What Are Alkenes
Alkenes are unsaturated hydrocarbons containing at least one carbon–carbon double bond. Because one double bond between two carbons costs the molecule two hydrogen atoms relative to the corresponding alkane, the general formula becomes $\ce{C_nH_{2n}}$. They are also called olefins (literally, oil-forming), a name that survives from the observation that the first member, ethene ($\ce{C2H4}$), forms an oily liquid on reaction with chlorine.
The first formal member, $\ce{CH2}$ (methene, obtained by setting $n=1$), has only a fleeting existence; the first stable member is ethene, $\ce{C2H4}$, known commonly as ethylene. Everything that follows in this subtopic — the geometry, the isomerism and the chemistry — flows from the single structural feature that defines the class: the $\ce{C=C}$ double bond.
Structure of the Double Bond
The carbon–carbon double bond is not two equivalent bonds. It consists of one strong sigma ($\sigma$) bond, formed by head-on overlap of two $sp^2$ hybridised orbitals, and one weaker pi ($\pi$) bond, formed by lateral (sideways) overlap of the two unhybridised $2p$ orbitals on the doubly bonded carbons. The $\sigma$ bond carries a bond enthalpy of about $397~\text{kJ mol}^{-1}$, while the $\pi$ bond is comparatively weak at about $284~\text{kJ mol}^{-1}$.
The combined double-bond enthalpy of roughly $681~\text{kJ mol}^{-1}$ exceeds that of a $\ce{C-C}$ single bond in ethane (about $348~\text{kJ mol}^{-1}$), and the double bond is shorter — about $134$ pm against $154$ pm for a single bond. Yet the loosely held, exposed $\pi$ electrons make alkenes behave as a source of mobile electron density, easily attacked by electron-seeking electrophilic reagents. This single fact governs almost all of alkene chemistry.
Orbital picture of the ethene $\ce{C=C}$ double bond.
The double bond = 1 $\sigma$ (in the molecular plane) + 1 $\pi$ (perpendicular $p$-overlap). The $\pi$ cloud is the reactive, easily attacked region.
Each $sp^2$ carbon is trigonal planar with bond angles close to $120^{\circ}$, and the four atoms attached to the two double-bonded carbons all lie in one plane. Crucially, the $\pi$ bond locks the geometry: free rotation about the $\ce{C=C}$ axis would require breaking the $\pi$ overlap, which does not happen at ordinary temperatures. This restricted rotation is what makes geometrical isomerism possible.
Nomenclature and Isomerism
In the IUPAC system, the longest carbon chain containing the double bond is chosen as the parent. Numbering starts from the end nearer the double bond, and the alkane suffix -ane is replaced by -ene, with a locant indicating the first doubly bonded carbon. Alkenes display both structural isomerism (chain and position) and, where the geometry permits, geometrical isomerism.
| Structure | IUPAC name | Isomerism note |
|---|---|---|
| $\ce{CH3-CH=CH2}$ | Propene | Single structure ($\ce{C3H6}$) |
| $\ce{CH2=CH-CH2-CH3}$ | But-1-ene | Position isomer of but-2-ene |
| $\ce{CH3-CH=CH-CH3}$ | But-2-ene | Shows cis–trans isomerism |
| $\ce{CH2=C(CH3)-CH3}$ | 2-Methylprop-1-ene | Chain isomer ($\ce{C4H8}$) |
| $\ce{CH2=CH-CH=CH2}$ | Buta-1,3-diene | Two double bonds (diene) |
Ethene ($\ce{C2H4}$) and propene ($\ce{C3H6}$) each have only one structure. From $\ce{C4H8}$ onward, multiple structures appear: but-1-ene and but-2-ene are position isomers (differing in the location of the double bond), while 2-methylprop-1-ene is a chain isomer of either.
Geometrical (Cis–Trans) Isomerism
Geometrical isomerism arises because rotation about the $\ce{C=C}$ bond is restricted. Each doubly bonded carbon completes its remaining two valences with two atoms or groups. If each of the two carbons carries two different groups (a $\ce{XYC=CXY}$ arrangement), two distinct spatial arrangements exist: the cis isomer, with identical groups on the same side of the double bond, and the trans isomer, with identical groups on opposite sides.
These are stereoisomers — same connectivity, different configuration — and they differ in physical properties such as melting point, boiling point, dipole moment and solubility. The cis form is generally more polar than the trans form. For but-2-ene, the cis isomer has a dipole moment of about $0.33$ D, whereas in trans-but-2-ene the two $\ce{C-CH3}$ bond dipoles point in opposite directions and cancel, making the molecule essentially non-polar ($\mu \approx 0$). For solids, the trans isomer typically has the higher melting point.
Cis and trans isomers of but-2-ene.
Same connectivity, different configuration. In cis the methyls are on the same side; in trans they are across the double bond, so their bond dipoles cancel.
No geometrical isomerism if one carbon has two identical groups
A common error is to assume every alkene shows cis–trans isomerism. It does not. $\ce{CH2=CBr2}$ and $\ce{(CH3)2C=CH-C2H5}$ have two identical groups on one of the doubly bonded carbons, so swapping sides changes nothing — no geometrical isomers exist. By contrast, $\ce{C6H5CH=CH-CH3}$ and $\ce{CH3CH=CClCH3}$ do show it.
Rule: geometrical isomerism requires each $\ce{C=C}$ carbon to carry two different groups.
Preparation of Alkenes
NCERT lists four standard routes to alkenes. Three of them are elimination reactions — they remove a small molecule (a hydrogen halide, a halogen acid as water, or a halogen) to create the double bond — while the first is a controlled partial reduction of an alkyne.
| Method | Reagents / conditions | Representative equation | Key point |
|---|---|---|---|
| From alkynes (partial reduction) | $\ce{H2}$, Lindlar's catalyst (Pd–C poisoned with quinoline / S); or Na in liq. $\ce{NH3}$ | $\ce{CH3-C#CH + H2 ->[Pd/C] CH3-CH=CH2}$ | Lindlar gives cis alkene; Na/liq. $\ce{NH3}$ gives trans |
| From alkyl halides (dehydrohalogenation) | Alcoholic KOH, heat ($\beta$-elimination) | $\ce{CH3CH2Br ->[alc.\,KOH][\Delta] CH2=CH2 + KBr + H2O}$ | Rate: I > Br > Cl; alkyl 3° > 2° > 1°. Follows Saytzeff |
| From vicinal dihalides (dehalogenation) | Zn metal | $\ce{CH2Br-CH2Br + Zn -> CH2=CH2 + ZnBr2}$ | Removes adjacent halogens as $\ce{ZnX2}$ |
| From alcohols (acidic dehydration) | Conc. $\ce{H2SO4}$, heat | $\ce{CH3CH2OH ->[conc.\,H2SO4][\Delta] CH2=CH2 + H2O}$ | $\beta$-elimination of water; follows Saytzeff with branched alcohols |
In dehydrohalogenation, the alkyl halide is heated with alcoholic potassium hydroxide (KOH dissolved in ethanol). One molecule of halogen acid is removed — a hydrogen leaves the $\beta$-carbon (the carbon next to the one bearing the halogen), hence it is a $\beta$-elimination. The rate increases with the polarisability of the halogen ($\ce{I} > \ce{Br} > \ce{Cl}$) and with the degree of substitution of the alkyl group (tertiary > secondary > primary).
Vicinal dihalides — those with halogens on two adjacent carbons — lose a molecule of $\ce{ZnX2}$ on treatment with zinc to regenerate the double bond. Acidic dehydration of alcohols with concentrated sulphuric acid likewise removes a $\beta$-hydrogen along with the $\ce{-OH}$ as water. In both elimination routes, when more than one alkene can form, the more substituted product dominates.
Saytzeff orientation: the more substituted alkene is the major product
When dehydrohalogenation (or dehydration) of an unsymmetrical substrate offers a choice of $\beta$-hydrogens, the hydrogen is preferentially removed so as to give the alkene with the greater number of alkyl substituents on the doubly bonded carbons. This more substituted alkene is more stable and is the major (preferred) product, while the less substituted alkene is minor.
Saytzeff: among possible alkenes, the most-substituted (most stable) $\ce{C=C}$ wins as the major product.
The orientation of $\ce{HX}$ and $\ce{H2O}$ addition is governed by carbocation stability. See Markovnikov and Anti-Markovnikov Addition for the full mechanism and the peroxide effect.
Physical Properties
As a class, alkenes resemble alkanes in physical properties, differing chiefly in isomerism and in their slight polar character. The first three members are gases, the next fourteen are liquids and higher members are solids. Ethene is a colourless gas with a faint sweet smell; the other alkenes are colourless and odourless, insoluble in water but fairly soluble in non-polar solvents such as benzene and petroleum ether.
Boiling point rises regularly with molecular size — each $\ce{-CH2-}$ unit added raises it — and, as with alkanes, straight-chain alkenes boil higher than their branched isomers because of greater surface contact between molecules.
Electrophilic Addition Reactions
The defining chemistry of alkenes is electrophilic addition: the exposed $\pi$ electrons attract electrophiles, the $\pi$ bond breaks, and two new $\sigma$ bonds form across the double bond. The product is a saturated addition compound. Several reagents add this way, and two of them double as classic tests for unsaturation.
| Reagent added | Conditions | Equation | Product / note |
|---|---|---|---|
| Dihydrogen, $\ce{H2}$ | Ni / Pd / Pt | $\ce{CH2=CH2 + H2 ->[Ni] CH3-CH3}$ | Gives the alkane (catalytic hydrogenation) |
| Halogen, $\ce{X2}$ (Br₂, Cl₂) | In $\ce{CCl4}$ | $\ce{CH2=CH2 + Br2 -> CH2Br-CH2Br}$ | Vicinal dihalide; reddish-orange Br₂ decolourised → test for unsaturation |
| Hydrogen halide, $\ce{HX}$ | Reactivity HI > HBr > HCl | $\ce{CH3-CH=CH2 + HBr -> CH3-CHBr-CH3}$ | Markovnikov: $\ce{X}$ to the carbon with fewer H atoms |
| Sulphuric acid, $\ce{H2SO4}$ | Cold conc. | $\ce{CH3-CH=CH2 + H2SO4 -> CH3-CH(OSO3H)-CH3}$ | Alkyl hydrogen sulphate; Markovnikov |
| Water, $\ce{H2O}$ | Few drops conc. $\ce{H2SO4}$ | $\ce{CH3-CH=CH2 + H2O ->[H+] CH3-CH(OH)-CH3}$ | Alcohol; Markovnikov (hydration) |
Addition of halogens proceeds through a cyclic halonium ion intermediate (studied in detail in higher classes). The disappearance of the reddish-orange colour of bromine in $\ce{CCl4}$ is the everyday laboratory test for a carbon–carbon double bond. Addition of $\ce{HX}$, $\ce{H2SO4}$ and $\ce{H2O}$ to unsymmetrical alkenes follows Markovnikov's rule: the negative part of the adding molecule attaches to the doubly bonded carbon bearing fewer hydrogen atoms, because that route passes through the more stable carbocation.
Q. Write the IUPAC names of the products formed when $\ce{HBr}$ adds to hex-1-ene (a) in the absence of peroxide and (b) in the presence of peroxide.
(a) Without peroxide, Markovnikov addition places $\ce{Br}$ on C-2 (fewer H atoms): the product is 2-bromohexane.
(b) With peroxide (Kharash effect, $\ce{HBr}$ only), addition is anti-Markovnikov, placing $\ce{Br}$ on the terminal C-1: the product is 1-bromohexane.
Oxidation and Ozonolysis
Alkenes are readily oxidised, and the outcome depends on the oxidant. With cold, dilute, aqueous potassium permanganate — Baeyer's reagent — the alkene is converted to a vicinal glycol (a 1,2-diol), and the purple permanganate colour is discharged. This decolourisation is a second standard test for unsaturation.
Under stronger conditions, acidic $\ce{KMnO4}$ or acidic $\ce{K2Cr2O7}$ cleaves the double bond, oxidising the alkene all the way to ketones and/or carboxylic acids depending on the substitution pattern. For example, but-2-ene is cleaved to two molecules of ethanoic acid:
$$\ce{CH3-CH=CH-CH3 ->[KMnO4/H+] 2\,CH3COOH}$$
Ozonolysis is the most diagnostically useful oxidation. Ozone first adds to the double bond to form an ozonide; the ozonide is then cleaved by $\ce{Zn-H2O}$ into smaller carbonyl compounds. Because the $\ce{C=C}$ bond breaks at exactly its original position, with each doubly bonded carbon becoming a carbonyl carbon, the carbonyl fragments reveal where the double bond was. This makes ozonolysis the standard tool for locating a double bond in an unknown alkene.
Reductive ozonolysis cleaves $\ce{C=C}$ into two carbonyl fragments.
Each doubly bonded carbon ends up as a carbonyl ($\ce{>C=O}$). Reading the carbonyl products backward reconstructs the parent alkene.
Reverse ozonolysis: rebuild the alkene from the carbonyl products
NEET frequently gives you the carbonyl products and asks for the parent alkene. The recipe: join the two carbonyl carbons by removing both oxygens and drawing a $\ce{C=C}$ in their place. If ozonolysis gives formaldehyde + 2-methylpropanal, joining $\ce{H2C=O}$ and $\ce{(CH3)2CH-CHO}$ at their carbonyl carbons gives $\ce{H2C=CH-CH(CH3)2}$, i.e. 3-methylbut-1-ene (NEET 2022). A methanal (formaldehyde) fragment always signals a terminal $\ce{=CH2}$ group.
Ozonolysis: $\ce{C=C}$ → two $\ce{C=O}$. To reverse, fuse the two carbonyl carbons with a double bond.
Polymerisation
Under high temperature, high pressure and a suitable catalyst, a large number of alkene (monomer) molecules combine into long-chain polymers. Ethene polymerises to polythene; propene polymerises to polypropene.
$$\ce{n\,CH2=CH2 ->[\text{high }T,\,P][\text{catalyst}] -(CH2-CH2)_n-}$$
These materials are used for plastic bags, squeeze bottles, pipes, toys, milk crates and moulded articles. Substituted dienes extend the same principle — for instance, chloroprene (2-chlorobuta-1,3-diene) polymerises to the synthetic rubber neoprene, a recurring NEET fact.
Alkenes in one screen
- General formula $\ce{C_nH_{2n}}$; at least one $\ce{C=C}$; also called olefins.
- $\ce{C=C}$ = 1 $\sigma$ ($sp^2$, ~397 kJ mol⁻¹) + 1 $\pi$ ($2p$, ~284 kJ mol⁻¹); shorter (~134 pm) and stronger than a single bond, but the loose $\pi$ electrons make it reactive toward electrophiles.
- Geometrical (cis–trans) isomerism needs each $\ce{C=C}$ carbon to bear two different groups; cis is more polar, trans usually higher-melting.
- Preparation: from alkynes (Lindlar → cis; Na/liq. $\ce{NH3}$ → trans), dehydrohalogenation of $\ce{R-X}$ (alc. KOH), dehalogenation of vicinal dihalides (Zn), and acidic dehydration of alcohols — eliminations follow Saytzeff.
- Electrophilic addition of $\ce{H2}$, $\ce{X2}$, $\ce{HX}$, $\ce{H2SO4}$, $\ce{H2O}$; $\ce{HX/H2O}$ follow Markovnikov. Br₂/CCl₄ and Baeyer's reagent both test for unsaturation.
- Ozonolysis ($\ce{O3}$ then $\ce{Zn/H2O}$) cleaves $\ce{C=C}$ to carbonyls, locating the double bond; polymerisation gives polythene, polypropene, neoprene.