Hydrocarbons
Alkanes, alkenes, alkynes and aromatic hydrocarbons for NEET
Alkanes, Alkenes and Alkynes
Alkanes: Saturated HydrocarbonsTopic 1
Hydrocarbons are compounds made of only carbon and hydrogen, and they are the framework of organic chemistry — the source of fuels, plastics and the carbon skeletons of biomolecules. They are classified as saturated (only single bonds, the alkanes), unsaturated (containing C=C or $\text{C}\equiv\text{C}$, the alkenes and alkynes), or aromatic (containing a benzene-type ring).
Alkanes have the general formula $\text{C}_n\text{H}_{2n+2}$, with every carbon $sp^3$ hybridised and tetrahedral. Because they have only strong, non-polar C–C and C–H single bonds, alkanes are relatively unreactive — their old name 'paraffins' means 'little affinity'. They are prepared by hydrogenation of alkenes/alkynes (with Ni/Pt catalyst), by reduction of alkyl halides, by Wurtz reaction (two alkyl halides + Na to give a symmetrical alkane), and by decarboxylation of carboxylic acid salts. NEET often asks you to predict the product of these named preparations.
The most important reaction of alkanes is free-radical substitution, especially halogenation. In sunlight or UV light, chlorine or bromine replaces hydrogen atoms one at a time through a chain mechanism with three stages: initiation (the halogen molecule splits homolytically into radicals), propagation (radicals abstract H and regenerate new radicals), and termination (two radicals combine). The reaction is not selective for chlorine and gives a mixture of mono- and polysubstituted products — a frequently tested point.
Alkanes also undergo combustion (the basis of their use as fuels, producing $\text{CO}_2$ and water with a large release of energy), controlled oxidation, isomerisation, and aromatisation (conversion of hexane and higher alkanes into benzene rings over a catalyst). Their physical properties follow clear trends: boiling point rises with chain length, and branching lowers the boiling point because branched molecules have less surface contact — a comparison NEET likes to test.
| Feature | Alkanes |
|---|---|
| Formula | $\text{C}_n\text{H}_{2n+2}$ |
| Hybridisation | $sp^3$, tetrahedral |
| Key reaction | free-radical substitution (halogenation) |
| Boiling point | rises with chain length, falls with branching |
Name the three steps in the free-radical chlorination of methane.
Show solution
Initiation ($\text{Cl}_2 \rightarrow 2\text{Cl}\cdot$ in UV light), propagation (Cl radical abstracts H, then the methyl radical reacts with $\text{Cl}_2$), and termination (two radicals combine).
Which has the higher boiling point: n-pentane or neopentane (2,2-dimethylpropane)?
Show solution
n-Pentane. It is unbranched, so its molecules have more surface contact and stronger van der Waals forces; the highly branched neopentane is more compact and boils lower.
The general formula of an alkane is:
Alkanes mainly undergo:
The Wurtz reaction is used to prepare:
Branching in an alkane causes its boiling point to:
The first step of free-radical halogenation is:
NEET tip: alkanes = substitution (free-radical, needs light, gives mixtures). Remember Wurtz (symmetrical alkane), decarboxylation, hydrogenation. Boiling point: longer chain higher, more branching lower.
Alkenes and Alkynes: Addition ReactionsTopic 2
Alkenes ($\text{C}_n\text{H}_{2n}$) contain one carbon–carbon double bond, with the doubly-bonded carbons $sp^2$ hybridised and planar. The pi electrons of the double bond are exposed and electron-rich, so alkenes are reactive towards electrophilic addition — the defining reaction of this topic. They are prepared by dehydrohalogenation of alkyl halides, dehydration of alcohols, and partial hydrogenation of alkynes.
Alkenes add many reagents across the double bond: hydrogen (catalytic hydrogenation to alkanes), halogens ($\text{Br}_2$ in $\text{CCl}_4$, whose decolourisation is the classic test for unsaturation), hydrogen halides (HX), and water (acid-catalysed hydration to alcohols). When an unsymmetrical reagent like HX adds to an unsymmetrical alkene, the product follows Markovnikov's rule: the hydrogen adds to the carbon already bearing more hydrogens, because this gives the more stable (more substituted) carbocation intermediate. Markovnikov problems are a NEET staple.
An important exception is the peroxide (Kharasch / anti-Markovnikov) effect: in the presence of organic peroxides, HBr (and only HBr) adds against Markovnikov's rule via a free-radical mechanism, placing bromine on the carbon with more hydrogens. This does not happen with HCl or HI — a detail NEET likes to test. Alkenes also undergo ozonolysis (cleaving the double bond to carbonyls, useful for locating the double bond) and oxidation with cold dilute $\text{KMnO}_4$ (Baeyer's reagent) to give diols.
Alkynes ($\text{C}_n\text{H}_{2n-2}$) contain a carbon–carbon triple bond with $sp$ hybridised, linear carbons. They undergo similar electrophilic additions (often twice, since there are two pi bonds), and add water in the presence of $\text{H}_2\text{SO}_4$/$\text{HgSO}_4$ to give carbonyl compounds. A distinctive feature is that terminal alkynes are weakly acidic: the hydrogen on the triply-bonded carbon can be removed by strong bases or by ammoniacal $\text{AgNO}_3$/$\text{Cu}_2\text{Cl}_2$, forming metal acetylides — a useful test to distinguish terminal from internal alkynes, and another common NEET question.
| Feature | Alkenes / Alkynes |
|---|---|
| Alkene formula | $\text{C}_n\text{H}_{2n}$ ($sp^2$) |
| Alkyne formula | $\text{C}_n\text{H}_{2n-2}$ ($sp$) |
| Key reaction | electrophilic addition |
| Unsaturation test | decolourises bromine water / Baeyer's reagent |
Predict the major product when HBr adds to propene ($\text{CH}_3\text{-CH=CH}_2$) in the absence of peroxide.
Show solution
By Markovnikov's rule, H goes to the terminal $\text{CH}_2$ (more H's) and Br to the middle carbon, giving 2-bromopropane ($\text{CH}_3\text{-CHBr-CH}_3$) via the more stable secondary carbocation.
How can you chemically distinguish 1-butyne (terminal) from 2-butyne (internal)?
Show solution
1-Butyne has an acidic terminal hydrogen, so it gives a white precipitate of silver acetylide with ammoniacal $\text{AgNO}_3$. 2-Butyne has no terminal H and gives no precipitate.
Alkenes characteristically undergo:
Markovnikov's rule predicts H adds to the carbon with:
The peroxide effect (anti-Markovnikov) is shown only by:
The general formula of an alkyne is:
A terminal alkyne gives a precipitate with:
NEET trap: Markovnikov gives the more substituted carbocation; only HBr + peroxide reverses it. Bromine-water decolourisation = unsaturation test. Terminal alkynes are acidic (acetylide test).
Aromatic Hydrocarbons
Benzene, Aromaticity and ResonanceTopic 3
Aromatic hydrocarbons (arenes) are built around the benzene ring. Benzene, $\text{C}_6\text{H}_6$, is a flat, regular hexagon of six $sp^2$ carbons, each carrying one hydrogen. All six C–C bonds are identical, intermediate in length between single and double bonds — a fact that puzzled early chemists and is explained by resonance: the three 'double bonds' are not fixed but delocalised as a ring of six pi electrons above and below the plane.
This delocalisation gives benzene unusual stability (its resonance/delocalisation energy is large), which is why benzene resists the addition reactions typical of alkenes and instead undergoes substitution, preserving the aromatic ring. The criteria for a compound to be aromatic are summarised by Hückel's rule: the molecule must be cyclic, planar, fully conjugated, and contain $(4n+2)$ pi electrons (where $n = 0, 1, 2, ...$). Benzene has $6$ pi electrons ($n=1$), so it is aromatic — a classification NEET often asks you to apply to other rings.
Benzene and its homologues are prepared by methods such as cyclic polymerisation of ethyne (three molecules of acetylene over hot tube give benzene), decarboxylation of aromatic acids, and reduction of phenol. The naming of substituted benzenes uses the ortho (1,2-), meta (1,3-) and para (1,4-) prefixes for disubstituted rings.
Aromatic compounds matter beyond the exam: many drugs, dyes and polymers contain benzene rings, and polynuclear (polycyclic) aromatic hydrocarbons such as benzo(a)pyrene — produced by incomplete combustion of tobacco, coal and fuels — are carcinogenic, a health link NEET sometimes references. Knowing the structure and stability of benzene, and being able to test aromaticity with Hückel's rule, sets up the substitution chemistry in the next topic.
| Property | Benzene |
|---|---|
| Formula / shape | $\text{C}_6\text{H}_6$, planar hexagon |
| Hybridisation | all $sp^2$ |
| Aromaticity (Hückel) | cyclic, planar, conjugated, $(4n+2)$ pi e⁻ |
| pi electrons | $6$ ($n=1$) — aromatic |
State Hückel's rule and use it to confirm benzene is aromatic.
Show solution
Hückel's rule: an aromatic ring is cyclic, planar, fully conjugated and has $(4n+2)$ pi electrons. Benzene is planar and conjugated with $6$ pi electrons, so $4n+2 = 6 \Rightarrow n = 1$ — it is aromatic.
Why does benzene undergo substitution rather than addition reactions?
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Addition would destroy the delocalised six-pi-electron system and its large stabilisation energy. Substitution replaces a hydrogen while keeping the aromatic ring intact, so it is energetically favoured.
The number of pi electrons in benzene is:
Hückel's rule for aromaticity requires:
All carbon atoms in benzene are:
Benzene's special stability is due to:
The 1,4-disubstituted benzene is named:
NEET tip: apply Hückel's $(4n+2)$ rule to test aromaticity — cyclic + planar + conjugated + $(4n+2)$ pi e⁻. Benzene resists addition (would break aromaticity), prefers substitution.
Electrophilic Substitution and Directing EffectsTopic 4
The characteristic reaction of benzene is electrophilic aromatic substitution (EAS), in which an electrophile replaces a ring hydrogen while the aromatic system is preserved. The general mechanism has two steps: the electrophile attacks the pi system to form a resonance-stabilised arenium ion (carbocation), then a proton is lost to restore aromaticity. Five named EAS reactions are essential for NEET.
These are: halogenation (with $\text{Cl}_2$ or $\text{Br}_2$ and a Lewis acid catalyst such as $\text{FeCl}_3$); nitration (with a mixture of concentrated $\text{HNO}_3$ and $\text{H}_2\text{SO}_4$, generating the electrophile $\text{NO}_2^+$); sulphonation (with fuming $\text{H}_2\text{SO}_4$, reversible); and the two Friedel–Crafts reactions — alkylation (with an alkyl halide and anhydrous $\text{AlCl}_3$) and acylation (with an acyl halide and $\text{AlCl}_3$, introducing a $\text{-COR}$ group). Knowing the reagent and the electrophile for each is a guaranteed NEET question.
When benzene already carries a substituent, that group controls where the next electrophile attacks — the directing effect. Ortho/para-directing groups are usually activating (they make the ring more reactive) and include $-\text{OH}, -\text{NH}_2, -\text{OR}, -\text{NHCOR}$, and alkyl groups; the halogens are an important exception — they are o/p-directing but deactivating. Meta-directing groups are deactivating and include $-\text{NO}_2, -\text{COOH}, -\text{CHO}, -\text{SO}_3\text{H}$, and $-\text{CN}$. Predicting the position of substitution from the existing group is one of the most frequently examined NEET skills.
The reason behind directing effects ties back to electronic effects from the previous chapter: activating o/p-directors donate electron density (by $+\text{M}$ or $+\text{I}$), stabilising the arenium ion when attack is at the ortho/para positions, while deactivating m-directors withdraw electron density, leaving the meta position relatively favoured. So the chemistry of substituted benzenes is a direct application of resonance and inductive reasoning — making this topic a high-yield, logic-based section rather than pure memorisation.
| Group type | Direction & activity |
|---|---|
| $-\text{OH}, -\text{NH}_2$, alkyl | ortho/para, activating |
| Halogens ($-\text{Cl}, -\text{Br}$) | ortho/para, deactivating |
| $-\text{NO}_2, -\text{COOH}, -\text{CHO}$ | meta, deactivating |
| Friedel–Crafts | alkylation / acylation with $\text{AlCl}_3$ |
Where will nitration occur on phenol, and why?
Show solution
The $-\text{OH}$ group is an activating, ortho/para director (it donates electron density by resonance). So nitration occurs mainly at the ortho and para positions, giving o- and p-nitrophenol.
Name the electrophile generated in the nitration of benzene.
Show solution
The mixture of concentrated $\text{HNO}_3$ and $\text{H}_2\text{SO}_4$ generates the nitronium ion $\text{NO}_2^+$, which is the attacking electrophile.
Benzene reacts mainly by:
The electrophile in nitration of benzene is:
Which is a meta-directing, deactivating group?
Friedel–Crafts alkylation uses the catalyst:
The $-\text{OH}$ group on benzene is:
NEET tip: learn the five EAS reactions + their electrophiles. o/p-directors are activating ($-\text{OH}, -\text{NH}_2$, alkyl) EXCEPT halogens (o/p but deactivating); m-directors ($-\text{NO}_2, -\text{COOH}$) are deactivating.
Quick Revision — Hydrocarbons
- Alkanes ($\text{C}_n\text{H}_{2n+2}$): saturated, $sp^3$, undergo free-radical substitution (halogenation); fairly unreactive ('paraffins').
- Alkenes ($\text{C}_n\text{H}_{2n}$): one C=C, $sp^2$; undergo electrophilic addition (of HX, $\text{H}_2\text{O}$, $\text{X}_2$).
- Markovnikov's rule: in HX addition, H adds to the carbon with more H's; the more stable carbocation forms. Peroxide (anti-Markovnikov) effect with HBr only.
- Alkynes ($\text{C}_n\text{H}_{2n-2}$): one $\text{C}\equiv\text{C}$, $sp$; addition + acidic terminal H.
- Aromatic hydrocarbons: benzene, planar, $6$ delocalised pi electrons (Hückel $4n+2$), extra stability.
- Benzene undergoes electrophilic substitution (nitration, halogenation, sulphonation, Friedel–Crafts).
- Directing groups: $-OH, -NH_2, -\text{alkyl}$ are o/p-directing & activating; $-NO_2, -COOH$ are m-directing & deactivating.
- Polynuclear aromatics (benzpyrene) in tobacco smoke are carcinogenic.
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