Hydrocarbons Formula Sheet — JEE Main Chemistry

Every key Hydrocarbons formula, definition and theorem for JEE Main Chemistry in one place — with common examiner traps and worked examples. Free to read; blurt from memory, then check your gaps.

Syllabus — topics coveredNTA · 12 sub-topics

Classification, isomerism, IUPAC nomenclature
General methods of preparation, properties and reactions
Alkanes - Conformations: Sawhorse and Newman projections
Mechanism of halogenation of alkanes
Alkenes - Geometrical isomerism, mechanism of electrophilic addition
Addition of hydrogen, halogens, water, hydrogen halides
Markownikoff's and peroxide effect, Ozonolysis and polymerization
Alkynes - Acidic character, addition of hydrogen, halogens, water and hydrogen halides
Aromatic hydrocarbons - Nomenclature, benzene structure and aromaticity
Mechanism of electrophilic substitution - halogenation, nitration
Friedel-Craft's alkylation and acylation
Directive influence of functional group in mono-substituted benzene

Alkanes — Preparation, Conformations & Halogenation

Alkanes: Saturated open-chain hydrocarbons, general formula

C_{n} H_{2n + 2}

, with only

s p^{3}

C–C and C–H

σ

bonds (tetrahedral,

109. 5^{\circ}

). Once called

paraffins

— chemically inert, so their main reactions are radical substitution and combustion.

Ethane in sawhorse and Newman projections: the staggered conformation (most stable) above and the eclipsed conformation (least stable, barrier about 12.5 kJ/mol) below. — Sawhorse and Newman projections of ethane: staggered (stable) versus eclipsed.

Preparation

▸ $Hydrogenation$ : $C = C$ or $C \equiv C + H_{2} Pt / P d / N i$ alkane.
▸ $Wurtz$ : $2 R - X + 2 Na dry ether R - R$ — gives $even$ -carbon alkanes.
▸ $Decarboxylation$ : $RCOONa + NaOH CaO RH$ (one C less).
▸ $Kolbe electrolysis$ : aqueous $RCOONa$ gives $R - R$ at the anode ( $2 C H_{3} COONa \to C_{2} H_{6}$ ). Methane cannot be made this way.
▸ $Reduction of alkyl halides$ : $R - X + H_{2} Zn, H^{+} R - H + HX$ .
▸ $Other reactions$ : $aromatisation$ (n-hexane $C r_{2} O_{3}, 773 K$ benzene), $isomerisation$ ( $AlC l_{3} /HCl$ ), $pyrolysis$ (cracking) and combustion.

★ Remember · Free-radical halogenation

A chain mechanism:

initiation

(

C l_{2} h ν 2 C l^{∙}

propagation

(

C l^{∙} + C H_{4} \to C H_{3}^{∙} + HCl

;

C H_{3}^{∙} + C l_{2} \to C H_{3} Cl + C l^{∙}

termination

(two radicals combine). Reactivity

F_{2} > C l_{2} > B r_{2} > I_{2}

; H abstraction

3^{\circ} > 2^{\circ} > 1^{\circ}

🎯 Exam · Conformations of ethane

Rotation about the C–C

σ

bond gives infinite conformers.

Staggered

(H's far apart) has the least torsional strain — most stable;

eclipsed

has the most. The barrier is only ~

12.5 kJ mo l^{- 1}

, so rotation is nearly free.

⚠️ Watch out · Branching lowers boiling point

b.p. rises with molecular mass (more van der Waals contact), but among isomers a

branched

alkane is more spherical, has less surface contact, and boils

lower

than its straight-chain isomer.

🚫 Examiner Trap · Examiner traps

(1)

Wurtz

gives only EVEN-carbon, symmetrical alkanes (R

-

R). (2) Halogenation reactivity F

_{2} > C l_{2} > B r_{2} > I_{2}

and H-abstraction

3^{\circ} > 2^{\circ} > 1^{\circ}

. (3)

Staggered

ethane is most stable (least torsional strain). (4) Among isomers, BRANCHING LOWERS the boiling point (less surface contact).

Alkenes I — Structure, Isomerism & Preparation

Alkenes (olefins): Unsaturated hydrocarbons with at least one

C = C

, general formula

C_{n} H_{2n}

. The double bond is 1

σ

+ 1

π

; the loosely-held

π

cloud makes alkenes electron-rich and attacked by

electrophiles

cis-but-2-ene (both methyls on one side, non-zero dipole) and trans-but-2-ene (methyls opposite, zero dipole). — Geometrical isomers of but-2-ene: cis ( $μ \neq = 0$ ) and trans ( $μ = 0$ ).

★ Remember · Geometrical (cis / trans) isomerism

Rotation about

C = C

restricted

, so when each doubly-bonded carbon carries two

different

groups,

cis

(like groups same side) and

trans

(opposite) differ. cis is more polar (

μ \neq = 0

); trans usually has the higher melting point.

Preparation (β-eliminations & reductions)

▸ $Dehydrohalogenation$ : alkyl halide $alc. KOH$ alkene (Saytzeff — the more substituted alkene predominates).
▸ $Dehalogenation$ : vicinal dihalide $+ Zn \to$ alkene $+ Zn X_{2}$ .
▸ $Dehydration$ of alcohol: conc. $H_{2} S O_{4}$ , $Δ$ (also β-elimination).
▸ $Partial reduction$ of an alkyne: Lindlar's catalyst $\to$ cis; $Na/liq. N H_{3} \to$ trans.

🎯 Exam · Structural isomerism too

C_{4} H_{8}

alkenes show both

chain

and

position

isomerism (but-1-ene, but-2-ene, 2-methylprop-1-ene) — count these before adding geometrical isomers.

🚫 Examiner Trap · Examiner traps

(1) cis/trans needs restricted C

=

C rotation AND each alkene carbon carrying two DIFFERENT groups. (2) cis is more polar (

μ \neq = 0

); trans usually has the higher m.p. (3) Dehydrohalogenation follows

Saytzeff

(more-substituted alkene). (4) Lindlar

\to

cis-alkene; Na/liq.N

H_{3}

\to

trans-alkene.

Alkenes II — Electrophilic Addition

★ Remember · Markovnikov's rule

In adding

HX

to an unsymmetrical alkene, the

H goes to the carbon with more H’s

and X to the other. The real reason: addition goes through the

more stable carbocation

—

C H_{3} CH = C H_{2} + HBr \to C H_{3} CHBrC H_{3}

(via a

2^{\circ}

cation).

🎯 Exam · Peroxide (Kharasch) effect

With a peroxide,

HBr

adds

anti-Markovnikov

(propene

\to

1-bromopropane) by a

free-radical

chain through the more stable radical. It works for

HBr

only

— not HCl or HI.

+ H_{2}

Ni/Pt/Pd

\to

alkane

+ X_{2}

vicinal dihalide (

B r_{2} /CC l_{4}

decolourises

=

test)

+ HX

>

HBr

>

HCl, Markovnikov

+ H_{2} O

dil.

H_{2} S O_{4} \to

alcohol (Markovnikov)

⚠️ Watch out · Why the carbocation matters

Markovnikov 'memorising H-to-more-H' can mislead — always check which intermediate carbocation is more stable (

3^{\circ} > 2^{\circ} > 1^{\circ}

). With rearrangement-prone substrates, the stable cation, not the rule of thumb, decides the product.

🚫 Examiner Trap · Examiner traps

(1) Markovnikov: H to the carbon with MORE H — really because it goes through the MORE STABLE carbocation. (2)

Peroxide effect

(anti-Markovnikov) works for

HBr ONLY

, not HCl/HI (free-radical chain). (3) B

r_{2}

/CC

l_{4}

decolourisation

=

unsaturation test. (4) Watch for carbocation REARRANGEMENT — the stable cation decides the product, not the rule of thumb.

Alkenes III — Oxidation, Ozonolysis & Polymerisation

★ Remember · Ozonolysis — locate the double bond

O_{3}

adds to the alkene to give an

ozonide

, which

Zn/ H_{2} O

cleaves into

two carbonyl

compounds. The carbonyls reveal exactly where the

C = C

was — e.g.

C H_{3} CH = C H_{2} \to C H_{3} CHO + HCHO

🎯 Exam · Baeyer's test (unsaturation)

Cold dilute

KMn O_{4}

(Baeyer's reagent) gives a

vicinal diol

(glycol) and its purple-pink colour

fades

— a standard test for

C = C

. Hot/acidic

KMn O_{4}

instead

cleaves

the double bond to acids/ketones.

Polymerisation

n C H_{2} = C H_{2} high T, P, cat. - (C H_{2} - C H_{2}) -_{n}

Many monomers join into a polymer — ethene → polythene, propene → polypropene.

Two oxidation outcomes

▸ $Mild$ (cold dil. $KMn O_{4}$ ): $C = C$ kept, adds two $- OH$ → glycol.
▸ $Vigorous$ (hot acidic $KMn O_{4}$ / $K_{2} C r_{2} O_{7}$ ): $C = C$ broken → ketones / carboxylic acids.
▸Ozonolysis is the structure-determination tool; combustion gives $C O_{2} + H_{2} O$ .

🚫 Examiner Trap · Examiner traps

(1)

Ozonolysis

(

O_{3}

then Zn/

H_{2}

O) cleaves C

=

C into TWO carbonyls — it locates the double bond. (2)

Baeyer’s

(cold dilute KMn

O_{4}

) gives a diol and the purple colour FADES; HOT/acidic KMn

O_{4}

instead CLEAVES to acids/ketones. (3) Don't confuse the two KMn

O_{4}

outcomes. (4) Addition polymerisation needs high T/P/catalyst.

Alkynes — Acidity, Preparation & Addition

Alkynes: Unsaturated hydrocarbons with a

C \equiv C

, general formula

C_{n} H_{2n - 2}

. The triple bond is 1

σ

+ 2

π

(sp, linear,

18 0^{\circ}

120

pm). Ethyne (acetylene) is the first member.

★ Remember · Acidic terminal hydrogen

The

\equiv C - H

carbon is sp (

50%

s-character), so it holds the bonding pair tightly and the H is

acidic

HC \equiv CH + Na \to HC \equiv C^{-} N a^{+} + \frac{1}{2} H_{2}

. Acidity order

HC \equiv CH > C H_{2} = C H_{2} > C H_{3} C H_{3}

— used to distinguish terminal alkynes.

Preparation

▸ $Ca C_{2} + 2 H_{2} O \to Ca (OH)_{2} + C_{2} H_{2}$ (industrial).
▸Vicinal dihalide $alc. KOH$ alkenyl halide $NaN H_{2}$ alkyne.
▸Cyclic polymerisation: $3 C_{2} H_{2} 873 K$ benzene (aliphatic → aromatic).

Tests for a terminal alkyne

▸Ammoniacal $AgN O_{3}$ (Tollens') gives a $white$ precipitate of silver acetylide.
▸Ammoniacal $C u_{2} C l_{2}$ gives a $red$ precipitate of copper(I) acetylide.
▸But-2-yne has no $\equiv C - H$ , so it gives neither — this separates it from but-1-yne.

+ H_{2}

Pt/Pd/Ni

\to

alkene

\to

alkane

+ X_{2}

+ HX

tetrahalide / gem-dihalide (Markovnikov)

+ H_{2} O

H g^{2 +} / H^{+}

, 333 K → carbonyl (Kucherov)

🎯 Exam · Kucherov hydration

Water adds across

C \equiv C

(

H g^{2 +} / H^{+}

) to give an unstable

enol

that tautomerises to a carbonyl: ethyne

\to

ethanal (

C H_{3} CHO

); propyne

\to

propanone.

🚫 Examiner Trap · Examiner traps

(1) The terminal (

\equiv C - H

) hydrogen is ACIDIC (sp,

50%

s); acidity

HC \equiv CH > C H_{2} = C H_{2} > C H_{3} C H_{3}

— used to distinguish terminal alkynes (acetylide test). (2) Only TERMINAL alkynes form acetylides. (3)

Kucherov

(

H_{2}

O, H

g^{2 +}

^{+}

): enol tautomerises to a carbonyl (ethyne

\to

ethanal). (4) HX adds Markovnikov to give a gem-dihalide.

Aromatic Hydrocarbons — Benzene, Aromaticity & Resonance

Benzene & arenes: Benzene (

C_{6} H_{6}

) is the parent

aromatic

hydrocarbon (arene): a flat ring of six

s p^{2}

carbons, six C–C

σ

and six C–H

σ

bonds, with each carbon's p orbital overlapping into a

delocalised

π

system.

Benzene as a resonance hybrid of two Kekulé forms (six delocalised pi electrons). — Benzene as a resonance hybrid of two Kekulé forms (six delocalised $π$ electrons).

★ Remember · Resonance & stability

Benzene is a

resonance hybrid

of the two Kekulé structures (drawn as a circle for the 6 delocalised

π

electrons). X-ray shows all six C–C bonds equal at

139 pm

(between single

154

and double

133

). This delocalisation gives extra stability, so benzene prefers

substitution over addition

🎯 Exam · Aromaticity — the Hückel rule

A ring is aromatic if it is (i)

planar

, (ii) fully

conjugated/delocalised

, and (iii) has

(4 n + 2)

π

electrons (

n = 0, 1, 2, \dots

). Benzene:

n = 1

6 π

. Naphthalene:

n = 2

10 π

. Anthracene:

n = 3

14 π

Preparation of benzene

▸ $Cyclic polymerisation$ of ethyne (red-hot iron, 873 K).
▸ $Decarboxylation$ of sodium benzoate with soda lime.
▸ $Reduction of phenol$ vapour over heated Zn dust.

⚠️ Watch out · Aromatic ≠ 'three double bonds'

Benzene's three 'double bonds' are not localised — that hypothetical cyclohexatriene would readily add. It is the

delocalised

(4 n + 2) π

system, not the formal double bonds, that confers aromatic stability.

Species	Ring $π$ e $^{-}$	Verdict
cyclopropenyl cation	2	aromatic ( $n = 0$ )
cyclopentadienyl anion	6	aromatic
tropylium (cycloheptatrienyl) cation	6	aromatic
cyclopentadienyl cation	4	antiaromatic
cyclobutadiene	4	antiaromatic
cyclooctatetraene	8	non-aromatic (tub-shaped, not planar)
cyclopentadiene / cycloheptatriene	—	non-aromatic (one $s p^{3}$ C breaks the ring)

🚫 Examiner Trap · Examiner traps

(1) Benzene is ONE resonance hybrid (all C–C

= 139

pm), NOT alternating single/double bonds. (2) Aromatic

=

planar

+

fully conjugated

+

(4 n + 2) π

electrons (benzene

6 π

, naphthalene

10 π

). (3) The delocalisation makes benzene prefer

substitution over addition

. (4) Count

π

electrons in the RING only for the Hückel rule.

Aromatic — Electrophilic Substitution & Directive Influence

Benzene plus E+ gives the arenium ion, whose top carbon is sp3 and carries both H and E, with the positive charge delocalised over the other five carbons. Loss of H+ gives the substituted benzene with E bonded to the ring. — Electrophilic aromatic substitution: $E^{+}$ adds to give the arenium ( $σ$ -complex), which loses $H^{+}$ to re-aromatise.

★ Remember · EAS mechanism (3 steps)

(a)

Generate the electrophile

E^{+}

(a Lewis acid like

AlC l_{3}

Fe X_{3}

makes

C l^{+}

R^{+}

, acylium;

H_{2} S O_{4}

makes

N O_{2}^{+}

). (b)

E^{+}

adds to the ring → resonance-stabilised

arenium ion

(

σ

-complex, one

s p^{3}

C). (c)

Lose H^{+}

to restore aromaticity.

Nitration

conc.

HN O_{3}

H_{2} S O_{4}

, 323–333 K (

N O_{2}^{+}

)

Halogenation

X_{2}

/anhyd.

Fe X_{3}

Sulphonation

fuming

H_{2} S O_{4}

(oleum)

F–C alkylation / acylation

RX

RCOCl

with anhyd.

AlC l_{3}

🎯 Exam · Ortho/para directors (mostly activating)

Electron-

donating

groups raise the

o / p

electron density and speed up substitution:

- OH

- N H_{2}

- NHR

- NHCOC H_{3}

- OR

- R

Exception

: halogens (

- Cl

- Br

) are

o / p

-directing but

deactivating

(strong

- I

★ Remember · Meta directors (deactivating)

Electron-

withdrawing

groups lower the ring density and slow substitution; the m-position is least deactivated, so the electrophile attacks there:

- N O_{2}

- CN

- CHO

- COR

- COOH

- COOR

- S O_{3} H

Friedel–Crafts limits

▸ $Alkylation$ over-reacts: the alkyl product is more activated, so polyalkylation follows. The carbocation can also $rearrange$ (1-chloropropane gives isopropylbenzene).
▸ $Acylation$ stops at one group, because the ketone deactivates the ring, and it does not rearrange.
▸It fails on strongly deactivated rings ( $- N O_{2}$ ) and on aniline, whose $- N H_{2}$ ties up $AlC l_{3}$ .

🚫 Examiner Trap · Examiner traps

(1) EAS goes via a resonance-stabilised

arenium

(

σ

-complex, one

s p^{3}

C); the slow step is its formation, then H

^{+}

is lost. (2)

o / p

-directors are EDG (activating) — but halogens are

o / p

-directing yet

deactivating

(strong

- I

). (3) m-directors are EWG (deactivating: N

O_{2}

, CN, CHO, COOH). (4) Friedel–Crafts fails on strongly deactivated (e.g. nitrobenzene) rings.

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What are the most important Hydrocarbons formulas for JEE Main?

This Hydrocarbons formula sheet covers all the high-yield Chemistry formulas, definitions and theorems you need for JEE Main, across Classification, isomerism, IUPAC nomenclature, General methods of preparation, properties and reactions, Alkanes - Conformations: Sawhorse and Newman projections, Mechanism of halogenation of alkanes, Alkenes - Geometrical isomerism, mechanism of electrophilic addition — each shown with the key result and, where useful, a worked example.

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