Naming of Alkenes

ChemistryHydrocarbonsFor NEET aspirants

Alkenes are unsaturated hydrocarbons that contain at least one carbon to carbon double bond, with the general formula C_nH_2n for a single double bond. Naming of alkenes follows the IUPAC pattern of picking the longest chain through the C=C, numbering from the end nearer the double bond, and adding $E$ / $Z$ or cis/trans where geometrical isomerism exists. The doubly bonded carbons are sp $^{2}$ hybridised and planar, which is exactly why alkenes are both reactive and capable of showing geometrical isomerism.

Key Formulas - Quick Reference

General formula (one double bond): C_nH_2n
Degree of unsaturation: $DU = \frac{2 n + 2 - H}{2}$ , where $n$ is the number of carbons
C=C bond length $= 1.34$ Å, bond energy $\approx 146$ kcal/mol (about 615 kJ/mol)
Hybridisation of alkene carbon: sp $^{2}$ , s-character $33%$ , bond angle near $12 0^{\circ}$
Stability order: $R_{2} C=CR_{2} > R_{2} C=CHR > R_{2} C=CH_{2} \approx RCH=CHR > RCH=CH_{2} > CH_{2} =CH_{2}$
Stability $\propto \frac{1}{Δ H _{hydrogenation}}$ for isomeric alkenes giving the same alkane

1. What Are Alkenes?

Alkenes are hydrocarbons containing at least one carbon to carbon double bond. They are also called olefins, from olefiant gas meaning oil forming gas, because the first member ethene forms an oily liquid when treated with halogens.

Alkenes are among the most important industrial organic compounds. Ethene is the largest volume industrial organic chemical in the world, used to make polyethylene and hundreds of downstream consumer chemicals. Many alkenes also occur naturally in plants and animals.

Definition: An alkene is an unsaturated hydrocarbon containing one or more C=C double bonds. With exactly one double bond and no ring, its molecular formula is C_nH_2n.

2. Structure and Bonding in Alkenes

Each doubly bonded carbon uses three sp $^{2}$ hybrid orbitals to make three $σ$ bonds, which point to the corners of an equilateral triangle. The leftover unhybridised p orbital on each carbon stands perpendicular to that triangle. When the two p orbitals overlap side-on, the $π$ bond forms.

Figure 1. Mixing one

s

with two

p

orbitals gives three sp

^{2}

hybrids at

12 0^{\circ}

, and leaves one untouched

p

orbital perpendicular to them. That leftover

p

orbital is what later forms the

π

bond.

Figure 2. The

σ

framework of ethene is planar (sp

^{2}

); side-on overlap of the leftover

p

orbitals gives the

π

bond.

The C=C bond consists of one strong $σ$ bond and one weaker $π$ bond.
All six atoms of ethene lie in a single plane; the molecule is flat.
The $π$ electron cloud lies above and below that plane, loosely held and easily attacked by electrophiles.
Rotation about the C=C is blocked, because rotating would break the side-on p overlap.

Figure 3. Geometry of ethene. The angle facing the double bond,

121. 2^{\circ}

, is larger than the H-C-H angle,

117. 6^{\circ}

, because the

π

cloud repels the C-H bonds more strongly.

Why the two angles are not equal

The H-C=C angle is $121. 2^{\circ}$ while the H-C-H angle is only $117. 2^{\circ}$ . By VSEPR reasoning, the fat $π$ cloud of the double bond repels the neighbouring C-H bonds more strongly than one C-H bond repels another, so the angle involving the double bond opens up.

Figure 4. Edge-on view. The

π

cloud sits above and below the plane of the molecule; the plane itself is a node where

π

density is zero.

Note: Because the

π

electrons are farther from the nuclei and more loosely held than

σ

electrons, the

π

bond is the nucleophilic site of the molecule. Almost every reaction of an alkene begins with an electrophile attacking this cloud.

3. General Formula and Degree of Unsaturation

A single double bond removes two hydrogens compared with the corresponding alkane, so C_nH_2n+2 becomes C_nH_2n. Each double bond or ring contributes one degree of unsaturation (DU), and a triple bond contributes two.

$DU = \frac{2 n + 2 - H}{2}$

For C₅H₁₀: $DU = \frac{2 ( 5 ) + 2 - 10}{2} = 1$ , so the compound is either an alkene or a cycloalkane.

4. IUPAC Nomenclature of Alkenes

Pick the parent chain. Choose the longest continuous chain that contains the double bond, even if a longer chain exists elsewhere in the molecule.
Number the chain. Start from the end that gives the double bond the lower locant. The double bond outranks alkyl branches.
Name the parent. Replace the -ane of the alkane with -ene, and place the locant of the first doubly bonded carbon just before it, as in pent-2-ene.
Add substituents. List them alphabetically with their locants as prefixes.
Add stereochemistry. Prefix $E$ or $Z$ (or cis/trans) in italics inside brackets when geometrical isomerism exists.

Figure 5. Number the chain from the end nearer the double bond. Here that gives pent-2-ene, not pent-3-ene.

Worked naming examples

Structure	IUPAC name	Common name
CH₂=CH₂	Ethene	Ethylene
CH₃CH=CH₂	Propene	Propylene
CH₃CH=CHCH₃	But-2-ene	-
(CH₃)₂C=CH₂	2-Methylprop-1-ene	Isobutylene
CH₂=CH-CH=CH₂	Buta-1,3-diene	Divinyl
CH₂=CH-CH₂-Cl	3-Chloroprop-1-ene	Allyl chloride

Naming tip: For dienes and trienes the ending becomes -adiene and -atriene, and the parent keeps its full stem: buta-1,3-diene, not but-1,3-diene.

5. Isomerism in Alkenes

(a) Structural isomerism

Chain isomerism: pent-1-ene and 3-methylbut-1-ene share C₅H₁₀ but differ in skeleton.
Position isomerism: but-1-ene and but-2-ene differ only in where the double bond sits.
Functional isomerism (ring-chain): propene and cyclopropane are both C₃H₆.

(b) Geometrical isomerism

Because the C=C cannot rotate, substituents are locked on one face or the other. Two conditions must both hold:

There must be restricted rotation, which the double bond supplies.
Neither doubly bonded carbon may carry two identical groups.

Figure 6. The parent chain must contain the C=C. The double bond locant is fixed first, so this is 4-methylpent-2-ene.

Figure 7. Rotating about C=C twists the two

p

orbitals out of alignment and destroys the side-on overlap that makes the

π

bond. The cost, about

264

kJ/mol, is far beyond room-temperature energy, so cis and trans forms are separate, isolable compounds.

Figure 8. Restricted rotation about C=C makes cis-but-2-ene and trans-but-2-ene separable compounds, not conformers. Note that cis boils higher but trans melts higher.

Figure 9. If either alkene carbon carries two identical groups, cis-trans isomerism disappears.

(c) E-Z nomenclature (CIP rules)

When all four substituents differ, cis and trans become meaningless. Rank the two groups on each alkene carbon by Cahn-Ingold-Prelog priority, which is decided by atomic number at the first point of difference.

Higher priority groups on opposite sides: E (from German entgegen, opposite).
Higher priority groups on the same side: Z (from zusammen, together).
Duplicate atoms for multiple bonds: a C=O counts as a carbon bonded to two oxygens.

Figure 10. Rank the two groups on each alkene carbon. Higher priorities on the same side gives

Z

; on opposite sides gives

E

6. Polyenes: Isolated, Conjugated and Cumulated

When a molecule holds more than one double bond, the spacing between them matters more than the count.

Figure 11. Apply the CIP rules strictly in order and stop at the first point of difference. Atomic numbers are compared, never added.

Figure 12. Stability order is conjugated

>

isolated

>

cumulated. Note that the central carbon of an allene is sp, so those three carbons are collinear.

Type	Arrangement	Example	Stability
Conjugated	Alternating single and double bonds	Buta-1,3-diene	Highest (delocalisation)
Isolated	Separated by two or more sp³ carbons	Penta-1,4-diene	Intermediate
Cumulated	Both bonds on one central carbon	Propa-1,2-diene (allene)	Lowest (strained, sp centre)

7. Relative Stability of Alkenes

Alkene stability rises steadily with the number of alkyl groups attached to the doubly bonded carbons. Two effects cooperate: the plus inductive effect of alkyl groups feeds electron density into the electron-poor sp $^{2}$ carbons, and hyperconjugation delocalises the $α$ C-H bonding electrons into the $π$ system.

Figure 13. More alkyl groups on the C=C means a more stable alkene, because of hyperconjugation and the electron-releasing inductive effect.

Figure 14. Stability tracks the number of

α

-hydrogens available for hyperconjugation with the

π

bond.

Measuring stability: heats of hydrogenation

Hydrogenation is exothermic, so $Δ H$ is negative. Isomeric alkenes that give the same alkane can be ranked directly: the one releasing the least heat began at the lowest energy and is therefore the most stable.

Figure 15. All three butenes give the same butane, so the alkene releasing the least heat,

Δ H = - 27.6

kcal/mol, started lowest and is the most stable.

Alkene	$Δ H_{hydrogenation}$ (kcal/mol)	Comment
Ethene	32.8	Least stable, no alkyl groups
Propene	30.1	Monosubstituted
But-1-ene	30.3	Monosubstituted
3-Methylbut-1-ene	30.23	Still monosubstituted at the C=C
cis-But-2-ene	28.6	Disubstituted, some steric strain
trans-But-2-ene	27.6	Disubstituted, strain relieved
2-Methylprop-1-ene	27.2	Disubstituted (1,1-pattern)
2-Methylbut-2-ene	26.9	Trisubstituted, most stable here

Remark: Heats of combustion can be compared the same way, but only for isomers, since all four butenes burn to the same 4CO₂ and 4H₂O. Comparing heats of combustion across different formulas is meaningless.

8. Alkanes, Alkenes and Alkynes Compared

Property	Alkane	Alkene	Alkyne
Bond length (Å)	1.54 (C-C)	1.34 (C=C)	1.20 (C $\equiv$ C)
Bond energy (kJ/mol)	415	615	835
Hybridisation	sp³	sp²	sp
s character	25%	33%	50%
p $K_{a}$	50	44	25
Shape at carbon	Tetrahedral	Planar	Linear
Rate of electrophilic addition	Does not react	Faster	Slower
General formula	C_nH_2n+2	C_nH_2n	C_nH_2n-2

Solved Examples

Solved Example 1

Write the IUPAC name of the compound in which a but-2-enyl group is attached to carbon 1 of a cyclohexene ring.

Solution:

The ring carries the double bond, so cyclohexene is the parent. The ring carbon bearing the substituent is numbered 1, and the double bond starts there. The substituent chain CH₂-CH=CH-CH₃ is a but-2-enyl group.

Name: 1-(but-2-enyl)cyclohex-1-ene.

Solved Example 2

Assign the configuration (

E

Z

) to 3-methylpent-2-ene, CH₃-CH=C(CH₃)-CH₂CH₃, drawn with the methyl on C2 and the ethyl on C3 on the same side.

Solution:

On C2 the groups are CH₃ and H, so CH₃ wins (C beats H). On C3 the groups are C₂H₅ and CH₃; at the first carbon both are (C, H, H), so move outward: ethyl gives (C, H, H) at the next atom while methyl gives (H, H, H). Ethyl wins.

With CH₃ and C₂H₅ on the same side, the answer is $Z$ . If they had been on opposite sides it would be $E$ .

Solved Example 3

Arrange the three C=C bonds marked a, b and c in increasing order of stability, where a is CH₃-CH=CH-CH₃ (disubstituted), b is (CH₃)₂C=CH-CH₃ (trisubstituted) and c is CH₂=CH-CH₃ (monosubstituted).

Solution:

Stability rises with the number of alkyl groups on the doubly bonded carbons, since each contributes hyperconjugation and a plus inductive effect.

Counting substituents: c has one, a has two, b has three.

Order: c $<$ a $<$ b.

Solved Example 4

Match each alkene with its heat of combustion: (i) pent-1-ene, (ii) cis-pent-2-ene, (iii) trans-pent-2-ene, against the values 804.03, 806.9 and 805.3 kcal/mol.

Solution:

All three are C₅H₁₀ isomers, so they burn to identical products and can be compared directly. The larger the heat released, the less stable the alkene.

Stability order is trans-pent-2-ene $>$ cis-pent-2-ene $>$ pent-1-ene, so heats of combustion run the other way.

(i) 806.9, (ii) 805.3, (iii) 804.03 kcal/mol.

Solved Example 5

How many stereocentres (sites of geometrical isomerism) are present in CH₃-CH=CH-CH=CH-CH₃?

Solution:

Check each double bond. On the first, one carbon carries CH₃ and H, the other carries H and a vinyl chain, so no carbon has two identical groups. The same holds for the second double bond.

Both double bonds qualify, giving $2^{2} = 4$ geometrical isomers in all.

Answer: two.

Solved Example 6

A hydrocarbon has the formula C₆H₁₀. What is its degree of unsaturation, and what structural possibilities does this allow?

Solution:

$DU = \frac{2 ( 6 ) + 2 - 10}{2} = \frac{4}{2} = 2$ .

Two degrees can be made up as: two double bonds (a diene), one triple bond, one ring plus one double bond (such as cyclohexene), or two rings. A bromine water test distinguishes the unsaturated options from the purely cyclic ones.

Common Mistakes to Avoid

Watch out

Choosing the longest chain in the molecule instead of the longest chain containing the double bond. The C=C always wins.
Numbering from the end nearer a methyl branch. The double bond takes priority over alkyl substituents for lowest locant.
Assuming every alkene shows cis-trans isomerism. Terminal alkenes and 1,1-disubstituted alkenes such as isobutene never do.
Assigning CIP priority by group size or by number of atoms. Only atomic number at the first point of difference counts, so -Br beats -C(CH₃)₃.
Assuming $Z$ always means cis. For CHBr=CHCl the higher priority groups may sit on the same side while the carbon chains do not.
Comparing heats of combustion of non-isomeric alkenes. That comparison only means something for isomers with the same molecular formula.
Forgetting that cumulated dienes are the least stable class, not the most stable, despite having two double bonds close together.

Frequently Asked Questions

What is the general formula of an alkene?

An acyclic alkene with one carbon to carbon double bond has the general formula CnH2n, for example C2H4 (ethene) and C3H6 (propene). Each extra double bond or ring removes two more hydrogens, so a compound with two double bonds follows CnH2n-2.

Why is the C=C bond length shorter than a C-C single bond?

The doubly bonded carbons are sp2 hybridised, so the sigma bond has 33 percent s character against 25 percent in an sp3-sp3 bond, and the extra pi overlap pulls the nuclei closer. The result is 1.34 angstrom for C=C against 1.54 angstrom for C-C.

Why can alkenes show cis-trans isomerism but alkanes cannot?

Rotation about a C=C would have to break the pi bond, which costs about 260 kJ per mole, so the two arrangements do not interconvert at room temperature and can be isolated separately. Single bonds rotate freely, so alkanes give conformers, not isomers.

When does an alkene fail to show geometrical isomerism?

When either doubly bonded carbon carries two identical groups. Isobutene, (CH3)2C=CH2, and every terminal alkene R-CH=CH2 fall in this class, because swapping the two identical groups regenerates the same molecule.

What is the difference between cis-trans and E-Z nomenclature?

Cis-trans compares two identical or obviously similar groups, so it fails when all four substituents differ. E-Z uses Cahn-Ingold-Prelog priorities on each carbon: higher priorities on opposite sides gives E, on the same side gives Z. Every alkene that shows geometrical isomerism can be labelled E or Z.

Which alkene is more stable, cis or trans?

Trans is normally more stable because the two bulky groups sit far apart, so steric strain is lower. The heats of hydrogenation confirm it: 27.6 kcal per mole for trans-but-2-ene against 28.6 for the cis isomer, a difference of about 1 kcal per mole.

How does heat of hydrogenation measure alkene stability?

Isomeric alkenes that give the same alkane can be compared directly. The one that releases less heat started lower in energy, so it was the more stable alkene. Among the butenes, trans-but-2-ene releases the least and is the most stable.

Why are more substituted alkenes more stable?

Alkyl groups release electron density into the pi system by hyperconjugation and by the plus inductive effect, spreading the charge and lowering the energy. So the order runs tetrasubstituted greater than trisubstituted greater than disubstituted greater than monosubstituted greater than ethene.

What is the degree of unsaturation and how is it used?

Degree of unsaturation counts rings plus pi bonds and equals (2n + 2 - H) divided by 2 for CnH compounds. One double bond or one ring gives one degree, a triple bond gives two. It narrows down possible structures before any chemical test is run.

Previous year questions on Naming of Alkenes

3 questions from past papers, each with a step-by-step solution.

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