Hydrocarbons Formula Sheet — NEET Chemistry

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

Syllabus — topics coveredNCERT · 5 sub-topics

Alkanes — conformations, preparation and reactions
Alkenes — geometrical isomerism, electrophilic addition, Markovnikov rule, ozonolysis
Alkynes — acidic character and addition reactions
Aromatic hydrocarbons — benzene, aromaticity and electrophilic substitution
Directive influence of functional groups in monosubstituted 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.

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).

★ 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

The C=C bond as one sigma plus one pi, cis and trans geometrical isomers of but-2-ene, and the preparation routes by dehydrohalogenation, dehalogenation, dehydration and partial reduction of alkynes — $C = C$ structure, cis/trans isomers and preparation routes.

★ 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).

+ 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 (

\equivC -

H) hydrogen is ACIDIC (sp,

50%

s); acidity HC

\equiv

>

_{2} = C H_{2}

>

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.

🚫 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

Electrophilic aromatic substitution: E^+ adds to give the arenium (sigma-complex), which loses H^+ to re-aromatise. — 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

🚫 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.

More NEET Chemistry formula sheets

Frequently Asked Questions

What are the most important Hydrocarbons formulas for NEET?

This Hydrocarbons formula sheet covers all the high-yield Chemistry formulas, definitions and facts you need for NEET, across Alkanes — conformations, preparation and reactions, Alkenes — geometrical isomerism, electrophilic addition, Markovnikov rule, ozonolysis, Alkynes — acidic character and addition reactions, Aromatic hydrocarbons — benzene, aromaticity and electrophilic substitution, Directive influence of functional groups in monosubstituted benzene — each shown with the key result and, where useful, a worked example.

Is this Hydrocarbons formula sheet free?

Yes — the full chapter formula sheet is free to read online, no login or payment required.

How should I revise Hydrocarbons for NEET?

Blurt the Hydrocarbons key points from memory, then check against this sheet to find your gaps — and practise a few previous-year questions on the chapter to make sure you can apply them under time pressure.

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