Organic Chemistry: Some Basic Principles and Techniques ISC Class 11

What Organic Chemistry: Some Basic Principles and Techniques is about

Weightage: Organic Chemistry: Some Basic Principles and Techniques belongs to unit 8 of the ISC Class XI Chemistry syllabus. CISCE weights units 8-9 together (Organic Chemistry) at 23 of the 70 marks in Paper I (Theory). Class XI is examined by the school, so your school sets the paper within this weightage.Source: CISCE ISC Chemistry syllabus, examination year 2028, Class XI unit weightage

Key concepts

Vital force theory, tetravalency and catenation

Vital force theory (Berzelius) held that organic compounds could be made only by living organisms. Wöhler (1828) heated ammonium cyanate (inorganic) and obtained urea, an organic compound, and later syntheses (Kolbe: acetic acid; Berthelot: methane) ended the theory. Carbon is tetravalent (ground state 2s²2p², excited 2s¹2p³, forming four bonds), directed tetrahedrally in sp³ carbon. Catenation is the ability of carbon atoms to bond to each other to form long chains, branches and rings; it is strongest for carbon because C–C bonds are strong (about 348 kJ mol⁻¹) and carbon is small.

Worked example

Give two reasons why carbon forms so many compounds.

Try it first, then show the solutionSolution

1. Catenation: strong C–C bonds give chains, branches and rings of any length
2. Tetravalency and the ability to form multiple bonds (C=C, C≡C, C=O, C≡N) with itself and other elements

Answer: Catenation and tetravalency (with multiple bonding)

Classification, homologous series and functional groups

Open-chain (acyclic, aliphatic) compounds have no rings. Closed-chain (cyclic) compounds are homocyclic (ring of carbon only), which may be alicyclic (cyclohexane) or aromatic (benzene, with delocalised π electrons), or heterocyclic (ring containing N, O or S: pyridine, furan). A homologous series is a family with the same functional group and general formula, in which successive members differ by CH₂ (14 u), with similar chemical properties and graded physical properties. A functional group is the atom or group that decides the characteristic reactions: –OH (alcohol), –CHO (aldehyde), >C=O (ketone), –COOH (carboxylic acid), –NH₂ (amine), –X (halide), –O– (ether), –COOR (ester).

Worked example

Classify benzene, cyclohexane, pyridine and butane, and give the next homologue of ethanol.

Try it first, then show the solutionSolution

1. Benzene: homocyclic aromatic; cyclohexane: homocyclic alicyclic
2. Pyridine: heterocyclic (N in the ring); butane: open-chain
3. Ethanol C₂H₅OH + CH₂ ⇒ propanol C₃H₇OH

Answer: Aromatic; alicyclic; heterocyclic; acyclic; propanol

IUPAC nomenclature

Steps: (1) find the longest chain containing the principal functional group (and multiple bonds); (2) number it to give the lowest locant to the principal group, then to multiple bonds, then to substituents; (3) write prefixes (substituents in alphabetical order with locants) + root (meth, eth, prop, but, pent, …) + primary suffix (-ane, -ene, -yne) + secondary suffix (-ol, -al, -one, -oic acid, -amine). Priority of principal groups: carboxylic acid > ester > aldehyde > ketone > alcohol > amine. Alicyclic names use 'cyclo-' (cyclohexanol); aromatic compounds are named as benzene derivatives with locants (1,2-, 1,3-, 1,4-) or by accepted names such as phenol, toluene (methylbenzene) and benzoic acid.

Worked example

Name (i) CH₃CH(CH₃)CH₂CH₃ (ii) CH₃CH₂CH(OH)CH₃ (iii) CH₂=CHCH₂CHO (iv) Cl–C₆H₄–OH with Cl para to OH.

Try it first, then show the solutionSolution

1. (i) Four-carbon chain with methyl on C2: 2-methylbutane
2. (ii) –OH on C2 of a four-carbon chain: butan-2-ol
3. (iii) CHO is C1, double bond between C3 and C4: but-3-enal; (iv) 4-chlorophenol

Answer: 2-methylbutane; butan-2-ol; but-3-enal; 4-chlorophenol

Structural isomerism

Same molecular formula, different structural formula. Chain: different carbon skeletons (butane and 2-methylpropane). Position: same skeleton and group, different position (propan-1-ol and propan-2-ol). Functional: different functional groups (ethanol and methoxymethane; propanal and propanone). Metamerism: different alkyl groups on either side of a polyvalent group (diethyl ether and methyl propyl ether). Tautomerism: rapidly interconverting isomers in dynamic equilibrium, usually by migration of a hydrogen, as in keto-enol tautomerism (CH₃COCH₃ ⇌ CH₂=C(OH)CH₃).

Worked example

Name the type of isomerism shown by (i) C₂H₅OC₂H₅ and CH₃OC₃H₇ (ii) CH₃CH₂CHO and CH₃COCH₃.

Try it first, then show the solutionSolution

1. (i) Both ethers; different alkyl groups on each side of –O–: metamerism
2. (ii) Aldehyde and ketone with the same formula C₃H₆O: functional isomerism

Answer: Metamerism; functional isomerism

Stereoisomerism: geometrical and optical

Geometrical (cis-trans) isomerism arises from restricted rotation about C=C when each doubly bonded carbon carries two different groups: in cis-but-2-ene the methyl groups are on the same side, in trans on opposite sides; cis isomers are usually more polar. For C=N (oximes) the terms syn and anti are used. Optical isomerism: a molecule with a chiral carbon (four different groups) is non-superimposable on its mirror image; the two enantiomers rotate plane-polarised light (produced by a Nicol prism) equally in opposite directions, measured in a polarimeter: dextrorotatory (d, +) or laevorotatory (l, −). Specific rotation [α] = α/(l × c), with l in dm and c in g mL⁻¹. A 1 : 1 mixture is racemic (no rotation). Lactic acid (one chiral C) has d and l forms. Tartaric acid (two similar chiral C) has d, l and an optically inactive meso form with an internal plane of symmetry.

Worked example

How many optical isomers does tartaric acid, HOOC–CH(OH)–CH(OH)–COOH, have, and why is the meso form inactive?

Try it first, then show the solutionSolution

1. Two chiral carbons with identical sets of groups: d, l and meso forms
2. Only d and l rotate light (2 optically active isomers); with the racemate, 4 forms in all
3. The meso form has a plane of symmetry: the rotation by one half cancels that by the other (internal compensation)

Answer: Two optically active forms (d, l) plus an inactive meso form

Methods of purification

Crystallisation: dissolve in hot solvent, filter, cool; impurities stay in solution. Sublimation: for substances that go directly from solid to vapour (camphor, naphthalene). Simple distillation: liquids with widely different boiling points; fractional distillation: close boiling points (crude oil); distillation under reduced pressure: liquids that decompose at their normal boiling point (glycerol); steam distillation: steam-volatile, water-immiscible liquids (aniline). Differential extraction: shaking with an immiscible solvent in a separating funnel. Chromatography: separation by different adsorption or partition between a stationary and a mobile phase; in paper or thin-layer chromatography, Rf = distance moved by substance/distance moved by solvent front.

Worked example

Choose a method to purify (i) camphor containing salt (ii) glycerol (iii) aniline from a reaction mixture.

Try it first, then show the solutionSolution

1. (i) Sublimation: camphor sublimes, salt does not
2. (ii) Distillation under reduced pressure: glycerol decomposes near its normal boiling point
3. (iii) Steam distillation: aniline is steam volatile and immiscible with water

Answer: Sublimation; reduced-pressure distillation; steam distillation

Qualitative analysis: Lassaigne's test

C and H: heat the compound with dry CuO; CO₂ turns lime water milky and H₂O turns anhydrous CuSO₄ blue. Lassaigne's (sodium fusion) test: fusing with sodium converts covalent N, S and X into ionic NaCN, Na₂S and NaX, which dissolve in water as the sodium fusion extract. Nitrogen: boil with FeSO₄, acidify with H₂SO₄ and add FeCl₃: Prussian blue (iron(III) hexacyanoferrate(II)). Sulphur: lead acetate gives a black PbS precipitate; sodium nitroprusside gives a violet colour. N and S together: NaSCN forms and gives a blood-red colour with Fe³⁺, $F{e}^{3+}+SC{N}^{-}\to [Fe(SCN){]}^{2+}$. Halogens: boil the extract with dilute HNO₃ (to decompose NaCN and Na₂S, which would interfere), then add AgNO₃: white AgCl (soluble in NH₃), pale yellow AgBr (sparingly soluble), yellow AgI (insoluble).

Worked example

Why is the sodium fusion extract boiled with dilute nitric acid before testing for halogens with silver nitrate?

Try it first, then show the solutionSolution

1. If N or S is present, the extract contains CN⁻ and S²⁻
2. These would give precipitates of AgCN (white) and Ag₂S (black) and mask the halide test
3. Boiling with HNO₃ expels them as HCN and H₂S

Answer: To remove cyanide and sulphide, which would interfere

Quantitative estimation of elements

Carbon and hydrogen (Liebig): burn a known mass with CuO; absorb H₂O in anhydrous CaCl₂ and CO₂ in KOH; weigh. %C = (12/44) × (mass CO₂/mass sample) × 100; %H = (2/18) × (mass H₂O/mass sample) × 100. Nitrogen (Kjeldahl): digest with conc. H₂SO₄ (N → (NH₄)₂SO₄), liberate NH₃ with NaOH, absorb it in a known volume of standard acid and back-titrate. %N = 1.4 × N × V/m, where V mL of acid of normality N is neutralised by the ammonia from m g of sample; not suitable for nitro, azo compounds or N in rings. Carius method: heat with fuming HNO₃ (with AgNO₃ for halogens) in a sealed tube. %X = (atomic mass of X/molar mass of AgX) × (mass AgX/m) × 100; %S = (32/233) × (mass BaSO₄/m) × 100; %P = (62/222) × (mass Mg₂P₂O₇/m) × 100. Oxygen is usually found by difference.

Worked example

0.30 g of urea is analysed by Kjeldahl's method; the ammonia formed neutralises 10.0 mL of 1.0 N H₂SO₄. Find %N.

Try it first, then show the solutionSolution

1. %N = 1.4 × N × V/m
2. = 1.4 × 1.0 × 10.0/0.30
3. = 46.7%

Answer: 46.7% (urea, CO(NH₂)₂, contains 28/60 = 46.7% N)

Electron displacement: inductive, electromeric, resonance and hyperconjugation

Inductive effect: permanent displacement of σ electrons along a chain due to an electronegative or electropositive group, fading with distance. −I groups withdraw electrons (–NO₂ > –F > –Cl > –Br > –I > –OH); +I groups donate (alkyl groups: (CH₃)₃C– > (CH₃)₂CH– > CH₃CH₂– > CH₃–). It explains why chloroethanoic acid is stronger than ethanoic acid. Electromeric effect: temporary, complete shift of a π pair to one atom of a multiple bond when an attacking reagent approaches; +E towards the reagent, −E away. Resonance (mesomeric) effect: permanent delocalisation of π or lone-pair electrons through a conjugated system; +R groups donate (–OH, –NH₂, –Cl), −R groups withdraw (–NO₂, –CHO, –COOH, –CN). Hyperconjugation: delocalisation of σ electrons of C–H bonds on the carbon next to a carbocation, radical or double bond ('no-bond resonance'); more α-hydrogens mean more stability.

Worked example

Explain why ClCH₂COOH is a stronger acid than CH₃COOH.

Try it first, then show the solutionSolution

1. Cl exerts a −I effect, pulling electron density through the chain towards itself
2. This weakens the O–H bond and, more importantly, disperses the negative charge of ClCH₂COO⁻, stabilising it
3. In CH₃COO⁻ the CH₃ group (+I) intensifies the charge, destabilising the anion

Answer: −I effect of Cl stabilises the conjugate base

Bond fission, reactive intermediates, electrophiles and nucleophiles

Homolytic fission: a bond breaks evenly, each atom taking one electron, giving free radicals (in light or heat, non-polar solvents). Heterolytic fission: one atom takes both electrons, giving a carbocation (positive carbon, sp², planar) or carbanion (negative carbon, sp³, pyramidal). Stability: carbocations and free radicals: 3° > 2° > 1° > methyl (+I and hyperconjugation), with benzylic and allylic ions extra stable by resonance; carbanions: methyl > 1° > 2° > 3° (alkyl groups intensify negative charge). Electrophiles are electron-seeking (positive or electron-deficient): H⁺, NO₂⁺, Cl⁺, carbocations, BF₃, AlCl₃. Nucleophiles are nucleus-seeking (electron-rich): OH⁻, CN⁻, Cl⁻, NH₃, H₂O, ROH, carbanions.

Worked example

Arrange the carbocations CH₃⁺, CH₃CH₂⁺, (CH₃)₂CH⁺ and (CH₃)₃C⁺ in increasing stability and explain.

Try it first, then show the solutionSolution

1. Each extra methyl group donates electrons (+I) and adds C–H bonds for hyperconjugation
2. Number of α-hydrogens: 0, 3, 6, 9

Answer: CH₃⁺ < CH₃CH₂⁺ < (CH₃)₂CH⁺ < (CH₃)₃C⁺

Types of organic reactions and SN1, SN2, E1, E2

Substitution: one atom or group replaces another (CH₃Br + OH⁻ → CH₃OH + Br⁻). Addition: atoms add across a multiple bond (CH₂=CH₂ + Br₂ → CH₂BrCH₂Br). Elimination: atoms are removed from adjacent carbons to form a multiple bond. SN2 (substitution, nucleophilic, bimolecular): one step; the nucleophile attacks from the back as the leaving group departs; rate = k[RX][Nu⁻]; inversion of configuration; favoured for methyl and 1° halides. SN1 (unimolecular): two steps through a carbocation; rate = k[RX]; partial racemisation; favoured for 3° halides and polar protic solvents. E2: one-step removal of H⁺ and X⁻ by a strong base; rate = k[RX][base]. E1: two steps via a carbocation; rate = k[RX]; competes with SN1.

Worked example

Predict the mechanism and rate law for (i) CH₃Br with aqueous OH⁻ (ii) (CH₃)₃CBr in water.

Try it first, then show the solutionSolution

1. (i) Methyl halide, little steric hindrance, strong nucleophile: SN2; rate = k[CH₃Br][OH⁻]; inversion
2. (ii) Tertiary halide forms a stable 3° carbocation; weak nucleophile: SN1; rate = k[(CH₃)₃CBr]

Answer: SN2 (second order); SN1 (first order)

Chemistry in real life: Organic Chemistry: Some Basic Principles and Techniques

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Topics in this chapter

• Electronic Effects, Intermediates and Reaction Mechanisms
• IUPAC Nomenclature and Types of Isomerism

Organic Chemistry: Some Basic Principles and Techniques formula sheet

Liebig's method

$\%C=\frac{12}{44}\times \frac{m_{C{O}_2}}{m}\times 100;\%H=\frac{2}{18}\times \frac{m_{H_{2}O}}{m}\times 100$

m = mass of sample.

Kjeldahl's method

$\%N=\frac{1.4\times N\times V}{m}$

V in mL of acid of normality N neutralised by the NH₃.

Carius: halogens

$\%X=\frac{\text{at. mass of }X}{\text{molar mass of }AgX}\times \frac{m_{AgX}}{m}\times 100$

AgCl 143.5, AgBr 188, AgI 235.

Carius: sulphur and phosphorus

$\%S=\frac{32}{233}\times \frac{m_{BaS{O}_4}}{m}\times 100;\%P=\frac{62}{222}\times \frac{m_{M{g}_2{P}_2{O}_7}}{m}\times 100$

Mg₂P₂O₇ contains 2P (62).

Specific rotation

$[\alpha ]=\frac{\alpha }{l\times c}$

l in dm, c in g mL⁻¹.

Retardation factor

$R_f=\frac{\text{distance moved by substance}}{\text{distance moved by solvent}}$

Always less than 1.

Rate laws

$S_{N}1:\text{rate}=k[RX];S_{N}2:\text{rate}=k[RX][N{u}^{-}]$

E1 and E2 follow the same patterns.

Organic Chemistry: Some Basic Principles and Techniques: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

How the school exam asks Organic Chemistry: Some Basic Principles and Techniques

The question patterns to expect, and the method that works for each.

Tricks to remember Organic Chemistry: Some Basic Principles and Techniques

Short memory aids. Say each one aloud once, then use it on a question the same day.

• Suffix shows the boss

  Principal group priority: acid > ester > aldehyde > ketone > alcohol > amine.

  Use it when: IUPAC naming.

• Prussian for N, blood for N + S

  Blue for nitrogen; blood red when nitrogen and sulphur come together as thiocyanate.

  Use it when: Lassaigne's test.

• 44, 18, 233, 222

  Divide by CO₂ 44, H₂O 18, BaSO₄ 233, Mg₂P₂O₇ 222 in estimation formulas.

  Use it when: Quantitative analysis.

• Three alkyls hold a plus

  Carbocations and radicals: 3° most stable; carbanions: the reverse.

  Use it when: Intermediate stability.

• SN2 inverts, SN1 racemises

  One step, back-side attack: inversion; carbocation, both faces: racemisation.

  Use it when: Mechanisms.

Organic Chemistry: Some Basic Principles and Techniques: exam-level solved questions

Ordered from exam level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section D style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 5-mark question split into 1 + 2 + 1 + 1 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in Organic Chemistry: Some Basic Principles and Techniques

The slips examiners see most often, and what to write instead.

• ✗Mistake: Numbering from the wrong end

  Students give lowest numbers to substituents rather than the principal group.

  Wrong working

  CH₃CH₂CH(CH₃)CH₂OH: 3-methylbutan-4-ol.

  Right working

  Number from the –OH carbon: 2-methylbutan-1-ol.

  ✓Do this: The principal functional group gets the lowest locant first.

  Usually costs: 1 mark.

• ✗Mistake: Forgetting to boil with HNO₃

  The halogen test is done directly.

  Wrong working

  Extract + AgNO₃ gives a black precipitate, so iodine is present.

  Right working

  Black Ag₂S comes from sulphur; remove S²⁻ and CN⁻ first.

  ✓Do this: Always mention boiling with dilute HNO₃.

  Usually costs: 1 mark.

• ✗Mistake: Using the wrong molar mass in estimation

  Students divide by the atomic mass instead of the compound's molar mass.

  Wrong working

  %S = (32/32) × (0.466/0.200) × 100.

  Right working

  %S = (32/233) × (0.466/0.200) × 100 = 32.0%.

  ✓Do this: Element mass ÷ molar mass of the weighed compound.

  Usually costs: Full marks for the part.

• ✗Mistake: Confusing inductive and electromeric effects

  Both are electron shifts.

  Wrong working

  The electromeric effect is permanent.

  Right working

  Electromeric: temporary, π electrons, only when a reagent attacks; inductive: permanent, σ electrons.

  ✓Do this: Compare by permanence, electrons involved and cause.

  Usually costs: 1 mark.

• ✗Mistake: Calling every molecule with a chiral carbon optically active

  Meso compounds are overlooked.

  Wrong working

  meso-tartaric acid is optically active because it has chiral carbons.

  Right working

  It has a plane of symmetry and is inactive.

  ✓Do this: Check for an internal plane of symmetry.

  Usually costs: 1 mark.

Practice questions on Organic Chemistry: Some Basic Principles and Techniques

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

1. Q1 · 1 mark · MCQExam level

   The IUPAC name of CH₂=CH–CH₃ is:
   (a) propane
   (b) propene
   (c) propyne
   (d) allyl

   Hint

   Three carbons, one C=C.

   Answer

   (b)

2. Q2 · 1 mark · MCQExam level

   A Prussian blue colour in Lassaigne's test shows the presence of:
   (a) sulphur
   (b) nitrogen
   (c) chlorine
   (d) phosphorus

   Hint

   From NaCN.

   Answer

   (b)

3. Q3 · 1 mark · Very short answerExam level

   Name the functional group present in ketones.

   Hint

   >C=O between two carbons.

   Answer

   Carbonyl (ketonic) group

4. Q4 · 1 mark · Very short answerExam level

   Which is the most stable free radical: CH₃•, CH₃CH₂•, (CH₃)₂CH• or C₆H₅CH₂•?

   Hint

   Resonance outweighs hyperconjugation.

   Answer

   C₆H₅CH₂• (benzyl), stabilised by resonance with the ring

5. Q5 · 2 marks · Short answerExam level

   State three characteristics of a homologous series.

   Hint

   Formula, difference, properties.

   Answer

   Same general formula and functional group; successive members differ by CH₂ (14 u); similar chemical properties with a gradation in physical properties.

6. Q6 · 2 marks · Short answerHarder exam level

   Distinguish homolytic and heterolytic fission, naming the intermediates formed.

   Hint

   Who keeps the electrons?

   Answer

   Homolytic: each atom keeps one electron, giving free radicals. Heterolytic: one atom keeps both, giving a carbocation and/or carbanion (or ions).

7. Q7 · 2 marks · Short answerHarder exam level

   On a thin-layer plate two spots move 1.8 cm and 4.5 cm while the solvent front moves 6.0 cm. Find their Rf values and say which compound is more strongly adsorbed.

   Hint

   Rf = spot ÷ solvent front.

   Answer

   1.8/6.0 = 0.30 and 4.5/6.0 = 0.75; the spot with Rf 0.30 is held more strongly by the stationary phase.

8. Q8 · 3 marks · Short answerHarder exam level

   0.300 g of an organic compound gave 0.444 g of Mg₂P₂O₇. Find the percentage of phosphorus.

   Hint

   62/222.

   Answer

   (62/222) × (0.444/0.300) × 100 = 41.3%

9. Q9 · 3 marks · Short answerHarder exam level

   Draw the geometrical isomers of but-2-ene and state which has the higher dipole moment.

   Hint

   Methyl groups same or opposite side.

   Answer

   cis: both CH₃ on the same side of C=C; trans: on opposite sides. cis has the higher dipole moment (in trans the bond dipoles cancel).

10. Q10 · 3 marks · Short answerHarder exam level

    Give IUPAC names: (i) CH₃CH(Cl)CH₂COOH (ii) C₆H₅CH₂OH (iii) the five-membered cyclic ketone C₅H₈O.

    Hint

    COOH carbon is C1.

    Answer

    (i) 3-chlorobutanoic acid (ii) phenylmethanol (iii) cyclopentanone

11. Q11 · 2 marks · Short answerHarder exam level

    A solution of 0.15 g mL⁻¹ in a 2.0 dm tube rotates plane-polarised light by +4.5°. Find the specific rotation.

    Hint

    [α] = α/(l × c).

    Answer

    4.5/(2.0 × 0.15) = +15°

12. Q12 · 3 marks · Short answerHarder exam level

    Classify as electrophile or nucleophile: OH⁻, NH₃, NO₂⁺, BF₃, CN⁻, H₂O, (CH₃)₃C⁺.

    Hint

    Electron-poor or electron-rich?

    Answer

    Electrophiles: NO₂⁺, BF₃, (CH₃)₃C⁺. Nucleophiles: OH⁻, NH₃, CN⁻, H₂O.

13. Q13 · 5 marks · Long answerHarder exam level

    0.180 g of a compound of C, H and O gave 0.264 g of CO₂ and 0.108 g of H₂O.
    (a) Find %C and %H. [2]
    (b) Find %O and the empirical formula. [2]
    (c) Its molar mass is 180 g mol⁻¹. Find its molecular formula. [1]

    Hint

    O by difference.

    Answer

    (a) %C = (12/44)(0.264/0.180)(100) = 40.0%; %H = (2/18)(0.108/0.180)(100) = 6.67%. (b) %O = 53.3%; ratio C 3.33 : H 6.67 : O 3.33 = 1 : 2 : 1, so CH₂O. (c) n = 180/30 = 6: C₆H₁₂O₆.

14. Q14 · 3 marks · Short answerHOTS / competency

    Explain keto-enol tautomerism in propanone, and use hyperconjugation to explain why propene is more stable than ethene.

    Hint

    Count α-hydrogens.

    Answer

    CH₃COCH₃ ⇌ CH₂=C(OH)CH₃: an α-H moves to O and the C=O becomes C=C (keto form predominates). Propene's CH₃ has three α-hydrogens that delocalise σ electrons into the π bond; ethene has none, so propene is more stable.

15. Q15 · 3 marks · Short answerHOTS / competency

    How many stereoisomers does butane-2,3-diol, CH₃CH(OH)CH(OH)CH₃, have? Explain.

    Hint

    Like tartaric acid.

    Answer

    Two similar chiral carbons: a d form, an l form and an inactive meso form (with a plane of symmetry): three stereoisomers, not 2² = 4.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1. 1.Who disproved vital force theory?Show answer

   Wöhler, by making urea from ammonium cyanate (1828).

2. 2.Difference between successive homologues?Show answer

   CH₂ (14 u).

3. 3.Colour for N and S together in Lassaigne's test?Show answer

   Blood red.

4. 4.Kjeldahl formula?Show answer

   %N = 1.4 × N × V/m.

5. 5.Specific rotation?Show answer

   [α] = α/(l × c).

6. 6.Which effect is temporary?Show answer

   Electromeric effect.

7. 7.Rate law of SN1?Show answer

   Rate = k[RX].

8. 8.Stereochemistry of SN2?Show answer

   Inversion of configuration.

Quick revision checklist

The night before the exam: can you do each of these without looking?

• ☐Can you explain why carbon forms so many compounds?
• ☐Can you name a branched alcohol, an unsaturated aldehyde and a substituted phenol?
• ☐Can you identify five kinds of structural isomerism?
• ☐Can you explain optical activity, racemic and meso forms?
• ☐Can you choose purification methods for given mixtures?
• ☐Can you describe Lassaigne's tests with colours and reasons?
• ☐Can you find %C, %H, %N, %X, %S and %P?
• ☐Can you compare SN1 and SN2 mechanisms?

Organic Chemistry: Some Basic Principles and Techniques: frequently asked questions