Alkenes: Structure and Isomerism

General formula and shape. Alkenes follow the general formula

where: $n$ is the number of carbon atoms in the chain, so an alkene with $n$ carbons carries exactly $2n$ hydrogen atoms — two fewer H atoms per molecule than the corresponding alkane, the "missing" pair used up in forming the extra bond of the double bond. Each doubly-bonded carbon is $sp^2$ hybridised: three $sp^2$ orbitals lie in one plane at $120^\circ$ to each other, each forming a sigma bond (two to substituents, one to the other alkene carbon), while a leftover, unhybridised $2p$ orbital on each carbon sits perpendicular to that plane.

Sigma plus pi. Those two perpendicular $p$ orbitals overlap sideways, above and below the sigma-bond plane, to form one pi bond. So C=C is not "two of the same kind of bond" — it is a strong sigma bond along the internuclear axis plus a weaker pi bond straddling it. The figure below shows both the electronic picture and its geometric consequence.

Why rotation is restricted. Because the pi bond's sideways overlap needs the two $p$ orbitals to stay parallel, twisting one carbon's substituents relative to the other around the C=C axis would pull the $p$ orbitals out of alignment and break that overlap. A C–C single bond in an alkane has no such pi bond, so its groups rotate freely; a C=C double bond does not.

The cis-trans rule. For a disubstituted alkene $\text{R–CH=CH–R}$, the cis isomer has the two R groups on the same side of the double bond and the trans isomer has them on opposite sides. This is a form of stereoisomerism — same connectivity, different arrangement in space — and it is only possible when each doubly-bonded carbon bears two different groups. It is also a separate phenomenon from structural isomerism (chain, position or functional-group isomerism, which alkenes can show too — for example but-1-ene and but-2-ene differ in where the double bond sits along the same four-carbon chain).

The goal here is to reason from the electronic structure of the C=C double bond all the way to the exact structural test for geometrical isomerism, using nothing beyond the sigma-pi picture already built above and simple counting of substituents on each doubly-bonded carbon, so that the structure of any alkene alone tells you whether cis and trans forms exist for it.

Step 1. Fix the electronic picture. Each alkene carbon holds three $sp^2$ sigma bonds in a plane plus one $p$ orbital standing perpendicular to that plane; the two perpendicular $p$ orbitals on the two double-bond carbons overlap sideways to make the pi bond, which lies above and below the sigma-bond plane.

Step 2. Ask what a rotation about the C=C axis would do to that overlap. Twisting one CH end relative to the other tilts its $p$ orbital out of parallel with the other carbon's $p$ orbital, shrinking and eventually destroying the sideways overlap that constitutes the pi bond — so a full rotation effectively requires breaking the pi bond rather than merely bending a single bond, and this does not happen under ordinary conditions.

Step 3. Because the molecule cannot freely rotate about C=C, whatever spatial arrangement the substituents are built in stays fixed — the two ends of the double bond are "frozen" relative to each other, unlike free rotation about a single bond.

Step 4. Turn that frozen geometry into a counting rule. If a doubly-bonded carbon carries two identical groups, swapping them produces the same molecule again — no new arrangement — so no cis or trans distinction arises at that carbon. Only when both doubly-bonded carbons carry two different groups does swapping produce a genuinely different, non-superimposable arrangement, giving distinct cis and trans isomers. This is exactly the rule applied in the worked examples below.

Q. But-2-ene, $\text{CH}_3\text{-CH=CH-CH}_3$, exists as two geometrical isomers. Identify them and explain, from the structure, why interconversion does not happen under ordinary conditions.

Step 1. Locate the doubly-bonded carbons: C2 and C3, each bearing one $\text{CH}_3$ group and one H atom.

Step 2. Apply the rule: each of C2 and C3 carries two different substituents (H and $\text{CH}_3$), so cis-trans isomerism is possible.

Step 3. Name the two arrangements. In cis-but-2-ene both $\text{CH}_3$ groups lie on the same side of the C=C bond (and both H atoms on the other, shared side). In trans-but-2-ene the two $\text{CH}_3$ groups lie on opposite sides, with the H atoms likewise opposite.

Step 4. Explain the stability. Converting cis to trans would require rotating one $\text{CH}_3$-CH= end by $180^\circ$ about the C2–C3 axis, which pulls the two $2p$ orbitals out of parallel alignment and breaks the sideways overlap forming the pi bond. Cis-but-2-ene and trans-but-2-ene are therefore distinct, isolable compounds, not interconverting conformations of one molecule.

Competitive stretch

Q. Does 1,2-dichloroethene, $\text{ClCH=CHCl}$, show geometrical isomerism?

Step 1. C1 carries one H and one Cl; C2 also carries one H and one Cl — both doubly-bonded carbons have two different substituents.

Step 2. By the same rule as above, this satisfies the condition for cis-trans isomerism, so cis-1,2-dichloroethene (both Cl atoms on the same side) and trans-1,2-dichloroethene (Cl atoms on opposite sides) are both valid, distinct compounds.

• Assuming every C=C bond gives cis-trans isomers — the rule needs each double-bond carbon to carry two different groups; symmetric alkenes like 2-methylpropene or ethene itself show none.
• Mixing up structural isomerism with geometrical isomerism — but-1-ene and but-2-ene are structural (position) isomers of each other, while cis- and trans-but-2-ene are geometrical isomers of each other.
• Treating the C=C bond as freely rotating like a C–C single bond — the pi bond's sideways $p$-orbital overlap specifically prevents this.
• Judging cis/trans from only one substituent instead of checking the arrangement on both carbons of the double bond.

Ask the tutor: Not sure if a particular alkene shows cis-trans isomerism, or want help drawing its cis and trans structures — ask the AI tutor to check your reasoning.