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NEET UG (Medical Entrance) · Chemistry (JEE & NEET)

Basic Principles of Organic Chemistry and Hydrocarbons

IUPAC nomenclature, isomerism, electronic effects, reaction mechanisms and the chemistry of alkanes, alkenes, alkynes and aromatics.

Thirteen concepts from NCERT Class 11 Units 8–9: electronic effects and carbocations, structural and geometrical isomerism, DU and IUPAC, Wurtz and ethane conformations, free-radical halogenation, Markovnikov addition, ozonolysis/Baeyer, terminal-alkyne acidity with Lindlar/trans reduction, and aromaticity with electrophilic substitution.

  • NEET UG (Medical Entrance)
  • Medium level
  • 13 concepts
  • 5 practice questions

1Inductive, mesomeric and hyperconjugative effects

Three through-bond effects decide most of the reactivity rankings in this chapter. The inductive effect (±I) is the permanent polarisation of a σ bond by electronegativity difference: −I groups (NO₂, CN, halogens) withdraw electron density through the chain, while +I groups (alkyl) push it. The mesomeric (resonance) effect (±M) moves π or lone-pair density through a conjugated system: +M donors (OH, OR, NH₂, halogen) put electron density into the ring or multiple bond; −M acceptors (NO₂, CHO, COR, CN) pull it out.

Hyperconjugation — no-bond resonance — is the delocalisation of C–H σ electrons from an alkyl group into an adjacent empty p orbital or π system. More α-hydrogens means more hyperconjugative structures, which is why alkyl substitution stabilises carbocations and alkenes, and why Saytzeff elimination prefers the more substituted alkene.

Figure. Three ways electron density moves without a full bond breaking: inductive along σ bonds, mesomeric through a π system, and hyperconjugation from α-C–H into an empty or π orbital. Classify the stem before drawing arrows.

Which effect is doing the work

  1. σ onlyIf the group is attached by a single bond and no π/lone-pair conjugation is available, read ±I.
  2. π or lone pairIf a lone pair or π bond can conjugate with an adjacent π system, ±M usually dominates the directing and acid–base story.
  3. Count α-HCarbocation, free-radical and Saytzeff rankings: count hydrogens on carbons adjacent to the empty or radical centre — that is hyperconjugation.
The three electronic effects
EffectWhat movesStabilisesClassic + / − groups
Inductive (±I)σ-electron densityCharges through the chain+I alkyl; −I NO₂, CN, F, Cl, Br, I
Mesomeric (±M)π or lone-pair densityResonance forms of ions and radicals+M OH, OR, NH₂, X; −M NO₂, CHO, COR, CN
HyperconjugationC–H σ into empty p / πCarbocations, radicals, substituted alkenesMore α-H → stronger; tert-butyl ≫ methyl
The tert-butyl cation is more stable than the methyl cation mainly because
  1. tert-butyl has a −I effect from three methyl groups that withdraw electrons
  2. nine α-hydrogens allow extensive hyperconjugation into the empty p orbital, plus +I from the alkyl groups
  3. tert-butyl is aromatic by Hückel's rule

Each methyl attached to the cationic carbon contributes three α-H atoms that hyperconjugate into the empty p orbital, and the alkyl groups also donate by +I. −I would destabilise a cation; aromaticity is irrelevant here.

2Carbocation stability

Carbocation stability rises with the number of alkyl groups on the charged carbon because each alkyl group donates by +I and opens more hyperconjugative structures. The alkyl order is therefore 3^\circ > 2^\circ > 1^\circ > \mathrm{CH}_3^+. Allylic and benzylic cations sit above that ladder: the empty p orbital conjugates with an adjacent π system, so the charge is shared over two or more carbons.

The same ranking controls the regiochemistry of electrophilic addition and of SN1: the path that builds the more stable carbocation wins. Counting α-hydrogens is the fastest comparison when the candidates are all alkyl cations with no resonance option.

Figure. α-Hydrogen counts on the four alkyl cations in the worked example. Bar length is the integer α-H count that feeds hyperconjugation — not a measured energy. Allyl and benzyl are off this chart on purpose: their extra stability is resonance, a different axis.

How to rank

  1. Resonance firstAllyl or benzyl beats any simple alkyl cation of the same substitution class.
  2. Then substitutionAmong alkyl cations: tertiary > secondary > primary > methyl.
  3. Count α-HFor close alkyl cases, more α-hydrogens means more hyperconjugation — tert-butyl has 9, isopropyl 6, ethyl 3, methyl 0.

Ranking carbocation stability

Arrange \mathrm{CH}_3^+, \mathrm{CH}_3\mathrm{CH}_2^+, (\mathrm{CH}_3)_2\mathrm{CH}^+ and (\mathrm{CH}_3)_3\mathrm{C}^+ in order of stability.

  • α-H on \mathrm{CH}_3^+0
  • α-H on \mathrm{CH}_3\mathrm{CH}_2^+3
  • α-H on (\mathrm{CH}_3)_2\mathrm{CH}^+6
  • α-H on (\mathrm{CH}_3)_3\mathrm{C}^+ → order9 → 3^\circ > 2^\circ > 1^\circ > \mathrm{CH}_3^+

Pro tip. α-H count is a hyperconjugation score for alkyl cations only. Once a candidate is allylic or benzylic, stop counting and put resonance above the alkyl ladder — benzyl beats tert-butyl in most JEE rankings that mix the two classes.

Which cation is the most stable?
  1. (\mathrm{CH}_3)_3\mathrm{C}^+
  2. \mathrm{CH}_2\mathrm{=CH{-}CH}_2^+ (allyl)
  3. \mathrm{CH}_3\mathrm{CH}_2^+

Allyl is resonance-stabilised over two carbons. Tert-butyl is the most stable simple alkyl cation, but resonance beats hyperconjugation here; ethyl is primary and least stable of the three.

3Structural isomerism: chain, position, functional

Structural isomers share a molecular formula but differ in which atoms are bonded to which. NCERT splits the common exam types as chain isomerism (different carbon skeletons — n-butane vs isobutane), position isomerism (same skeleton, functional group or multiple bond at different carbons — but-1-ene vs but-2-ene), and functional-group isomerism (different functional groups — ethanol vs dimethyl ether for C₂H₆O). Metamerism is the ether/amine special case of unequal alkyl groups on either side of a heteroatom.

Degree of unsaturation does not choose among these — it only constrains the formula. Naming a pair as chain versus position is an atom-connectivity question: redraw both, then ask whether the longest chain changed or only a locant.

Figure. Structural isomers share a molecular formula but differ in connectivity: chain branching, functional-group position, or a different functional group entirely. Name the class before naming the compound.

Classify the pair

  1. Same formula?If not, they are not isomers. If yes, compare connectivity.
  2. Same functional group?No → functional-group isomerism (or metamerism for R–X–R'). Yes → check the carbon skeleton.
  3. Skeleton vs locantDifferent branching → chain. Same chain, different position of group/π bond → position.
Structural isomer types
TypeWhat changesClassic pair
ChainCarbon skeletonButane / 2-methylpropane
PositionLocant of group or π bondBut-1-ene / but-2-ene
Functional groupFunctional group identityEthanol / methoxymethane
MetamerismAlkyl groups about a heteroatomDiethyl ether / methyl propyl ether
But-1-ene and but-2-ene are
  1. Chain isomers, because both are C₄H₈
  2. Position isomers — same chain, different C=C locant
  3. Functional-group isomers — one is an alkene and one an alkyne

Both are butenes (same functional group and C₄ chain); they differ only in where the double bond sits. That is position isomerism.

4Geometrical (cis–trans) isomerism

Alkenes can show geometrical isomerism when each doubly bonded carbon carries two different substituents. Rotation about C=C is restricted, so the relative disposition of groups is fixed: identical groups on the same side give the cis isomer; on opposite sides, the trans isomer. But-2-ene is the textbook pair.

Cis-but-2-ene is polar (NCERT quotes dipole moment ≈ 0.33 D) while the trans form is essentially non-polar because the bond dipoles cancel. For solids, the trans isomer typically melts higher. If either doubly bonded carbon has two identical groups, cis–trans isomerism vanishes.

Figure. cis–trans needs each doubly bonded carbon to carry two different groups. Same-side substituents (cis) often leave a net dipole; opposite (trans) can cancel — as in but-2-ene. Propene cannot form the pair.

Does cis–trans apply?

  1. Check each =CEach doubly bonded carbon needs two different attachments. Same two on one carbon → no geometrical isomers.
  2. Name cis/transSame side of C=C for the identical reference groups → cis; opposite → trans.
  3. Property cueCis often more polar; trans often higher melting point for solids — do not invent boiling-point rules beyond that.

But-2-ene geometries

State whether propene and but-2-ene can show cis–trans isomerism, and give the dipole-moment contrast NCERT quotes for but-2-ene.

  • Propene (CH₃CH=CH₂)one =C has two H → no cis–trans
  • But-2-ene (CH₃CH=CHCH₃)each =C has H and CH₃ → cis and trans
  • μ(cis-but-2-ene) vs μ(trans)≈0.33 D vs ≈0 (NCERT)

Pro tip. Geometrical isomerism is about configuration at a restricted π bond — it is not another name for position isomerism.

Which alkene shows cis–trans isomerism?
  1. Propene
  2. 2-Methylpropene
  3. But-2-ene

But-2-ene has two different groups on each doubly bonded carbon. Propene and 2-methylpropene each have a =CH₂ or =C(CH₃)₂ end that blocks geometrical isomerism.

5Degree of unsaturation

The degree of unsaturation (index of hydrogen deficiency) counts how many rings and π bonds a formula hides. For a molecule \mathrm{C}_c\mathrm{H}_h\mathrm{N}_n\mathrm{O}_o\mathrm{X}_x the formula is \mathrm{DU} = (2c + 2 + n - h - x)/2. Oxygen is ignored; each halogen counts as a hydrogen; each nitrogen adds one to the hydrogen count in the saturated parent.

A DU of 1 is one double bond or one ring; DU of 2 can be a triple bond, two double bonds, a ring plus a double bond, and so on. Benzene's \mathrm{C}_6\mathrm{H}_6 gives DU = 4 (the ring plus three double bonds of the Kekulé picture), which is why an aromatic formula jumps out of an elemental analysis long before you draw it.

Figure. Degree of unsaturation counts rings plus π bonds: DU = (2C+2−H−X+N)/2. Each missing H₂ pair is one unit — benzene's four is the classic check.

Computing DU

  1. Read the formulaPull c, h, n and x (halogens). Drop oxygen and other divalent atoms from the count.
  2. Plug in\mathrm{DU} = (2c + 2 + n - h - x)/2. A non-integer means the formula is wrong or ionic.
  3. InterpretEach unit is one ring or one π bond. DU ≥ 4 with six carbons is a benzene-ring alarm.

DU for three common formulae

Find the degree of unsaturation of \mathrm{C}_4\mathrm{H}_8, \mathrm{C}_3\mathrm{H}_6\mathrm{O} and \mathrm{C}_6\mathrm{H}_6.

  • \mathrm{C}_4\mathrm{H}_8: (2\times 4 + 2 - 8)/21 (one C=C or one ring)
  • \mathrm{C}_3\mathrm{H}_6\mathrm{O}: O ignored → (2\times 3 + 2 - 6)/21 (e.g. acetone's C=O)
  • \mathrm{C}_6\mathrm{H}_6: (2\times 6 + 2 - 6)/24 (ring + three π)
  • Check \mathrm{C}_3\mathrm{H}_7\mathrm{Br}: (2\times 3 + 2 - 7 - 1)/20 (saturated alkyl halide)

Pro tip. The bromine check is the halogen rule in action: X is treated as H, so \mathrm{C}_3\mathrm{H}_7\mathrm{Br} is saturated just like \mathrm{C}_3\mathrm{H}_8. If you forget to subtract X you invent a false double bond.

The degree of unsaturation of \mathrm{C}_5\mathrm{H}_9\mathrm{N} is
  1. 1
  2. 2
  3. 3

(2\times 5 + 2 + 1 - 9)/2 = (10 + 2 + 1 - 9)/2 = 4/2 = 2. The nitrogen contributes +1 in the numerator; two units mean some mix of rings and π bonds (for example a ring plus a double bond, or a nitrile).

6IUPAC chain and locant rules

IUPAC naming for simple open-chain compounds is a short checklist, not a drawing problem. Choose the longest continuous carbon chain that contains the principal functional group (the parent). Number that chain to give the principal functional group the lowest possible locant; if that does not decide, minimise the locants of the substituents as a set.

Name substituents as prefixes (methyl, chloro, nitro), cite them alphabetically, and write the parent with its functional-group suffix (alkene -ene, alkyne -yne, alcohol -ol, aldehyde -al, ketone -one, acid -oic acid). Multiple identical substituents take di-, tri-, tetra-; those multipliers do not affect alphabetical order.

Figure. IUPAC naming is a checklist: pick the principal functional group, take the longest chain that includes it, number for lowest set of locants, then list substituents alphabetically.

Naming checklist

  1. Pick the parentLongest chain that includes the principal functional group; for alkenes/alkynes the multiple bond must sit in the parent.
  2. NumberLowest locant to the principal group; then lowest set of substituent locants.
  3. AssembleAlphabetical prefixes + locants + parent stem + unsaturation + principal suffix.
Principal-group priority (common JEE set)
ClassSuffixPrefix if not principal
Carboxylic acid-oic acidcarboxy-
Aldehyde-aloxo- / formyl-
Ketone-oneoxo-
Alcohol-olhydroxy-
Amine-amineamino-
Alkene / alkyne-ene / -yne—
For \mathrm{CH}_3\mathrm{CH}(\mathrm{OH})\mathrm{CH}_2\mathrm{CH}_3, the correct IUPAC name is
  1. 1-methylpropan-1-ol
  2. butan-2-ol
  3. 2-hydroxybutane

The longest chain containing the OH has four carbons, so the parent is butane with suffix -ol. Numbering from the end nearer OH gives locant 2 → butan-2-ol. '2-hydroxybutane' treats OH as a substituent when it is the principal group; '1-methylpropan-1-ol' chose the wrong parent.

7Wurtz reaction for alkanes

Alkyl halides treated with sodium metal in dry ether couple to a higher alkane: 2\,\mathrm{RX} + 2\,\mathrm{Na} \rightarrow \mathrm{R–R} + 2\,\mathrm{NaX}. That is the Wurtz reaction. It is used to build alkanes with an even number of carbons when both halide molecules are the same.

Two different alkyl halides give a mixture of three alkanes (R–R, R'–R', R–R'), so the method is a poor route to a single unsymmetrical product. Methane cannot be made by Wurtz — there is no C₀ partner. Dry ether matters because water would destroy the sodium and the organosodium path.

Figure. Wurtz couples two alkyl halides with sodium in dry ether to R–R. Mixed halides scramble into three alkanes — count carbons before you promise a single product.

Predict the Wurtz product

  1. Same RX twice2 RBr + 2 Na → R–R. Count carbons: product has twice the carbon count of R.
  2. Mixed halidesRBr + R'Br → R–R + R'–R' + R–R' — three products.
  3. Dry etherMoisture kills the reaction. Do not confuse with alcoholic KOH elimination.

Bromoethane coupling

What alkane forms when bromoethane reacts with sodium in dry ether?

  • HalideCH₃CH₂Br (2 carbons)
  • CouplingCH₃CH₂–CH₂CH₃
  • Productbutane (4 carbons, even)

Pro tip. Wurtz doubles the alkyl fragment — even carbon count when starting from one RX.

Wurtz reaction on a mixture of CH₃Br and C₂H₅Br in dry ether gives
  1. Only propane
  2. Ethane, butane and propane as a mixture
  3. Only methane, because methyl always wins

Cross-coupling of two different alkyl halides yields R–R, R'–R' and R–R': ethane, butane and propane.

8Conformations of ethane

Rotation about the C–C single bond in ethane gives conformations (rotamers). The extremes are the eclipsed conformation (C–H bonds on the two carbons aligned) and the staggered conformation (each front C–H bisects a rear H–C–H angle). Intermediate arrangements are skew. Infinite conformations exist; exams ask the two extremes and their relative stability.

Staggered ethane is more stable than eclipsed because eclipsed C–H bonds raise torsional strain. NCERT quotes the barrier as about 12.5 kJ mol⁻¹ — small enough that rotation is rapid at room temperature, so conformational isomers of ethane are not separately isolable. Sawhorse and Newman projections are the two standard drawings; this vocabulary cannot render Newman’s circle, so the stability contrast is taught in words and a 1D energy sketch.

Figure. Qualitative torsional energy along the ethane dihedral: minima at staggered, maxima at eclipsed. Barrier height is schematic — NCERT’s 12.5 kJ mol⁻¹ is the verbal number, not a plotted scale.

Read a conformation question

  1. Name the extremeAligned C–H bonds → eclipsed. Bisecting arrangement → staggered.
  2. StabilityStaggered lower energy (less torsional strain). Barrier ~12.5 kJ mol⁻¹ — free enough at RT.
  3. Not geometrical isomersConformers interconvert by σ-bond rotation; cis–trans isomers do not.

Barrier size

NCERT quotes the eclipsed–staggered energy difference for ethane as about 12.5 kJ mol⁻¹. Is that large enough to isolate the eclipsed form at room temperature?

  • Barrier (NCERT)≈12.5 kJ mol⁻¹
  • Thermal energy scale at ~300 Kcollisions readily supply this
  • Isolable eclipsed ethane?no — rotation is essentially free

Pro tip. Quote the barrier to explain rapid rotation — do not memorise it as a stoichiometry input.

The most stable conformation of ethane is
  1. Eclipsed, because C–H bonds align for maximum bonding
  2. Staggered, because torsional strain is minimised
  3. A frozen cis geometrical isomer about the C–C bond

Staggered ethane has lower torsional strain than eclipsed. Geometrical isomerism requires a restricted π bond, not a freely rotating σ bond.

9Free-radical halogenation of alkanes

Alkanes do not undergo polar electrophilic or nucleophilic substitution under ordinary conditions; they react with \mathrm{Cl}_2 or \mathrm{Br}_2 under light or heat by a free-radical chain. The three stages are initiation (homolytic cleavage of \mathrm{X}_2 to two halogen atoms), propagation (the chain: halogen atom abstracts H to give R·, then R· abstracts X from \mathrm{X}_2 to give RX and a new X·), and termination (any two radicals combine and stop the chain).

Chlorination is fast and relatively unselective; bromination is slower and much more selective for the most stable radical (tertiary ≫ secondary ≫ primary), because the H-abstraction step is endothermic for bromine and feels the radical-stability difference. Reactivity of H–X abstraction follows the same hyperconjugation order as carbocations: 3^\circ > 2^\circ > 1^\circ > \mathrm{CH}_3.

Figure. Free-radical chain as a flow of labels, not structures: initiation makes X·, the two propagation boxes recycle X· while converting RH + X₂ into RX + HX, and termination removes radicals. No skeletal formulae — only the three stage names the exam asks you to assign.

The chain

  1. Initiation\mathrm{X}_2 \xrightarrow{h\nu} 2\mathrm{X}\cdot — one photon starts many later turnovers.
  2. Propagation\mathrm{X}\cdot + \mathrm{RH} \rightarrow \mathrm{HX} + \mathrm{R}\cdot, then \mathrm{R}\cdot + \mathrm{X}_2 \rightarrow \mathrm{RX} + \mathrm{X}\cdot. Net: \mathrm{RH} + \mathrm{X}_2 \rightarrow \mathrm{RX} + \mathrm{HX}.
  3. TerminationAny two of \mathrm{X}\cdot, \mathrm{R}\cdot combine (X₂, R₂ or RX) and kill the chain.

Selectivity of bromination

Propane has six primary and two secondary hydrogens. Relative rates of H-abstraction by Br· are about 1600 (2°) : 1 (1°) at ordinary temperature. Find the approximate product ratio of 2-bromopropane to 1-bromopropane.

  • Statistical factor, 2° sites2 hydrogens
  • Statistical factor, 1° sites6 hydrogens
  • Rate weight 2° = 2 × 16003200
  • Ratio 2-bromo : 1-bromo = 3200 : (6 × 1)3200 : 6 ≈ 533 : 1

Pro tip. Chlorination numbers are far flatter (roughly 4–5 : 1 for 2° vs 1° per hydrogen), so propane chlorination gives a messy mixture while bromination is almost cleanly 2-bromopropane. Selectivity is a kinetic isotope of radical stability, not a different mechanism.

In the free-radical chlorination of methane, the step \mathrm{Cl}\cdot + \mathrm{CH}_4 \rightarrow \mathrm{HCl} + \mathrm{CH}_3\cdot is
  1. Initiation, because a new radical appears
  2. Propagation, because one radical is consumed and another is produced
  3. Termination, because HCl is a stable molecule

Propagation keeps the radical count at one: Cl· in, CH₃· out. Initiation is the light-driven split of Cl₂ with no organic radical yet; termination removes radicals in pairs.

10Electrophilic addition and Markovnikov's rule

Alkenes and alkynes undergo electrophilic addition: the π bond attacks an electrophile, a carbocation forms, and a nucleophile captures it. Markovnikov's rule is the regiochemical consequence of that mechanism — the electrophile (H⁺ from HX, or the equivalent) adds so that the more stable carbocation intermediate forms. For HBr on propene that means H on C-1 and Br on C-2, giving 2-bromopropane via the secondary cation.

The peroxide (Kharasch) effect reverses the orientation for HBr only: peroxides start a radical chain in which Br· adds first, and the radical intermediate prefers the more substituted carbon, so H ends up on the more substituted carbon and the product is anti-Markovnikov 1-bromopropane. HCl and HI do not show a useful peroxide reversal under the same conditions.

Figure. Energy up the page. Both paths start and finish at the same levels; the dashed primary-cation path climbs higher at the first TS than the solid secondary path. Markovnikov selectivity is that barrier difference — the 2° cation route is lower. Schematic heights, not kJ values from a table.

Reading the addition

  1. No peroxideH⁺ adds first; pick the carbocation that is more stable; X⁻ captures. Markovnikov product.
  2. HBr + peroxideBr· adds first; the carbon radical prefers the more substituted site; anti-Markovnikov alkyl bromide.
  3. ScopeKharasch reversal is a teaching fact for HBr. Do not invent it for HCl or HI in a JEE stem.

Markovnikov addition of HBr

Predict the major product when HBr adds to propene (\mathrm{CH}_3\mathrm{CH}{=}\mathrm{CH}_2) in the absence of peroxides.

  • Protonation at C-1 → cation\mathrm{CH}_3\mathrm{CH}^+\mathrm{CH}_3 (2°)
  • Protonation at C-2 → cation\mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}_2^+ (1°)
  • More stable cation2° wins
  • Br⁻ capture at C⁺\mathrm{CH}_3\mathrm{CHBrCH}_3 (2-bromopropane)

Pro tip. With peroxides the same alkene gives 1-bromopropane: Br· adds to C-1 so the radical sits on C-2, then H is delivered to C-2 from HBr. Same atoms, opposite regiochemistry, different intermediate.

HBr adds to propene in the presence of benzoyl peroxide. The major product is
  1. 2-bromopropane (Markovnikov)
  2. 1-bromopropane (anti-Markovnikov)
  3. 1,2-dibromopropane

Peroxides divert HBr into the radical Kharasch path. Br· adds to the less substituted carbon so the secondary radical forms; hydrogen abstraction then places H on that carbon, giving 1-bromopropane. Dibromide would require Br₂, not HBr.

11Baeyer's test and ozonolysis

Two classical alkene tools answer different questions. Baeyer's test — cold dilute alkaline \mathrm{KMnO}_4 — is a qualitative presence test: the purple permanganate decolourises as the alkene is oxidised to a vicinal diol (and \mathrm{MnO}_2 may brown-out). Alkanes leave the purple colour standing. It does not locate the double bond.

Ozonolysis does. Ozone cleaves the C=C; reductive workup (\mathrm{Zn}/\mathrm{H}_2\mathrm{O} or dimethyl sulfide) delivers the two carbonyl fragments that were joined by the double bond — aldehydes and ketones whose carbon skeletons are the two sides of the original alkene. Oxidative workup (\mathrm{H}_2\mathrm{O}_2) further oxidises aldehydes to carboxylic acids. Reading the carbonyl products backwards reconstructs where the π bond sat.

Figure. Baeyer's cold alkaline KMnO₄ browns on an alkene (test). Ozonolysis cleaves the C=C into two carbonyls whose structures locate the double bond — stitch the fragments on paper from the products.

Which tool

  1. PresencePurple \mathrm{KMnO}_4 fades → unsaturation present (Baeyer). No positional information.
  2. PositionOzonolysis + reductive workup → two carbonyls; stitch them at the carbonyl carbons to rebuild the alkene.
  3. WorkupReductive keeps aldehydes; oxidative turns them into acids. Ketones survive either workup.
Alkene diagnosis
ToolWhat you seeWhat you learn
Baeyer (cold dil. KMnO₄)Purple fades; diol formsC=C (or C≡C) is present
Br₂ in CCl₄Brown-red fadesUnsaturation present (addition)
O₃, then Zn/H₂OAldehydes / ketonesWhere the C=C was
O₃, then H₂O₂Ketones / carboxylic acidsWhere the C=C was (aldehydes oxidised)

Rebuilding an alkene from ozonolysis

Reductive ozonolysis of an alkene yields equimolar acetone and acetaldehyde. Identify the alkene.

  • Acetone fragment(\mathrm{CH}_3)_2\mathrm{C}= end
  • Acetaldehyde fragment\mathrm{CH}_3\mathrm{CH}= end
  • Join at the two carbonyl carbons(\mathrm{CH}_3)_2\mathrm{C}{=}\mathrm{CHCH}_3
  • Name2-methylbut-2-ene

Pro tip. Equimolar 1:1 means one alkene, not a mixture of two symmetrical ones. If both carbonyls had been acetone, the alkene would have been 2,3-dimethylbut-2-ene — each carbon of the double bond carrying two methyls.

Cold dilute alkaline KMnO₄ decolourises with an unknown, and reductive ozonolysis of the same unknown gives only propanone. The unknown is
  1. propene
  2. 2,3-dimethylbut-2-ene
  3. but-1-ene

Baeyer confirms a C=C. A single ketone product means the alkene was symmetrical with both ends identical to the acetone fragment — so each doubly bonded carbon carries two methyl groups: 2,3-dimethylbut-2-ene. Propene and but-1-ene would give aldehyde-containing mixtures.

12Terminal alkyne acidity and Lindlar reduction

Hydrogen on a triply bonded carbon is acidic enough to react with Na or NaNH₂, giving an acetylide salt and H₂. Ethyne and terminal alkynes (RC≡CH) show this; internal alkynes (RC≡CR') do not have that hydrogen. Acidity falls CH≡CH > CH₂=CH₂ > CH₃–CH₃, and among alkynes HC≡CH > CH₃C≡CH ≫ CH₃C≡CCH₃ — a distinction test for terminal versus non-terminal triple bonds.

Partial hydrogenation of an alkyne over Lindlar’s catalyst (palladised charcoal poisoned with sulfur compounds or quinoline) stops at the cis-alkene. Sodium in liquid ammonia instead gives the trans-alkene. Full hydrogenation (Pt/Pd/Ni, excess H₂) goes to the alkane.

Figure. Only a terminal alkyne loses ≡C–H to NaNH₂. Partial reduction of an internal alkyne is stereoselective: Lindlar gives cis; Na in liquid ammonia gives trans.

Terminal test and partial reduction

  1. Na / NaNH₂Gas or acetylide formation → acidic ≡C–H present → terminal (or ethyne). No reaction → internal alkyne or alkene/alkane.
  2. Lindlar + H₂Alkyne → cis-alkene; do not over-reduce.
  3. Na / liq. NH₃Alkyne → trans-alkene — the complementary stereochemical path.
Alkyne reductions and acidity
ReagentOn RC≡CH / RC≡CROutcome
Na or NaNH₂Terminal onlyAcetylide + H₂
H₂ / LindlarBoth (partial)cis-Alkene
Na / liq. NH₃Both (partial)trans-Alkene
H₂ / Pt (excess)Both (full)Alkane

Butynes with NaNH₂

Which of but-1-yne and but-2-yne reacts with NaNH₂ to liberate dihydrogen?

  • But-1-yneCH₃CH₂C≡CH — has ≡C–H
  • But-2-yneCH₃C≡CCH₃ — no ≡C–H
  • Reacts with NaNH₂but-1-yne only

Pro tip. The acidity test sorts terminal from internal; Lindlar sorts cis partial reduction from full hydrogenation.

Lindlar’s catalyst with H₂ converts but-2-yne mainly into
  1. Butane, because all π bonds are reduced
  2. cis-But-2-ene by partial hydrogenation
  3. trans-But-2-ene, which is Lindlar’s stereochemical specialty

Lindlar stops at the alkene and gives the cis isomer. Trans comes from Na/liq. NH₃; excess H₂ on active metal gives the alkane.

13Aromaticity and electrophilic aromatic substitution

A species is aromatic when it is cyclic, planar, fully conjugated around the ring, and holds (4n+2)\pi electrons (Hückel's rule) — benzene's six π electrons are the n=1 case. Antiaromatic systems are planar, cyclic and conjugated with 4n π electrons and are sharply destabilised; non-aromatic systems fail one of the geometric requirements and behave like ordinary polyenes.

Benzene's chemistry is electrophilic aromatic substitution (EAS), not the electrophilic addition alkenes prefer: addition would destroy the aromatic sextet, while substitution (nitration, halogenation, sulfonation, Friedel–Crafts) replaces H and restores aromaticity after the σ-complex. Activating +M/+I groups direct ortho/para; deactivating −M/−I groups (except halogens) direct meta.

Figure. Hückel aromaticity needs a cyclic, planar, fully conjugated ring with (4n+2) π electrons — benzene and C₅H₅⁻ both hold six. Four or eight π electrons fail the count (antiaromatic when planar and conjugated).

Hückel checklist

  1. GeometryCyclic, planar, and a p orbital on every atom of the ring (full conjugation).
  2. Count π electronsEach fully conjugated double bond contributes 2; a lone pair in a p orbital may contribute 2 if it is part of the conjugated circuit (furan, pyrrole).
  3. Match 4n+26, 10, 14… → aromatic. 4, 8, 12… with geometry OK → antiaromatic. Geometry fails → non-aromatic.
π counts you must recognise
Speciesπ electronsVerdict
Benzene6Aromatic (n=1)
Cyclobutadiene4Antiaromatic
Cyclooctatetraene8Non-aromatic (tub-shaped, not planar)
Cyclopentadienyl anion6Aromatic
Tropylium cation C₇H₇⁺6Aromatic
Naphthalene10Aromatic (n=2)

Hückel count for the cyclopentadienyl anion

Show that \mathrm{C}_5\mathrm{H}_5^- is aromatic.

  • Ring geometrycyclic, planar, every C conjugated
  • π from two C=C bonds4 electrons
  • Lone pair on the anionic carbon in the p system+2 electrons
  • Total π = 6 = 4(1)+2aromatic (n=1)

Pro tip. The cyclopentadienyl cation has only 4 π electrons and is antiaromatic — same ring, opposite charge, opposite verdict. Always count electrons after fixing the charge, not before.

Benzene reacts with electrophiles primarily by
  1. Electrophilic addition across one double bond, like an alkene
  2. Electrophilic aromatic substitution, preserving the aromatic sextet
  3. Free-radical substitution at a side-chain carbon only

Addition would give a non-aromatic cyclohexadiene adduct and costs the aromatic stabilisation. EAS proceeds through a transient σ-complex but expels H⁺ to restore the sextet. Side-chain radical substitution is toluene chemistry under UV, not the ring reaction under electrophilic conditions.

Notes

  • Electronic effects: the inductive effect (−I/+I through sigma bonds), resonance/mesomeric effect (+M/−M), and hyperconjugation (no-bond resonance) together govern stability and reactivity.
  • Carbocation stability: 3^\circ>2^\circ>1^\circ>CH_3^+ (hyperconjugation and +I), while allylic and benzylic cations are extra-stabilised by resonance.
  • IUPAC nomenclature: choose the longest chain containing the principal functional group, then assign the lowest locants to substituents.
  • Reaction types: alkanes undergo free-radical substitution (chlorination: initiation, propagation, termination), while alkenes and alkynes undergo electrophilic addition (Markovnikov's rule).
  • Aromaticity: a species is aromatic if it is cyclic, planar, fully conjugated and has (4n+2)\pi electrons (Hückel's rule); benzene undergoes electrophilic aromatic substitution.

Formulas

  • Degree of unsaturation =\frac{2C+2+N-H-X}{2}
  • Markovnikov: H adds to the carbon with more H's; peroxide effect (Kharasch) reverses it for HBr only
  • Hückel's rule: aromatic if (4n+2)\pi electrons
  • Carbocation stability: allyl/benzyl >3^\circ>2^\circ>1^\circ>CH_3^+
  • Baeyer's test: an alkene decolourises cold dilute KMnO_4 to a diol

Exam traps & shortcuts

  • More alpha-hydrogens means more hyperconjugation, so Saytzeff's rule favours the more substituted alkene in elimination.
  • Ozonolysis pinpoints the C=C position: reductive workup gives aldehydes/ketones showing where the double bond was.
  • +I groups (alkyl) stabilise carbocations but destabilise carbanions; −I groups do the reverse.

Reference tables

Three rankings that share hyperconjugation / resonance logic. Read across a row when a stem mixes classes (alkyl against allyl); read down a column when only substitution level changes.

Stability and selectivity ladders
LadderOrderWhat decides it
Carbocationallyl/benzyl > 3° > 2° > 1° > CH₃⁺Resonance first, then +I / hyperconjugation
Carbon radicalallyl/benzyl > 3° > 2° > 1° > CH₃·Same electronic logic; controls Br· selectivity
Alkene (Saytzeff)more substituted C=C favoured in eliminationMore α-H → more hyperconjugation

Same alkene, same HBr, opposite regiochemistry once peroxides appear. The intermediate type is the whole difference.

Markovnikov against Kharasch
ConditionsFirst addendIntermediateMajor product from propene
HBr, dark, no peroxideH⁺2° carbocation2-bromopropane
HBr, ROOR / lightBr·2° carbon radical1-bromopropane
HCl or HI + peroxidestill ionic in practicecarbocationMarkovnikov (no useful reversal)

Organic basics and hydrocarbons share this topic.

NCERT Unit 8–9 map
NCERT focusConcept
8.6 Structural isomerismstructural_isomerism_types
Geometrical isomerism (alkenes)geometrical_isomerism_cis_trans
9.2 Conformations of ethaneethane_conformations
9.2 Wurtzwurtz_reaction
9.4 Acidic alkynes; Lindlaralkyne_acidity_and_lindlar

Recap

Read only this the night before.

Three effects
±I through σ bonds, ±M through conjugation, hyperconjugation from α-C–H into an empty p or π. More α-H means more hyperconjugation — that single count ranks alkyl cations, radicals and Saytzeff alkenes.
Cation ladder
Allyl/benzyl > 3° > 2° > 1° > CH₃⁺. Count α-H only inside the alkyl series (9 / 6 / 3 / 0 for tert-butyl through methyl).
DU
(2C + 2 + N - H - X)/2. Oxygen silent; halogens count as H. C₆H₆ = 4 is the benzene alarm.
Name
Longest chain with the principal group; lowest locant to that group; alphabetical prefixes. Multipliers do not affect alphabetisation.
Radical chain
Initiation (X₂ → 2X·), propagation (X·/R· recycle), termination (radicals pair). Bromination ≫ chlorination in selectivity for 3°/2°.
Markovnikov
H⁺ first → more stable cation → Markovnikov. HBr + peroxide → Br· first → anti-Markovnikov. HCl/HI do not usefully reverse.
Locate C=C
Baeyer says a π bond is there; ozonolysis says where. Reductive workup keeps aldehydes; stitch carbonyl carbons to rebuild the alkene.
Aromatic
Cyclic, planar, conjugated, (4n+2)\pi. Benzene does EAS, not addition. C₅H₅⁻ and C₇H₇⁺ are the six-electron ions to recognise cold.
Structural isomers
Chain = skeleton; position = locant; functional = different group; metamerism about a heteroatom.
cis–trans
Needs two different groups on each =C. Cis more polar for but-2-ene; restricted rotation about C=C.
Conformations
Staggered ethane below eclipsed; barrier ~12.5 kJ mol⁻¹ — not isolable rotamers.
Wurtz
2 RX + Na (dry ether) → R–R. Mixed RX → three alkanes. No methane.
Terminal alkyne
≡C–H acidic (Na/NaNH₂). Lindlar → cis-alkene; Na/liq. NH₃ → trans.

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