a carbon–carbon double bond (alkene, C=C) and the phenyl (benzene) ring.

Extra depth. A frequent mistake is to call the OH of vanillin ‘alcohol’; because it is attached directly to the benzene ring it is a phenol (more acidic than an alcohol). When naming such a molecule, the highest-priority group decides the suffix: vanillin is an aldehyde (benzaldehyde), procaine an ester (benzoate), the nitrostyrene a nitro-substituted alkene. These three molecules are drawn from real compounds: vanillin (flavouring), procaine (local anaesthetic) and β-nitrostyrene (a reagent in synthesis).

Conclusion: (a) –CHO, ether (OMe), phenolic –OH; (b) –NH₂, ester, tertiary amine; (c) –NO₂, C=C, benzene ring.

Concept. The stability of an anion depends on how well its negative charge is dispersed. An electron-withdrawing group (−I effect) pulls electron density away from the charged atom and disperses (stabilises) the negative charge; an electron-releasing group (+I effect) pushes electron density towards the charged atom and intensifies (destabilises) it.

1. In O₂NCH₂CH₂O⁻: the nitro group is strongly electron-withdrawing (−I); it pulls electrons through the CH₂–CH₂ bonds away from O⁻, so the negative charge on oxygen is delocalised/reduced.
2. In CH₃CH₂O⁻: the ethyl group is electron-releasing (+I); it pushes electrons towards the already negative oxygen, increasing the charge density on O and making the ion less stable.
3. Conclusion: O₂NCH₂CH₂O⁻ is more stable than CH₃CH₂O⁻.

Extra depth. The same reasoning explains acidity: the more stable the conjugate base, the stronger the parent acid, so 2-nitroethanol (O₂NCH₂CH₂OH) is a stronger acid than ethanol. The inductive effect weakens with distance (becoming negligible after about three bonds); here the nitro group is only two carbons away from O, so the effect is still substantial. The same logic orders the acidity of chloroacetic acids in Exercise 8.17.

Conclusion: O₂NCH₂CH₂O⁻ is more stable because the −I effect of NO₂ disperses the negative charge on oxygen, whereas the +I effect of the ethyl group intensifies it.

Concept. Alkyl groups attached to a π system (C=C, C≡C, benzene ring) behave as electron donors mainly through hyperconjugation (a permanent effect) and, to a smaller extent, through the +I (inductive) effect of the sp³ carbon, which is less electronegative than the sp²/sp carbon it is attached to.

1. Hyperconjugation: the C–H σ bonds of the alkyl group that are in the same plane as the p orbitals of the π system can overlap sideways with the π orbital. The σ electrons of such C–H bonds therefore become partially delocalised into the π system (‘no-bond resonance’: H⁺ … CH₂=CH–CH₂⁻ ↔ H–CH₂–CH=CH₂).
2. Result: the π system gains electron density from the C–H σ bonds — the alkyl group has donated electrons. The greater the number of C–H bonds (the CH₃ group has three), the greater the effect.
3. Inductive contribution: the sp²/sp carbon is more electronegative than the sp³ carbon of the alkyl group (more s-character), so the sp³ carbon also pushes electrons towards the π system (+I).
4. Evidence: alkyl groups stabilise carbocations and alkenes: tertiary > secondary > primary carbocations; more substituted alkenes are more stable; alkylbenzenes are more reactive than benzene towards electrophiles (alkyl groups are ortho–para directing activators).

Extra depth. Hyperconjugation is often called ‘Baker–Nathan effect’ and the number of hyperconjugative structures equals the number of α-hydrogens (C–H on the carbon directly attached to the π system): propene has three, which explains why propene is more stable than ethene (heat of hydrogenation lower), and why 2-methylbut-2-ene (nine α-hydrogens) is the most stable of the isomeric pentenes.

Conclusion: Alkyl groups donate electrons to a π system by hyperconjugation (delocalisation of C–H σ electrons into the π orbital) together with a +I effect.

Concept. Resonance structures differ only in the position of electrons (π electrons and lone pairs); the atoms stay in the same positions. Each curved arrow starts at an electron pair (a lone pair or a π bond) and ends where that pair goes (to form a new π bond or to become a lone pair). Draw the arrows one at a time and check octets and charges after each step. In the descriptions below, ring positions are numbered from the carbon carrying the substituent (C-1): C-2 and C-6 are ortho, C-4 is para.

1. (a) Phenol. Arrow 1: a lone pair on O moves into the C-1–O bond (it becomes C=O⁺H). Arrow 2: the C-1=C-2 π pair moves onto C-2 (a negative charge). Result: structure with O⁺ and C-2⁻. Repeating the shifts around the ring puts the negative charge on C-4 and then on C-6. So phenol is a hybrid of two Kekulé structures and three charge-separated structures (O⁺ with C⁻ at ortho, para, ortho). The high electron density at ortho/para explains their reactivity towards electrophiles.
2. (b) Nitrobenzene. The nitro group itself is a hybrid: Ph–N⁺(=O)–O⁻ ↔ Ph–N⁺(–O⁻)=O (arrow from the O⁻ lone pair to form a new N=O, and the old N=O π pair onto the other O). In addition the ring is conjugated with the group: arrow 1 from a ring π bond to the C-1–N bond (so C-1=N⁺), arrow 2 from the N=O π pair onto oxygen (O⁻). Result: structure with the exocyclic C-1=N⁺(O⁻)₂ and a positive charge on C-2; further shifts place it on C-4 and C-6. Ortho and para carbons become electron-deficient, so the meta position is the least deactivated.
3. (c) CH₃CH=CH–CH=O (crotonaldehyde). Arrow 1: the C=O π pair moves onto O (O⁻). Arrow 2: the C=C π pair moves to form a new C=C between the two middle carbons, leaving C⁺ on the carbon bearing CH₃: CH₃CH=CH–CH=O ↔ CH₃C⁺H–CH=CH–O⁻. The neutral structure is the major contributor (no charge separation).
4. (d) Benzaldehyde. Arrow 1: the C=O π pair moves onto O (O⁻). Arrow 2: a ring π pair moves to form a C-1=CH exocyclic double bond, leaving C-2⁺ in the ring: ring(C-2⁺)=CH–O⁻; similarly the positive charge appears at C-4 and C-6. The CHO group withdraws electrons by resonance (−R), so ortho/para carbons carry partial positive charge.
5. (e) Benzyl cation C₆H₅–C⁺H₂. Arrow from a ring π bond (C-1=C-2) to the C-1–CH₂ bond: the exocyclic C=CH₂ forms and the positive charge moves to C-2; repeated shifts put it on C-4 and C-6. The cation is a hybrid of the benzylic form (two Kekulé versions) and three ring-cation forms, so the positive charge is spread over four carbons — one reason benzyl cations are unusually stable.
6. (f) CH₃CH=CH–C⁺H₂ (an allylic cation). Arrow from the C=C π bond to the adjacent C–C bond: the double bond shifts and the positive charge moves two carbons away: CH₃CH=CH–C⁺H₂ ↔ CH₃C⁺H–CH=CH₂. The second contributor is a secondary cation, so it is slightly more stable than the primary one.

Extra depth. The patterns are general. A substituent with a lone pair on the atom attached to the ring (OH, NH₂, halogens) donates electrons (+R) and makes the ortho and para positions electron-rich; a substituent with a multiple bond to an electronegative atom (NO₂, CHO, COOH, CN) withdraws electrons (−R) and makes the ortho and para positions electron-poor. That is why phenol and aniline undergo electrophilic substitution at ortho/para positions whereas nitrobenzene is attacked at the meta position. Charge-separated structures are less stable contributors than the neutral Kekulé structures, but they show how electrons are delocalised.

Conclusion: (a) O⁺ with C⁻ (ortho, para); (b) N⁺–O⁻ hybrid and ring C⁺ (ortho, para); (c) CH₃C⁺H–CH=CH–O⁻; (d) ring C⁺ (ortho, para) with CH–O⁻; (e) benzyl cation with C⁺ on ortho, para carbons; (f) CH₃C⁺H–CH=CH₂.

Concept. In a polar (ionic) organic reaction an attacking reagent seeks out the reactive site of the substrate. If the reagent is electron-rich it is a nucleophile; if it is electron-deficient it is an electrophile. The names come from Greek-based meaning: nucleophile = ‘nucleus-seeking’ (it seeks a positive centre), electrophile = ‘electron-seeking’.

1. Nucleophile (Nu:): a reagent that brings an electron pair to the reactive site (it donates a pair; it is a Lewis base). Examples: negatively charged ions with lone pairs — hydroxide HO⁻, cyanide ⁻CN, halide ions, carbanions R₃C⁻, HS⁻, RO⁻; and neutral molecules with lone pairs — H₂O:, NH₃, R₃N:, ROH. The reaction it causes is nucleophilic substitution/addition.
2. Electrophile (E⁺): a reagent that takes away an electron pair from the reactive site (it accepts a pair; it is a Lewis acid). Examples: carbocations (CH₃⁺), NO₂⁺, Cl⁺, H⁺, BF₃, AlCl₃ (six-electron species), and neutral molecules with polar functional groups, e.g. the carbonyl carbon of >C=O or the carbon of R–X (alkyl halide). Its reaction is electrophilic substitution/addition.
3. How they react: a nucleophile attacks an electrophilic centre of the substrate (an electron-deficient atom, e.g. the carbon of C–Br or C=O); an electrophile attacks a nucleophilic centre (an electron-rich atom or bond, e.g. the π electrons of a double bond). The movement of the electron pair from nucleophile to electrophile is shown by a curved arrow.

Extra depth. Typical illustration: in CH₃CH₂Br + HO⁻ → CH₃CH₂OH + Br⁻, HO⁻ (nucleophile) attacks the carbon bearing δ+ (the electrophilic centre of the substrate); in CH₂=CH₂ + Br₂, the π electrons of ethene (a nucleophilic centre) attack the electrophile Br^δ+. A species can be a nucleophile in one reaction and an electrophile in another only by playing different parts — for example H₂O is a nucleophile (lone pair on O) but H⁺ is an electrophile. Nucleophilicity (rate) and basicity (equilibrium) are related but not identical.

Conclusion: Nucleophiles are electron-pair donors (HO⁻, CN⁻, NH₃); electrophiles are electron-pair acceptors (H⁺, CH₃⁺, BF₃, >C=O).

Concept. Ask: does the bold species bring an electron pair (nucleophile) or accept one (electrophile)? Look for lone pairs/negative charge (nucleophile) and for positive charge or an incomplete octet (electrophile).

1. (a) HO⁻ is a nucleophile: it has lone pairs and a negative charge; here it uses a lone pair to remove a proton (acts as a base) from the acid CH₃COOH giving CH₃COO⁻ + H₂O.
2. (b) ⁻CN is a nucleophile: the carbon of cyanide carries a lone pair and a negative charge; it attacks the electrophilic carbonyl carbon of propanone (C=O, δ+) forming a C–C bond (cyanohydrin formation: (CH₃)₂C(OH)CN).
3. (c) CH₃C⁺O (the acylium ion) is an electrophile: the carbon has a positive charge and an incomplete octet; it accepts the π electrons of benzene, giving acetophenone C₆H₅COCH₃ (a Friedel–Crafts acylation).

Extra depth. In (a) the same species HO⁻ acts as a base (it takes H⁺) because the acid proton is attacked; a base is simply a nucleophile that attacks hydrogen. In (b) cyanide is a classic carbon nucleophile that lengthens the carbon chain by one carbon; in (c) benzene supplies the electrons: it is the nucleophile and the acylium ion is the electrophile — an example of electrophilic aromatic substitution that is discussed again in Unit 13.

Conclusion: (a) nucleophile; (b) nucleophile; (c) electrophile.

Concept. The four reaction types are substitution (one atom/group replaced by another), addition (two species add to a multiple bond), elimination (two atoms/groups removed from adjacent carbons with formation of a multiple bond) and rearrangement (a group migrates within the molecule, so the carbon skeleton or position of a group changes). Compare reactants and products to decide.

1. (a) Br is replaced by SH: the nucleophile HS⁻ displaces the leaving group Br⁻ → nucleophilic substitution.
2. (b) H and Cl add across the C=C of 2-methylpropene to give 2-chloro-2-methylpropane (H goes to the carbon with more H’s — Markovnikov’s rule); the attacking species H⁺ is an electrophile → electrophilic addition.
3. (c) HBr is removed (H from one carbon, Br from the adjacent one) and a C=C forms → elimination (dehydrohalogenation, E2: HO⁻ acts as a base).
4. (d) OH is replaced by Br, but the product has a rearranged carbon skeleton: neopentyl alcohol (CH₃)₃CCH₂OH gives 2-bromo-2-methylbutane (CH₃)₂CBrCH₂CH₃, because the primary carbocation formed after loss of water rearranges (a CH₃ shifts) to a more stable tertiary cation which then adds Br⁻ → substitution accompanied by rearrangement (nucleophilic substitution through a rearranged carbocation).

Extra depth. Note how the same substrate type can follow different paths depending on the reagent: CH₃CH₂Br with HS⁻ (a good nucleophile, weakly basic) undergoes substitution (a), whereas with HO⁻ in alcoholic solution (a strong base) it undergoes elimination (c); in aqueous solution HO⁻ gives substitution (the same reaction as Exercise 8.40). In (d) the 1,2-methyl shift converts an unstable primary carbocation into a stable tertiary carbocation — a classic rearrangement.

Conclusion: (a) nucleophilic substitution; (b) electrophilic addition; (c) elimination; (d) substitution with rearrangement.

Concept. Isomers are different compounds with the same molecular formula (differing in the arrangement of atoms), whereas resonance contributors are different electron-distribution drawings of the same structure (same atom positions). Structural isomers differ in connectivity; geometrical isomers have the same connectivity but differ in arrangement about a double bond.

1. (a) The two drawings are CH₃CH₂COCH₂CH₃ (pentan-3-one) and CH₃COCH₂CH₂CH₃ (pentan-2-one): both C₅H₁₀O, same functional group but the C=O at different positions → structural (position) isomers.
2. (b) Both structures are CHD=CHD: connectivity is identical, but in one the two D atoms are on opposite sides of the C=C (trans) and in the other on the same side (cis) → geometrical (cis–trans) isomers.
3. (c) Both have the same atoms in the same positions (H, C, two OH-type oxygens); they differ only in where the positive charge and the π bond are placed (C=O⁺H versus C⁺–OH) → resonance contributors (of protonated formic acid).

Extra depth. A quick test: if you must move an atom to convert one drawing into the other, they are isomers; if you only move electrons (π bonds or lone pairs), they are resonance contributors. Geometrical isomerism requires restricted rotation (the C=C π bond) and two different groups on each double-bonded carbon (here H and D, which are isotopes but are different groups). Resonance forms are not separate molecules and cannot be isolated; isomers can be isolated and have different physical properties.

Conclusion: (a) structural (position) isomers; (b) geometrical isomers; (c) resonance contributors.

Concept. Bond cleavage is homolysis if each atom keeps one electron of the shared pair (half-headed fish-hook arrows, free radicals formed) and heterolysis if the shared pair goes to one atom (full-headed arrows, ions formed). The intermediate is a free radical (unpaired electron), a carbocation (C⁺, sextet) or a carbanion (C⁻, lone pair).

1. (a) CH₃O–OCH₃ → 2CH₃O•: two half-headed arrows, one from the middle of the O–O bond to each O: homolysis; reactive intermediate: the methoxy free radical (an oxygen radical).
2. (b) Acetone + HO⁻: HO⁻ removes a proton from CH₃; the C–H bonding pair stays on carbon (full arrow from the C–H bond to carbon, another from the HO⁻ lone pair to H): heterolysis; intermediate: a carbanion (⁻CH₂COCH₃, stabilised as an enolate by resonance with C=O).
3. (c) (CH₃)₃C–Br: full arrow from the C–Br bond to Br (the pair goes to the more electronegative Br): heterolysis; intermediate: the tertiary butyl carbocation, (CH₃)₃C⁺, together with Br⁻.
4. (d) C₆H₆ + E⁺: a full arrow from a ring π bond to E⁺ forms the C–E σ bond and leaves a positive charge in the ring: heterolytic (polar) process (a pair of electrons moves); intermediate: a carbocation (the cyclohexadienyl cation or arenium ion, resonance-stabilised).

Extra depth. In (b) the carbanion is stabilised because the negative charge can delocalise onto the oxygen of the C=O (enolate), which is why the hydrogens next to a carbonyl group are acidic. In (c) the tertiary carbocation is the most stable simple alkyl cation (nine hyperconjugating C–H bonds), explaining why tert-butyl bromide ionises easily. In (d) the cation retains aromatic stability after loss of H⁺ — the basis of electrophilic aromatic substitution.

Conclusion: (a) homolysis, free radical; (b) heterolysis, carbanion; (c) heterolysis, carbocation; (d) heterolysis (pair movement), carbocation.

Concept. Inductive effect: the permanent polarisation of a σ bond caused by the polarisation of an adjacent σ bond (a shift of σ electrons along a chain due to the electronegativity difference between atoms). It is transmitted through σ bonds, falls off rapidly with distance (negligible after three bonds) and is classified as −I (electron-withdrawing groups: halogens, NO₂, CN, COOH) or +I (electron-donating: alkyl groups). Electromeric effect: the temporary complete transfer of a shared pair of π electrons to one of the atoms of a multiple bond under the influence of an attacking reagent; it exists only while the reagent is present (+E if the electrons go to the atom to which the reagent attaches, −E if they go to the other atom).

1. Acidity depends on the stability of the carboxylate anion RCOO⁻: anything that disperses the negative charge makes the anion more stable and the acid stronger.
2. (a) The inductive (−I) effect of chlorine. Cl is more electronegative than C and pulls electrons through the σ bonds towards itself, dispersing the negative charge on the carboxylate oxygens. The effect increases with the number of Cl atoms: Cl₃C– (three −I groups) > Cl₂CH– > ClCH₂–. Hence Cl₃CCOOH > Cl₂CHCOOH > ClCH₂COOH in acidity.
3. (b) The inductive (+I) effect of alkyl groups. The CH₃ group pushes electrons towards the carboxyl carbon, increasing charge density on the carboxylate oxygens and destabilising the anion (making the acid weaker). The more methyl groups on the α-carbon, the stronger the +I effect: CH₃CH₂– < (CH₃)₂CH– < (CH₃)₃C–, so the acidity falls in the order CH₃CH₂COOH > (CH₃)₂CHCOOH > (CH₃)₃CCOOH.

Extra depth. Typical pK_a values show the strength of the inductive effect: CH₃COOH 4.76, ClCH₂COOH 2.86, Cl₂CHCOOH 1.29, Cl₃CCOOH 0.65 — each Cl lowers the pK_a by about one unit. The inductive effect is transmitted only through σ bonds and weakens with distance (4-chlorobutanoic acid is only slightly stronger than butanoic acid). The electromeric effect, by contrast, is a temporary effect that operates in π bonds in the presence of a reagent and does not explain these orders.

Conclusion: Both orders of acidity are explained by the inductive effect: (a) −I of Cl; (b) +I of alkyl groups.

Concept. All three are physical methods of purification/separation that exploit a difference in a physical property of the components: solubility (crystallisation), boiling point (distillation) and differential adsorption or partition (chromatography).

1. (a) Crystallisation — principle: difference in solubility of the compound and its impurities in a suitable solvent. The impure solid is dissolved in a minimum of hot solvent (in which it is sparingly soluble in the cold but quite soluble hot) to get a nearly saturated solution; on cooling the pure compound crystallises out and is filtered; the impurities remain in the mother liquor. Coloured impurities are removed with activated charcoal. Example: purification of impure sugar (or benzoic acid from hot water; aspirin from ethanol–water).
2. (b) Distillation — principle: liquids with different boiling points vaporise at different temperatures; the vapour of the more volatile liquid is condensed and collected. It is used for (i) separating a volatile liquid from non-volatile impurities and (ii) liquids with a sufficient difference in boiling points; fractional distillation (using a fractionating column) is used when the difference is small. Example: chloroform (b.p. 334 K) and aniline (b.p. 457 K) are separated by simple distillation; crude oil fractions by fractional distillation.
3. (c) Chromatography — principle: different components of a mixture are distributed differently between a stationary phase (solid or liquid) and a mobile phase (solvent or gas) so they move at different rates. In adsorption chromatography (column, TLC) the components are adsorbed to different degrees on silica gel or alumina; in partition chromatography (paper) they partition between water held in the paper and the moving solvent. Example: separation of the coloured pigments of a plant extract (chlorophylls and carotenoids) on an alumina column or on a TLC plate; amino acids on paper detected with ninhydrin.

Extra depth. Choosing between the methods: for a solid contaminated with a different solid use crystallisation; for liquids use distillation (simple, fractional, under reduced pressure or steam depending on the data); for small quantities of closely similar substances or for testing purity use chromatography. A purified compound shows a sharp melting/boiling point. The R_f value (distance moved by the spot/distance moved by the solvent) identifies a compound in TLC or paper chromatography.

Conclusion: Crystallisation: solubility difference; distillation: boiling-point difference; chromatography: differential adsorption/partition between stationary and mobile phases.

Concept. When two solid compounds have different solubilities in a solvent S, they can be separated by fractional crystallisation (an application of crystallisation based on the difference in solubility).

1. Dissolve the mixture in the minimum amount of hot solvent S so as to obtain a nearly saturated solution of both compounds.
2. Cool the solution slowly: the less soluble compound reaches saturation first and crystallises out, while the more soluble compound stays in the solution (the mother liquor).
3. Filter the crystals of the less soluble compound and wash with a little cold solvent; concentrate the filtrate and cool again to obtain the more soluble compound (in successive crops, the ‘fractions’).
4. Repeat (recrystallise) each fraction until the compound is pure, as shown by a sharp melting point.

Extra depth. If the compounds differ greatly in solubility, a single crystallisation is enough; if the solubilities are comparable, repeated crystallisation is needed — each cycle enriches the less soluble component in the crystals. A mixture of solvents can also be used if the compound is very soluble in one solvent and little in another. Charcoal is used to remove coloured impurities. The success of the method is checked by melting point or TLC.

Conclusion: Fractional crystallisation: dissolve in hot S, cool; the less soluble compound crystallises first and is filtered; the filtrate is concentrated to get the second compound.

Concept. All three separate liquids by vaporisation and condensation, but they differ in the pressure/medium at which the liquid boils and in the type of liquids they suit.

1. Distillation (simple): the liquid is boiled at atmospheric pressure at its normal boiling point and the vapour condensed; used for volatile liquids from non-volatile impurities or for liquids whose boiling points differ considerably (chloroform and aniline).
2. Distillation under reduced pressure (vacuum distillation): the pressure over the liquid is lowered with a water/vacuum pump so that the liquid boils (vapour pressure = external pressure) at a temperature below its normal boiling point; used for liquids with very high boiling points or which decompose at or below their boiling point (glycerol from spent lye).
3. Steam distillation: steam is passed through the liquid mixture; the liquid boils when the sum of the vapour pressures of water and of the organic liquid equals atmospheric pressure (p = p₁ + p₂), i.e. below 373 K, which is lower than the boiling point of the organic compound; used for steam-volatile compounds immiscible with water (aniline, essential oils).

Extra depth. In steam distillation the distillate contains water and the organic compound, which are later separated with a separating funnel; the composition of the vapour depends on the ratio of vapour pressures (p₁/p₂ = n₁/n₂), so a compound with a low molar mass and a high vapour pressure distils more effectively. In reduced pressure distillation the apparatus has a receiver connected to the pump (Fig. 8.8), a thermometer in the vapour and often a capillary for smooth boiling.

Conclusion: Simple: atmospheric pressure, normal b.p.; reduced pressure: lowered pressure for high-boiling or unstable liquids; steam: boils below 373 K for steam-volatile, water-immiscible liquids.

Concept. Lassaigne’s test detects nitrogen, sulphur, halogens (and phosphorus) in an organic compound by converting them from the covalent form into the ionic form. The compound is fused with sodium metal in a fusion tube and plunged into distilled water; the filtered solution is the sodium fusion extract, which is tested for CN⁻, S²⁻ and X⁻.

1. Fusion with sodium: Na + C + N → NaCN; 2Na + S → Na₂S; Na + X → NaX (X = Cl, Br, I); (C, N, S and X come from the organic compound).
2. Test for nitrogen: boil the extract with FeSO₄ and acidify with conc. H₂SO₄: 6CN⁻ + Fe²⁺ → [Fe(CN)₆]⁴⁻; on heating with the acid some Fe²⁺ is oxidised to Fe³⁺, which reacts with hexacyanoferrate(II): 3[Fe(CN)₆]⁴⁻ + 4Fe³⁺ → Fe₄[Fe(CN)₆]₃·xH₂O (Prussian blue) — blue colour confirms N.
3. Test for sulphur: (a) acidify the extract with acetic acid and add lead acetate: S²⁻ + Pb²⁺ → PbS (black precipitate); (b) with sodium nitroprusside: S²⁻ + [Fe(CN)₅NO]²⁻ → [Fe(CN)₅NOS]⁴⁻ (violet colour).
4. When N and S both are present: sodium thiocyanate forms (Na + C + N + S → NaSCN), giving a blood red colour with Fe³⁺ (Fe³⁺ + SCN⁻ → [Fe(SCN)]²⁺) and no Prussian blue (no free CN⁻). With excess sodium, thiocyanate breaks down: NaSCN + 2Na → NaCN + Na₂S, and the usual tests are obtained.
5. Test for halogens: acidify with HNO₃ and add AgNO₃: X⁻ + Ag⁺ → AgX; white precipitate soluble in NH₄OH (Cl); pale yellow precipitate sparingly soluble in NH₄OH (Br); yellow precipitate insoluble in NH₄OH (I). If N or S is present, first boil the extract with conc. HNO₃ to destroy CN⁻ and S²⁻.

Extra depth. Why sodium? It is a strong reducing metal that breaks the covalent bonds and converts the non-ionic N, S and X into their water-soluble sodium salts (NaCN, Na₂S, NaX). The test fails for compounds in which N is not converted easily (for example diazo compounds) unless the fusion is done at high temperature; and in the absence of carbon (hydrazine N₂H₄) no cyanide is formed. Phosphorus is detected separately by oxidation to phosphate followed by the ammonium molybdate test.

Conclusion: Fusion with sodium converts N, S and halogens into NaCN, Na₂S and NaX; these are detected as Prussian blue (N), black PbS/violet (S), blood-red thiocyanate (N + S) and AgX precipitates (halogens).

Concept. Both methods convert the nitrogen of an organic compound into a measurable form, but the chemistry and the measured quantity differ: the Dumas method converts nitrogen to free N₂ gas whose volume is measured, while the Kjeldahl method converts nitrogen to ammonia (via ammonium sulphate) whose amount is found by titration.

Extra depth. The Dumas method is the more general but needs a precise gas measurement and temperature/pressure correction (including the aqueous tension); Kjeldahl’s method is a volumetric (titration) method that is convenient for routine analysis of large numbers of samples, e.g. protein content of foods. Both give the percentage of nitrogen, from which the empirical formula can be derived. The copper gauze in the Dumas tube is there to reduce any nitrogen oxides to N₂; KOH absorbs CO₂ so that only N₂ is measured.

Conclusion: Dumas: N converted to N₂ gas, measured by volume; Kjeldahl: N converted to NH₃ via (NH₄)₂SO₄, determined by back-titration of the acid.

Concept. In each estimation the element is quantitatively converted into a weighable inorganic compound by oxidation (heating in a sealed Carius tube with fuming nitric acid, plus AgNO₃ for halogens); the mass of the compound is determined (gravimetric analysis) and the percentage of the element is calculated from its fraction in the compound.

1. Halogens (Carius method). A known mass (m g) of the compound is heated with fuming HNO₃ in the presence of AgNO₃ in a sealed hard-glass Carius tube in a furnace; C and H are oxidised to CO₂ and H₂O and the halogen forms silver halide AgX, which is filtered, washed, dried and weighed (m₁ g). % halogen = (atomic mass of X × m₁ × 100)/(molecular mass of AgX × m).
2. Sulphur. A known mass of compound is heated in a Carius tube with sodium peroxide or fuming nitric acid: S is oxidised to H₂SO₄, which is precipitated as BaSO₄ by adding excess BaCl₂; the precipitate is filtered, washed, dried and weighed. 233 g BaSO₄ contains 32 g S: % S = (32 × m₁ × 100)/(233 × m).
3. Phosphorus. A known mass of compound is heated with fuming HNO₃: P → H₃PO₄, precipitated as ammonium phosphomolybdate, (NH₄)₃PO₄·12MoO₃ (M = 1877 g mol⁻¹) by adding ammonia and ammonium molybdate: % P = (31 × m₁ × 100)/(1877 × m). Alternatively H₃PO₄ is precipitated as MgNH₄PO₄ by magnesia mixture and ignited to Mg₂P₂O₇ (M = 222): % P = (62 × m₁ × 100)/(222 × m).

Extra depth. These are worked examples from the chapter: 0.15 g of a bromo compound giving 0.12 g AgBr has 34.04 % Br; 0.157 g of a sulphur compound giving 0.4813 g BaSO₄ has 42.10 % S (Problems 8.23 and 8.24). The word ‘2P’ (mass 62) appears for magnesium pyrophosphate because each formula unit contains two phosphorus atoms. Oxygen is not estimated in this way; it is found by difference or by conversion to CO₂ via CO and I₂O₅.

Conclusion: Halogens as AgX (Carius), sulphur as BaSO₄, phosphorus as ammonium phosphomolybdate or Mg₂P₂O₇; percentage = (mass fraction of the element in the precipitate × m₁ × 100)/m.

Concept. Paper chromatography is a form of partition chromatography: it is based on the continuous differential partitioning of the components of a mixture between two liquid phases — a stationary phase and a mobile phase.

1. Stationary phase: the water trapped in the fibres of the special chromatography paper (cellulose fibres hold a layer of adsorbed water). Mobile phase: a suitable solvent or mixture of solvents.
2. Procedure: a spot of the mixture is placed near one end of a strip of chromatography paper; the strip is suspended in a closed jar so that its lower end dips in the solvent (the spot is above the solvent level). The solvent rises up the paper by capillary action and flows over the spot.
3. Separation: each component partitions between the water on the paper and the moving solvent according to its relative solubility in the two phases. A component more soluble in the mobile phase moves up faster and further; one more soluble in the stationary phase (water) lags behind. After some time the components appear as separate spots at different heights.
4. The chromatogram: the developed paper strip. Coloured spots are visible directly; colourless ones are detected under UV light or by spraying a reagent (e.g. ninhydrin for amino acids). The R_f value = distance moved by the substance/distance moved by the solvent is characteristic of a compound.

Extra depth. Contrast with adsorption chromatography (column, TLC) where the stationary phase is a solid adsorbent and separation depends on different degrees of adsorption; in paper chromatography the separation depends on partition between two liquid phases. Paper chromatography is simple, needs only very small amounts and is widely used to separate and identify amino acids, sugars and plant pigments. A closed jar saturated with solvent vapour prevents evaporation of the solvent from the paper, so the run is reproducible.

Conclusion: Components partition differently between the water held in the paper (stationary phase) and the rising solvent (mobile phase), so they travel different distances and give separate spots with characteristic R_f values.

Concept. The sodium fusion extract may contain CN⁻ (if N is present) and S²⁻ (if S is present) along with the halide ions. Both cyanide and sulphide ions react with Ag⁺ to give precipitates (AgCN and Ag₂S) that would be mistaken for silver halides.

1. What nitric acid does: on boiling with concentrated HNO₃ the sodium cyanide and sodium sulphide are decomposed (converted to volatile HCN and H₂S, and oxidised): NaCN + HNO₃ → NaNO₃ + HCN↑; Na₂S + 2HNO₃ → 2NaNO₃ + H₂S↑. The gases escape on boiling.
2. Result: only halide ions remain in the solution, so the precipitate formed with AgNO₃ is genuinely AgX (AgCl white, AgBr pale yellow, AgI yellow) and not AgCN or Ag₂S.
3. Another role: the acid also keeps the solution acidic so that hydroxide or carbonate ions in the (basic) extract do not give a precipitate of Ag₂O or Ag₂CO₃.

Extra depth. Always carry this out in a fume cupboard because the HCN and H₂S evolved are toxic. If neither N nor S is present, the boiling step is not needed but the acidification is still done. Nitric acid is chosen (not HCl or H₂SO₄) because HCl would supply chloride ions and H₂SO₄ can give the sparingly soluble Ag₂SO₄; AgNO₃ and nitrates are all soluble.

Conclusion: HNO₃ decomposes NaCN and Na₂S (as HCN and H₂S) that would otherwise precipitate AgCN and Ag₂S and interfere with the halide test.

Concept. The nitrogen, sulphur and halogens in an organic compound are held by covalent bonds, so they do not give ions in water and cannot be detected by the ordinary ionic tests. Fusion with sodium converts them into ionic compounds that are soluble in water and respond to the usual qualitative tests.

1. Reaction: on fusion, sodium (a strong reducing agent) reacts with the elements of the compound: Na + C + N → NaCN; 2Na + S → Na₂S; Na + X → NaX (X = Cl, Br, I).
2. Extraction: boiling the fused mass with distilled water dissolves these sodium salts as the sodium fusion extract, which contains CN⁻, S²⁻ and X⁻ ions.
3. Detection: the ions are then identified: CN⁻ (Prussian blue test), S²⁻ (lead acetate / nitroprusside), X⁻ (AgNO₃).

Extra depth. Sodium is chosen because it is very reactive, melts at low temperature (about 98 °C) and forms sodium salts that are all freely soluble in water. Carbon must be present in the compound to form cyanide for nitrogen; if both N and S are present, thiocyanate is formed. The fusion is carried out in an ignition (fusion) tube by heating the compound with a small piece of sodium until red hot, then plunging the hot tube into distilled water — a dangerous step that should be done behind a screen.

Conclusion: Sodium converts covalently bonded N, S and halogens into water-soluble ionic salts (NaCN, Na₂S, NaX) that can be identified by ionic tests.

Concept. The mixture contains one sublimable substance (camphor sublimes on heating) and one non-sublimable substance (calcium sulphate does not). The technique that separates them is sublimation.

1. Principle: some solids change directly from solid to vapour on heating, without passing through the liquid state, and the vapour solidifies again on cooling.
2. Procedure: place the mixture in a china dish covered with an inverted funnel (the stem plugged with cotton wool) and heat gently. Camphor vapourises and condenses as white crystals on the cool inside surface of the funnel; CaSO₄ remains in the dish.
3. Alternative: because camphor is soluble in organic solvents and CaSO₄ is not, one can dissolve camphor in ether/alcohol, filter and evaporate (solvent extraction), but sublimation is the simplest and the technique indicated by the question.

Extra depth. Other substances purified by sublimation include naphthalene, anthracene, benzoic acid, iodine and ammonium chloride. The method is applicable only when the substance has a sufficiently high vapour pressure below its melting point and the impurity is non-volatile. For camphor sublimation is a standard laboratory demonstration.

Conclusion: Sublimation: camphor sublimes and is collected on the cold funnel; calcium sulphate remains.

Concept. In steam distillation the liquid mixture contains two immiscible liquids (water and the organic liquid). Each exerts its own vapour pressure independently, and the mixture boils when the sum of their vapour pressures becomes equal to the atmospheric pressure.

1. Condition for boiling: p = p₁ + p₂, where p is the atmospheric pressure, p₁ the vapour pressure of the organic liquid and p₂ that of water at the temperature of the mixture.
2. Consequence: since p₂ contributes a considerable part of the total (at 370 K, water’s vapour pressure is close to atmospheric), the organic liquid needs a vapour pressure of only p₁ = p − p₂, which is less than the atmospheric pressure p. A liquid boils on its own only when its vapour pressure equals p; so it has already begun to distil at a temperature where its own vapour pressure is less than p.
3. Hence the mixture (and therefore the organic liquid) distils at a temperature below the normal boiling point of the organic liquid — close to but below 373 K, the boiling point of water.

Extra depth. This is exactly why heat-sensitive substances that boil above 373 K (aniline b.p. 457 K) can be distilled at about 371 K without decomposition. The same idea is used to extract essential oils (eucalyptus, lavender, turpentine oil). The mass ratio of the two components in the distillate is given by w₁/w₂ = (p₁M₁)/(p₂M₂), so substances of high molar mass with a reasonable vapour pressure distil efficiently with a small amount of steam.

Conclusion: The mixture boils when p₁ + p₂ = atmospheric pressure; as p₁ is less than p, the organic liquid distils at a temperature lower than its own boiling point.

Concept. A precipitate of AgCl forms only when free chloride ions are present to combine with Ag⁺ ions: Ag⁺ + Cl⁻ → AgCl↓. The question is therefore whether CCl₄ produces Cl⁻ ions in solution.

1. CCl₄ is a covalent compound: all four C–Cl bonds are covalent (non-polar overall, tetrahedral molecule); it does not ionise in water.
2. No Cl⁻ ions are available in the solution, because the covalently bound chlorine does not dissociate; AgNO₃ has nothing to react with.
3. Answer: No, CCl₄ will not give a white precipitate of AgCl on heating with AgNO₃. (CCl₄ is also insoluble in water and does not hydrolyse under these conditions.)
4. To test for chlorine in CCl₄ the compound must first be fused with sodium (Lassaigne’s test): CCl₄ + 4Na → C + 4NaCl; the extract then contains Cl⁻ ions and gives a white precipitate of AgCl on adding AgNO₃ (after acidifying with HNO₃).

Extra depth. This is the reason that Lassaigne’s fusion is needed for organic halogen compounds: alkyl and aryl halides (CH₃Cl, C₆H₅Cl) do not give an immediate precipitate with aqueous AgNO₃. In contrast, some reactive halides (tertiary alkyl halides, acyl chlorides) hydrolyse readily in water to give halide ions and precipitate AgX — CCl₄ is very resistant to hydrolysis because the C atom has no vacant orbital or lone-pair path for attack by water and the carbon is shielded by four large chlorine atoms.

Conclusion: No: CCl₄ is covalent and does not give Cl⁻ ions, so no AgCl precipitate forms (the chlorine can be detected only after sodium fusion).

Concept. In the estimation of carbon, the compound is burnt in excess oxygen with CuO and the CO₂ formed is absorbed in a weighed U-tube; the increase in mass of that tube gives the mass of CO₂, from which % C = 12 × m₂ × 100/(44 × m). For this to work the absorbent must take up CO₂ completely and chemically.

1. CO₂ is an acidic oxide; potassium hydroxide is a strong base, so they react readily and completely: CO₂ + 2KOH → K₂CO₃ + H₂O (and, with excess CO₂, K₂CO₃ + CO₂ + H₂O → 2KHCO₃).
2. Quantitative absorption: a concentrated KOH solution absorbs all the CO₂ passing through, so none escapes uncounted, and the rise in mass of the tube equals the mass of CO₂ produced.
3. Order of the tubes: the CaCl₂ tube (which absorbs only water, and not CO₂) comes first and the KOH tube second, so the two products (H₂O and CO₂) are determined separately in the same experiment.

Extra depth. The same absorption is used in the Dumas method for nitrogen: the gas mixture is collected over aqueous KOH, which absorbs CO₂ so that the unabsorbed gas is nitrogen alone. In the CH estimation the CaCl₂ tube must come first: an aqueous KOH tube would absorb water vapour as well as CO₂ and would also hand moisture back to the gas stream, so neither mass would be reliable. Soda lime or Ascarite could also be used for CO₂.

Conclusion: KOH, a strong base, reacts quantitatively with the acidic oxide CO₂ (forming K₂CO₃), so the increase in mass of the KOH tube equals the mass of CO₂ produced.

Concept. In the lead acetate test the sodium fusion extract is acidified and lead acetate is added; sulphide ions give a black precipitate of PbS: S²⁻ + Pb²⁺ → PbS↓. The acid used for acidification must not itself produce a precipitate with lead(II) ions.

1. With sulphuric acid: H₂SO₄ supplies sulphate ions, which react with lead acetate to give a white precipitate of lead sulphate: (CH₃COO)₂Pb + H₂SO₄ → PbSO₄↓ + 2CH₃COOH. The white PbSO₄ would appear whether sulphur is present or not, so it would mask or be confused with the black PbS and give a false result.
2. With acetic acid: CH₃COOH only provides acetate ions, and lead acetate itself is soluble; lead acetate does not form any precipitate with it. So a black precipitate (PbS) appears only if S²⁻ is present.
3. Conclusion: acetic acid (a weak acid with a common ion, acetate, of the reagent) is used so that the test is specific for sulphide.

Extra depth. Likewise hydrochloric acid is unsuitable (it would give white PbCl₂ with cold, concentrated solutions); nitric acid is also avoided because it can oxidise S²⁻ to sulphur or sulphate. Acetic acid is mild enough not to affect the sulphide but acidic enough to liberate the S²⁻ from any contaminating species and to avoid the precipitation of lead hydroxide in the basic extract.

Conclusion: H₂SO₄ would precipitate white PbSO₄ with lead acetate and interfere with the black PbS; acetic acid gives no precipitate, so the test is specific for sulphide.

Concept. All the carbon of the compound ends up as CO₂ and all its hydrogen as H₂O, so mass of CO₂ = (mass of C) × 44/12 and mass of H₂O = (mass of H) × 18/2. Oxygen is not needed (it is the remainder: 100 − 69 − 4.8 = 26.2 %).

1. Mass of carbon in 0.20 g: 0.20 × 69/100 = 0.138 g.
2. Mass of CO₂: 12 g C gives 44 g CO₂, so 0.138 g C gives 0.138 × 44/12 = 0.506 g of CO₂.
3. Mass of hydrogen in 0.20 g: 0.20 × 4.8/100 = 0.0096 g.
4. Mass of H₂O: 2 g H gives 18 g H₂O, so 0.0096 g H gives 0.0096 × 18/2 = 0.0864 g of water.

Extra depth. This is the reverse of Problem 8.20 (finding % C and % H from the masses of CO₂ and H₂O). Check: % C = 12 × 0.506 × 100/(44 × 0.20) = 69.0 %; % H = 2 × 0.0864 × 100/(18 × 0.20) = 4.8 %. Remember that 44/12 = 3.67 and 18/2 = 9 are the conversion factors from C to CO₂ and from H to H₂O — worth memorising.

Conclusion: Mass of CO₂ = 0.506 g; mass of H₂O = 0.0864 g.

Concept. Kjeldahl method: N → NH₃ absorbed in a known excess of H₂SO₄; the unreacted acid is titrated with NaOH. The acid used by NH₃ = total acid − acid left over. Since H₂SO₄ is dibasic and NH₃ reacts as 2NH₃ + H₂SO₄ → (NH₄)₂SO₄, 1 mol H₂SO₄ = 2 mol NH₃ = 2 mol N.

1. Total acid taken: 50 mL of 0.5 M H₂SO₄ = 50 × 0.5 = 25 mmol H₂SO₄.
2. Acid left over (from the NaOH titration): 60 mL of 0.5 M NaOH = 30 mmol NaOH; H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O, so 30 mmol NaOH neutralises 15 mmol H₂SO₄ (= 30 mL of 0.5 M H₂SO₄).
3. Acid used by NH₃: 25 − 15 = 10 mmol H₂SO₄ (i.e. 20 mL of 0.5 M).
4. NH₃ and N: 10 mmol H₂SO₄ reacts with 20 mmol NH₃ = 20 mmol N = 20 × 14 mg = 280 mg = 0.28 g.
5. Percentage of N = (0.28/0.50) × 100 = 56.0 %. (By the formula: % N = 1.4 × M × 2(V − V₁/2)/m = 1.4 × 0.5 × 2(50 − 30)/0.50 = 56.0 %.)

Extra depth. The common error is to forget the factor 2 in NaOH → H₂SO₄ (60 mL of 0.5 M NaOH equals only 30 mL of 0.5 M H₂SO₄) or in NH₃ → H₂SO₄. A nitrogen content of 56 % is unusually high (urea has 46.7 %), but this is a textbook data set meant for practice. Kjeldahl’s method assumes the nitrogen is in the amine, amide or similar state (not nitro, azo or ring).

Conclusion: % N = 0.28 g/0.50 g × 100 = 56.0 %.

Concept. In the Carius method the chlorine of the compound is converted into AgCl, which is weighed. The mass of chlorine in the precipitate is (atomic mass of Cl/molar mass of AgCl) × mass of AgCl, and % Cl = (35.5 × m₁ × 100)/(143.5 × m).

1. Molar mass of AgCl = 108 + 35.5 = 143.5 g mol⁻¹.
2. Mass of Cl in 0.5740 g AgCl = (35.5/143.5) × 0.5740 = 0.2474 × 0.5740 = 0.1420 g.
3. Percentage of Cl = (0.1420/0.3780) × 100 = 37.57 %.

Extra depth. Check using the general formula: % Cl = 35.5 × 0.5740 × 100/(143.5 × 0.3780) = 2037.7/54.243 = 37.57 %. The fraction of chlorine in AgCl is 24.7 %, a number worth remembering for gravimetric calculations. Silver chloride is used because it is quantitatively precipitated, stable on drying (in the dark, as it is photosensitive) and has a large molar mass, which reduces the relative error of weighing.

Conclusion: % Cl = 35.5 × 0.5740 × 100/(143.5 × 0.3780) = 37.57 %.

Concept. In the Carius method the sulphur of the compound is oxidised to H₂SO₄ and precipitated as BaSO₄, which is weighed. 1 mol BaSO₄ (233 g) contains 1 mol S (32 g), so % S = (32 × m₁ × 100)/(233 × m).

1. Molar mass of BaSO₄ = 137 + 32 + 4(16) = 233 g mol⁻¹.
2. Mass of sulphur in 0.668 g BaSO₄ = (32/233) × 0.668 = 0.1373 × 0.668 = 0.0917 g.
3. Percentage of sulphur = (0.0917/0.468) × 100 = 19.6 %.

Extra depth. By the direct formula: 32 × 0.668 × 100/(233 × 0.468) = 2137.6/109.04 = 19.60 %. (Answers between 19.59 and 19.61 % arise from rounding of atomic masses.) The fraction of S in BaSO₄ is 13.7 % — the ‘gravimetric factor’ for sulphur. As a check on the sense of the answer: 0.468 g of a compound containing 19.6 % S contains 0.0917 g of sulphur — consistent with the 0.668 g of BaSO₄ obtained.

Conclusion: % S = 32 × 0.668 × 100/(233 × 0.468) = 19.6 %.

Concept. Number the chain from the double-bond end: C₁H₂=C₂H–C₃H₂–C₄H₂–C₅≡C₆H. The hybridisation of each carbon follows from its multiple bonds.

1. C₁ and C₂: carbons of the C=C double bond → sp².
2. C₃ and C₄: CH₂ groups with four single bonds → sp³.
3. C₅ and C₆: carbons of the C≡C triple bond → sp.
4. The C₂–C₃ bond is therefore formed by the overlap of an sp² orbital of C₂ with an sp³ orbital of C₃: option (c) sp²–sp³.

Extra depth. Hence the bond lengths differ along the chain: the C₂–C₃ single bond (sp²–sp³) is shorter than the C₃–C₄ bond (sp³–sp³, 154 pm), and the C₄–C₅ bond (sp³–sp) is shorter still, because the higher the s-character of the hybrid orbitals, the shorter the bond. Wrong options (a) and (b) include an sp carbon, which is C₅ or C₆, not C₃.

Conclusion: (c) sp²–sp³.

Concept. In the test, sodium cyanide of the extract reacts with iron(II) sulphate to give sodium hexacyanoferrate(II), Na₄[Fe(CN)₆]; on heating with concentrated H₂SO₄ some Fe²⁺ is oxidised to Fe³⁺, which combines with the hexacyanoferrate(II) to give the blue compound.

1. Step 1: 6CN⁻ + Fe²⁺ → [Fe(CN)₆]⁴⁻ (sodium hexacyanoferrate(II), Na₄[Fe(CN)₆] — this is colourless/yellow, option (a), not the blue compound).
2. Step 2: oxidation by the hot acid: Fe²⁺ → Fe³⁺.
3. Step 3: 3[Fe(CN)₆]⁴⁻ + 4Fe³⁺ → Fe₄[Fe(CN)₆]₃ · xH₂O — iron(III) hexacyanoferrate(II) (ferriferrocyanide), the Prussian blue: option (b). Charge balance: 4 × (+3) = +12 and 3 × (−4) = −12.

Extra depth. Check each option by charge balance: Fe₄[Fe(CN)₆]₃ has 4 Fe³⁺ (+12) and 3 [Fe(CN)₆]⁴⁻ (−12). The compound is the same as that of the ‘Prussian blue’ pigment used in inks and blueprints (the ‘Turnbull’s blue’ with Fe²⁺ and [Fe(CN)₆]³⁻ is the same substance). The presence of both sulphur and nitrogen gives instead the blood-red [Fe(SCN)]²⁺.

Conclusion: (b) Fe₄[Fe(CN)₆]₃.

Concept. Carbocation stability depends on the number of alkyl groups on the positively charged carbon: alkyl groups stabilise the cation by the +I (inductive) effect and by hyperconjugation (more C–H bonds adjacent to C⁺ → more delocalisation). The order is tertiary > secondary > primary > methyl.

1. (a) (CH₃)₃C–C⁺H₂: the positive carbon is attached to only one carbon (the quaternary C): primary (neopentyl cation); only 0 α-hydrogens on the neighbouring carbon.
2. (b) (CH₃)₃C⁺: the positive carbon is attached to three carbons: tertiary; nine α-C–H bonds for hyperconjugation.
3. (c) CH₃CH₂C⁺H₂: one carbon attached: primary.
4. (d) CH₃C⁺HCH₂CH₃: two carbons attached: secondary.
5. Order of stability: (b) tertiary > (d) secondary > (a), (c) primary. The most stable is (b) (CH₃)₃C⁺ (tert-butyl cation).

Extra depth. The neopentyl cation (a) is unusually unstable even among primary cations, and it rearranges (a methyl shift) to the tertiary 2-methylbut-2-yl cation — as seen in Exercise 8.14(d). The α-hydrogen count: propyl cation 2 (on the adjacent CH₂), sec-butyl cation 5 (3 on CH₃ + 2 on CH₂), tert-butyl 9. A larger number of such bonds means more hyperconjugation and a more stable cation.

Conclusion: (b) (CH₃)₃C⁺ (tertiary) is the most stable.

Concept. Crystallisation, distillation and sublimation are classical methods based on solubility, boiling point and volatility; each works only for particular types of mixtures and generally needs a reasonable quantity of material. Chromatography is the modern, versatile technique.

1. Wide applicability: it separates solids, liquids and gases, coloured and colourless substances, even closely similar compounds (isomers) that cannot be separated by crystallisation or distillation.
2. Small quantities: works with micro- to nanogram amounts; one technique serves for isolation, purification, identification and purity testing (R_f values).
3. Variants: column, thin layer, paper chromatography, gas chromatography (GC) and high-performance liquid chromatography (HPLC), often coupled with spectrometers (GC–MS).
4. Answer: (d) Chromatography.

Extra depth. Chromatography is the basis for modern analysis in drug testing, forensic science, food and environmental analysis. Its two basic categories studied in this chapter are adsorption (column, TLC) and partition (paper) chromatography; the more advanced forms (GC, HPLC) follow the same principle of differential distribution between a stationary and a mobile phase.

Conclusion: (d) Chromatography.

Concept. Decide by comparing the reactant and the product: the iodine atom of iodoethane has been replaced by the OH group, so it is a substitution reaction. The attacking species is HO⁻ (an electron-rich nucleophile), and the substrate carbon (C–I, δ+) is the electrophilic centre. So it is nucleophilic substitution.

1. Substitution? I is replaced by OH; no multiple bond is formed (so not elimination) and no multiple bond is consumed (not addition).
2. Which reagent attacks? HO⁻ has lone pairs and a negative charge — a nucleophile; it attacks the carbon of the polar C–I bond (C^δ+–I^δ−) and displaces I⁻ (the leaving group).
3. Answer: (b) nucleophilic substitution.

Extra depth. The medium decides the outcome: with aqueous KOH (HO⁻ acts as a nucleophile) iodoethane gives ethanol (substitution); with alcoholic KOH (a strong base) it gives ethene by dehydrohalogenation (elimination, as in Exercise 8.14c). The iodine is a good leaving group because C–I is the weakest of the carbon–halogen bonds and I⁻ is a stable ion.

Conclusion: (b) nucleophilic substitution.