Alkene Structure and Pi Bonding

Introduction

An alkene has at least one carbon-carbon double bond. The two bonded carbons are approximately trigonal planar, and their double bond contains one sigma and one pi component. Side-by-side p-orbital overlap makes rotation around C=C very different from ordinary rotation around an alkane C–C single bond.

Core explanation

In ethene, each double-bond carbon is attached to two hydrogens and the other carbon, giving three sigma-bond directions approximately 120° apart in a plane. The common sp² hybridization model uses three in-plane orbitals for sigma bonding and one unhybridized p orbital perpendicular to that plane. The two p orbitals overlap sideways above and below the carbon framework, forming a π bond. C=C therefore consists of one sigma and one pi bond, not two equivalent sigma bonds.

Twisting one alkene carbon around the C=C axis disrupts the p-orbital alignment required for the π bond. Ordinary thermal motion does not freely rotate a stable alkene double bond as it does many alkane single bonds. If each double-bond carbon has two different substituents, two distinct arrangements may be possible, giving geometric isomerism. The simplest ethene lacks this because each carbon bears two identical hydrogen atoms.

The π component is more accessible to many reagents than the C–C sigma framework. Electrophilic addition often uses π electrons to form new sigma bonds to the two alkene carbons. The double bond becomes a single C–C bond as two new attachments form. This explains why an alkene can add bromine or hydrogen under suitable conditions whereas an alkane normally requires a different reaction pathway. Yet not every reagent reacts spontaneously; conditions, mechanism, and thermodynamics still matter.

An acyclic monoalkene has formula CₙH₂ₙ, two hydrogens fewer than the analogous acyclic alkane. The formula reflects one degree of unsaturation but is shared by simple cycloalkanes. Double-bond position and branching create constitutional isomers, while fixed substituent orientation can create stereoisomers. A drawing with a double bond as two flat parallel lines indicates connectivity; for stereochemical questions the substituents on each carbon must also be placed explicitly.

Step-by-step reasoning

1. Locate the C=C and count three sigma directions around each bonded carbon. 2. Identify one sigma and one pi component of the double bond. 3. Test whether each carbon has two different substituents. 4. Predict restricted rotation and possible geometric isomers.

Visual explanation

Draw two trigonal-planar carbons with in-plane sigma bonds. Show p lobes above and below the plane overlapping sideways to form the π electron region.

Real-world analogy

Two boards linked by a single round hinge can turn, but adding a broad second connection resists twisting. A double bond similarly has an additional overlap that must be disrupted for rotation.

Real-world example

Ethene is polymerized into polyethylene by reactions that consume its π bond and form new carbon-carbon sigma links. Its double-bond geometry explains why it acts as a monomer.

Why?

Why is alkene rotation restricted? Twisting makes the two p orbitals cease to overlap effectively, costing the stabilization supplied by the π bond between them.

Common misconception

“C=C contains two identical bonds.” It contains a sigma framework and a distinct pi overlap with different geometry and reaction behavior.

Worked example

Consider propene, CH₃–CH=CH₂. Its double-bond carbons each have three sigma-bond directions and one perpendicular p orbital. The C=C has one sigma plus one pi bond. The terminal CH₂ carbon has two identical H substituents, so propene cannot have a pair of E/Z geometric isomers. Its formula C₃H₆ also fits the one-degree-of-unsaturation count but does not alone prove double-bond location.

Quick check

1. How many sigma and pi bonds make up one ordinary C=C double bond? Answer: One sigma bond and one pi bond.

Exam focus

Use bond geometry, not just a formula, to test stereoisomerism. Both double-bond carbons must carry distinguishable substituent pairs for E/Z configurations.

Advanced insight

Pi bonding is delocalized in conjugated systems, so localized double-bond drawings may be only one resonance contributor. Ethene is the simpler isolated case for understanding the basic overlap.

Summary

An alkene C=C combines a sigma bond with a perpendicular pi bond. Its carbons are roughly trigonal planar, and disrupting pi overlap makes rotation restricted.

Practice questions

1. What geometry surrounds each double-bond carbon in ethene? Answer: Approximately trigonal planar. 2. Can propene exhibit E/Z isomerism about its double bond? Answer: No. Its terminal double-bond carbon has two identical hydrogens. 3. What bond component is lost when an alkene undergoes simple addition? Answer: The pi component, while the C–C sigma bond remains.