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Pyramidal alkene

Pyramidal alkene is a chemistry topic covered in the lgStudy science library. This page brings together a partial reference excerpt, illustrations, worked examples, real-world applications and a short study plan, so you can understand Pyramidal alkene rather than just read about it. In short: Pyramidal alkenes are alkenes in which the two carbon atoms making up the double bond are not coplanar with their four substituents. This deformation results from geometric constraints.

Pyramidal alkene — main illustration
Pyramidal alkene — illustration

Key takeaways

  • Pyramidal alkene belongs to chemistry; place it in that map before memorising details.
  • Learn the definition first, then one example that makes the definition concrete.
  • Connect Pyramidal alkene to a quantity you can measure, compute or draw — that is where exam questions come from.
  • Reproduce the core statement of Pyramidal alkene from memory before moving on to harder problems.

Reference excerpt

Pyramidal alkenes are alkenes in which the two carbon atoms making up the double bond are not coplanar with their four substituents. This deformation results from geometric constraints. Pyramidal alkenes are of interest because much can be learned from them about the nature of chemical bonding.

Energetics Twisting to a 90° dihedral angle between two of the groups on the carbons requires less energy than the strength of a pi bond, and the bond still holds. The carbons of the double bond become pyramidal, which allows preserving some p orbital alignment—and hence pi bonding. The other two attached groups remain at a larger dihedral angle. This contradicts a common textbook assertion that the two carbons retain their planar nature when twisting, in which case the p orbitals would rotate enough away from each other to be unable to sustain a pi bond. In a 90°-twisted alkene, the p orbitals are only misaligned by 42° and the strain energy is only around 40 kcal/mol. In contrast, a fully broken pi bond has an energetic cost of around 65 kcal/mol.

Examples In cycloheptene (1.1) the cis isomer is an ordinary unstrained molecule, but the heptane ring is too small to accommodate a trans-configured alkene group resulting in strain and twisting of the double bond. The p-orbital misalignment is minimized by a degree of pyramidalization. In the related anti-Bredt molecules, it is not pyramidalization but twisting that dominates.

Pyramidalized cage alkenes also exist where symmetrical bending of the substituents predominates without p-orbital misalignment.

The pyramidalization angle φ (b) is defined as the angle between the plane defined by one of the doubly bonded carbons and its two substituents and the extension of the double bond and is calculated as: cos ⁡ φ = − cos ⁡ ( ∠ R C C ) cos ⁡ ( 1 2 ∠ R C R ) {\displaystyle \cos \varphi =-{\frac {\cos(\angle \mathrm {RCC} )}{\cos({\frac {1}{2}}\angle \mathrm {RCR} )}}}

the butterfly bending angle or folding angle ψ (c) is defined as the angle between two planes and can be obtained by averaging the two torsional angles R1C=CR3 and R2C=CR4. In alkenes 1.2 and 1.3 these angles are determined with X-ray crystallography as respectively 32.4°/22.7° and 27.3°/35.6°. Although stable, these alkenes are very reactive compared to ordinary alkenes. They are liable to dimerization creating cyclobutane rings, or react with oxygen to epoxides. The compound tetradehydrodianthracene, also with a 35° pyramidalization angle, is synthesized in a photochemical cycloaddition of bromoanthracene followed by elimination of hydrogen bromide.

This compound is very reactive in Diels–Alder reactions due to through-space interactions between the two alkene groups. This enhanced reactivity enabled in turn the synthesis of the first-ever Möbius aromat. In one study, the strained alkene 4.4 was synthesized with the highest pyramidalizion angles yet, 33.5° and 34.3°. This compound is the double Diels–Alder adduct of the diiodocyclophane 4.1 and anthracene 4.3 by reaction in presence of potassium tert-butoxide in refluxing dibutyl ether through a diaryne intermediate 4.2. This is a stable compound but will slowly react with oxygen to an epoxide when left standing as a chloroform solution.

In one study, isolation of a pyramidal alkene is not even possible by matrix isolation at extremely low temperatures unless stabilized by metal coordination:

A reaction of the diiodide 5.1 in Figure 5 with sodium amalgam in the presence of ethylenebis(triphenylphosphine)platinum(0) does not give the intermediate alkene 5.2 but the platinum stabilized 5.3. The sigma bond in this compound is destroyed in reaction with ethanol. Isobenzvalene (tricyclo[3.1.0.02,6]hex-1(6)-ene), an isomer of benzene, has been synthesized and trapped by a Diels–Alder reaction with anthracene.

Dodecahedrane dehydrogenates to C20 fullerene, as well as a variety of partially-unsaturated clusters. The latter are thermodynamically quite stable despite extremely pyramidalized alkenes (φ ≈ 40°). In 2024, Neil Garg and Kendall Houk reported a general way to make pyramidalized anti-Bredt olefins, violating Bredt's Rule, and later described chemistry of cubene and quadricyclene. They introduced the term "hyperpyramidalized" to describe the severe pyramidalization of cubene and quadricyclene carbons and showed correlations between pyramidalization and non-integer bond orders.

References

Illustrations

Pyramidal alkene: Figure 2. Angle definitions
Figure 2. Angle definitions
Pyramidal alkene: Figure 3. Tetradehydrodianthracene synthesis
Figure 3. Tetradehydrodianthracene synthesis
Pyramidal alkene: Figure 4. Cyclophane anthracene adduct
Figure 4. Cyclophane anthracene adduct
Pyramidal alkene: Figure 5. (Ph3P)2Pt complex of 3,7-dimethyltricyclo[3.3.0.03,7]oct-1(5)-ene
Figure 5. (Ph3P)2Pt complex of 3,7-dimethyltricyclo[3.3.0.03,7]oct-1(5)-ene
Pyramidal alkene illustration

Worked examples

Example 1 — a first encounter with Pyramidal alkene

Start with the simplest possible case. Write down what Pyramidal alkene claims or describes in one sentence, then invent the smallest concrete situation in which that sentence is true. In chemistry, the smallest case is usually a single object, a single equation or a single measurement. Check that every symbol or term in your sentence has a meaning in that case.

Example 2 — changing one variable

Take the situation from Example 1 and change exactly one quantity: double it, halve it, or set it to zero. Predict what should happen to Pyramidal alkene before you calculate. Comparing your prediction with the result is the fastest way to find out whether you understand the idea or only the words.

Example 3 — an exam-style question

Typical questions about Pyramidal alkene ask you to (a) state it precisely, (b) apply it to given data, and (c) explain a limitation. Practise writing all three answers in under five minutes; the third part is what separates a full-mark answer from an average one.

Applications of Pyramidal alkene

In research
Pyramidal alkene appears in chemistry research whenever the underlying quantities have to be modelled precisely. Papers usually cite it as a starting assumption and then explore where it breaks down.
In technology and industry
Engineering practice reuses Pyramidal alkene in design rules, simulations and safety margins. Knowing the idea lets you read a specification sheet and understand why the numbers look the way they do.
In the classroom
Pyramidal alkene is common in secondary-school and first-year university syllabi. It links to neighbouring topics Alkenes, Chemical bonding, so understanding it makes those chapters shorter.
In everyday life
Look for Pyramidal alkene outside the textbook — in sport, cooking, traffic, electronics or the sky above you. An example you found yourself is remembered far longer than one you were given.
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How to study Pyramidal alkene in 20 minutes

  1. Read the reference excerpt below once, without taking notes.
  2. Close the page and write down what Pyramidal alkene means in your own words.
  3. Compare your version with the excerpt and mark what you missed.
  4. Work through the three examples above with pen and paper.
  5. Explain Pyramidal alkene out loud to somebody else — or to Teacher Smith in the lgStudy chat.

Frequently asked questions

What is Pyramidal alkene in simple terms?

Pyramidal alkenes are alkenes in which the two carbon atoms making up the double bond are not coplanar with their four substituents. This deformation results from geometric constraints.

Why does Pyramidal alkene matter?

Because it connects several chemistry ideas at once: it gives you a definition you can apply, a quantity you can calculate, and a way to check whether a result is plausible.

How should I study Pyramidal alkene?

Read the excerpt, restate it from memory, then work through the examples and applications listed on this page. The five-step study plan above takes about twenty minutes.

What does this page cover?

It gives you a compact reference excerpt plus original lgStudy explanations, examples, applications and study material on Pyramidal alkene.

Tags

  • Alkenes
  • Chemical bonding

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