Fundamental Principles of Organic Chemistry

Unit 12 taught you what organic compounds are. This unit teaches you the working grammar of the subject: how to give every compound a name that no two chemists can disagree about, how to prove by experiment which elements a compound contains, how the same set of atoms can build several different substances, and why some bonds break one way and some the other. Ten hours of the syllabus are spent here because everything in Units 14 and 15 — and the whole of the Grade 12 organic course — is written in this language.

13.1 IUPAC Nomenclature of Organic Compounds (up to chain having 6 carbon atoms)

Before 1892 every compound had a trivial name — acetic acid, glycerol, urea — which told you nothing about its structure. The IUPAC system (International Union of Pure and Applied Chemistry) builds a name out of pieces, so that the name and the structure can always be converted into each other.

An IUPAC name has up to four parts:

secondary prefix + word root + primary suffix + secondary suffix

Part	What it gives	Examples
Secondary prefix	substituents that are not the principal group	methyl-, ethyl-, chloro-, nitro-, hydroxy-, amino-
Word root	number of carbons in the principal chain	meth (1), eth (2), prop (3), but (4), pent (5), hex (6)
Primary suffix	saturation of the chain	-ane, -ene, -yne
Secondary suffix	the principal functional group	-ol, -al, -one, -oic acid, -amine

The naming rules, in the order you apply them

1. Find the principal chain. It is the longest continuous chain of carbon atoms that contains the principal functional group, and then the greatest number of multiple bonds. If two chains are equally long, take the one with more substituents.
2. Find the principal functional group. If a molecule has more than one, only the senior one gets a suffix; the rest become prefixes. The seniority order is

—COOH > —COOR (ester) > —COCl > —CONH₂ > —CN > —CHO > >C=O > —OH > —NH₂ > C=C > C≡C

3. Number the chain from the end that gives the lowest locant to the principal functional group. If there is no functional group, number to give the lowest locant to the multiple bond; if there is neither, number to give the lowest set of locants to the substituents (first point of difference rule).
4. Name the substituents and put them in alphabetical order, each with its locant. The multiplying prefixes di-, tri-, tetra- are not counted when alphabetising, so 4-ethyl-2-methylhexane has ethyl before methyl, and 2,3-dimethylpentane files under "m".
5. Assemble and elide. Drop the final "e" of -ane/-ene/-yne when the next letter is a vowel: hexan-2-ol (not hexane-2-ol), but hexane-1,2-diol (the "e" survives before a consonant).

Class	Group	Prefix	Suffix	Example (IUPAC name)
Carboxylic acid	—COOH	carboxy-	-oic acid	CH₃COOH ethanoic acid
Ester	—COOR	alkoxycarbonyl-	alkyl …oate	CH₃COOC₂H₅ ethyl ethanoate
Acid chloride	—COCl	chlorocarbonyl-	-oyl chloride	CH₃COCl ethanoyl chloride
Amide	—CONH₂	carbamoyl-	-amide	CH₃CONH₂ ethanamide
Nitrile	—C≡N	cyano-	-nitrile	CH₃CN ethanenitrile
Aldehyde	—CHO	oxo- / formyl-	-al	CH₃CH₂CHO propanal
Ketone	>C=O	oxo-	-one	CH₃COCH₃ propanone
Alcohol	—OH	hydroxy-	-ol	C₂H₅OH ethanol
Amine	—NH₂	amino-	-amine	CH₃NH₂ methanamine
Ether	—OR	alkoxy-	—	CH₃OC₂H₅ methoxyethane
Halide	—X	fluoro-, chloro-, bromo-, iodo-	—	CH₃Cl chloromethane
Nitro	—NO₂	nitro-	—	CH₃NO₂ nitromethane

The common alkyl groups up to C4 are worth memorising, because NEB questions still quote the trivial names:

Structure	IUPAC name	Trivial name
CH₃—	methyl	methyl
CH₃CH₂—	ethyl	ethyl
CH₃CH₂CH₂—	propyl	n-propyl
(CH₃)₂CH—	propan-2-yl	isopropyl
CH₃CH₂CH₂CH₂—	butyl	n-butyl
CH₃CH₂CH(CH₃)—	butan-2-yl	sec-butyl
(CH₃)₂CHCH₂—	2-methylpropyl	isobutyl
(CH₃)₃C—	2-methylpropan-2-yl	tert-butyl

13.2 Qualitative analysis: detection of N, S and halogens by Lassaigne's test

In an organic compound nitrogen, sulphur and the halogens are joined by covalent bonds, so the ordinary inorganic tests — which need free ions — give nothing. Lassaigne's test solves this by first converting the covalent element into an ionic sodium salt.

During the fusion sodium reduces the elements into ionic salts:

Na + C + N --fuse--> NaCN

2Na + S --fuse--> Na₂S

Na + X --fuse--> NaX (X = Cl, Br, I)

If nitrogen and sulphur are both present and sodium is not in excess, they combine instead as sodium thiocyanate:

Na + C + N + S --fuse--> NaSCN

Detection of nitrogen

The extract is boiled with freshly prepared iron(II) sulphate solution and then acidified with dilute sulphuric acid. A Prussian blue colour or precipitate confirms nitrogen.

6NaCN + FeSO₄ → Na₄[Fe(CN)₆] + Na₂SO₄

3Na₄[Fe(CN)₆] + 2Fe₂(SO₄)₃ → Fe₄[Fe(CN)₆]₃↓ + 6Na₂SO₄ (Prussian blue)

Some of the Fe²⁺ is oxidised to Fe³⁺ by air during the boiling; the acid then dissolves the greenish iron hydroxides so that the blue colour can be seen.

Detection of sulphur

Two independent tests are used.

1. Sodium nitroprusside test. A few drops of sodium nitroprusside give a deep violet or purple colour:

Na₂S + Na₂[Fe(CN)₅NO] → Na₄[Fe(CN)₅NOS] (violet)

2. Lead acetate test. Acidify with acetic acid and add lead acetate; a black precipitate of lead sulphide appears:

Na₂S + (CH₃COO)₂Pb → PbS↓ + 2CH₃COONa (black)

Detection of nitrogen and sulphur together

If both are present as SCN⁻, adding iron(III) chloride gives a blood-red colour:

FeCl₃ + 3NaSCN → Fe(SCN)₃ + 3NaCl (blood red)

Fusing with excess sodium destroys the thiocyanate (NaSCN + 2Na → NaCN + Na₂S), after which the ordinary nitrogen and sulphur tests both work.

Detection of halogens

The extract is first boiled with dilute nitric acid. This is essential: any cyanide or sulphide left in solution would precipitate AgCN and Ag₂S and give a false positive.

NaCN + HNO₃ → NaNO₃ + HCN↑

Na₂S + 2HNO₃ → 2NaNO₃ + H₂S↑

Silver nitrate solution is then added:

NaX + AgNO₃ → AgX↓ + NaNO₃

Halogen	Precipitate	Colour	In excess NH₄OH
Chlorine	AgCl	white, curdy	completely soluble
Bromine	AgBr	pale yellow	sparingly soluble
Iodine	AgI	bright yellow	insoluble

For completeness: carbon and hydrogen are detected by heating the compound with dry copper(II) oxide. The carbon dioxide turns lime water milky and the water condensing on the cool part of the tube turns anhydrous copper sulphate from white to blue.

13.3 Isomerism: definition and classification

Isomerism exists because carbon is tetravalent and catenates: for a given formula there is usually more than one way to join the atoms up. The number of possibilities rises very fast — C₄H₁₀ has 2 isomers, C₅H₁₂ has 3, C₆H₁₄ has 5, C₁₀H₂₂ has 75 and C₂₀H₄₂ has 366 319.

13.4 Structural isomerism: chain, position, functional, metamerism and tautomerism

Structural isomers have the same molecular formula but a different connectivity — the atoms are bonded to different partners. Five types are examined.

Chain isomerism

The isomers differ in the length or branching of the carbon skeleton. It begins at C₄.

• C₄H₁₀: CH₃CH₂CH₂CH₃ (butane) and (CH₃)₃CH (2-methylpropane)
• C₅H₁₂: pentane, 2-methylbutane, 2,2-dimethylpropane
• C₆H₁₄: hexane, 2-methylpentane, 3-methylpentane, 2,2-dimethylbutane, 2,3-dimethylbutane — five in all

Position isomerism

The skeleton is the same; the functional group or multiple bond sits on a different carbon.

• C₃H₈O: CH₃CH₂CH₂OH (propan-1-ol) and CH₃CH(OH)CH₃ (propan-2-ol)
• C₄H₈: CH₂=CHCH₂CH₃ (but-1-ene) and CH₃CH=CHCH₃ (but-2-ene)
• C₆H₄Cl₂: 1,2-, 1,3- and 1,4-dichlorobenzene

Functional isomerism

The isomers contain different functional groups altogether, so they belong to different families and behave completely differently.

Formula	Isomer 1	Isomer 2
C₂H₆O	ethanol CH₃CH₂OH (liquid, b.p. 78 °C)	methoxymethane CH₃OCH₃ (gas, b.p. −24 °C)
C₃H₆O	propanal CH₃CH₂CHO	propanone CH₃COCH₃
C₂H₄O₂	ethanoic acid CH₃COOH	methyl methanoate HCOOCH₃
C₃H₉N	propan-1-amine (1°)	N-methylethanamine (2°)

Metamerism

• C₄H₁₀O ethers: C₂H₅—O—C₂H₅ (ethoxyethane) and CH₃—O—C₃H₇ (1-methoxypropane)
• C₅H₁₀O ketones: CH₃—CO—C₃H₇ (pentan-2-one) and C₂H₅—CO—C₂H₅ (pentan-3-one)

Metamerism is really a special case of chain/position isomerism, but NEB lists it separately, so learn the definition word for word.

Tautomerism

The common type is keto–enol tautomerism, which needs at least one hydrogen on the carbon next to the carbonyl (an α-hydrogen):

CH₃—CO—CH₃ ⇌ CH₃—C(OH)=CH₂

(keto form) ⇌ (enol form)

At room temperature propanone is 99.99 % keto; ethyl acetoacetate, CH₃COCH₂COOC₂H₅, is about 93 % keto and 7 % enol, because its enol is stabilised by conjugation and an internal hydrogen bond. The keto form normally dominates because the C=O bond (≈ 749 kJ mol⁻¹) is much stronger than the C=C bond (≈ 614 kJ mol⁻¹).

13.5 Geometrical isomerism (cis and trans) and optical isomerism (d and l form)

Stereoisomers have the same connectivity. They differ only in the three-dimensional arrangement of the atoms.

Geometrical (cis–trans) isomerism

A C=C double bond cannot rotate: rotating it would mean breaking the π bond, which costs about 250 kJ mol⁻¹. The groups on the two carbons are therefore locked in place.

• but-2-ene CH₃CH=CHCH₃ — each carbon carries CH₃ and H → yes, cis and trans
• but-1-ene CH₂=CHCH₂CH₃ — C1 carries two H → no
• 2-methylbut-2-ene (CH₃)₂C=CHCH₃ — C2 carries two CH₃ → no

In the cis isomer the two like groups are on the same side of the double bond; in the trans isomer they are on opposite sides.

Property	cis isomer	trans isomer
Position of like groups	same side	opposite sides
Dipole moment	bond dipoles add — polar	dipoles cancel — zero or small
Boiling point	higher (stronger dipole–dipole forces)	lower
Melting point	lower (packs badly, less symmetric)	higher (packs well)
Stability	lower (the two groups crowd each other)	higher
Solubility in water	greater	smaller

The standard NEB illustration is the pair of butenedioic acids. Maleic acid (cis) melts at 130 °C, is very soluble in water and loses water on heating to give maleic anhydride. Fumaric acid (trans) melts at 287 °C, is much less soluble and forms no anhydride, because its two —COOH groups are too far apart to reach each other.

Optical isomerism

The usual cause is a chiral (asymmetric) carbon — a carbon atom joined to four different groups. Such a molecule has no plane of symmetry, so its mirror image cannot be laid on top of it, just as a left hand cannot be laid on a right hand.

Key facts about enantiomers:

• They have identical melting point, boiling point, density, solubility and ordinary chemical reactivity. They differ only in the direction in which they rotate polarised light, and in how they react with other chiral things (enzymes, other enantiomers, polarised light).
• A racemic mixture is an equimolar (50:50) mixture of the d and l forms. It is written (±) or dl and is optically inactive by external compensation — the rotation of one half exactly cancels the other.
• A meso compound has chiral carbons but also an internal plane of symmetry, so it is optically inactive by internal compensation. Tartaric acid therefore exists in three forms: d, l and meso.
• A molecule with n different chiral carbons has at most 2^{n} optical isomers (van't Hoff's rule).

The rotation is measured in a polarimeter and reported as the specific rotation

where \alpha is the observed rotation in degrees, l the length of the tube in decimetres and c the concentration in g cm⁻³.

13.6 Reaction Mechanism: homolytic and heterolytic fission

Every organic reaction starts by breaking a covalent bond, and a shared pair can be split in only two ways.

Homolytic fission (homolysis)

The two bonded atoms take one electron each. Each fragment keeps an unpaired electron and is a free radical. The movement of a single electron is shown by a half-headed "fish-hook" arrow.

Cl—Cl --hν--> Cl• + Cl•

CH₃—CH₃ --Δ, 500 °C--> CH₃• + CH₃•

Homolysis is favoured by high temperature, ultraviolet light, peroxide initiators, non-polar bonds and non-polar solvents such as CCl₄. The reactions it starts are called free-radical reactions; substitution in alkanes (Section 14.2) is the standard example.

Heterolytic fission (heterolysis)

Both electrons go to one atom — the more electronegative one. Two oppositely charged ions result, and the movement of the electron pair is shown by a full curved arrow.

CH₃—Br → CH₃⁺ + Br⁻ (carbocation)

CH₃—MgBr → CH₃⁻ + MgBr⁺ (carbanion)

Heterolysis is favoured by polar bonds, polar solvents (which solvate and stabilise the ions) and the presence of a catalyst such as AlCl₃ that can pull the leaving group away. It gives ionic (polar) reactions.

	Homolytic fission	Heterolytic fission
Sharing of the pair	one electron to each atom	both electrons to one atom
Arrow	half-headed (fish-hook)	full curved arrow
Products	two free radicals	a cation and an anion
Favoured by	UV light, heat, peroxides, non-polar solvent	polar bond, polar solvent, catalyst
Energy needed	bond dissociation energy	usually less, if the ions are solvated
Reaction type produced	free-radical reactions	ionic (electrophilic / nucleophilic) reactions

13.7 Electrophiles, nucleophiles and free radicals

The species produced by fission are the reaction intermediates; the species that attack them are the reagents. Organic reagents are sorted by what they do with electrons.

	Electrophile (E⁺ or E)	Nucleophile (Nu⁻ or Nu)
Electron demand	deficient — accepts a pair	rich — donates a pair
Lewis character	acid	base
Attacks	the electron-rich site (π bond, benzene ring, C of C=O)	the electron-poor site (C of C—X, C of C=O)
Charged examples	H⁺, Cl⁺, Br⁺, NO₂⁺, NO⁺, R⁺, R—C≡O⁺	OH⁻, CN⁻, Cl⁻, Br⁻, RO⁻, CH₃COO⁻, HSO₃⁻, R⁻
Neutral examples	BF₃, AlCl₃, FeCl₃, ZnCl₂, SO₃, carbene :CCl₂	H₂O, NH₃, ROH, RNH₂, R₃P, alkenes, benzene

Two points examiners like:

• Every carbocation is an electrophile and every carbanion is a nucleophile, but the reverse is not true — BF₃ is an electrophile with no charge at all.
• Ambident nucleophiles such as CN⁻ and NO₂⁻ have two donor atoms and can attack through either, which is why KCN gives nitriles while AgCN gives isocyanides.

A free radical is the third kind of reactive species: an atom or group with an unpaired electron, electrically neutral, paramagnetic and extremely reactive. It is neither an electrophile nor a nucleophile, although it is electron deficient.

Intermediate	Electrons on C	Hybridisation	Shape	Stability order
Carbocation R₃C⁺	6	sp²	trigonal planar, 120°	3° > 2° > 1° > CH₃⁺
Carbanion R₃C⁻	8 (incl. lone pair)	sp³	pyramidal, ≈107°	CH₃⁻ > 1° > 2° > 3°
Free radical R₃C•	7	sp²	planar (shallow pyramid)	3° > 2° > 1° > CH₃•

The reason for the carbocation order is that alkyl groups release electrons (+I effect) and also share the C—H electrons of the neighbouring carbon with the empty p orbital (hyperconjugation); the more alkyl groups, the more the positive charge is spread out and the more stable the ion. For a carbanion the charge is negative, so electron-releasing alkyl groups make it worse, and the order reverses.

13.8 Inductive effect (+I and −I); Resonance effect (+R and −R)

Substituents change how a molecule reacts by pushing or pulling electron density. Two permanent electronic effects are on the syllabus.

Inductive effect

Hydrogen is taken as the reference, so an effect is measured relative to a C—H bond.

• −I groups (electron-withdrawing) pull electrons away from the chain. In decreasing order of strength:

—NO₂ > —CN > —SO₃H > —CHO > —COOH > —F > —Cl > —Br > —I > —OR > —OH > —C₆H₅

• +I groups (electron-releasing) push electrons towards the chain:

—O⁻ > —COO⁻ > —C(CH₃)₃ > —CH(CH₃)₂ > —CH₂CH₃ > —CH₃ > —H

The effect dies away quickly: its strength falls to roughly one-third at each successive carbon, so it is negligible beyond the third carbon from the substituent. That is why 4-chlorobutanoic acid is barely stronger than butanoic acid, while 2-chlorobutanoic acid is much stronger.

Application 1 — strength of carboxylic acids. An acid is strong when its conjugate base is stable. A −I group drains negative charge away from the carboxylate ion and stabilises it, so the acid is stronger:

Acid	CH₃CH₂COOH	CH₃COOH	HCOOH	ClCH₂COOH	Cl₂CHCOOH	Cl₃CCOOH
pKa	4.87	4.76	3.75	2.87	1.29	0.66

Formic acid beats acetic acid because it has no +I methyl group at all. Among the monohalo acids, the more electronegative halogen wins: FCH₂COOH (2.59) > ClCH₂COOH (2.87) > BrCH₂COOH (2.90) > ICH₂COOH (3.17).

Application 2 — carbocation stability. Alkyl groups are +I, so they feed electron density into the electron-poor carbon: (CH₃)₃C⁺ > (CH₃)₂CH⁺ > CH₃CH₂⁺ > CH₃⁺. This single fact explains Markovnikov's rule in Unit 14.

Resonance (mesomeric) effect

• +R groups carry a lone pair on the atom attached to the conjugated system and push it in: —OH, —OR, —NH₂, —NHR, —NR₂, —NHCOR, —OCOR, —F, —Cl, —Br, —I.
• −R groups have a multiply bonded electronegative atom and pull electron density out: —NO₂, —CN, —CHO, —COR, —COOH, —COOR, —CONH₂, —SO₃H.

Two properties distinguish resonance from induction and are worth a mark each:

	Inductive effect	Resonance effect
Electrons involved	σ electrons	π electrons or a lone pair
Needs conjugation?	no	yes — a planar, conjugated system
Range	dies out after 2–3 σ bonds	transmitted right along the conjugated chain
Charge produced	partial (δ+, δ−)	full formal charges in the contributing structures
Magnitude	weaker	stronger

For resonance to be possible the contributing structures must have the same positions for all nuclei, the same number of unpaired electrons, and the system must be planar and conjugated. The real molecule is a single resonance hybrid — never one structure "changing into" another — and it is always more stable than the most stable contributing structure. That extra stability is the resonance energy (152 kJ mol⁻¹ for benzene, Unit 15).

Application — acidity of phenol versus ethanol. Phenol (pKa 10.0) is far more acidic than ethanol (pKa 15.9), because the phenoxide ion C₆H₅O⁻ spreads its negative charge over the ortho and para carbons of the ring by resonance, while the ethoxide ion has nowhere to put it.

Application — basicity of aniline versus ammonia. In aniline the nitrogen lone pair is delocalised into the ring (+R). It is therefore less available for protonation, which is why aniline (pKb 9.4) is a much weaker base than ammonia (pKb 4.75) or methylamine (pKb 3.4).

Chapter summary

• An IUPAC name is secondary prefix + word root + primary suffix + secondary suffix. Choose the longest chain containing the senior functional group, number for the lowest locant to that group, list substituents alphabetically, and elide the final "e" before a vowel.
• Lassaigne's test fuses the compound with sodium to make NaCN, Na₂S and NaX. Nitrogen gives Prussian blue Fe₄[Fe(CN)₆]₃; sulphur gives a violet nitroprusside colour and a black PbS precipitate; halogens give AgCl (white), AgBr (pale yellow) or AgI (yellow) after the extract is boiled with dilute HNO₃.
• Isomers share a molecular formula. Structural isomerism = chain, position, functional, metamerism, tautomerism. Stereoisomerism = geometrical and optical.
• Geometrical isomerism needs restricted rotation and two different groups on each doubly bonded carbon. Trans is the more stable and higher-melting; cis is the more polar and higher-boiling.
• Optical isomerism needs a chiral carbon (four different groups). Enantiomers rotate polarised light equally and oppositely; a 50:50 racemic mixture and a meso form are both optically inactive; n chiral centres give up to 2^{n} isomers.
• Homolysis splits the pair one electron each and gives free radicals; heterolysis gives both electrons to one atom and produces a carbocation or a carbanion. Carbocations and radicals are sp² and planar, 3° > 2° > 1°; carbanions are sp³ and pyramidal, with the order reversed.
• Electrophiles accept an electron pair (Lewis acids); nucleophiles donate one (Lewis bases); free radicals have one unpaired electron.
• The inductive effect works through σ bonds, is short-range and gives partial charges; the resonance effect works through a conjugated π system, is long-range and gives full charges. Both are permanent, and together they explain acid strength, base strength and carbocation stability.

Practice questions

Group A — Multiple choice (1 mark each)

1. The IUPAC name of (CH₃)₂CHCH₂CH₂OH is [1]a2-methylbutan-1-olb3-methylbutan-1-olc2-methylbutan-4-oldpentan-1-ol
2. Prussian blue in Lassaigne's test confirms the presence of [1]asulphurbnitrogencchlorinedphosphorus
3. Before testing for halogens the sodium extract is boiled with dilute HNO₃ in order to [1]aoxidise the halidebremove cyanide and sulphide ionscneutralise the sodiumdincrease the solubility of AgX
4. Ethoxyethane and 1-methoxypropane are [1]achain isomersbtautomerscmetamersdfunctional isomers
5. Which compound shows geometrical isomerism? [1]a1,1-dichloroetheneb2-methylprop-1-enecpent-2-enedpropene
6. A racemic mixture is optically inactive because of [1]ainternal compensationbexternal compensationcthe absence of a chiral carbonda plane of symmetry in each molecule
7. Heterolytic fission of a covalent bond gives [1]atwo free radicalsba cation and an anionctwo cationsdtwo atoms
8. Which of the following is the strongest −I group? [1]a—CH₃b—OHc—Id—NO₂
9. The number of optical isomers of a compound with three different chiral carbons is [1]a3b4c6d8

Answers to Group A

1. (b) — the longest chain is four carbons with OH on C1 and a methyl on C3.

2. (b) — the CN⁻ from fused nitrogen ends up as Fe₄[Fe(CN)₆]₃.

3. (b) — otherwise AgCN and Ag₂S precipitate and give a false positive.

4. (c) — same functional group (ether) but different alkyl groups on the oxygen.

5. (c) — in CH₃CH=CHC₂H₅ each doubly bonded carbon carries two different groups.

6. (b) — the d and l halves rotate the plane equally and oppositely and cancel.

7. (b) — both electrons go to the more electronegative atom.

8. (d) — the nitro group heads the −I series.

9. (d) — 2^{n} with n = 3 gives 8.

Group B — Short answer (5 marks each)

1. State the rules of IUPAC nomenclature and use them to name (i) CH₃CH₂CH(CH₃)CH₂CH₃ (ii) (CH₃)₂C=CHCH₃ (iii) CH₃CH(OH)CH₂CH₂CHO (iv) CH₃CH₂COOCH₃ [5]
2. How is Lassaigne's extract prepared? Describe with equations how nitrogen and sulphur are detected in it. [5]
3. Define isomerism and classify it with one example of each type. [5]
4. What are the necessary conditions for geometrical isomerism? Draw the cis and trans forms of but-2-ene and give three ways in which their properties differ. [5]
5. What is a chiral carbon? Explain optical isomerism with lactic acid as an example, and distinguish between a racemic mixture and a meso compound. [5]
6. Distinguish between homolytic and heterolytic fission. Give the shape, hybridisation and stability order of a carbocation and a carbanion. [5]
7. Define the inductive effect. Explain why trichloroacetic acid is a much stronger acid than acetic acid, and why the effect of the chlorine in 4-chlorobutanoic acid is almost negligible. [5]
8. Write the seven structural isomers of C₄H₁₀O, name them, and state the type of isomerism between (i) isomers 1 and 2 and (ii) an alcohol and an ether. [5]

Answers to Group B

1. (i) 3-methylpentane (ii) 2-methylbut-2-ene (iii) 4-hydroxypentanal (iv) methyl propanoate. In (iii) the —CHO carbon is C1, the chain is five carbons, and the OH becomes the prefix hydroxy-.

2. Extract: fuse the compound with clean sodium to redness, plunge the hot tube into distilled water, boil and filter. Nitrogen — boil with FeSO₄, acidify with dilute H₂SO₄: 6NaCN + FeSO₄ → Na₄[Fe(CN)₆] + Na₂SO₄, then 3Na₄[Fe(CN)₆] + 2Fe₂(SO₄)₃ → Fe₄[Fe(CN)₆]₃↓ + 6Na₂SO₄, Prussian blue. Sulphur — Na₂S + Na₂[Fe(CN)₅NO] → Na₄[Fe(CN)₅NOS], violet; and Na₂S + (CH₃COO)₂Pb → PbS↓ + 2CH₃COONa, black.

4. Conditions: restricted rotation about a C=C (or a ring) and two different groups on each of the doubly bonded carbons. cis-but-2-ene has both methyls on the same side (b.p. 3.7 °C, m.p. −139 °C, dipole moment 0.33 D); trans-but-2-ene has them on opposite sides (b.p. 0.9 °C, m.p. −106 °C, dipole moment 0). The trans form is more stable and melts higher; the cis form is more polar and boils higher.

7. Definition as in Section 13.8. In Cl₃CCOOH three strongly −I chlorines drain electron density from the carboxylate oxygen, spreading the negative charge and stabilising the anion, so the acid ionises far more completely (pKa 0.66 against 4.76). In 4-chlorobutanoic acid the chlorine is three carbons away; the inductive effect falls to roughly one-third of its strength at each carbon, so almost nothing reaches the —COOH group.

8. butan-1-ol, butan-2-ol, 2-methylpropan-1-ol, 2-methylpropan-2-ol, 1-methoxypropane, 2-methoxypropane, ethoxyethane. (i) butan-1-ol and butan-2-ol are position isomers; (ii) any alcohol and any ether are functional isomers.

Group C — Long answer (8 marks each)

1. (a) State the rules used to select and number the principal chain in IUPAC nomenclature, and give the seniority order of the common functional groups. [4](b) Name: (i) CH₃CH(CH₃)CH₂CH(OH)CH₃ (ii) HOOC—CH₂—CH(CH₃)—COOH (iii) CH₃CH₂CH₂CHO (iv) (CH₃)₃CCl. [4]
2. (a) Define isomerism and classify it fully, giving one example of each of the seven types. [4](b) What is optical isomerism? State the conditions for a compound to be optically active, define enantiomer, racemic mixture and meso compound, and work out how many optical isomers tartaric acid, HOOC—CH(OH)—CH(OH)—COOH, actually has. [4]
3. (a) Define the inductive effect and the resonance effect and give three points of difference between them. [4](b) Using these effects, explain: (i) the order HCOOH > CH₃COOH > CH₃CH₂COOH in acid strength; (ii) why phenol is more acidic than ethanol; (iii) why aniline is a weaker base than ammonia. [4]

Answers to Group C

1. (b) (i) 4-methylpentan-2-ol — the chain is five carbons, the OH takes the lower locant 2, so the methyl lands on C4. (ii) 2-methylbutanedioic acid — both —COOH carbons are in the chain, giving a four-carbon diacid with a methyl on C2. (iii) butanal. (iv) 2-chloro-2-methylpropane.

2. (b) A compound is optically active if it is chiral, i.e. its mirror image is not superimposable on it; the usual cause is a carbon carrying four different groups, and the molecule must have no plane or centre of symmetry. Enantiomers are the two non-superimposable mirror images; a racemic mixture is a 50:50 blend of them, inactive by external compensation; a meso compound contains chiral carbons but has an internal plane of symmetry and is inactive by internal compensation. Tartaric acid has two chiral carbons, so 2^{2} = 4 is the maximum — but the two chiral carbons carry the same four groups, so one of the four is a meso form and its mirror image is identical to it. Tartaric acid therefore has only three forms: d, l and meso.

3. (b) (i) The methyl group in CH₃COOH is +I and pushes electron density onto the carboxylate, destabilising the anion; the ethyl group in CH₃CH₂COOH is a still stronger +I group. Formic acid has no alkyl group at all, so its anion is the most stable and it is the strongest of the three. (ii) In the phenoxide ion the negative charge is delocalised over the ortho and para carbons of the ring by the −R effect of the ring, so the anion is strongly resonance stabilised. The ethoxide ion has no such delocalisation, so ethanol is a far weaker acid (pKa 10.0 against 15.9). (iii) In aniline the nitrogen lone pair is conjugated with the ring (+R effect of —NH₂) and is delocalised into it, so it is much less available to accept a proton. In ammonia the lone pair is entirely localised on nitrogen. Hence aniline (pKb 9.4) is a much weaker base than ammonia (pKb 4.75).